Talk:Photon/Photons and Mass Debate4

Photons do not have rest mass, but they contribute to inertia and gravitate

Hi all,

Someone asked at the Physics Wikiproject to clarify the relationship of photons and mass. I'll do my best! :) I hope everyone will be patient with my explanations.

First, photons do contribute to inertia. That's stated explicitly by Einstein in his 1909 article, The Development of Our Views on the Composition and Essence of Radiation, which I translated recently. You can find the original German text on the web, but I'm not allowed to post it at the German Wikisource because of copyright issues.

Second, photons do gravitate. Since they have energy, they contribute to the stress-energy tensor and, thus, to the curvature of space time. Hence, they act gravitationally on other objects. Conversely, they respond to a gravitational field of other objects; light's path is curved in a gravitational field, as shown (for example) by gravitational lensing.

The "weighing" of a box of photons was used by Einstein in a famous thought-experiment to refute the Heisenberg uncertainty principle, $\Delta E\Delta t\geq \hbar /2$. His idea was that a hole in a box of photons could be opened for an arbitrarily short time, allowing one photon to escape. The energy of that photon could be measured to arbitrary accuracy by weighing the box before and after the hole opened. Thus, it seemed that the uncertainty in time and energy could be made as small as possible. Bohr's refutation (delivered the next morning) was a watershed moment for quantum mechanics.

All that being said, light does not have rest mass (to the best of our present knowledge). That's not a problem, though; the rest mass of an object is its mass in a coordinate frame where the object is at rest. There is no such frame for light moving through the vacuum; it moves at c in all reference frames.

Hope this helps, Willow 12:06, 9 August 2006 (UTC)

COMMENT

It would help if you'd clarify. A photon's inertia is undefined (because it's energy is undefined) and cannot be measured quantitatively, until it's part of a system. Same for its weight. These two constitute a mass of E/c^2, which a photon adds to a system in which it is trapped (and in reference to which its energy is E, which now CAN be defined because we have defined a reference frame). And similarly a photon subtracts E/c^2 invariant mass from a system (seen in the COM frame) which emits it. SBHarris 15:28, 9 August 2006 (UTC)

Another comment written concurrently

As you can see by Ati's reply below, it doesn't help much. What you've stated is pretty much what we all know and agree on. What's left is the question of whether adding inertia and gravitation to an object is the same as adding to its mass. Most of us think it is. We all agree that light does not have rest mass. Can you address his response below in terms of how it relates to standard modern physics as you know it? Dicklyon 15:32, 9 August 2006 (UTC)

His response has nothing to do with physics as we know it, so there's not much place to compromise on the issue. If you increase the energy, inertia, and weight of a resting compound object, whether by heating it or adding one or more photons to it (which of course can be the same process-- try an infrared broiler), you have increased its proper mass, invariant mass, rest mass, and just plain old everyday mass as we and modern physicists use the word. The end. It's not complicated. Only cranks believe otherwise. SBHarris 15:50, 9 August 2006 (UTC)

Photons have energy, momentum and spin but no mass

The views that the photon contributes E/c^2 to the inertia of a system are not the view of modern physics. While it is true that Einstein demonstrated in september 1905 [1] that a body radiating em would loose a mass $\Delta m$ where that math is related to the energy variation $\Delta E=\Delta m*c^{2}$ , the converse is paradoxically not true, the naive view that adding a photon to a system would increase its mass by E/c^2 is refuted by modern experiments. One can only tell that the total energy of the system has increased by E and its momentum has increased by p (proper mass is not additive in relativity, the so called "relativistic" mass is, but that is an obsolete term ). Turns out that the photon has zero mass as proven by modern experiments. Modern experiments [2] look for a violation of the Maxwell's laws predicted by A. Proca in his 1937 seminal paper. The violation is of the order of $mu_{g}=m_{g}*c/\hbar$ where $m_{g}$ is the "photon mass". Any $mu_{g}>0$ would result into an observable phenomenon , confirming indirectly that $m_{g}$ is greater than zero. This would bring about a violation of gauge invariance. As it can be seen from the many other experiments [3]-[11], this is not the case.Ati3414 14:05, 9 August 2006 (UTC)

References:

[1] http://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

[2] http://silver.neep.wisc.edu/~lakes/mu.pdf

[3] Eric Adelberger, Gia Dvali, Andrei Gruzinov, "Photon Mass Bound Destroyed by Vortices", arXig.org

[4] Goldhaber, Alfred S., and Nieto, Michael Martin, "Terrestrial and Extraterrestrial Limits on The Photon Mass", Rev. Mod. Phys. vol.43 #3 pp.277–296, 1971 [1]

[5] E. Fischbach et al., Physical Review Letters, 73,514-517 25 July 1994.

[6] Official particle table http://pdg.lbl.gov/2005/tables/gxxx.pdf

[7] L. Davis, A. S. Goldhaber, and M. M. Nieto, Phys. Rev. Lett. 35, 1402 (1975)

[8] Roderic Lakes, "Experimental Limits on the Photon Mass and Cosmic Magnetic Vector Potential", Phys. Rev. Lett. 80, 1826 (1998) [2]

[9] J. Luo et al., Phys. Rev. D 59, 042001 (1999)

[10] B. E. Schaefer, Phys. Rev. Lett. 82, 4964 (1999)

[11] J.Luo et al., Physical Review Letters, (28 February 2003)

COMMENT

You actually write:

The views that the photon contributes E/c^2 to the inertia of a system are not the view of modern physics. While it is true that Einstein demonstrated in september 1905 [1] that a body radiating em would loose a mass Δm where that math is related to the energy variation ΔE = Δm * c2 , the converse is paradoxically not true...

I wonder if you know how completely crazy that idea is? You're suggesting conservation of energy and mass in one direction for a process, but not the other direction. The same process. Give it up. You've lost this one.

