Talk:Photon/Photons and Mass Debate3

message: 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)
replies: text: 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. SBHarris 05:08, 6 August 2006 (UTC)
replies: text: 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)
replies: text: 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.SBHarris 05:33, 6 August 2006 (UTC)
thread_title: In a box?  With a fox?

message: 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)
replies: text: 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? SBHarris 05:10, 6 August 2006 (UTC)
replies: text:  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)

text: 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)
replies: text: 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)
replies: text: 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.SBHarris 05:20, 6 August 2006 (UTC)
replies: text: 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)
replies: text: 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. SBHarris 05:43, 6 August 2006 (UTC)
replies: text: 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)
thread_title: Photon in the box is very poorly written

message: :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)
replies: text: 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)

text: 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)
replies: text: 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)

text: 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)

text: 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)

text: 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)
replies: text: "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)

text: 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)
replies: text: 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)
replies: text: 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.  SBHarris 22:23, 8 August 2006 (UTC)
replies: text:  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)
replies: text: 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.SBHarris 00:52, 9 August 2006 (UTC)

text: 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)
replies: text: 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)
replies: text: 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)

text: 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)

text: 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)
replies: text: As any student of Newton will tell you, F = dp/dt is how Newton defined it also.SBHarris 20:53, 8 August 2006 (UTC)
thread_title: compromise