Talk:Photon/Photons and Mass Debate3

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1 In a box? With a fox?
2 Photon in the box is very poorly written
3 Weight/mass of single photon in a box bouncing up and down
4 Ati3414's reversions
5 Single photon trapped transversely in a rocket; inertia and weight change

[edit] In a box? With a fox?

What's up with the mass of a photon in a box? It says "Although a single photon has zero mass, multi-particle objects including photons may collectively have mass. For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater mass (by Δm = E/c²) than if the box did not contain photon(s)." Now if you wait a second until the photons are absorbed, is the mass the same, or different? What does it mean to have a photon in a box anyway? Photons are emitted and absorbed. So who put them there? If they were inserted from an external source, then the added mass is the same as the mass of the photons themselves, is it not? Reflection absorbs photons and radiates new ones. Do some of these have zero mass? Can someone explain how to make sense of this section? Dicklyon 04:44, 6 August 2006 (UTC)

We have a mirrored box so we don't have to talk about absorption, because it just makes things more complicated. But no, absorption doesn't change the mass of the system.

. In a box with a single photon bouncing back and forth, the COM frame is the center-of-mass frame, the same as if the box had a superball or gas molecule bouncing from side to side. The box in free space will wiggle back and forth about a point, and that point is the center of mass. An inertial frame with observer stationary with respect to that point, is the COM frame. Okay?

. Now the total box mass in the COM frame is the mass of the empty box plus E/c^2 where E is the photon energy in this frame. So in that sense the trapped photon has mass which is E/c^2.<g>But what if the photon isn't trapped, you ask? Suppose somebody puts a hole in the box ahead of it, or it's just flying through? Answer is that the answer does not change. There is a still a COM frame, and in that frame the system of photon-plus-box still has mass equal to mass of empty box plus E/c^2 for the photon. The difference is this mass isn't easily accessible by putting the thing on a scale, which is why we trapped the photon. However, if you calculate the invariant mass of photon+box system, you get what the mass WOULD BE, if you weighed it "at rest" (ie weighed it in the COM frame). So doing that calc can give you the rest mass of a particle before it decays-- you just figure out the energy of the decay products in the COM frame and the mass of the whole shebang, including rest mass of the particle which was the origin of it all, is E/c^2, where E is the COM total energy. S B H arris 05:08, 6 August 2006 (UTC)

So the explanation for the photon without a box not have a mass is that you can't be in its center-of-mass system, and that if you approach it the photon energy goes to zero? Dicklyon 05:12, 6 August 2006 (UTC)

Exactly right. It's the presence of the box (which we have to assume has SOME energy or mass itself) that gives you a 2-particle system, so that there's a certain SYSTEM energy you can't get rid of by choice of frame (you can chase the photon, but you'll find your box gaining energy when you do it-- a losing game). The (minimal) energy for the whole system is your (invariant mass) * c^2.S B H arris 05:33, 6 August 2006 (UTC)

[edit] Photon in the box is very poorly written

Please :

a. delete

b. correct in order to contain a correct explanation of what is going on:

The correct explanation of the "photon in a box" issue is that the trajectory of the photon inside the box gets slightly curved in the gravitational field such that there is a vertical component of the photon momentum p_tangential=p*sin(theta) where p=E/c is the momentum and theta is a very small angle. At each collision with the vertical walls of the box the photon transmits a downward force dp=dp_tangential/dt to the box that can be (mis)construed as an increase of mass equal to dp/g where g is the gravitational constant. Please remove any misleading references to Δm = E/c². Such ridiculous stuff can only further the misconception that the photon has non-zero mass that can be measured.

Thank youAti3414 05:03, 6 August 2006 (UTC)

COMMENT: I see you don't read your talk page. The reason somebody reverted your edit, is that it indeed misses the point. You have the MECHANISM for how the photon energy in a box shows up as an extra weight g* (E/c^2). But this extra weight is not at all "pseudo" weight. It's as real as any weight and it's due to the mass of the trapped photon(s) bouncing back and forth. Single free photons have no mass, but single trapped ones do, since they are forced into a 2-object system, which has a COM frame, which gives the photon energy an invariant mass.

You'll notice, BTW, that your explanation applies to any particle in a box, not just a photon. This is the mechanism by which all trapped particles show their mass or weight in a system. Note that for a relativistic massive particle, this mass is proportional to relativistic momentum γmv, which means that the "mass" or "weight" measured for it is not just the rest energy, but the total energy. So the kinetic energy of the particle gets "weighed" too, and shows up as mass of the box. Interesting stuff, yes? S B H arris 05:10, 6 August 2006 (UTC)

I think the original explanation was terrible, where is this gedankexperiment coming from anyway? What is its relevance? Is there any such experiment that has ben reported in a peer refereed journal? If not, why is it included? I also think that the explanation based on g* (E/c^2) makes no sense whatsoever. In effect, it encourages the people that claim that the photons have an equivalent mass m=E/c^2 which is totally wrong. Come to think of it, the whole page misses the fact that current experiments put a limit of 6*10^-17 eV on the photon mass (with newer experiments set to lower this limit). This limit is in total contradiction with any type of E/c^2 derivation. If you don't like my explanation of "photon in the box" could you please remove the entry altogether? It is very misleading and badly written as originally set. Thank you Ati3414 05:21, 6 August 2006 (UTC)

I was looking for an explanation in the absense of gravity, hopefully relating to when the energy mass is considered to be a mass and when it is not. I think the gravity effects you are describing are beside the point. Dicklyon 05:09, 6 August 2006 (UTC)

Right, they are beside the point, and so is the box, for that matter. Two photons, not going in exactly the same direction, constitute a system having nonzero invariant mass. --Trovatore 05:14, 6 August 2006 (UTC)

Yep. The box is only used to illustrate this fact, since in a g-field, the invariant mass of a closed system can be simply determined by weighing the thing in the COM frame.S B H arris 05:20, 6 August 2006 (UTC)

Then why aren't you writing an entry on systems of photons instead of the very badly written "Photon in the box". Two of us already complained about the shoddy story, almost simultaneously. You cleaned it up a little by removing some errors and now the entry is almost incomprehensible. Someone reading the entry might conclude that the mass of the photon is E/c^2 a classical rookie mistake by antirelativists. Please write a scholarly piece or take the "photon in the box" out. altogether.Ati3414 05:34, 6 August 2006 (UTC)

The entry uses one photon to illustrate a point, because the box then becomes the other mass in the system. One photon has no mass. One photon which is part of ANY system in which there is ANY other mass, contributes a photonic mass to the system, E/c^2. Just as a particle with rest mass M and kinetic energy T contributes (M+T)/c^2 mass to such systems (in the COM frame). The point is that ANYTHING in there gets its energy counted as mass. Okay? Part of the reason I'm doing this, is because it's poorly understood out there. I'm having problems with you, looks like. So you should be grateful. S B H arris 05:43, 6 August 2006 (UTC)

No need to resort to personal attacks. By insisting on the posting, you simply reinforce the belief of many crackpots who think that m_photon=E/c^2 (or, expressed in eV, m=hf), notwithstanding that there are experiments that clearly refuted that idea. You can have your entry, try to make things a little clearer and don't take it personally. As it is currently written, is very bad , though I must say that after my interventions you cleaned it up quite a bit. So, I will put in an entry, that is sorely missing on the mass of the photon, such that there is no possibility of confusion.Ati3414 06:38, 6 August 2006 (UTC)

[edit] Weight/mass of single photon in a box bouncing up and down

ONE photon has a mass (and a weight) in relativity, if you trap it in a box so that it bounces around, and put the box on a scale. The weight and mass turn out to be due to the gravitational Doppler shift. Here, I'll do the calc (excuse my editing):

Put one (well localized) gamma photon of ground energy E_o = hf_o in a perfectly gamma mirrored box (okay, it's Gedanken) of height x, in a gravitational field g. Let it bounce up and down, hitting the top and bottom of the box, each bounce after time t=2x/c. Each bounce transfers 2 times the photon momentum E/c. Now the force that the photon exerts on box top (or bottom) is dP/dt = Force = F = (2E/c)/(2x/c) = E/x.

