User talk:JDoolin

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Contents 1 Contribution to "Lorentz transformation" 2 Lorentz transformation image 3 Milne Model 4 Reply to comment on Bell's problem 4.1 Welcome Argument Rod Ball 4.2 Saturday, August 26, 2006 4.3 Lorentz Pre-Relativity Theory? 5 Big Bang article NPOV dispute 6 IP Block

[edit] Contribution to "Lorentz transformation"

Please clarify your contribution to "Lorentz tranformation"! (Actually I think what you wrote is completely wrong and should be removed.)

What are "passing event" and "measured event"? What is \tau, \chi, \tau', \chi'? What means "origins of the two reference frames make their closest approach"?

The Lorentz Transformation is a transformation around a specific event. Just like a rotation cannot be done without a center to rotate around, a Lorentz Transformation can't be done without an event to transform around. In general this is the Origin of the Lorentz transformation. The event at (t=0,x=0,y=0,z=0). Presumably, there is something AT that event, or else there is little of interest about the transformation. i.e. an object is changing velocity, thereby changing reference frames, thereby requiring a Lorentz Transformation. Another way of thinking of it is if two objects have passed each other at that event, or will pass each other at a certain place and time, this also gives an appropriate position around which to perform a Lorentz Transformation.

The following sentence makes no sence: "If the variables are referenced from this passing event, the equations will properly map the coordinates (x,t) of a set of events in the first reference frame to the coordinates (x',t') of the same set of events in the second frame."

I was fumbling for words, I suppose. I meant to make clear that the Lorentz Transformation works as a bijection that maps the Euclidian space-time of one observer into the Euclidian space-time of another observer.

You should not say things like "If the origins of the two reference frames are approaching". The origins of the space-time coordinate systems are single events, that do not "approach".

Reference frames are not space-time coordinate systems. They are space-space-space. They have moving origins.

I don't understand what you want to say about Andromeda. What the heck is a "passing event"?

It's a hypothetical future event where Andromeda and the Milky way pass one another, or make their closest approach. Presumably the origin for an observer in Andromeda galaxy is at their location, which we are approaching, while the origin for an observer here is here. The passing event is where those two origins make their closest approach.

What you say about the big bang is complete nonsense. Special Relativity and Lorentz transformation do not apply to cosmology. We need General Relativity for that!

If you could come up with one good reason that Lorentz transformation does not apply to cosmology, that might be of some interest. However, no one has ever done so. It's just repeated ad nauseam, similar to the assumption of universal homogeneity. I've recently discovered that this debate has been in the graveyard since about 1933, when A. E. Milne developed a model for the universe which was nonhomogeneous and satisfied the cosmological principle. His arguments in "Relativity, Gravitation, and World Structure" are such that utterly demolish all the modern cosmology that has been done in the 70 years since.

I read the paragraph "These four relations..." again and I know now what you mean. Please remove this paragraph, it is more confusing than explanatory! You seem to have a poor understanding of the subject.

The following is nonsense for sure:

"If the origins of the two reference frames are approaching when the measured event occurs, then t is negative, while x and v have opposite signs."

I'm confused about where you're getting this. My statements were immediately deleted as crackpottery almost two years ago. You want me to edit an old revision?

I agree that I was unclear. I did not distinguish very clearly between origin events of a Lorentz Transformation and origin positions of a reference frame. The origin of a reference frame is a non-accelerating (i.e. inertial) observer. If the two observers are approaching one another, then the most appropriate event around which to perform a Lorentz transformation is the event at which they will pass. Since the time at this event will be zero, the current time at either of the two observers must be negative. As far as x and v having opposite signs, it depends on which x and which v you're thinking of, and as such, I can see why you thought I didn't have any idea what I was talking about. Sorry 'bout that.

A "measured event" can have any coordinates t and x. Again: Don't say "two reference frames are approaching"! Two observers can approach. Being in different reference frames means using different space-time coordinate systems, that are related by lorentz transformation. Reference frames do not approach!

