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Posts Tagged ‘Green’s function’

PHY450H1S. Relativistic Electrodynamics Lecture 18 (Taught by Prof. Erich Poppitz). Green’s function solution to Maxwell’s equation.

Posted by peeterjoot on March 12, 2011

Covering chapter 8 material from the text [1].

Covering lecture notes pp. 136-146: continued reminder of electrostatic Greens function (136); the retarded Greens function of the d’Alembert operator: derivation and properties (137-140); the solution of the d’Alembert equation with a source: retarded potentials (141-142)

Solving the forced wave equation.

See the notes for a complex variables and Fourier transform method of deriving the Green’s function. In class, we’ll just pull it out of a magic hat. We wish to solve

\begin{aligned}\square A^k = \partial_i \partial^i A^k = \frac{4 \pi}{c} j^k\end{aligned} \hspace{\stretch{1}}(2.1)

(with a $\partial_i A^i = 0$ gauge choice).

Our Green’s method utilizes

\begin{aligned}\square_{(\mathbf{x}, t)} G(\mathbf{x} - \mathbf{x}', t - t') = \delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t')\end{aligned} \hspace{\stretch{1}}(2.2)

If we know such a function, our solution is simple to obtain

\begin{aligned}A^k(\mathbf{x}, t)= \int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t') G(\mathbf{x} - \mathbf{x}', t - t')\end{aligned} \hspace{\stretch{1}}(2.3)

Proof:

\begin{aligned}\square_{(\mathbf{x}, t)} A^k(\mathbf{x}, t)&=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\square_{(\mathbf{x}, t)}G(\mathbf{x} - \mathbf{x}', t - t') \\ &=\int d^3 \mathbf{x}' dt' \frac{4 \pi}{c} j^k(\mathbf{x}', t')\delta^3( \mathbf{x} - \mathbf{x}') \delta( t - t') \\ &=\frac{4 \pi}{c} j^k(\mathbf{x}, t)\end{aligned}

Claim:

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }\end{aligned} \hspace{\stretch{1}}(2.4)

This is the retarded Green’s function of the operator $\square$, where

\begin{aligned}\square G(\mathbf{x}, t) = \delta^3(\mathbf{x}) \delta(t)\end{aligned} \hspace{\stretch{1}}(2.5)

Proof of the d’Alembertian Green’s function

Our Prof is excellent at motivating any results that he pulls out of magic hats. He’s said that he’s included a derivation using Fourier transforms and tricky contour integration arguments in the class notes for anybody who is interested (and for those who also know how to do contour integration). For those who don’t know contour integration yet (some people are taking it concurrently), one can actually prove this by simply applying the wave equation operator to this function. This treats the delta function as a normal function that one can take the derivatives of, something that can be well defined in the context of generalized functions. Chugging ahead with this approach we have

\begin{aligned}\square G(\mathbf{x}, t)=\left(\frac{1}{{c^2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} - \Delta\right)\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi c^2 {\left\lvert{\mathbf{x}}\right\rvert} }- \Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }.\end{aligned} \hspace{\stretch{1}}(2.6)

This starts things off and now things get a bit hairy. It’s helpful to consider a chain rule expansion of the Laplacian

\begin{aligned}\Delta (u v)&=\partial_{\alpha\alpha} (u v) \\ &=\partial_{\alpha} (v \partial_\alpha u+ u\partial_\alpha v) \\ &=(\partial_\alpha v) (\partial_\alpha u ) + v \partial_{\alpha\alpha} u+(\partial_\alpha u) (\partial_\alpha v ) + u \partial_{\alpha\alpha} v).\end{aligned}

In vector form this is

\begin{aligned}\Delta (u v) = u \Delta v + 2 (\boldsymbol{\nabla} u) \cdot (\boldsymbol{\nabla} v) + v \Delta u.\end{aligned} \hspace{\stretch{1}}(2.7)

Applying this to the Laplacian portion of 2.6 we have

\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)\Delta\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}+\left(\boldsymbol{\nabla} \frac{1}{{2 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\right)\cdot\left(\boldsymbol{\nabla}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \right)+\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\Delta\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right).\end{aligned} \hspace{\stretch{1}}(2.8)

Here we make the identification

\begin{aligned}\Delta \frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }} = - \delta^3(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.9)

This could be considered a given from our knowledge of electrostatics, but it’s not too much work to just do so.