And by the way, if your (true) arguments that photons have no rest mass were relevant here (which they aren't), a system wouldn't lose mass when emiting a photon, either. So you shot yourself in the foot there by even admiting that. But I understand why you don't want to call Einstein wrong, since they you'd be claiming he said something incorrect a century ago, and nobody caught him for a century until YOU came along. LOL. SBHarris 15:34, 9 August 2006 (UTC)

What else is new, you obviously can't read (or you twist whatever you are reading). Keep on reading : "One can only tell that the total energy of the system has increased by E and its momentum has increased by p (proper mass is NOT additive in relativity, the so called "relativistic" mass IS, but that is an OBSOLETE term)." This was the argument from the beginning but it doen't register with you. Try reading the modern papers and the modern way of teaching the subject, I put them on the website for you. Maybe you learn and you stop "manufacturing" mass from gedank experiments.Ati3414 15:43, 9 August 2006 (UTC)

Actually, proper mass (aka invariant mass) is additive in special circumstances, so long as you measure mass in the COM inertial frame for each object or submass, and measure the total mass in a common COM frame for the whole. Put two boxes on a scale in relativity and you get the summed mass of two boxes. Put one box on a scale and keep it closed, and its mass will stay the same no matter what goes on inside it. You could generate heat, have a nuclear reaction, blow up a bomb-- even an atomic bomb if the box was superstrong. So long as you let nothing out of the box, all the heat, light, and mass of whatever was inside, would continue to have the same mass and weight, before, during and after the reaction. SBHarris 16:08, 9 August 2006 (UTC)

A simple exercise

Let's try something really simple. Imagine your empty box, with no momentum. It has the mass:

$m={\frac {E}{c^{2}}}$ [1]

Now, we add a single photon (not a pair), exactly as in the "photon in a box" wiki entry having energy e and momentum p=0 (how could that be, the single photon has nonzero momentum) to the box as in your example. The mass of the system becomes:

$m'={\frac {E+e}{c^{2}}}=m+{\frac {e}{c^{2}}}$ [2]

It is correct to write the formula

$m'=m+{\frac {e}{c^{2}}}$ [3]

but one SHOULD NOT call the term ${\frac {e}{c^{2}}}$ "mass", "mass increase", "mass increase due to photon". All you can say is that the mass of the system has increased by ${\frac {e}{c^{2}}}$ . You cannot attribute mass to the term ${\frac {e}{c^{2}}}$ because if you do, the gauge invariance in QED is violated. This is purely a pedagogical issue but it is very important.

Now let's try the real case, when the photon has a nonzero momentum. You must now modify [2] to accomodate the photon's momentum. You get something quite different:

$m'={\frac {\sqrt {(E+e)^{2}-(pc)^{2}}}{c^{2}}}$[4]

Not easy to relate m' to m anymore.....What is the "photon's contribution"? Does it make sense to even mention it anymore? What about the most general case:

$m'={\frac {\sqrt {(\Sigma E)^{2}-(\Sigma pc)^{2}}}{c^{2}}}$[5]

Ati3414 22:33, 9 August 2006 (UTC)

A lovely string of muddled assertions and nonsequiturs. I was particularly impressed by

You cannot attribute mass to the term ${\frac {e}{c^{2}}}$ because if you do, the gauge invariance in QED is violated. This is purely a pdagogical[sic] issue but it is very important.

which is what happens when you fail to distinguish between rest mass and relativistic mass. Gauge invariance of QED is violated if the photon has a nonzero rest mass, and your first reference to mass is really relativistic mass. So your sentence reduces to:

You cannot attribute relativistic mass to the term ${\frac {e}{c^{2}}}$ because if you do, rest mass of the photon becomes non-zero. This is purely a pdagogical[sic] issue but it is very important.

and we see what a load of garbage it is. --Michael C. Price ^talk 23:03, 9 August 2006 (UTC)

PS it's quite easy to add zero-momentum photons to a box. Pair them up so their momenta cancel. --Michael C. Price ^talk 23:06, 9 August 2006 (UTC)

There is no such thing as rest/relativistic mass for the photon. So your counter is a practical zero. Try reading the Proca paper and try retaking the relativity classes, you are stuck in the past. Ati3414 23:09, 9 August 2006 (UTC)

"...one SHOULD NOT call the term ${\frac {e}{c^{2}}}$ "mass", "mass increase", "mass increase due to photon". All you can say is that the mass of the system has increased by ${\frac {e}{c^{2}}}$." ...Okay: I add a momentum-less photon (yes, I know; a pair of photons with equal but opposite momentums would be the better system, but I want to maintain the original situation) that has an energy ${e}$. The mass of the system increases by ${\frac {e}{c^{2}}}$. However, I cannot attribute that increase ($m'-m>0$) to the introduction of the photon into the system. Um, lol? If only mass and energy were equivalent... Regardless of whether you (or a Harvard textbook) approve of the term 'relativistic mass' for all cases or not doesn’t change the fact that we can consider the system as a whole to have gained mass by the addition of a photon. If the photon adds mass to the system, then it must have had a mass prior (in the form of energy). // Sidebar: I do not see how you intend to have this community accept this so-called modern take on physics, if it truly is the modern view, by insulting people. The negative responses that you are getting are not because people think that your information is flawed. (And before responding, I would suggest reviewing the wiki-formatting of your posts (Show Preview) as well as running them through an MSWord spelling/grammar check, as your credibility tends to decrease with the increase of such errors) --HantaVirus 17:37, 10 August 2006 (UTC)

Try the above exercise with one photon. Then try it with 3 photons, one going in one direction, the other two going at 180 degrees. Come back when you have the answers.Ati3414 20:04, 10 August 2006 (UTC)

Actually, when trying to prove your point, it works best if you prove it. As you've yet to provide persuading evidence, I believe that the burden of proof still lies with you. Moreover, if, in performing the suggested calculations using your equations, I was to get a nonsensical answer (which is what I assume you are trying to get at), then that means that those equations do not accurately represent the system as stated. --HantaVirus 20:51, 10 August 2006 (UTC)

Actually, I did (see above calculations). How about you calculated something? Let's see the 3 photon arrangement, couldn't be too complicated. Ati3414 21:04, 10 August 2006 (UTC)

Don't you mean 3 equal energy photons going at 120 degrees, so their momenta add to zero? What's the point of having one go at 180 degrees from TWO others?? SBHarris 22:29, 10 August 2006 (UTC)

No , precisely 180 degrees, such that you cannot get the "COM" frame triviallyAti3414 00:03, 11 August 2006 (UTC)

On second thought, I'm pulling out. I lack both the interest and subject matter expertise to continue in this debate. If it matters to anyone, I, as a layman when it comes to relativistic physics, I understood the article as is. Good luck resolving this, guys. --HantaVirus 22:43, 10 August 2006 (UTC)

At least you admit it. Ati3414 00:03, 11 August 2006 (UTC)

Separation of two distinct issues

Hi all,

There seem to be two issues; perhaps disentangling them and solving them separately will help?