First error: I took another look, this is obviuosly wrong, the "dt" in the calculation of force has nothing to do with the time taken by the photon to travel up and down the box , t=2x/c. The "dt" is actually the very short time during which the photon bounces of the wall and reverses momentum. F=dP/dt where dt is the short (and unknown) interval of momentum reversal. You are confusing a finite time interval , t with an infinitesimal one , dt. So, your derivation seems totally bogus, this is why it comes up with E/c^2.Ati3414 04:11, 8 August 2006 (UTC)

Sign you comments, please. If you don't like dP/dt note that this is average force (pressure time area) averaged over the time of many photon impacts, so you can just as well use ΔP/Δt because we're averaging the momentum transfered over long periods of time and many photon impacts (that is what the ratio I use does-- do I have to spell it out?). Your sniping is ridiculous here, as you have to do exactly the same thing to get an average downward momentum transfer with time in the side-to-side force = weight, in the example you wanted to put in the article. S B H arris 17:19, 6 August 2006 (UTC)

No, I didn't , I used the proper dt. I did not mix in an abitrary time interval as you do in your derivation. I think that your entry has no business in the page since it is clearly based on personal research and it is wrong to boot. Ati3414 18:50, 6 August 2006 (UTC)

It's not an arbitrary time. It's the average time over which the momentum transfer takes place. You get the same momentum transfer rate per second as you do per hour. Which means the same force. Only if you go to very small times < x/c does it get lumpy. S B H arris 20:22, 8 August 2006 (UTC)

The difference in photon force (ΔF) between the top and bottom of the box is the "weight" of the photon, as measured by the scale the box is on, and THAT is due to the Doppler change in energy, (ΔE)/x.

Second Error: Nothing gives you the right to attribute "wight" to the photon. A resulting downward force, yes, but "weight"? Nothing gives you the right to assign it weight. You are preparing the ground to assign mass to the photon and I will expose this trick here. We all know that there is such a thing as a pressure radiation (see Einstein 1905).Ati3414 04:11, 8 August 2006 (UTC)

Hello? Yes, the weight is here the difference between radiation pressures. And bouyancy force in water is due to the difference in water pressures with height. Here's a shock: Did you know you can calculate the weight of a cube of gas (never mind a photon gas) by integrating the difference in gas pressure on a bottom vs. the top of a vessel? Same with a container of water, for that matter. S B H arris 20:22, 8 August 2006 (UTC)

Nothing permits you to mix in the Newtonian notion Weight=m*g , you are mixing relativistic calculations with Newtonian ones. You are attributing a the force due to the variation of radiation pressure to weight, with no justification whatsoever. Ati3414 02:14, 9 August 2006 (UTC)

Any permanent increase in force on a scale in a g field, we call "weight." The justification is the definition of the word. The scale says the box weighs more due to its content of photons, and so it does. What, do you think this is some kind of magical rocket motor? S B H arris 02:42, 9 August 2006 (UTC)

Now ΔE is just h*(Δf) where f is the photon frequency. From Einstein's formula, the Δf from a gravitational shift through a constant field g and height x, is given by Δf = f_o[gx/c²]. Here f_o is the frequency at ground level. Substituting gives: Weight = ΔE/x = h(Δf)/x = h(f_o)[gx/c²]/x. Cancel x's and note that hf_o = E_o which is the energy of the the photon at the ground, and you have:

Weight = (E_o) g/c²

If you equate the weight as mg, then the "mass" you weigh for the photon is m = E_o/c².

Third and most fatal error: You are now mixing relativistic calculations with the Newtonian ones. You are saying : Weight=g*"something" so "something" must be the "mass" of the photon. Sloppines and sleigh of hand can produce any result (especially if you already know the result you are after). Ati3414 04:11, 8 August 2006 (UTC)

Some things are the same for both Newton and Einstein. The fact that mass is proportional to weight is one of them. Also the fact that mass is proportional to inertia. There's no way to increase the passive weight of a box, with nothing but gravity acting on it, except by increasing its mass and inertia. That's what weight is: m*g. By suggesting not, you're embarrassing yourself S B H arris 20:22, 8 August 2006 (UTC)

For a photon going from side to side you have to figure the g field as an acceleration producing a relative box-side photon lateral velocity, then do the calc from relativistic aberration. But the answer comes out the same. S B H arris 02:42, 9 August 2006 (UTC)

BS, I challenge you to do it and you tried to weasel out by asking me to do it. I am not the one supporting the crackpot idea that m=E/c^2. So, for the third time, show the proof or cease and desist on your claims Ati3414 02:14, 9 August 2006 (UTC)

Okay, a calculation for mass contribution of a single lateral photon in a rocket and a box, showing increase in ineria, therefore weight, of E/c^2. See last section of this page. All for you. Afterall, I used your suggested method, even though you don't believe its result. S B H arris 02:42, 9 August 2006 (UTC)

This may not be true. If you really wanted to prove that the photon trapped in a box imparts "mass" to the system, you would need to demonstrate mathematically that this happens INDEPENDENT of the direction of photon motion. Since you botched the simpler case (up/down) motion, the chances are slim that you will be able to make your point for the side to side case. But , please try (and try to do it without sneaking a peek at my previous solution to this situation)67.170.224.36 04:00, 8 August 2006 (UTC)

You gave no previous formal solution-- just a few handwaves. If you'd actually worked the problem, you'd have had to deal with the things you just criticized me over, namely having to integrate over many impacts. So let's see you do it. Post it at the end of the TALK here, as a section. I'll give you a week, and if you can't, I will.S B H arris 20:38, 8 August 2006 (UTC)

THe point is that you botched your solution and you twisted the math such that it gives an answer (E/c^2) that is 17-19 orders of magnitude larger than the experimental data. You refuted your own calculations, remember? Try taking care of your errors, then we;ll talk. Especially the mish-mash of relativistic and newtonian mechanics. Also the fact that your solution is for one direction only, it should be direction invariant. Be a good sport, take out the embarassing E/c^2 "answer" , it makes the post ridiculous. Ati3414 20:50, 8 August 2006 (UTC)

A photon may not have mass in and of itself, but photon energy contributes mass to all systems it's part of. S B H arris 05:39, 6 August 2006 (UTC)

A photon contributes energy and momentum to the system into which is being incorporated. The extra energy manifests itself as radiation pressure which in turn may manifest itself as a force. No need to go past this point and venture into the murky waters of "relativistic mass", rest mass", etc. Change the post to drop any mention of "mass contribution" and I will stop objecting to your posts.