An aside. Proving the Laplacian Green’s function.

If $-1/{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }$ is a Green’s function for the Laplacian, then the Laplacian of the convolution of this with a test function should recover that test function

\begin{aligned}\Delta \int d^3 \mathbf{x}' \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right) f(\mathbf{x}') = f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.10)

We can directly evaluate the LHS of this equation, following the approach in [2]. First note that the Laplacian can be pulled into the integral and operates only on the presumed Green’s function. For that operation we have

\begin{aligned}\Delta \left(-\frac{1}{{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} \right)=-\frac{1}{{4 \pi}} \boldsymbol{\nabla} \cdot \boldsymbol{\nabla} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(2.11)

It will be helpful to compute the gradient of various powers of ${\left\lvert{\mathbf{x}}\right\rvert}$

\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert}^a&=e_\alpha \partial_\alpha (x^\beta x^\beta)^{a/2} \\ &=e_\alpha \left(\frac{a}{2}\right) 2 x^\beta {\delta_\beta}^\alpha {\left\lvert{\mathbf{x}}\right\rvert}^{a - 2}.\end{aligned}

In particular we have, when $\mathbf{x} \ne 0$, this gives us

\begin{aligned}\boldsymbol{\nabla} {\left\lvert{\mathbf{x}}\right\rvert} &= \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &= -\frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}^3}} &= -3 \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^5}.\end{aligned} \hspace{\stretch{1}}(2.12)

For the Laplacian of $1/{\left\lvert{\mathbf{x}}\right\rvert}$, at the points $\mathbf{e} \ne 0$ where this is well defined we have

\begin{aligned}\Delta \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} &=\boldsymbol{\nabla} \cdot \boldsymbol{\nabla} \frac{1}{{{\left\lvert{\mathbf{x}}\right\rvert}}} \\ &= -\partial_\alpha \frac{x^\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - x^\alpha \partial_\alpha \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} - \mathbf{x} \cdot \boldsymbol{\nabla} \frac{1}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= -\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}^3} + 3 \frac{\mathbf{x}^2}{{\left\lvert{\mathbf{x}}\right\rvert}^5}\end{aligned}

So we have a zero. This means that the Laplacian operation

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \Delta \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}},\end{aligned} \hspace{\stretch{1}}(2.15)

can only have a value in a neighborhood of point $\mathbf{x}$. Writing $\Delta = \boldsymbol{\nabla} \cdot \boldsymbol{\nabla}$ we have

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla} \cdot -\frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.16)

Observing that $\boldsymbol{\nabla} \cdot f(\mathbf{x} -\mathbf{x}') = -\boldsymbol{\nabla}' f(\mathbf{x} - \mathbf{x}')$ we can put this in a form that allows for use of Stokes theorem so that we can convert this to a surface integral

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^3 \mathbf{x}' \boldsymbol{\nabla}' \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &=\lim_{\epsilon = {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert} \rightarrow 0}f(\mathbf{x}) \int d^2 \mathbf{x}' \mathbf{n} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= \int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\mathbf{x}' - \mathbf{x}}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \cdot \frac{\mathbf{x} - \mathbf{x}'}{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^3} \\ &= -\int_{\phi=0}^{2\pi} \int_{\theta = 0}^\pi \epsilon^2 \sin\theta d\theta d\phi \frac{\epsilon^2}{\epsilon^4}\end{aligned}

where we use $(\mathbf{x}' - \mathbf{x})/{\left\lvert{\mathbf{x}' - \mathbf{x}}\right\rvert}$ as the outwards normal for a sphere centered at $\mathbf{x}$ of radius $\epsilon$. This integral is just $-4 \pi$, so we have

\begin{aligned}\Delta \int d^3 \mathbf{x}' \frac{1}{{-4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} }} f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.17)

The convolution of $f(\mathbf{x})$ with $-\Delta/4 \pi {\left\lvert{\mathbf{x}}\right\rvert}$ produces $f(\mathbf{x})$, allowing an identification of this function with a delta function, since the two have the same operational effect

\begin{aligned}\int d^3 \mathbf{x}' \delta(\mathbf{x} - \mathbf{x}') f(\mathbf{x}') =f(\mathbf{x}).\end{aligned} \hspace{\stretch{1}}(2.18)

Returning to the d’Alembertian Green’s function.

