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 2022-09-25 Answered
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?
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Answers (1)

ululatonh
Answered 2022-09-26 Author has 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.
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