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Why can't photons have a mass? Could you explain this to me in a short and mathematical way?

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This is on an entirely different question, does the photon have a mass. The question is different, the answers are different. In fact, the "duplicate" question has NO answer that tells why it is believed that photons are massless. And the question is one of some interest. In addition to the current literature (i.e. ), J. D. Jackson has a discussion of this question and the question here has attracted 5 answers, none from a less than 400 rep author, and with a total of 9,000 reputation. Reopen the question. – Carl Brannen Feb 6 '11 at 22:56
Jackson's book on Electrodynamics starts off in the introduction on the experimental evidence for why we believe a photon does not have a mass. It is not an axioma as far as I know. – Gerard Feb 7 '11 at 15:22

The other answers explain that there's no paradox but they don't explain why the particular particle called photon is massless.

It's massless because it is the messenger particle responsible for electromagnetism which is a long-range force. Its range is infinite so the mass has to be zero. One may view the Coulomb potential as the zero-mass limit ($m\to 0$) of the Yukawa potential $$V(r) = \frac{\exp(-mr)}{r} $$ OK, so why is it massless and why the range is infinite? It's because of the unbroken $U(1)$ gauge invariance for the electromagnetic field that acts on the electromagnetic gauge potential as $$A_\mu\to A_\mu+\partial_\mu \lambda$$ The mass term (in the Lagrangian) for a gauge field would have the form $m^2 A_\mu A^\mu/2$ and it is not invariant under the gauge invariance above. The gauge invariance is needed to make the time-like mode $A_0$ unphysical - otherwise it would produce quanta with a negative norm (because of the opposite sign in the signature for the timelike direction) which would lead to negative probabilities.

However, gauge fields may consistently become massive via the Higgs mechanism - like the W-bosons and Z-bosons. Then they lead to short-range forces. Beta-decay is mediated by the W-bosons.

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Neutrinos also travel at the speed of light but have non-zero masses. – Murod Abdukhakimov May 26 '13 at 10:01
@MurodAbdukhakimov: That's not quite correct. Neutrinos always travel at some significant fraction of $c$ ;-) – Waffle's Crazy Peanut May 26 '13 at 16:29
@ϚѓăʑɏβµԂԃϔ: How do you know that? – Murod Abdukhakimov May 27 '13 at 6:58
@MurodAbdukhakimov: Well, I believe in observations ;-) – Waffle's Crazy Peanut May 27 '13 at 7:18
Why the strong force has a short-range but gluons are massless? – Hakim Feb 18 '14 at 12:41

According to the special theory of relativity, any particle with a finite mass requires an infinite amount of energy to reach the speed of light. Therefore no particles with any intrinsic mass can travel with the speed of light. The energy required to attain a speed $v$ is given by $E$ = $\frac{mc^2}{\surd1-v^2/c2}$ - $mc^2$ As $v$ approaches $c$, $E$ approaches $\infty$.

Only massless particles are allowed to travel at the speed of light. Photon is massless, hence it can travel with the speed of light. The energy of a photon is given by $E = pc$ where p is the momentum of the photon.

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Good answer, just minor points: "are allowed" -> "are forced/required", "it can travel" -> "it has to travel". Because otherwise it seems like they can also decide not to :) – Marek Feb 6 '11 at 19:01
This doesn't explain why photons don't have (rest) mass - very small rest mass, yes. – John McVirgo Feb 7 '11 at 3:16
@John McVirgo: Are you really voting on the basis of the merit of the answer? BTW your comment is absolute nonsense. – user1355 Feb 7 '11 at 3:24
@sb1 "why can't photons have mass?" was the question. It could be that photons have a finite rest mass and so don't reach this limiting velocity. How can we know that the speed of a photon is this limiting velocity? – John McVirgo Feb 7 '11 at 5:24
It is a nice precise explanation but it's wrong, unless you make the circular definition "the c used in Maxwell's field equations is the same c used in special relativity". If there were some proof that it was impossible in the manner of 2+2=4 people wouldn't be getting paid to set limits on the mass of the photon as in Phys.Rev.D82:065026,2010, "Upper Bounds on the Photon Mass", Antonio Accioly, José Helayël-Neto, Eslley Scatena, Google Proca electrodynamics. – Carl Brannen Feb 10 '11 at 3:18

There is nothing special about the photon having zero mass. Although zero is the smallest mass any particle can have, it is as good as any other value. In this sense, there is no mathematical proof that the photon has to have zero mass, this is a purely experimental fact. And, to our best knowledge, the photon mass is consistent to zero.

