# how is shown that photon speed is constant using QED?

In Feynman's simple QED book he talks about the probability amplitude P(A to B) ,where A and B are events in spacetime, and he says that it depends of the spacetime interval but he didn't put the expression. I would like to know what expression was he referring to. He said then that the amplitudes for a photon to go faster or slower than c are very smaller than the contribution from speed c. How is that this contributions cancel out for long distances?

I would like also that you explain how it is related to the path integral of the form exp(iKL) of the question How is the path integral for light explained, or how does it arise? I think this path integral is very strange because it doesn't have spacetime fixed points but space fixed points. Its paths doesn't have the same time if we assume speed constant so they wont reach the same event.

• I have to run (literally) so just a note: $K \cdot L$ is an invariant. If you transform to a different frame you both blueshift $K$ and contract $L$ and these effects compensate. As for the main answer: as always, for nice theories (meaning weakly interating) classical limit of a path integral recovers the classical theory with variational principle. This is a general fact in any quantum mechanical theory and QED is no exception. So in this case you'll recover the motion of light along null geodesics. Jun 13 '11 at 16:26

This is a very good question, but it is really two completely different questions in one.

### Feynman's propagator

The probability amplitude for a photon to go from x to y can be written in many ways, depending on the choice of gauge for the electromagnetic field. They all give the same answer for scattering questions, or for invariant questions involving events transmitted to a macroscopic measuring device, but they have different forms for the detailed microscopic particle propagation.

Feynman's gauge gives a photon propagator of:

$P(k) = {g_{\mu\nu} \over w^2 - k^2 + i\epsilon}$

And it's Fourier transform is

$2\pi^2 P(x,t) = {g_{\mu\nu} \over {t^2 - x^2 + i\epsilon}}$

This is the propgation function he is talking about. It is singular on the light cone, because the denominator blows up, and it is only this singularity which you can see as propagating photons for long distances. For short distances, you see a $1/s^2$ propagation where s is the interval or proper time, between source and sink.

To show that you recover only physical light modes propagating, the easiest way is to pass to Dirac gauage. In this gauge, electrostatic forces are instantaneous, but photons travel exactly at the speed of light. It is not a covariant gauge, meaning it picks a particular frame to define instantaneous.

The issues with the Feynman gauge is that the propagator is not 100% physical, because of the sign of the pole on the time-time component of the photon propagator. You have to use the fact that charge is conserved to see that non-physical negative-coefficient-pole states are not real propagating particles. This takes thinking in the Feynman picture, but is not a problem in the Dirac picture. The equivalence between the two is a path integral exercise in most modern quantum field theory books.

### Fermat's principle and Lagrangians

Fermat's principle, as you noted, is not a usual action principle because it doesn't operate at fixed times. The analog of the Fermat principle in mechanics is called the principle of Maupertuis. This says that the classical trajectory is the one which minimizes

$$J = \int p dx = \int \sqrt{2m(E-V(x))} dx$$

between the endpoints. This principle is also timeless, and it can be used to construct an approximate form for the time Fourier transform of the propagator, and this is called the Gutzwiler trace formula.

the Gutzwiller trace formula is the closest thing we have to a proper quantum analog of the Maupertuis principle at this time.

### Lagrangian for light

The analog of the Lagrangian principle for light is just the principle of that light travels along paths that minimize proper time, with the additional constraint that these proper times are zero.

The Lagrangian is

$m\int ds = m\int \sqrt{1-v^2} dt$

but this is useless for massless particles. The proper transformation which gives a massless particle propagator is worked out in the early parts of Polyakov's "Gauge Fields and Strings" as a warm-up to the analogous problem for string theory. The answer is:

$S= \int {\dot{x}^2\over 2} + m^2 ds$

The equivalence between this form and the previous one is actually sort of obvious in Euclidean space, because of the central limit theorem you must get falling Gaussians with a steady decay rate. Polyakov works it out carefully because the anlogous manipulations in string theory are not obvious at all.

The second form is not singular as m goes to zero, and gives the proper massless propagator. Transitioning between the two introduces an "einbein" along the path, a metric tensor in one dimension.