User talk:David Tombe

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Hi David,

I'm moving the discussion here, in order to leave Talk:Centrifugal force for what it is about, namely discussing the article.

You have been on Wikipedia for a while now, so you should be familiar with how it works. The main principles for resolving disputes is to find consensus, and referring to published work. Before you came to the page, there was a consensus on the two meanings of the term "centrifugal force". You want to unilaterally change this. Don't.

I am telling you that you have misunderstood the subject matter. If you don't want to believe me, that's fine. But remember, that, just as one year ago, when everybody is else is disagreeing with you, there's a chance they may be right.

The reason why I don't want to argue the physics with you is that I tried that last year, but you did not seem to listen, and continued inserting nonsense into the article until you were blocked by an administrator. As long as you believe that you are right and everybody else is wrong, it would be a waste of my time and energy to try to reason with you.

If you are willing to accept that some of that which you put into the article is wrong, I'll be happy to work with you on a better introduction, on the condition that we keep that work on a subpage of the talk page until it's done, so as to not disturb the main article.

In either case I won't revert your edits to the article for now. The introduction still contains a few errors. For example, you say that the fictitious centrifugal force only applies to objects that co-rotates with the frame. This is wrong, because the centrifugal force applies to any object in the frame, just as gravity affects any object, not just objects that are falling. (But it the object is not stationary in the rotating frame, then it is also affected by the Coriolis force, which may offset the centrifugal force.)

--PeR (talk) 13:44, 23 April 2008 (UTC

PeR,

Imagine a man standing still on the ground. Viewed from a rotating frame of reference, he will trace out a circle in that frame. There will be no centrifugal force. David Tombe (talk) 15:09, 23 April 2008 (UTC)

Because he is moving in the rotating frame, there is a Coriolis force, which is twice as large as the centrifugal force, and in the opposite direction. The net result is an inwards force causing a circular motion, as seen from the rotating frame.

Since my first message here, where I pointed out that there were errors in your edits to the centrifugal force article, those edits were reverted by another editor, as I predicted. You then reverted him with the message "No Rracecarr. There were no errors. If there were, lets' hear them". Did you not read my message? I will now revert your edits again. Please do not re-revert, as this will be a violation of the three revert rule. --PeR (talk) 16:51, 23 April 2008 (UTC)

Hi again, David,

I read what you just wrote on Talk:Centrifugal force. If you where to ask the question "What is it that I have misunderstood?" I would try to help you, but with your "I'm right and the rest of the world is wrong" attitude, I'm afraid I can't waste my time writing long explanations that you will ignore anyway. Sorry. ~~Please don't edit the article anymore. Your edits will likely contain inaccuracies and get reverted quickly.~~ I repeat my offer to work with you on a new introduction on a talk page. Otherwise, I am sure there are other areas of Wikipedia where you can contribute in a more productive way. --PeR (talk) 07:38, 24 April 2008 (UTC)

1 Your definition of acceleration?
2 Vandalism warning, April 2008
3 More vandalism
4 ANI-notice
- 4.1 Other accounts ?
5 Enough
6 Centrifugal force
7 Neutral Point of View
8 My reverting you on Centrifugal Force
9 Verifiability policy
- 9.1 Warning
- 9.2 Discussion
- 9.3 Blocked
- 9.4 Warning again
- 9.5 Discussion again
10 Have you noticed...?
11 Blocked again, as per warning above
12 Evading the block.
13 Blocked again
14 A word of caution
15 Centrifugal force
16 Blocked for one week
17 Reply to Wolfkeeper
18 Reply to SCZenz
19 Quote for possible comment
20 Request to post
21 Discussion
22 Continuation of discussion
23 I'll respond here, since you are blocked again
24 Links now work
25 Centripetal force
26 So, which is it?
27 Finding an expert
28 Questions for experts
29 Reply to Anome's Questions
30 The Artificial Circle
31 Fictitious Force article
32 Reply to SBharris on Circular Motion
33 Reply to Anome
34 Reply to Brews ohare
35 Reply to Anome
36 Reply to Timothy Rias
37 Reply to SBharris
38 About orbital theory
39 Having fun yet?
40 The Three Body Problem
41 Proposed addition to Centrifugal Force
42 Unresponsive
43 Induction
44 Reply to SBharris on Circular Motion (2)
45 An intuitive approach
46 Personal attacks
- 46.1 Blocked for personal attacks
47 Regrets
48 Airplane banking
49 Kinematics
50 Reply to SBharris
51 Editor controversy

[edit] Your definition of acceleration?

Hi again, David,

I asked this question before, on talk:Centrifugal force, but it was answered by an anon, and I'd like to hear your answer. You say:

...it is possible to have an obect moving in a curved path with no net acceleration if that curved path forms part of a circle. [1]

Now, most textbooks define acceleration as "rate of change in velocity". Do you use a different definition? Or do you propose that an object moving in a circle never changes its velocity?

Just to be clear: those same textbooks also define velocity as a vector, that includes both speed and direction.

--PeR (talk) 14:53, 26 April 2008 (UTC)

[edit] Vandalism warning, April 2008

Please stop. If you continue to vandalize Wikipedia, as you did to Rollerblade, you will be blocked from editing. Just because we disagree on a different article, you are not allowed to wikistalk me, and undo edits that are obviously noncontroversial. --PeR (talk) 07:53, 28 April 2008 (UTC)

[edit] More vandalism

Please do not vandalize pages, as you did to Volleyball here. Rracecarr (talk) 16:49, 28 April 2008 (UTC)

[edit] ANI-notice

Hello, David Tombe. This message is being sent to inform you that there currently is a discussion at Wikipedia:Administrators' noticeboard/Incidents regarding an issue with which you may have been involved. Yours, --PeR (talk) 21:24, 28 April 2008 (UTC)

[edit] Other accounts ?

David, Are you also editing as User:George Smyth XI, as you seem to imply on User talk:PhySusie? Then you should say so on your user page, to avoid confusion. See WP:SOCK. --PeR (talk) 14:27, 30 April 2008 (UTC)

[edit] Enough

I have, Solomon-like, decided what the title of the thread is going to be. Please do not change the title of the thread again. You are disruptively edit warring. --barneca (talk) 18:49, 30 April 2008 (UTC)

[edit] Centrifugal force

David, you seem to find yourself in vigorous disagreement with a number of other contributors to the centrifugal force articles. Since Wikipedia has no way of directly determining the WP:TRUTH in any matter, the usual procedure in these circumstances is to cite reliable sources such as peer-reviewed papers and physics textbooks, in order to support controversial views. No matter how sure you are of the correctness of your views, and the incorrectness of other editors, this is likely to be the only way that you will be able to make any progress in this matter. -- The Anome (talk) 10:34, 1 May 2008 (UTC)

I've just reverted your removal of the term "fictitious force", and, in the spirit of the above, added a cite to support its usage. Please could you do the same, and supply cites to reliable sources to support your edits? -- The Anome (talk) 10:48, 1 May 2008 (UTC)

Following up my earlier comment, you might want to read the essay on THE TRUTH, and see the Neutral Point of View policy for guidance on how to proceed when there is controversy regarding an issue. -- The Anome (talk) 10:57, 1 May 2008 (UTC)

[edit] Neutral Point of View

David, could you please take another look at Wikipedia's Neutral Point of View policy and attribution, verifiability and reliable sources principles? I look forward to your providing suitable cites to support your contributions from now on. -- The Anome (talk) 11:18, 1 May 2008 (UTC)

[edit] My reverting you on Centrifugal Force

David, I explained the reasons for my revert in the edit summary[2], but because you asked, I will elaborate: I deleted your text because it seemed incorrect to me, and it failed to cite a reliable source. (See WP:V.) In the spirit of what the Anome has already said above, I have also added a cite to support the sentence that I restored. Please could you now do the same, and supply cites to reliable sources to support your edits? --PeR (talk) 12:29, 2 May 2008 (UTC)

[edit] Verifiability policy

David, you have been repeatedly asked to provide verifiable references to reliable sources to back up your assertion that centrifugal force is a real force that is independent of reference frames. So far, you have not given any references to reliable sources to back support this opinion, even as being a significant minority opinion, yet alone as being a fundamental truth, as you assert.

After hundreds of edits and many pages of discussion with other editors who have repeatedly tried to broker a compromise with you, you have continued to make these edits without any justification from reliable sources, in direct opposition to Wikipedia's basic principles of NPOV and verifiability, and have repeatedly deleted the contributions of other editors whose contributions are in accordance with the multiple reliable sources referenced in the article, all of which agree with one another that centrifugal force is an apparent, reference-frame-dependent pseudoforce.

This pattern of behavior seems to have gone on for some time, so it is hard to conclude that you are unaware of Wikipedia's policies: see this earlier discussion. Edits like this one, in which you seem to explicitly reject the requirement for verifiable references, do not help your case. Unfortunately, I find myself drawn to the conclusion that your edits are now becoming actively disruptive. Wikipedia's policies are not negotiable on these matters: please abide by them. -- The Anome (talk) 18:13, 5 May 2008 (UTC)

[edit] Warning

Regarding this edit, I would like to remind you that providing references for disputed material upon request is required by Wikipedia policy. Continuing to add such material without references repeatedly constitutes disruptive editing. Please do not continue to do this. If you do, it will become necessary to block your account in order to avoid continued disruption to the article.

Please feel free to write a message on my talk page if you have questions about Wikipedia's verifiability policy. -- SCZenz (talk) 21:59, 5 May 2008 (UTC)

[edit] Discussion

If you are referring to this edit in your comment on my talk page, it would appear that Plvekamp left an edit summary that appears entirely accurate. The quotes you gave do not support your statement, and your inference violates our policy on original research. -- SCZenz (talk) 08:32, 6 May 2008 (UTC)

SCZenz, The entry was in response to a question 'Is centrifugal force real?'. I gave those quotes from Maxwell and Bernoulli which clearly show that they believed it to be real.

How could you possibly interprete those quotes in any other way? They talk about outward pressure being caused by centrifugal force. In fact, it is the whole basis of Maxwell's theory of magnetic repulsion.

The article wasn't about actually pushing Maxwell's theories. It was merely drawing attention to the fact that Maxwell believed centrifugal force to be real, and a valid citation was given.

If you accept Plvekamp's reasoning then it is clear that you are not genuine, because there was absolutely no logic behind what Plvekamp said.

In normal circumstances, interesting historical quotes would gladly be accepted into such an article. But at the moment, we have a group who not only want to push the idea that centrifugal force is absolutely fictitious, but they want to hide all knowledge of the fact that the top physicists in the past believed it to be real. David Tombe (talk) 08:41, 6 May 2008 (UTC)

The quotes mention centrifugal force, but do not state that they "believed it to be real" in the sense you assert. Plvekamp explained that point further in this edit. Others have explained both the physics issues and the issues of Wikipedia policy to you. I commend them for their patience, and see no reason to repeat their effort. I can only repeat: do not repeatedly add statements for which you are unable to provide a source, and do not add material which reflects your personal interpretation of sources. If you persist in this type of disruptive editing, such edits may be reverted on sight and your account may be blocked; please consider this your final warning. -- SCZenz (talk) 09:00, 6 May 2008 (UTC)

SCZenz, I'm sorry but if you read Maxwell's 1861 paper, you will see that he utilzed centrifugal force to account for magnetic repulsion. The quote which I supplied is unambiguous in its meaning. It refers to vortices pushing against each other with centrifugal force. If Maxwell did not believe centrifugal force to be real how could he have believed in that statement?

I don't believe that you are being altogether genuine in your denial of this obvious interpretation of the statement. David Tombe (talk) 09:08, 6 May 2008 (UTC)

[edit] Blocked

You have been blocked for three hours, in order to prevent continued disruptive editing in violation of Wikipedia's Verifiability and No Original Research policies. You are welcome to appeal this decision, but I hope it is now clear that continued edit warring will not be an effective strategy. -- SCZenz (talk) 09:22, 6 May 2008 (UTC)

[edit] Warning again

Please cease and desist from adding claims that centrifugal force is not a fictitious force, in contradiction to sources cited in the article and without sources of your own. Instead, find sources that support your claims and discuss your concerns on the talk page. You will be blocked again if you persist. -- SCZenz (talk) 06:33, 7 May 2008 (UTC)

[edit] Discussion again

Actually, I am a professional physicist; my mastery of the facts is quite adequate to the task at hand. And while my professional skills are certainly an asset when I compare claims with sources, in this matter I am firmly on the "side" of Wikipedia policy, which I will continue to enforce. You are removing accurate material supported by sources, and replacing it with innacurate material that cannot be supported by sources, and policy is clear on how to handle this. -- SCZenz (talk) 11:38, 7 May 2008 (UTC)

David; the issue is not whether article material is orthodox or heterodox; the issue is whether it is attributable to third parties as attested to by verifiable, reliable sources while avoiding synthetic arguments. This has been explained to you repeatedly, by many, many different editors. If, after extensive discussion and multiple warnings, you still persist in ignoring Wikipedia's core editorial policies, blocking is unfortunately the only remaining option available to stop you doing so. -- The Anome (talk) 12:22, 7 May 2008 (UTC)

[edit] Have you noticed...?

That this edit also added some rubbish to the top of the article? Please be more careful. -- SCZenz (talk) 16:13, 7 May 2008 (UTC)

I see. It looked to be very much in keeping with your edit style and interests, but if you say it wasn't you then I'll not press you further. -- SCZenz (talk) 16:43, 7 May 2008 (UTC)

A number of other IP addresses in the 72.64.0.0/11 block also appear to have edited articles on the same topics, and with a very similar writing style. -- The Anome (talk) 16:59, 7 May 2008 (UTC)

Automated analysis of the contribution logs show that the 72... editor is User:Electrodyanmicist. I believe this may truly be a person unconnected with User:David Tombe (David does have at least one alias, but I have not seen any evidence of sockpuppetry). Rracecarr (talk) 19:22, 7 May 2008 (UTC)

[edit] Blocked again, as per warning above

David, I can only repeat what I have said in my previous replies. The issue is not whether article material is orthodox or heterodox; the issue is whether it is attributable to third parties as attested to by verifiable, reliable sources while avoiding synthetic arguments. This has been explained to you repeatedly, by many, many different editors. The "vandals" you are complaining of are editing according to Wikipedia's policies, and the material you are complaining about is clearly supported by the multiple references cited in the article. Most of the changes made by other editors have simply been attempts to express these ideas in such as way as to be persuasive to even the most skeptical reader, driven by your apparent difficulty in accepting that this is the mainstream scientific view of centrifugal force. As far as I can tell, you have refused to provide a single reference that backs up your assertions; unfortunately, your inferences based the quotes from Newton and Bernoulli do not count, since they are based on your personal interpretation of these quotes, something clearly dealt with in the WP:SYN and WP:OR policies.

If you can abide by Wikipedia's ground rules, and can demonstrate that your point of view is attested to -- even as a significant minority viewpoint -- by multiple reliable sources, we can make progress. If, however, after extensive discussion and multiple warnings, you still persist in ignoring Wikipedia's core editorial policies, blocking unfortunately appears to be the only remaining option available to stop you doing so.

This discussion from last year suggests that this is not your first exposure to Wikipedia's policy enforcement processes. Since the recent block of this account for 3 hours has had no apparent effect, and given your subsequent re-insertion of the same unreferenced assertions that you have been warned about before (see this diff), I am therefore now blocking you from editing for a period of 31 hours. You are welcome to make constructive contributions again when the block expires. -- The Anome (talk) 10:55, 8 May 2008 (UTC)

[edit] Evading the block.

You are not allowed to evade a block by not logging in, as you did here [3]. When you do that, an admin may reset your block, and the IP that you are editing from may also be blocked. --PeR (talk) 07:21, 9 May 2008 (UTC)

[edit] Blocked again

David, after many, many requests not to add material without cites, you inserted the following text without a supporting citation:

As regards trajectories however, the matter becomes somewhat more complex. Providing that an object begins on the floor and doesn't move too far away from the floor, the result will be a good approximation to gravity. For example, when playing catch, a ball that is thrown upwards will come down again due to centrifugal force. For longer range trajectories that go closer to the center of rotation, a noticeable deflection will occur due to the rotation of the space station.

This is [a] contrary to the predictions of the classical treatment of rotational motion (which is fully cited within the article), [b] uncited. You have persistently and consistently refused to provide cites for your idiosyncratic theories, and this is just one more example. Accordingly, I am blocking you again, this time for 48 hours.

You are welcome to appeal this decision if you think you have been wrongly blocked. -- The Anome (talk) 15:32, 12 May 2008 (UTC)

David, this is not a game. You are using you Ip address to evade your block. I am re-setting your block for 48 hours from now. If you continue to evade blocks that have gained consensus on WP:ANI, you will soon find yourself indefinitely blocked. --barneca (talk) 18:20, 13 May 2008 (UTC)

[edit] A word of caution

David, you might want to take a look at the current discussion concerning you at WP:ANI. I think you are under the impression that it is just this username that risks getting blocked. That is not the case. You risk being banned from Wikipedia. That means that anything you write, from anywhere, will be reverted without discussion, and regardless of its merits. --PeR (talk) 12:52, 14 May 2008 (UTC)

Yes, do read that thread before posting anything more while blocked, PeR's interpretation is correct. --barneca (talk) 13:30, 14 May 2008 (UTC)

I don't know who the IP server is that begins with 71. So Wolfkeeper had no right to remove those comments claiming them to be mine.

In fact, bearing in mind that the block totally broke wikipedia's rules, since it was totally without justification, one wonders what makes Wolfkeeper and Itub presume to police the block. David Tombe (talk) 03:41, 15 May 2008 (UTC)

If you don't want people to be making that kind of mistake, then - please - only edit from your account in the future. --PeR (talk) 06:31, 15 May 2008 (UTC)

Well of course if you hadn't blocked my username on a dishonest pretext, then it wouldn't have come to this.

In fact, I'm wondering is the IP server beginning with 71 the same one that put confusing stuff into the centripetal force page. I was objecting to some vector derivation that a 71 put unto the centripetal force pages on the grounds that they were in fact the Cartesian/polar conversion equations and don't explicitly refer to centripetal force. David Tombe (talk) 07:36, 15 May 2008 (UTC)

[edit] Centrifugal force

Please do not remove valid cites from articles. Please see Wikipedia's verifiability policy, which explains the need for citations. -- The Anome (talk) 14:56, 16 May 2008 (UTC)

Please do not accuse other editors of "talking nonsense". Please see Wikipedia's civility policy. -- The Anome (talk) 15:01, 16 May 2008 (UTC)

[edit] Blocked for one week

Hi Mr. Tombe. In this edit, you removed cited material from Centrifugal force. I understand from the talk page that your interpretation of various sources was such that the material was incorrect—but you know very well that all the other editors of the page read them in a different way than you. In both the view of the community editing the article, and in my own personal judgment, your readings of sources are inconsistent with what they say and constitute original research, while the other editors have bent over backward to provide multiple sources for their statements. (And to explain them to you personally on the talk page.) You know this, yet you have continued to add unsupported material, and remove verifiable material, despite many warnings. You must seek consensus before making such edits in the future. You are blocked for one week; if you persist on your current path, an indefinite block is very likely very soon. -- SCZenz (talk) 06:11, 17 May 2008 (UTC)

[edit] Reply to Wolfkeeper

Wolfkeeper, You asked me for a proof. The proof lies in the bit which SCZenz has just tried to conceal.

Check out my last edit to the main page before SCZenz reverted it and blocked me from editing for a week under a totally dishonest pretext.

I put in the correct expression for the centrifugal force. Somebody else had fudged the sign by taking it out of its vector format.

You will notice that both the Coriolis force and the centrifugal force terms have exactly the same sign.

Yet your fringe theory about the artificial circle depends on the Coriolis force and the centrifugal force having opposite signs. David Tombe (talk) 06:33, 17 May 2008 (UTC)

[edit] Reply to SCZenz

SCZenz, I gave a citation which unambiguously contradicted the sentence which I removed. The citations which were in that sentence did not address the issue in question. You have totally abused your administrative authority.

You are not playing straight. David Tombe (talk) 06:33, 17 May 2008 (UTC)

If you believe that is the case, please see Wikipedia:Appealing a block. -- SCZenz (talk) 10:29, 17 May 2008 (UTC)

[edit] Quote for possible comment

Below is a reply to an anonymous contributor with some of the same concerns you have voiced. I wonder if you find my response satisfactory? How about the article revisions?

You have suggested that an error in sign has been made in evaluating the vector cross-products. This matter is somewhat subtle, because the velocity entering the Coriolis force is the velocity seen from within the rotating frame. The centrifugal term is straightforward: −Ω × (Ω × r ). Vector r is radially outward,(Ω × r ) is tangential to the orbit, and Ω × (Ω × r ) is normal to the orbit and inward. Hence, with the − sign the centrifugal force is outward, as perhaps you agree. The Coriolis term is −2(Ω × v_rot ) =+2(Ω × (Ω × r )! The plus sign comes because the velocity of a body in the rotating frame is the negative of (Ω × r ). Consequently, (Ω × (Ω × r )) points inward, and the sum 2m(Ω × (Ω × r )) –mΩ × (Ω × r ) =mΩ × (Ω × r ) is the necessary inward centripetal force.

An example is the dropping ball problem. Here the centrifugal force is outward (as always). However, (Ω × r ) points oppositely to the observed velocity in Figure 3. Consequently the Coriolis force is inward.

It's confusing, I agree. I've edited the article to make the evaluation clearer. Thank you. Brews ohare (talk) 03:51, 18 May 2008 (UTC)

Brews, the answer that I would have given to your statement These unit vectors are attached to the rotating frame, not to the object under observation. would have been different. I would simply have said that you have proved my very point. The equations do not apply to objects that are not co-rotating.

This is confirmed by the GD Scott statement which I cited. Scott says that centrifugal force only occurs on objects at rest in the rotating frame.

"The centrifugal forces must be recognized for what they are --- Non-Newtonian forces acting on masses at rest in a rotating frame."

The existing cited references do not even address the issue and SCZenz's decision to block me for one week for making this correction has surely got to be one of the most corrupt administrative decisions in the history of wikipedia.

On the point raised by the anonymous, anonymous is absolutely correct. Actually yesterday I corrected the signs in the main article, but SCZenz reverted them again to their fudged state in which one was in vector format and the other not.

We are both agreed that both the Coriolis term and the centrifugal term have negative signs. Lets say for the sake of argument that ω refers to a right hand grip rule anti-clockwise rotation and that ω is upwards. In that case, the v in the Coriolis force for a staionary particle, under your way of looking at it (because it actually can't be radial) must be in a clockwise direction. Hence ωXv will be inwards.

In your centrifugal term, the ωXr term, where r is outwards, will lead to a v in the exact same direction as the v in the Coriolis term. So both your terms, if applied radially, will be in the exact same direction. David Tombe (talk) 06:29, 18 May 2008 (UTC)

Brews, Wolfkeeper replied to the anonymous with a question. I can answer that question. Can you please do me a favour and put this on the main page under Wolfkeeper's question.

Wolfkeeper, the answer to your question is threefold.

(i) The centrifugal force and Coriolis force in hydrodynamics are mutually perpendicular components of a parent term vXω which can be obtained by direct differentiation of the velocity vector without reference to polar or Cartesian coordinates. This parent force is just like the convective term in the Lorentz force. It will act symmetrically all the way around a cyclone.