First issue: does the photon have rest mass? The answer at present (to the best of our experimental knowledge) is "no". The references you cite above (e.g., [2]) discuss the issue in detail. I don't have my books here, but I recall that Feynman also recounts several arguments for limits on the photon rest mass in his first autobiography Surely you're joking, Mr. Feynman!. Moreover, if we believe the standard electroweak unification theory, the photon is exactly massless, since its associated U(1) symmetry is not broken; given that the other vector bosons (W and Z) were discovered with the correct properties, there's good reason to believe in the predictions of that theory for photons as well.

Second issue: does the absorption or emission of a photon change the inertia or gravitational interaction of the absorbing/emitting body? The answer is "yes", if we believe the theory of special relativity; Einstein himself says so in the last sentence of reference [1] that you cite: "The photon conveys inertia between the emitting and absorbing bodies." If inertial mass were lost upon emission but not gained upon absorption, that would violate the principle of time reversibility; moreover, any material system would be continually losing mass with every emission, which is not observed.

Hopefully, this separation of topics clarifies the difficulties. Willow 15:48, 9 August 2006 (UTC)

PS. I wrote this reply before all the other responses; I'll try to read those now and see what needs to be added.

Definitions of "mass"; Helpful thought experiment?

Hi, it's Willow again.

From what I can tell, the remaining difficulties seems to lie in the definition/usage of the word "mass" — they're semantic difficulties.

One definition of mass (for a massive particle) is its rest-mass, i.e., the coefficient $m_{0}$ in front of the energy $E=\gamma m_{0}c^{2}$ and momentum $\mathbf {p} =\gamma m_{0}\mathbf {v}$. By contrast for a massless particle like the photon, the energy-momentum 4-vector is given by $(E,\mathbf {p} )=\hbar (\omega ,\mathbf {k} )$. If we all agree that the photon is massless, there's no need to argue over this definition.

Another (more experimental/practical) definition of mass is by inertia or gravitation: how difficult is it to move? How strongly does it attract other bodies? I hope we now agree that the photon conveys inertia between the emitting and absorbing bodies; and that the photon is capable of gravitational interactions, since it contributes to the stress-energy tensor.

Perhaps the following thought experiment will clarify matters. Consider a very large, hollow cavity bounded by perfectly reflecting, perfectly rigid but extremely thin walls; inside the cavity are free electrons, atoms and also charges bound by springs of every resonant frequency. At thermal equilibrium, this cavity will be filled with the normal Planck distribution of blackbody radiation, no? If the cavity volume is large enough, the energy in the blackbody radiation will be much larger than that of the walls.

Now assume that the cavity is prepared in a non-equilibrium state such that it has no photons within at time $t=0$, i.e., all the energy is in the matter. As the cavity equilibrates, the energy in the photons will increase over time and that of the matter will decrease over time, until the thermal equilibrium distribution of blackbody radiation (given by Planck's formula) is achieved. Still with me?

The pivotal question is: does the inertia of the box (as seen from outside the box) change as the energy distribution within the cavity gradually equilibrates? For example, does the box gradually become easier to accelerate? The answer is "no", according to modern physics. The photons within the cavity contribute to the inertia and gravitation of the box. Expressing the pivotal question in another way: does the effective $m_{0}$ of the box vary with time as the energy in its interior equilibrates? The answer again is "no".

Hoping again that this helps, Willow 16:32, 9 August 2006 (UTC)

Thank you, this is much better. Now, how do we reflect this in the much beleagured entry "Photon in the box" that claims that: "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" ? can you clean it up ? This misbegotten sentence conjures the idea that one could see a measurable photon mass of about 3eV when, in reality all modern experiments put a limit at about 6*10^-17 or smaller. I tried to explain (in vain) that for the photon the notions of "rest" and "relativistic" mass do not exist in modern physics. Because of that the two notions have been eliminated altogether from the way relativity is being taught (see links). The current experiments, based on the Proca violation extensions to the Maxwell's equations talk about "mass", period. The difference is subtle but very important: when someone comes in with a badly written gedankexperiment and tells us that "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" and people see that the whole rest of the wiki page says clearly "the photon is a massless particle" is it to wonder that people reading the "photon in the box" get confused? To counter all the confusion I wrote the whole entry on "Photon mass, experimental limits" but I still hope that someone (maybe Willow?) can clean up the "Photon in the box". I fully agree with what you wrote above, this is the view of modern (I should say contemporary ) physics, now can you convey it by cleaning up all the refences to "relativistic/rest mass" and the ill begotten "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)"?

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.

Thank you Ati3414 16:54, 9 August 2006 (UTC)

Hope so, too. I thought we had pretty much agreement on these points, but now Ati is saying that the energy of the photon does NOT add to the inertia of the system (in the view of modern physics). I'm perplexed as to why he would make such a claim, as nothing in his references that I've been able to find address that. Dicklyon 16:39, 9 August 2006 (UTC)

Yes, that's a good example. If the photon gas the cavity fills with, doesn't have mass, then the whole thing will get lighter as the photons leave the walls and lower the mass by E/c^2 (per Einstein) but then don't contribute EQUALLY to the externally measured mass of the object, while they bounce around inside.

And it will be medically interesting to see the response from those who wish to argue otherwise. SBHarris 16:42, 9 August 2006 (UTC)

Please, please refrain from that sort of talk; it makes me sick at heart. I'm sure that we're all here to improve Wikipedia and our own understanding by conversing. No one, living or dead, has a perfect understanding of nature. Let Physics speak for Herself. Willow 16:56, 9 August 2006 (UTC)

I see you're new here. In this discussion we've long gone past the point where physics is allowed to speak for itself. As you will see from the replies you are about to get to your example. SBHarris 17:29, 9 August 2006 (UTC)

No need to sling mud and get personal, I am just trying to explain thigns to you that come at odds with what you learned many years ago.There is no more such notion of "rest/relativistic" mass. Neither of them makes any sense in the case of the photon and both have been banished from modern teaching of physics. Here it is, the same explanation again:

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 17:36, 9 August 2006 (UTC)

Please stop quoting these cites. They don't discuss mass of systems except in the examples (which aren't fully referenced), and thus they really are irrelevant.SBHarris 19:36, 9 August 2006 (UTC)