True, because they coexist in the same space. But I didn't say the two reference frames--I specified the Origins of the two reference frames.

Please revert the artice to the previous version: http://en.wikipedia.org/wiki/Wikipedia:How_to_revert_a_page_to_an_earlier_version

If you don not react, then I will do it!

Sorry I guess I didn't see this in time! Wasn't that a couple of years ago?

[edit] Lorentz transformation image

Could you consider changing the license for Image:Animated_Lorentz_Transformation.gif to CC-By rather than CC-By-SA? The image was included in Lorentz transformation, but I think it's way to big to put in an article. I thought of taking putting one frame rather than the whole animation sequence into the article, but the CC-By-SA license does not allow this. Han-Kwang 20:40, 12 July 2006 (UTC)

I'm pretty sure that CC-By-SA allows you to do this. Let me know if I'm wrong. Anyway, you've got my permission!

Yes, I was wrong. I thought SA means no changes to the file, but it applies to the licensing conditions. Han-Kwang 12:50, 9 August 2006 (UTC)

May I see the Mathematica code that makes that image if it is yours and you have it to give out? Alejandr013 22:20, 23 August 2006 (UTC)

I have located the program. Do you know an easy way to get Mathematica Source Code into a text document?

To my great surprise this worked! Just copy and paste this code directly into Mathematica 5.1. There may be some extra stuff in here, too, and I don't think I have the line in here that turns it into a gif image.

