Photon Propagator with Retarded Green’s Function

Some prerequisites: complex analysis, Fourier transform, Laplace transform (optional if you are a physicist).

1. Introduction

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

Before directly working with Maxwell’s equations, let us consider an important toy model. Consider a field {\phi(t,\vec{x}).} Suppose that it is influenced by the source {{\rho}(t,\vec{x})} through the inhomogeneous differential equation

\displaystyle  (-{\partial}_t^2 + \nabla^2){\phi}(t,\vec{x}) = {\rho}(t,\vec{x}), \ \ \ \ \ (1)

where {(\vec{x}) = (x,y,z),} and

\displaystyle  {\partial}_t^2\equiv \frac{{\partial}^2}{{\partial} t^2},\qquad \nabla^2\equiv \frac{{\partial}^2}{{\partial} x^2}+\frac{{\partial}^2}{{\partial} y^2} + \frac{{\partial}^2}{{\partial} z^2}. \ \ \ \ \ (2)

In fact, I have made use of the convention by setting speed of light {c=1.} Roughly interpreting eq.(1), the source {{\rho}(t,\vec{x})} produces the field {\phi(t,\vec{x}).} In more details but still rough, the field {\phi(t,\vec{x})} is modified in response to the presence of the source {{\rho}(t,\vec{x}).}

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 {t\rightarrow-\infty}, the source is absence {{\rho}(-\infty,\vec{x})=0} and the field is identically zero everywhere {{\phi}(-\infty,\vec{x})=0}. 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 {t = t_0.} During the time when the source is still off, the field remains zero, i.e. {{\phi}(t,\vec{x}) = 0} for {t<t_0.} But once the source is turned on, the field in influenced by the source through eq.(1): {(-{\partial}_t^2 + \nabla^2){\phi}(t,\vec{x}) = {\rho}(t,\vec{x}).} We would like to describe the values of the field after the source is turned on, i.e. {{\phi}(t,\vec{x}) = ?} for {t>t_0.}

The first step in studying this problem is to solve eq.(1) by Fourier expanding the source and the field as

\displaystyle  {\phi}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (3)

\displaystyle  {\rho}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (4)

where the integral range for {{\omega}} is {(-\infty,\infty)} and the integral range for {\vec{k}} is {(-\infty,\infty)^3.} But in the calculation, we will see that {{\omega}} 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 {{\omega}}) 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 {e^{{\epsilon} t}} with small {{\epsilon}>0} to the field and source:

\displaystyle  {\phi}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}, \ \ \ \ \ (5)

\displaystyle  {\rho}(t,\vec{x}) =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}. \ \ \ \ \ (6)

This extra factor vanishes at initial time {t\rightarrow-\infty,} but is finite at any time of measurement (we always measure the value of the field at finite {t.}). Then substituting eq.(5)(6) into eq.(1) gives

\displaystyle  \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} (({\omega}+i{\epsilon})^2 - \vec{k}^2)\tilde{\phi}({\omega},\vec{k})e^{-i{\omega} t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t} = \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \tilde{\rho}({\omega},\vec{k})e^{-i({\omega}+i{\epsilon}) t+i\vec{k}\cdot\vec{x}}e^{{\epsilon} t}. \ \ \ \ \ (7)

This implies

\displaystyle  \tilde{\phi}({\omega},\vec{k}) =\frac{\tilde{\rho}({\omega},\vec{k})}{({\omega}+i{\epsilon})^2 - \vec{k}^2}. \ \ \ \ \ (8)

Substituting into eq.(3) and using Fourier transform [Until I can find a rigorous or satisfying explanation, let me be sloppy with the factor {e^{{\epsilon} t}} by intentionally forgetting it from time to time.]

