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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.

"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]"

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 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 , 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

## 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 , 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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