(ii) In terms of centrifugal force, the east-west currents on opposite sides of the cyclone will represent a greater and a lesser tangential speed in relation to the larger body of entrained and co-rotating atmosphere, hence we will get a differential centrifugal force which will yield a torque.

(iii) If those answers aren't good enough for Wolfkeeper, then he must remember that official theory has it that the Coriolis force only sets the direction for the cyclones, but that once they are started, other inter molecular forces and conservation of angular momentum take over. It would only take an east-west deflection of north-south air currents to trigger it all off. David Tombe (talk) 06:50, 18 May 2008 (UTC)

[edit] Request to post

Hi David: I will not post this material for several reasons. The main one is that I still have not reached agreement with you over what is going on. If that agreement occurs, I will be happy to engage the changes on my own initiative. For the moment, we disagree about the statement "the ωXr term, where r is outwards, will lead to a v in the exact same direction as the v in the Coriolis term". I believe the dropping ball example refutes this remark, as clarified by addition of a figure in this example, as does the discussion you can find on line in John Robert Taylor (2004). pp. 349ff. ISBN 189138922X. Brews ohare (talk) 16:07, 18 May 2008 (UTC)

Brews, we need to look at the full equation and examine the physical meaning of the terms,

$\mathbf{a}_\mathrm{rot}=\mathbf{a} - 2\mathbf{\Omega \times v_\mathrm{rot}} - \mathbf{\Omega \times (\Omega \times r)} \,$

Going back to the derivation of the above equation, the $\mathbf{(\Omega \times r)}\,$ term in the centrifugal force term is one of the components of the velocity of the moving particle that is considered at the beginning of the derivation. In fact, it is the tangential component. In the above equation it effectively refers to the tangential speed of a point on the rotating frame. It can only take on that mathematical form if it is the tangential speed.

That is the essence of the argument. The centrifugal force only applies to a particle which is stationary at that point in the rotating frame, or more accurately to a particle which actually possesses that tangential velocity. And that is backed up by the GD Scott citation,

"The centrifugal forces must be recognized for what they are --- Non-Newtonian forces acting on masses at rest in a rotating frame."

The $\mathbf{v}_\mathrm{rot}$ term in the Coriolis force, as per the derivation, is the other component of the moving particle in question, and so it has to be radial. The Coriolis force and the centrifugal force are mutually perpendicular components of the velocity of a particle which has been routed through a point on a rotating frame of reference.

So there is no point in even discussing the hypothetical question of whether these two mutually perpendicular components of velocity would be in the same direction if they were to suddenly become parallel to each other.

In your example of a particle that is stationary in the inertial frame, its tangential velocity cannot be linked to either the $\mathbf{v}_\mathrm{rot}$ term in the Coriolis force or the $\mathbf{(\Omega \times r)}\,$ term in the centrifugal force.

There are no forces acting on the particle that is stationary in the inertial frame, and the artificial circle which it appears to trace out as viewed from the rotating frame, is not even worthy of discussion as a topic in physics. It's just like you yourself said "These unit vectors are attached to the rotating frame, not to the object under observation".

This so-called equivalence principle that denies the Bucket argument and claims that the artificial circle is equivalent to the real radial centrifugal force which occurs in real rotation is in fact a case of the modern view of centrifugal force being extrapolated to its ridiculous conclusion.

Your dropping of the ball demonstration illustrates a tangential artifact as viewed from the rotating frame which is similar to Coriolis force, but it is not Coriolis force. To get Coriolis force we need to go into a field where Kepler's law of areal velocity doesn't hold.David Tombe (talk) 16:39, 18 May 2008 (UTC)

David: Nothing in what you have said "drops the penny" for me. I am not unaware of how blind one can be when seeing things from one mind set, but I simply can't say that your arguments are moving me off center. I think you can see from the changes I have made in the various articles that you have precipitated a complete review of the equations and arguments on my part, and I haven't turned up anything more than imprecision in some of the presentation. Brews ohare (talk) 19:28, 18 May 2008 (UTC)

[edit] Discussion

David: I'll intersperse my reactions among your comments:

Brews, your Coriolis $\mathbf{v}_\mathrm{rot}$ ostensibly applies to the particle at rest in the inertial frame. And in your own words, you said that the equations don't apply to the particle under observation. Remember,

"These unit vectors are attached to the rotating frame, not to the object under observation"

The other velocity in the centrifugal force term $\mathbf{(\Omega \times r)}\,$ applies at a point fixed in the rotating frame.

Hence how can we sum together two forces that aren't even acting on the same point?

The above is not what I understand the formalism to say. Rather, both forces are ubiquitous in the rotating frame and apply to everything with mass. My reaction to your remarks is that you have not read the articles closely. If you have, it would be helpful if you pointed out how your interpretation of the articles is possible, so they can be changed to eliminate this possibility.

Why not pay heed to the GD Scott quote and realize that the transformation equation only applies at points that are rotating with the rotating frame. It does not apply to particles that are at rest in the inertial frame.

I find this isolated quote hard to interpret. Before attempting interpretation, I haven't looked into why GD Scott should be definitive. In my opinion, there are plenty of documented sources cited in the article that support its take on the matter. As I understand matters, you do not believe these sources have been correctly interpreted, which leads to a conundrum: we will be unable to agree on what the sources say, regardless of who the sources are or what they actually say. That is a predicament.

We need to have actual real rotation, relative to the background stars, in order to induce centrifugal force. And when it is induced we can see that it is a real outward radial effect. It can create hydrostatic pressure in a bucket of water.

There is no dispute here.

But the example that you have given involving a stationary object being observed to trace out a circle from a rotating reference frame is quite simply not an example of centrifugal force. A circular motion involves a zero radial acceleration. That is a geometrical fact. Your theory is trying to argue that a net inward centripetal force will yield a circular motion. I explained to you that centripetal force acts in opposition to outward centrifugal force and when they are balanced we get circular motion.

I am sorry about this David, but this statement is simply dead wrong and contradicts everybody's approach to this topic.

Your Cartesian example neglects to consider that the inward centripetal force is relative to a straight line motion which already contains a centrifugal force. From the radial perspective, they are both present and if one of them is real, then they are both real. David Tombe (talk) 19:54, 18 May 2008 (UTC)

Here again, you are entertaining an analysis of the situation that does not agree with customary terminology in today's textbooks on mechanics. It may be feasible to redefine all the ordinary terms so that your approach leads to the correct equations. However, the standard, ubiquitous approach has become standard and ubiquitous because it works. It is unlikely that we will return to the Ptolemaic approach (I'm trying to be funny here), even if it works too. The objective of the articles is to present the standard view of matters, not an historical view (assuming that is where your ideas come from). However, there is always room in an article for some historical development, such as you provided for Maxwell's equations. Would you care to try that avenue? Brews ohare (talk) 15:04, 19 May 2008 (UTC)

[edit] Continuation of discussion

Again, I'll intersperse my remarks:

Brews, in circular motion, the radial acceleration is zero. There is no dispute about that. However, some people here have been confused by the high school derivation of centripetal force in circular motion, which I can do standing on my head, as I have taught it myself to students on the blackboard. I know fine well about the vector triangle and the inward pointing dv vector in the limit as it tends to zero. But that is all relative to the Cartesian frame.

I do not agree with this remark, so count me with the confused. How do you square this comment with Newton's law of inertia? If the path is curved, if the direction of the velocity changes, there is a force. I'm entering a state of amazed disbelief at this point.

Brews, it's Newton's law of inertia that gives rise to the centrifugal force that counterbalances the centripetal force in circular motion.

In the radial direction, the acceleration will be zero. It is the radial direction itself that is rotating in the Cartesian frame.

Hence, the artificial circle theory is wrong because it tries to explain a circular motion with a net radial inward centripetal force.

Again, amazed disbelief.

On the other matter, which also all boils down to a vector triangle, you should look at the components of that vector triangle and see what they are referring to. The $\mathbf{(\Omega \times r)}\,$ term is the important term because it is the one in the centrifugal force. It is the component of the particle velocity leading between the fixed point in the inertial frame and a point on the rotating frame. It takes on that mathematical form in the limit as the triangle tends to zero (just like in your centripetal force derivation). It then physically refers to the tangential speed of a fixed point on the rotating frame.

You cannot apply those transformation equations to a point that is at rest in the inertial frame because such a particle has got zero velocity and hence both components of its velocity will be zero and hence the Coriolis force and the centrifugal force will be zero. There will be no force acting on the stationary particle in the inertial frame. It is not a centrifugal force affair in any respect.

Think about the Lorentz force. We need a real vXB to kick it off. We can't get it by rotating a magnet on its magnetic axis. Likewise with centrifugal force. We need a real vXω.

You were asking about the higher meaning behind the A vector. If you read Maxwell's stuff, you'll soon discover that EM is steeped in centrifugal force.David Tombe (talk) 15:30, 19 May 2008 (UTC)

Sorry, David. I do not think any further discussion is useful. Brews ohare (talk) 16:49, 19 May 2008 (UTC)

Brews, in a circular orbit, the inward gravity force (centripetal force) is exactly balanced by the outward centrifugal force. There is a net zero radial acceleration. This is an undisputed fact amongst anybody who has ever studied orbital mechanics.

On the transformation equations, never loose sight of what the mathematical symbols mean. The Coriolis force and the centrifugal force in those equations are the components of the actual polar acceleration of a particle in the inertial frame. If that particle is at rest, then those two components must exactly cancel.

The artificial circle theory is original research by an editor here and it is wrong. It is an over extrapolation of the transformation equations to a particle to which they don't apply. David Tombe (talk) 06:18, 20 May 2008 (UTC)

[edit] I'll respond here, since you are blocked again

Since you were blocked again I'll respond to you here for the moment

Timothy, G David Scott of the University of Toronto wrote about centrifugal force. He said,

"The centrifugal forces must be recognized for what they are --- Non-Newtonian forces acting on masses at rest in a rotating frame."

I'll try and get you the full name of the publication. David Tombe (talk) 16:36, 16 May 2008 (UTC)

I read your reference, and the phrasing is a bit awkward. The quoted statement is made in the context of refuting the approach describing the centrifugal force as a reaction to a centripetal force. (He very correctly notices that the reaction to a centripetal force is also centripetal force.) The main message of the quote is that that is not the centrifugal force. The specification that centrifugal force act on objects at rest in a rotating frame serves mainly to distinguish them from Coriolis forces which act on masses moving in a rotating frame.

What is clear is that Scott is not arguing that centrifugal forces do not apply to masses moving in a rotating frame, since he provides no further argument supporting this statement. Whereas he thoroughly argues the other points he is trying to make. (TimothyRias (talk) 10:09, 19 May 2008 (UTC))

I'll also respond to this:

Brews, I'll summarize my points of disagreement and then we'll have to go into to them each in more detail.

(1) The circular motion as viewed from the rotating frame is only an artifact. There is nothing real about it. No forces are acting at all. There is no inertia. There is no centripetal force.

(2) Coriolis force never acts in the radial direction. Coriolis force and centrifugal force are mutually perpendicular components of a parent vXω force. The former causes an east-west deflection on north-south motion. The latter causes a radially outward deflection on tangential motion.

(3) Even if we were to accept that the Coriolis force could act radially, we still end up with a net inward acceleration. That is not circular motion. The radial acceleration for circular motion must be zero.

This is borne out by orbital theory. The radial acceleration is the sum of the inward acting gravity force, which is the centripetal force, and the outward acting centrifugal force. When we have a circular orbit, these two cancel and the radial acceleration is zero.

(4)The signs in the transformation equations are based purely on the vector notation. We cannot know the actual signs until we know the real physical situation. And in this case, there is no real physical situation.

(5) I don't know where you found that theory about a fictitious centripetal force being composed of a fictitious outward centrifugal force and a fictitious inward Coriolis force twice as strong. Did you find it here on wikipedia? I can assure you that it is badly wrong.

(6) the transformation equations only apply to objects that possess the angular velocity in question. The derivation insists upon that.David Tombe (talk) 18:38, 16 May 2008 (UTC)

(1) This is true in a certain sense. But: a particle stationary in an inertial frame will really makes a circular motion in that frame. (I don't think you are denying this, so we probably agree thus far.) The consequence is that in this frame there must be an inward pointing centripetal force causing the circular motion. (At least when insisting on Newton's second law to descibe the kinemetics. (2) This statement is just plain wrong. I challenge you to provide either a source of a derivation of this claim. (3) This statement is wrong. When using Newton's second law to describe motion a circular orbit requires an inward pointing force. (This simply a consequence of the fact that the direction of the velocity is constantly changing, but the magnitude is not.)

Regarding the orbital equation in this context. The orbital equation is a special case of a co-rotating frame (with non-constant radial postion dependent angular velocity!) It can be obtained in variety of ways, most of which circumvent the use of forces altogether and derive the orbital equation straight from the action, using symmetry arguments the simplify the resulting EoM. Standing still in in this co-rotating frame requires that is this frame the sum of the forces on a object must zero. Of course standing still in this frame is the same thing as circular motion in the inertial frame. This has lead some authors in applied maths (Kreysig for example) to make the somewhat confusing statement that circular motion requires the net forces on a mass to be zero. This sentence commits a major physical sin by combining physical statements about different reference frames in a single statement. Physically more correct would be circular motion in the inertial frame requires the net forces- in the co-rotating frame- on a mass to be zero. Circular motion in the rotating frame (which would be non-circular motion in the inertial frame!) still requires a centripetal force. (4) The signs in the transformation equation follow directly from the derivation. (You must be careful to use consistent convention for the sign of the angular velocity though) (5)It is quite trivial fact, easily derived frame the transformation equation. Which is probably how most people here obtained it. Which might make it an OR issue in some sense. Finding a proper reference should be possible, although it might be hard to find because the example is to easy to use in most textbooks. But there must be at least one undergrad classical mechanics book that does this either as an example or as an exercise. (6) This is a statement that you have frequently repeated, but have never provided an arguments or references for.

It seems that your determination is based on a couple of very fundamental misconceptions. (TimothyRias (talk) 10:09, 19 May 2008 (UTC))

Timothy, you seem to be under the impression that I have never seen the high school derivation for centripetal force in circular motion. I am quite familiar with the derivation and the arguments that go with it about the centripetal force causing a change in direction relative to the Cartesian frame. But there is absolutely no need to cloud the issue with Cartesian coordinates. The entire issue can be understood in polar coordinates. In fact, polar coordinates are really the only viable system for analyzing central force situations.

At the end of the day, when an object is moving in a circle, its radial acceleration is zero. And when we do orbital theory, we see that this state of affairs occurs when the inward centripetal force (gravity) is exactly counterbalanced by the outward centrifugal force.

In the artificial circle example, which is being pushed as the flagship piece of original research in the article, you have a net inward radial force and so that does not correspond to circular motion.

The next point is that you all seem to have lost track of the physical meaning behind the maths symbols in the transformation equations.

What in your mind is the physical meaning of the term $\mathbf{(\Omega \times r)}\,$ ?

Where did it come from?

In my mind, it refers to the tangential speed of a fixed point in the rotating frame. That would make GD Scott's quote correct. Centrifugal force applies to objects that are at rest in a rotating frame of reference.

Finally, the artificial circle thing is indeed original research by one of the editors here. You cannot get off the hook on the citations requirement by arguing that it is too trivial to be in any textbooks. This original research, which is false, is dominating the entire article at the expense of any mention of the real radial effects that are induced by actual rotation. David Tombe (talk) 12:55, 19 May 2008 (UTC)

About polar coordinates. Yes they are very useful when dealing with central forces. But you must realize that Newton's second law takes a very unconventional form when written in these coordinates. This makes the concept force in this situation very awkward. Anyway, looking at any coordinate system clouds the issue. Physics is properly expressed in coordinate free quantities. If you do this it is clear in circular motion the velocity is not constant. (in technical terms: parallel transporting a tangent vector at some point p along a circular path to some other point q, does not yield the tangent vector in q). Even in polar coordinates it is easy to show that there must be a centripetal force when the motion is circular. Assuming a conservative force the radial part of the EoM is (easily derived from the action via the Euler-Lagrange equations):

$F_r = m\ddot{r} - m r \dot{\phi}^2$

The phi part of the EoM will generally contain all sorts of nasty stuff, but when all forces are conservative and central (i.e. the corresponding potentials are independent of phi.) Then it takes the very simple form:

$F_\phi = mr^2\ddot{\phi},$

expressing conservation of angular momentum. It is easy to sea that for circular motion (r is constant phi = w t) these EoMs imply that $F r = - m r ω 2$ and $F φ = 0$ . That is there must exist an inward pointing force that cause the circular motion.

I finally come to your question:

What in your mind is the physical meaning of the term $\mathbf{(\Omega \times r)}\,$ ?

Where did it come from?

Indeed $(\vec\Omega \times \vec{r})\,$ is the speed of the the point $\vec{r}$ w.r.t. the inertial frame. (Where does it come from? Technically, when it arises in equations it tends to come from the non-trivial part of then covariant derivative in rotating coordinates.) Adding $(\vec\Omega \times \vec{r}(t))$ to the velocity $\vec{v}(t)$ of a particle in the rotating frame yields the velocity of the particle in the inertial frame. This holds for any path $\vec{r}(t)$ in the rotating frame, including those that already have tangential speed. So, yes, Scott's statement is true, but please note that he does not say that centrifugal force only applies to particles at rest in the rotating frame. He does not say so, because in fact centrifugal force applies to all objects in a rotating frame. (TimothyRias (talk) 14:22, 19 May 2008 (UTC))

Timothy, let's deal with you last point first. GD Scott DOES say that centrifugal force only applies to particles at rest in the rotating frame. How can you possibly deny that when it is written down clearly for us all to see? Your logic was unbelievable. You said "He does not say so, because in fact centrifugal force applies to all objects in a rotating frame.".

In other words, because you disagree with what GD Scott says, you are now telling us that that means that he didn't say it?

On the other points, the mathematical expression $(\vec\Omega \times \vec{r})\,$ only applies when that component of the velocity of the particle is tangential. It is basic calculus. That is the expression which applies only in the limit as the vector triangle tends to zero. Hence it is the tangential component. Hence the other component must be the radial component.

Finally, on orbital theory, the radial force is the sum of the outward $m r \dot{\phi}^2$ centrifugal force and the inward centripetal acceleration (which is gravity). When these two forces are balanced all the time, we will have a circular motion and the radial acceleration will be zero. There is no net inward radial centripetal force acting in circular motion. Kreysig is right on that point. High school students don't get it emphasized to them that the changing velocity due to changing direction, is relative to the Cartesian frame, but that in polar coordinates, the centripetal force is balanced by a centrifugal force in the radial direction. David Tombe (talk) 15:11, 19 May 2008 (UTC)

It has nothing to do with polar coordinates! So long as the frame is inertial (whether polar, cylindrical, cartesian, spherical-- doesn't matter what coordinate system you use), moving in a circle requires a radial acceleration, and this requires a radial force. That's provided by a rope or gravity, and in this inertial frame, that's the ONLY force there is. Like all forces it has two ends (one end acts on the central body, the other end acts on the rotating body), but that's it. A free-body diagram on just the body moving in a circle shows just ONE force acting on it, end of story. IN THE ROTATING FRAME (again what coordinates you use doesn't matter) now the same body has two forces on it, balanced. Thus, it does not accelerate and in this frame, it doesn't MOVE. S B H arris 00:46, 20 May 2008 (UTC)

SBharris, it has got everything to do with radial motion. And in a circular orbit there is a net zero radial acceleration. The inward gravity force is exactly balanced by the outward centrifugal force. Anybody who has studies orbital mecahnics knows this. David Tombe (talk) 06:12, 20 May 2008 (UTC)

It has nothing to do with orbital mechanics, which in the limit of a circular orbit, is just one more kind of circular motion. All circular motions are accelerated motions. The fact that the radius doesn't change, does NOT mean the object isn't being accelerated radially! Of course it is, but the change in radius due to not moving tangentially exactly makes up the radial change, so the residual is zero. Astronauts in Earth orbit are still inside 90% of the Earth's g field. The reason they feel weightless is that they're falling (accelerated) all the time. An astronaut is moving toward the horizon so fast that the Earth drops out from underneath him/her as it curves away. In order to keep a constant distance, the astronaut must fall toward the Earth by just the amount to adjust that tangent distance so it's not increasing, but constant. It takes a fall, an acceleration, to do that. That acceleration is no different than your first second as a skydiver, except in space you're not just going DOWN, but also horizontally. IF you go horizontally fast enough that the distance to the Earth never decreases, even though you continue to fall toward it, you're in circular orbit. Work it out. Even easier to see is the case of a bug on a record turntable. If there's no radial force inward on the bug, all the time, it will not move in a circle, but will slide outward and off. S B H arris 06:31, 20 May 2008 (UTC)

SBharris, I know all that. I taught it on the blackboard many years ago. The centripetal force changes the direction of an object away from its inertial path. If there was no centripetal force the object would continue in a straight line.

Yes. And that's all there is too it. In the inertial frame, the centripetal force causes it to go in a circle instead of a line, and that it.S B H arris 21:32, 20 May 2008 (UTC)

But now examine that straight line with respect to the same origin. You will find that there is an outward centrifugal acceleration. The centripetal force that makes it go in a circle is merely cancelling out with the already existing centrifugal force.

I have no idea what you mean "with respect to the same origin". There is no centrifugal acceleration (or force) unless you move to a rotating coordinate system. Any inertial system gives you just the one force. You do not "cancel out" two forces without resulting in zero acceleration, and an object moving in a circle is accelerated, as seen in an inertial frame. It's only when you mentally tranfer to the object's frame that a second force appears, and that's because in THAT frame, the object isn't accelerating.S B H arris 21:32, 20 May 2008 (UTC)

In fact, in an elliptical orbit, it doesn't even cancel it in general. Sometimes the centrifugal force is greater and sometimes the centripetal force is greater. But I can assure you that when the two are balanced, the net radial acceleration will be zero. David Tombe (talk) 16:32, 20 May 2008 (UTC)

In an elliptical orbit you STILL only have ONE force (the force of gravity), UNLESS you choose to view it from a rotating frame. But stop doing that! S B H arris 21:26, 20 May 2008 (UTC)

Question: We have a particle moving with trajectory $\vec{r}(t)$ in a frame rotating with angular velocity $\vec\Omega$ . What is, according to you David, the velocity of this particle in the inertial frame? My answer has been clear on this point, the velocity in the inertial frame is $\dot\vec{r}(t)+\vec\Omega\times\vec{r}(t)$ , the velocity of in the inertial frame at time t is the velocity of particle in the rotating frame at time t plus the velocity of the rotating frame at the postion of the particle at time t. According to you this is only true if the particle is co-rotating (without any argument of why this would be), so my question to you is, what is the expression for the general case. (TimothyRias (talk) 08:02, 20 May 2008 (UTC))

Timothy, the velocity of the particle in the inertial frame is the sum of the velocity of a fixed point in the rotating frame and the velocity of the particle relative to that fixed point in the rotating frame. The centrifugal force acts at the fixed point in the rotating frame which is obviously co-rotating. David Tombe (talk) 13:08, 21 May 2008 (UTC)

And this is where you are wrong. Velocity of the particle is the sum of the velocity of its current position (which need not be fixed) in the rotating frame (wrt to the inertial frame) and the velocity of the particle in the rotating frame (wrt this point). (TimothyRias (talk) 14:11, 21 May 2008 (UTC))

Timothy, it has to be fixed for the velocity term in the Coriolis force to mean the velocity relative to the rotating frame. David Tombe (talk) 09:49, 22 May 2008 (UTC)

[edit] Links now work

Sorry that the links on my TALK page, now fixed, didn't work. They contained a pipe which I had to remove. Anyway, here are two derivations: [4] is the arXive Euclidian, and [5] is a standard one. In neither one is the approximation you mention made. The velocity vector is completely general and free. BTW, the centrifugal force in both of these is seen to be a vector quantity also, but this is usually not a problem since the direction of the radial vector to the rotating particle is generally defined to be the shortest one orthogonal to the rotation axis, so that ω X (ω X r) = ω²r (unit vector in the radial direction). The last paper has a good analysis which should benefit you, including a merry-go-round example: a quote: The radial Coriolis force associated with azimuthal motion is much like an increase or slackening of the centrifugal force and so is not particularly difficult to compensate. However, the azimuthal Coriolis force associated with radial motion is quite surprising, even assuming that you are the complete master of this analysis. Heh. Which apparently you aren't, yet. S B H arris 00:06, 20 May 2008 (UTC)

SBharris, I have just looked at both of your links. In both cases, contrary to what you say above, the derivation requires that the term $\mathbf{(\Omega \times r)}\,$ only applies in the limit. In the former paper (arXive) it occurs at equation (13). In the latter paper (J price) this is clearly stated in the last sentence of the wording beside fig. 5 on page 14 of the pdf file.

