# Getting stuck trying to solve electromagnetic wave equation using Green's function

I've recently learned about Green's function and am trying to derive an equation similar to that of the Biot-Savart law but for the electric field around a wire of changing current using the electric component of the electromagnetic wave equation:

$$\frac{\partial^2 E}{\partial t^2} = c^2\nabla^2E -c^2\mu_0 \frac{\partial J}{\partial t}$$

However, I am getting stuck on a step involving a dirac delta function in an integral and I'm not sure how to proceed next. Here are my steps:

Rewrite the wave equation so that we can use a Green's function for the 3-D wave equation:

$$\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}-\nabla^2E=-\mu_0 \frac{\partial J(t)}{\partial t}$$

From Wikipedia, the green's function is:

Resulting in the following equation to find $$E$$,

$$E=\int\limits_{\mathbb{R}^3}G(x)f(x) dr^3=-\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\frac{\delta^{3}(t-r/c)}{r}\cdot\frac{\partial J(t)}{\partial t}dr^3$$

At this point I'm a bit stuck, I'm not sure how to deal with the dirac delta inside the integral. One step I have tried is rewriting the current density of a thin wire as $$J=I\cdot\delta_2(r)$$,

$$E=-\frac{\mu_0}{4\pi}\int\limits_{\mathbb{R}^3}\frac{\delta_3(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t}\delta^{2}(r)\cdot dr^3$$

Letting us rewrite the equation as a line integral, similar to how the Biot-Savart law is written,

$$E=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \left[ \iint \frac{\delta_3(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t}\delta^{2}(r)\cdot dr^2 \right] dl$$

$$=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \frac{\delta^{3}(t-r/c)}{r}\cdot\frac{\partial I(t)}{\partial t} dl$$

But after this I'm completely lost. Based on Jefimenko's equations I need to get something like,

$$=-\frac{\mu_0}{4\pi}\int\limits_{a}\limits^{b} \frac{1}{r} \cdot \frac{\partial I(t-r/c)}{\partial t} dl$$

Which does seem to give the correct behavior, but I'm not exactly sure how multiplying $$I(t)$$ with a dirac-delta results in $$I(t-r/c)$$. I've tried finding a property that results in this, but I can't find anything.

I think I'm missing a step or doing something wrong. Because I'm a newbie in Green's functions and just this area of math in general, I would really appreciate an easy to understand answer. But really any help would be appreciated.

• try the potential formulation maybe? Nov 17, 2021 at 13:41
• also to obtain this relation you take the derivative of the equations, I could be wrong but this loses information contained in maxwells equation as using this new equation. many functions satisfy this new equation due to the "+c" in transforming back. like when solving the freespace equation, taking the curl of the function require you to use the other maxwell equations to prove orthoganality, Nov 17, 2021 at 13:45
• your transformation from volume current to line current is also only valid of a straight wire, as if its e.g I $\delta (x,y)$ this represents a wire located at x=0 y=0 and length in the z plain only. the general transformation to any wire is much much more complicated Nov 17, 2021 at 13:48
• @jensenpaull For the line current part, r is the distance from the infinitesimal piece to the point in space you want to calculate for. So assuming this, the same technique works when trying to derive the Biot-Savart law so I believe it would work here. I realize however that my notation isn't very precise, sorry about that.
– nreh
Nov 17, 2021 at 19:26
• It feels to me that your problem here is more "integrals of delta functions" than it is "Greens functions". For one thing, it is really important to note that that $\delta^{3}$ really means the product of three different delta functions in your three variables, which is why it is written with a superscript, and not a subscript. Nov 19, 2021 at 19:09

General considerations

I think it will help to carefully write out all the expressions. The PDE should be read component-wise

$$\partial_{tt}\psi_j(x,t)-\nabla^2\psi_j(x,t)=F_j(x,t)$$

Where $$x\in\mathbb{R}^3$$. I've replaced $$J$$ with a generic source $$\mathbf{F}$$, $$\mathbf{E}$$ with a generic field $$\boldsymbol{\psi}$$, and set all constants to unity. The causal Green's function is

$$G(r,t)=\frac{\delta(t-r)}{4\pi r}$$

The solution for $$\psi_j$$ is

$$\psi_j(x,t)=\int dt' \int d^3x' \ G(x-x',t-t')F_j(x',t') \ \ + \ \ \text{surface terms}$$

The surface terms are to match initial conditions which we can specify to vanish (see eg. Zangwill Modern Electrodynamics chapter 20). The delta collapses the $$t'$$ integral and we are left with

$$\psi_j(x,t)=\frac{1}{4 \pi}\int d^3x' \ \frac{F_j(x',t-|x-x'|)}{|x-x'|}$$

Note the arguments of the source appearing in the integrand.