\!\(\(SetOptions[ParametricPlot, AspectRatio -> 1\/GoldenRatio, Axes -> Automatic, AxesLabel -> None, AxesOrigin -> Automatic, AxesStyle -> Automatic, Background -> Automatic, ColorOutput -> Automatic, Compiled -> True, DefaultColor -> Automatic, DefaultFont \ :> $DefaultFont, DisplayFunction :> $DisplayFunction, Epilog -> {}, \ FormatType :> $FormatType, Frame -> False, FrameLabel -> None, FrameStyle -> Automatic, FrameTicks -> Automatic, GridLines -> None, ImageSize -> Automatic, MaxBend -> 10.`, PlotDivision -> 30.`, PlotLabel -> None, PlotPoints -> 25, \ PlotRange -> Automatic, PlotRegion -> Automatic, PlotStyle -> Automatic, Prolog -> {}, RotateLabel -> True, TextStyle :> $TextStyle, Ticks -> Automatic];\)\ \[IndentingNewLine] \(Clear["\<Global`*\>"];\)\[IndentingNewLine] \(setScale[x0_, x1_, y0_, y1_, h_] := SetOptions[ParametricPlot, PlotRange -> {{x0, x1}, {y0, y1}}, AspectRatio -> \(y1 - y0\)\/\(x1 - x0\), ImageSize -> {h/\(\(y1 - \ y0\)\/\(x1 - x0\)\), h}];\)\[IndentingNewLine] \(setColor[r_, g_, b_, t_] := SetOptions[ParametricPlot, PlotStyle -> {RGBColor[r/8, g/8, b/ 8], Thickness[t]}];\)\[IndentingNewLine] \(SetOptions[ParametricPlot, DisplayFunction -> Identity];\)\ \[IndentingNewLine] \(out = 11;\)\[IndentingNewLine] setScale[\(-out\), out, \(-out\), out, 300]\[IndentingNewLine] \(l1 = ParametricPlot[{t, t}, {t, \(-10\), 10}];\)\[IndentingNewLine] \(\(l2 = ParametricPlot[{\(-t\), t}, { t, \(-10\), 10}];\)\(\[IndentingNewLine]\) \)\[IndentingNewLine] \(wl[x0_, t0_, v_] := ParametricPlot[{x0 + v*t, t0 + t}, {t, \(-10\), 10}, PlotStyle -> {RGBColor[1/8, 1/8, 4/8], Thickness[ \ .01]}];\)\[IndentingNewLine] \(wr[x0_, t0_, v_] := ParametricPlot[{x0 + x, t0 + v*x}, { x, \(-10\), 10}, PlotStyle -> {RGBColor[1/8, 4/8, 1/ 8], Thickness[ .01]}];\)\[IndentingNewLine] future[n_] := ParametricPlot[{ x, \@\(x\^2 + n\^2\)}, {x, \(-10\), 10}, PlotStyle -> { RGBColor[6/8, 0/8, 6/8], Thickness[ .01]}]\[IndentingNewLine] past[n_] := ParametricPlot[{x, \(-\@\(x\^2 + n\^2\)\)}, {x, \(-10\), 10}, PlotStyle -> { RGBColor[6/8, 0/8, 6/8], Thickness[ .01]}]\[IndentingNewLine] right[n_] := ParametricPlot[{\@\(t\^2 + n\^2\), t}, {t, \(-10\), 10}, \ PlotStyle -> {RGBColor[6/8, 0/8, 0/8], Thickness[ .01]}]\[IndentingNewLine] left[n_] := ParametricPlot[{\(-\@\(t\^2 + n\^2\)\), t}, {t, \(-10\), 10}, PlotStyle -> {RGBColor[6/8, 0/8, 0/8], \ Thickness[ .01]}]\[IndentingNewLine] futureEvent[x0_, y0_, v_, t_] := {x0 + \(v*t\)\/\@\(1 - v\^2\), y0 + t\/\@\(1 - v\^2\)}\[IndentingNewLine] rightEvent[x0_, y0_, v_, x_] := {x0 + x\/\@\(1 - v\^2\), y0 + \(v\ x\)\/\@\(1 - v\^2\)}\[IndentingNewLine] \(futureSpace[v_, t_] := \(v*t\)\/\@\(1 - v\^2\);\)\[IndentingNewLine] \(futureTime[v_, t_] := t\/\@\(1 - v\^2\);\)\[IndentingNewLine] \(rightSpace[v_, x_] := x\/\@\(1 - v\^2\);\)\[IndentingNewLine] \(\(rightTime[v_, x_] := \(v\ x\)\/\@\(1 - v\^2\);\)\(\[IndentingNewLine]\) \)\[IndentingNewLine] "\<(a) World line of O's clock at x=1m; red:\>"\[IndentingNewLine] \(setColor[8, 0, 0, .001];\)\[IndentingNewLine] \(\(p3a = ParametricPlot[{1, t}, {t, \(-10\), 10}];\)\(\[IndentingNewLine]\) \)\[IndentingNewLine] "\<(b) World line of Particle Moving at .1c from (.5,0); green:\>"\ \[IndentingNewLine] \(setColor[0, 6, 0, .001];\)\[IndentingNewLine] \(\(p3b = wl[0, 0, .1];\)\(\[IndentingNewLine]\) \)\[IndentingNewLine] "\<(c) World line and world region of particle moving at .5c from (0,0); blue:\>"\[IndentingNewLine] \(setColor[0, 0, 8, .001];\)\[IndentingNewLine] \(p3cwl = wl[0, 0, .5];\)\[IndentingNewLine] \(setColor[0, 0, 6, .001];\)\[IndentingNewLine] \(p3cwr = wr[0, 0, .5];\)\[IndentingNewLine] \(\(Show[l1, l2, p3a, p3b, p3cwl, p3cwr, DisplayFunction -> \ $DisplayFunction];\)\(\[IndentingNewLine]\) \)\[IndentingNewLine] \*"\"\<(d) Locus of events whose interval from the origin is 1 secom. One secom is the amount an object would age if it moved at a steady pace from the origin to the location of the event. Since the author insists on making c unitless, I am creating the unit 'secom' to be the time it takes light to travel one meter. It \ is tiny, 3*\!\(10\^8\)secom = 1 second. (blue)\>\""\[IndentingNewLine] "\<(e) Locus of events whose interval from the origin is 1 meter. A 1 \ meter interval between two events means there is some observer where the two \ events occur simultaneously and one meter apart. (red)\>"\[IndentingNewLine] \(p3df = future[1];\)\[IndentingNewLine] \(p3dp = past[1];\)\[IndentingNewLine] \(setColor[7, 0, 0, .001];\)\[IndentingNewLine] \(p4el = left[1];\)\[IndentingNewLine] \(p4er = right[1];\)\[IndentingNewLine] \(Show[l1, l2, p3df, p3dp, p4el, p4er, DisplayFunction -> $DisplayFunction];\)\[IndentingNewLine] \(SetOptions[ListPlot, DisplayFunction -> Identity];\)\[IndentingNewLine] \(tTicks[ v_] := ListPlot[Table[futureEvent[0, 0, v, t], {t, \(-10\), 10, \ 1}], PlotStyle -> PointSize[0.03]];\)\[IndentingNewLine] \(xTicks[v_] := ListPlot[Table[rightEvent[0, 0, v, x], {x, \(-10\), 10, 1}], PlotStyle -> PointSize[0.03]];\)\[IndentingNewLine] Layers := Join[Table[future[n], {n, \(-10\), 10}], Table[past[n], { n, \(-10\), 10}], Table[left[n], {n, \(-10\), 10}], Table[right[n], {n, \(-10\), 10}]]\[IndentingNewLine] \(Show[l1, l2, xTicks[ .5], tTicks[ .5], p3cwl, p3cwr, Layers, DisplayFunction -> $DisplayFunction];\)\[IndentingNewLine] wRegions[v_] := Table[wr[futureSpace[v, t], futureTime[v, t], v], {t, \(-10\), 10}]\[IndentingNewLine] wLines[v_] := Table[wl[rightSpace[v, x], rightTime[v, x], v], {x, \(-10\), 10}]\[IndentingNewLine] \(vel[p_] := p\/\@\(p\^2 + 1\);\)\[IndentingNewLine] "\<Hello!\>"\[IndentingNewLine] \(Table[Show[wLines[vel[p]], wRegions[vel[p]], l1, l2, xTicks[vel[p]], tTicks[vel[p]], Layers, DisplayFunction -> $DisplayFunction], {p, \(-3\), 3, .05}];\)\[IndentingNewLine] \)