Mass of systems of photons

If we could keep things off the repeated personal attack, the only thing I have been saying all along is that you cannot infer that "For example, a mirrored box containing a gas of photons, or even a single photon, with total energy E will have greater rest mass (by Δm = E/c²)" because it conveys the notion of photon having mass. And this comes to odds with the whole rest of the wiki page (and with mainstream physics). Try to accept this: since there is no "rest" vs. "relativistic" mass for the photon, it makes no sense to talk about the "photon mass contribution". It also makes no sense to talk about "rest/relativistic mass" anymore. The modern teachers have moved on. That's allAti3414 17:36, 9 August 2006 (UTC)

No, try and accept this (it's quite simple):there is no photon "rest mass", there IS a photon "relativistic mass" . Ok? --Michael C. Price ^talk 17:50, 9 August 2006 (UTC)

Just a quibble: There is a photon rest mass; it just happens to be zero. (Dilbert: "I've got a personality." Dogbert: "Let's not have that 'is zero a number' argument again.") --Trovatore 17:56, 9 August 2006 (UTC)

Just a quibble-quibble: Debatable, since the photon can never be at rest... :-) --Michael C. Price ^talk 17:59, 9 August 2006 (UTC)

OK, but the serious part of my point is that we should always phrase things in terms of whether things "have nonzero (foo) mass" rather than whether they "have (foo) mass". --Trovatore 18:01, 9 August 2006 (UTC)

There is no more such notion of "rest/relativistic" mass. Neither of them makes any sense in the case of the photon and both have been banished from modern teaching of physics. Here it is, the same explanation again:

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 18:15, 9 August 2006 (UTC)

That POV is already explained at relativistic mass. Go there. --Michael C. Price ^talk 18:33, 9 August 2006 (UTC)

It is very poorly explained there. The photon is an exception, you cannot apply to it the notions of "relativistic/rest mass" . I have been trying to get the new way of teaching physics in, would you please have a look at the referenced curses from Cornell and Harvard. I know the subject very well, I do not need more refences to the way wiki botches it. This is not only my POV , it is the POV of leading universities in the US:

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 18:52, 9 August 2006 (UTC)

What part of "relativistic mass" $=E/c^{2}$ don't you understand? --Michael C. Price ^talk 19:04, 9 August 2006 (UTC)

The same one that you seem unwilling to read on, the one that is no longer used in teaching physics because of the confusion it generates, the one that does not apply to photon:

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 19:09, 9 August 2006 (UTC)

Can't you even figure out the indent thing? BTW I have read those links, as I've already told you on my talk page. Their POV is explained at relativistic mass. --Michael C. Price ^talk 19:14, 9 August 2006 (UTC)

Sorry, this is a nonsense argument. There is no "rest mass" for the kinetic part of the energy of a particle (I'm talking about JUST the kinetic energy). However, despite that, the kinetic energy of particles in a system (2 particles or more) DOES HAVE MASS. And it can be rest mass, if the system is at rest. This kinetic energy ADDS mass to all systems which contain the moving particles. Not only their rest mass, but their kinetic energy shows up as mass in the COM frame. Kinetic energy is part of the mass of "heat." It's half of heat in solids, and it's ALL of heat in ideal gases. And yet this kinetic energy has no rest mass when you look at the particles one at a time. So where is the mass it represents? Well, the mass of the kinetic energy of particles is part of the system of particles, even though you cannot connect it with any particular individual particle (indeed it hops around from place to place as you change your reference frame). That confuses some people when they study relativity, but our job in these articles is dig out this confusion and confront it, not bury it and pretend it doesn't exist. Kinetic energy has no rest mass for single particles but it DOES have rest mass for SYSTEMS of particles when THEY are at "rest" (rest in systems means we are looking at the system from the COM frame). Get used to it. Get over it. It's modern physics.

Photons are merely a special case of the same problem, since in a sense they're ALL kinetic energy and nothing else. None has ANY definable rest mass when viewed one at a time (you can make their energy anything you like by choice of reference frame), any more than massive particles have any particular mass associated with THEIR kinetic energy, when viewed one at at time (again, you can make it anything you like). BUT the kinetic energy of particles AND photons (which for photons we conventionally call merely their energy) adds mass to any system they're part of.

All of this is ordinary mass. It has inertia, gravity, weight. It's not only relavistic mass, but also proper and invariant mass in the COM frame. We count it as heat whether it's kinetic energy or photons. And as heat, it HAS mass and ADDS mass, no matter which kind of energy it is. SBHarris 19:55, 9 August 2006 (UTC)

Perhaps the problem is merely pedagogy

It does seem like more than one person here could use a nap. ;) Let's all remain true to the physics!

The last remaining difficulty — which seems perfectly valid — is pedagogical, i.e., not with the physics per se, but with how to explain it to lay people (i.e., non-physicists), which will include most people reading Wikipedia. There will naturally be at least a few readers who will be confused by the apparent paradox that the photon is massless and yet is able to gravitate and to increase the inertial mass of an object that absorbs a photon. As Ati seems to be pointing out, we should try to explain to these readers as carefully as possible why this is not contradictory.

The two web-pages cited by Ati do not seem to contradict what I said earlier. The second one (in its section 11.8) argues that physics teachers should replace the term "rest mass" with "mass" (and the symbol $m_{0}$ with $m$) for pedagogical reasons, since inexpert students might try to take over other elements of Newtonian physics that are not valid relativistically. That's a valid viewpoint, but it is not universal in the physics community, since there are other confusions that can arise from using just the term "mass". Working physicists, however, do generally stick with the briefer term "mass" when speaking among themselves. Regardless of the teaching method, the essential physics has not changed, merely its nomenclature and symbols.

