- #1
Lindsayyyy
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Hi everyone
Give is a generall gauge transformation [tex] \Phi \rightarrow \Phi ' =\Phi -\frac {\partial \chi}{\partial t}[/tex]
and
[tex]\vec A \rightarrow \vec A' = \vec A + \nabla \chi[/tex]
first task for now is the following: How do I have to choose Chi in order to fulfill the lorenz gauge condition.
[tex] {\rm div} \vec A + \frac{1}{c^2} \frac{\partial}{\partial t}\phi = 0[/tex]
FIrst of all I'm not even sure if I have to discuss phi and A as if they are linked to each other or not. But let's take a look at my A
I tried to use the divergence on my A'
[tex]div \vec A' = div \vec A + div \nabla \chi[/tex] then I use the Lorenz gauge condition for div a and I finally get
[tex] \nabla ^2 \chi +\mu_0 \epsilon_0 \frac {\partial^2 \chi}{\partial t^2}=0[/tex]
Is this the right approach ? I'm stuck here though I don't know how I have to choose my chi now and I still haven't taken a look at my potential.
Thanks for your help in advance.
Homework Statement
Give is a generall gauge transformation [tex] \Phi \rightarrow \Phi ' =\Phi -\frac {\partial \chi}{\partial t}[/tex]
and
[tex]\vec A \rightarrow \vec A' = \vec A + \nabla \chi[/tex]
first task for now is the following: How do I have to choose Chi in order to fulfill the lorenz gauge condition.
Homework Equations
[tex] {\rm div} \vec A + \frac{1}{c^2} \frac{\partial}{\partial t}\phi = 0[/tex]
The Attempt at a Solution
FIrst of all I'm not even sure if I have to discuss phi and A as if they are linked to each other or not. But let's take a look at my A
I tried to use the divergence on my A'
[tex]div \vec A' = div \vec A + div \nabla \chi[/tex] then I use the Lorenz gauge condition for div a and I finally get
[tex] \nabla ^2 \chi +\mu_0 \epsilon_0 \frac {\partial^2 \chi}{\partial t^2}=0[/tex]
Is this the right approach ? I'm stuck here though I don't know how I have to choose my chi now and I still haven't taken a look at my potential.
Thanks for your help in advance.