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

  • $\begingroup$ Stokes theorem is relating a surface integral to a line integral. Gauss' Law relates a surface integral (flux) to a volume integral (total charge/source). You are confusing the two fundamental theorems. $\endgroup$
    – jerk_dadt
    Apr 7, 2014 at 0:31
  • $\begingroup$ @jerk_dadt: Gauss's law is a special case of Stokes' theorem. $\endgroup$
    – Kyle Kanos
    Apr 7, 2014 at 0:53
  • $\begingroup$ Oh my bad. Neglect my original comment then. $\endgroup$
    – jerk_dadt
    Apr 7, 2014 at 0:58
  • $\begingroup$ Yeah, but I'm talking about it in terms of differential form. $\endgroup$
    – user44056
    Apr 7, 2014 at 1:05
  • 1
    $\begingroup$ Huh - I've never heard it called "Gauss's Law for magnetism," and it's one of the most referenced equations in my field. $\endgroup$
    – user10851
    Apr 7, 2014 at 4:51

3 Answers 3


Maxwell's equations in curved spacetime are written in the form $$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with $F$ the Faraday two-form, $J^a$ the current four-vector, $\nabla$ the covariant derivative and $[]$ denotes antisymmetrization of the indices. In terms of exterior calculus they become: $$ \begin{split} d\star F &= 4\pi \star J\\ dF&=0,\end{split}$$ 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, $\Sigma$, with normal time-like vector $t^a$, then Stokes' theorem yields $$\int_\Sigma d\star F = \int_S \star F,$$ with $S$ the boundary of $\Sigma$ with normal $n_a$. Since $\Sigma$ is space-like and $(\star F)_{cd} = \frac{1}{2} F^{ab} \epsilon_{abcd}$, $S$ 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 $$\int_S \star F = -\int_S F^{ba} t_a n_b d S = -\int_S E_b n^b d S.$$ Now we integrate the right side, $$\int_\Sigma \star J = \int_\Sigma J^a \epsilon_{abcd}=\int_\Sigma J^a t_a d \Sigma = -q,$$ and after combining this and the previous one, we obtain: $$\int_S E_a n^a d S = 4\pi q,$$ Gauss's law applies also in curved spacetime. Note that $\epsilon_{abcd}$ 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, $B_a = - \frac{1}{2} \epsilon_{abcd} F^{cd} t^b$, only the space-like components of $F_{ab}$ are used, and the magnetic field part of the Faraday tensor is $$F = F_{12} dx^1\wedge dx^2 + F_{23} dx^2\wedge dx^3 + F_{31} dx^3\wedge dx^1,$$ and the components of the field are $B^1 = - |\det g_{\mu\nu}|^{-1/2} F_{23}$ etc, therefore the integrals become $$0=\int_S F = -\int_S \sqrt{|\det g_{\mu\nu}|} (B^1 dx^2\wedge dx^3 + B^2 dx^3\wedge dx^1 + B^3 dx^1\wedge dx^2) = - \int_S B^a n_a dS.$$


So on in three-dimensional Euclidean space we have an isomorphism between vectors and 1-forms, the usual way $$\eta_\mu = g_{\mu\nu} \eta^\mu.$$ We also have an isomorphism between 1-forms and 2-forms, given by $\star : dz\mapsto dx\wedge dy$ and cyclically. This isomorphism has a fancy name, the Hodge dual, if you want to know about it in general. Then if $B^\mu$ is the magnetic field, we can make a 3-form -- something that can be integrated over a volume -- out of it by (i) lowering the index to get a 1-form (ii) taking the Hodge dual to get a 2-form (iii) using $d$ to get a 3-form. More explicitly, $$B_\mu = B_x dx + B_y dy + B_z dz$$ $$(\star B)_{\mu\nu} = B_x dy\wedge dz + B_y dz\wedge dx + B_z dx\wedge dy$$ $$(d\star B)_{\mu\nu\rho} = \frac{\partial B_x}{\partial x} dx\wedge dy \wedge dz + \frac{\partial B_y}{\partial y} dy\wedge dz\wedge dx + \frac{\partial B_z}{\partial z} dz\wedge dx\wedge dy$$ But this is $$(d\star B)_{\mu\nu\rho} = \left(\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} \right) dx\wedge dy\wedge dz$$ which is what those poor souls who don't know about differential forms call $\nabla \cdot \mathbf B$. Now you can just apply Stokes's wonderful theorem!

This is what they should teach you in multivariable calculus but don't!


Without specifying any particular scenario, and ignoring any proportionality constants, simply consider some general differential form $\omega$, and let this represent the electric flux through a closed surface which bounds some volume V. In classical electromagnetism, the Gauss law tells us that the flux through a closed surface is proportional to the amount of charge enclosed within that surface; in other words, visualise the flux as "flux tubes" which terminate inside the volume V. If $\omega$ represents the flux tubes, then $d\omega$ represents their end points, and we can write the Gauss law simply and intuitively as

$\displaystyle{\int_{\partial V}\omega =\int_{V}d\omega}$

This is just precisely the generalised Stokes Theorem - the amount of flux tubes ending inside the volume equals the amount of flux tubes crossing the surface.


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