This means that the $\mathbf{(\Omega \times r)}\,$ term must be exclusively a tangential term. Hence the other v term in the Coriolis force must be a radial term. In other words, these transformation equations only apply to co-rotating radial motion such as would occur if an object were to be constrained to move in a radial groove on a rotating turntable.

As regards circular motion, it appears that both you and Brews are unfamiliar with planetary orbital theory. A circular motion occurs when the inward gravity force (centripetal force) always exactly cancels with the outward centrifugal force.

You both insist that a circular motion involves a net inward centripetal force. If that were so, the radius would shrink and we wouldn't have a circular motion.

You are basing your beliefs in this regard upon the restricted high school derivation of centripetal force in circular motion. The force needed to rotate the velocity vector inwards shows up to be the standard expression for centripetal force. However, if we examine the exact same vector diagram for a straight line motion measured relative to the same origin, we will obtain the exact same expression, only radially outwards. The centripetal force in circular motion is therefore merely cancelling with the centrifugal force. This fact is totally overlooked in high school physics courses.

You must never lose track of the physical meaning behind the maths symbols which you use. These transformation equations cannot be applied to a particle which is stationary in the inertial frame. The entire derivation of those equations is based on the actual components of its actual acceleration. One component leads to the Coriolis force and the other component leads to the centrifugal force. If there is no actual acceleration, then these two components will both be either zero, or equal and opposite. There will be no net inward centripetal force no matter what frame of reference we view it from.

As regards the merry-go-round example, it actually undermines your point and bolsters the very point that I have been trying to make. The quote The radial Coriolis force associated with azimuthal motion is much like an increase or slackening of the centrifugal force tends to confirm my view that the so-called radial Coriolis force is in fact a centrifugal force and that in meteorology it would invoke Archimedes' principle and invoke a torque on a cyclone, due to differential centrifugal force on the east-west wind at one side, and the west-east wind at the other side. David Tombe (talk) 06:01, 20 May 2008 (UTC)

The $\mathbf{\Omega \times r}$ term ( $\mathbf{\Omega \times X'}$ in the notation used in Price's paper), representing the contribution to the motion of that point generated by the relative rotation of the frames, is indeed, as you say, tangential at any given instant. But its time derivative is not: this is because, although it is tangential at any given moment, that tangent is continuously rotating. Taking the time derivative generates two acceleration terms, corresponding to the Coriolis and centrifugal accelerations respectively. Note that the two terms do not correspond to the radial and tangential components of force; they are instead the terms corresponding to X' and its time derivative respectively, and that there is no requirement for the Coriolis acceleration to be tangential. See equations 19 to 23 of Price's paper for the derivation. You're also right that Price says that "The radial Coriolis force associated with azimuthal motion is much like an increase or slackening of the centrifugal force" (page 24) -- this is indeed how it feels. But Price says "much like" for a reason; saying that something is "much like" something else is not at all the same thing as that it is the same thing. -- The Anome (talk) 11:00, 20 May 2008 (UTC)

Anome, I didn't get your point. The argument was about whether or not $\mathbf{\Omega \times r}$ is tangential. Now we are agreed that it is tangential. And if it is the tangential component of the velocity of the particle, then the other component which leads away from that fixed point on the rotating frame, can only be the radial component. If a velocity has two components and one is tangential, then the other must be radial.

Hence, Coriolis force only acts on radial motion.

And now you can see that the centrifugal force acts at that fixed point on the rotating frame whose velocity is $\mathbf{\Omega \times r}$ . So how are you going to deny the GD Scott reference which clearly says that the centrifugal force is a force that acts on stationary points within a rotating frame of reference?

And remember, it was for that citation which trumps all your other irrelevant citations, that I got blocked.

I think we need to start getting real about what is actually stated in all these citations. And we need to start getting real about the application of the transformation equations. They don't apply to particles that don't have a component of velocity that is connected to the angular velocity in question.

You are pushing original research which is false. No textbook talks about an artificial circle being caused by a summation of fictitious forces. David Tombe (talk) 11:30, 20 May 2008 (UTC)

David, there is a world of difference between stating that a component ( $\mathbf{\Omega \times r}$ ) is tangential and that it is the tangential component. The first is obviously true (by definition of the cross product)and nobody is denying this. The second is not. A velocity vector may have more than one component that is tangential. If I write $\vec{a}=\vec{b}+\vec{c}$ this does not imply that $\vec{b}$ and $\vec{c}$ are perpendicular. For all we care they could be co-linear. (TimothyRias (talk) 11:39, 20 May 2008 (UTC))

Timothy, It is the tangential component. At any rate, even if the two components were colinear, in your artificial circle example involving the stationary paticle in the inertial frame, the two components would cancel out. They wouldn't lead to a net inward force. David Tombe (talk) 12:12, 20 May 2008 (UTC)

In the artificial circle example the result is a net tangential velocity in the rotating frame. Taking the cross product with Omega then (by definition of the cross product) produces a radial term. (and no it is not the tangential velocity. If you think it is prove it!) (TimothyRias (talk) 12:29, 20 May 2008 (UTC))

Timothy, that net tangential velocity fits into neither the Coriolis force term nor the centrifugal force term. To fit into the former, it needs to be radial. To fit into the latter, it needs to be stationary in the rotating frame.David Tombe (talk) 09:53, 22 May 2008 (UTC)

[edit] Centripetal force

You also state that "You both insist that a circular motion involves a net inward centripetal force. If that were so, the radius would shrink and we wouldn't have a circular motion." Does this mean that you disagree with the proposition, universal to all treatments of classical mechanics,^[1]^[2]^[3]^[4]^[5]^[6] that a body in circular motion with constant radius and angular velocity is subject to a constant inward acceleration, and thus, according to Newton's second law of motion, it must therefore be subject to an inward force? -- The Anome (talk) 11:20, 20 May 2008 (UTC)

That is exactly what David has said at least once: see User_talk:David_Tombe#Continuation_of_discussion. 11:44, 20 May 2008 (UTC)Brews ohare (talk)

Anome, no I'm not denying that at all. The constant inward force cancels the outward centrifugal force that is inherent in inertia. There is no net radial acceleration in circular motion. Just consider the circular gravity orbit. There is an inward gravity force in the radial direction which is exactly counterbalanced by an outward centrifugal force.

Have you ever done orbital mecahnics? Have you seen the equation? There is an inward gravity force added to the outward centrifugal force and the result is the radial acceleration. The radial acceleration is zero when the two forces cancel.

How could you possibly have circular motion with an actual radial acceleration? If r double dot was non-zero, then the radius would be changing. David Tombe (talk) 11:36, 20 May 2008 (UTC)

Just to repeat the standard derivation, taking orthogonal components of x, the position vector relative to the center of rotation:

x = cos(omega t)
y = sin(omega t)
  implies (by differentiation)
xˈ = - omega sin(omega t)
yˈ = omega cos(omega t)
  implies (by differentiation)
xˈˈ = - omega^2 cos(omega t) = - omega^2 x
yˈˈ = - omega^2 sin(omega t) = - omega^2 y

Therefore, if x = (x, y), xˈˈ = - omega^2 x, which is (a) nonzero, and (b) radially inwards.

If you disagree with the above, am I therefore correct in my understanding that what you are saying is that everyone else involved in editing this article is wrong about the standard interpretation of classical mechanics, and unanimous in their misinterpretation of all the references they have cited to support their position, and you alone are right? Have you considered writing a paper about this, and putting it up for publication in a peer-reviewed physics journal? -- The Anome (talk) 11:41, 20 May 2008 (UTC)

Anome, that derivation is absolutely correct. I can do it standing on my head. I have taught it to students on the blackboard. Yes, we need such a inward centripetal force in order to have circular motion.

If we didn't have that centripetal force, we would have straight line motion.

Now do the derivation over again but this time for straight line motion referenced to exactly the same origin. This time the change in direction points outwards.

There is an outward radial centrifugal force built into the straight line motion. That is inertia. The centripetal force cancels it in circular motion and we are left with a net zero radial acceleration.

Bzzt. What are you talking about?? There is a gonzo idea, and it doesn't even come from moving in a circle. If you have some idea that there is an opposing "force" built into straight line acceleration, that can only mean you've (perhaps unconsciously) translated to the accelerated coordinate system of the accelerated body. Don't do that. A body moving in a straight line being pushed on, has ONE force acting on it, from the inertial frame. "Inertia" does not generate force. In this frame, it merely sets the acceleration F = m a. It's given by the mass.S B H arris 21:57, 20 May 2008 (UTC)

You would have known all of this if you had studied planetary orbital theory. Gravity is the centripetal force and it works in opposition to a centrifugal force. You cannot solve the planetary orbit differential equation with gravity alone. You need that outward radial centrifugal term. David Tombe (talk) 12:07, 20 May 2008 (UTC)

Strangely enough, I do remember studying classical orbital mechanics, in the "catch up with what you learned at high school" bit right at the start of my physics degree course. No doubt my memory is confused by all the piles of stuff about bras and kets and tensors and Feynman diagrams and symmetries and fields and cosmology that came later and, you know, more stuff about Schwarzschild this and Hawking that, that probably blew all that real physics knowledge clean out of my head... -- The Anome (talk) 12:16, 20 May 2008 (UTC)

David, the radial acceleration in polar coordinates is NOT $\ddot{r}$ ! It is a vector quantity $(\ddot{r}-\dot\phi^2r)\hat{r}$ , where $\hat{r}$ is the radial pointing unit vector. So even though $\ddot{r}$ is zero for circular motion, the radial acceleration is not! Basic physics, really. (TimothyRias (talk) 12:10, 20 May 2008 (UTC))

Timothy, I am fully aware of that expression. The centrifugal force term in that expression adds to the gravity force to obtain the r dot dot term. The r dot dot term is the second time derivative of the radial distance. If you don't want to refer to r dot dot as the radial acceleration, that is fine. We don't need to play on words. The r dot dot term is zero in a circular orbit.

In your artificial circle example, the r dot dot term is not zero. David Tombe (talk) 12:30, 20 May 2008 (UTC)

Sure it is. It is a circle, thus the radial distance is not changing. Artificial or not a circle is a circle. (or are you claiming that the artificial circle is not a circle?) (TimothyRias (talk) 12:50, 20 May 2008 (UTC))

Timothy, that's the whole problem. It should be zero. But if you have a net inward radial r dot dot, then it will not be zero. Sow me your equation for the artificial circle.David Tombe (talk) 13:21, 20 May 2008 (UTC)

Here we go. In polar coordinates the EoMs for a free particle moving in a rotating frame (angular momentum) are:

$\begin{align} 0 &= \ddot{r} - r(\dot\phi+\omega)^2 = \ddot{r} -r\dot\phi^2 -r\omega^2 -2r\omega\dot\phi \\ 0 &= \ddot\phi \end{align}$

A particle that starts out as stationary in the inertial frame, begins with angular velocity $\dot\phi(0) = -\omega$ in the rotating frame. So the second equation implies that

φ(t) = - ω t

and $\dot\phi(t) = -\omega$ . Plugging this into the first equation yields:

$0 = \ddot{r} - r\dot(\phi+\omega)^2 = \ddot{r} -r\omega^2 -r\omega^2 +2r\omega^2 = \ddot{r}$

Hence

r (t) = r (0)

. The particle at rest in the inertial frame moves in a clockwise circle in the rotating frame with angular velocity

ω

against the direction of rotating. This all o so simple.

In the first equation the $-2r\omega\dot\phi$ term is the Coriolis term. It is this term that cancels the centrifugal terms.(TimothyRias (talk) 14:05, 20 May 2008 (UTC))

(note: there was a minor slip up in the above derivation the EoM for phi of course is $0 = r^2\ddot\phi + 2r\dot{r}(\dot\phi+\omega)$ . It is not hard to check that the above solution also solves this equation. (the extra term vanishes when r is constant.) So, no harm was done. But it was sloppy. (TimothyRias (talk) 07:41, 22 May 2008 (UTC)))

Timothy, now you are contradicting your colleagues. They are arguing that the Coriolis force is twice as strong as the centrifugal force and that the combination results in a net inward radial acceleration. That is what I am objecting too. David Tombe (talk) 17:20, 20 May 2008 (UTC)

You are again confusing polar coordinates with vectors. The translation to vectors from polar coordinates is always a bit obscure. The acceleration (vector!) of a particle in polar coordinates is $(\ddot{r}-r\dot\phi^2)\hat{r}$ . In the above problem this thus gives a net acceleration of

- r ω 2) h a t r

which is constantly pointing inward, as everybody here accept you has been claiming. (This has nothing to do with Cartesian coordinates as you seem to think.) (TimothyRias (talk) 07:41, 22 May 2008 (UTC))

David, as far as I can see it, you are interpreting the tangential velocity of the "co-rotating" object from the viewpoint of an inertial frame in which the centre of the circular path is stationary (in which it is is moving in a circle at a constant speed), and its radial acceleration from the viewpoint of a rotating frame rotating at the same rate and direction as the object's circular motion (in which appears to be stationary, but is apparently subject to pseudoforces). You can't mix coordinate frames like that: it's unphysical; you have to pick one coordinate frame at any given time, and stick to it. -- The Anome (talk) 12:56, 20 May 2008 (UTC) [Updated 13:24, 20 May 2008 (UTC)]

Anome, there was no question of me mixing coordinate frames. The centrifugal force expression contains a wXr term. That term refers to the fixed point on the rotating frame. Check the original derivation. The derivation begins with a vector triangle for two components of the actual velocity of a particle relative to the inertial frame.

The first component wXr leads from the origin in the inertial frame to the fixed point on the rotating frame. It is at that point that the centrifugal force acts, just as per GD Scott's citation. David Tombe (talk) 13:29, 20 May 2008 (UTC)

Vectors don't point from one point in one frame to some other point in an other frame. That statement is pure gibberish. Most certainly it doesn't point from the origin in any frame since it is an element of the tangent space at the point r. The map v->v + wXr is the pullback to the tangent space at the point r of the map sending the inertial to the rotating frame. This type of confusion is very common among people that don't understand the physically significant difference between a space and its tangent space. (especially because in classical mechanics the two are mathematically isomorphic is vector spaces. (TimothyRias (talk) 13:45, 20 May 2008 (UTC))

Timothy, in the derivation, there are three important points that make up a triangle.

(1) the origin in an inertial frame. (2) the point on the rotating frame. (3) the point where the particle is.

(1) to (2) becomes the tangential velocity wXr in the limit. (2) to (3) then becomes the radial velocity.

That is all there is to it. There is no need to involve Lagrangian, or Hamiltonian, or matrix algebra, or three body problems in celestial mechanics, or words like isomorphic or vector space.

The tangential component is the one that is relevant for the centrifugal force and it applies at the fixed point on the rotating frame. David Tombe (talk) 13:59, 20 May 2008 (UTC)

Your statement is a bit limited as it is not able to describe the general situation. Let write it down a little more generally (I'll use the somewhat flawed language confusing position space and its tangent spaces you seem to be used to). We have a particle following some trajectory. Act an given time t we have the following vectors.

We express its position in the inertial frame by $\vec{r}_{in}(t)$ a vector pointing from the origin in the inertial frame to the current position of the particle in the inertial frame.
The current position in the rotation frame is expressed by $\vec{r}_{rot}(t)$ a vector pointing from the origin in the rotating frame to the current position in the rotating frame.
The current velocity of the particle in the inertial frame is $\vec{v}_{in}(t)$ , is a vector point from the current position of the particle in the inertial frame to the position a infinitesimal time $δ t$ later. This means that $\vec{v}_{in}(t)=\frac{\vec{r}_{in}(t+\delta t)-\vec{r}_{in}(t)}{\delta t}$ .
The current velocity of the particle in the rotating frame is $\vec{v}_{rot}(t)$ , is a vector point from the current position of the particle in the rotating frame to the position a infinitesimal time $δ t$ later. This means that $\vec{v}_{rot}(t)=\frac{\vec{r}_{rot}(t+\delta t)-\vec{r}_{rot}(t)}{\delta t}$ .

Now how do we relate $\vec{v}_{in}(t)$ to $\vec{v}_{rot}(t)$ . For this we need one more vector:

The "transformation" vector $\vec{\rho}(t)= \vec{r}_{rot}(t)-\vec{r}_{in}(t)$ the vector pointing from the current position of the particle in the inertial frame to its current position in the rotating frame.

We can relate the two velocities in the following way

$\begin{align} \delta t\vec{v}_{rot}& =\vec{r}_{rot}(t+\delta t)-\vec{r}_{rot}(t)\\ &= \vec{r}_{in}(t+\delta t)-\vec{r}_{in}(t) - (\vec{r}_{in}(t+\delta t)-\vec{r}_{in}(t)) + \vec{r}_{rot}(t+\delta t)-\vec{r}_{rot}(t)\\ &= \delta t\vec{v}_{in} + (\vec\rho(t+\delta t)-(\vec\rho(t)) \end{align}$

The term $\frac{(\vec\rho(t+\delta t)-(\vec\rho(t))}{\delta t}$ is the velocity of the rotating frame wrt to inertial frame. In is not hard to show that in the limit that

δ t

goes to zero, it is expressed by the crossproduct $\vec\omega\times \vec{r}_{in}$ . It is clear that this quantity is completely of the value of $\vec{v}_{in}(t)$ and only depends on the current position of the particle and the angular velocity of the frame. In particular it is not the tangential component of $\vec{v}_{in}(t)$ . This can be easily seen if we choose the particle to be stationary in the inertial frame, in which case $\vec{v}_{in}(t)=0$ but $\vec\omega\times \vec{r}_{in}$ is not. (TimothyRias (talk) 15:23, 20 May 2008 (UTC))

Timothy, if we choose the particle to be stationary in the inertial frame, then we don't have a derivation. The whole derivation is based on the components of the particle if it is moving. David Tombe (talk) 17:17, 20 May 2008 (UTC)

Why? (TimothyRias (talk) 17:46, 20 May 2008 (UTC))

The derivation works just fine when the object is stationary in the inertial frame. $\vec{r}_{in}(t)$ is then just a constant vector. Consequently, $\vec{v}_{in}(t)$ is zero. This is perfectly fine. What I see here, is you grasping at straws when confronted with actual arguments that completely refute your claims. Any time this happens, you either respond with total nonsense or you start complaining about something totally irrelevant like the formalism being complicated. Has it never crossed your mind dead you actually may be just dead wrong. People make mistakes, it is natural. (TimothyRias (talk) 08:14, 21 May 2008 (UTC))

No Timothy, if the particle is stationary then it will have no components of acceleration to constitute the centrifugal force and the Coriolis force. And we certainly won't have two components that sum to give a net radial acceleration as you need for your artificial circle. David Tombe (talk) 13:22, 21 May 2008 (UTC)

You are changing the subject. The above derivation show that the term $\vec\omega\times \vec{r}_{in}$ is not the tangential component of the velocity. It is merely the difference between the velocity in the inertial frame and the velocity in the rotating frame. Since it is the difference it is completely independent of the actual value of the velocity in either frame. (TimothyRias (talk) 13:58, 21 May 2008 (UTC))

No Timothy, it is the velocity of a fixed point in the rotating frame measured relative to the inertial frame. David Tombe (talk) 10:41, 22 May 2008 (UTC)

[edit] So, which is it?

David, since everyone else here appears to disagree with you about, well, just about everything regarding this, can you tell us whether, in your opinion, everyone else in this discussion is wrong and you are right? Or can you suggest another possible explanation for their disagreement with your ideas? -- The Anome (talk) 12:34, 20 May 2008 (UTC)

Anome, they are wrong in key points.

(1)The transformation equations don't apply to particles that are not involved in the angular velocity in question. The centrifugal force applies to particles fixed in the rotating frame that possess the tangential speed given by the wXr term in the centrifugal force.

This is backed up by the GD Scott reference which is the only citation which has directly addressed the issue, and which has been totally swept under the carpet.

The theory about applying the transformation equations to a stationary particle in the inertial frame and accounting for a centripetal force on the artificial circle is the original research of somebody amongst the editors here and it is not to be found in any textbook.

(2) Most people here seem to be unfamiliar with orbital theory. If they knew about orbital theory, they would know that gravity is an inward acting centripetal force that acts in opposition to an outward acting centrifugal force. When the two are balanced, the second time derivative of the radial distance (call it what you like) vanishes.