I'm rather pre-occupied now with family matters, but I can try to find a compromise wording, if everyone is patient enough with me. I was thinking of working more on photons, anyway; I'm in the middle of translating Planck's 1900 paper and Einstein's 1905 "photoelectric" effect paper for Wikisource. I'm kind of busy for the rest of today, though. Willow 18:42, 9 August 2006 (UTC)

I'm against "compromising" with Ati3414; that will not improve the article. We can't allow articles to be skewed just because editors with, shall we say, unusual enthusiasms show up and make a lot of noise. --Trovatore 18:47, 9 August 2006 (UTC)

I absolutely agree. You can't comprise with a crank either on the talk pages or in the article. His beef is with conventional terminology -- it has no physical insight or merit. He simply can't accept that "relativistic mass" $=E/c^{2}$. That's his problem, not ours, and articles should not suffer because of it. --Michael C. Price ^talk 18:53, 9 August 2006 (UTC)

Sure, let's leave things as they are. Ignorance is bliss. Ati3414 18:54, 9 August 2006 (UTC)

As pointed out above, the kinetic energy of particles has no "rest mass" and yet it contributes to the mass of resting systems full of such particles (like a can of heated gas). If you can deal with that paradox, you can deal with photons. Perhaps our problem is we didn't make people deal with it before they got to photons. Here, in an article on photons, it returns to bite us. Had it been thoroughly dealt with by the people arguing here in relativistic kinematics before they ever heard of photons, they'd have an easier time seeing photons as just "bare" kinetic energy. SBHarris 20:03, 9 August 2006 (UTC)

Mass of an empty box

This article fails to mention that the zero photon state also has energy and that this gives rise to the well known Casimir effect. Another consequence is that the mass of an empty box with reflecting walls is not zero (plus the mass of the walls). Count Iblis 21:08, 9 August 2006 (UTC)

E = m c²

A while back, I finished up writing at the end of the photon in a box section "consistent with the system's total energy and mass at rest satisfying E = m c²." I later took it out as redundant, but MichaelCPrice put it back modified as "each component of the composite system, including the total composite system itself, obeys E=mc², where m is the relativistic mass." Now, I've been trying to be good and avoid the concept of relativistic mass. Is it really correct here? Is it to emphasize that in the system of the box, the photon can indeed be attributed a relativistic mass, as an independent component? What are the components of the system? Is kinetic energy a component? Or it kinetic energy to associated with some unit of mass? Does the notion of components really make sense? Are we sure we want to go this far? Obviously Ati will object; anyone else? Dicklyon 04:31, 10 August 2006 (UTC)

Look up "A simple exercise" entry above. Try getting the "photon mass contribution" from formula [5] Would be good to read the older controversy "Photons carry mass?" from further above as well. Looks like this group has been gone thru this before Ati3414 04:59, 10 August 2006 (UTC)

I don't see why you'd want to muddle the issue by adding to a system in a way that changes its momentum. It's unclear to me whether your equations demonstrate that it is not in general easy to assign mass to components such as photons, but suppose for the sake of argument that I accept that. You might be right. What I'm trying to ask is whether anybody else sees such a problem. If so, let's leave the more general statement that all of the energy in the system counts in its mass, and not try to split it up over "components". Dicklyon 05:07, 10 August 2006 (UTC)

It's not "muddling", it is the (most) general case you asked for, remember? You asked about the "components". I gave you the "components". Now that you got an "inconvenient answer" you are trying to weasel out of the situation. How does the most general formula applies to the "photon(s)in a box" now? What is the "mass contribution" in the general case? Have you read the section "Photons carry mass?". Yes or no? Don't let all this "inconvenient questions" rattle your (mis)conceptions about what is really mainstream Ati3414 13:53, 10 August 2006 (UTC)

I said I can accept it. I'm wondering if anyone else does. Dicklyon 15:55, 10 August 2006 (UTC)

Well, you can add relativistic masses and energies, so it's correct that the total relativistic mass and relativistic energy of the box is the sum of these things for the components. But that's one reason we don't use relativistic mass, because we do this job using energy, then just dividing by c^2 where we like (and after we're done). In the COM frame (as for a box) the rest mass and invariant mass happen to equal the relativistic energy and relativistic mass, SO that fact (in this special frame) allows you to get invariant mass of the box by adding up all the RELATIVISTIC energies. Which seems a bit weird to many people because you certainly can't get invariant mass by adding the invariant masses of components. SBHarris 20:38, 10 August 2006 (UTC)

Resolving physics disputes; and a consensus nomenclature?

Hi all,

My impression is that the confusion here has persisted because people argue past one another using different definitions of "mass", without proposing a concrete physical experiment that discerns the two points of view. I recommend that we all do this when we have a substantial disagreement about the physics.

Even where there's agreement about the physics per se, there also seem to be disagreements about the best method for presenting it to lay readers. A first simple step towards solving several of our difficulties may be to agree on a common nomenclature. Here's a proposed nomenclature that may help us clarify the debates and make articles more intelligible to Wikipedia's readers:

(1) I propose that we not use the unqualified term "mass", since it does not have a single definition and can be confusing.

(2) I propose that we use only the term rest mass (not an unqualified "mass") to represent the coefficient $m_{0}$ in the energy momentum equations $E=\gamma m_{0}c^{2}$ and $\mathbf {p} =\gamma m_{0}\mathbf {v}$. The term "rest mass" is still in common use among physicists and is found in modern physics textbooks (see below). "Rest mass" has exactly one physics meaning and, therefore, seems preferable to "mass", which has several. Moreover, "rest mass" is an intrinsic property of a particle, the one being referred to when we say that a photon is massless, or that the electron's mass is 9.1 x 10^-31 kg (0.511 MeV). Being shorter, "rest mass" seems preferable to the equivalent term "invariant mass", also in common use.

(3) I propose that we not use the term relativistic mass, since that term is not used by practicing physicists, is not taught to physics students at major universities, is generally not used in solving relativistic mechanics problems, and may confuse the readership of Wikipedia. To support this third proposal, I went to a friend's house and surveyed commonly used physics textbooks that cover relativistic mechanics. The following 10 textbooks made no mention of "relativistic mass" at all

1. Halzen F and Martin AD (1984) Quarks and Leptons, Wiley.
2. Wald RM (1984) General Relativity, University of Chicago Press.
3. Goldstein H (1980) Classical Mechanics, 2nd ed., Addison-Wesley.
4. Barut AO (1980) Electrodynamics and Classical Theory of Fields and Particles, Dover.
5. Rindler W (1977) Essential Relativity, Springer Verlag.
6. Misner CW, Thorne KS and Wheeler JA (1973) Gravitation, W. H. Freeman.
7. Weinberg S (1972) Gravitation and Cosmology, Wiley.
8. Landau LD and Lifshitz EM (1975) The Classical Theory of Fields, 4th ed., Pergamon
9. Symon KR (1971) Mechanics, 3rd ed., Addison-Wesley.
10. Weyl H (1952) Space-Time-Matter, Dover.