[edit] Milne Model

Who is Milne and where is his model published? Gazpacho 20:37, 16 August 2006 (UTC)

Milne is Edward Arthur Milne and the book I refer to is Relativity, Gravitation, and World Structure. In great part the first several chapters of the book are largely his personal take on his model vs. that of Einstein, Eddington, Russel, etc. and their comparitive merit. JDoolin 21:21, 16 August 2006 (UTC)

I see. I encourage you to write about this physicist and his theories, but this article seems to be largely your personal take on the model and its comparative merit. Articles on Wikipedia are supposed to be neutral and not represent particular editors' views. Could you revise the article with this in mind? Gazpacho 20:52, 16 August 2006 (UTC)

I could post the article on Usenet and see if anyone has any suggestions for a way to make it seem more neutral. Can you set Galileo's model of the solar system and Ptolemy's model of the solar system side by side and take a neutral stance? Not likely. The best I can do is take all references of the Standard Model OUT of this description, so they are not directly compared. I would be willing to do that. Although I would really prefer to have a supporter of the Standard Model make a side-by-side alternative view of Milne's Model, or pointing out the flaws. Removing the information seems a little drastic. How can anyone make an informed decision without hearing both sides? JDoolin 21:21, 16 August 2006 (UTC)

Actually, Galileo's model was more complex that the Ptolemaic model of the day, due to his use of circular orbits and epicycles. So yes, it is possible and necessary to treat such questions neutrally. If Milne, or another published author who liked his model, pointed out a problem in another physicists model, you can cite that. But original (e.g. first-person) commentary should be left out. Gazpacho 21:40, 16 August 2006 (UTC)