\displaystyle  \tilde{\rho}({\omega},\vec{k}) =\int dt'\int d\vec{x}'{\rho}(t',\vec{x}') e^{i{\omega} t' - i\vec{k}\cdot\vec{x}} \ \ \ \ \ (9)


\displaystyle  {\phi}(t,\vec{x}) =\int dt'\int d^3\vec{x}' \int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{{\rho}(t',\vec{x}')}{({\omega}+i{\epsilon})^2 - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}. \ \ \ \ \ (10)

This equation is in the form

\displaystyle  {\phi}(t,\vec{x}) =\int dt'\int d^3\vec{x}' G(t-t',\vec{x}-\vec{x}'){\rho}(t',\vec{x}'), \ \ \ \ \ (11)

where {G(t-t',\vec{x}-\vec{x}')} is called Green’s function. In our case, it is given by

\displaystyle  G(t-t',\vec{x}-\vec{x}') =\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{({\omega}+i{\epsilon})^2 - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}. \ \ \ \ \ (12)

Furthermore, it satisfies

\displaystyle  (-{\partial}_t^2 + \nabla^2)G(t-t',\vec{x}-\vec{x}') ={\delta}(t-t'){\delta}^{(3)}(\vec{x}-\vec{x}'). \ \ \ \ \ (13)

Our task now is to simplify eq.(12). So

\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{2|\vec{k}|}\left(\frac{1}{{\omega}+i{\epsilon} - |\vec{k}|}-\frac{1}{{\omega}+i{\epsilon} + |\vec{k}|}\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle{\cal P}\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{2|\vec{k}|}\left(\frac{1}{{\omega} - |\vec{k}|}-\frac{1}{{\omega} + |\vec{k}|}\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\qquad\displaystyle+\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{i{\pi}}{2|\vec{k}|}\left(-{\delta}({\omega}-|\vec{k}|)+{\delta}({\omega}+|\vec{k}|)\right) e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}, \end{array} \ \ \ \ \ (14)

where we have made use of the Sokhotski-Plemelj formula

\displaystyle  \frac{1}{x+i{\epsilon}} = {\cal P}\frac{1}{x}-i{\pi}{\delta}(x) \ \ \ \ \ (15)

with {{\cal P}} being the principal part. Let us now proceed by considering

\displaystyle  \begin{array}{rl} \displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega} - |\vec{k}|} e^{-i{\omega} (t-t')} &=\displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega}} e^{-i({\omega} + |\vec{k}|) (t-t')}\\ &\\ &=\displaystyle-\frac{i}{2}sign(t-t')e^{-i|\vec{k}|(t-t')}, \end{array} \ \ \ \ \ (16)

where we used the identity [I might come back and add how to prove this identity]

\displaystyle  \displaystyle{\cal P}\int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x} =i{\pi}\ sign(s). \ \ \ \ \ (17)


\displaystyle  \begin{array}{rl} \displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega} + |\vec{k}|} e^{-i{\omega} (t-t')} &=\displaystyle{\cal P}\int_{-\infty}^{\infty} \frac{d{\omega}}{2{\pi}} \frac{1}{{\omega}} e^{-i({\omega} - |\vec{k}|) (t-t')}\\ &\\ &=\displaystyle-\frac{i}{2}sign(t-t')e^{i|\vec{k}|(t-t')}. \end{array} \ \ \ \ \ (18)


\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d^3\vec{k}}{(2{\pi})^3}\left(-\frac{i}{2}\right)sign(t-t') \frac{1}{2|\vec{k}|}\left(e^{-i|\vec{k}|(t-t')}-e^{i|\vec{k}|(t-t')}\right) e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\qquad\displaystyle+\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{i}{4|\vec{k}|}\left(-e^{-i|\vec{k}| (t-t')}+e^{i|\vec{k}| (t-t')}\right) e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int \frac{d^3\vec{k}}{(2{\pi})^3}(sign(t-t')+1) \frac{\sin(|\vec{k}|(t-t'))}{2|\vec{k}|} e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int \frac{d^3\vec{k}}{(2{\pi})^3}\Theta(t-t') \frac{\sin(|\vec{k}|(t-t'))}{|\vec{k}|} e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}, \end{array} \ \ \ \ \ (19)

where {\Theta(t-t')} is Heaviside step function. Let us use spherical polar coordinates in the space of {\vec{k}} such that the “{z-}axis” is aligned with {\vec{x}-\vec{x}'.} So the integral above simplifies as follows