We need two additional computations to finish the job. The first is the gradient of the delta function

\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ? \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= ?\end{aligned}

Consider $\boldsymbol{\nabla} f(g(\mathbf{x}))$. This is

\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))&=e_\alpha \frac{\partial {f(g(\mathbf{x}))}}{\partial {x^\alpha}} \\ &=e_\alpha \frac{\partial {f}}{\partial {g}} \frac{\partial {g}}{\partial {x^\alpha}},\end{aligned}

so we have

\begin{aligned}\boldsymbol{\nabla} f(g(\mathbf{x}))=\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g.\end{aligned} \hspace{\stretch{1}}(2.19)

The Laplacian is similar

\begin{aligned}\Delta f(g)&= \boldsymbol{\nabla} \cdot \left(\frac{\partial {f}}{\partial {g}} \boldsymbol{\nabla} g \right) \\ &= \partial_\alpha \left(\frac{\partial {f}}{\partial {g}} \partial_\alpha g \right) \\ &= \left( \partial_\alpha \frac{\partial {f}}{\partial {g}} \right) \partial_\alpha g +\frac{\partial {f}}{\partial {g}} \partial_{\alpha\alpha} g \\ &= \frac{\partial^2 {{f}}}{\partial {{g}}^2} \left( \partial_\alpha g \right) (\partial_\alpha g)+\frac{\partial {f}}{\partial {g}} \Delta g,\end{aligned}

so we have

\begin{aligned}\Delta f(g)= \frac{\partial^2 {{f}}}{\partial {{g}}^2} (\boldsymbol{\nabla} g)^2 +\frac{\partial {f}}{\partial {g}} \Delta g\end{aligned} \hspace{\stretch{1}}(2.20)

With $g(\mathbf{x}) = {\left\lvert{\mathbf{x}}\right\rvert}$, we’ll need the Laplacian of this vector magnitude

\begin{aligned}\Delta {\left\lvert{\mathbf{x}}\right\rvert}&=\partial_\alpha \frac{x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} + x_\alpha \partial_\alpha (x^\beta x^\beta)^{-1/2} \\ &=\frac{3}{{\left\lvert{\mathbf{x}}\right\rvert}} - \frac{x_\alpha x_\alpha}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \\ &= \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned}

So that we have

\begin{aligned}\boldsymbol{\nabla} \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &= -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ \Delta \delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) &=\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \end{aligned} \hspace{\stretch{1}}(2.21)

Now we have all the bits and pieces of 2.8 ready to assemble

\begin{aligned}\Delta \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }&=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \\ &\quad +\frac{1}{{2\pi}} \left( - \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3} \right)\cdot-\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}} \\ &\quad +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }}\left(\frac{1}{{c^2}} \delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) -\frac{1}{{c}} \delta'\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \frac{2}{{\left\lvert{\mathbf{x}}\right\rvert}} \right) \\ &=-\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) +\frac{1}{{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2 }}\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \end{aligned}

Since we also have

\begin{aligned}\frac{1}{{c^2}} \partial_{tt}\frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\frac{\delta''\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} c^2}\end{aligned} \hspace{\stretch{1}}(2.23)

The $\delta''$ terms cancel out in the d’Alembertian, leaving just

\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.24)

Noting that the spatial delta function is non-zero only when $\mathbf{x} = 0$, which means $\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c) = \delta(t)$ in this product, and we finally have

\begin{aligned}\square \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} }=\delta(t) \delta^3(\mathbf{x}) \end{aligned} \hspace{\stretch{1}}(2.25)

We write

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert} },\end{aligned} \hspace{\stretch{1}}(2.26)

Elaborating on the wave equation Green’s function

The Green’s function 2.26 is a distribution that is non-zero only on the future lightcone. Observe that for $t < 0$ we have

\begin{aligned}\delta\left(t - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)&=\delta\left(-{\left\lvert{t}\right\rvert} - \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right) \\ &= 0.\end{aligned}

We say that $G$ is supported only on the future light cone. At $\mathbf{x} = 0$, only the contributions for $t > 0$ matter. Note that in the “old days”, Green’s functions used to be called influence functions, a name that works particularly well in this case. We have other Green’s functions for the d’Alembertian. The one above is called the retarded Green’s functions and we also have an advanced Green’s function. Writing $+$ for advanced and $-$ for retarded these are

\begin{aligned}G_{\pm} = \frac{\delta\left(t \pm \frac{{\left\lvert{\mathbf{x}}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.27)

There are also causal and non-causal variations that won’t be of interest for this course.