If you want to describe a theory with a zero mass vector in a manifestly relativistic way, you have to have gauge invariance. This is a mathematical fact. As is the fact that if you force this symmetry to be quantum mechanically exact, the mass will not receive quantum corrections (perturbatively, at least). Gauge theories can be shown to have all sorts of other nice features (like IR finiteness, if you sum enough virtual and real diagrams) and that makes us believe that at low energies they are the right theories.

But one would be inverting the logical order within physics if one says that the mass of the photon is zero because EM is described by a gauge theory. EM is described by a gauge theory because the photon has zero mass. There would be no problem with special relativity either. The fact that the maximal velocity is the same as the velocity of light in the vacuum is, again, an experimental fact (equivalent to the one we are discussing here) but by no means necessary by any mathematical theorem.

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If Maxwell's equations are to be relativistically covariant, then light must travel at some constant c' in all frames. If it wasn't the same as the universal limiting velocity c, you'd end up with a different velocity for a photon and hence light from the velocity addition formula. From this it must have zero rest mass, correct? – John McVirgo Feb 9 '11 at 17:43
Sure, any zero mass field equation will have this property, not only Maxwell's equation. I know that historically it was not done this way, but teaching in the historical sequence is usually the worst thing to do. Today, the way people understand is the following: you measure the mass and the spin of the particle. From these values and special relativity you can get a well define lagrangian with which you calculate all the rest. – Rafael Feb 9 '11 at 18:36
"If you want to describe a theory with a zero mass vector in a manifestly relativistic way, you have to have gauge invariance.". This is wrong. Maxwell equations allow for magnetic charges, in wich case no $U(1)$ gauge invariance is possible. But source-free EM waves are still possible, even with magnetic charges. – Murod Abdukhakimov May 26 '13 at 10:05

In the context of special relativity, anything that travels at the speed of light can't have a nonzero rest mass. One way to see this is that the kinetic energy of an object of mass $m$ moving at speed $v$ is $$ mc^2\left({1\over\sqrt{1-v^2/c^2}}-1\right), $$ which tends to infinity as $v\to c$. Physically, this means that it would cost an infinite amount of energy to raise a massive particle up to speed $c$.

As far as special relativity is concerned, it's logically possible that photons do have mass and travel at speeds (slightly) less than $c$. (This would mean that the quantity $c$ that occurs in special relativity should not be called "the speed of light.") The experimental limits on this possibility are extremely severe, though.

I may have misguessed the level of your question and the sort of answer you're looking for. For instance, there are separate reasons for believing the photon to be strictly massless based on gauge invariance.

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I think the central issue is the invariant interval and invariant mass. A particle moving in spacetime has the interval $ds^2~=~dt^2~-~dx^idx_i$. There is the corresponding invariant mass $m^2~=~E^2~-~p^2$, which is the momentum spacetime interval. So let us consider the plane wave $\psi~=~exp(-i{\vec k}\cdot{\vec x}~+~i\omega t)$ $=~exp(-ik^\mu x_\mu)$. The Laplacian operator $\Delta~=~\nabla^2~-~\partial^2/\partial t^2$ applied to $\psi$ is $$ (\nabla^2~-~\partial/\partial t)~=~(\omega^2~-~k^2)\psi~=~\hbar^{-2}(E^2~-~p^2)\psi. $$ This is an eigenvalued problem with $\Delta\psi~=~\lambda\psi$. If the particle has mass this eigenvalue is the mass squared. This means there is dispersion as $|k|~=~\sqrt{\omega^2~-~m^2} $ and for $\omega~=~2\pi/\lambda$ we then have that $$ |k|~=~c\sqrt{2\pi/\lambda^2~-~m^2} $$ The velocity of a wave is then only exactly $c~=~1$, or $|k|~=~2\pi/\lambda$, with of course $kc~=~\omega$, for $m~=~0$, and otherwise we contradict our assumption of the wave existing on a null interval $ds^2~=~0$. If the invariant interval in the spacetime is zero, then the corresponding invariant mass in the momentum-spacetime must be zero.

The above Laplacian in the case of a photon is predicted as the wave equation operator by Maxwell's equations.

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put simply - mass terms for photons break gauge invariance.

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In its current form this is a very poor answer. Please look at some of the more highly voted answers on the site for a better model. – dmckee May 25 '13 at 23:47
You need to elaborate or state WHY a mass term breaks guage invariance or this answer is useless. – Brandon Enright May 26 '13 at 2:02
useless is a tad strong - should i elaborate and state why gauge symmetries hold in standard model too?????? this was meant to be a 'short' and 'mathematical' answer as requested. – phaedrus May 26 '13 at 15:03

protected by Qmechanic May 25 '13 at 22:55

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