NepanitaNesg3a

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?

Brianna Sims

Beginner2022-05-03Added 19 answers

Maxwell's equations in curved spacetime are written in the form

$\begin{array}{rl}{\mathrm{\nabla}}_{a}{F}^{ab}& =-4\pi {J}^{b},\\ {\mathrm{\nabla}}_{[a}{F}_{bc]}& =0,\end{array}$

with $F$ the Faraday two-form, ${J}^{a}$ the current four-vector, $\mathrm{\nabla}$ the covariant derivative and [] denotes antisymmetrization of the indices. In terms of exterior calculus they become:

$\begin{array}{rl}d\star F& =4\pi \star J\\ dF& =0,\end{array}$

with $\star $ 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, $\mathrm{\Sigma}$, with normal time-like vector ${t}^{a}$, then Stokes' theorem yields

${\int}_{\mathrm{\Sigma}}d\star F={\int}_{S}\star F,$

with S the boundary of $\mathrm{\Sigma}$ with normal ${n}_{a}$. Since $\mathrm{\Sigma}$ is space-like and $(\star F{)}_{cd}=\frac{1}{2}{F}^{ab}{\u03f5}_{abcd}$, S is also space-like and one component of F must be time-like, therefore $\star F={F}^{ab}{t}_{a}{n}_{b}dS$. 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 ${E}_{a}={F}_{ab}{t}^{b}$, hence

${\int}_{S}\star F=-{\int}_{S}{F}^{ba}{t}_{a}{n}_{b}dS=-{\int}_{S}{E}_{b}{n}^{b}dS.$

Now we integrate the right side,

${\int}_{\mathrm{\Sigma}}\star J={\int}_{\mathrm{\Sigma}}{J}^{a}{\u03f5}_{abcd}={\int}_{\mathrm{\Sigma}}{J}^{a}{t}_{a}d\mathrm{\Sigma}=-q,$

and after combining this and the previous one, we obtain:

${\int}_{S}{E}_{a}{n}^{a}dS=4\pi q,$

Gauss's law applies also in curved spacetime. Note that ${\u03f5}_{abcd}$ is the Levi-Civita (volume) tensor, not the symbol. In local coordinates its components are the product of the symbol with $\sqrt{|det{g}_{\mu \nu}|}$

For the case of the magnetic field, ${B}_{a}=-\frac{1}{2}{\u03f5}_{abcd}{F}^{cd}{t}^{b}$, only the space-like components of ${F}_{ab}$ are used, and the magnetic field part of the Faraday tensor is

$F={F}_{12}d{x}^{1}\wedge d{x}^{2}+{F}_{23}d{x}^{2}\wedge d{x}^{3}+{F}_{31}d{x}^{3}\wedge d{x}^{1},$

and the components of the field are ${B}^{1}=-|det{g}_{\mu \nu}{|}^{-1/2}{F}_{23}$ etc, therefore the integrals become

$0={\int}_{S}F=-{\int}_{S}\sqrt{|det{g}_{\mu \nu}|}({B}^{1}d{x}^{2}\wedge d{x}^{3}+{B}^{2}d{x}^{3}\wedge d{x}^{1}+{B}^{3}d{x}^{1}\wedge d{x}^{2})=-{\int}_{S}{B}^{a}{n}_{a}dS.$

$\begin{array}{rl}{\mathrm{\nabla}}_{a}{F}^{ab}& =-4\pi {J}^{b},\\ {\mathrm{\nabla}}_{[a}{F}_{bc]}& =0,\end{array}$

with $F$ the Faraday two-form, ${J}^{a}$ the current four-vector, $\mathrm{\nabla}$ the covariant derivative and [] denotes antisymmetrization of the indices. In terms of exterior calculus they become:

$\begin{array}{rl}d\star F& =4\pi \star J\\ dF& =0,\end{array}$

with $\star $ 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, $\mathrm{\Sigma}$, with normal time-like vector ${t}^{a}$, then Stokes' theorem yields

${\int}_{\mathrm{\Sigma}}d\star F={\int}_{S}\star F,$

with S the boundary of $\mathrm{\Sigma}$ with normal ${n}_{a}$. Since $\mathrm{\Sigma}$ is space-like and $(\star F{)}_{cd}=\frac{1}{2}{F}^{ab}{\u03f5}_{abcd}$, S is also space-like and one component of F must be time-like, therefore $\star F={F}^{ab}{t}_{a}{n}_{b}dS$. 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 ${E}_{a}={F}_{ab}{t}^{b}$, hence

${\int}_{S}\star F=-{\int}_{S}{F}^{ba}{t}_{a}{n}_{b}dS=-{\int}_{S}{E}_{b}{n}^{b}dS.$

Now we integrate the right side,

${\int}_{\mathrm{\Sigma}}\star J={\int}_{\mathrm{\Sigma}}{J}^{a}{\u03f5}_{abcd}={\int}_{\mathrm{\Sigma}}{J}^{a}{t}_{a}d\mathrm{\Sigma}=-q,$

and after combining this and the previous one, we obtain:

${\int}_{S}{E}_{a}{n}^{a}dS=4\pi q,$

Gauss's law applies also in curved spacetime. Note that ${\u03f5}_{abcd}$ is the Levi-Civita (volume) tensor, not the symbol. In local coordinates its components are the product of the symbol with $\sqrt{|det{g}_{\mu \nu}|}$

For the case of the magnetic field, ${B}_{a}=-\frac{1}{2}{\u03f5}_{abcd}{F}^{cd}{t}^{b}$, only the space-like components of ${F}_{ab}$ are used, and the magnetic field part of the Faraday tensor is

$F={F}_{12}d{x}^{1}\wedge d{x}^{2}+{F}_{23}d{x}^{2}\wedge d{x}^{3}+{F}_{31}d{x}^{3}\wedge d{x}^{1},$

and the components of the field are ${B}^{1}=-|det{g}_{\mu \nu}{|}^{-1/2}{F}_{23}$ etc, therefore the integrals become

$0={\int}_{S}F=-{\int}_{S}\sqrt{|det{g}_{\mu \nu}|}({B}^{1}d{x}^{2}\wedge d{x}^{3}+{B}^{2}d{x}^{3}\wedge d{x}^{1}+{B}^{3}d{x}^{1}\wedge d{x}^{2})=-{\int}_{S}{B}^{a}{n}_{a}dS.$

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