This arms us now to solve any problem in the Lorentz gauge

\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' dt' \frac{\delta\left(t - t' - \frac{{\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}}{c}\right)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}j^k(\mathbf{x}', t')+\text{An arbitrary collection of EM waves.}\end{aligned} \hspace{\stretch{1}}(3.28)

The additional EM waves are the possible contributions from the homogeneous equation.

Since $\delta(t - t' - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$ is non-zero only when $t' = t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)$, the non-homogeneous parts of 3.28 reduce to

\begin{aligned}A^k(\mathbf{x}, t) = \frac{1}{{c}} \int d^3 \mathbf{x}' \frac{j^k(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(3.29)

Our potentials at time $t$ and spatial position $\mathbf{x}$ are completely specified in terms of the sums of the currents acting at the retarded time $t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c$. The field can only depend on the charge and current distribution in the past. Specifically, it can only depend on the charge and current distribution on the past light cone of the spacetime point at which we measure the field.

Example of the Green’s function. Consider a charged particle moving on a worldline

\begin{aligned}(c t, \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.30)

($c$ for classical)

For this particle

\begin{aligned}\rho(\mathbf{x}, t) &= e \delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ \mathbf{j}(\mathbf{x}, t) &= e \dot{\mathbf{x}}_c(t) \delta^3(\mathbf{x} - \mathbf{x}_c(t))\end{aligned} \hspace{\stretch{1}}(4.31)

\begin{aligned}\begin{bmatrix}A^0(\mathbf{x}, t)\mathbf{A}(\mathbf{x}, t)\end{bmatrix}&=\frac{1}{{c}}\int d^3 \mathbf{x}' dt'\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\begin{bmatrix}c e \\ e \dot{\mathbf{x}}_c(t)\end{bmatrix}\delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\ &=\int_{-\infty}^\infty\frac{ \delta( t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c }{{\left\lvert{\mathbf{x}_c(t') - \mathbf{x}}\right\rvert}}\begin{bmatrix}e \\ e \frac{\dot{\mathbf{x}}_c(t)}{c}\end{bmatrix}\end{aligned}

PICTURE: light cones, and curved worldline. Pick an arbitrary point $(\mathbf{x}_0, t_0)$, and draw the past light cone, looking at where this intersects with the trajectory

For the arbitrary point $(\mathbf{x}_0, t_0)$ we see that this point and the retarded time $(\mathbf{x}_c(t_r), t_r)$ obey the relation

\begin{aligned}c (t_0 - t_r) = {\left\lvert{\mathbf{x}_0 - \mathbf{x}_c(t_r)}\right\rvert}\end{aligned} \hspace{\stretch{1}}(4.33)

This retarded time is unique. There is only one such intersection.

Our job is to calculate

\begin{aligned}\int_{-\infty}^\infty \delta(f(x)) g(x) = \frac{g(x_{*})}{f'(x_{*})}\end{aligned} \hspace{\stretch{1}}(4.34)

where $f(x_{*}) = 0$.

\begin{aligned}f(t') = t - t' - {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}/c\end{aligned} \hspace{\stretch{1}}(4.35)

\begin{aligned}\frac{\partial {f}}{\partial {t'}}&= -1 - \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \sqrt{ (\mathbf{x} - \mathbf{x}_c(t')) \cdot (\mathbf{x} - \mathbf{x}_c(t')) } \\ &= -1 + \frac{1}{{c}} \frac{\partial {}}{\partial {t'}} \frac{(\mathbf{x} - \mathbf{x}_c(t')) \cdot \mathbf{v}_c(t_r)}{{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}}\end{aligned}

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

[2] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

Desai Chapter 10 notes.