It strikes me that there are too many people here that are keen to write about this subject without having a full comprehension of it. David Tombe (talk) 13:06, 20 May 2008 (UTC)

David, please read the reference by Burgel, B. (1967). "Centrifugal Force". American Journal of Physics 35: 649. It talks quite explicitly about what you call the "artificial circle" example and how it is explained using the sum of the centrifugal and Coriolis forces. Regarding the Scott reference--yes, he says that centrifugal force acts on objects that are at rest in a rotating reference frame. But he doesn't say only on objects at rest, because that wouldn't make any sense. Otherwise, when you are standing on the Earth, you are subject to a centrifugal force, but when you start walking West you suddenly don't feel the centrifugal force anymore because you are "disconnected" from the Earth's rotating frame of reference? I suppose you would create a new frame of reference where you are no longer rotating at the Earth's speed but slightly slower, with a smaller centrifugal force. But no one else does that. Everyone else keeps using the Earth's frame of reference and the same centrifugal force, which does not involve velocity in the equation, only the distance to the axis of rotation and the angular velocity of the frame of reference. --Itub (talk) 13:08, 20 May 2008 (UTC)

Bingo. It's quite difficult, because of course the component of Coriolis force which is in the same direction the centrifugal force, can be SEEN as a new centrifugal force, just in a different rotating system (one with a different rotational rate). We choose to define it differently, keeping the old rotational system rate, mostly because it's convenient for places like planets. If I walk around the Earth's equator in the direction of longitude (or along any line of longitude, actually), I experience a new force which makes me lighter or heavier. It's convenient to call that the radial Coriolis force component, since we're already assigning "Coriolis" to any forces that I feel as a result of moving about on the planetary surface with a velocity with respect to land or water. If I get a force for moving along lines of latitude (which pushes me longitudinally, due to my change in distance from the Earth's center) then it's easy mathematically to say a similar walk longitudinally pushes me "up" (away from the ground), or "down" (makes me heavier). A combination of the two velocities gives me a combination of the forces, and that's the cross product. S B H arris 22:23, 20 May 2008 (UTC)

SBharris, the Coriolis force and the centrifugal force never double up. They are both mutually perpendicular components of a parent force vXω. David Tombe (talk) 10:43, 22 May 2008 (UTC)

David, if we were to get an actual professor of theoretical physics to check the article out, like oh, say, John Baez [6], would that help? -- The Anome (talk) 13:13, 20 May 2008 (UTC)

David, you are taking the GD Scott reference out of its context. (Which makes me wonder if you have actually read the article) The article is basically an essay explaining why the fictitious centrifugal force is the only useful concept worth teaching in high school. The one sentence quote you posted is refuting that the reactive centrifugal force is a centrifugal force. The at rest remark in this context is there to avoid discussing Coriolis forces which are completely out of context in that article. If we were to take the quote in the literal way you are taking it (i.e. that the centrifugal force works only on particle at rest in a rotating frame), then you should also conclude that it does not work on particles moving radially in a rotating frame. This would spell disaster for the orbital equation you are so fond of. (TimothyRias (talk) 13:30, 20 May 2008 (UTC))

Timothy, the centrifugal force acts on the actual tangential speed of a particle irrespective of reference frames.David Tombe (talk) 13:37, 20 May 2008 (UTC)

David, are you saying that tangential speed is independent of coordinate frames? If so, how does this square with your belief that centrifugal force is a real, coordinate-system independent force? Again: David, if we were to get an actual professor of theoretical physics to check the article out, like oh, say, John Baez [7], would that help? -- The Anome (talk) 13:44, 20 May 2008 (UTC)

David, there is no such thing as an actual speed independent of the frame of reference. Or are you claiming that there is no such thing a Galilei transformation in classical physics (or Lorentz transformations in relativistic physics.) The simple fact is that changing the reference frame changes the velocity of particle. To find the velocity in the new frame you simply subtract the velocity of the new frame (wrt to the old frame) at the current position of the particle from the current velocity in the old frame. How hard is this to understand.

Also the centrifugal force is completely independent of the speed of a particle. It only depends on the current position and the current angular velocity of the rotating frame. Again this is so very simple. (TimothyRias (talk) 14:36, 20 May 2008 (UTC))

Anome, a tangential speed only needs a point origin. David Tombe (talk) 13:52, 20 May 2008 (UTC)

I get the impression that David believes the following: that centrifugal force is (a) a real, Newtonian, coordinate-system-independent force, and (b) applies only to objects which are stationary from the viewpoint of a rotating coordinate frame. Apparently, if I read his comments correctly, you have to choose the correct "co-rotating" rotating frame for each object; effectively, each object gets to obey its own individual set of physical laws. Again: David, if we were to get an actual professor of theoretical physics to check the article out, like oh, say, John Baez [8], would that help? -- The Anome (talk) 13:42, 20 May 2008 (UTC)

Anome, that is a reasonably accurate assessment of my views apart from your statement effectively, each object gets to obey its own individual set of physical laws. There is one general explanation for all induced centrifugal force.

These views are borne out by orbital theory, the centrifuge and the GD Scott reference. Also by Maxwell's use of the concept in explaining magnetic repulsion.

In reply to your suggestion, you need to find somebody who knows the importance of not neglecting the physical significance behind the mathematical symbols. I don't know enough about John Baez to make any judgement on your proposal. He might be fine for the purpose. David Tombe (talk) 13:50, 20 May 2008 (UTC)

Good. If you'd like to see his credentials, you might want to look here: http://math.ucr.edu/home/baez/vita.pdf -- The Anome (talk) 13:57, 20 May 2008 (UTC)

Again you are misrepresenting GD Scott. He says quite explicitly that the centrifugal force is non-Newtonian or inertial. He recognizes its "fictitious" character although he doesn't like calling fictitious a force that may black out the pilot of and aircraft. He also says that these forces are introduced when using a rotating reference frame. Nothing about the forces "being there" independently of the reference frame. --Itub (talk) 15:47, 20 May 2008 (UTC)

Itub, how can I misrepresent somebody just by quoting them? It wasn't as if the quote was out of context. It was a clear unequivocal statement that centrifugal force occurs on objects that are at rest in a rotating frame of reference. David Tombe (talk) 16:58, 20 May 2008 (UTC)

I was referring to your statement that your views are based on GD Scott's, where one of them (worded by Anome, but you said it was a reasonably accurate assessment), is that centrifugal force is a real, Newtonian, coordinate-system-independent force. Which is exactly the opposite of what GD Scott says. --Itub (talk) 17:08, 20 May 2008 (UTC)

Well if I ever made a statement saying that my views are based on those of GD Scott, then I retract it because I only heard of GD Scott for the first time about four days ago. I don't recall making such a statement. All I know is that he clearly stated that centrifugal force occurs on objects at rest in a rotating frame of reference. David Tombe (talk) 17:14, 20 May 2008 (UTC)

[edit] Finding an expert

Sorry folks, but I cannot see how the expert would help. First, already we have expert testimony in texts. David has refused to really go through these texts and point out how they support him. And he has claimed that contributors have misinterpreted these texts. Second, there is an impeccable coordinate-free derivation at Fictitious force that David refuses to sift through and point out its misconceptions (according to him), even though it completely disagrees with him. BTW, this derivation also is backed up with expert texts.

What will happen when you find an expert? David has several recourses. First, he can say the expert is not expert in "physica obscura" or whatever area is most important according David (right now it's orbital mechanics, which has absolutely nothing to do with this argument which is a purely kinematic argument; or it requires expertise in elliptic coordinates, etc.). Second, he can say that the expert is confused, just like us, because, of course, the expert will agree with us. Or, third, he can say we are misinterpreting the expert.

What is NOT going to happen is David's saying: "Oh golly, now I see the light. You guys should accept my deepest apologies for involving you in hours of effort to change my mind. I see now that you were right, but somehow the light just did not turn on until the expert said the magic words "abracadabra"." I make that assertion based upon the manner in which David approaches all this discussion, which indicates an unwillingness to really take any contributor's argument seriously, but to latch onto some detail and knit a sweater around it oddly similar to his previous design.

The "find an expert" routine boils down to this: if the expert agrees with David, David will take that as proof of his views. If the expert does not agree with David, either David will find a way to say he actually does agree (it's just that the expert's meaning for "inward" corresponds to our "outward") or that the expert is off the beam, expert though they may be.

So what is the final result? We'll have to say: "David, we've given it our best shot. Everyone in the world that we could find disagrees with you, and they all are unanimous on what is happening here. So unfortunately, having exhausted all known means of human persuasion, we will have to adopt the unanimous view." That could be said right now, in somewhat less flamboyant terms. Brews ohare (talk) 15:24, 20 May 2008 (UTC)

Brews, your coordinate free derivation at fictitious forces doesn't begin at the beginning. At the beginning, we have a vector triangle and that's where all the physics is. One component is ωXr and it is the component that is relevant for the centrifugal force term. It refers to the fixed point on the rotating frame.David Tombe (talk) 17:04, 20 May 2008 (UTC)

This is where you are confused. The r in ωXr does not refer to a fixed point in the rotating frame. It refers to the current position of the object. This position may, of course, change with time. (TimothyRias (talk) 08:21, 21 May 2008 (UTC))

No Timothy, the ωXr term is the velocity of the fixed point in the rotating frame. David Tombe (talk) 13:16, 21 May 2008 (UTC)

This seems to be at the core of your misunderstanding of this subject. r in ωXr does not refer to any magical fixed point. It is precisely the velocity of the rotating frame (wrt the inertial frame) at the position r, which denotes the current position of the particle. Any derivation of these matters makes this instantly clear. (TimothyRias (talk) 14:16, 21 May 2008 (UTC))

No Timothy, if it didn't refer to a fixed point, then the velocity term in the Coriolis force would no longer refere to the velocity relative to a fixed point in the rotating frame.

It does refer to a fixed point as this is the anchor point upon which the whole theory hinges. David Tombe (talk) 09:45, 22 May 2008 (UTC)

I agree. I wouldn't bug poor John Baez about this, personally. Frankly, I am under the impression that several of the editors involved in this already have ample academic credentials to be considered "experts" in undergraduate-level physics. -- SCZenz (talk) 15:21, 20 May 2008 (UTC)

Yes, I'm already aware that other editors already here have far more than the level of expertise needed to deal properly with this topic; asking for a review by someone who is both a leading professional physicist and a Wikipedia contributor is a bit of a last resort, bordering on overkill. However, since Professor Baez has already said (see User talk:John Baez) that he may take a look at the article at some point when he has some free time, I certainly wouldn't refuse the offer. Of course, since he's already rather busy doing real physics, and has a life as well, that might take some time. -- The Anome (talk) 16:05, 20 May 2008 (UTC)

[edit] Questions for experts

To calm down a bit, I am aware that it is easy to believe one is right, even when wrong. An expert can be useful in some circumstances. What are those circumstances? I'd say the expert should be consulted only after an agreed upon list of questions has been constructed, and it is agreed in advance what the implications of the answers to these questions are for the argument. For example, if a list of questions with "yes" or "no" answers can settle the matter, that would be pretty satisfactory.

I am dubious that construction of this list will prove feasible, and that David would accept the results in the end, even if he agreed in the beginning. He'd want to "dig into" the answers to see whether the expert really understood what their "yes" or "no" implied, especially, was the expert aware of xyz when they made their "off-the-cuff" response? Possibly, these responses by David could be ruled out by proper preamble to each question that includes all the relevant factors the expert should consider. The agreement is made up front that no other "second thoughts" are permitted.

Here is a possible list of questions for consideration:

1. Is the presentation of Fictitious_force#Mathematical_derivation_of_fictitious_forces correct? (y or n)?

2. Is the presentation of Fictitious_force#Rotating_coordinate_systems correct? (y or n)?

3. Is the presentation of Centripetal_force#General_planar_motion correct? (y or n)?

4. Are the examples in Centrifugal_force#Examples correctly analyzed? (y or n)?

5. Are the examples in Centrifugal_force#Examples properly interpreted?" (y or n)?

In the event of "no" answers, can you kindly provide some suggestions that would bring these articles into compliance with expert opinion? Brews ohare (talk) 16:10, 20 May 2008 (UTC)

I'd add these questions for David:

6. "Does the conventional interpretation, regardless of its inherent rightness or wrongness, make numerical predictions about the motion of bodies that are in agreement with experiment?" (y or n)?

7. "If not, can you give an example of an experiment which might be used to distinguish between the physical predictions of your theory and those of the conventional theory?" (y or n)?

8. "Have you or anyone else, performed this experiment?" (y or n)?

9. "Have the results of the experiment been replicated by multiple investigators, and published in peer-reviewed journals?" (y or n)?

-- The Anome (talk) 16:12, 20 May 2008 (UTC)

And David's questions for the experts are:?? Brews ohare (talk) 16:56, 20 May 2008 (UTC)

[edit] Reply to Anome's Questions

Anome, I think that you have missed the point. My aim was never to insert any unorthodox theory into the main article.

My aim was (1) to remove the nonsense about the artificial circle, (2) To simplify the article and make it more coherent and readable, (3) To draw attention to scenarios such as the centrifuge and planetary orbital theory and emphasize the fact that real effects are being induced as a direct consequence of actual rotation, and (4) to draw attention to the changing attitudes about centrifugal force over the last few hundred years.

The subsequent debate seems to have now focused on a fierce defence of the artificial circle. That in turn has split into two key arguments. (i) The issue of ωXr meaning that centrifugal force only applies at stationary points in the rotating frame, and (ii) the issue of whether or not a circular motion involves a radial acceleration.

I maintain that the artificial circle is not only wrong, but that the entire article is absorbed in the fictitious implications of it.

The edit war itself was mainly over my attempts to insert sentences about the real effects in a centrifuge being caused by rotation.

Anyway, since Timothy Rias, Brews ohare, and SBharris are all adamant that a circular motion does involve a net inward radial acceleration, I think that we ought to clear this matter up once and for all.

There has been a tendency for some of them to lecture me on some basic high school stuff which I myself taught to students.

We all know that a centripetal force is needed to change the direction of an object that is moving in a straight line in order to make it move in a circle.

But they all overlook the fact that there is already an outward centrifugal force acting on the object. When the circular motion occurs, then the two are balanced and there is no net radial acceleration.

I drew attention to planetary orbital theory since it was the best example to illustrate the point.

The orbital equation unequivocally contains an outward radial centrifugal force and an inward radial centripetal force (gravity). When the two are balanced (all the time) we will have a circular orbit and a net zero radial acceleration.

In the artificial circle example, we can't have a net zero radial acceleration if we have a net inward centripetal force and so we can't have a circle. The theory is a nonsense. And it's a nonsense on at least two more counts as well.David Tombe (talk) 16:49, 20 May 2008 (UTC)

Not just them, I'm afraid: me too, and also see [9][10][11][12][13][14]. Just because the distance from the center does not get smaller does not mean there is no inward acceleration. (And yes, that's acceleration from the viewpoint of an inertial frame, the same one that the circular motion is occurring in: if we talk about circular motion, and thus velocity. in terms which are relative to a particular frame, we must also consider any accelerations from the viewpoint of the same frame for them to be consistent.)

The requirements for a physical theory to be regarded as valid are:

it must be logically self-consistent
it must make physically testable predictions
it must not have been falsified by the observations of experimental results that conflict with its predictions

Where two equally good theories exist, it is also generally assumed the the simpler theory is "better", based on the principle Occam's Razor. The classical theory of mechanics meets all of these requirements for nearly all natural phenomena, to very high precision; the only exceptions to this appear to be on energy, size, or velocity scales where the effects of quantum mechanics or relativistic mechanics take hold.

Unfortunately, this theory also uses words like "force", "velocity", "acceleration", "coordinate frame", "inertial" and "fictitious" in ways that do not fit your own conceptions of those quantities. This does not matter: only explanatory power and falsifiability matter. (If you don't like this sort of usage, you really aren't going to like quantum physics or general relativity, which do further violence to even more basic common-sense ideas like "space", "time" and "matter" in their formal conceptions of those entities.)

By the way, when you say your aim "was never to insert any unorthodox theory into the main article.", does this mean that your answer to question #6 above is "yes"? If it's "no", could you please also tell me your answer to question #7? -- The Anome (talk) 17:59, 20 May 2008 (UTC)

Mr Anome, by definition of acceleration towards the center, there is a change in velocity towards the center, in this case from zero to some finite value. So if there is acceleration there must be movement towards the center. But there is none. So you are confused. —Preceding unsigned comment added by 72.64.49.79 (talk) 19:07, 20 May 2008 (UTC)

Could it be that in high school you did not learn the difference between speed and velocity? As viewed from an inertial frame, an object moving in a circle does not change speed, either tangentially or radially. However, it does change velocity, and this means it has an acceleration, and this requires an unbalanced force. Okay? Do I really need to start citing basic texts? [15]. If you're in an inertial frame, there are no centrifugal or Coriolus forces. There's just the one force which causes the object to deviate from straight-line motion. S B H arris 21:07, 20 May 2008 (UTC)

Again, for the benefit of User:72.64.49.49, here's the derivation of the acceleration involved for circular motion in a unit circle in the plane. Using an intertial frame with its origin at the center of rotation, taking orthogonal components of x, the position vector relative to the center of rotation:

x = cos(omega t)
y = sin(omega t)
  implies (by differentiation)
xˈ = - omega sin(omega t)
yˈ = omega cos(omega t)
  implies (by differentiation)
xˈˈ = - omega^2 cos(omega t) = - omega^2 x
yˈˈ = - omega^2 sin(omega t) = - omega^2 y

Therefore, if x = (x, y), xˈˈ = - omega^2 x, which is (a) nonzero, and (b) radially inwards, in spite of the fact that |x| = 1 throughout. Seemingly paradoxically, the body accelerates towards the center of motion forever, without ever getting any closer to it. -- The Anome (talk) 21:50, 20 May 2008 (UTC)

Yes, but less of a paradox when you think about velocity rather than speed. At 12 O'Clock on a circle the clockwise object is going TO THE RIGHT at speed s, whereas by 6 O'Clock it is going TO THE LEFT at the same speed. So something has completely slowed it to a stop, turned it around, and sent it going in the opposite direction just as fast. That's a lot of change in velocity, and it takes an unbalanced force to cause this acceleration. There are no "balanced forces" when viewed from this frame. If there were, there'd be no explanation for the direction and velocity change. S B H arris 22:39, 20 May 2008 (UTC)

[edit] The Artificial Circle

I maintain that the artificial circle is not only wrong, but that the entire article is absorbed in the fictitious implications of it.

– David Tombe

My understanding of the "artificial circle":

In the dropping ball example, the ball has no force upon it and falls vertically at constant speed. Thus, in the rotating frame the ball appears to spiral downward, and a projection of the ball's path on the rotating plane obeys a description in the rotating plane as uniform circular motion. David's view is that there is really nothing happening here (hence the adjective "artificial"). As a result (possibly involving some further argument) there is no force on the ball from the rotating observer's viewpoint.

In contrast to David's view, the article adopts the view of the rotating observer, that the path is an example of uniform circular motion, and therefore according to this observer, a centripetal force is necessary to cause the circular motion.

The argument can then go on to discuss other matters, but already David is in disagreement.

I've deleted some unnecessary remarks of mine. Brews ohare (talk) 18:49, 20 May 2008 (UTC)

This would seem easy to settle experimentally. Does anyone have access to a ball, a rotating platform, and a video camera? (I've just carried out this experiment with the aid of a dropped ball and a swivelling office chair, and to within human observational error, the results were in accordance with the theories expounded in the article, just as they were with my earlier held-and-then-released-object demonstration of the reality of the Coriolis force. But I don't have a video camera.)-- The Anome (talk) 18:40, 20 May 2008 (UTC)

The matter is not one of observing circular motion, where David agrees about what is seen. The argument is how one should go about calculating the observed motion based on Newton's laws. Personally, I don't see how David's approach can do that, and to date he hasn't. Brews ohare (talk) 18:49, 20 May 2008 (UTC)

In many ways, it doesn't matter. If the conventional theory is self-consistent and predicts the same observables as experimentally observed in all known reproduceable, peer-review-published experiments to date, it is valid. But actually, we don't have to prove even that: for the purposes of Wikipedia, we only need to demonstrate that, rightly or wrongly, this is the view believed by most physicists, and that there is no substantially notable group (see WP:UNDUE) that thinks otherwise.

I'm afraid that when faced with intransigence, a bureaucratic approach like this is the only possible way to resolve the matter. As with most bureaucratic approaches, it has no regard for logic, justice, accuracy, good sense or compassion. But it does settle matters. Brews ohare (talk) 19:00, 20 May 2008 (UTC)

By the way, David, when you say your aim "was never to insert any unorthodox theory into the main article.", does this mean that your answer to question #6 above is "yes"? If it's "no", could you please also tell me your answer to question #7?-- The Anome (talk) 18:52, 20 May 2008 (UTC)

[edit] Fictitious Force article

Copied from above, David's comment:

Hi David: Not surprisingly, I disagree. The derivation begins precisely at the beginning, with the displacement vector. It then proceeds to find the acceleration according to the standard rules for differentiation. The final result is a direct consequence. There is not much that can go wrong here.

BTW, I recall your saying in earlier commentary that you agreed with the result (which happens to be exactly duplicated in at least three cited references). However, you quarreled with how the cross-product was evaluated, leading to my insertion of a Figure to show exactly how that was done. At that point you broke off, and began a different line of discussion. Brews ohare (talk) 19:36, 20 May 2008 (UTC)

Brews, the derivation begins with a vector triangle. The position vector of an actual particle is considered and its displacement is then split into two components.

The first component is the displacement to a fixed point on a rotating frame.

The second component is the displacement from that fixed point on the rotating frame to the particle.

Hence, the first component which eventually leads to the centrifugal force at the end of the derivation, relates to a fixed point on the rotating frame.

The second component leads to the velocity of the particle relative to the rotating frame. But in the limit in which the first component becomes equal to ωXr then the first component necesarily becomes the tangential component and hence the second component can only be the radial component of the velocity.

The conclusion is that a Coriolis force causes tangential deflections on radial motion in a rotating frame. The centrifugal force and the Coriolis force are two mutually perpendicular aspects of the same thing. But they can never double up or cancel each other out.

We are not free to apply the Coriolis force term to velocities that are in any direction.

You should know this simply by looking at the derivation for acceleration in polar coordinates which uses exactly the same principles. The Coriolis term is exclusively a tangential effect and so we cannot talk about a radially directed Coriolis force.David Tombe (talk) 05:44, 21 May 2008 (UTC)

Apart from the arbitrariness of your proposed method of analysis, see Figure 9.12 in Taylor for a counter example to your final remark. You did not reply to the paragraph in my comment beginning: "BTW, I recall your saying in earlier commentary that you agreed with the result (which happens to be exactly duplicated in at least three cited references)." Brews ohare (talk) 06:14, 21 May 2008 (UTC)

Brews, I've always agreed with the final mathematical form. But there are alot of people, including Taylor, who have lost track of the restrictions on the applicability of those equations.

You must go to the beginning of the derivation and work through it line by line, paying attention to the physical significance of each line. If you do that, you will see that v(rot) has to be radial.David Tombe (talk) 06:30, 21 May 2008 (UTC)

Hi David: So now we are down to the significance of the result and the limitations on applicability. It is clear that Taylor and Arnol'd and others have determined that the equations and their derivation imply a Coriolis force with properties different from yours. As a minority of one, without credentials that match an Arnol'd, all demos of validity and physical implication fall on your shoulders, nobody else's. That requires a focus and eloquence of argument that is hard to accomplish.

In particular, as you have agreed that the final mathematical form is OK, it's a very tight corner to argue that the interpretation of the results has ambiguity, and to find some definitive example where there are multiple interpretations with a clear advantage for your own.

For example, it is improbable that application of the final mathematical form allows multiple interpretations of the dropping ball example. Likewise for the rotating sphere examples, which will have to be explained your way with success equal to the present explanation. Brews ohare (talk) 19:00, 21 May 2008 (UTC)

Brews, see the reply in the special section below. Also why not consider a real scenario in which the transformation equations are applicable.

Consider a rotating turntable with a radial groove. There is a marble in the groove. Centrifugal force will accelerate the marble radially outwards. The ensuing co-rotating radial motion will experience a tangential Coriolis force and it will in turn apply a reactive Coriolis force on the turntable in the opposite direction.

