How can I solve for A in the Lorenz Gauge with a static source?

In summary, the conversation discusses how, in the Lorenz Gauge, A(x,t) satisfies a wave equation with the current density J(x,t) as a source. For static sources, this reduces to a Poisson-like equation. The attempt at a solution involves substituting the Lorenz Gauge into Maxwell's equations and using the Coulomb potential to solve for A.
  • #1
ApeXaviour
35
0
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" 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, [tex]\vec{A}(x,t)[/tex] satisfies a wave equation with the current density [tex]\vec{J}(x,t)[/tex] as source, and that for static sources this reduces to a Poisson-like equation.
Calulate [tex]\vec{A}(x,t)[/tex] for [tex]\vec{J}(x,t)=\vec{J}_0\delta(x-x_0)[/tex]"

Homework Equations


Lorentz guage: [tex] \vec{\nabla}\cdot\vec{A}=-\mu_0\epsilon_0\frac{dV}{dt}
[/tex]
delta function:[tex]\int_{I}f(x)\delta(x-x_0)dx=f(x_0)[/tex]
(once [tex]I[/tex] includes the point [tex]x_0[/tex])
Otherwise [tex]\delta(x-x_0)=0[/tex]
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:
[tex]-\vec{\nabla}^2\vec{A}+\mu_0\epsilon_0\frac{\partial^2 \vec{A}}{\partial t^2}=\mu_0\vec{J}[/tex]For static sources
[tex] \vec{\nabla}\cdot\vec{A}=0
[/tex]?
So the poisson like equation that you get comes up as: [tex] \vec{\nabla}\cdot V^2=-\frac{\rho}{\epsilon_0}
[/tex]

Now for the last bit :frown: , since [tex]\vec{A}=\vec{A}(x,t)[/tex] then the wave equation can be simplified down to:
[tex]-\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)[/tex]
(with the [tex]J_0[/tex] term substituted in)

So how can I solve for A? My idea was to isolate [tex]d^2\vec{A}[/tex] and integrate to solve it but that gets too messy. Also over what limits would I integrate? +/- infinity? For the [tex]dx[/tex] that will give [tex]\vec{A}(x_0,t)[/tex]? What happens when I integrate the delta function in terms of [tex]dt[/tex]? 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
 
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  • #2
maybe, you can try [tex]A(x,t)[/tex] as a plane wave solution,
then you can get a solution
 
  • #3
"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.
 
  • #4
Each Cartesian component of A satisfies the scalar Poisson equation for a point charge, so just use the Coulomb potential for A_x, etc.
 

What is the Lorenz gauge in electromagnetism?

The Lorenz gauge is a mathematical condition used in electromagnetism to simplify the equations of motion for electromagnetic fields. It ensures that the equations are consistent with the principles of energy and momentum conservation.

How is the Lorenz gauge different from other gauges?

The Lorenz gauge is a special case of the more general Coulomb gauge. It differs from other gauges in that it explicitly takes into account the time derivative of the potential in addition to the spatial derivatives.

Why is the Lorenz gauge important?

The Lorenz gauge is important because it allows for a more simplified and elegant formulation of Maxwell's equations, which govern the behavior of electromagnetic fields. It also helps to ensure that the equations are consistent with the fundamental laws of physics.

What are the implications of using the Lorenz gauge?

Using the Lorenz gauge has several implications in the study of electromagnetism. It helps to simplify and unify the equations, making them easier to solve and analyze. It also has implications for the quantization of the electromagnetic field and the development of quantum field theories.

Are there any limitations to using the Lorenz gauge?

While the Lorenz gauge is a useful tool in electromagnetism, it does have limitations. In some cases, it may lead to unphysical solutions or difficulties in certain boundary conditions. In these cases, other gauges may be more appropriate to use.

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