The following two textbooks mention "relativistic mass", but treat it as an intermediate quantity used merely to arrive at the 4-vector formulation of energy and momentum:

1. Feynman RP, Leighton RB and Sands ML (1989) The Feynman Lectures on Physics, vol. I, Addison-Wesley.
2. Eyges L (1972) The Clasical Electromagnetic Field, Dover.

For example, the latter reference says of "relativistic mass" that it is not fundamental and "merely introduces new terminology". Two other textbooks follow the same path for deriving the 4-vector, but do not use the term "relativistic mass"

1. Jackson JD (1975) Classical Electrodynamics, 2nd. ed., Wiley.
2. Schwartz M (1972) Principles of Electrodynamics, Dover.

Finally, two older textbooks introduce "relativistic mass" for completeness (along with longitudinal and transverse mass) but state that the energy-momentum 4-vector approach is better

1. Sommerfeld A. (1942) Mechanics, Academic Press.
2. Pauli W (1921) Theory of Relativity, Dover.

Summarizing, I have not identified a commonly used textbook that treats "relativistic mass" as an important concept in its own right. In 2/16 instances, it is used as an intermediate concept to derive the energy-momentum 4-vector, which is subsequently treated as fundamental. In two other cases, it is presented as a historical relic on an equal footing with transverse/longitudinal mass (which I hope we agree that we don't want to cover here).

On the basis of these data, it seems fair to say that "relativistic mass" is not in current use among practicing physicists. One might argue that it has pedagogical value but, speaking for myself, it seems more likely to be confusing than helpful to the majority of lay readers.

I have restricted myself to published, paper-copy textbooks on relativity, because those seem to be the most reliable sources. Lecture notes posted on the web and similar sources are often made by people who lack expertise or who may be a little careless in their wording, such as graduate students or harried professors whose research is in a different field.

Please forgive the long post, but I think many of our disagreements will be resolved if we agree on a common nomenclature and insist upon describing concrete experiments that discern different models of the physics. Willow 19:40, 10 August 2006 (UTC)

Thank you, Willow , a very sensible post.Devoid of personal attacks and pettiness.There are multiple references to "relativistic mass" throughout wiki. Now, how do we get from what you wrote to removing those references and the m_photon=E/c^2 from the same several wiki posts? Ati3414 20:09, 10 August 2006 (UTC)

Sorry, relativistic mass has to be discussed because that is how it appears in the equation $E=mc^{2}$. [Added by Michael Price]

Answer, is that this is not true for resting particles, and also for systems in the COM frame. There, you can still use E = mc^2 and the energy is the system rest energy (which also is the total relativistic energy) and m is the system rest mass. The main problem for this approach is that people like Ati3414 don't really believe what I just said, because it implies that relativistic energy (and mass) adds system rest mass. Including when the energy comes from a photon (or from system heat or kinetic energy). So the issue is not pedagogical, it's an issue of reality. Basically Ati3414 will reject the physicists like Baez or whoever he needs to, because he doesn't believe it. In spite of what the equations say. Look, you just SAW him say that Einstein is right that a body which emits a photon of energy E has less rest mass, by E/c^2. But he doesn't believe that a body absorbing a photon E grows in rest mass by E/c^2: That would violate his fixed belief, so he denies it. I'm afraid that at some point you have stop looking for reasons why unreasonable people have unreasonable beliefs. SBHarris 20:24, 10 August 2006 (UTC)

Sbharris wrote: "....you just SAW him say that Einstein is right that a body which emits a photon of energy E has less rest mass, by E/c^2. But he doesn't believe that a body absorbing a photon E grows in rest mass by E/c^2"

Ati3414 answers: No, this is not what I am saying. You have been told multiple times that, while the above is true and I do not dispute it, you cannot and should not infer from it that the equivalent mass of the photon is E/c^2. This has always been the issue. I have said this to you and to the others maybe 20 times by now. There is a subtle difference but an important one. Go over all the previous exchanges, it was always there Ati3414 21:01, 10 August 2006 (UTC)

No, $E=mc^{2}$ is still true for COM/resting particles, it's just that in those circumstances the relativistic mass happens to equal rest mass. Of course we both know and understand what we both mean and that we only have a minor linguistic quibble here -- it's just Ati who has a screw loose. --Michael C. Price ^talk 20:39, 10 August 2006 (UTC)

Repeated personal attacks only show that you can't argue something scientifically and only reflect on you. Since you are attacking with such violence a couple of questions:

COM frame calculation

1. what is the "COM" of a single photon?

That's easy; you can say either that it hasn't got one, and hence has no uniquely defined energy, mass, or momentum, or you can say that its COM frame moves at the speed of light, so the mass, energy, and momentum are zero. I believe the latter answer is too far stretched to be meaningful, and we've all agreed already on the former, so let's just go with that. Dicklyon 21:28, 10 August 2006 (UTC)

2. what is the "COM" of a system of 3 photons , two flying at 180 degress wrt the third one?

That's a more interesting case. If you specify the conditions more exactly, it should be easy to find the frame with respect to which the total momentum is zero. That's the COM frame. Dicklyon 21:28, 10 August 2006 (UTC)

You have enough info, two photons fly to the right, the third one flies to the left, along the same lineAti3414 23:24, 10 August 2006 (UTC)

3. For both you and/or Sbharris, go back to "A simple exercise" and calculate the "mass of the photon" for one of the more complicated cases (like the 3 photon system above). Ati3414 21:01, 10 August 2006 (UTC)

The contribution of mass due to the photon will be E/c2, for the E evaluated in the COM frame. No magic there. The (proper) mass of the photon is still zero. Dicklyon 21:28, 10 August 2006 (UTC)

Can you do some calculations? English composition doesn't count in physics, math does. Ati3414 23:22, 10 August 2006 (UTC)

COMMENT

The only people doing calculation here so far is us. Haven't see you do one yet.

Presuming all 3 of your photons to each have energy E in the lab frame, then 2 going one direction and 1 going in the other, is equivalent to one photon of energy 2E going one way, and another of energy E going oppositely (reduction to a problem of 2 photons, one twice the energy and frequency of the other, which has the same answer). The COM frame is the one in which these 2 photons have momenta (and therefore energies, since they ARE photons) summing to zero. In one direction momentum is 2E/c, and in the other it's E/c. So our COM frame is the one which chases the 2E photon so that by Doppler shift it's energy drops to equal that of E photon. Of course, the Doppler shift will increase E for one photon while decreasing it for the other.

