Some prerequisites: complex analysis, Fourier transform, Laplace transform (optional if you are a physicist).
In this post, I will solve Maxwell’s equation with source term. This would provide an idea on what it means by “propagation”. Of course, as in other posts for this blog, the material is standard, but I present it in the way that I prefer. Note that at some points of this post, the presentations are not rigorous as I still do not completely understand this subject. Furthermore, I am stuck at the final step where I have no idea if the final integral can be integrated out.
2. Green’s function for d’Alembertian operator
In fact, I have made use of the convention by setting speed of light Roughly interpreting eq.(1), the source produces the field In more details but still rough, the field is modified in response to the presence of the source
Let us suppose that initially, the source is absence and the field is identically zero everywhere. One normally waits for quite a relatively long while from the initial time until one turns on the source. During the time when the source is still off, the field remains zero. But once the source is turned on, the field in influenced by the source through eq.(1). We would like to describe the values of the field after the source is turned on.
Having read the above sentence once let me repeat it again, but this time I include symbols and more details:
“ Let us suppose that initially , the source is absence and the field is identically zero everywhere . One normally waits for quite a relatively long while from the initial time until one turns on the source. This means that one turns on the source at a finite During the time when the source is still off, the field remains zero, i.e. for But once the source is turned on, the field in influenced by the source through eq.(1): We would like to describe the values of the field after the source is turned on, i.e. for ”
The first step in studying this problem is to solve eq.(1) by Fourier expanding the source and the field as
where the integral range for is and the integral range for is But in the calculation, we will see that is extended to a complex number. I believe that in order to do so properly, one should have used inverse Laplace transform like in the case for mathematicians and engineers. [One difference between Laplace transform and Fourier transform is that for Fourier transform, the transformed variable (e.g. angular frequency ) is a real number whereas the one for Laplace transform is a complex number. But physicists remedy this by saying something like `well, in order to describe the real world, let us allow the frequency in Fourier transform to be complex. The imaginary part will instruct how the amplitude of the wave decays’. I think this is not necessary because one should have used Laplace transform in the first place, and the decaying of the amplitude would already be taken into account.] However, being trained as a physicist for so long, I am not so familiar with Laplace transform. So let me instead keep using physicist’s convention by working non-rigorously with Fourier transform and extends the variable to be complex when necessary.
In fact eq.(3)–(4) are still not suitable with the subsequent calculation. In order to do so, needs to impose an extra real-world condition into account. Suppose that when turning on the source, its value does not jump up at instantly. Instead, its value gradually increases. This is known as adiabatic approximation. In order to model this, we introduce a factor with small to the field and source:
Substituting into eq.(3) and using Fourier transform [Until I can find a rigorous or satisfying explanation, let me be sloppy with the factor by intentionally forgetting it from time to time.]
Furthermore, it satisfies
Our task now is to simplify eq.(12). So
where we have made use of the Sokhotski-Plemelj formula
with being the principal part. Let us now proceed by considering
where In the above equation, each step is a simple manipulation and is left as an exercise for interested readers, especially students. [In fact, I do not feel like explaining the calculations in each step.] Next, note that due to the presence of Green’s function vanishes unless This condition makes which in turns implies Therefore, the second term in the last line of eq.(20) does not contribute. This leaves us with
Substituting into eq.(11) gives
The interpretation is that is the value of the field measured at and is obtained from the superposition of the response from the source. The sourse at will spend the time to reach and influence the field situated at [the speed of light is restored to make the argument clearer]. So one could say that the signal from the source travels with the speed of light.
3. Photon Propagator
Let us now consider Maxwell’s equations
where and indices are raised and lowered by It is often convenient to impose Coulomb’s gauge It is usually useful to separate the components into transverse and longitudinal parts. To be more precise, let us expand in Fourier modes:
with fixed and We see that in fact each mode has components labelled by index The longitudinal part is defined to be the projection along the direction of By a very simple vector analysis, the longitudinal part of the mode (25) is given by
So we define the longitudinal part of to be the superposition of the longitudinal part of is mode. That is
Now, Coulomb gauge demands
The component is
The interpretation for this result is that is only influenced by the th component of the source. Furthermore, once the source is turned on, is influenced directly. The signal does not need to spend time travelling. So is non-physical and is not counted towards degrees of freedom of
Let us now consider the component of eq.(29). This is
Then by using the result of the previous section, one obtains
This form tells us that the transverse part (recall that the longitudinal part vanishes because of the Coulomb gauge) of is propagating. The source at needs to spend time travelling to with the speed of light Next, in order to read off Green’s function, let us slightly rewrite this equation as
From this, one can read off Green’s function as
I have no idea whether the integral in the final stage can be integrated out.
In fact, whenever propagator is mentioned in physics, one almost always mean the Green’s function in (energy-)momentum space. The presentation in this post however concerns how to write the Green’s function in coordinate space. So I will stop this post here but might start a new one in the future concerning Green’s function in momentum space in order to catch up with the standard notion.