\displaystyle  \begin{array}{rl} G(t-t',\vec{x}-\vec{x}') &=\displaystyle-\int_{0}^{2{\pi}}\int_0^{\pi}\int_{0}^\infty \frac{dk d{\theta} d{\phi}}{(2{\pi})^3}\ k^2\sin{\theta}\Theta(t-t') \frac{\sin(k(t-t'))}{k} e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{-1}^1\int_{0}^\infty \frac{dk \ d\cos{\theta}}{(2{\pi})^2}\ k\Theta(t-t') \sin(k(t-t')) e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{(2{\pi})^2}\ k\Theta(t-t') \sin(k(t-t')) \frac{1}{ikR}\left(e^{ikR}-e^{-ikR}\right)\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \sin(k(t-t')) \frac{2}{R}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \sin(k(t-t')) \frac{1}{R}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{(2{\pi})^2}\ \Theta(t-t') \frac{1}{2R}(\cos(k(R-t+t'))-\cos(k(R+t-t')))\\ &\\ &=\displaystyle-\frac{1}{(2{\pi})}\ \Theta(t-t') \frac{1}{2R}({\delta}(R-t+t')-{\delta}(R+t-t')) \end{array} \ \ \ \ \ (20)

where {k\equiv|\vec{k}|, R\equiv |\vec{x}-\vec{x}'|.} 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 {\Theta(t-t'),} Green’s function vanishes unless {t-t'>0.} This condition makes {R+t-t'>0,} which in turns implies {{\delta}(R+t-t')=0.} Therefore, the second term in the last line of eq.(20) does not contribute. This leaves us with

\displaystyle  G(t-t',\vec{x}-\vec{x}') =-\dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}\ \Theta(t-t'){\delta}(|\vec{x}-\vec{x}'|-t+t'). \ \ \ \ \ (21)

Substituting into eq.(11) gives

\displaystyle  \begin{array}{rl} \displaystyle{\phi}(t,\vec{x}) &=\displaystyle-\int dt'\int d^3\vec{x}' \dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}\ \Theta(t-t'){\delta}(|\vec{x}-\vec{x}'|-t+t'){\rho}(t',\vec{x}')\\ &\\ &=\displaystyle-\int d^3\vec{x}' \dfrac{1}{4{\pi} |\vec{x}-\vec{x}'|}{\rho}(t-|\vec{x}-\vec{x}'|,\vec{x}'). \end{array} \ \ \ \ \ (22)

The interpretation is that {{\phi}(t,\vec{x})} is the value of the field measured at {(t,\vec{x}),} and is obtained from the superposition of the response from the source. The sourse at {(t',\vec{x}')} will spend the time {|\vec{x}-\vec{x}'|/c} to reach and influence the field situated at {\vec{x}} [the speed of light {c} 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

\displaystyle  {\partial}_{\mu} F^{{\mu}{\nu}} = j^{\nu}, \ \ \ \ \ (23)

where {F_{{\mu}{\nu}} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu},} and indices are raised and lowered by {({\eta}^{{\mu}{\nu}}) = diag(-1,1,1,1)= ({\eta}_{{\mu}{\nu}}).} It is often convenient to impose Coulomb’s gauge {{\partial}_i A^i = 0.} It is usually useful to separate the components {A_i} into transverse and longitudinal parts. To be more precise, let us expand {A_i(t,\vec{x})} in Fourier modes:

\displaystyle  A_i(t,\vec{x}) =\int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}\tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}, \ \ \ \ \ (24)

Here, a `mode’ means a plane wave

\displaystyle  \tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (25)

with fixed {{\omega}} and {\vec{k}.} We see that in fact each mode has {3} components labelled by index {i.} The longitudinal part is defined to be the projection along the direction of {\vec{k}.} By a very simple vector analysis, the longitudinal part of the mode (25) is given by

