By oneat

2011-02-06 17:15:38 8 Comments

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


@phaedrus 2013-05-25 22:50:43

put simply - mass terms for photons break gauge invariance.

@dmckee --- ex-moderator kitten 2013-05-25 23:47:44

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.

@Brandon Enright 2013-05-26 02:02:48

You need to elaborate or state WHY a mass term breaks guage invariance or this answer is useless.

@phaedrus 2013-05-26 15:03:36

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.

@Rafael 2011-02-09 16:33:57

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.

@John McVirgooo 2011-02-09 17:43:27

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?

@Rafael 2011-02-09 18:36:45

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.

@Murod Abdukhakimov 2013-05-26 10:05:18

"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.

@Lawrence B. Crowell 2011-02-06 22:04:07

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.

@Luboš Motl 2011-02-06 19:05:25

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.

@Murod Abdukhakimov 2013-05-26 10:01:06

Neutrinos also travel at the speed of light but have non-zero masses.

@Waffle's Crazy Peanut 2013-05-26 16:29:22

@MurodAbdukhakimov: That's not quite correct. Neutrinos always travel at some significant fraction of $c$ ;-)

@Murod Abdukhakimov 2013-05-27 06:58:41

@ϚѓăʑɏβµԂԃϔ: How do you know that?

@Waffle's Crazy Peanut 2013-05-27 07:18:10

@MurodAbdukhakimov: Well, I believe in observations ;-)

@Hakim 2014-02-18 12:41:44

Why the strong force has a short-range but gluons are massless?

@Luboš Motl 2014-02-18 13:36:54

Because the strong force is confining, so only color-neutral ("uncharged") particles are allowed to exist in isolation. Correspondingly, the force among such neutral particles decreases quickly with the distance. In fact, the decreases is faster than the power law because the gluons self-interact so they are confined, too. The mass of colored objects is a subtle thing - it depends on the RG scale and the masslessness is only relevant at very short distances, much shorter than the proton radius (QCD scale) where the confinement starts to matter.

@Isomorphic 2015-03-13 20:26:20

Can the answer be because it is impossible to accelerate them ?

@Xiaoyi Jing 2015-08-17 20:27:16

Hi Professor Motl. Is that also related with the mass gap of Yang-Mills fields?

@Luboš Motl 2015-08-18 04:39:37

Hi @XiaoyiJing - most things are related in some way. The mass gap of Yang-Mills fields is just a different statement of the fact that there exist no massless - or arbitrarily light - allowed (color-neutral) bound states in QCD and similar theories. For QCD, the conclusion is the opposite one - a massless particle does not exist.

@HolgerFiedler 2016-02-01 12:11:41

Luboš That the electrostatic force is a infinite force seems to be a postulate. Please pay attention to my answer here.

@HolgerFiedler 2016-02-01 12:17:37

The related elaboration shows how with a postulate about finite electrostatic fields the exchange between positive and negative charges happens without virtual photons.

@HolgerFiedler 2016-02-01 12:19:23

And the EM radiation is not a force. But of course it's range is infinite because photons are indivisible particles, travelling until they hit something.

@Jimmy G. 2016-03-14 15:16:11

@MurodAbdukhakimov: that nothing with mass can travel at the speed of light is an axiom of physics. Einstein's equations show that in order for mass to travel at the speed of light would take an infinite amount of energy. Also, results show that neutrinos ALWAYS travel sub-C. So, you asked Crazy Peanut how he knew that neutrinos did not travel at the speed of light. I am going to reverse that on you. How do you know that neutrinos travel at the speed of light? Your profile lists theoretical particle physics as your specialty, so you must have some basis for your stance.

@AccidentalFourierTransform 2017-02-17 12:37:46

@LubošMotl This answer is wrong. If you add the mass term $\frac12 m^2 A^2$ to the QED Lagrangian, you break gauge invariance but there are no negative norm quanta. The theory (Proca Lagrangian) is well defined, unitary and finite (there are problems with naïve power-counting renormalisation, but as long as the current is conserved, all measurable quantities are finite). The most correct and general treatment of massive photons is through the Stückelberg mechanism (which becomes the Proca theory in the unitary gauge). The S. theory is perfectly well defined, unitary, covariant and finite.

@Luboš Motl 2017-02-18 06:12:16

OK, I actually agree. I would still bet anything that the photon is exactly massless - and there are also reasons that will be understood beyond the QFT framework why it has to be so. I think it's not an accident that the light but massive gauge bosons we know get their mass from the Higgs mechanism, not via Stueckelberg, and in quantum gravity, the Stueckelberg masses have to be high enough - comparable to the fundamental scale - for consistency, for some reasons similar to the weak gravity conjecture if not exactly that one.

@user1355 2011-02-06 17:35:49

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.

@Marek 2011-02-06 19:01:56

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 :)

@John McVirgooo 2011-02-07 03:16:34

This doesn't explain why photons don't have (rest) mass - very small rest mass, yes.

@user1355 2011-02-07 03:24:39

@John McVirgo: Are you really voting on the basis of the merit of the answer? BTW your comment is absolute nonsense.

@John McVirgooo 2011-02-07 05:24:22

@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?

@user1355 2011-02-07 08:18:07

@john McVirgo: Photons are particles of the electromagnetic field. Electromagnetic field is described by Maxwell's field equations. The requirement, that these equations are invariant under inertial transformations forced us to accept a limiting velocity $c$ which is nothing but the velocity of the electromagnetic waves. Hence photons which are particles of em waves move with this limiting velocity $c$. Do you understand now why light moves with the speed of light? ;)

@John McVirgooo 2011-02-08 03:21:22

@sb1, that's a nice, precise explanation.

@Carl Brannen 2011-02-10 03:18:56

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.

@user1355 2011-02-10 14:50:10

@Carl Brannen: Are you sure the above explanation is wrong? Are you sure you understand the context of the paper you cited? I am not too sure. Since you have gone to a more generalized context than Maxwell's electrodynamics, namely proca theory. By this you have brought QED quite unnecessarily. Yes, under proca theory a massive photon is allowed and in the appropriate limit the proca theory produces the classical Maxwell's theory. There is nothing circular about the above point. The c used in Maxwell's theory is indeed the same c in special relativity. I've already explained why and..

@user1355 2011-02-10 14:56:07

..nothing can obscure this simple logic. You certainly didn't provide any logic except referring to a paper which, as anticipated, is based on proca theory. If you change the context of the discussion then many things can change. If a new TOE is discovered then many current ideas may change accordingly but within their appropriate context the ideas will remain safe for ever.

@Carl Brannen 2011-02-10 18:25:02

I am quite sure. I'm not going to argue the point, but you might consider the possibility that the editors at Phys. Rev. D, the authors of the paper, the reviewers, and me are right and you're wrong. You might try doing a further literature search, I gave you a 2010 reference but this is an old problem.

@Ted Bunn 2011-02-06 17:27:08

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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