That pretty well sums the whole situation up. Why not put in that example instead of the artificial circle which involves neither centrifugal force nor Coriolis force. David Tombe (talk) 10:30, 22 May 2008 (UTC)

[edit] Reply to SBharris on Circular Motion

SBharris, You keep emphasizing the change in velocity that occurs in circular motion. That change is indeed caused by the centripetal force. Nobody is denying it. We are all agreed that the centripetal force deflects the object from its inertial path.

But you are totally ignoring the centrifugal force. Even when the object is moving in a straight path, it still has centrifugal force relative to the exact same origin that we measure the centripetal force from. It is known as inertia.

The straight line path is what results when only centrifugal force acts and no centripetal force acts.

Consider the gravity orbit. In the case of zero gravity, the centrifugal force alone will yield a hyperbola of infinite eccentricity. That is a straight line.

When the centripetal force accelerates the object in a circle, the centripetal force will then exactly cancel the centrifugal force and we will have a net zero acceleration in the radial direction.David Tombe (talk) 05:56, 21 May 2008 (UTC)

I am NOT ignoring the centifugal force. As long as you stay OFF the orbiting or accelerated object, there IS NO CENTRIFUGAL force. STAY OFF IT! S B H arris 18:41, 21 May 2008 (UTC)

SBharris, I looked at some of your other comments. It seems that you only want to consider this issue from the Cartesian coordinate frame.

However, the centripetal force is a radial force and so also is the centrifugal force. If we are to consider the radial direction, then we have to consider both the centripetal force and the centrifugal force.

We cannot talk about the centripetal force and then ignore the centrifugal force on the grounds that we are looking at things from a Cartesian perspective.

I notice also that you deny the idea that centrifugal force is contained within inertia. What I then suggest is that you solve the planetary orbital equation for a case of negligible gravity. That will leave only the centrifugal force in the equation. Then tell me what the solution will be. Circle? Ellipse? Parabola? oe Hyperbola? And if it is a hyperbola, what will the eccentricity be? And if the eccentricity is infinite, will the result look anything like a straight line?

Of course it is a hyperbola (what you get when an object passes by another with faint gravity and experiences a minor deflection), and yes, it approaches a straight-line path, as the force of gravity tapers to zero (which happens far away from the primary, both incoming and outgoing). Happy, now? S B H arris 19:06, 21 May 2008 (UTC)

I also notice that you had difficulty with the concept of analyzing a straight line motion with reference to a point which is not on its path. It's simple. Draw a line between the point and the moving particle. As the particle moves in a sraight line, the radial distance between the point and the particle will change, and the radial direction will change too. If you do the exact same analysis that you did to derive centripetal force, you will find that the result is exactly the same but that this time, you will get a centrifugal force outwards from the point. David Tombe (talk) 06:10, 21 May 2008 (UTC)

No, indeed. Physics doesn't change, nor do its predictions (paths, forces, accelerations) change when you merely change coordinate systems, so long as you remain in the same reference frame. In a given frame, Cartesian and cylindrical and polar coordiate systems all give the same answers. It's only when you unconsiously skip to another accelerated or displaced frame at the same time you apply the new coordinates, that you get different answers.

Yes, an object moving in a straight line at speed V (think of a bowling ball), then being deflected to another direction by application of a momentary exactly lateral/transverse force (say a billiard ball bounces off the side of it), is the simplest case of our disagreement. In the original frame, you see the object moving at V in one direction, and a transverse force simply causes it to deflect to the same speed V in another off-direction. There is only one force. There no mysterious counterforce. Velocity (the vector) changes here, since direction changes, but speed does not change. Now, transfer to the frame where you're moving with the object, no matter what. In THAT frame, every time an external force acts on the object, a mysterious counterforce must appear, because in our frame the object never moves, so all forces must always be balanced at all times. This opposing force is what you call the "inertial force." It's not a standard physics idea. It's just a ficticious force (one of many) that appear when you insist on inhabiting an accelerated frame, as you do when riding a bowling ball when exterior forces act on it. Please realize what you're mentally doing. There is no inertial "force" so long as you stay OFF the flipping ball! S B H arris 18:33, 21 May 2008 (UTC)

SBharris, a centripetal force causes an inward radial acceleration, yet in circular motion, the net inward radial acceleration is still zero. That can only mean one thing. There must be an outward centrifugal force acting in the radial direction.David Tombe (talk) 10:24, 22 May 2008 (UTC)

The inward acceleration is only zero in a rotating frame where the rotating object doesn't move. In the frame of the central point, or (if you will) the inertial frame (put yourself on the center of the merry-go-round, but don't rotate), then objects moving in a circle DO have a radial acceleration. Why? Because they're changing velocity all the time. Velocity is speed and direction. The speed doesn't change, but in a non-rotating frame, the direction does, and thus the velocity does. Halfway around a circle, an object is going at the same speed, but in a totally opposite direction. Thus, it's been stopped, turned around, and sent the other way. That requires an unbalanced force, acting all the time. Objects with balanced forces on them do not change velocity. You're claiming two balanced forces acting on such objects, but if you were right, you'd see no velocity change, and you do. S B H arris 21:14, 22 May 2008 (UTC)

[edit] Reply to Anome

Anome, this argument doesn't come down to Popperism. It comes down to the objectives of you, PeR, Henning Makholm, Itub, SZCenz, Wolfkeeper and others.

Those objectives are to write an article on centrifugal force while trying to deny that centrifugal force exists.

That is why you have steadfastly refused to allow references to the fact that the real effects of a centrifuge are caused by actual rotation.

That is why you have been introducing 'newspeak' into the article, such as 'The centrifugal tendency'.

That is why, a few days ago, you deleted the reference to rotation in the last line of the introduction.

That is why you went to the Bucket argument page and inserted the word 'attempts' in relation to the assertion that Newton's bucket experiment proves absolute rotation.

And to that end, you want to focus the article on a nonsense theory that involves falsely extrapolating the transformation equations to objects that are at rest in the inertial frame with the view to falsely demonstrating the fact that centrifugal force is merely a fictitious artifact.

In fact, I'm surprised that you have even allowed the last line in the introduction to remain because it actually lets everybody know that there is a real centrifugal force.

The last line effectively reads, "If you came here wanting to read about centrifugal force, you have come to the wrong page. We don't have any page on centrifugal force".David Tombe (talk) 06:23, 21 May 2008 (UTC)

Yes, "extrapolating the transformation equations to objects ... in the inertial frame with the view to ... demonstrating the fact that centrifugal force is merely a fictitious artifact" is pretty much a correct summary of what these other editors are saying. This is the conventional viewpoint of the modern treatment of classical mechanics. Yes, centrifugal force feels real, and for most everyday purposes it can be treated as a real force. But it does not transform under coordinate transformation like a real force, and in every case the effects it appears to generate can be explained completely by using the less intutitive but mathematically simpler inertial frame interpretation. Quite simply, the modern formulation of classical mechanics has no need for it as an intellectual contstruct, except as a d'Alembert force term when considering motion from the viewpoint of a rotating reference frame.

Whether you like it or not, this is the approach used in modern textbooks of classical mechanics, and used from day-to-day by physicists in their mathematical calculations, even if this means that they are using terms like "force", "fictitious force", "acceleration" and "inertia" in ways that appear to you to be nonsense. "Centrifugal force" is as useful and "real" a concept as suction in the sense that for everyday, informal, purposes these are perfectly useful and sensible concepts that explain real-world phenomena in an intuitive way; but in physics, these phenomena are more rigorously explained by mathematically simpler, albeit perhaps counter-intuitive, theories which do not require these concepts at all.

As I have said before, Wikipedia is not an arbiter of truth, it is a reporter of usage. All that matters here is that the article is based on the common usage among experts, as attested to by large numbers of direct cites to verifiable reliable sources. If you can demonstrate that your alternative view that centrifugal force is a real coordinate-system-independent force is held by a significant number of experts (and if it is, you should easily be able to cite significant numbers of named adherents to this view explicitly saying this via reliable sources: see WP:NPOV and WP:UNDUE for guidelines), it can be also be added to the article as a report of a minority view. If you cannot do so, these opinions cannot be placed in the article, and, as advised elsewhere by multiple editors, you will find yourself blocked if you persist in doing so. The same applies to removing cite-supported material from the article to the same end.

-- The Anome (talk) 08:37, 21 May 2008 (UTC)

Mr Anome. It is clearly obvious that what you say is true from all of the erronous information contained in wikipedia. It is certainly no surprise to me that wikipedia is not concerned with truth. Now that you have officially confirmed this, which I have long suspected, I am greatful. So now we can proceed to understand why what I read in wikipedia is in most cases false and incorrect. How can we correct that? Maybe by getting new editors and a new management to run things differently. I am voting for that solution.Boggled (talk) 12:17, 21 May 2008 (UTC)

No Anome, your analysis of the edit war is completely wrong. The edit war is about the fact that gravity acts in the radial direction as does centrifugal force.

However, while gravity is accepted as a real force which is independent of reference frames, centrifugal force has been the subject of a deceitful switching on and off at will, as the editors here play conjuring tricks between polar coordinates and Cartesian coordinates. David Tombe (talk) 13:12, 21 May 2008 (UTC)

This might be part of your problem. The modern vision on gravity is that it is not a real force. Gravity is just as fictitious as centrifugal or Coriolis forces. All are the result of describing physics in a coordinate frame where straight lines are not geodesics. A minor difference is, that it is generally not possible to choose a coordinate frame in which gravity vanishes at all points. (TimothyRias (talk) 14:03, 21 May 2008 (UTC))

David, please re-read the paragraph emphasized in red above. -- The Anome (talk) 15:01, 21 May 2008 (UTC)

I think I've located the problem, which is that David Tombe is simply unable not to translate his viewpoint to the accelerated frame of any accelerated object. So even when an object moving in a straight line is deflected by a tangential force to another straight line path, he thinks that an opposing "inertial force" appears (momentarily) to offset the deflecting force, so long as it exists. That means he's put his frame on the object, where a mysterious counterforce appears for EVERY outside force, because in that frame, the object does not move, so cannot experience unblanced forces. So forget gravity. Forget circles. Forget the math, which simply confuses the issue. The very simplest case where we come down to disagreement (and perhaps this should begin a general Fictious Force wiki) is a single transverse deflection from straight line motion, where Tombe STILL cannot keep from switching reference frames. I think in this be may be unteachable, like the person who never does manage to see the other figure in one of those double-drawings where you can switch back and forth from one view to the other. S B H arris 18:54, 21 May 2008 (UTC)

SBharris, you haven't located the problem yet. I don't involve reference frames at all. The problem is that you can't reconcile the idea that an inward centripetal force results in a zero net inward radial acceleration. The problem is resolved when you realize that there is also an outward centrifugal force cancelling the centripetal force in circular motion. That centrifugal force is already there and exists independently of whether we have a centripetal force or not.David Tombe (talk) 07:09, 22 May 2008 (UTC)

This simply false. Even though the second derivative of the radial coordinate is zero this doesn't mean the acceleration (which is defined as the instantaneous change of the velocity) in the radial direction is zero. You seem to have a hard time with dealing with vector quantities in polar coordinates. You also seem to fail to realize that the position vector, the velocity vector and acceleration vector all live in different vector spaces. This point is often glossed over in simple treatments of classical mechanics (and is often delayed to the teaching of General relativity where this difference becomes very important) which has caused confusion in many students. The ground for this confusion is that all vector spaces are just standard 3-dimensional vector spaces and hence isomorphic, and you can sometimes use this fact to represent all vectors in the same vector space, but in the end it is important the realize that physically these object belong to completely different families. The problem you are running into is that changing coordinates in the position vector space leads to no trivial changes in the velocity and acceleration vector spaces. In high school physics this is all treated as a whole bunch of hocus pocus, but once you understand what is really going on these facts become really simple. You seem to be stuck in the high school hocus pocus phase, probably because you have been teaching just that. (TimothyRias (talk) 08:03, 22 May 2008 (UTC))

Timothy, I'm not the one stuck in high school physics. I'm the one that was pointing out the limitations of the high school teaching. They teach there that an inward radial centripetal force causes the particle to keep changing its direction.

But they don't teach why the radial acceleration is still zero nevettheless. They omit to teach that the centripetal force is working against an outward centrifugal force.David Tombe (talk) 10:14, 22 May 2008 (UTC)

[edit] Reply to Brews ohare

Brews, I didn't say that the final maths expression had multiple interpretations. I said that its application is restricted. As regards the centrifugal force article, the most important restriction is that the centrifugal force acts at a fixed point in the rotating frame.

There has only been one single citation produced which actually addresses this issue. It is the GD Scott citation which confirms exactly what I say. David Tombe (talk) 06:13, 22 May 2008 (UTC)

[edit] Reply to Anome

Anome, I read your bit in red. There has only been one citation so far presented that directly addresses the controversy, and it agrees with what I have been saying. It is the GD Scott citation. He says that the centrifugal force only applies at stationary points in a rotating frame of reference. Hence it doesn't apply to the artificial circle. All the many citations that you have lined up have got nothing whatsoever to do with the controversy. You have come along with a large hand of bogus citations.

And the artificial circle theory appears in no textbook. David Tombe (talk) 06:19, 22 May 2008 (UTC)

But the "artificial circle" does appear in a reliable source: Burgel, B. (1967). "Centrifugal Force". American Journal of Physics vol 35: pp.649-650. Burgel shows (example 1) that for an object that is stationary in an inertial frame, in the rotating frame the sum of the centrifugal force (radially outward) and the Coriolis force (radially inward) gives the required centripetal force necessary for the observed circular motion. --FyzixFighter (talk) 06:47, 22 May 2008 (UTC)

FyzixFighter, I'm not at all surprised to hear that. Somebody was bound to put the idea in print at some stage. But you've got to ask yourself whether or not you believe it. I never wanted this to come down to a cheap game of citations and counter citations. GD Scott says that the centrifugal force only applies at fixed points in the rotating frame. I was trying to draw your attention to the derivation of the transformation equations. It all comes down to a simple vector triangle and identifying what the two components mean. The first component, which ultimately develops into centrifugal force, is the the displacement to a fixed point on the rotating frame. The second is the displacement from that fixed point to the particle. Hence the second can be interpreted as the velocity of the particle relative to the rotating farme. In the limit, these two components become the tangential and radial components and we can write the tangential component as ωXr. You are the very one who liked to start with the velocity vector. Can you not see that the direction changing acceleration on a velocity vector is ωXv and that this is the parent force for both the Coriolis force and the centrifugal force? Can you not see that Coriolis force and centrifugal force are two mutually perpendicular components of this parent force when we write it in polar coordinates? If you could see that, you would know that the Coriolis force only ever acts on radial motion. What appears to you to be a Coriolis force acting on tangential motion is actually only a tightening up or slackening of centrifugal force. This was confirmed in one of the references suplies by SBharris. David Tombe (talk) 07:04, 22 May 2008 (UTC)

The first remark is a cheap shot. First, you claim that there are no citations, then when somebody notes that perfectly fine citations have been given, you reply that you don't care.

Also, have you actually read the GD Scott reference, you seem to be so fond of? It appears not. Scott in that article is not discussing the point that is at issue here. The quote you give from that article does not explicitly say that centrifugal force does not apply to objects moving in a rotating frame. Nowhere in the article does Scott argue that it shouldn't. He merely notes that centrifugal force is a force that applies to object stationary in a rotating frame. (As opposed to a reaction on a centripetal force) (TimothyRias (talk) 08:20, 22 May 2008 (UTC))

"What appears to you to be a Coriolis force acting on tangential motion is actually only a tightening up or slackening of centrifugal force." Exactly, this is what happens if you keep changing your rotating frame of reference so that the particle is always fixed on it regardless of its angular velocity. This is what happens in the walking-on-the-Earth example I gave above. But if you keep a consistent frame of reference which keeps rotating at the same speed (i.e., the Earth), then you need a radial Coriolis force. Here's another paper which mentions radial Coriolis forces explicitly, when dealing with the case of walking in a rotating space station.[16] It talks about how when walking in the direction of rotation or against it, one feels heavier or lighter, due to the Coriolis force. --Itub (talk) 09:24, 22 May 2008 (UTC)

Timothy and Itub, You both need to get real about this. Instead of looking at abstract artificial circles where no forces act, let's look at a real example which is what the transformation equations are all about.

Consider a real turntable with a radial groove in it. Consider a marble in that groove. When the turntable rotates, a centrifugal force will act radially outwards on the marble and the marble will then be constrained to a co-rotating radial path.

A Coriolis force will act tangentially on the moving marble and the marble will exert an equal and opposite reactive Coriolis force on the turntable.

That is the kind of situation that the transformation equations are designed for.

If you would put stuff like that in the main article instead of nonsense about artificial circles that involve neither Coriolis force nor centrifugal force, then the article would be alot better.

You both need to pay attention to the physical meaning behind the maths symbols. You are both far too engrossed in abstract extrapolations of the maths and you have both lost touch with the actual underlying physics. David Tombe (talk) 09:40, 22 May 2008 (UTC)?

So, basically naive physics rules and mathematical physics sucks. Everything that has been done since Newton is BS, and we shouldn't use math in science because we lose sight of the "real physics". That's not the way it works. The reason math is used in physics is that if you start with the right assumptions and do the math correctly, the result will be physically correct. There is no need for adding ad hoc restrictions at the end and saying that, although the equations allow you to have the Coriolis force pointing in any direction, it was "not meant to be that way" and thus can only be radial. That sounds more like religion than science to me. --Itub (talk) 09:47, 22 May 2008 (UTC)

Itub, the point here is that these equations are being applied beyond what the derivation permits. The derivation is anchored on a fixed point in a rotating frame and that is where the centrifugal force acts. David Tombe (talk) 10:10, 22 May 2008 (UTC)

You are the one denying physical reality. If Coriolis forces dont act radially on particle that are not co-rotating Lagrangian points would not be stable. Real life physics has shown that they are. Who is losing sight of physical reality? (TimothyRias (talk) 10:02, 22 May 2008 (UTC))

Timothy, I'm not going to get drawn into the controversies of the three body problem and the various attempts to solve it. David Tombe (talk) 10:17, 22 May 2008 (UTC)

There is nothing controversial about the reduced 3-body problem, which is just a single body moving in a non-central potential. (TimothyRias (talk) 10:39, 22 May 2008 (UTC))

Timothy, I see. So it's the reduced three body problem that you are talking about now? And what's your point?David Tombe (talk) 10:48, 22 May 2008 (UTC)

That in the reduced three body problem, the Coriolis force acting radially is responsible for the stability of the L4 and L5 points and makes the other point dynamically stable. If your ridiculous claim that the Coriolis force cannot act radially were true, there could be no such stability. (TimothyRias (talk) 11:59, 22 May 2008 (UTC))

Timothy, which radial direction are you talking about? Their are two radial directions leading from L4 and L5. And it's not so much the Coriolis force that is involved so much as it's a hybrid velocity dependent term that is the parent of both the Coriolis force and the centrifugal force. See http://www.physics.montana.edu/faculty/cornish/lagrange.pdf David Tombe (talk) 16:42, 22 May 2008 (UTC)

[edit] Reply to Timothy Rias

Timothy, the derivation involves routing the vector for the actual particle through a fixed point in the rotating frame. I don't know why you suddenly decided that the point in the rotating frame can be moving. It can't. The derivation needs it to be fixed. David Tombe (talk) 06:22, 22 May 2008 (UTC)

Because it can? And because if you do this you get an equation that is more general than the restricted interpretation you have been promoting? And because this more general interpretation agrees with all sorts of empirical observations, such as dynamical stability of lagrangian points in orbital systems?

Only in your head those it need to be fixed. This seems to be some weird limitation of your imagination. (TimothyRias (talk) 08:10, 22 May 2008 (UTC))

Timothy, if we make the point on the rotating frame dynamic, then the velocity term in the Coriolis term no longer means the velocity of the particle relative to the rotating frame. It then comes to mean the velocity of the particle relative to the moving point on the rotating frame. The whole thing then becomes a moving farce. David Tombe (talk) 09:11, 22 May 2008 (UTC)

It is perfectly fine to measure the velocity of a particle with respect to its current position. Namely $\vec{v}_{rot}(t) = \lim_{dt\rightarrow 0} \frac{\vec{r}_{rot}(t+dt)-\vec{r}_{rot}(t)}{dt}$ . In the same way you can also talk about the velocity of the rotation frame wrt to the inertial frame at the current position of the particle. (please note that the velocity difference between the two frame is position dependent, you cannot measure it using only a fixed point.) (TimothyRias (talk) 09:55, 22 May 2008 (UTC))

Timothy, no need for the maths. If the point on the rotating frame becomes dynamic, then you lose the anchor upon which the entire theory hinges. The velocity term in the Coriolis force will no longer refer to the velocity relative to the rotating frame. It will become an arbitary velocity measured relative to some arbitary moving point. David Tombe (talk) 09:59, 22 May 2008 (UTC)

Let's put it this way. It is perfectly fine to express the velocity of the particle in the rotating frame wrt to the fixed point in the rotating frame defined by the current position of the particle at time t. Physically, the velocity of a particle can only be defined locally at its current position. Doing anything else is losing track of the actual physical meaning of velocity. Something you have been excusing me and others of but are in habit of doing yourself. (TimothyRias (talk) 10:38, 22 May 2008 (UTC))

Timothy, it's either a fixed point or it's not. I don't know what you are talking about above. All I know is that the entire theory is anchored on a fixed point in the rotating frame. David Tombe (talk) 10:46, 22 May 2008 (UTC)

That shows what you know, and what you understand about physics. (TimothyRias (talk) 11:56, 22 May 2008 (UTC))

[edit] Reply to SBharris

SBharris, Here's how you look at the picture.

You see that an acceleration occurs by virtue of the direction of the particle constantly changing.

You see that there must therefore be an inward force (the centripetal force) acting radially on the particle.

But that's all that you can see.

I was taught it that way at high school too, and indeed in first year university physics. But when I did orbital theory at university in secoind year applied maths I then saw the remainder of the picture.

In a circular motion, there is no net inward radial acceleration. That is an obvious geometrical fact. It is indisputable. The second time derivative of radial length is zero. There is no net radial acceleration.

But there is. The radial vector you have to choose to have it NOT shorten, is a moving one. It's complosed of x and y components which change continuously to keep the length of r the same. Which means you're now in the rotating system of coordinates, which I told you to stay out of! But if you pick ONE radial vector, say the one pointing at 12 O'Clock for a moving clock hand at the moment it points to 12, the next instant the vector pointing at the clock hand tip (now a bit past 12) IS shorter in the y direction, and longer in the x direction. So there is acceleration with regard to the original position, due to a change in velocity. The position changes also, just as expected. No acceleration means straight-line motion, or else no motion, and this isn't either of those. Thus, there is acceleration. S B H arris 21:46, 22 May 2008 (UTC)

But you have gone into total denial about this obvious geometrical fact because you have been taught that there is an inward force causing an acceleration inwards in the radial direction and so you think that there must be a net inward radial acceleration even though it is obvious that there isn't.

It's obvious that there is! From my center, not rotating, I see that I must apply a force to keep anything moving in a circle around me, and that force provides the one and only acceleration that object feels.S B H arris 21:46, 22 May 2008 (UTC)

The answer to the conundrum is that the inward radial centripetal force is counterbalanced by an outward centrifugal force. This outward centrifugal force forms a very real part of the central force orbital equation. Both the centripetal force and the centrifugal force act in the radial direction so we cannot pretend that one of them doesn't exist by playing games with reference frames.