I guess it's not at all clear whether circular orbits and epicycles are more or less complex than elliptical orbits that we now expect to be the case. The law is simpler, but the three-body-problem is infamous for it's difficulty. JDoolin 16:47, 22 August 2006 (UTC)

[edit] Reply to comment on Bell's problem

Although your "outside user" comment on my RfC perhaps missed the point slightly, in that it concerned the problem itself rather than the conduct of argument, which was what the RfC was about, if you don't mind I can't help disagreeing with the point you make. I honestly think you gave up unnecessarily and that Russel is wrong. If you begin by saying they start accelerating simultaneously ,then that is a matter of assumed fact in the setup and common to both launch and s'ship frames. It makes no sense to say they "suddenly" find themselves in a frame where it's not simultaneous. Presented graphically in the usual x-t diagram, the x' axis of the s'ships starts horizontal and smoothly tilts w.r.t. launch x axis with increasing v. This does not "change the past" by making the forward one have started first. Both s'hips see the other one start slightly behind by just about the light time of their launch separation. As I explain in the talk page, the blue shift of front to back signals and red shift of rear to front signals are not due to any "Doppler" increasing separation but simply due to the increase in v of both s'ships during signal transit, which involves no change in distance nor difference in clock rates. Rod Ball 13:37, 24 August 2006 (UTC)

Let's try to define a very explicit problem so that there is no confusion. Let us say that two ships accelerate by the same amount. Can we pretend at least as an approximation that instantaneous acceleration is possible? (I have had some arguments about this.) Also, let's pretend we have an external observer watching. So both ships are seen to accelerate at the same moment by the external observer. These events are at x=t=0, and x=-1,t=0

Both ships accelerate to 50% of the speed of light, instantly. To the external observer, are the ships any closer or further from one another? I would say definitely no.

However, if you assume the validity of the Lorentz Transformation, you can calculate the position of the x=-1,t=0 event in the new frame.

 t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
 x'=(x-vt)    /sqrt(1-v^2/c^2)

The (0,0) event remains at (0,0)

However, the (-1,0) event becomes

  t'=0-.5(-1)/sqrt(.75)=0.577 (*later than t'=0*)
  x'=-1/sqrt(.75)=-1.154 (*further away that x'=-1*)

In effect, they "suddenly" find themselves in a reference frame where the events are no longer simultaneous.

Now I don't want to say that history has in any way changed because of this transformation. It is only that the observer's velocity affects both his perception of history and his perceptions of the present. All of the events are still in place, and no cause-effect relationships have reversed. It is merely a whee little change in when and where things happened, but not how.

This argument will of course not hold up if you don't regard the Lorentz Transformations as valid, and it may be greatly complicated if you wish to apply a more calculus based approach where the acceleration is a function of time rather than instantaneous.

The confusion that I had in my earlier argument was that I knew that a second acceleration would result in length contraction in the original frame. And since I was considering the case of a constant acceleration, the before-and-after acceleration reference frames were more muddled.

Finally, if the two ships were actually connected by a rope, the rope might hold up to a constant acceleration but it would also cause the two ships clocks to become desynchronized. The clock in the rear would lose time to the one in front, just as clocks lower down in gravity wells lose time to those on high mountains (so I've heard). That's a problem for another day, though.

Oh dear, I'm afraid I have to point out that you are mis-applying the Lorentz transformations. This is quite a common problem. What you have calculated are the coordinates of two different time separated events. To achieve a meaningful separation, both events must be recorded at the same time in the observer's frame. It is as if you are passing a stationary bus of rest length one unit in a car and saying "the forward end (of the bus) is now at (0,0) and in .577 time units the rearward end will be at (-1.154,0.577), by my coordinates".

