Electromag - Lorenz gauge

[ApeXaviour]

I've gotten out most of this question, it's really just the last part that's getting to me at this stage. I'd never seen the http://mathworld.wolfram.com/DeltaFunction.html" [Broken] before so it might be because of that. I've an idea how to do it but I just end up in a mess of partial derivatives. I'd say it's something simple I just can't see.

Homework Statement

"Demonstrate that in the Lorenz Gauge, $$\vec{A}(x,t)$$ satisfies a wave equation with the current density $$\vec{J}(x,t)$$ as source, and that for static sources this reduces to a Poisson-like equation.
Calulate $$\vec{A}(x,t)$$ for $$\vec{J}(x,t)=\vec{J}_0\delta(x-x_0)$$"

Homework Equations

Lorentz guage: $$\vec{\nabla}\cdot\vec{A}=-\mu_0\epsilon_0\frac{dV}{dt}$$
delta function:$$\int_{I}f(x)\delta(x-x_0)dx=f(x_0)$$
(once $$I$$ includes the point $$x_0$$)
Otherwise $$\delta(x-x_0)=0$$
and Maxwell's equations.

The Attempt at a Solution

The wave equation was relatively easy. Substituting the lorenz gauge into maxwell's equations and getting:
$$-\vec{\nabla}^2\vec{A}+\mu_0\epsilon_0\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J}$$


For static sources
$$\vec{\nabla}\cdot\vec{A}=0$$?
So the poisson like equation that you get comes up as: $$\vec{\nabla}\cdot V^2=-\frac{\rho}{\epsilon_0}$$

Now for the last bit , since $$\vec{A}=\vec{A}(x,t)$$ then the wave equation can be simplified down to:
$$-\frac{\partial^2 \vec{A}}{\partial x^2}+\mu_0\epsilon_0\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J_0}\delta(x-x_0)$$
(with the $$J_0$$ term substituted in)

So how can I solve for A? My idea was to isolate $$d^2\vec{A}$$ and integrate to solve it but that gets too messy. Also over what limits would I integrate? +/- infinity? For the $$dx$$ that will give $$\vec{A}(x_0,t)$$? What happens when I integrate the delta function in terms of $$dt$$? I'd say there is something about the delta function that makes this fairly simple but I'm just not accustomed to it... :grumpy:

Thanks in advance for any hints you can give me...

Dec

 

[PRB147]

maybe, you can try $$A(x,t)$$ as a plane wave solution,
then you can get a solution

 

[dextercioby]

"Static sources' means no time dependence of the current vector, hence no time dependence of the potential vector.

To solve the eqn, Fourier expand all terms of the equation.

 

[Meir Achuz]

Each Cartesian component of A satisfies the scalar Poisson equation for a point charge, so just use the Coulomb potential for A_x, etc.