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ApeXaviour
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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" 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, \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)"
Lorentz guage: \vec{\nabla}\cdot\vec{A}=-\mu_0\epsilon_0\frac{dV}{dt}<br />
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 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<br />?
So the poisson like equation that you get comes up as: \vec{\nabla}\cdot V^2=-\frac{\rho}{\epsilon_0}<br />
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...
Thanks in advance for any hints you can give me...
Dec
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}<br />
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<br />?
So the poisson like equation that you get comes up as: \vec{\nabla}\cdot V^2=-\frac{\rho}{\epsilon_0}<br />
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...
Thanks in advance for any hints you can give me...
Dec
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