You got this idea from an ill-inspired entry in the John Baez FAQ. The FAQ is known to have errors, I have personally convinced him in the past to change something that was wrong in the test theories of relativity paragraph, I contacted him again asking to correct the paragraph about the mass of the photon. I see no reason why the FAQ will not be corrected. Ati3414 04:11, 8 August 2006 (UTC)

Looks very similar to my original post. Would you care to put it on the front page such that people don't get the mistaken idea that photons have mass? We are only talking about photons getting WEIGHTED, correct? Which was my point. Also, I think that the case with the photon moving side to side was treated correctly in my post that you removed.Come to think of it, what is the point of the whole "Photon in a box"? It is a gedankexperiment, no real application (we already know about radiation pressure). Why confuse people with it? Come to think of it, after looking at your explanation, I believe that it is wrong. You calculate the radiation pressure on the bottom of the box (2E/c)/(2x/c). But there is EQUAL pressure on the TOP of the box, so in reality the box is "jumping" up and down on the balance platter, so on average, you should see ZERO extra weight. This is a very bad entry for an encyclopedia, would you please remove it? Thank you Ati3414 05:46, 6 August 2006 (UTC)

There is no equal pressure on the top of the box for Doppler shift reasons explained.

OK, I see that.Ati3414 06:17, 6 August 2006 (UTC)

I didn't remove your example. Somebody else who agreed with me did (yep, another crank here). And since we're not only talking about photons being weighed, we don't put the mechanism for this in the article (though it's fun for the talk page, so I did it). Photon energy DOES contribute E/c^2 mass to systems, in the COM frame. Not just weight. Actual, really real mass.S B H arris 06:25, 6 August 2006 (UTC)

How could it? You are only "weighting" your system in the gedankexperiment. You cannot infer that mass has been added based on the fact that in reality you have a differential in radiation pressure between the top and the bottom of the box.To say that the photons are contributing "Actual,reallly real mass" is philosophically abhorrent. They are contributing energy and momentum to the open system into which they have been injected. The resultant system has a higher energy and a higher momentum. We can WEIGH the new system and we find the weigh to be higher than the weigh prior to photon injection. This does not mean that mass has been created, that the photons have contributed "Actual, really real mass". I believe that the entry is very bad and that it should be removed, notwithstanding your very interesting explanation. Ati3414 06:18, 6 August 2006 (UTC)

I was under the impression that in standard physics, the three kinds of mass, total energy mass, inertial mass, and gravitational mass, are regarded as being always equal for all closed systems. Are you saying that's not so? Dicklyon 06:27, 6 August 2006 (UTC)

Look, the same explanations inevitably arise with gluons, which have no more mass than photons (ie, they are massless). But most of the mass of nucleons is kinetic energy of quarks and energy of massless gluons. IOW, a hadron is just a box, being weighed. So shake any ordinary object. 1% is rest mass of quarks and electrons. The rest of the mass (that other 99%) is the sort of stuff we're discussing here. So it's important as a topic. S B H arris 06:25, 6 August 2006 (UTC)

I agree it is important, this is why it needs to be written properly or not included at all.Ati3414 06:47, 6 August 2006 (UTC)

We have written it properly. Massless gluons in a box (nucleons) contribute mass to the system. Massless photons when trapped in the same way, contribute mass to THEIR system. How much shorter, cleaner, and clearer could be it be? Massless particles can, and do, contribute mass to systems. Period. S B H arris 07:47, 6 August 2006 (UTC)

You mean that you cleaned it up after yesterday's criticisms? This is a more correct statement. Even so, read by a neophyte is still bad: it implies a photon mass of 3eV , in clear disagreement with the current experiments (and with the official .gov site) that place a limit of 6*10^-17 eV on it. There must be an error in the calculations of your gedankexperiment. 67.170.224.36 14:10, 6 August 2006 (UTC)

It implies nothing of the sort, and there is no error. The mass measured in systems due to trapped photons is not the mass of the photon but the mass of their energy contribution. Same as for gluons. S B H arris 17:21, 6 August 2006 (UTC)

I took another look, this is obviuosly wrong, the "dt" in the calculation of force has nothing to do with the time taken by the photon to travel up and down the box , t=2x/c. The "dt" is actually the very short time during which the photon bounces of the wall and reverses momentum. F=dP/dt where dt is the short (and unknown) interval of momentum reversal. You are confusing a finite time interval , t with an infinitesimal one , dt. So, your derivation seems totally bogus, this is why it comes up with E/c^2. You should do the right thing and eliminate the entry. It has no place in an encyclopedia since it apperas to be the outgrowth of your personal interpretation of a college exercise. Ati3414 18:52, 6 August 2006 (UTC)

If not so, the local gravity field would disappear POOF every time an electron met a positron. Do you really think it does? That was a serious question. S B H arris 06:25, 6 August 2006 (UTC)

[edit] mention relativistic mass

I think the basic underlying problem is this: What the section really is, is an attempt to explain that photons have nonzero relativistic mass, without referring to that unfashionable concept by name. I understand that contemporary physicists don't find the concept particularly useful, and I understand (at least reasonably well) why they don't, and that's fine. Nevertheless, from a historical point of view, the concept was used rather recently, and it's not as though it's been refuted; it's just been found not to be the most convenient way of organizing information.

Since many readers probably have heard explanations that implicitly used relativistic mass, and want to know how those relate to the descriptions they're reading in this article, I think at least a brief summary (giving a link to the excellent discussion in the mass in special relativity article) would be appropriate. The summary should also mention that the concept of relativistic mass is not much used these days, but should not imply that it's a bad, evil concept. --Trovatore 18:19, 6 August 2006 (UTC)

It is a bad, evil concept. Probably it's behind the rising tide of global terrorism. I hate relativistic mass because it's frame-dependent. We have relativistic energy to do its job. But it has been noted that the special thing about the COM frame, is that it's where the relativistic mass of the system, and the ordinary "rest mass" (or really invariant mass), are the same. So E=mc^2 holds no matter which kind of mass you're talking about. Which means the "relativistic mass" of photons counts HERE in this frame as rest mass (of the system = box), but not so in other frames. Perhaps I need to just SAY something like that, here. S B H arris 18:25, 6 August 2006 (UTC)

Just say they have nonzero relativistic mass and be done with it; detailed discussion isn't really about photons per se and can go in other articles. It doesn't matter whether you or I think relativistic mass is the right way of thinking about things. It's the way workers in the field did think of it, until, I don't know, 1950 or so? and it's still in plenty of popular literature. It shouldn't be implied to readers that those writers were spouting gibberish, except in the cases that they were (which I'm sure are abundant, but this isn't one of them). It's just a different way of describing the same facts. --Trovatore 18:33, 6 August 2006 (UTC)

Ati3414, please desist from removing references to relativistic mass. Your hostility to the usage notwithstanding, the point needs to be addressed, descriptively rather than as an advocate for contemporary usage. It would certainly be correct to note that the usage is not very current, but it isn't wrong, and readers who are used to it need to know what has happened to it. --Trovatore 04:53, 7 August 2006 (UTC)

The usage of "rest" and "relativistic" mass has been abandoned in all modern texts. It has been replaced with "proper" mass, the only mass that current relativistic textbooks and papers deal with. Relativistic mass has been expunged. "Photon relativistic mass" is even worse, there is no such thing.

