Classical Electromagnetism

Chapter 4: Electromagnetism Green's Functions

By Mark Lawrence

Please help support this web site

  • If you need a windshield, consider ours.
  • Contribute to our site maintenance fund:
  • Support our advertisers. Thanks, Mark

Electrostatics is the theory of electricity when nothing is changing in time. This means that all the time derivatives are zero, and we won't consider any magnetic fields. In this case, all we need is Poisson's equation,

Ñ2Φ = -ρ

or if there are no charges present, LaPlace's equation

Ñ2Φ = 0

Let's suppose we have only one charge in the entire universe. We'll locate that charge at x'. The charge will be a dimensionless point particle. Now let's ask what the potential is at any other location, x. Poisson's equation is now:

Ñ2Φ(x) = -δ( x, x' )

We already know the solution to this equation, it's

Φ(x) = 
  | x - x' |

There's an interesting thing about this equation: it's symmetric in x and x'. If you and the charge trade places, there's no difference in the potential where you are sitting. We expect this: if you and the charge trade places, then you're the same distance from the charge that you were before, and distance is all that matters.

What if instead of one point charge, we had several point charges. These charges would be located at x'i. Now, since the electrostatic field is linear, since it obeys the laws of superposition, all we need to do is add up the potential from each of the charges:

Φ(x) = 
  | x - x'i |

Since we're working on a classical theory, we'll ignore the idea of point charges and instead we'll presume we have a continuous charge distribution, ρ(x). Now the potential from a charge distribution in a particular volume is,

Φ(x) = 
  | x - x'i |  

If we lived in empty space and there were only a few charges in the universe, this would be it: electrodynamics would be a completely solved problem. But, things are just a little more complicated. We're going to be writing this little function, 1 / | x - x' | a lot, so we're going to invent a little short hand for it: G( x, x' ). As we noted above, G( x, x' ) = G( x', x ), that is G is symmetric in its two arguments. G, by the way, stands for Green's Function. We'll look a bit more closely at this function now.

The first thing we notice is that Ñx2G( x, x' ) = -δ( x, x' ). We put a subscript on Ñ to help us remember that we're differentiating with respect to x, not x'. This equation, Ñx2 G( x, x' ) = -δ( x, x' ), is the defining equation for G. Now we can ask, is G = 1 / | x - x' | a unique solution? Immediately we see that it is not. Imagine we have some function F(x) which is a solution to Laplace's equation Ñ2F = 0. Now, Ñx2( G( x, x' )+F(x) ) = -δ( x, x' ). So, we see that we can add any solution to Laplace's equation to G and G is still a solution to Poisson's equation. We will use this freedom a lot. This freedom is called a Gauge freedom, for historic reasons. Probably the gauge freedom most familiar to physics students is the choice of a voltage for ground.

Green's First and Second Identities:

We start with Gauss' Divergence theorem,

ÑA d3x = AdS = An dS

Where S is a surface that bounds our volume of interest and n is an outward pointing unit vector on S.

Now, choose A = φÑψ.

nA = n⋅φÑψ = φn Ñψ = φ ∂ψ/∂n

ÑA = Ñ⋅(φÑψ) = φÑ2ψ + Ñφ⋅Ñψ

ÑA d3x = φÑ2ψ + Ñφ⋅ Ñψ d3x = φ ∂ψ/∂n dS        Green's First Identity

Now, take Green's first identity, exchange φ and ψ, and subtract. This gives us:

φÑ2ψ - ψÑ2φ d3x = φ ∂ψ/∂n - ψ ∂φ/∂n dS       Green's Second Identity