In order to find the moving distance in terms of the "stationary" undashed coordinates you must have t(1)=t(2) so that using delta(x')=gamma[delta(x)- v.delta(t)] and delta(t)=t(2)-t(1)=0 gives delta(x)=delta(x')/gamma,where delta(x) means difference between coords of each s'ship in the x-t frame etc.. Thus moving distances appear contracted from the undashed frame by the factor 1/gamma where gamma=1/sqrt(1-v^2) which is >1. This is what we would expect and is in exact accordance with the measured contraction of other moving lengths such as the nominally "rigid" rod. [For instance consult length contraction in a standard textbook such as d'Inverno's.]

I also say the two ships clocks do not, indeed cannot become desynchronised (they are after all in identical situations). Acceleration is not the same thing as gravitation (remember space-time curvature). The accelerating s'ships are in Euclidean space. The "equivalence principle" only says that the effects of each are equal and opposite for a system in gravitational free fall. Thus the redshift-blueshift effect I described earlier is exactly cancelled by the gravitational time dilation (difference in clock rates) in a freely falling system provided we neglect "tidal" effects, so that it becomes effectively "inertial" for the comoving experimenter falling with it. Rod Ball 15:06, 25 August 2006 (UTC)

[edit] Welcome Argument Rod Ball

Let's start with where I agree.

If you have two ships connected by a rope, and they stay connected by that rope, they will stay approximately the same distance from one another in "their" frame, and the rope will appear shorter and shorter in any inertial reference frame.

I disagree about mis-application of the Lorentz Transformation. I can't see any other way to apply it than to map the coordinates of events from one reference frame to another.

I disagree that the space coordinates of events are less meaningful just because they don't happen at the same time. Each observer considers an event to have a definable location and time.

I see that you wish to talk about "moving distance" and "stationary in quotes." which is an interesting problem, but it is much easier to deal with the events rather than the objects. Don't try to define the position of a vehicle. Just the events where it left, and where it arrived, and maybe where it wrecked. From these events, the whole route (world-line) of the car can be easily determined. And as an added bonus, the Lorentz Transformation is specifically designed to deal with the events, but it does not do so good to try to apply it to the paths.

I don't understand what you mean by "The accelerating ships are in Euclidian Space." Once the two ships begin accelerating they cannot synchronize their clocks until they both return to an intertial state.JDoolin 03:23, 26 August 2006 (UTC)

I'm, sorry, you seem to be wandering off into other areas of speculation. Originally you said the clicher was that onceboth s'ships were at a v, the front one would suddenly be in a frame where the front started first - but I pointed out that it would also be a frame where the launch sites were suddenly going backwards at -v so the would appear at 0.866, not affecting an assertion that at any v the s'ships would remain at exactly the same distance in their own frame. This is the crucial question - one can do without the string altogether, it's not essential to the problem. What counts is whether the gap between the identially accelerating identical s'ships behaves in SR in the same way as the proverbial nominally "rigid". I am saying that it does. But I agree that in Lorentz's pre-relativity theory, where c was not posulated constant for all observers, and where contraction happens even for observers at rest w.r.t. object - then for such theory the string would break. Bell preferred Lorentz's theory [���� l�4GET http://k.b5z.net/i/u/6038535/i/menu/(UTC)

[edit] Saturday, August 26, 2006

First of all I must apologise for rather missing your points about simultaneity in my hasty addition, which was over-concerned with showing that the s'ship distance should measure contracted from original launch frame whilst, as I'm glad you seem to agree, it would be constant in the accelerating frame. The argument I usually have to contra is that it would be constant from the launch frame so by "reverse Lorentz contraction" it is argued they must be moving apart in the s'ship frame.

Reminds me of three blind men trying to describe an elephant, and coming to an argument because none of them see the whole thing.

Only given a single instantaneous burst of acceleration would the ship's distance be constant in the "launch pad" frame. So in this area, I agree with the people you have argued with. But if the acceleration continues, the two s'ships will come closer together in the "launch pad" frame. So in this area, I agree with you.

However, in each acceleration event, the two ships will find each other further apart in their own new reference frame than they were in the previous reference frame. (unless they are attached by a rope or other structure.)

