# Is there a connection between spontaneous symmetry breaking and massless photons?

I haven't studied a lot about these topics to put it that way. But I wonder if there is a connection between spontaneous symmetry breaking and the fact that photons are massless?

The spontaneous symmetry breaking breaks the symmetry of the universe predicted by the force and particle equations and thus, makes other particles acquire mass. Is that right? Then, is this symmetry breaking that causes mass to be in some way related to the at speed of light traveling photons that do not have mass?

The SSB of the electroweak Lagrangian occurs, when you first have 4 massless gauge fields: one for the photon, and three for the W+,W- and Z boson. Through the process of symmetry breaking, the Lagrangians symmetry is no longer preserved in the physical reality. In the U(1) case eg one has two degrees of freedom for the complex Higgs field and two dof for the massless gauge field. After SSB, the Higgs field loses one dof, becomes massive (the massive Higgs boson) and the gauge field gains the reminescent dof, carrying now 3 dof and therefore becoming massive, since the additional dof adds a longitudinal mode to the gauge field.

When you look at the famous sombrero potential, you should see that adjacent states in the pan of the hat share the sane energy and are therefore connected by a massless mode of the gauge field: the photons.

A comprehensive paper on the topic was published here

My three cents:

1. Zero rest mass of photons is simply a consequence of the fact that the EM Lagrangian is only a function of the derivatives of the (gauge) field, $$A$$. One can choose the vacuum or the ground state (state w/o any excitations/waves) to be $$A=\text{constant}$$, where the "constant" could be anything. However, if you look for waves, you've to fix this constant to something $$-$$ thereby breaking a continuous symmetry of the Lagrangian and resulting in massless Goldstone excitations (photons in this case). This is a general characteristic of Lagrangians that only depend on the derivatives of the field, e.g. waves on a rope, sound waves etc. (we're not even using the fact that we're dealing with a gauge field here!)

Now if we want to count the number of massless excitations we need to look at our gauge field $$-$$ it is a four-vector and one would naively conclude that there would be 1 massless mode for breaking translational symmetry for each of its components, i.e. a total of 4 modes. However, enter gauge symmetry and that's not the case. Why? Because gauge symmetry is a kind of redundancy in our description and the true degrees of freedom can only be seen after a gauge is fixed. For example, choosing the radiation gauge: $$A^0 = 0, \partial_\mu A^\mu =0$$, one can see that there are only two degrees of freedom, leading to two massless photons corresponding to two choices of polarization for every $$\vec{k}$$.

2. The fact that these excitations travel with a phase velocity equal to the speed of light can be traced back to Einstein's argument: If it were moving at anything lesser than $$c$$, you could catch up with it, resulting in a static EM field, which has to be zero in the absence of sources (assuming field at spatial infinity is zero and invariance of Maxwell's equations). If it were moving faster than $$c$$, it would violate causality. In fact, $$c$$ is the only invariant speed. (Lorentz invariance)

For a particular rigorous/mathematical application/example, check my answer to this question. TL;DR: $$(\omega,\vec{k})$$ is a 4-vector, $$\implies \omega^2 - k^2 = \text{constant}$$. But Goldstone's theorem forces $$k\to 0$$ for $$\omega \to 0$$, meaning that $$\omega^2-k^2=0$$, aka phase velocity is $$c$$.

3. An argument of classical relativists that supports the above claim is an elevation of the rule $$E^2 = m_{o}^2c^4+p^2 c^2$$ where $$m_{o}$$ is the rest mass and $$p$$ is the momentum for material bodies to all "particles", including those with rest mass $$0$$. These particles then have relativistic dispersion and can carry a non-zero energy/momentum only if $$v=c$$, otherwise $$p=\frac{m_ov}{\sqrt{1-\big(\frac{v}{c}\big)^2}}$$ would be zero. As Griffiths says in his famous textbook on Electrodynamics, "personally I would take this argument as a joke, were it not for the fact that at least one massless particle is known to exist in nature: the photon".

In QED this fits very nicely with the picture that we're trying to derive a quantum field theory of the electromagnetic field whose excitations are particles with zero rest mass, which travel at the speed of light and obey the relativistic dispersion formula (which is exactly what we want in order to be able to describe the electromagnetic field).

There are cases eg. superconductors/Higgs mechanism, wherein a gauge field can eat a zero mode of the field it couples to, resulting in massive photons. This is a separate story and tells us how massless photons become massive by conspiring with another field, as pointed out already by @Peter Sanctus.