In addition to the embarassing terminology, the section is plain wrong, looks like someone's favorite college book exercise. Wiki does not allow for personal exercises to be turned into postings. To make matters even worse, there is no credible reference as to where the posting came from (textbook , paper published in a peer refereed journal, etc) To make matters even worse, SBharris' attempt to prove the statement can be easily refuted , his proof is not correct. Either way, why have a college book EXERCISE as a post in an encyclopedia?

So, why use wiki to perpetrate incorrect notions, take out the posting and we are done.Thank you for your cooperation. I think that the two of us have worked together in the past removing some crackpot statements from other areas of relativity (Luminiferous Aether). Why does it take so much effort in removing wrong posts? Ati3414 06:22, 7 August 2006 (UTC)

It's not my exercise. It's part of the USENET physics FAQ and has been for many years. This is written by professional physicists. Do read and learn, please. [1] S B H arris 17:07, 7 August 2006 (UTC)

By the same author: "You should not use this to justify the statement that light has mass in general." I agree with you, read and learn.

Gedankexperiments with botched explanations do not constitute science. And they have no room in a work that aspires to be an encyclopedia. Ati3414 17:20, 7 August 2006 (UTC)

Can you guys tell what is the "modern" alternative to this concept? I don't understand what is motivating this dispute, since I haven't see Ati3414 provide an alternative accounting except in terms of "weight" which seems to only confuse the question of mass. Dicklyon 05:13, 7 August 2006 (UTC)

I suppose the contemporary approach is just to describe the momentum, energy, etc, using the formulas that work when you plug in rest mass instead of relativistic mass. I am not arguing against that approach; I kind of see why contemporary physicists prefer it.

However Ati3414 is simply wrong when he refers to relativistic mass as an "incorrect notion". It's not even a different notion from the contemporary one—it not only predicts exactly the same outcome to every conceivable experiment (which should satisfy instrumentalists), it even has exactly the same underlying noumena, for the hardcore realists. It's purely a difference in language and bookkeeping.

Indeed, but as pointed out, it's a concept which is redundant. We have total energy/c^2 to do the job of "relavistic mass". Meanwhile, what we measure as mass in our daily lives and work with as "mass" is actually invariant mass. It only HAPPENS to be the same as relativistic mass in the COM frame, but otherwise it's not the same thing at all. Since this quantity is separately conserved and invariant for all observers, and is usable for all the old concepts of mass, let's use it for "mass" (relativistic mass is not usable for mass in many cases where you have motion == for example you can't make a black hole from relativistic mass unless you're in the COM frame, and then it's also invariant mass as well).S B H arris 17:07, 7 August 2006 (UTC)

Look, I'm not arguing with you on this point. Physicists prefer to use invariant mass and that's fine with me. But that does not erase the historical notion of relativistic mass, and it doesn't make it wrong. I don't know that we need as elaborate an explanation as you've provided; I'd be happy with a brief footnote, after the text saying that the photon mass is zero, pointing out that there is a historical notion of relativistic mass, which is not zero for the photon, and pointing the reader to mass in special relativity. --Trovatore 17:21, 7 August 2006 (UTC)

The statement that the photon has nonzero relativistic mass is a correct statement. Ati3414 may wish that the notion of relativistic mass had never been formulated, but it was formulated, and it is not WP's role to rewrite history to conform to someone's prescriptive notions. Some neutral statement about the photon's relativistic mass must be made; I don't insist on SBHarris' version, but something needs to be there. --Trovatore 15:33, 7 August 2006 (UTC)

What has to be is something that is supported by experiment and references, not a botched grad school exercise as the "Photon in a box". Wiki tries to be an encyclopedia, not a collection of exercises and answers sheet. As to your statement that "the photon has nonzero relativistic mass" , you will need to back it up with published experimental references or retract it. 12.36.122.2 17:13, 7 August 2006 (UTC)

Obviously you don't understand the definition of relativistic mass, which is relativistic energy/c^2. And READ the *&%%$ing reference, please. [2] S B H arris 17:23, 7 August 2006 (UTC)

You are resorting to personal attacks again, which means that you are loosing the scientific argument. You mean your own personal interpretation of the term relativistic mass based on a FAQ? Since when are YOUR PERSONAL interpretations of FAQ entries scientific references? Besides, your definition comes at odds with wiki's. See here: http://en.wikipedia.org/wiki/Relativistic_mass Same reference you gave above, have you read it? Here is what wiki says:

The relativistic mass M is then formulated as:

$M = \gamma m \!$

where

m is the rest mass, and

$\gamma = {1 \over {\sqrt{1 - u^2/c^2}}} \!$ is the Lorentz factor,

u is the relative velocity between the observer and the object, and

c is the speed of light.

Nothing to do with YOUR INTERPRETATION, actulally at ODDS with YOUR INTERPRETATION No wonder that wiki is such a mess. Ati3414 17:40, 7 August 2006 (UTC)

Your math is a mess. The equations you're using do not apply to photons, which have no rest mass.S B H arris 17:53, 7 August 2006 (UTC)

It is not MY math, it is lifted off the wiki page that YOU referenced as the definition of the antiquated term called relativistic mass (or mess?). You never anwered my challenge: "You mean your own personal interpretation of the term relativistic mass based on a FAQ? Since when are YOUR PERSONAL interpretations of FAQ entries scientific references?" Ati3414 18:11, 7 August 2006 (UTC)

Your notation is a mess, and you didn't lift it off the Wiki, or the equations would come out properly. See those symbols

m a t h

? They exist for a reason. I'm not going to fix it for you. It's educational, so you do it yourself.

You are using equations which do not apply. The wiki on relavivistic mass specifically states: Another downside of this approach is that since γ depends on velocity, observers in different inertial reference frames will measure different values, which can be complicated. It should also be noted that these equations apply to matter and they are unsuited for photons: with v=c and m=0, γm of a photon is undefined. Got that? No disagreement, no contradiction. Now, as for the other, what are you asking for? The invariant mass of systems of photons is a topic of several published literature cites. Is that what you want? The relativistic mass of a photon is a concept not used much these days, but for a very long time was a common term in the physics literature. If I find you a cite, will you zip it? There's a reason you were blocked for a month. Your behavior is obnoxious. S B H arris 18:27, 7 August 2006 (UTC)

I was blocked because I challenged cranks that pass themselves as "editors" by editing out their nonsense. The point is that you send the wiki page referring to "relativistic mass". I pointed out that the wiki term m_relativistic=m_rest*gamma comes at ODDS with your personal interpretation inferred from the John Baez FAQ. So your post on the "photon in the box" has only the scientific support that is inferred from a FAQ for beginners, right? No experiment, no peer reviewed paper? As such, it has no place in wiki. Ati3414 19:00, 7 August 2006 (UTC)