You are in denial of this centrifugal force even though it is obviously there. If you do orbital theory, you can't possibly deny it. There are two forces that add together to yield the radial acceleration. They are both radial. One is the gravitational (centripetal force) and the other is the centrifugal force. It has got nothing to do with frames of reference.

One other point. You have been misrepresenting me above. I didn't say anything about forces that suddenly switch on when directions are changed. I was saying that a centrifugal force is there all the time when a particle moves in a straight line. This can easily be confirmed by considering straight line motion referenced to the centre of a circle. There will be an acceleration in the identical form of the centripetal force, but it will always be directed outwards, even if the particle is approaching the origin.

This is nonsense! Saying that some force is always "there all the time" for an object moving inertially in a straight line (as seen from an inertial frame), contradicts Newton's second law of motion. Moving in a straight line is the way objects move when they have NO forces on them. Where the devil did you get the idea otherwise? How do YOU think objects move when they have no forces on them??? I have no idea what you mean by straight line motion "referenced" to the center of a circle. Are you talking about a rotating frame or not? If you just draw a circle in space and look at an object approaching its origin, the circle many as well not be there. The path of the object is a line of constant velocity. No acceleration is seen, or needed. S B H arris 21:46, 22 May 2008 (UTC)

The conclusion is that the artificial circle theory is a nonsense because it involves a circular motion with a net centripetal force that is caused by the Coriolis force. It is an absolute total nonsense on three counts.

And this nonsense is the flagship example in the article. David Tombe (talk) 06:49, 22 May 2008 (UTC)

[edit] About orbital theory

David you are misinterpreting orbital theory in the way you try to extraploate its results to this discussion.

In orbital theory the rotational symmetry of the central force problem is used to reduce the inherently 3-dimensional problem to a single one dimensional equation for the radial distance as it changes with time. Effectively this has placed us in a frame of reference that is always co-rotating with the particle.

We can obtain the same result by taking the 3-dimensional problem in transforming directly to this co-rotating frame. Because this frame is co-rotating with the object the Coriolis and Euler forces always point tangentially, while the centrifugal force always points radially. The Coriolis and Euler forces conspire in the tangential equation to force conservation of angular momentum, allowing the co-rotating condition to be expressed purely in terms of the current radius and conserved angular momentum. The result is a radial equation that is independent of the other coordinates.

In this equation we can find stationary points. At these points the particle doesn't move at all in the co-rotating frame, it is stationary, all the forces working on it are balanced. In particular there is no circular motion in the co-rotating frame. Of course, of we now translate back to the original inertial frame, this stationary motion in the co-rotating frame becomes circular motion in the co-rotating frame. Conversely, stationary motion in the inertial frame implies circular motion in the co-rotating frame. (This doesn't require any fancy talk about force at all).

In short, the orbital equation deals with co-rotating frames and with co-rotating frames only. It cannot say anything about what happens with non co-rotating objects. (TimothyRias (talk) 08:53, 22 May 2008 (UTC))

Timothy, I'm not misinterpreting orbital theory. An outward radial centrifugal force is working in opposition to an inward radial gravitational centripetal force. In the special case of circular orbits, there is no net radial acceleration. David Tombe (talk) 09:23, 22 May 2008 (UTC)

When the centrifugal en centripetal forces cancel in orbital theory, there in deed is no acceleration. In fact, there is no motion. The object is standing still in the frame in which the orbital equation is written, the co-rotating frame. There is no circular motion in THIS FRAME. Only when you translate back to an inertial frame do you get circular motion, but this transformation also kills the centrifugal force leaving us with a net inward force. (TimothyRias (talk) 09:45, 22 May 2008 (UTC))

No Timothy, both the gravitational force and the centrifugal force are in the radial direction. Reference frames have got nothing to do with it. You cannot make the centrifugal force disappear and yet keep the gravitational force. David Tombe (talk) 09:55, 22 May 2008 (UTC)

David, consider this: let's add one more body to the system, so we now have three objects, all orbiting one another under their mutual gravitational attraction (the "three-body problem"), can you tell me how you would select an appropriate "co-rotating" frame within which to consider the motion of any particular body? How would you then consider the motion of the other objects in the system, from the viewpoint of that frame?

Is it possible that, in this case, the system might be more simply considered in terms of solving the equations of motion within an inertial frame? If so, you might want to consider setting the mass of one of the particles in the 3-body problem to zero... -- The Anome (talk) 10:00, 22 May 2008 (UTC)

Anome, I don't recall involving frames of reference in this at all. The radial direction is absolute. If its real for gravity, then its real for centrifugal force too.

There is no need to cloud the issue by introducing the three body problem which to this day has never been satisfactorily reseolved. David Tombe (talk) 10:06, 22 May 2008 (UTC)

See the link above:

"In 1912, the Finnish mathematician Karl Fritiof Sundman proved that there exists a series solution in powers of

t 1 / 3

for the 3-body problem."

Consider this: in the case of the 3-body problem, which of the radial distances (or both) would you consider, and what rate and direction of rotation would you use, to compute the centrifugal force on a particular object? Can you see how this might be linked to the 2-body problem? -- The Anome (talk) 10:12, 22 May 2008 (UTC)

Anome, we can see it all clearly in the two body problem. The centrifugal force and the gravitational force are both radial.

You are only introducing the three body problem to try and cloud the issue. It is an old trick. If somebody is faced with an undesirable truth in a simple situation, then bring in extra complications to try and mask out that truth.

I see that Timothy Rias has been trying the same game.David Tombe (talk) 10:21, 22 May 2008 (UTC)

No, my point is that the 2-body problem is a special case of the n body problem. The inertial-frame approach allows the equations of motion to be written down in a remarkably simple form:

$\ddot{\mathbf{q}}_j = G \sum\limits_{k\neq j } \frac{ m_k(\mathbf{q}_k-\mathbf{q}_j)}{|\mathbf{q}_k-\mathbf{q}_j|^3}, j,k \in \{1,\ldots,n\}$

...and can be easily integrated by numerical methods, or solved analytically for the cases of n=1, 2 or 3. All of this is accomplished without introducing any consideration of rotating frames or pseudoforces. For example, it can be used to predict the motions of objects in the solar system with high precision. Are you denying that this is the case? -- The Anome (talk) 10:26, 22 May 2008 (UTC)

Anome, what's it got to do with the argument that if both gravity and centrifugal force are radial, then they are either both real or both fictitious?David Tombe (talk) 10:35, 22 May 2008 (UTC)

What is has to do with it is that the inertial-frame model makes correct physical predictions, in spite of completely lacking any consideration of centrifugal force or rotating frames, and moreover does this in situations where n > 2, and there is no clear or simple way of defining a "co-rotating" frame, because there is no clear center of rotation to choose (you've got a choice of n-1 at any time), or clear expectation of uniform rotation about any axis. Nevertheless, the inertial frame solution works. Clearly, we don't need to consider centrifugal terms when n > 2.

But now consider that the physical situation is the same for any n, and that this also applies to the 2-body problem (in particular, consider the 3-body problem in the limit where one of the masses tends to zero). And yet we can continue to make testable physical predictions about the system without the concept of centrifugal force. Now, if it was a real force, that is preserved by coordinate frame transformations, how could we ignore it, and still make correct predictions? The only conclusion that can be drawn is that, from the viewpoint of an inertial frame, it doesn't exist, and we don't need it as a concept. -- The Anome (talk) 10:47, 22 May 2008 (UTC)

Anome, when you write out the full equation of motion, the centrifugal force term will be present because it comes automatically with the general expression for radial acceleration. Each of the gravity terms in the summation above will be joined by a centrifugal force term in the full equation of motion. When you expand the $\ddot{\mathbf{q}}_j$ terms, you will see the centrifugal terms inside.

One minute Timothy Rias is trying to tell me that the Coriolis force is involved in the three body problem and the next minute you are trying to tell me that neither centrifugal force nor Coriolis force is involved. David Tombe (talk) 11:10, 22 May 2008 (UTC)

And both are correct, depending on the frame you use. If you use the inertial frame, there are no fictitious forces. The only forces present are gravity. (note: gravity is also fictitious, but can be treated as a real force in the classical limit.) You can also choose a frame that is co-rotating with two of the bodies, the third body obviously will not be co-rotating and will have to deal with the fictitious Coriolis and centrifugal forces. The presence of fictitious forces depends on the chosen frame, this is what makes them fictitious. (TimothyRias (talk) 11:54, 22 May 2008 (UTC))

Timothy has it exactly right. The two are just alternative ways of writing effectively the same set of equations, just transformed to apply to different reference frames. Centrifgual force and other pseudoforce terms appear in the rotating-frame representation, but are completely absent in the inertial-frame representation. From the viewpoint of the inertial frame representation, which physicists choose by convention because it makes the equations simplest (for example, $\ddot{\mathbf{q}}_j = (\ddot{x_j}, \ddot{y_j}, \ddot{z_j})$ , with no centrifugal or other complicating terms), they don't exist. Like Laplace, we "have no need of that hypothesis." This is what is meant by "fictitious" in the term "fictitious force". -- The Anome (talk) 12:02, 22 May 2008 (UTC)

Anome, you cannot be serious. If the centrifugal force is present in the equation of motion for the two body problem, then it will be present when we sum it over n bodies. The centrifugal force does occur in your equation above. It is contained within the $\ddot{\mathbf{q}}_j$ terms. Have you never seen how that term expands when we use it in orbital theory? Check out the radial expression for acceleration and you will see a centrifugal force term.

If centrifugal force is not present in the n-body problem then all the particles would simply collapse into each other.

You are clearly not being serious, and you are pathetically trying to maintain a united front with Timothy Rias who has just said something different from you.David Tombe (talk) 13:42, 22 May 2008 (UTC)

Gentlemen, in your attempt to argue that Mr Tombe is wrong, you have deliberately changed the point and purpose of the discussion. His point was a valid example of a physical interpretation which you changed into a discussion about mathematical computations which have little to do with physical interpretation. Your arguments are as usual not to the point and are invalid as a result. Please address the issues and do not insist upon changing the subject to irrevalent red herrings. The main point of this discussion is as Mr Tombe pointed out that there is a valid physical interpretation for the concept of centrifugal force, which is used in the practical applications of centrifugal machines. You refused to acknowledge these technological applications as valid uses of centrifugal force and now persist in a pointless discussion involving terminology. You should simply admit that Mr Tombe's point was valid and move on to fix other problems which plague your poorly written articles. But you persit in this silly nonsense of insisting Mr Tombe is wrong, by unfairly blocking him, and now by continuing to harass, him even though he has been unfairly blocked by you for one week. Please confirm his edits and get on with other necessary work, and stop this silly bickering over something that is obviously a fine point of interpretation of no real consequence to the typical user of wikipedia. The more I read of this, the more I am convinced that this is all motivated by personal animosity directed towards Mr Tombe.

P.S. Mr Anome dont erase this as you have done with other of my comments.Boggled (talk) 18:31, 22 May 2008 (UTC)

A look at Special:Contributions/Boggled, and this page, reveals that no comments made by User:Boggled have ever been erased by anyone. If your comments were erased while you were using another account, can you please indicate which account(s) you were using? It helps to understand which people are contributing, without having the issue muddle by the use of sockpuppets. -- SCZenz (talk) 21:17, 22 May 2008 (UTC)

Mr SCZenz. This is just another example of the nasty character of wikipedia editors. It demonstrates the typical behavior of wikipedia editors towards those editors for whom they have a particular animosity. A good example is the way you have personally behaved towards Mr. Tombe. In this case, Sir, your comment is so rude and snotty that I can certainly understand why someone whould develope an animosity towards you as your behavior is certainly rude, and uncivil. Your bad behavior is clearly a good example of why wikipedia has a bad reputation, and objective users, such as myself, consider it a rather bad joke run by nasty persons. As I understand you are an administrator, your behavior is certainly reflective of wikipedia management policy, and as such reflects rather more badly on its management. Now reflecting upon your behavior, I can certainly understand why you personally blocked Mr Tombe, as your attitude and behavior is certainly an example of what is undesirable about the management of wikipedia. I now think that both you and Mr Anome should offically apologise to Mr Tombe, since I have personally observed your bad behavior on numerous occasions and it does reflect badly on wikipedia management. P.S. Dont erase this as that would be vandalism. I understand that you would like to do this as it does give you a black mark on your records.Boggled (talk) 13:19, 23 May 2008 (UTC)

[edit] Having fun yet?

All this chatter is going nowhere. David refuses to engage in real discussion of the articles, or even to read what they say beyond mining a few items out of context, and wants to keep everything in the domain of rhetoric. And you all are just enjoying it no end. Brews ohare (talk) 12:42, 22 May 2008 (UTC)

Brews, I gave you a very good practical example for which the transformation equations would actually apply. It involved a marble rolling along a radial groove on a rotating turntable. It clarified both centrifugal force and Coriolis force.

When my block expires, I intend to add it to the main article. It will be an all in one that even explains reactive centrifugal force and reactive Coriolis force.

It will explain how centrifugal force causes the marble to roll outwards in a co-rotating radial path which will cause a tangential action-reaction Coriolis force with the turntable.

I'm fully expecting it to be instantly erased and for me to be blocked indefinitely, because the editors here don't like references to real effects and the administrators have shown that they have no respect for either the truth or the wikipedia rules. They much prefer the extrapolation of maths beyond its applicability to abstract scenarios about artificial circles.

I don't agree with you that I refuses to engage in real discussion of the articles, or even to read what they say beyond mining a few items out of context, and wants to keep everything in the domain of rhetoric.

I have been trying to explain to you the key points in the derivation that limit it to fixed points in the rotating frame for centrifugal force and to co-rotating radial motion for Coriolis force. That constitutes engaging in relavant discussion.

The key points lie in a vector triangle. The first component is displacement to a fixed point on the rotating frame. That leads to centrifugal force. The second component leads from that point to the moving particle. In the limit, these become the tangential and radial components and so Coriolis force is always mutually perpendicular to centrifugal force. David Tombe (talk) 15:32, 22 May 2008 (UTC)

[edit] The Three Body Problem

Timothy, the two stable Lagrange points are definitely a very interesting subject. But to debate it here would result in us rapidly drifting off topic.

Your main point in introducing this topic was to point out that it is claimed that the Coriolis force comes into play whereas I have been claiming that the Coriolis force doesn't occur naturally in free space.

In my books, the centrifugal force and the Coriolis force are two mutually perpendicular aspects of the same thing. The latter is a tangential deflection on a radial motion in the simple case. However, if we have two radial directions, then the centrifugal force and the Coriolis force can get all mixed up to the extent that we have an effect which we cannot definitely state to be exclusively a Coriolis force or a centrifugal force.

In the simple two body problem, a Coriolis force is exclusively a tangential deflection on a radial motion and it never doubles up with centrifugal force. In fact, Kepler's law of areal velocity rules out Coriolis force for the two body problem.

But in the three body problem, the distinction between the two begins to become somewhat vague. So let's agree that a convective effect does occur. How would we decide if it were a Coriolis force or a centrifugal force?

Which radial line would we choose in order to try and rationalize it to be a Coriolis force as opposed to a centrifugal force?

One thing is sure. We are both agreed that the convective forces vXω are very much present in the multi-body problems.

Anome was trying to tell us that we don't need the convective forces in the n-body problem. He was quite wrong on that point. It's a pity that you had to agree with him to show a united front even though it is obvious to everybody that you don't actually agree with him. David Tombe (talk)

[edit] Proposed addition to Centrifugal Force

When my block expires, I intend to add it to the main article. It will be an all in one that even explains reactive centrifugal force and reactive Coriolis force.

– D. Tombe

Hi David: Well, I'd guess your forecast about erasure is pretty accurate. You could try posting it in small sections now on your User Page here to see what the reaction is and maybe iron out some kinks. That could begin by examining your remark "Coriolis force is always mutually perpendicular to centrifugal force". If we agree that Coriolis force is directed along v × Ω and centrifugal force along ( Ω × r ) × Ω what happens when v happens to be along (Ω × r)? That is a rhetorical question, but one you might address in your posting. Brews ohare (talk) 17:37, 22 May 2008 (UTC)

Brews, when v happens to be along (Ω × r) it will either increase or decrease the centrifugal force according to whether it is going east or west. In meteorology, this would invoke Archimedes' principle and yield an effect virtually indistinguishable from the Coriolis force.

Interestingly Timothy Rias has drawn attention to the Lagrange points in the three-body problem. At first I thought that this was off-topic. But on closer examination, I now realize that this is additional evidence of the reality of centrifugal/Coriolis force because the stability is dependent on them. They can hardly be fictitious illusions if they create real stability.

I supplied an interesting link above which I want to study in more detail. It subsumed the centrifugal and Coriolis forces into a single potential term. Once we have two radial directions in the problem then the distinction between Coriolis and centrifugal becomes obscured.

But what is important is that they are there and very real and serving the purpose of stability at the Lagrange points.

Did you consider the rotating turntable with the radial groove? Does it make a good all round example of the two mutually perpendicular aspects of the parent convective force?David Tombe (talk) 19:24, 22 May 2008 (UTC)

In my mind, the subject of fictitious forces is one of kinematics. By that I mean: Throw me a trajectory r (t) and I'll tell you what accelerations are a consequence of such a trajectory. I will not undertake to tell you how to generate these accelerations, nor whether it is in fact possible to achieve a trajectory like r (t), nor under what circumstances.

Next, switch to a different reference frame, and ask how the description of the original trajectory looks in the new frame. Ask: what happens to the accelerations? Answer: in inertial frames - no change. In non-inertial frames - fictitious accelerations. Do I care whether these accelerations are realizable -no. Do I care about how they are to be obtained - no. Are there any attributes of these accelerations that matter - yes : v = dr / dt and a = dv / dt. The r (t) in frame A and frame B are related by a mathematical transformation of coordinates. If this transformation itself varies in time, well of course the time derivatives of r (t) in frame A will differ from the time derivatives in frame B.

$\mathbf{r_A} = \mathfrak{M} (t) \mathbf{r_B}$

$\frac{d}{dt} \mathbf{r_A} = \frac{d}{dt}\left(\mathfrak{M} (t)\right) \mathbf{r_B} + \mathfrak{M} (t)\frac{d}{dt} \mathbf{r_B}$

$\frac{d}{dt} \mathbf{v_A} = \frac{d^2}{dt^2}\left(\mathfrak{M} (t)\right) \mathbf{r_B}+2\frac{d}{dt}\left(\mathfrak{M} (t)\right) \frac{d}{dt}\left(\mathbf{r_B}\right) + \mathfrak{M} (t)\frac{d^2}{dt^2} \mathbf{r_B}$

$\mathbf{a_A} = \frac{d^2}{dt^2}\left(\mathfrak{M} (t)\right) \mathbf{r_B}+2\frac{d}{dt}\left(\mathfrak{M} (t)\right) \mathbf{v_B} + \mathfrak{M} (t)\mathbf{a_B} \ .$

Call them centrifugal, Coriolis or whatever (what's in a name?), $\mathfrak{M}$ and its derivatives are all that they are. The transformation $\mathfrak{M}$ can be anything you like: rotating, expanding, twisting, you pick it. (What if $\mathfrak{M}= \mathfrak{M} (\mathbf{r} (t) )$ , what then, eh? An example of this situation is Centripetal_force#General_planar_motion).)

This is actually a very limited topic. Physical intuition be damned. Conservation laws be damned. Kepler's laws be damned. And rotating turntables with grooves - well, you tell me. Brews ohare (talk) 23:42, 22 May 2008 (UTC)

Brews, Have a look at Maxwell's original works, in particular his 1861 paper which is available on-line. In particular look at part I and see how he uses the third and fourth terms on the right hand side of equation (5) to explain the force on a current carrying wire. He explains magnetic repulsion in terms of centrifugal pressure in the equatorial plane of a solenoidally aligned sea of molecular vortices. That centrifugal force acts in the radial direction of each of the individual vortices. It is a real effect which we cannot make disappear simply by viewing the whole situation from a Cartesian frame.

Your mistake is that you have failed to realize that the radial and tangential directions are the only reality on both the microscopic scale and the cosmological scale. The Cartesian frame is only a close up section in which hyperbolic motion is viewed as a straight line when the eccentricity is high.

You will find by studying Maxwell's 1861 paper that he explains the F = qvXB force in terms of differential centrifugal pressure acting on a wire from the equatorial plane of the molecular vortices.

The centrifugal force is as much an induced and real effect as the forces of electromagnetic induction because that's exactly what the forces of EM induction ultimately come down to. Centrifugal force is not an illusion which only occurs in rotating frames of reference.

Hence, when we write an article about centrifugal force, we should put an emphasis on the real effects of centrifugal force. The maths, which you like to quote so much, is ideal for the purposes.

But we are dealing with a group of editors who are falsely extrapolating the maths to situations where it doesn't apply and then engaging in specious arguments about fictitious forces combining together to produce illusions.

They are deliberately trying to write centrifugal force out of the centrifugal force page.

As regards the other page on 'reactive centrifugal force' it is in a much better state. However, it totally ommits to mention the real centrifugal force which is being cancelled by the centripetal force. The reactive centrifugal force comes from the real centrifugal force. The reactive centrifugal force is to the real centrifugal force what weight is to gravity.

Reactive centrifugal force is a topic which could be legitimately duplicated on both a centripetal force page and a single centrifugal force page. David Tombe (talk) 06:24, 23 May 2008 (UTC)

[edit] Unresponsive

Hi David: I regret to say I find your above remarks unresponsive, ignoring everything I said, and in no way building upon it. In particular, no engagement of the coordinate transformation, kinematic perspective and its connection to fictitious forces. The red herring of Cartesian coordinates is brought up again, although it does not enter the discussion (neither above, nor in the articles). Brews ohare (talk) 13:32, 23 May 2008 (UTC)

Brews, a coordinate transformation only translates the way we look at something in one frame of reference to how we look at it in another frame of reference. It doesn't tell us any physics.

I have been trying to draw attention to the very real induction effect which is independent of coordinate frames ie. when an object undergoes tangential motion relative to a point, an outward radial acceleration will be induced. That is a real effect. It is centrifugal force.

No maths is necessary. All we need is examples.

I wasn't the one to keep bringing up the Cartesian red herring. It was the others. I have been trying to say that it is irrelevent. David Tombe (talk) 14:25, 23 May 2008 (UTC)

We agree about coordinate transformations being physics independent from the analyst's point of view. Exactly my point. And I'd agree that if you live in a non-inertial frame the artificial constructs of fictitious force are as real a way of looking at things to you as reality can get. I don't know if the reality to the non-inertial observer is what you mean by "real induction effect", however. It appears that you may mean the "real induction effect" (whatever that is) is real to the inertial observer as well. Brews ohare (talk) 14:59, 23 May 2008 (UTC)

Brews, That's it exactly. The induction effect is real no matter how we look at it.

Let's use the example of the marble rolling in the radial groove on the rotating turntable. It experiences a radially outward acceleration that is inducd by the rotation.

Anybody can observe that reality. We can observe it by sitting in a seat in the corner of the room. We don't need to be on the rotating turntable to observe it.