\displaystyle  \frac{k^j \tilde A_j({\omega},\vec{k})}{|\vec{k}|^2}\ k_i e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (26)

So we define the longitudinal part of {A_i(t,\vec{x})} to be the superposition of the longitudinal part of is mode. That is

\displaystyle  A_i^{\parallel}(t,\vec{x}) \equiv\int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}\frac{k^j \tilde A_j({\omega},\vec{k})}{|\vec{k}|^2}\ k_i e^{-i{\omega} t + i\vec{k}\cdot\vec{x}}. \ \ \ \ \ (27)

Now, Coulomb gauge demands

\displaystyle  \int\frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3}{\partial}^i\tilde A_i({\omega},\vec{k}) e^{-i{\omega} t + i\vec{k}\cdot\vec{x}} =0, \ \ \ \ \ (28)

which implies {{\partial}^i\tilde A_i({\omega},\vec{k})=0.} This in turn gives {A_i^{\parallel}(t,\vec{x}) = 0.} That is, Coulomb gauge condition demands that the longitudinal part of {A_i} automatically vanishes. Furthermore, in this gauge Maxwell’s equations read

\displaystyle  {\partial}_{\mu}{\partial}^{\mu} A^{\nu} - {\partial}^{\nu}{\partial}_0 A^0 = j^{\nu}. \ \ \ \ \ (29)

The component {{\nu} = 0} is

\displaystyle  \nabla^2 A_0 = j_0. \ \ \ \ \ (30)

By following a similar analysis to the one in the previous section, one obtains, in parallel to eq.(11)(12),

\displaystyle  A_0(t,\vec{x}) =\int dt'\int d^3\vec{x}' G_0(t-t',\vec{x}-\vec{x}')j_0(t',\vec{x}'), \ \ \ \ \ (31)


\displaystyle  \begin{array}{rl} \displaystyle G_0(t-t',\vec{x}-\vec{x}') &=\displaystyle\int \frac{d{\omega}}{2{\pi}}\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{ - \vec{k}^2} e^{-i{\omega} (t-t')+i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle\int \frac{d^3\vec{k}}{(2{\pi})^3} \frac{1}{ - \vec{k}^2}{\delta}(t-t') e^{i\vec{k}\cdot(\vec{x}-\vec{x}')}\\ &\\ &=\displaystyle-\int_{0}^{2{\pi}}\int_0^{\pi}\int_{0}^\infty \frac{dk d{\theta} d{\phi}}{(2{\pi})^3}k^2\sin{\theta} \frac{1}{k^2}{\delta}(t-t') e^{ikR\cos{\theta}}\\ &\\ &=\displaystyle-\int_{0}^\infty \frac{dk}{2{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{4{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\int_{-\infty}^\infty \frac{dk}{4{\pi}^2} {\delta}(t-t')\frac{1}{kR}\sin(kR)\\ &\\ &=\displaystyle-\frac{1}{4{\pi}^2R}{\delta}(t-t')\ \textrm{Im}\int_{-\infty}^\infty dk \frac{e^{ikR}}{k}\\ &\\ &=\displaystyle-\frac{1}{4{\pi} R}{\delta}(t-t'), \end{array} \ \ \ \ \ (32)

where in the last step, we used eq.(17). Substituting eq.(32) into eq.(31) gives

\displaystyle  \begin{array}{rl} \displaystyle A_0(t,\vec{x}) &=\displaystyle-\int dt'\int d^3\vec{x}' \frac{1}{4{\pi} |\vec{x}-\vec{x}'|}{\delta}(t-t')j_0(t',\vec{x}'),\\ &\\ &=\displaystyle-\int d^3\vec{x}' \frac{j_0(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}. \end{array} \ \ \ \ \ (33)

The interpretation for this result is that {A_0} is only influenced by the {0}th component {j_0} of the source. Furthermore, once the source is turned on, {A_0} is influenced directly. The signal does not need to spend time travelling. So {A_0} is non-physical and is not counted towards degrees of freedom of {A_{\mu}.}

