The use of Green’s functions was really the one big piece of insight I gained from a graduate class in electrodynamics. The rest was all machinery.

Start with some arbitrary linear differential operator D (could be something like $\partial^2+m^2$), and we want to solve the equation $D f (\vec{r}) = \rho(\vec{r})$

Here’s Green’s trick: First solve the equation $D G (\vec{x,x_0}) = \delta^3 (\vec{x-x_0})$, where $G (\vec{x,x_0})$ is called a Green’s function for the equation and $\delta$ is the dirac delta .

Then the solution to the equation ‘proper’ is just given by:

$f (\vec{x}) = \chi (\vec{x})+\int_{R^3} G (\vec{x,x_0}) \rho (\vec{x_0})d^3 x_0$

where if we set $\rho = 0$, we see that $\chi$ is just the solution to $D \chi (\vec{x}) = 0$

And that’s all there is to it! One hard equation to solve has been reduced to an easier equation for G and a boundary condition (the $\rho = 0$ condition). You can check that $f (\vec{x})$ is the correct solution by operating on it with D.

The reason all this is significant is this: if you call $f$ the field and $\rho$ the source, then once you know the green’s function for a particular equation, you just plug in the source $\rho$ into the above equation, and out pops the field $f$ at any other point in space and time (as we’ll see). In other words , G tells you how the field responds to the presence of a source. Since the Green’s function tells you how the field changes from point to point, it’s also known as a propagator. It’s used everywhere that fields show up, from radiation to Feynman diagrams. This technique is so powerful that when Schwinger applied it to Quantum Field Theory, his method was known as ‘sourcery’.

Now let’s put it to work in some physicsy situations…

Here are three equations that are crucial to electrodynamics but also show up in all sorts of different situations.

Case 1: Poisson’s equation $\nabla^2 \phi(\vec{x})= - \frac{\rho}{\epsilon}$

(This follows from the maxwell’s equation $\nabla \cdot E = \frac{\rho}{\epsilon}$ and $E = - \nabla \phi$)

We need to find the Green’s function G in $\nabla^2 G (\vec{x,x_0}) = \delta^3 (\vec{x-x_0})$

Here it is: $G = \frac{-1}{4 \pi |\vec{x-x_0}|}$ does the trick. Chap 1 of Griffiths E&M book has a good explanation of why this is true. (Good point to remember, $\nabla^2 \frac{1}{r}$ is not equal to zero, as you might calculate on first glance, but is equal to $-4 \pi \delta^3 (\vec{r})$)

Say you didn’t remember this result and had to derive it. Here’s the trick (and it’s a good one): you solve the equation by solving it’s fourier transform. Let’s see how that would work.

You can write a delta function in k-space as $\delta^3(\vec{x-x_0}) = \frac{1}{(2 \pi)^3}\int_{R^3} e^{i \vec{k} \cdot (\vec{x-x_0})} d^3 k$

So for $\nabla^2 G (\vec{x,x_0}) = \frac{1}{(2 \pi)^3}\int_{R^3} e^{i \vec{k} \cdot (\vec{x-x_0})} d^3 k$,

G has to be $G = -\frac{1}{(2 \pi)^3}\int_{R^3}\frac{e^{i \vec{k} \cdot (\vec{x-x_0})}}{k^2} d^3 k$

We can do this integral by writing the spherical volume element $d^3 k$ as $2 \pi k^2 dk d(cos\theta)$ .

So, writing out the dot product and defining $\vec{r}=\vec{x-x_0}$ and $u=cos\theta$, the integral becomes

$-\frac{2\pi}{(2\pi)^3} \int_{0}^{\infty}\int_{-1}^{1}\frac{e^{i k r u}}{k^2}k^2 du dk$

First do the u integral, giving $\frac{-1}{2\pi^2}\int_0^{\infty}\frac{sin(kr)}{kr} dk$

The k integral is equal to $\frac{\pi}{2}\frac{1}{r}$ so $G = -\frac{1}{4\pi r}$, as expected.

So the solution to our equation is $f (\vec{x}) = \chi (\vec{x})+ \frac{1}{4 \pi \epsilon}\int_{V} \frac{\rho (\vec{x_0})}{|\vec{x-x_0}|}$

So we get back the good old electrostatic result, where $\chi (\vec{x})$ represents the potential due to all the charges outside of V.

#### Case 2: The Helmholtz Equation $(\nabla^2 + k^2) \phi(\vec{x})= - \frac{\rho(\vec{x})}{\epsilon}$

First we want to solve $(\nabla^2 + k^2) G (\vec{x,x_0}) = \delta^3 (\vec{x-x_0})$.

It turns out that $G_{\pm} = -\frac{e^{\pm i k r}}{4 \pi r}$ does the trick. (In the limit $k \rightarrow 0$, you get back the solution to case 1).