The wave equation for $$E$$

Because $$J$$ and $$\rho$$ form part of a continuity equation, they cannot be specified completely independently. For $$\psi=E$$, the source should be $$F_j=-(\partial_t J_j+\partial_j \rho)$$, not just $$\partial_t J$$. You can check this by deriving it from Maxwell's equations. The expression for $$E$$ is

$$E_j(x,t)=-\frac{1}{4\pi} \int d^3x' \ |x-x'|^{-1} \bigg[\partial'_{j} \rho(x',t-|x-x'|) \ + \ \partial'_{t}J_j(x',t-|x-x'|) \bigg]$$

Note the primes on partial derivatives within the integral: the retarded time is $$t_r:=t-|x-x'|$$. You can see why it's preferable to write the integral as

$$E_j(x,t)=-\frac{1}{4\pi} \int d^3x' \ |x-x'|^{-1} \bigg[\partial'_{j} \rho(x',t') \ + \ \partial'_{t}J_j(x',t') \bigg]_{t'=t_r}$$

With $$R_j:=x_j-x_j'$$, and using the multivariate chain rule we find (Jackson chapter 6.5)

$$\big[\partial'_j \rho(x',t') \big]_{t'=t_r} = \partial'_j \big[ \rho(x',t')\big]_{t'=t_r} - \hat{R} \big[ \partial'_t \rho(x',t')\big]_{t'=t_r}$$

On integrating the term $$|x-x'|^{-1}\partial'_j \big[ \rho(x',t')\big]_{t'=t_r}$$ by parts, we find Jefimenko's equation for $$E_j$$.

Finally, if you want to specify a thin wire of arbitrary shape, the expression is not as simple as you may think. Start by parametrizing in $$\tau$$ such that the wire is the set of points $$(x,y,z)=(p_1(\tau),p_2(\tau),p_3(\tau))$$, then (up to a proportionality constant)

$$J_j(\mathbf{x})= \int d\tau \ \frac{dp_j}{d\tau} \delta^{(3)}(\mathbf{x}-\mathbf{p}(\tau))$$

• This is a very good answer. One edit I would suggest is to clarify that OP's first and second equations are not correct in general. (That is, OP's starting point itself is wrong). You state this implicitly by noting that the correct source term has a gradient of the charge density, but it might be good to state it explicitly.
– hft
Nov 19, 2021 at 18:56
• @hft Thank you. Good point: I've made it (I think) clearer now
– Sal
Nov 19, 2021 at 19:24
• Thank you so much for such a clear answer, I have a question though, you say that the delta collapses the t' integral, could you elaborate a little bit? I'm also confused why there's a t' integral, I thought the Green's function was just the integral over all of space?
– nreh
Nov 19, 2021 at 19:31
• @nreh You're welcome. A Green's function $G(\mathbf{x}-\mathbf{x}',t-t')$ for the wave equation is a solution to $\partial_{tt}G-\nabla^2G=\delta^{(3)}(\mathbf{x}-\mathbf{x}')\delta(t-t')$. Note the $\delta(t-t')$ on the RHS. Formally, you can see that multiplying by $F(\mathbf{x}',t')$ and integrating with respect to both $x'$ and $t'$ will yield your inhomogenous equation, hence there must be an integration wrt $t'$
– Sal
Nov 19, 2021 at 19:48
• @nreh Regarding the wire: yes $\tau$ is continuous, have a look at arc length parametrization. Heuristically, the first term in the integrand, $\frac{d \mathbf{p}}{d\tau}$ gives the tangent vector to the curve (at some given point $\tau$), this endows $\mathbf{J}$ with the proper vectorial direction, and the delta picks out only those points in space which really do lie on the curve
– Sal
Nov 19, 2021 at 19:59