Posted by peeterjoot on December 7, 2010

Motivation.

Chapter 10 notes for [1].

Notes

In 10.3 (interaction with a electric field), Green’s functions are introduced to solve the first order differential equation

\begin{aligned}\frac{da}{dt} + i \omega_0 a = - i \omega_0 \lambda(t)\end{aligned} \hspace{\stretch{1}}(2.1)

A simpler way is to use the usual trick of assuming that we can take the constant term in the homogeneous solution and allow it to vary with time.

Since our homogeneous solution is of the form

\begin{aligned}a_H(t) = a_H(0) e^{-i\omega_0 t},\end{aligned} \hspace{\stretch{1}}(2.2)

we can look for a specific solution to the forcing term equation of the form

\begin{aligned}a_S(t) = f(t) e^{-i\omega_0 t}\end{aligned} \hspace{\stretch{1}}(2.3)

We get

\begin{aligned}f' = -i \omega_0 \lambda(t) e^{i \omega_0 t}\end{aligned} \hspace{\stretch{1}}(2.4)

which can be integrate directly to find the non-homogeneous solution

\begin{aligned}a_S(t) = a_S(t_0) e^{-i \omega_0 (t - t_0)} - i \omega_0 \int_{t_0}^t \lambda(t') e^{-i \omega_0 (t-t')} dt'\end{aligned} \hspace{\stretch{1}}(2.5)

Setting $t_0 = -\infty$, with a requirement that $a_S(-\infty) = 0$ and adding in a general homogeneous solution one then has 10.92 without the complications of Green’s functions or the associated contour integrals. I suppose the author wanted to introduce this as a general purpose tool and this was a simple way to do so.

His introduction of Green’s functions this way I didn’t personally find very clear. Specifically, he doesn’t actually define what a Green’s function is, and the Appendix 20.13 he refers to only discusses the subtlies of the associated Contour integration. I didn’t understand where equation 10.83 came from in the first place.

Something like the following would have been helpful (the type of argument found in [2])

Given a linear operator $L$, such that $L u(x) = f(x)$, we search for the Green’s function $G(x,s)$ such that $L G(x,s) = \delta(x-s)$. For such a function we have

\begin{aligned}\int L G(x,s) f(s) ds &= \int \delta(x-s) f(s) ds \\ &= f(x)\end{aligned}

and by linearity we also have

\begin{aligned}f(x) &=\int L G(x,s) f(s) ds \\ &= L \int G(x,s) f(s) ds \\ \end{aligned}

and can therefore identify $u(x) = \int G(x,s) f(s) ds$ as the desired solution to $L u(x) = f(x)$ once the Green’s function $G(x,s)$ associated with operator $L$ has been determined.

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] Wikipedia. Green’s function — wikipedia, the free encyclopedia [online]. 2010. [Online; accessed 20-November-2010]. http://en.wikipedia.org/w/index.php?title=Green%27s_function&oldid=3911%86019.

Reader notes for Jackson 12.11, Retarded time solution to the wave equation.

Posted by peeterjoot on September 20, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

Motivation

In [1] I blundered my way towards the retarded time Green’s function solution to the 3D wave equation. Jackson’s [2] (section 12.11) covers this in a much more coherent fashion. It is however somewhat terse, and some details that were not immediately obvious to me were omitted.

Here are my notes for this section in case I want to refer to it again later.

Guts

The starting point is the electrodynamic wave equation

\begin{aligned}\partial_\alpha F^{\alpha\beta} = \frac{4 \pi}{c} J^\beta\end{aligned} \quad\quad\quad(1)

A substitution of $F^{\alpha \beta} = \partial^\alpha A^\beta - \partial^\beta A^\alpha$ gives us

\begin{aligned}\partial_\alpha F^{\alpha\beta} = \partial_\alpha \partial^\alpha A^\beta - \partial_\alpha \partial^\beta A^\alpha= \square A^\beta - \partial^\beta (\partial_\alpha A^\alpha)\end{aligned} \quad\quad\quad(2)

Thus with the Lorentz condition $\partial_\alpha A^\alpha = 0$ we have

\begin{aligned}\square A^\beta = \frac{4 \pi}{c} J^\beta\end{aligned} \quad\quad\quad(3)

A set of four non-homogeneous wave equations to solve. It is assumed that a Green’s function of the form

\begin{aligned}\square_x D(x - x') = \delta^4(x - x')\end{aligned} \quad\quad\quad(4)

can be found. Jackson states that this is possible in the absense of boundary surfaces, which seems to imply that the more general case would require $\square_x D(x, x') = \delta^4(x - x')$, where $D$ is not neccessarily a function of the four vector difference $x - x'$.