Let us now consider the component {{\nu}=i} of eq.(29). This is

\displaystyle  \begin{array}{rl} \displaystyle (-{\partial}_t^2 + \nabla^2)A_i &=\displaystyle j_i + {\partial}_i{\partial}_0 A^0. \end{array} \ \ \ \ \ (34)


\displaystyle  \begin{array}{rl} \displaystyle {\partial}_i{\partial}^0 A_0(t,\vec{x}) &=\displaystyle-{\partial}_i\int d^3\vec{x}' \frac{{\partial}^0 j_0(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}\\ &\\ &=\displaystyle{\partial}_i\int d^3\vec{x}' \frac{{\partial}'^j j_j(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|}\\ &\\ &=\displaystyle-\int d^3\vec{x}' \frac{{\partial}'^j j_j(t,\vec{x}')}{4{\pi} |\vec{x}-\vec{x}'|^3}(x_i-x_i'). \end{array} \ \ \ \ \ (35)


\displaystyle  j_i(t,\vec{x}) + {\partial}_i{\partial}^0 A_0(t,\vec{x}) =\int d^3\vec{x}'\left({\delta}^{(3)}(\vec{x}-\vec{x}'){\delta}_i^j -\frac{x_i-x_i' }{4{\pi} |\vec{x}-\vec{x}'|^3}{\partial}'^j\right)j_j(t,\vec{x}'). \ \ \ \ \ (36)

Then by using the result of the previous section, one obtains

\displaystyle  \begin{array}{rl} \displaystyle A_i(t,\vec{x}) &=\displaystyle\int d^3\vec{x}'\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{x_i'-x_i'' }{4{\pi} |\vec{x}'-\vec{x}''|^3}{\partial}''^j\right)j_j(t-|\vec{x}-\vec{x}'|,\vec{x}'')\\ &\\ &=\displaystyle\int d^3\vec{x}'\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\times\\ &\qquad\displaystyle j_j(t-|\vec{x}-\vec{x}'|,\vec{x}''). \end{array} \ \ \ \ \ (37)

This form tells us that the transverse part (recall that the longitudinal part vanishes because of the Coulomb gauge) of {A_i} is propagating. The source at {\vec{x}''} needs to spend time travelling to {\vec{x}} with the speed of light {c.} Next, in order to read off Green’s function, let us slightly rewrite this equation as

\displaystyle  \begin{array}{rl} \displaystyle A_i(t,\vec{x})&=\displaystyle\int d^3\vec{x}'\int dt''\int d^3\vec{x}''\frac{1}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\times\\ &\qquad\displaystyle {\delta}(t''-t+|\vec{x}-\vec{x}'|)j_j(t'',\vec{x}''). \end{array} \ \ \ \ \ (38)

From this, one can read off Green’s function as

\displaystyle  \begin{array}{rl} \displaystyle G_i^j(t,\vec{x};t'',\vec{x}'') &=\displaystyle \int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j -\frac{{\delta}_i^j }{4{\pi} |\vec{x}'-\vec{x}''|^3} +\frac{3(x_i'-x_i'')(x'^j-x''^j) }{4{\pi} |\vec{x}'-\vec{x}''|^5}\right)\\ &\\ &=\displaystyle \int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|}\left({\delta}^{(3)}(\vec{x}'-\vec{x}''){\delta}_i^j +{\partial}'_i{\partial}'^j\frac{1 }{4{\pi} |\vec{x}'-\vec{x}''|}\right)\\ &\\ &=\displaystyle \frac{{\delta}(t''-t+|\vec{x}-\vec{x}''|)}{4{\pi}|\vec{x}-\vec{x}''|}{\delta}_i^j +\int d^3\vec{x}'\frac{{\delta}(t''-t+|\vec{x}-\vec{x}'|)}{4{\pi}|\vec{x}-\vec{x}'|} {\partial}'_i{\partial}'^j\frac{1 }{4{\pi} |\vec{x}'-\vec{x}''|}. \end{array} \ \ \ \ \ (39)

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.

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