How can the integral form of Gauss's law for magnetism

NepanitaNesg3a

NepanitaNesg3a

Answered question

2022-05-02

How can the integral form of Gauss's law for magnetism be described as a version of general Stokes' theorem? How does it follow?

Answer & Explanation

Brianna Sims

Brianna Sims

Beginner2022-05-03Added 19 answers

Maxwell's equations in curved spacetime are written in the form
a F a b = 4 π J b , [ a F b c ] = 0 ,
with F the Faraday two-form, J a the current four-vector, the covariant derivative and [] denotes antisymmetrization of the indices. In terms of exterior calculus they become:
d F = 4 π J d F = 0 ,
with the Hodge dual, which sends p-forms to 4 p-forms in dimension 4. If we integrate the left side of the first equation over a space-like hypersurface of dimension 3, Σ, with normal time-like vector t a , then Stokes' theorem yields
Σ d F = S F ,
with S the boundary of Σ with normal n a . Since Σ is space-like and ( F ) c d = 1 2 F a b ϵ a b c d , S is also space-like and one component of F must be time-like, therefore F = F a b t a n b d S. This is easy to see also if one takes the restriction of the dual on S in local coordinates, all the 2-forms are space-like. But we know that E a = F a b t b , hence
S F = S F b a t a n b d S = S E b n b d S .
Now we integrate the right side,
Σ J = Σ J a ϵ a b c d = Σ J a t a d Σ = q ,
and after combining this and the previous one, we obtain:
S E a n a d S = 4 π q ,
Gauss's law applies also in curved spacetime. Note that ϵ a b c d is the Levi-Civita (volume) tensor, not the symbol. In local coordinates its components are the product of the symbol with | det g μ ν |
For the case of the magnetic field, B a = 1 2 ϵ a b c d F c d t b , only the space-like components of F a b are used, and the magnetic field part of the Faraday tensor is
F = F 12 d x 1 d x 2 + F 23 d x 2 d x 3 + F 31 d x 3 d x 1 ,
and the components of the field are B 1 = | det g μ ν | 1 / 2 F 23 etc, therefore the integrals become
0 = S F = S | det g μ ν | ( B 1 d x 2 d x 3 + B 2 d x 3 d x 1 + B 3 d x 1 d x 2 ) = S B a n a d S .

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