What is really meant by this Green’s function? It only takes meaning in the context of the convolution integral. Namely

\begin{aligned}A^\beta = \int d^4 x' D(x, x') \frac{4 \pi}{c} J^\beta(x') \end{aligned} \quad\quad\quad(5)

So that

\begin{aligned}\square_x A^\beta &= \int d^4 x' \square_x D(x, x') \frac{4 \pi}{c} J^\beta(x') \\ &= \frac{4 \pi}{c} \int d^4 x' \delta^4(x - x') J^\beta(x') \\ &= \frac{4 \pi}{c} J^\beta(x) \\ \end{aligned}

So if a function with this delta filtering property under the Delambertian can be found we can find the non-homogeneous solutions directly by four-volume convolution.

It is implied in the text (probably stated explicitly somewhere earlier) that the asymmetric convention for the Fourier transform pairs is being used

\begin{aligned}\tilde{f}(k) &= \int d^4 z f(z) e^{i k \cdot z} \\ f(z) &= \frac{1}{{(2\pi)^4}} \int d^4 k \tilde{f}(k) e^{-i k \cdot z} \end{aligned} \quad\quad\quad(6)

where $d^4 k = dk_0 dk_1 dk_2 dk_3$, and $d^4 z = dz^0 dz^1 dz^2 dz^3$, and $k \cdot z = k_\mu z^\mu = k^\mu z_\mu$.

Assuming the validity of this transform pair, even for the delta distribution, we can find an integral representation of the delta using the transform pairs. For the Fourier transform of delta we have

\begin{aligned}\tilde{\delta^4}(k) &= \int d^4 z \delta^4(z) e^{i k \cdot z} \\ &= e^{i k \cdot 0} \\ &= 1\end{aligned}

Performing the inverse transformation provides the delta function exponential integral representation

\begin{aligned}\delta^4(z) &= \frac{1}{{(2\pi)^4}} \int d^4 k \tilde{\delta^4}(k) e^{-i k \cdot z} \\ &= \frac{1}{{(2\pi)^4}} \int d^4 k e^{-i k \cdot z} \\ \end{aligned}

Just as a Fourier representation of the delta can be found, we can integrate by parts to find an integral representation of the Green’s function that we seek. Taking Fourier transforms

\begin{aligned}\mathcal{F}(\square_x D(z))(k) &= \int d^4 z \partial_\alpha \partial^\alpha D(z) e^{i k \cdot z} \\ &= -\int d^4 z \partial^\alpha D(z) \partial_\alpha e^{i k_\beta z^\beta} \\ &= -\int d^4 z \partial^\alpha D(z) i k_\alpha e^{i k_\beta z^\beta} \\ &= \int d^4 z D(z) i k_\alpha \partial^\alpha e^{i k^\beta z_\beta} \\ &= -\int d^4 z D(z) k_\alpha k^\alpha e^{i k \cdot z } \\ &= - k^2 \tilde{D}(k)\end{aligned}

Usign the assumed delta function property of this Green’s function we also have

\begin{aligned}\mathcal{F}(\square_x D(z))(k) &= \int d^4 z \delta^4(z) e^{i k \cdot z} \\ &= 1\end{aligned}

This completely specifies the Fourier transform of the Green’s function

\begin{aligned}\tilde{D}(k) &= - \frac{1}{{k^2}}\end{aligned} \quad\quad\quad(8)

and we can inverse transform to complete the task of finding an initial representation of the Green’s function itself. That is

\begin{aligned}D(z) = -\frac{1}{{(2\pi)^4}} \int d^4 k \frac{1}{{k^2}} e^{-i k \cdot z} \end{aligned} \quad\quad\quad(9)