I see you've learned something about the math editor. Good. Now continue in student mode, please. The invariant mass of pairs of photons is such a common concept in physics that it's textbook material. Therefore it's rather perverse to demand "experimental" or "peer reviewed" discussion of a student level concept. Peer reviewed papers are long past this concept, and take it for granted. I gave you a reference on another page, but here's another example from dozens of abstrants and papers you'll find in the physics arXive: [3]. Are you really suggesting that pairs of photons have an invariant mass but a system of a photon and a box does NOT? A pair of photons from a positron-electron anihilation obviously has the same invariant mass as the electron and positron, since invariant mass is conserved. Or are you arguing against that, also? Frankly, it's difficult for me to tell what your problem is. You don't like relativistic mass in connection with photons, and you don't like invariant mass in connection with photons, and one gets the impression you don't think photons have mass in any circumstances at all. But since mass turns into photons, that would mean that mass would have to disappear. Since the equations of general relativity do not allow for that (no discontinuities in the G field), it would be an obvious problem. Fortunately, it's one we don't have. S B H arris 01:26, 8 August 2006 (UTC)

I restored the section again, then made some minor changes. In particular, I wrote it to make sense for the case of what happens after the photons are absorbed. The energy is still in the box, so the mass doesn't change. Dicklyon 01:37, 8 August 2006 (UTC)

Thank you. After Ati3414 deleted the section yet again, I restored it yet again. That's revert #2 for me. One more delete from him and he runs up against WP:3R. I leave it to the community overlooking this article to deal with him on this matter from here. S B H arris 01:54, 8 August 2006 (UTC)

[edit] compromise

p.31 of Concepts of Modern Physics, by Arthur Beiser, has a pretty good explanation that doesn't invoke relativistic mass, but rather assigns the energy (of the photon in our case) to the rest energy of the system, which therefore becomes part of the rest mass of the system. Maybe such wording would go down better with him? The result is the same, if I understand correctly. You can "search inside the book" on Amazon for energy mass. Dicklyon 02:03, 8 August 2006 (UTC)

It explains the observations fine, but doesn't solve the main problem, which is that it doesn't point out to readers confused by the claim that photons have zero mass, that this doesn't mean they have zero relativistic mass. That needs to be pointed out. Once it is, I have no great attachment to the whole "photons in a box" section, and I don't really care if it goes away. --Trovatore 02:14, 8 August 2006 (UTC)

I didn't find the term "relativistic mass" in the books I looked in, which is why I looked for "energy mass" and settled for what I found. Can you recommend a book that explains it more like you prefer? I'm not sure what it is that you think is the main problem that is not explained well now. Dicklyon 03:49, 8 August 2006 (UTC)

You won't find "relativistic mass" and you won't find "rest mass" anymore.They are outdated, incorrect, misleading. This was one of my points.

My second point is that it is plain wrong to talk about "photon mass". QED mandates that the photon is a massless particle and there are many experiments that confirm it. So, talking (or even worse, writing) about "photon mass" encourages all the crackpots that think that the "mass" of the photon is hf (i.e. about 3eV) when in reality the current experiments set an upper limit at 6*10^-17. I edited the post in such a way that it avoids any of the pitfalls and any of the tell-tale signs of old terminology. Hopefully this will be an acceptable compromise to all parties. Ati3414 04:28, 8 August 2006 (UTC)

I made another edit, to try to make it more clear, and put the term "relativist mass" in quotes for what it is called. And I took out the complicated bits about the inertial and weight, as they really are beside the point; I just summarized the point that energy mass, inertial mass, and gravitational mass are the same (so far as standard physics knows, I should have added). Dicklyon 03:58, 8 August 2006 (UTC)

And then I added the example of the system of two photons that come from electron–positron annihilation. If anyone disagrees, please speak up; I might have it not quite right. Dicklyon 04:18, 8 August 2006 (UTC)

He has perverted the section again to a very different view, while ignoring my pleas for discussing it here. I reverted it, since it seem unreasonable for a person to cite something that he is disparaging. It appears that his problem is with the word "mass" for all the usual manifestations that everyone else calls "mass". Or is there some other explanation that anyone knows, since he's not saying? Dicklyon 04:29, 8 August 2006 (UTC)

"He" has not perverted anything. "He" has been trying to get you to understand that physics has changed since you learned it and since someone made an ill fated post in 1997 on the John Baez FAQ. "He" has explained this to you numerous times and has shown the errors in the derivations of such effects. Read above my answer to SBharris derivation. "He" has tried to compromise and phrase things in such a way such that the entry doesn't fly in the face of current understanding and such that it doesn't get laughed by current grad students. Ati3414 04:53, 8 August 2006 (UTC)

Oops, I see I did miss a talk entry above. Sorry. Still, there's nothing dangerous about the old terminology that associated energy and mass via E=mc^2. Dicklyon 04:31, 8 August 2006 (UTC)

E=mc^2 is not even correct. Why don't you let it be? You can't have decisions over posts that you only have partial and obsolete understanding. Leave things alone, let others decide.Ati3414 04:53, 8 August 2006 (UTC)

E=mc^2, using m for invariant mass or rest mass, still continues to be perfectly correct in the CM or COM frame, which is the frame we usually work in, and is certainly the frame we always are in when we're using scales to measure masses. So we note that and pass on. Heat an object, and E goes up and m goes up and E=mc^2 remains true. Very handy. E(total) is directly weighable in the COM frame. That's sort of the point of this whole discussion. And why it needs to go in the Wiki. S B H arris 22:23, 8 August 2006 (UTC)

So, you are saying that E=mc^2 is correct for photons. this is how you got your naive "solution" m=E/c^2, a classical rookie mistake. Any self-respecting textbook will tell you that E=pc for photons (http://www.lassp.cornell.edu/~cew2/P209/part11.pdf). Good diversion, though, got an insight how you "manufactured" the result for your "proof". Ati3414 22:41, 8 August 2006 (UTC)

E=mc^2 is correct for systems of photons in the COM frame, yes. It's correct for single photons and one other particle, so long as the COM frame is used. It's not correct for single free photons, but I never said it was. One rarely meets single free photons anyway.S B H arris 00:52, 9 August 2006 (UTC)

There is nothing like persisting in a rookie mistake. E=mc^2 does not apply to the photon. E=pc is what applies. Crack open a more modern book, you are showing your ageAti3414 02:14, 9 August 2006 (UTC)

OK, I'll back off for now and see who else cares about this. Can you recommend a good book to explain why (and when) E=mc^2 got to be wrong, or incomplete, as an expression of the equivalent of mass and energy? Maybe I'm still stuck in teh 20th century. I need a source besides you to explain how the old view is no longer acceptable. Dicklyon 05:02, 8 August 2006 (UTC)

Try this , it is free and comes from Harvard, it is pretty much the norm for modern teaching of relativity . Thank you for being so nice to me.

