In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential: A^alpha = nu_0 j^alpha, to get A in terms of J, however, we have to use a considerably uglier formula

tarjetaroja2t

tarjetaroja2t

Answered question

2022-09-25

In the Lorenz gauge, we have a beautiful relation between the four-current and the four-potential:
A α = μ 0 J α
To get A in terms of J, however, we have to use a considerably uglier formula; or, at least, this is the formula presented in textbooks:
A α ( t , r ) = μ 0 4 π J α ( t 1 c r r , r ) r r d 3 r
The first equation is evidently Lorentz covariant. The second one, on the other hand, doesn't look covariant at all. The integrand has some messy dependence on ( t , r ) and the integration goes over only the spatial dimensions.
Can we rewrite the second equation in a covariant form? If not, then why not?

Answer & Explanation

ululatonh

ululatonh

Beginner2022-09-26Added 4 answers

The integral expression is Lorentz-covariant, too, and it may be made manifestly Lorentz-covariant, too.
The integral measure d 3 r / | | r r | | is equal to and may be rewritten as the four-dimensional integral with a delta-function (and step function) added:
2 d 4 x δ [ ( x x ) 2 ] θ ( t t )
It's understood that J is substituted at the point J ( x ).
Note that the step function θ (equal to one for positive arguments and zero otherwise) is Lorentz-covariant assuming that the points x , x aren't spacelike-separated (because the ordering of a cause and its effect is frame-independent), and they're not spacelike-separated as guaranteed by the delta-function that is only non-vanishing near/at the null separation of x , x
The step function guarantees that the cause precedes its effect.
The argument of the delta-function is a Lorentz invariant, ( x x ) 2 which means ( x x ) μ ( x x ) μ . The sign convention for the metric doesn't matter becase this invariant is the argument of an even function (delta-function).
Finally, the equivalence of the two integrals may be shown by performing the integral over t . The theta-function implies that we only integrate over the semi-infinite line t < t. The delta-function implies that the integral is only sensitive on the value of J ( t ) where c | t t | = | r r | where the delta-function vanishes.
Finally, the delta-function also automatically generates the 1 / | r r | factor because
δ ( y 2 ) = δ ( y 0 2 | y | 2 ) = δ [ ( y 0 + | y | ) ( y 0 | y | ) ] =
which is equal, because δ ( k X ) = δ ( X ) / | k | , to
= δ ( y 0 | y | ) y 0 + | y | = δ ( y 0 | y | ) 2 | y |
You see that the factor of two was needed, too.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Relativity

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?