It's easiest for me here to simply pretend we're IN the COM frame where both photons have the same energy E and frequency hv. We're looking for a velocity v to produce a Doppler shift from this frame which is large enough that one photon now has twice the frequency of the other. Let us define β = b = v/c, which I'll use just so I don't have to use the Greek letter all the time, then the finally Doppler shifted frequency UP for one photon (1+b)/(1-b) is going to equal 2 times the shifted frequence DOWN for the other: (1-b)/(1+b). Writing that out and getting rid of fractions gives

(1+b)^2 = 2 (1-b)^2

Solving for b gives b = v/c = [sqrt2 - 1]/[sqrt2 + 1]. This is not an easily resolvable fraction. It's also 1/2 [(sqrt2)-1] = about 0.2071. Thus, the COM frame is the frame which is moving at 0.2071 c in the direction of the 2 photons. In that frame, the two photons have the same energy and momentum as the other photon headed in the opposite direction. I will leave what the total energy and invariant mass for this system, to the student. Ati, here's your chance to show you can do algebra.

Now, Ati, a much easier problem for you, which doesn't involve photons so you might be able to get around the mental block. Two objects of mass M move away from each other with relative velocity V. What is the kinetic energy of one mass, as seen by the other? What is the total kinetic energy as seen in their COM frame? What is the system invariant mass? If we put these objects in a can, how much increase in mass could we expect from the can if we put it on a scale in the COM frame? SBHarris 00:15, 11 August 2006 (UTC)

No calculation of the "photons equivalent mass"? Ati3414 00:20, 11 August 2006 (UTC)

It's easily done, now that we have the COM frame. I'm leaving it up to you. Show us your stuff SBHarris 00:43, 11 August 2006 (UTC)

Finish what you started please, you are not a professor though you put on the airs . Let's see your final result. Let us know when you stop changing your answers. Ati3414 01:12, 11 August 2006 (UTC)

The final answer is total energy is 2*sqrt2*E in the COM frame, for 3 photons of E in the lab frame, with invariant mass of the system therefore = 2sqrt2*E/c^2. Now your turn with the simple kinetic energy problem.SBHarris 01:35, 11 August 2006 (UTC)

SB, thanks for bailing me out here, but are you sure that the ratio (1-b)/(1+b) is in general correct, for system velocities that are getting into a relativistic range? Is it really that simple? And is it really in general OK to treat the two photons as one of double the energy? Does the doppler shift on that do the same thing to energy and momentum as it would to two photons of the original energy? Looks like it does, but I never thought of that trick. If that's all true, you just need to solve for the frequencies of the momentum-matched photons in terms of the original photons, which are just 2*(1-b)/(1*b) and 1*(1+b)/(1-b) and should be equal. But, with b = 1/3 (your original answer) or b = 0.2071 (your new answer) these are not equal. Either your original ratio needs a sqrt in it, or your solution is in error, since the correct solution of your equation is b = 3-sqrt(8) = 0.1715 according to my scratch paper. Using that, the two photon energies are sqrt(2) times the original one photon's frequency or energy. There are two of those, so the total energy in the COM system if this is right is 2sqrt(2)E, and the rest mass of the system is that over c^2, as always. It's somewhat less than 3E, as expected. I don't have confidence that the Doppler formula is exactly correct, so if this result is wrong, patch it accordingly. Dicklyon 00:57, 11 August 2006 (UTC)

ANSWER Arggh! Yes, you've caught me squaring both sides and not showing it, and then forgetting to do it to the factor of 2. The relativistic Dopper factors have square roots. Let's see if I can get it right.

Sqrt [(1+b/1-b)] = 2 Sqrt [(1-b)/(1+b)]

(1+b)^2 = 4 (1-b)^2

1+b = 2 (1-b)

So it looks like v = c/3 was right the first time.

So re-inserting you get sqrt ( [4/3] / [2/3]) = 2 * Sqrt([2/3] /[4/3])

Which is Sqrt(2) = 2 Sqrt(1/2) so this works.

Because these are square roots, the E photon gets bumped up to (Sqrt2)E and the other big 2E one (or if you like the 2 E ones going the same direction) gets taken DOWN to 2E/(Sqrt2), but because Sqrt2 = 2/Sqrt2, the energies of both sides are now equal in the COM frame and so are the momenta, as advertized. And the total COM energy and therefore rest energy of the system is 2*(sqrt2)E. Divide by c^2 for invariant mass. So Ati got us to do it anyway, blast. SBHarris 01:29, 11 August 2006 (UTC)

---

Yes, getting the correct relativistic Doppler formula can always get you, professor. My turn:

$m'={\frac {\sqrt {(\Sigma E)^{2}-(\Sigma pc)^{2}}}{c^{2}}}$ [1]

is an invariant, i.e. one should get the same result in any frame, no need for the pesky COM frames that can trip the best.

$\Sigma E=3E$ [2]

$\Sigma p=p+p-p=p$ [3]

Substitute [2],[3] in [1] remembering E=pc and you get

$m'={\frac {{\sqrt {8}}E}{c^{2}}}$ [4]

At this point the humble student gives the self-appointed professor a D :-) Ati3414 02:01, 11 August 2006 (UTC)

I don't see why I should get a D for getting the correct answer, since of course your sqrt8 is my 2sqrt2. And IF you didn't get your problem and your final answer out of a textbook. Yes, doing invariant mass in the lab frame CAN be a shortcut if something cancels, and so it is here. But remember, you haven't done the whole problem. You didn't originally ask for the invariant mass or energy of the system--- you asked to calculate the COM frame velocity. Which I did. It might be faster to do the invariant mass first, but when you back-calculate the velocity, you're still stuck with the Doppler formula (or the equivalent) for photons, so there are no royal roads. Anyway, now your turn, on the kinetic energy problem. Show us where that kinetic energy is, in the system. How much mass does it contribute? How can it contribute any mass, if it has no rest mass? SBHarris 02:38, 11 August 2006 (UTC)

Too bad, SB, you'll have to live with getting a D from him who claimed that photons don't add to the mass of a system, even if he just showed you exactly that they do. That's a grade you can frame (in a COM frame?). Dicklyon 02:45, 11 August 2006 (UTC)

Naturally a COM frame. This is the invariant mass of a system of nothing but photons, as described, good in any frame--- and we're being told it's not a REAL mass. Proof positive you can calculate something you don't understand. SBHarris 02:58, 11 August 2006 (UTC)