Nope. Already you're wrong. If you can just understand this one point, you'll be fine. From the point of view of somebody sitting in the corner of the room, there is no outward acceleration of the marble. The marble wants to go in a straight line, and there is only one thing keeping it from doing so: a single unblananced inward force on it from the rim, providing a radially inward acceleration. The direction of this force and acceleration change, but their magnitude doesn't. The marble's x and y velocities change due to it, but their vector sum doesn't. It is called a centripital force. It's all there is, from the view of the corner-person. Of course, by Newton's 3rd law the marble also exerts an outward force on the rim, but that doesn't result in any acceleration of the rim, because it's tacked down mechanically. If we confine ourselves to the freebody diagram of marble-only, there's just one force, and it's inward.

Now, translate to riding with the marble, so it's motionless in our new frame. If it's motionless it's not accelerating, and thus any forces on it must be balanced, like two people pushing on opposite sides of a block. Obviously, the rim is pushing inward on one side of the marble-- you can still feel that centripital force, even riding with the marble. So why isn't the marble being accelerated? Which it is NOT, in this frame. Because there is now another NEW force, pushing it outward, exactly balancing the force inward. That new force feels like gravity in your frame, and it's called the centrifugal force. It appears in the rotating frame and the man in the corner doesn't see it. But he doesn't need to, because for him the marble is accelerated. Only the man on the rim sees it, because for him, the marble is NOT accelerated, yet is being pushed by the rim. Thus we require some mysterious new force to push it back toward the rim. And there you are. S B H arris 00:02, 24 May 2008 (UTC)

I think the fictitious idea comes about because if we are on the rotating turntable, since we don't theoretically observe the rotation, then we don't see the cause of the centrifugal force. And so from the perspective of a rotating frame, a centrifugal force seems to be something that just naturally occurs on all stationary objects in the frame.

The Coriolis force also acts tangentially on the marble as it rolls out because the marble is being constrained to follow a co-rotating radial path. The marble also exerts a reactive Coriolis force in the opposite tangential direction to that which the groove exerts on the marble.

When the marble reaches the wall at the edge, the wall exerts a centripetal force on the marble which cancels the centrifugal force. The marble then exerts a reactive centrifugal force on the wall.

But now imagine a marble rolling around the inside of an elliptical enclosure in the horizontal plane.

The reactive centrifugal force will always equal the centripetal force. But The reactive centrifugal force will not always equal the actual centrifugal force. When the radius is expanding, the centripetal force will be less than the actual centrifugal force.

There is an analogy with an accelerating elevator. When the elevator floor accelerates upwards, the normal reaction is greater than the downward gravity. But the weight will always be equal to the normal reaction. Hence the person in the lift will feel heavier.

Accelerating elevators are like elliptical motion.

Uniformly moving elevators or stationary elevators are like circular motion.

(1) Normal reaction is to centripetal force. (2) Gravity is to centrifugal force, and (3) Weight is to reactive centrifugal force. David Tombe (talk) 16:15, 23 May 2008 (UTC)

We've got a semantic difficulty here, because I don't know what "inductive effect" means. Your above remarks do seem to indicate that you don't believe Newton's law of inertia. You probably don't agree with Newton's Principia either. (Click cancel when asked for a password). Brews ohare (talk) 16:49, 23 May 2008 (UTC)

Brews, you know what electromagnetic induction means. The convective case is when a particle moves in a magnetic field and a force is induced on it at right angles to its direction of motion as per F = qvXB. According to Maxwell who derived this expression, B is vorticity.

Centrifugal force is the same. The expression F = mvXω is the parent term for both centrifugal force and Coriolis force.

In other words, tangential motion of a particle INDUCES a force at right angles to it.

Regarding Newton, it's something I need to read more about. My understanding is that he may well be the source of the problem. I believe in both centripetal force and centrifugal force. But I understand there was some kind of dispute between Newton and Huygens and that Huygens was pushing centrifugal force whereas Newton wanted to emphasize centripetal force and the law of inertia. I would need to read exactly what Newton said.

I would probably lean towards Huygenism on the limited amount that I know about their disagreement.

I have written an article stating that inertia IS centrifugal force. Newton's law of inertia strikes me as semantics that mask out the reality of centrifugal force.

The tendency of a body to continue in its straight line path at uniform speed unless acted upon by an external force is merely, in my opinion, an archaic way of saying that the hyperbola of infinite eccentricity is the solution to the central force problem when the centripetal force is negligible.

It's possible that Newton was indeed the source of the problem. David Tombe (talk) 17:41, 23 May 2008 (UTC)

[edit] Induction

In other words, tangential motion of a particle INDUCES a force at right angles to it.

– D. Tombe

Hi David: That is a neat use of the word "induction", and I'd apply it to Coriolis force. As for Huygens vs. Newton, I haven't got the stomach to sort out what these gentlemen were arguing over.

However, whatever the more transparent or intuitive picture may be, its hard to argue against the two points that Newton's formulation (i) works, and (ii) is the currently adopted viewpoint. So an argument that the law of inertia isn't the best approach is better put into an historical comment with citations of the ancient texts.

I don't think there is much to be gained in trying to persuade the everyday engineer that there is a more intuitive approach dating back 4 centuries. Once you have learned a method, there is too much overhead in learning a new one. Reminds me of the adage about "FORTRAN spoils one as a programmer" or "The real programmer can write FORTRAN in any language". Here's another one: "Anyone could learn Lisp in one day, except that if they already knew Fortran, it would take three days." Brews ohare (talk) 18:35, 23 May 2008 (UTC)

Brews, yes Coriolis force too. But Coriolis force only gets naturally induced in a vortex. The example of the marble on the turntable only involves applied and reactive Coriolis force. There is no induced Coriolis force on the turntable like there is an induced centrifugal force.

Newton's method is fine as regards many everyday situations. But it is nevertherless good to be explicity aware of induced centrifugal force as a concept. It is needed in planetary orbital theory as the induced effect that keeps the planets up.

Since you are interested in electromagnetism, I'm sure you'll be interested to see how Maxwell extrapolated the induced centrifugal force into his tiny molecular vortices in order to derive the F = qvXB force.

In the main article, induced centrifugal force has been relegated to a 'colloquial centrifugal force' that is briefly mentioned in the last sentence of the introduction. Anome even took the precaution of removing the reference to the fact that it is induced by rotation. I'm expecting to see that entire sentence erased soon because it is essentially saying that if you have come here expecting to read about centrifugal force then you have come to the wrong page. We don't even have such a page.

The other article deals with reactive centrifugal force. That is to centrifugal force what weight is to gravity. It is the reaction effect when a centripetal force counteracts a centrifugal force totally or partially.

The main article itself is in a bit of a mess. It starts off trying to imply that centrifugal force is only an artifact of coordinate transformation. This is quite wrong. But further down, it gives examples of real centrifugal force devices. It even does an interesting section on centrifugal potential energy in a rotating bucket of liquid, which steers close to the missing fourth term in the Lorentz force. I would expect that that section will eventually be deleted by the fictitiousists. David Tombe (talk) 19:12, 23 May 2008 (UTC)

"induced centrifugal force ... is needed in planetary orbital theory as it is the induced effect that keeps the planets up" Hey, David, the way I was taught, the moon's motion is two-fold: it falls to earth under gravity, but fortunately it also is moving tangentially just fast enough that the distance it drops is exactly the distance it has traveled away from earth on a straight line: ergo we have circular motion. This is Newton's explanation. It doesn't require a "real" centrifugal force. I'd argue that this approach works just fine. I won't argue that there aren't other approaches that use a "real" centrifugal force. However, having alternatives doesn't mean that one of them is "better". And it would seem that "induced centrifugal force" is not needed: Newton's approach doesn't need it. Brews ohare (talk) 20:09, 23 May 2008 (UTC)

Brews, Newton's approach is useless when it comes to dealing with the central force orbital equation. In that equation, need to explicitly state the centrifugal force which is a very active component in the analysis. Anyway, we'll now go over to the centrifugal force talk page and continue on the crucial issue of centrifugal force as a real radial force. David Tombe (talk) 07:09, 24 May 2008 (UTC)

[edit] Reply to SBharris on Circular Motion (2)

SBharris, I am not denying that an inward radial centripetal force causes the direction of the particle to continually change. We all know that its velocity is continually changing by virtue of its direction changing.

But you are quite mistaken when you then deduce that this means there must be a net inward radial acceleration. Yes, there is an inward radial acceleration caused by the centripetal force. But there is no NET inward acceleration.

If the second time derivative of the radial distance is zero, then there is no radial acceleration. That state of affairs exists in circular motion. It is an obvious geometrical fact that there is no net inward radial acceleration in circular motion.

So there must also be an outward centrifugal force acting which cancels the centripetal force when circular motion occurs. The centripetal force merely deflects the particle from its inertial path which already contains centrifugal force.

We can see this centrifugal force even if the centripetal force were to disappear. Cut the string that is holding the object in circular motion. The object will fly off both tangentially and radially. There will be an outward radial acceleration with respect to the origin.

Nope. Cut the sting and the object moves off on a staight line tangent. There is NO acceleration of an object on such a line, with respect to a point off the line (such as the origin). It's no different then if you were hanging in space and somebody threw a baseball at you, and missed. You'd see it approach on a straight line, draw toward a minimal distance (the impact parameter), and then recede from you, again. But no acceleration would you ever see.

Now, if you sit on a point off the line and ROTATE, then things are different. But in that case, even objects at rest to you at a distance, experience circular motion and acceleration (with regard to you). S B H arris 00:45, 24 May 2008 (UTC)

Look at it another way. Consider a rotating turntable with a marble in a radial groove. The centrifugal force will cause the marble to accelerate to the edge. (A tangential Coriolis force will also act on the co-rotating radial motion).

When the marble is restrained at the wall at the edge of the turntable, there will be an inward centripetal force exactly counterbalancing the outward centrifugal force and the result will be that there will be zero net radial acceleration.David Tombe (talk) 06:00, 23 May 2008 (UTC)

Acceleration is a vector quantity, defined to be the rate of change in the velocity vector. So when you say "its velocity is continually changing by virtue of its direction changing," but also "there is no NET inward acceleration," you are simply not using these words in the normal way. This could be part of the general difficulty in communication. -- SCZenz (talk) 13:19, 23 May 2008 (UTC)

SCZenz, the centripetal force causes an inward radial acceleration which has the effect of changing the direction of the particle.

But there is no NET radial acceleration in circular motion because there is also an equal and opposite outward centrifugal force in the radial direction.

It is quite obvious that the second time derivative of the radial distance in circular motion is zero. If it wasn't zero, then we wouldn't have a circular motion.

You can see that centrifugal force simply by considering a radial line from an origin to a particle undergoing straight line motion. Inertia IS centrifugal force. A straight line is the hyperbolic solution to the orbital equation in the special case of zero gravity. David Tombe (talk) 13:36, 23 May 2008 (UTC)

When you are using the radius as one of your coordinates you are using a rotating frame. Hence the centrifugal force. --Itub (talk) 13:39, 23 May 2008 (UTC)

As I just said, the acceleration that appears in Newton's laws is the second time derivative of the position, as a vector quantity, independent of reference frame. The second time derivative of the radial coordinate is not the same as this quantity, and is therefore not the acceleration. -- SCZenz (talk) 13:42, 23 May 2008 (UTC)

There is a parallel here with normal reaction and gravity. The normal reaction of the floor of an elevator causes an upward acceleration. But the net acceleration depends on the degree to which the normal reaction counteracts gravity.

When the normal reaction is equal and opposite to gravity, there will be no net acceleration.

So you can have a normal reaction causing acceleration (deceleration) yet still have no net acceleration. Likewise, you can have a centripetal force causing acceleration yet still have zero net acceleration.

Both the normal reaction and the centripetal force are acting to alter the object's inertial path. David Tombe (talk) 13:36, 23 May 2008 (UTC)

You don't have to call the second order time derivative of the radial distance "acceleration" if you don't want to. I do however. The terminolgy doesn't alter the material point of the argument in any way. The second order time derivative of the radial distance is zero in circular motion and so the net force in the radial direction will be zero.

If it were a gravity orbit circular motion and the centrifugal force suddenly stopped, the centripetal force would cause a net inward second order time derivative of the radial distance. David Tombe (talk) 13:52, 23 May 2008 (UTC)

If you choose to change the definition of "acceleration," then of course the definition of "force" is also changed, and it's not surprising that fictitious forces under the standard definition might become real, or vice versa. However, the standard definitions are, well, standard for the following reason: the F in F = ma coincides with the forces that are actually produced by an identifiable physical mechanism. You can make up whatever new meanings of words you like—you just can't put them in Wikipedia articles. -- SCZenz (talk) 14:04, 23 May 2008 (UTC)

SCZenz, Reality can't be dependent on definitions. In circular motion, there is no net displacement along the radial line. Hence, the centripetal force acting along that line must be counterbalanced by a centrifugal force. Definitions will not allow us to escape from that reality.

You are also ignoring the fact that this centrifugal force can be observed even in the absence of a centripetal force. David Tombe (talk) 14:12, 23 May 2008 (UTC)

Read what I just wrote again. I very specifically connected the standard definition of acceleration to reality; it's your assertion that a lack of "net displacement along the radial line" requires a balance of radial forces which isn't justified, or supported by anything but your own flawed arguments. I began this dialog because I thought there was a very small, clear point of fact upon which I could communicate something new; it's now devolving into the same old argument, so I doubt it will continue past your next comment. -- SCZenz (talk) 14:28, 23 May 2008 (UTC)

SCZenz, Nobody has been interfering with the standard definition of acceleration. The discussion is focused on the radial component of acceleration. If that radial component is zero in circular motion, then all the radial forces must be balanced.

The centripetal force clearly acts inwards and cancels the already existing outward centrifugal force, just as the normal reaction on the Earth's surface cancels out the gravity force. David Tombe (talk) 15:56, 23 May 2008 (UTC)

Mr SCZenz. In your definition a force can only be a force if it results in an action. So this means that forces which do not cause actions can not be referred to in wikipedia articles. Do you want me to check just how many actually are and delete references to them based on your poorly phrased statement? You need to acquire more clarity of thought. As a force does not require an acceleration be produced to be real. Your definition is just a pointless excerise in a pointless logomachy with Mr Tombe based upon your personal animonsity towards him. This is demonstrated by the fact that your argument is not a valid argument at all. Please stop your pointless arguments based on your personally nasty animonsity towards Mr Tombe and admit that he actually does have a valid point.Boggled (talk) 18:39, 23 May 2008 (UTC)

Newton's second law requires that unbalanced forces produce accelerations. Thus, lack of acceleration is evidence of either 1) no force, or 2) balanced forces. An object moving in a straight line, according to Newton's first law, generally continues to do so, and requires no forces to do so. You may postulate pairs of force acting on all free bodies moving in straight lines, but unless there is a reason to do so (such as being able to feel one of them), it is better to assume no forces at all are acting. None of this has anything to do with animosity toward Mr. Tombe. S B H arris 00:40, 24 May 2008 (UTC)

SBharris, the point is that we CAN feel the centrifugal force. Anyway, it's time to go back to the main centrifugal force talk page. I want to deal with this very topic more generally. The issue of net radial acceleration in circular motion is one of three crucial points which relate to my argument that the transformation equations are being wrongly over-extrapolated to fictitious situtions. David Tombe (talk) 07:04, 24 May 2008 (UTC)

Yes, you can feel it, as you can the force of gravity. But of course, only when in the accelerated frame. If you move to an inertial (free fall) frame, it disappears (as does the force of gravity). In any case, even with gravity you don't really feel it as a force. What you feel is the floor pushing up on you, but but that's not gravity-- that's the floor. You feel the floor pushing up, and you see you're not moving up, and you ASSUME the opposing force of gravity. Gravity is indeed much like an inertial force in an accelerated frame, and Einstein points out that it actually is just that. S B H arris 23:59, 24 May 2008 (UTC)

SBharris, see the centrifugal force talk page for a reply. David Tombe (talk) 08:09, 25 May 2008 (UTC)

[edit] An intuitive approach

Hi David: As suggestion made once by User_talk:The_Anome in different words perhaps, is that the article does not adequately represent the "naive" point of view. The natural examples are those of the cornering car, the centrifuge, the graviton amusement park ride, and the centrifugal dryer. The problem with the article is not technical inaccuracy (I'd say) but visceral pallor. Maybe even David would be happy with the article if it were more even-handed in its treatment of non inertial frames? Brews ohare (talk) 20:39, 23 May 2008 (UTC)

Brews, we can now go back to the proper centrifugal force talk page. On your point above, I wouldn't call practical examples naive. I would call the over extrapolation of the maths to fictitious situations naive. David Tombe (talk) 07:11, 24 May 2008 (UTC)

Well, then you're calling Richard Feynman naive, since in section 19 of book I in his lectures on physics, he defines Coriolis force the same way we do, give the two independent components of it, including the one that acts like centrifugal force, and even uses the artificial circle example. Guess he was fooled, too, and you're the only one who's figured this out, eh? S B H arris 02:40, 25 May 2008 (UTC)

SBharris, see the centrifugal force talk page for a reply. David Tombe (talk) 08:10, 25 May 2008 (UTC)

[edit] Personal attacks

In your recent edit[17], you call a fellow editor "a wikistalker and a hypocrite and an absuser of administrative authority". You have been reminded of Wikipedia policies on civility countless times, most recently here: [18].

You may not realize it, but edits like these, besides being likely to get you blocked or even banned, from Wikipedia, also make it even less likely that anything you write will stay in the article. People (including me) get fed up with your aggressive style and, rather than trying to compromise, simply revert the things that you write that we don't agree with. Is this unfair? - Perhaps, but that's the way people tend to behave, and we are within our rights to remove anything that we disagree with unless it is backed up by reliable sources. (And so far you haven't provided any citations supporting the statements that you want to put in the article.)

My advice to you is to take some time off from the fictitious force related articles. (It is likely that your last edit will get you blocked, forcing you to do this, but in either case) The article will still be here a year from now, and then you might feel differently about it. And when you do come back, try finding an authoritative text that says, more or less, what you want to say. Paraphrase the statements from that text into the article, and include a reference to the source, and I am certain that you will see a different result.

--PeR (talk) 10:44, 29 May 2008 (UTC)

Hi David. There is good advice above. I'm sorry that you believe that the world is ignorant, wrong, and evil, but there's nothing I can do about that. I know you are frustrated, but your increasingly-shrill name-calling is creating an unpleasant environment for everyone and violates our civility policy. If you don't want your additions reverted, then you can back them up with reliable sources, as you know. Otherwise it's a wiki, and people get to edit without constantly being called hypocrites and liars. I will not warn you again before blocking you if your incivility continues. -- SCZenz (talk) 11:21, 29 May 2008 (UTC)

[edit] Blocked for personal attacks

Calling other editors mindless, pathetic, and deceitful is an unacceptable violation of our civility policy, regarding which you have been warned extensively. You are blocked for one week. Further blocks will extend very rapidly in length. If you won't change your ways in response to warnings, then you're a hindrance rather than an asset to Wikipedia. -- SCZenz (talk) 06:25, 30 May 2008 (UTC)

[edit] Regrets

David: I'm sorry that the centrifugal force article has proved such a thorn in your side. Response to your arguments has improved the article. However, you have been unable to gain acceptance for some of your ideas. I'd submit that going into melt-down is not an effective response. In the cold light of another day, perhaps you will see that resistance is based upon a logical viewpoint (logical to those in opposition), not animosity, and be better able to approach matters. I hope that is so. Brews ohare (talk) 14:13, 30 May 2008 (UTC)

Brews, I wasn't going to bother coming back until next week. But since SBharris decided to take advantage of me being blocked to talk about the source of my confusion, as if such a source actually exists, can you tell him that I have never been confused between the Coriolis force and the Euler force. I know exactly what the difference is between them and I have even written articles on the issue. They do however both have in common the fact that they are tangential accelerations.

They can both be invoked when the physical set up so demands. But we need to know what that set up is before we can see the role of either of them.

I gave you one example of a rotating turntable with a particle rolling radially outwards along a groove. At constant angular velocity we would have a Coriolis force but no Euler force. The Coriolis force would be applied, and it would be causing the direction of the motion to change.

However, if the turntable were to be angularly accelerated, an Euler force would even act on a particle that is stationary on the turntable.

In a gravitational field, Kepler's law of areal velocity rules out all tangential accelerations, hence it rules out both the Euler force and the Coriolis force.

It was totally specious of SBharris to suggest that I was getting Coriolis force and Euler force mixed up, especially knowing that I wouldn't be able to answer back.

SBharris always seems to assume that he is dealing with a form 1 boy who still hasn't learned the difference between a speed and a velocity. SBharris is clearly unknowledgeable about the role of Kepler's law of areal velocity in planetary orbital theory. The geometrical formula for areal velocity is differentiated and equated to zero. The equation is then multiplied through by r and expanded to obtain exactly the Coriolis force and the Euler force, meaning that Kepler's law of areal velocity eliminates these two forces from the gravitational field.

The Euler force and the Coriolis force both act tangentially. The Euler force changes the magnitude of a tangential speed and the Coriolis force changes the direction of a radial velocity. In that respect, the Coriolis force is more closely related to the centrifugal force than is the Euler force. But neither of them really need to be mentioned in a centrifugal force article. Neither of them occur naturally in gravitational space whereas centrifugal force does occur naturally in gravitational space.

The Euler force and the Coriolis force both need to be physically applied. In the case of the Euler force, that essentially amounts to a torque.

In EM induction, the (partial)dA/dt term is an applied Euler force.

SBharris needs to learn to check up on the existing state of training and knowledge that a person possesses about a topic before assuming that he is dealing with someone of inferior knowledge to himself. He may be a medical doctor but that doesn't put him in the position of being able to lecture applied maths graduates on the difference between a speed and a velocity. David Tombe (talk) 06:24, 31 May 2008 (UTC)

[edit] Airplane banking

David: To divert you from planetary motion, I'd like to engage you about a banking airplane. For the sake of simplicity, let's forget about gravity and lift, and simply imagine the plane flying an elliptical path about some origin of coordinates. Of course, just how the pilot manages this maneuver is a question, but I suppose for now that he has accomplished this trajectory somehow.

Elliptic coordinate system

Copying from the article elliptic coordinates, the most common definition of elliptic coordinates $(μ,ν)$ is

$x = a \ \cosh \mu \ \cos \nu$

$y = a \ \sinh \mu \ \sin \nu$

where $μ$ is a nonnegative real number and $\nu \in [0, 2\pi)$ .

These definitions correspond to ellipses and hyperbolae. The trigonometric identity

$\frac{x^{2}}{a^{2} \cosh^{2} \mu} + \frac{y^{2}}{a^{2} \sinh^{2} \mu} = \cos^{2} \nu + \sin^{2} \nu = 1$

shows that curves of constant $μ$ form ellipses, whereas the hyperbolic trigonometric identity

$\frac{x^{2}}{a^{2} \cos^{2} \nu} - \frac{y^{2}}{a^{2} \sin^{2} \nu} = \cosh^{2} \mu - \sinh^{2} \mu = 1$

shows that curves of constant $ν$ form hyperbolae.

Suppose an airplane travels an elliptical path given by a curve μ = constant = μ₀. We ask, what is the centripetal force acting upon the plane to maintain its path, and what is the tangential acceleration necessary, as a function of ν as viewed from the frame with origin at x = y = 0?

