Why can't photons have a mass?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

put simply - mass terms for photons break gauge invariance.