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8..... http://www.courses.fas.harvard.edu/~phys16/Textbook/ch12.pdf

Ati3414 05:24, 8 August 2006 (UTC)

Thanks for the references. But I don't buy your point. That book clearly states that "Some treatments of relativity refer to the mass of a motionless particle as the “rest-mass” m0, and the mass of moving particle as the “relativistic mass” mrel = γ m0." So it admits what is "conventional", which is what the rest of us have been saying. Then it goes on to argue for a different way that it claims is better, and supports that argument by a ridiculous strawman about Newtonian stuff, without even recognizing that Newton's law is usually stated as the more sensible F=dp/dt. And it never really addresses the question of whether the total rest mass of a system that has internal energy needs to account for that energy. The guy who wrote it is "Lecturer on Physics & Assistant Head Tutor, Assistant Director of Undergraduate Studies," with a Ph.D. just 10 years ago; not exactly the kind of authority who is likely to change the conventions in the field of physics by sheer force of personality. I have no problem with the concept of redefining mass to only have rest mass, and phase out relativistic total mass, and to redefine the meaning of E=mc^2 accordingly. More power to him; and to you; but not here, not now, not by using Wikipedia as the podium. Write a paragraph that describes this as a new alternative view, reference the book, but leave the other part in the conventional form. Dicklyon 05:55, 8 August 2006 (UTC)

This is just Harvard, do you want to see the MIT or Stanford website on the subject? I understand how difficult it is to admit that your views are outdated. The newer generation has a new view, wiki needs to reflect the most up to date view. (A small correction F=dp/dt is not Newtonian, it is SR. F=ma is the Newtonian view). Ati3414 06:05, 8 August 2006 (UTC)

As any student of Newton will tell you, F = dp/dt is how Newton defined it also.S B H arris 20:53, 8 August 2006 (UTC)

[edit] 328 new books, many with "outdated" view

I'm OK admitting my views are outdated if it can be shown to me. But when I search recent books I find many that don't seem to have gone this way. Maybe your view is just too "with it" to be encyclpedic? Why not write it up as a paragraph about the new view, instead of denying the conventional view? Books 2003-2006 with "relativistic mass" mentioned

Pleading ignorance is not a valid argument. Books change much later,they take a lot longer to edit, reissue, etc. I think that your argument is specious and you know it. Look at how things are being taught today, books will follow suit:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf

and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8....Ati3414 06:20, 8 August 2006 (UTC)

I don't dispute that some people are teaching it your way. That's no reason to deny the conventional way, which is still being put into current books, when writing a pedia article on it. So give it up; your radical approach is inappropriate here. It's fine to talk about the new way, but not to exclude the conventional way. If you are really "though with this BS" as you said on my talk page, then so long. Dicklyon 06:38, 8 August 2006 (UTC)

Because an encyclopedia is supposed to convey the most up to date view of science. It is supposed to correct older, outdated views. If one wilfully conveys the outdated view, in the context of such an encyclopedia being easy to update instantaneously, it turns the encyclopedia into a joke. If a joke is what you want, then continue to wallow in the blissful ignorance. After all, all that matters is that you have your way, scientific truth is secondary. Ati3414 06:47, 8 August 2006 (UTC)

There isn't any difference on "scientific truth"; this is language and bookkeeping. It's like the way people are taught to subtract by borrowing. My generation was told that, when you have to borrow, you take one away from the digit on top; an earlier generation was taught to add one to the digit on the bottom. (Tom Lehrer has a song about this, among other things.) That's what relativistic mass/invariant mass is like: Do you multiply by γ before or after you do something else?

Teachers of elementary courses (including first-year university courses) often simplify the story to save time. Algebra teachers teach their classes that "you can't divide by zero"; there's no need for them to bother with the Riemann sphere, which would just confuse their students anyway. That may be what's going on with these texts. Or maybe the professors really believe that one bookkeeping method is right and another is wrong; as silly as that would seem to me, we can certainly report that view. But we should not expunge the historically well-attested account in terms of relativistic mass, simply because it's inconvenient for a few instructors' lesson plans. --Trovatore 07:01, 8 August 2006 (UTC)

Sorry, it goes much deeper than that. From a professional physicist poit of view, the "photon in the box" is a joke, trying to convey what's not there, a mass increase of the system. As scientists , or wanna-be scientists, the people that put it together should know. But heck, you are all happy with the outcome, your collective POV prevailed, this is all that counts. I no longer give a s####Ati3414 07:09, 8 August 2006 (UTC)

I believe you have it wrong. What about the case of the massless gluons in nuclei? When you fission a U235 nucleus, and get a set of other particles where a lot of gluon energy has gone into kinetic energy, if you catch those particles and weigh them they add up to less than the U235 you started with. Isn't that because the U235 mass includes the energy of the massless gluons that are part of it? Isn't that just like the photons in the box? One of us is still confused. Dicklyon 15:11, 8 August 2006 (UTC)

If Ati seriously believes that the box plus photon doesn't have a higher invariant mass than the box without the photon, then he's simply in error. I had thought he had given up on that point and was now arguing about what should be called "mass"; looks like I was wrong about that. --Trovatore 15:30, 8 August 2006 (UTC)

And if you're waiting for Ati to explain how massless gluons contribute to the mass of the gluon-container that is a nucleon (or any hadron) then you're going to be waiting awhile. This contradicts his worldview, therefore he's going to ignore it. Modern physics or not. That is how you tell reasonable from unreasonable people in a discussion. When the man with the fixed delusion is brought up against a perfect counterexample, he just shuts down like a locked-up computer running Windows. But his mind is not changed. The reasonable person, on the other hand says: "Wups, looks like I was wrong." S B H arris 15:39, 8 August 2006 (UTC)

Hmm, you may be crediting me with more reasonableness than I really intended. I was saying I was wrong about what Ati3414 was arguing, not about what should be called mass. My position on that remains that, yes, as I knew, physicists no longer like to describe things in terms of relativistic mass, and they have good reasons and that's fine, but that doesn't make the descriptions in terms of relativistic mass wrong, and we should mention those descriptions, neutrally and briefly, for the benefit of readers who would otherwise be confused. --Trovatore 21:17, 8 August 2006 (UTC)

On the idea of mentioning relativistic mass as a previous notation, sure. That's why there's a section about it in mass in special relativity. Relativistic mass as a notion, should be mentioned, then the advantages and disadvantages of the notation noted, and finally the fact that the physics field is generally moving away from it, tending to use energy instead. Although not without carnage. :) You know, one of the worst problems with "relativistic mass" is that when people decided to discard it, some people seem to have been left with the impression that heat and radiation, which had relativistic mass but no rest mass, somehow now are left massless in all ordinary senses of the word mass. But that's not true. All energy continues to have rest mass in the system COM frame, and that includes all energies of the things which were composed wholey or in part of "relativistic mass". Heat, light, and kinetic energy in systems continues to exhibit mass when the system as a whole is at rest, and it remains just plain old ordinary mass of the kind you weigh. According to E = mc^2, end of story. Doing away with "relativistic mass" hasn't changed this remarkable fact. So thanks, Dr. Einstein. It's been more than a century now since you clued us, and some people, by god, still don't understand it. And apparently this Ati3414 is one of them. S B H arris 01:10, 9 August 2006 (UTC)

Please carry on with the personal attacks, they form a very good scientific argument. In between personal attacks maybe you find a few minutes to have a critical look at your mathematical "proof". I added some notes yesterday, you may have missed them.Ati3414 18:08, 8 August 2006 (UTC)

OK, I'll be the Mac, like in those fun commercials, since you brought it up. But seriously, I added one more example, of the mass of the "virtual photons" holding an atom together. Again, I seek feedback if I got this wrong, but more feedback from Ati will only serve to make me want to further belabor the point that he disagrees with. Is there any reasonable physicist out there who can tell me if I've captured the essense of truth, at least as accepted in modern physics, with these little examples? Dicklyon 17:12, 8 August 2006 (UTC)