The point being that both rope and s'ship distance measure shorter and shorter from launch pad frame. Yes your Lorentz calculation was essentially the same as mine - I was merely looking at it from the other end. Instead of saying the starting distance is 1 and becomes 1.154 at speed v, I am saying it's 1 at speed v and 0.866 when measured from the starting frame.

If the two ships were attached by a rope and somehow it managed to keep its proper length without stretching or breaking, I would agree with you. However, this is a big "if." The two ships are starting off simultaneously. So there is a shock applied to both ends of the rope, and let's pretend that the rope reacts at the speed of light (though the speed of sound is what would actually apply here.) Consider this from the lab frame and see where the shock waves would meet, relative to the two ships.

The rope won't necessarily appear shorter in any inertial frame. Consider a passing 3rd. s'ship at constant v who observes the launch. From that POV the launch pads will already appear to be 0.866 apart, flying in backward direction at -v. The front s'ship will appear to start first, reducing its -v speed until the second s'ship starts, the end result being that both rope and s'ship distance become 1 when at speed v w.r.t. launch site ie. zero w.r.t. passing observer.

So when you say the s'ships suddenly find themselves in a frame where the front one started first, they also, it should be noted, find themselves in a frame where the lauch pad distance is suddenly 0.866, so that the "starting first" does not mean they are now more than 1 apart but rather bridges the gap between the 0.866 they suddenly appear to have started from and the 1 that they now are at v. The rope of course, undergoes the same apparent change so there is no reason or mechanism for breakage.

The launch pads will appear to be .866 apart, but they will be moving in this frame. The front s'ship starts first, and the back s'ship leaves from the length contracted launchpad later. There is not just a delay in time, but motion in space. The rear ship not only starts at a later time, but it's starting line has moved back during that delay. Since the launchpad is moving at 50% of the speed of light for .577 seconds, it moves .2885 light seconds before the rear ship takes off. .866 + .2885=1.1545. So even though the launchpad appears length contracted, the ships are going to be further apart in their new reference frame.

About the coordinates, I simply want to emphasise that for any observer to be able to say what another moving length or distance is from his frame, he naturally must record the endpoints simultaneously as also defined in his frame. So as SR says the moving length/distance appears shorter from launch frame and the launch pad distance appears shorter from the moving frame - as would be the case by swapping t and t'.

Essentially I agree. One method for doing this is to state the motions of the particles in terms of functions of time, such as x_a(t) and x_b(t). These functions of time will vary based on the velocity of the observer. Now, if you take two stationary objects, and associate two events with each of them, it is not difficult to create functions x1(t) and x2(t). This creates two parallel lines. When you perform Lorentz transformations on thes events, in general the lines may lean over and get closer together, even though the events get further and further apart.

I said Euclidean space because it's assumed no gravitational fields are present, and they cannot be conjured into existance by any acceleration - only significant mass can do that. It's sometimes argued that because clocks differ in rate at different heights in a Gr. field, that this should also happen between two identically accelerating s'ships. I disagree.

I also find that to be a weak argument, though I am not sure I have ever heard anyone explain the whole argument. I think it actually works better in the opposite direction. Instead of going to experiment in gravitational field, to unfounded assumption of equivalence, to description of acceleration, I find that it works better to go from gedanken of acceleration, to unfounded assumption of equivalence, to surprising agreement with experiment in gravitational fields.

I guess what I'm saying is that the equivalence between gravity and acceleration, is still, in my mind a surprising correlation, rather than a proven principle.

I find that a careful analysis of an accelerating body 
from an outside inertial frame, taking into consideration 
the constant speed of light and the Lorentz-fitzgerald contraction 
gives great insight into how acceleration should affect clocks.  
And since gravity seems to give roughly the same phenomena as 
acceleration, we can predict that the same things should happen.  