With an explicit spacetime split we have our integral prepped for the contour integration

\begin{aligned}D(z) = -\frac{1}{{(2\pi)^4}} \int d^3 k e^{i \mathbf{k} \cdot \mathbf{z}} \int_{-\infty}^\infty dk_0 \frac{1}{{k_0^2 - \mathbf{k}^2}} e^{-i k_0 z_0} \end{aligned} \quad\quad\quad(10)

Here $\kappa = {\left\lvert{\mathbf{k}}\right\rvert}$ is used as in the text. If we let $k_0 = R e^{i\theta}$ take on complex values, integrating over a semicircular arc, we have for the exponential

\begin{aligned}{\left\lvert{e^{-i k_0 z_0}}\right\rvert}&= {\left\lvert{e^{-i R (\cos\theta + i \sin\theta) z_0} }\right\rvert} \\ &= {\left\lvert{e^{ z_0 R \sin\theta} e^{ -i z_0 R \cos\theta } }\right\rvert} \\ &= {\left\lvert{e^{ z_0 R \sin\theta} }\right\rvert}\end{aligned}

In the upper half plane $\theta \in [0,\pi]$, so $\sin\theta$ is never negative, and the integral on an upper half plane semi-circular contour can only vanish as desired for $z_0 0$ for a lower half plane contour. This is mentioned in the text but I felt it more clear just writing out the exponential as above.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourBoth}
\caption{Contours strictly above the $k_0 = 0$ axis}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourAroundPole}
\caption{Contour around pole}
\end{figure}

Having established the value on the loop at infinity we can now integrate over the contour $r_1$ as depicted in figure (\ref{fig:jacksonRet:retardedContourBoth}). The problem is mainly reduced to an integral of the form figure (\ref{fig:jacksonRet:retardedContourAroundPole}) around the simple poles at $\alpha = \pm \kappa$

\begin{aligned}I_\alpha = \oint \frac{f(z)}{z - \alpha} dz\end{aligned} \quad\quad\quad(11)

With $z = \alpha + R e^{i\theta}$, and $\theta \in [\pi/2, 5\pi/2]$, we have

\begin{aligned}I_\alpha = \int \frac{f(z)}{R e^{i\theta}} R i e^{i\theta} d\theta\end{aligned} \quad\quad\quad(12)

with $R \rightarrow 0$, we are left with

\begin{aligned}I_\alpha = 2 \pi i f(\alpha)\end{aligned} \quad\quad\quad(13)

There are six arcs on the contour of interest. For the first two around the poles lets lable the integral contributions $I_\kappa$ and $I_{-\kappa}$. Along the infinite semicircular contour the integral vanishes with the right sign choice for $z_0$. For the remainder lets write the integral contributions $I$.

Summing over the complete contour, specially chosen to enclose no poles, we have

\begin{aligned}I + I_\kappa + I_{-\kappa} + 0 = 0 \end{aligned} \quad\quad\quad(14)

For this $z_0 > 0$ integral we are left with the residue sum

\begin{aligned}\int_{-\infty}^\infty dk_0 \frac{1}{{k_0^2 - \mathbf{k}^2}} e^{-i k_0 z_0} &= - 2 \pi i \left( {\left. \frac{1}{{k_0 - \kappa}} e^{-i k_0 z_0} \right\vert}_{k_0 = -\kappa}+{\left. \frac{1}{{k_0 + \kappa}} e^{-i k_0 z_0} \right\vert}_{k_0 = \kappa}\right) \\ &= \frac{2 \pi i^2}{\kappa} \sin(\kappa z_0)\end{aligned}

Since I can never remember the signs and integral orientations for the residue formula so I’ve always done it “manually” as above picking a zero valued contour.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{retardedContourOnAxis}
\caption{Contour exactly on the $k_0 = 0$ axis?}
\end{figure}

Now, the issue of where to place the contour wasn’t really discussed mathematically. Physically this makes the difference between causal and acausal behaviour, but why put the contour strictly above or below the axis and not right on it. If we put the contour exactly on the $k_0 = 0$ axis as in (\ref{fig:jacksonRet:retardedContourOnAxis}), then our integrals around the two half circular poles gives us a result off by a factor of two? There is also an (implied) limiting procedure required to place the contour strictly above the axis, and the details of this aren’t mentioned (and I also haven’t thought them through). Some of these would be worth thinking through in more detail, but for now lets ignore these. We are left with

\begin{aligned}D(z) = \frac{\theta(z_0)}{(2\pi)^3} \int d^3 k e^{i \mathbf{k} \cdot \mathbf{z}} \frac{1}{{\kappa}} \sin(\kappa z_0)\end{aligned} \quad\quad\quad(15)