Well, alas, Mac, I had to revert you, before you remembered (in embarrassment) that the atom has less mass (not more) as a bound system. Just as the nucleus does. But you're not wrong about the virtual photons (or for the nucleus, virtual pions too). Static electromagnetic field is DESTROYED when opposite charges are brought together, as in an atom. So the missing mass is missing virtual photons. In a bound nucleus you gain virtual photons, but you lose even more virtual pions, so the net comes out lost mass. Which of course must appear as mass someplace else (the mass of kinetic energy, radiation, etc) when the bound object is bound. Missing mass is only "missing" if the system wasn't closed and it was allowed to leak out (which is generally the case in nuclear physics, of course). That's one more oddity of how physics is taught, which is historical. Mr. PC. S B H arris 18:08, 8 August 2006 (UTC)

Yes, even Macs have their flaky moments. I stand corrected and humbled. Dicklyon 19:43, 8 August 2006 (UTC)

[edit] Ati3414's reversions

By my count, the following edits of Ati3414 are reversions within the meaning of WP:3RR :

[4] (restoration of slightly reworded note about section being "embarrassing" or "a disaster")
[5] (removal of the section altogether)
[6] (removal of the section altogether)
[7] (removal of the section altogether)
[8] (removal of the section altogether)
[9] (partial revert; removal of reference to relativistic mass)
[10] (partial revert; removal of reference to relativistic mass)

These are all within 24 hours (from 05:11, 07 August 2006 UTC to 04:40, 08 August 2006). The following edit misses being within the 24 hours by less than half an hour:

[11] (full revert to a previous version of Ati3414)

I think someone (preferably someone not a party to the debate, with admin powers) should warn him and block him on the next revert, or even without warning given the magnitude of the violation. If this behavior continues an RFC should be opened. --Trovatore 06:09, 8 August 2006 (UTC)

Disclosure: On reviewing the record, it looks like I did it too — four reverts in the same time period. So, person I've invited above, you may wish to block me as well; that would be fair. --Trovatore 06:52, 8 August 2006 (UTC)

Certainly not, you did everything within your powers to maintain the status quo of "design by committee" so pervasive in wiki. No point in penalizing you for maintaining backward views. After all, this is what wiki is all about. Ati3414 07:00, 8 August 2006 (UTC)

Dear Il Trovatore

I understand your frustration, you don't like to be corrected. The truth is that physics has moved on since you last looked at it and the concept of mass has suffered some serious changes. So, instead of admitting that you have outdated knowledge why not ask for the banning or even better, execution of the person that does not conform to your views. See here:

http://www.lassp.cornell.edu/~cew2/P209/part11.pdf

and here:

http://www.courses.fas.harvard.edu/~phys16/Textbook/ch11.pdf Have a look at chapter 11.8.....

Heck, have it your way, it isn't worth it. Let's keep wiki a joke, the most important thing is that "editors" are never proven wrong, scientific truth is secondary. Ati3414 06:16, 8 August 2006 (UTC)

For the record there's nothing in this chapter which contradicts our basic point that the mass of a system in the COM or CM frame, is system total energy/c^2. And that includes a system of photons, which add energy and therefore mass to the system. Or a box that contains photons. Photon E adds to system mass in the ordinary E/c^2 way. Any kind of energy added to the system does, whether it's rest mass of system particles, the kinetic energy of system particles, photons injected, whatever. Anything in a box contributes to its mass. Also, any part of a system contributes to its mass, which in the COM frame is the system mass or system rest mass or system invariant mass, or whatever you want to call it. But it's what you weigh, and it's the inertia the system has in the COM frame (equivalence principle). Things get heavier when heated, and they get heavier when they absorb photons. And they get heavier even if the photons go into them and just bounce around. Which part of this don't you understand? S B H arris 16:34, 8 August 2006 (UTC)

I took a second pass at your "proof" a few days ago. have a look at it, a few sections above. as to the personal attacks that I shut down and all this stuff, you should have had a look at the answers before that. Anyways, I decided to add a non-crackpot section on the experimental mass of the photon, you must have "missed" that too.Ati3414 18:02, 8 August 2006 (UTC)

1. The experimental mass of the photon is talking about single photons, not photons which are part of systems, and so is not relevant. Kinetic energy also has mass in systems, but not in single particles, because it can be made to go to zero (or be any value) by choice of frame. This is exactly the situation with photons, except they have no residual rest mass. However, their energy contributes mass to systems.

2. As for my derivation, you still have not come to grips with the fact that you can replace dp/dt with p/t if you like (and as I did), since this momentum transfer is constant over time (just as is mean time-averaged pressure exerted by an ordinary molecular gas). So take out the calculus, use p/t, where you pick a p and a t that goes with it. And get over it. S B H arris 18:37, 8 August 2006 (UTC)

1. But your "photon in a box" talks exacly about adding E/c^2 by virtue of adding one photon to the box. You can't have the cake and eat it too.In other words, you cannot claim that adding one photon adds E/c^2 to the "mass" of the box. Looks like you provided the best argument against your own post. Finally.

Whaterver you do, never admit to error, doc. So, after you botched the calculations by mixing Newtonian concepts (no, dt/dt is not Newtonian, is SR, and you are "calculating" and oddball dp/t) with SR concepts in order to get your desired answer E/c^2 which is many orders of magnitude off you repeatedly "forget" that you contradicted your toy experiment by the posting comparing it with the recognised experiments in the field. You are 17-19 orders of magnitude off, doc!Ati3414 21:28, 8 August 2006 (UTC)

The photon does not fall as m*g in the gravitational field, so you cannot infer the "mass" of the photon from its downwards force, and your derivation needs to be direction invariance. Try the experiment with the photon going horizontally and let's see how you "manufacture" the E/c^2 "solution" again. Don't ask me, I am just a spectator for your magic acts 21:26,Ati3414 21:28, 8 August 2006 (UTC)

The photon does fall as m*g in a gravitational field. For a uniform gravity field (note the caveat) a photon falls as you'd expect for any object traveling at its speed (deflections are larger when traveling into a gravity field and then back out, but those don't concern us here-- we're weighing starlight in a box, not deflecting starlight into and out of a gravity well). And yes, while bouncing around inside a box, a photon will add an effective mass m to whatever it is that contains it, whether traveling "up and down," or "transversely" in the field. I asked you to do the calculation so you could demonstrate your superior knowledge of physics, not stand on the side and snipe at others. Come on, Mr. Critic.S B H arris 01:19, 9 August 2006 (UTC)

2.To the dp/dt replacement with p/t, I don't think so. You have several other errors as well that you are conveniently not addressing. Have a second look. Ati3414 18:45, 8 August 2006 (UTC)

How many different ways do I have to say it? See those dt's and dp's? Integrate over time and get rid of them. IOW, integrate over MANY bounces. Just as you would for a box containing ONE gas molecule. Which would exert an average pressure, and by virtue of this, an average mass/weight over time. Of course, over shorter times, a box with one molecule or one photon jumps like a jumping bean. So does an elevator with a man jumping up and down in it, for that matter. But in the end, over time, exactly the right extra weight shows up on the cable. Or the scale. S B H arris 20:01, 8 August 2006 (UTC)