Then, the next question is, if we are the pilot (and therefore in a frame where the plane is apparently at rest) what are the fictitious forces present according to the pilot?

Can you help me with this problem? Brews ohare (talk) 15:58, 31 May 2008 (UTC)

Here is where I find myself at the moment

The plane is located at position r(t) = ( x(t), y(t) ). In time the plane's velocity is:

$\mathbf{v} (t) = \frac{d}{dt} \left( x(t),\ y(t)\right) = \frac{d}{dt} \left( a \ \cosh {\mu}_0 \ \cos \nu (t) ,\ a \ \sinh {\mu}_0 \ \sin \nu (t) \right) \$

$= \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ \frac{d}{dt} \nu (t) \ ,$

which is tangential to the ellipse as can be seen by looking at the element of arc length along the trajectory. d s:

$d \boldsymbol{s} = \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ d \nu (t) \ .$

The acceleration of the plane is:

$\mathbf{a} = \frac{d}{dt}\mathbf{v} = \frac{d^2}{dt^2}\boldsymbol{s} \$

$= \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ \frac {d^2}{dt^2} \nu (t) + \left( -a \ \cosh {\mu}_0 \ \cos \nu (t) ,\ - a \ \sinh {\mu}_0 \ \sin \nu (t) \right) \ \left( \frac {d}{dt} \nu (t) \right)^2$

$= \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ \frac {d^2}{dt^2} \nu (t) - \boldsymbol{r} (t) \ \left( \frac {d}{dt} \nu (t) \right)^2 \ .$

My problem here is that the interpretation of centripetal force didn't just drop out of the equations, as I expected, but has to be worked at. For example, the normal to the trajectory appears to be in the direction:

$\boldsymbol{n} = \left( a \ \sinh {\mu}_0 \ \cos \nu (t) ,\ a \ \cosh {\mu}_0 \ \sin \nu (t) \right) \ .$

I'd expect the centripetal accelertation to be

$\boldsymbol{a_c} = \left( \mathbf{a} \cdot \boldsymbol{n} \right) \frac {\boldsymbol{n}}{\boldsymbol{n} \cdot \boldsymbol{n}} = a\ \left( \frac {d}{dt} \nu (t) \right)^2 \ \left(\sinh {\mu}_0 \cosh {\mu}_0 \right)\frac{ \left( \ \sinh {\mu}_0 \ \cos \nu (t) ,\ \ \cosh {\mu}_0 \ \sin \nu (t) \right)}{\left( \ \sinh^2 {\mu}_0 \ \cos^2 \nu (t)\ +\ \ \cosh^2 {\mu}_0 \ \sin^2 \nu (t) \right)}$

$= a^3\ \left( \frac {d}{dt} \nu (t) \right)^4 \ \left(\sinh {\mu}_0 \cosh {\mu}_0 \right)\frac{ \left( \ \sinh {\mu}_0 \ \cos \nu (t) ,\ \ \cosh {\mu}_0 \ \sin \nu (t) \right)}{{\mathbf{v}}^2}\ ,$

with magnitude:

$|\boldsymbol{a_c}|=\frac{ a^2}{{|\mathbf{v}|}}\ \left( \frac {d}{dt} \nu (t) \right)^3 \ \left(\sinh {\mu}_0 \cosh {\mu}_0 \right)\ .$

Although this result agrees with the result for a circular path in that limit, I have no appealing interpretation of this result.

There also is a tangential contribution to acceleration from the second term:

$\boldsymbol{a_t}=a\left( \frac {d}{dt} \nu (t) \right)^2\ \sin \nu (t)\cos \nu (t) \frac{\left( - \ \cosh {\mu}_0 \ \sin \nu (t) ,\ \ \sinh {\mu}_0 \ \cos \nu (t) \right)}{\left( \ \sinh^2 {\mu}_0 \ \cos^2 \nu (t)\ +\ \ \cosh^2 {\mu}_0 \ \sin^2 \nu (t) \right)} \ ,$

which contributes zero at ν = 0, π/2, π, 3π/2. The total tangential acceleration is then

$\boldsymbol{a_t}=\mathbf{v}\left[\frac{\frac {d^2}{dt^2} \nu (t)}{ \frac {d}{dt} \nu (t) } + \left( \frac {d}{dt} \nu (t) \right)\ \frac{\sin \nu (t)\cos \nu (t)}{\left( \ \sinh^2 {\mu}_0 \ \cos^2 \nu (t)\ +\ \ \cosh^2 {\mu}_0 \ \sin^2 \nu (t) \right)} \right]\ .$

The interpretation of this acceleration? Perhaps the second derivative contribution is the acceleration leading to the Euler force in the pilot's frame of reference? Brews ohare (talk) 18:22, 31 May 2008 (UTC)

Brews, the way to tackle this problem is to set up polar coordinates with an origin, perhaps at the centre of the ellipse. If it were a Keplerian ellipse we would choose the focus, but this is not a Keplerian ellipse.

David: I'll intersperse my remarks among yours. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

The next thing that we must do is establish what radial forces are acting and what tangential forces are acting.

I believe that has been accomplished above: the tangential and normal components of acceleration have been found. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

There will definitely be both a centripetal force and a centrifugal force always acting in the radial direction but they will not in general be equal in magnitude.

I'd interpret the centripetal force as the force normal to the ellipse. I do not find a centrifugal force in the inertial frame, but in the frame of the pilot, the centripetal force becomes the centrifugal force from his viewpoint. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

The centripetal force will be applied inwards from the aerodynamics. The centrifugal force will be induced (as in EM induction) outwards from the tangential motion.

Your terminology for induced implies a velocity dependence - at least in the case of circular motion, neither the cetripetal nor the centrifugal force has velocity dependence. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

There may or may not be a tangential acceleration. If there is a tangential acceleration, it will come from the traction force of the aircraft's engines and it will indeed constitute an Euler force.

A tangential acceleration has been found; I'm not sure that it can entirely be viewed from the pilot's perspective as an Euler force. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

There will be no Coriolis force involved.

Why do you say that? I don't disagree, I just don't know how to argue the point. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

If there is an Euler force present, the problem will be very difficult to analyze.

If there are only radial forces present, the problem may still be very difficult to analyze because the mathematical expression for the aerodynamically induced centripetal force may make it non-analytical.

I think that the best that we could do is a qualitative treatment.

I'd like to be able to do that in a convincing manner. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

The pilot would simply be controlling the elliptical motion on an ongoing basis. When he is flying out to the long ends of the ellipse, he will reduce his banking in order to reduce the centripetal force and let the centrifugal force dominate. Then when he is flying back in again to the minor axis, he will bank more sharply in order to switch on more centripetal force.

I can't see the pilot ever considering himself in anything other than the inertial frame. I couldn't see him bothering to consider himself to be in a rotating frame of varying angular velocity.

David Tombe (talk) 09:22, 1 June 2008 (UTC)

I agree with you about this. Brews ohare (talk) 17:04, 1 June 2008 (UTC)

Brews, You need a second order differential equation in the relevant variable. You then need to sum the forces on the other side of the equation. Centripetal force will always be an applied force so its mathematical expression cannot fall out of vector calculus. We need to physically model it. Off hand, I don't know how you would physically model a centripetal force based on aerodynamic pressure under the wings of a banking aircraft. I suspect that the problem is non-analytical.

On Coriolis force, I may have been slightly wrong. The problem with Coriolis force and the Euler force is that in the natural state of affairs, they always mutually cancel each other.

But if you apply a torque, then that will upset the balance. So in that case, a Coriolis force probably would show through. I'd need to think more about that.

In your particular example, the aircraft engine is indeed likely to be applying a torque occasionally, and so Coriolis force and Euler force may well be coming and going.

Could I say to you that in general you are making the same mistake as FyzixFighter.

Classical mechanics at advanced level is all about solving equations of motion. We use the expressions from the polar coordinates, but those expressions alone tell us no physics. We need a physical scenario upon which to build a differential equation. We have to examine which forces are present and do our best to construct a mathematical expression for those forces. You and FyzixFighter seem to be under the impression that those polar expressions alone are telling us about some particular physical scenario.

I'll return once again to the gravity orbit as it is the best example of a complex analysis which is fully analytical. Newton's law of gravity (the inverse square force law) goes into the equation for the centripetal force. The centrifugal force is a naturally induced velocity dependent force and we use the standard expression. These two are then summed and equated to the second order derivative of the radial distance.

We will have a very difficult equation to solve, but it can be done and it checks out perfectly to an ellipse, hyperbola, or parabola.

However your aircraft example is much more difficult since it involves more variables, which are somewhat arbitrary or which need to be aproximated. David Tombe (talk) 07:33, 2 June 2008 (UTC)

[edit] Kinematics

Hi David: Thanks for your thoughts. I believe you are engaging a problem more difficult than the one I am attempting. Rather than consider actual flight with the problems of lift, gravity, and engine thrust, I am attempting only to find what accelerations an elliptical path entails. That problem needs only differentiation of the displacement r ( t ), considered to be a realized trajectory (somehow) rather than the actual calculation of r ( t ) from first principles. Brews ohare (talk) 11:12, 2 June 2008 (UTC)

Brews, if you are only looking for the requirements for elliptical motion, then all you need is that the inward centripetal force and the outward centrifugal force alternate cyclically between which one is of the greater magnitude. David Tombe (talk) 12:48, 2 June 2008 (UTC)

Hi David: For convenience, I've reprised the math below:

The plane is located at position r(t) = ( x(t), y(t) ). In time the plane's velocity is:

$\mathbf{v} (t) = \frac{d}{dt} \left( x(t),\ y(t)\right) = \frac{d}{dt} \left( a \ \cosh {\mu}_0 \ \cos \nu (t) ,\ a \ \sinh {\mu}_0 \ \sin \nu (t) \right) \$

$= \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ \frac{d}{dt} \nu (t) \ ,$

The acceleration of the plane is:

$\mathbf{a} = \frac{d}{dt}\mathbf{v} = \frac{d^2}{dt^2}\boldsymbol{s} \$

$= \left( -a \ \cosh {\mu}_0 \ \sin \nu (t) ,\ a \ \sinh {\mu}_0 \ \cos \nu (t) \right) \ \frac {d^2}{dt^2} \nu (t) - \boldsymbol{r} (t) \ \left( \frac {d}{dt} \nu (t) \right)^2 \ ,$

leading to a centripetal (inward) acceleration:

There also is a tangential contribution

These are the results in an inertial frame, watching the plane. They are not fictitious accelerations. Other than the results being indigestible, do you have a basic quarrel with the logic of the approach? If so, please explain. Brews ohare (talk) 16:58, 2 June 2008 (UTC)

Brews, there is no physics in it. All you have done is expanded general Cartesian expressions. That maths doesn't describe any particular motion.

Hi David

The math does describe motion in an elliptical path set by μ₀ and parameter a. The rate of travel on the path is chosen as you wish by setting the time dependence of r ( t ), but the path must be elliptical. It may be only terminology, but I'd call this a "particular motion".Brews ohare (talk) 14:35, 3 June 2008 (UTC)

Any centripetal force will have to be physically applied, and before we can obtain any useful differential equation, we need to know an expression for the centripetal force as it arises from aerodynamic pressure. It would probably be some function of the bank angle of the wings and also probably of the speed of the aeroplane, and there would have to be a magnitude coefficient in it determined from experiment.

It seems to me that you are totally locked into the maths, to the extent that you are forgetting about the fact that these problems involve applied physical forces that cannot be calculated from maths alone. David Tombe (talk) 06:19, 3 June 2008 (UTC)

Hi David

I'm aware that physics is a great subject, and that the universe is complex. However, here we have a tiny slice of all that. But not a tiny as you believe. In particular: the forces that are needed to achieve a specified trajectory are determined by the trajectory alone, as done above: what is needed is only the path and how rapidly it is traveled. As I'm trying to say to you, kinematics does not treat the issues you are dying to discuss, namely, how is the trajectory achieved. Be the path traversed by a meteor or by a motor scooter, the kinematics is identical, and the accelerations necessary also are identical. However, certainly, I agree, the origin of the forces (another matter, please) is totally different.

Perhaps we need to discuss the differences between kinetics and kinematics? These two fields are traditionally separated, and their subject matters are clearly distinguished from each other in the literature. Brews ohare (talk) 14:39, 3 June 2008 (UTC)

Brews, the difference between kinematics and kinetics was one of the first things that I learned in applied maths. But even in kinematical problems we need an expression that relates to the source of the acceleration, or else we need the actual acceleration itself.

For example, in alot of those S = ut + 1/2at^2 type problems we would be given a value of 9.81 for the acceleration. There always has to be some input as regards the value of the acceleration or the cause of the acceleration. We can't just solve a general kinematical expression without any input information about the cause of the acceleration.

And in your example above, we need the expression for the centripetal force in terms of banking angle, velocity, and a coefficient. David Tombe (talk) 20:45, 3 June 2008 (UTC)

David: I am not getting it. In the math above we have the acceleration in tangential and perpendicular directions based upon traveling an elliptical orbit at a rate given by the coordinate ν ( t ). We don't need banking angle etc. unless we wish to determine just how a specific type of plane can duplicate the assumed circling of the path, and that question is about that specific plane, a separate matter. If instead we traversed the path in a motor boat on Lake Erie, we'd have different expressions relating the physics of turning boats in polluted water, but the required accelerations in both problems would be the same. I took the trouble to add several citations to the articles on kinetics and kinematics that expound on these points. Please take a look. Brews ohare (talk) 21:31, 3 June 2008 (UTC)

Questions and answers

Here's a yes or no question: Do the following equations express the acceleration for the cited problem? The component normal to the elliptical orbit is:

and there is also a tangential contribution:

Here's a follow-up question: If the answer to the above question is "no", do you agree that acceleration as a matter of definition is:

$\boldsymbol{a} (t) = \frac{d^2}{dt^2} \boldsymbol{r} (t) \ ,$

where r ( t ) locates the moving object relative to a selected point of origin in an inertial reference frame?

Here's the next follow-up question: if the answer to the above is "no", what is the definition? Brews ohare (talk) 22:18, 3 June 2008 (UTC)

No Brews, that's not how dynamics is done. We set up a differential equation and substitute the terms for what we physically know about the applied forces.

I'll now give you a very simple example. Circular motion as in a weight being flung around on a string.

We begin with the standard radial scalar equation,

centripetal force + centrifugal force = radial force.

We know that the net radial force will be zero. Hence, the centrifugal force mv^2/r wil be equal to the centripetal force which will be T for the tension in the string. In other words, we will have centripetal force = T = mv^2/r.

In your aeroplane example, we need to know what the expression is for centripetal force in terms of bank angle and velocity. If we were to get that, I guarantee that the final differential equation would be so complicated that it would be non-analytical.

Gravity however is a well known case that is very complicated but nevertheless analytical. David Tombe (talk) 10:34, 4 June 2008 (UTC)

Hi David: Well, you have ducked answering my "yes" or "no" questions, and have not looked at the references I put up about kinematics. So, you have not addressed the key issue about the role of:

$\boldsymbol{a} (t) = \frac{d^2}{dt^2} \boldsymbol{r} (t) \ .$

As you see above, this equation has been used to find the acceleration, and there is no doubt in my mind that it has been found correctly using this definition of acceleration. That settles the issue about what the accelerations are, except for (maybe) some algebraic errors. It is an entirely mathematical (not physics) exercise.

The matters you bring up, about bank angle or gravity, are not pertinent to simply finding the derivative

$\boldsymbol{a} (t) = \frac{d^2}{dt^2} \boldsymbol{r} (t) \ .$

I feel you are refusing to have a dialog, and simply making unresponsive assertions that fall in the realm of (as you say yourself) dynamics. Brews ohare (talk) 15:08, 4 June 2008 (UTC)

Brews, we'll go over to the proper centrifugal force talk page and discuss it there. David Tombe (talk) 12:07, 6 June 2008 (UTC)

[edit] Reply to SBharris

SBharris, I have given further thought to the matter. I was never in any doubt that there are no net tangential forces involved in the simple two body planetary orbit. It follows directly from Kepler's law of areal velocity. And there was never a time when I was confused between the Euler force and the Coriolis force. Kepler's law of areal velocity gets rid of the entire package of tangential forces.

David: Kepler's law of areal velocity has no place in the articles on centrifugal force, fictitious force, which are strictly kinematical discussions. Brews ohare (talk) 18:25, 4 June 2008 (UTC)

Brews, If I was allowed to write he article, Kepler's law of areal velocity would not be appearing in it other than in passing. I am only talking about it in the discussion pages in order to help the likes of you get a better comprehension of the overall topic of central forces. The importance of Kepler's law of areal velocity is that only radial forces are involved unless an external torque is applied to the system. David Tombe (talk) 07:23, 5 June 2008 (UTC)

I did however review the situation on further reflecton, as I told Brews above. I considered that the net zero tangential force might be an ongoing summation of an Euler force and a Coriolis force that always cancels out, just as a centripetal and a centrifugal force always cancel out in the radial direction in the special case of circular motion.

David: The Euler force is zero in the case of uniform angular velocity, and so cannot possible cancel the Coriolis force in a general case. Brews ohare (talk) 18:25, 4 June 2008 (UTC)

Brews, if you had read on, you would have seen that I also concluded that the Euler force doesn't cancel with the Coriolis force.

The case for making such an assertion rests with exclusively focusing on the radial component of the motion which does appear to be subjected to a side on Coriolis force.

But I have now rejected the idea on two counts. We cannot look at a direction change in relation to one scalar component of a vector. Any direction changes must be considered in relation to the vector itself, in which case any direction change is always in the radial direction.

There is also the issue of the ridiculousness of having two naturally induced forces that always mutually cancel.

So I concluded finaly that in he simple Keplerian two body problem there is unequivocally no Coriolis force involved.

David: Again, the Keplerian two body problem is simply a fascination of your own, and has no place in the articles on centrifugal force, fictitious force.

I wouldn't say that at all. It is an excellent case scenario that deals with centrifugal force in a very general way in the context of a topic of general interest.David Tombe (talk) 07:23, 5 June 2008 (UTC)

I'll be back on the main page on Saturday and I think that you ought to seriously reconsider your position on the artificial circle. The Coriolis force never swings into the radial direction and in circular motion we absolutely need to have a net radial acceleration of zero.

David: The "artificial circle" is a kinematical example that naturally appears in the articles on centrifugal force, fictitious force. None of your arguments regarding planetary motion and inverse square law systems has any bearing upon it at all. Brews ohare (talk) 18:25, 4 June 2008 (UTC)

Brews, they do. The planetary orbital equation is a practical demonstration of the fact that in circular motion, the radial acceleration must be zero. In the artificial circle, the equation is not balanced due to an excess of inward centripetal force which is ridiculously ascribed to the Coriolis force. David Tombe (talk) 07:23, 5 June 2008 (UTC)

David

Well, I'm pretty frustrated with your answers, which invariably fail to address any issue or any citation, or any example that is raised to try to open a real dialog. Brews ohare (talk) 14:25, 5 June 2008 (UTC)

Brews, we can carry on the discussion on the centrifugal force talk pages. David Tombe (talk) 12:11, 6 June 2008 (UTC)

[edit] Editor controversy

And finally, I've been watching events on the discussion page and I can't intervene. There is an anonymous editor who begins with 72. It is not me. He appears to be supporting me but I would need to clear up a few things with him about that Newton reference. For example, what was Newton's attitude to centrifugal force when it doesn't balance with centripetal force?

Anyway, I see that some new guy "Akroterion" has arrived on the scenes who hasn't been involved in this debate up until now and his record indicates that he has no interest in this topic. His arrival seems to be for the sole purpose of deleting the comments of the anonymous 72. And he has even gone to the extent of making the inuendo that it is me.

Whoever he is, it would appear that he is totally guilty of the very things that he is accusing the anonymous 72 of. I'd be grateful if you might somehow inform he to reconsider his involvement in this discussion because he has clearly arrived as biased and prejudiced as anybody could ever be. David Tombe (talk) 10:34, 4 June 2008 (UTC)

Actually, I saw page blanking and reverted it, insisting that the editor play by the rules and add comments at the end, not blanking other peoples' edits. You're right, I have little interest (and nothing to add) in this subject, but as an admin I do expect everybody to follow the rules. When a new anonymous editor appears after a named account is blocked voicing support for the blocked editor and alleging a conspiracy, it raises concerns. If I was convinced of your involvement, I would have extended your block. I have not. Acroterion _(talk) 18:03, 4 June 2008 (UTC)

Acroterion, the point is that the anonymous editor in question has been active for quite some time. Whoever he is, he didn't just appear when I was blocked. He agrees with some things that I say but I need to discuss with him about his Newton reference.David Tombe (talk) 07:13, 5 June 2008 (UTC)

David, I've been looking through the discussions on Talk:Centrifugal force as you requested. I don't see anything "malicious" on either side. I do see a degree of frustration on both sides: I've seen far worse, though, and some editors appear to be enjoying the debate. As I stated, I won't judge the content: my closest personal relevant education is in statics and I simply don't have the right background to sort the content. I don't see any reason for administrator intervention: administrators don't resolve content disputes, and the tools to deal with misbehavior are blunt instruments. I arrived here because an IP was blanking other people's comments with his statements, which is an admin-related matter.

My reading of the discussion is that most editors disagree with your positions. They also sometimes disagree with each other, as in the example you cite. A fair number of people have passed through, and apart from the IP that was demanding apologies, you haven't successfully convinced them. There is always an opportunity for a Request for Comment or even Requests for Mediation if you feel that broader participation is needed.

The episodic nature of this kind of discussion makes it a poor substitute for a seminar room full of people arguing/discussing. Something that would be hashed out in half an hour in person (with perhaps a few minutes of shouting) takes days of wrangling and leaves hurt feelings. It's the nature of the medium. People take offense, express frustration, and it's all permanently recorded.

In summary, I see frustration, but not malice. Regards, Acroterion _(talk) 02:15, 10 June 2008 (UTC)

User talk:David Tombe

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[edit] Your definition of acceleration?

[edit] Vandalism warning, April 2008

[edit] More vandalism

[edit] ANI-notice

[edit] Other accounts ?

[edit] Enough

[edit] Centrifugal force

[edit] Neutral Point of View

[edit] My reverting you on Centrifugal Force

[edit] Verifiability policy

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[edit] Discussion

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[edit] Warning again

[edit] Discussion again

[edit] Have you noticed...?

[edit] Blocked again, as per warning above

[edit] Evading the block.

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[edit] A word of caution

[edit] Centrifugal force

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[edit] Reply to Wolfkeeper

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[edit] Quote for possible comment

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[edit] I'll respond here, since you are blocked again

[edit] Links now work

[edit] Centripetal force

[edit] So, which is it?

[edit] Finding an expert

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[edit] Reply to Anome's Questions

[edit] The Artificial Circle

[edit] Fictitious Force article

[edit] Reply to SBharris on Circular Motion

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[edit] Reply to Brews ohare

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[edit] Reply to SBharris

[edit] About orbital theory

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[edit] The Three Body Problem

[edit] Proposed addition to Centrifugal Force

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[edit] Induction

[edit] Reply to SBharris on Circular Motion (2)

[edit] An intuitive approach

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