How to reduce this to the single variable integral in $\kappa$ was not immediately clear to me. Aligning $\mathbf{z}$ with the $\mathbf{e}_3$ axis, and using a spherical polar representation for $\mathbf{k}$ we can write $\mathbf{z} \cdot \mathbf{k} = R \kappa \cos\theta$. With this and the volume element $d^3 k = \kappa^2 \sin\theta d\theta d\phi d\kappa$, we have

\begin{aligned}D(z) = \frac{\theta(z_0)}{(2\pi)^3} \int_0^\infty d\kappa \sin(\kappa z_0) \int_0^{2\pi} d\phi \int_0^\pi d\theta \kappa e^{i R \kappa \cos\theta} \sin\theta\end{aligned} \quad\quad\quad(16)

This now happily submits to a nice variable substitution, unlike an integral like $\int e^{i \mu \cos\theta} d\theta = J_0({\left\lvert{\mu}\right\rvert})$ which can be evaluated, but only in terms of Bessel functions or messy series expansion. Writing $\tau = \kappa \cos\theta$, and $-d\tau = \kappa \sin\theta d\theta$ we have

\begin{aligned}\int_0^\pi d\theta \kappa e^{i R \kappa \cos\theta} \sin\theta&=-\int_{\kappa}^{-\kappa} d\tau e^{i R \tau} \\ &=\frac{e^{i R \kappa}}{i R} -\frac{e^{-i R \kappa}}{i R} \\ &=2 \frac{1}{{R}} \sin(R \kappa)\end{aligned}

Our Green’s function is now reduced to

\begin{aligned}D(z) = \frac{\theta(z_0)}{2 \pi^2 R} \int_0^\infty d\kappa \sin(\kappa z_0) \sin(\kappa R)\end{aligned} \quad\quad\quad(17)

Expanding out these sines in terms of exponentials we have

\begin{aligned}D(z) &= -\frac{\theta(z_0)}{8 \pi^2 R} \int_0^\infty d\kappa ( e^{i\kappa(z_0+R)} + e^{-i\kappa(z_0+R)} -e^{i\kappa(R-z_0)} - e^{i\kappa(z_0-R)} ) \\ &= -\frac{\theta(z_0)}{8 \pi^2 R} \left(\int_0^\infty d\kappa \left( e^{i\kappa(z_0+R)} -e^{i\kappa(R-z_0)} \right) +\int_0^{-\infty} -d\kappa \left( e^{i\kappa(z_0+R)} - e^{-i\kappa(z_0-R)} \right) \right)\\ &= \frac{\theta(z_0)}{8 \pi^2 R} \int_{-\infty}^\infty d\kappa \left( e^{i\kappa(R-z_0)} -e^{i\kappa(z_0+R)} \right) \\ \end{aligned}

the sign in this first exponential differs from what Jackson obtained but it won’t change the end result. Did I make a mistake or did he? Wonder what the third edition shows? Using $\delta(x) = \int e^{-ikx} dk/2\pi$ we have

\begin{aligned}D(z) = \frac{\theta(z_0)}{4 \pi R} \left( \delta(z_0 -R) - \delta(-(z_0 + R)) \right)\end{aligned} \quad\quad\quad(18)

With $R = {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert} \ge 0$, and $z_0 = c(t - t') > 0$, this second delta cannot contribute, and we are left with the retarded Green’s function

\begin{aligned}D_r(z) = \frac{\theta(z_0)}{4 \pi R} \delta(c(t -t') - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}) \end{aligned} \quad\quad\quad(19)

Very slick. I like the procedure, despite a few magic steps (like the choice to offset the contour).

References

[1] Peeter Joot. {Poisson and retarded Potential Green’s functions from Fourier kernels} [online]. http://sites.google.com/site/peeterjoot/math2009/poisson.pdf.

[2] JD Jackson. Classical Electrodynamics Wiley. 2nd edition, 1975.