Selective answering, congratulations.What you are doing in your hack is calculating dp/t. No matter: there are three other errors in your "proof", like your continous mixing of relativistic mechanics with newtonian mechanics. You are also conveniently missing the fact (point 1) that you have refuted your own calculations. Just take the E/c^2 of the post, would you? It looks ridiculous, you are 17-19 orders of magnitude off.Ati3414 20:06, 8 August 2006 (UTC)

No, for force I calculated p/t (or Δp/Δt if you like). There are no differentials, because with a differential left in you cannot get an answer. The mean force is a number. It has a value. Had you bothered to get a value from your dP/dt you'd have had to do the same average that I did. So go for it. S B H arris 20:59, 8 August 2006 (UTC)

[edit] Single photon trapped transversely in a rocket; inertia and weight change

A rocket of mass M flies free in space, firing its engine to give it exactly an acceleration of 1 g (as seen by the passengers). Now a photon of light with energy E (as seen by the passengers of the rocket) flies in through a porthole and begins to bounce around inside the rocket transversely, in a direction orthogonal to the acceleration of the rocket. Does the engine suddenly now need to fire a bit harder, to maintain g? Does the photon act as though it has MASS, requiring extra rocket power to now accelerate a rocket with mass M + m, where m is the effective mass of the photon? And how much would that mass be? Small (to be sure), but since photons have no mass, would it be zero? Einstein told us in 1905 that if the photon where absorbed, the rocket would gain mass equal to E/c^2. But what if it is not absorbed, but merely bounces around inside, before exiting through another porthole? Does it take extra push to accelerate it? Does the photon have inertia? And if so, how much?

Answer: It does. And it's the same as if the photon was bouncing in any direction. In this case, we call the direction toward the engine Y, and track the photon starting from the centerline axis, to a bounce at the far wall of the rocket, then return to the centerline. This is half a bounce cycle. The mean force which these cycles exert on the rocket in the Y direction (the photon weight in this system) Fy, is therefore F= P/t, where P is Py, or the momentum transferred in the Y direction by ONE photon bounce off the far wall, and t is the time for the photon to go from centerline, bounce, and return to the centerline. This answer does not change if the photon completes a hundred bounces, because P and t increase by the same amount in fixed ratio. The component Py won't be large, because the photon starts out headed exactly at the far wall, and not at all in the direction of the engine. But between the centerline and the wall, the rocket moves, so that by the time the photon hits, it's no longer moving exactly straight into the wall, but rather at an aberration angle sin(θ)= v/c (exercise left for the student, particularly why this is sin(θ) and not tan (θ), but remember the wall is moving and there is some Lorentz contraction to consider). Here v is component of velocity in the Y direction the photon picks up while going from centerline to wall, as a result of the acceleration. This process continues, with every bounce off a wall by the photon transfering force downward in the direction opposite rocket acceleration.

Now if P is the photon's total momentum to begin with, then Py = P sin(θ). We already know that sin(θ) = v/c. And v will be g * (t/2) where t is the time for the photon to go from centerline axis to wall, and back to centerline (it's t/2 because the rocket only gets half this time to accelerate while the photon goes from axis to wall). So:

Py = P(v/c) = P*g*(t/2)/c = (P*g*t) / (2c)

The average force downward toward Y due to the photon bouncing through a half cycle is P/t and in this case P is 2Py because a total reflection causes twice the momentum transfer. Putting in our expression for Py:

Thus F = 2Py/t = Pg/c

Remembering that in relativity photon P = E/c:

F = (E/c^2) * g

This is the force the photon exerts on the rocket, which is the same as the rocket exerts on the photon. The extra force the rocket motor needs to use due to the photon the rocket picked up, is E/c^2 *g, which is the same as if something with mass E/c^2 had come in through the porthole. So that is our answer. This is the effective mass of the trapped photon in this situation, until it gets out through another porthole.

What if the rocket is sitting on the pad on Earth, being weighed by a scale at 1 g. The rocket's mass is M, and its weight is M*g. If a single photon now comes in through a window and is trapped transversly in the same manner, the equivalence principle tells us that the gravitational field of the Earth will act exactly like the 1 g acceleration of the motor, and cause the photon to exert the same extra force on the rocket wall by bouncing back and forth. Thus, the rocket now weighs more due to the photon of energy E, by an amount E/c^2. The single photon not only has inertia in a moving rocket, but weight in a landed rocket. And inertia and weight is mass. If the rocket is shaped like a box, the answer does not change :) . S B H arris 02:21, 9 August 2006 (UTC)

Very clearly stated. That is in agreement with special relativity and other principals of physics as I learned it, and so far as I can tell it hasn't changed since then. Dicklyon 03:35, 9 August 2006 (UTC)

Very clever, you finished what I started and got erased. Now, the issues:

[1] same as in the previous solution, a gratuitous F=P/t which sets things in the desired direction but not is not justifiable

[2] This does not mean that the rocket picked up an extra mass E/c^2 , it simply means that there is a debatable (see [1]) resistence type force F = (E/c^2) * g exerted in the direction of rocket travel and with a sense opposite to rocket travel. Nothing in QED allows you to attribute mass to a photon, even worse, to attribute it weight.

[3] Even if objections [1],[2] were removed by some miracle , you are now stuck with your own refutation of the derivation. The well respected experiments put a constraint on the mass of ONE photon (as you well observed somewhere in a previous post) of 6*10^-17 eV . But your formula produces an equivalent em mass equal to E , which is about 3eV. So, your personal research is 17 orders of magnitude off the accepted experimental data (actually 19 orders of magnitude considering the latest experimental results). Interestingly enough, you are off by a factor of about (v/c)^2.

I liked your transposition of the problem using the EP. Ati3414 03:49, 9 August 2006 (UTC)

Answer

1) If you don't like my method, then YOU finish what you started. Do not criticize somebody else's physics unless you're prepared to demonstrate better technique. If not, you're just blowing smoke.

Well, you get it wrong repeatedly. And you keep mixing in Newtonian with SR mechanics <shrug> Ati3414 17:08, 9 August 2006 (UTC)

2) Inertia is mass. If you believe in the equivalence principle, the rocket also picks up weight. Are you really arguing that it picks up weight and inertia equal to E/c^2, but not mass? How else would you measure its mass?

Sure, I believe in EEP. This doesn't give you the right to infer (again , repeatedly) that the term in front of "g" is the photon "mass". The Proca equations on which the current experiments are based express the violation in "photon mass" , no mention of "rest mass". Have you read the papers? Ati3414 17:08, 9 August 2006 (UTC)

3) As has been carefully explained to you many times now, you are arguing apples and oranges when you argue that photons have no rest mass. We're not measuring their rest mass. A single gas molecule of rest mass m trapped in a rocket would add the inertia and weight m/c^2, but if it was moving with kinetic energy T, it would add (m+T)/c^2. A photon has no rest mass, so it only adds its kinetic energy T. Which is E/c^2. S B H arris 15:23, 9 August 2006 (UTC)

"...so it only adds its kinetic energy T. Which is E/c^2" I learn new physics from you every post. Now, seriously speaking there is no such notion of rest/relativistic mass for the photon. It was explained to you repeatedly. As such, modern teaching of relativity did away with the notion of rest/relativistic mass altogether. The paper I pointed you to talks about constraining the "photon mass", no reference to the "rest" mass. Ati3414 17:22, 9 August 2006 (UTC)

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