Scalar and vector potential

  • Thread starter Petar Mali
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  • #1
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Main Question or Discussion Point

[tex]\vec{B}=rot\vec{A}[/tex]

[tex]\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi[/tex]


If I define

[tex]\varphi=\widetilde{\varphi}-\frac{\partial f}{\partial t}[/tex]

[tex]\vec{A}=\widetilde{\vec{A}}+gradf[/tex]

where

[tex]f=f(x,y,z,t)[/tex]

I will get

[tex]\vec{B}=rot\vec{A}=rot\vec{\widetilde{\vec{A}}}[/tex]

[tex]\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi=-\frac{\partial\widetilde{\vec{A}}}{\partial t}-grad\widetilde{\varphi}[/tex]

But if I say

[tex]\varphi=\widetilde{\varphi}+\frac{\partial f}{\partial t}[/tex]

[tex]\vec{A}=\widetilde{\vec{A}}+gradf[/tex]

I wouldnt get that result. How I know how to take minus sign in this relations!
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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The transformation for A is clear, because you want B to stay the same and you simply exploit the fact that grad(rot F) = 0 for any vector field F.
So then you can simply define [itex]\tilde\phi = \phi + \delta[/itex], write out
[tex]-\frac{\partial \tilde{\vec A}}{\partial t} - \nabla \tilde\phi[/tex]
and see what [itex]\delta[/itex] has to be to cancel the extra term from the [tex]\tilde A[/tex]-derivative so you get
[tex]-\frac{\partial \vec A}{\partial t} - \nabla \phi[/tex]
back.
 
  • #3
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How I know how to take minus sign in this relations!
Another way to see it is with four vectors:

Define the (contravariant) four potential putting together the scalar and vector potential:

[tex]A^\mu=(\phi,\mathbf{A})[/tex]

and define the (covariant) four derivative putting together the time derivative and the gradient:

[tex]\partial_{\mu}=(\partial_t,\nabla)[/tex]

Then the gauge transformation is simply

[tex]A^\mu\longmapsto A^\mu+\partial^\mu f[/tex]

and the fact that you have to raise the index of [tex]\partial_\mu[/tex] yields the minus sign.
 

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