(To derive this, one way would be to fourier transform to k space as before, and do the integral $G = \frac{1}{(2 \pi)^3}\int_{R^3}\frac{e^{ i w r}}{-w^2+k^2} d^3 w$ Incidently, this is very similar the integral that Yukawa did to predict the mass of the pi meson and win the Nobel Prize in 1949. If you send $k \rightarrow i m$ you get the Yukawa integral. The solution $E = -\frac{e^{- m r}}{4 \pi r}$ is the Yukawa potential that gives the force of attraction between the nucleons. This is one of the most celebrated results of 20th century physics! See Zee, Quantum Field Theory in a Nutshell, chapter I.4 for a nice discussion of this derivation.)

In Electrodynamics, $G_{+}$ and $G_{-}$ represent incoming and outgoing spherical waves, and are apparently used in studying radiation.

The next case is probably the most important equation in all of electrodynamics.

#### Case 3: The Wave Equation$\partial^2 \phi(\vec{x},t) = - \frac{\rho(\vec{x},t)}{\epsilon}$

Or, expanding the dels, $(\nabla^2 -\frac{1}{c^2} \frac{\partial^2}{\partial t^2}) \phi(\vec{x},t)= - \frac{\rho(\vec{x},t)}{\epsilon}$, where we now need to allow time dependence, since we have a time dependent equation.

This equation is different from the previous two cases, since it depends on time. To deal with this, we need to modify our formalism slightly.

If we have a equation of the form $D \phi(\vec{x},t)=\rho(\vec{x},t)$

Then the green’s function is defined by $D G(\vec{x,x_0},t,t_0)=\delta^3(\vec{x-x_0}) \delta(t-t_0)$

And the solution to the equation is then

$f (\vec{x},t) = \chi (\vec{x},t)+\int_{R^3}\int_{-\infty}^{t} G (\vec{x,x_0},t,t_0) \rho (\vec{x_0})d t_0 d^3 x_0$

where we’ve included an integral over $t_0$ to pop the time delta function. To see that this solution works, just act on it with D.

To solve for the Green’s function, we use the same trick as above: fourier transform, this time in $\omega$ space.

So $G(\vec{x,x_0},t-t_0) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G_\omega(\vec{x,x_0},\omega)e^{- i \omega (t-t_0)}d\omega$

And $\delta(t-t_0) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{- i \omega (t-t_0)}d\omega$

So the Green’s function equation becomes $(\nabla^2 + k^2) G_\omega(\vec{x,x_0},\omega) = \delta^3(\vec{x-x_0})$ where $k=\frac{\omega}{c}$

But we’ve already solved this equation in part 2. The solution is just $G_\omega(\vec{x,x_0},\omega) = -\frac{e^{\pm i k r}}{4 \pi r}$

To get G, we just need to fourier transform back.

$G(\vec{x,x_0},t,t_0) = -\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{4 \pi r}e^{\pm i \frac{\omega}{c} r}e^{-i \omega (t-t_0)} d\omega$

The integral gives a delta function.

What you get is $G_{\pm}(\vec{x,x_0},t,t_0) = - \frac{1}{4 \pi r}\delta(t_0-(t\mp \frac{r}{c}))$

Now our Green’s function G is not only a function of space, but also time. To see it’s significance, remember that the solutions to our equation are given by:

$f (\vec{x},t) = \chi (\vec{x},t)+\int_{R^3}\int_{-\infty}^{t} G (\vec{x,x_0},t,t_0) \rho (\vec{x_0})d t_0 d^3 x_0$

So the Green’s function really is a propagator! You plug in the source at some point $(\vec{x_0},t_0)$ and you get the field everywhere in space and time.)

Plugging things in, the solution to the wave equation is:

$f (\vec{x},t) = \chi (\vec{x},t)+ \frac{1}{4 \pi \epsilon}\int_{V} \int_{-\infty}^{t} \frac{\rho (\vec{x_0},t_0)}{|\vec{x-x_0}|} \delta(t_0-(t\mp \frac{r}{c})) d t_0 d^3 x_0$

This is totally neat: we’ve stumbled upon one of the most celebrated results of 19th century physics! Electromagnetic signals are propagating through space at the speed of light and only at the speed of light (it’s also enforcing that the fields fall off as $\frac{1}{r}$). What the delta function is saying is this: if I flash a light for an instant of time (say at $t_0 = 0$) somewhere in space a distance r away from you, you could only have seen it at time $t = \frac{r}{c}$, or at $t = -\frac{r}{c}$. The second solution clearly violates causality, and unfortunately you have to throw it out by hand. The equations don’t seem to care, either way. Wikipedia calls this the problem of causality. Some attempts to attack this problem were made by Wheeler and Feynman (Link).

References:

In understanding and deriving these results, I’ve relied heavily on the excellent E&M lecture notes in Classical Electrodynamics by Robert Brown at Duke. In particular, I followed closely the bit on Green’s functions under the section on radiation.

1. In linear control systems you can always get the answer once you know the impulse response; i.e., the response to the delta function input.

My instructor said it’s the same principle as finding the fundamental tones of a bell by hitting it with a hammer.

The reason for that is that it’s all about Laplace transforms and delta has an L.t of 1.

I wonder if anyone’s proved that Green’s functions and impulse response are equivalent.

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