Yep, too bad you can't read: "invariant" . Not mass. But you did well, you told Sbharris that relativistic Doppler formulas are different from the Newtonian ones so he stopped doing his standard mish-mash of SR and Newtonian mechanics. Carry on guys, maybe you can collaborate on some scientific project. One more thing, pervect cleaned up a little mass in special relativity. Maybe you two guys can clean up "photon in a box". You look good together. Ati3414 03:16, 11 August 2006 (UTC)

Invariant means invariant mass. See the "m'" in your formula? The m stands for "mass" SBHarris 04:47, 11 August 2006 (UTC)

Yes, it is an invariant and it does have dimensions of mass. It should not be misconstrued that the 3 photons have the mass shown by the formula.One of these days it will sink in: the photon is a singular particle , you cannot apply your ideas of massive particles because you will get weird results. Same way you cannot mishmash SR formulas with Newtonian ones. Ati3414 05:58, 11 August 2006 (UTC)

So this system has an invariance with units of mass, and it has a center of mass that we all agree on. And the invariance that has units of mass was computed from the energies of the photons that make up the system. But the system doesn't have rest mass? What am I missing here? Dicklyon 06:11, 11 August 2006 (UTC)

Correct, it doesn't. If it had, it would violate the Proca gauge. Try reading the Roderic Lakes paper I sent you, maybe it will sink in. Look at the changes that Pervect made to mass in special relativity (not profound enough, but still) Ati3414 06:27, 11 August 2006 (UTC)

Answer to WillowW

Hi WillowW,

I appreciate your well-intended and thoughtful work, but I can't fully agree. The term "relativistic mass" is historically well-attested, and the mere fact that it is not currently fashionable to use it (even if for good reason) does not mean it should be wiped from the record. As I mentioned earlier (before you came in, probably) I would be happy with a brief footnote pointing out that while the "mass" in the contemporary sense -- that is, the invariant mass -- of the photon is zero, there is an older notion of relativistic mass, which is not zero for the photon, and pointing the reader to the discussion in the relevant article. With such a footnote, neutrally worded, I would be happy to see the "photons in a box" section just go away, since (as an earlier contributor pointed out) it isn't really about photons specifically. --Trovatore 21:17, 10 August 2006 (UTC)

How do you know : "which is not zero for the photon" ? Have you run any experiment that proves that? Can you name some experiments that prove that? I concur with you on one thing: the "photon in a box" gotta go. Delete it, please. Ati3414 21:38, 10 August 2006 (UTC)

It's definitional. You might as well ask for an experiment that proves that left is the opposite direction from right. --Trovatore 21:48, 10 August 2006 (UTC)

Please refrain from using names, this goes counter wiki privacy policy. So, your answer is that you have no answer, right? Ati3414 23:21, 10 August 2006 (UTC)

My answer is, it's true by definition. Arguing with well-attested historical definitions doesn't make much sense; you can use them or not, but they are what they are. Asking for an experiment to "prove" them is a complete non-sequitur.

As for this "privacy policy"—I don't exclude out of hand that it might exist; I haven't made a systematic study of all the policies. But you're the first person I've heard mention it. Perhaps you can point me to it. --Trovatore 23:33, 10 August 2006 (UTC)

"True by definition" 1. There is only one type of mass, proper mass. 2. In QED the mass of the photon is defined to be zero. From 1 and 2 you can easily conclude that the mass of the photon is NOT E/c^2. On the other subject look here, under harassment : http://en.wikipedia.org/wiki/Wikipedia:Harassment , you have done quite a few things that qualify as harassment, you can read for yourselfAti3414 23:37, 10 August 2006 (UTC)

Pervect has put references in the mass in special relativity article attesting the term "relativistic mass". You're entitled not to use the term; you're not entitled to insist that the rest of the world forget it. The claim is in fact true by definition. --Trovatore 23:55, 10 August 2006 (UTC)

I know, I am having a sidebar discussion with him and he's mulling over what I told him. Are you bowing out? No calculations from you? I set a few problemms in front of the opposing side, would you care to tackle one? You know, physics is about math, not about English compositionAti3414 00:07, 11 August 2006 (UTC)

The calculations are trivial and beside the point. Physics is not about English usage, but there is no physics issue here; everyone with a basic, minimal understanding will get the same final answers. The only issue is linguistic usage, what you call the various quantities involved. And on that point you are not entitled to dictate that the historical usage be ignored. --Trovatore 00:28, 11 August 2006 (UTC)

That is the key point: this is a linguistic debate only -- there is no physics here. The terms are defined by common and historical usage. End of story. --Michael C. Price ^talk 01:02, 11 August 2006 (UTC)

Would you care to help Sbharris with his calculations? He seems to need help.Ati3414 01:20, 11 August 2006 (UTC)

Worth repeating. Energy with no rest mass in single particles, DOES have rest mass in systems

The "odd" fact that relativistic energy can be added to get rest mass of systems, is what allows energies which have NO rest mass (like photons and the kinetic energies of moving particles) to be counted (indeed summed up to be) the regular mass (invariant mass) of bound systems. Energy with no rest mass DOES have rest mass in a system! We've made this point as many places as we could, in part because it seems to be widely misunderstood. It's a special property of systems which does not appear for individual particles. The kinetic energy of a (single) free gas molecule doesn't WEIGH anything, because as soon as you go to a frame where it can be weighed, it goes away. And if you trap the gas molecule, you're no longer talking about a single object, but a system of molecule plus can. The kinetic energy of a particle in a can of gas, DOES weigh something. Kinetic energy is a SYSTEM property. It doesn't reside in single objects, but in systems of two or more objects. The MASS associated with kinetic energy IS THE SAME KIND OF THING. It's not really "in" a single moving object, but rather is a property of a system of 2 or more objects in motion relative to each other.

And much the same is true of photons. Single photons have no definable energy, anymore than massive objects have a definable kinetic energy--- it's all frame dependent. The kinetic energy of a single particle or a single photon can be anything you like, depending on your reference frame chosen. But get a system of two or more objects, or photons, and now you have some residual minimal energy which you cannot get rid of by choice of frame. THAT energy has mass. THAT energy contributes invariant mass to systems.

When you get this, you'll be fine. SBHarris 20:36, 10 August 2006 (UTC)