Derivation of Lienard-Wiechart

  • #1
716
9

Main Question or Discussion Point

Can anyone point me to a derivation of the Lienard-Wiechart potential formulas? I assume that they can be derived from Maxwell's equations alone.

Thanks.
 

Answers and Replies

  • #3
716
9
That is very helpful, thank you.

A follow-up question, if I may:

What makes

[tex]

( \frac {\phi }{c} , A^x , A^y , A^z )

[/tex]

a four-vector? That is, why is applying a Lorentz transformation to it vaild? Is the electric potential somehow the time-component of the magnetic potential? Is

[tex]

\vec B = \nabla \times \vec A

[/tex]

still valid if A is four dimensional? Is the curl of a four-dimensional vector field even defined?
 
  • #4
dx
Homework Helper
Gold Member
2,011
18
Is

[tex]

\vec B = \nabla \times \vec A

[/tex]

still valid if A is four dimensional? Is the curl of a four-dimensional vector field even defined?
The analog of that equation is Fαβ = ∂αAβ - ∂βAα. The thing on the right is the four dimensional curl. The thing on the left is the Farady tensor.
 
  • #5
clem
Science Advisor
1,308
15
That is very helpful, thank you.
What makes
[tex]
( \frac {\phi }{c} , A^x , A^y , A^z )
[/tex]
a four-vector? That is, why is applying a Lorentz transformation to it valid? Is the electric potential somehow the time-component of the magnetic potential?
[tex](\phi,{\vec A})[/tex] is chosen to be a 4-vector in a LT so that the continuity equation
[tex]\partial_\mu A^\mu=0[/tex] will hold in any LT so charge conservation will hold in any Loentz system.
They are no longer called the electric and magnetic potential, but just the 4-vector potential.
 
  • #6
716
9
The thing on the right is the four dimensional curl.
OK, good. Thanks.

[tex](\phi,{\vec A})[/tex] is chosen to be a 4-vector in a LT so that the continuity equation [tex]\partial_\mu A^\mu=0[/tex] will hold in any LT so charge conservation will hold in any Loentz system.
I'm sorry - could you re-word this? I'm not sure I get your meaning.
 

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