No resynchronisation (between the s'ships) would be necessary because they are in identical situations throughout. I you imagine n equally spaced identical such s'ships in a line, isotropy dictates that they must all have the same time, as otherwise it would be impossible to define a time for any given one without knowing how many others there were and what time they had elsewhere, leading to a vicious circle. Anyway the equivalence principle would apply only in that the redshift-blueshift effect I mentioned before is "equivalent" to the Gr. time dilation so that the compensate in an equal and opposite way if the s'ships were free-falling in a Gr. field.

I still disagree. Even in an infinite line of accelerating ships, no two of them are in an identical situation. Any two adjacent ships can talk to each other and in NO situation can *both* say to the other "I am accelerating toward you, and you are accelerating away from me."

Also in this hypothetical string of identically accelerating space ships, those you are accelerating toward will appear blue-shifted, while those behind will appear red-shifted. This will not just affect the color of the lights but the measurement of time itself.

Unlike inertial relative motion, where both observers see the other person's watch going slow, this group will all agree on whose clocks are faster than whose.

If you still think that the rope/string would break, I'd be keen to hear more of a specific argument, or also if you disagree with my stuff just written. Regards. Rod Ball 13:46, 26 August 2006 (UTC)

Now it's easy to cop out on this argument and describe the magnitude of acceleration we're talking about, and simply note that no known material can stand up to it.

But this argument runs a little deeper than that; No one doubts that a string would break if it were long enough and you pulled on it hard enough. If you have a twenty mile long string and accelerate both ends of it to ten times the speed of sound in less than a second, the inertia of the center of the string would cause a tension which would break the front end.

However, if you idealize the string, and make it inertialess, and assume that signals can pass through it at the speed of light, or even instantaneously, accelerate all portions of the string at the same time, this is the deeper question.

Is it even possible to describe a synchronizable non-inertial frame? I'm sure we have not done so here, and I don't think it is possible to construct such a frame. JDoolin 18:00, 26 August 2006 (UTC)

[edit] Lorentz Pre-Relativity Theory?

Rod Ball said: Originally you said the clicher was that onceboth s'ships were at a v, the front one would suddenly be in a frame where the front started first - but I pointed out that it would also be a frame where the launch sites were suddenly going backwards at -v so the would appear at 0.866, not affecting an assertion that at any v the s'ships would remain at exactly the same distance in their own frame. This is the crucial question - one can do without the string altogether, it's not essential to the problem. What counts is whether the gap between the identially accelerating identical s'ships behaves in SR in the same way as the proverbial nominally "rigid". I am saying that it does. But I agree that in Lorentz's pre-relativity theory, where c was not posulated constant for all observers, and where contraction happens even for observers at rest w.r.t. object - then for such theory the string would break. Bell preferred Lorentz's theory Rod Ball 19:56, 27 August 2006 (UTC)

In the launchpad frame there are two distinct modes. In the first, the launchpad is stationary and the spaceships are stationary.

In the second, the launchpad is stationary, and the two ships are moving.

In the moving frame there are three distinct modes.

The first mode: The launchpad and the two spaceships are moving at .5c to the left.

The second mode: The launchpad and the rear spaceship are moving at .5c to the left. (It is during this time that the discrepancy between what you think should be and what I know is the case is occurring.)

The third mode: The launchpad is moving to the left and the two spaceships are moving to the right.

The proverbial nominally "rigid" rod is a question of greater complexity than this, which I've tried to address above, but you called it "speculation." Whether or not it is speculation, it is the product of months of my own arguments on Usenet, where I stopped at some level of consensus with people who could apply the Lorentz Transformations with some skill.

At no point did they argue that the Lorentz Transformations were pre-relativity, although many did argue that these equations were not applicable to the universe as a whole.

[edit] Big Bang article NPOV dispute

Hi you started a long discussion (with continuing follow-ups!) on the Big bang Talk page, but lately you didn't comment. As a result, the discussion is becoming one-sided. When you have time, please comment. Harald88 13:47, 9 September 2006 (UTC)

[edit] IP Block

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