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I have read about a generalization of Maxwell equation on manifolds that employs differential forms and Hodge duality that goes as follow: $$dF = 0\qquad \text{and}\qquad d \star F = J.\tag{1}$$ As I understand exterior derivative is defined to be the differential on 0-forms and to be an antiderivation (roughly speaking). The Hodge dual of a $p$-form $w$ is defined as the unique $(n-p)$-form $\star w$ that satisfy $\eta \wedge \star w = \langle \eta , w \rangle \mathrm{vol}$ for all $p$-forms $\eta$. We can express the Faraday tensor in coordinates such that the metric tensor is trivialized at one point i.e. $g_{\mu \nu} = \eta_{\mu \nu}$. Then one should compute the hodge dual using the general formula in a basis: \begin{equation} \star \alpha = \frac{1}{k!(n-k)!} \epsilon_{i_1,\dots,i_n} \sqrt{|\det(g)|} \alpha_{j_1,\dots,j_k} g^{i_1,j_1} \cdots g^{i_k,j_k} e^{i_{k+1}} \wedge \cdots \wedge e^{i_n} \end{equation} That formula can be used to obtain $\star F$ in a single point if one put $g_{\mu \nu} = \eta_{\mu \nu}$. But due to the vanishing of first derivative of the metric tensor once differentiate with the exterior derivative one gets the correct expression in the point where the metric was trivialized. By decomposing the effect of the derivative as spatial and temporal parts one gets Maxwell equation. Here I'm following John Baez and Javier P. Muniain "Gauge fields, knots and gravity". Thus one sees that in an local inertial coordinate reference usual Maxwell equation are obtained. This formulation uses the fact that there is a Lorenzian metric and that the manifold is orientable, but doesn't rely on any addiction structure like a connection.

However I am aware of another way to generalize maxwell equation: the minimal coupling which is the substitution of usual partial derivative with the covariant derivative leading to $$\nabla_a F^{ab} = J^{b}\qquad \text{and}\qquad\nabla_{[a} F_{bc]} = 0.\tag{2}$$ With the Levi-Civita connection.

I don't understand how this two are connected. Are those two generalization the same? How can it be possible if the differential form formulation is not aware of Levi-Civita connection? It will be remarkable if the "right" connection popped out from the differential forms version of the Maxwell equation!

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Maxwell's equations in differential form notation reads $$ dF = 0~, \qquad d\ast F = \ast J~. $$ We now show that these equations are equivalent to $\nabla_{[a} F_{bc]} = 0$, $\nabla_a F^{ab} = J^b$.

First, by definition \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} + \Gamma^d_{[ab}F_{c]d} - \Gamma^d_{[ac} F_{b]d} \end{align} If the connection is torsion-free (not necessarily the Levi-Civita connection) then $\Gamma^a_{[bc]} = 0$ so that \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} = \frac{1}{3} (dF)_{abc} \end{align} The last equality is true by definition. Thus, $\nabla_{[a} F_{bc]} \implies dF = 0$.

Next, consider the second equation \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a_{ac} F^{cb} + \Gamma^b_{ac} F^{ac} \end{align} Again, if the connection is torsion free, then the last term is zero. To simplify the second term, we have to assume that $\Gamma$ is the Levi-Civita connection so that $$ \Gamma^a_{ac} = \frac{1}{2} g^{ab} ( \partial_a g_{cb} + \partial_c g_{ab} - \partial_b g_{ac} ) = \frac{1}{2} g^{ab} \partial_c g_{ab} = \frac{1}{2} \partial_c \log \det g = \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} $$ Then, we have \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} F^{cb} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} F^{ab} \right)~. \end{align} So we can write Maxwell's equation as $$ \partial_e \left( \sqrt{\det g} F^{ed} \right) = \sqrt{\det g} J^d $$ Now, contract both sides with the Levi-Civita symbol (not tensor), ${\tilde \varepsilon}_{abcd}$ to get $$ \partial_e \left( \sqrt{\det g} {\tilde \varepsilon}_{abcd} F^{ed} \right) = \sqrt{\det g}{\tilde \varepsilon}_{abcd} J^d $$ Now, recall the the Levi-Civita tensor is ${\varepsilon}_{abcd} = \sqrt{\det g} {\tilde \varepsilon}_{abcd}$ $$ \partial_e \left( {\varepsilon}_{abcd} F^{ed} \right) = {\varepsilon}_{abcd} J^d = (\ast J)_{abc} \tag{1}$$ The last equality is true by definition. Finally, we wish to write the LHS in terms of $\ast F$. To do this, we write $$ F^{ed} = -\frac{1}{2} \varepsilon^{edmn} (\ast F)_{mn} $$ Then, $$ (\ast J)_{abc} = \frac{1}{2} \partial_e \left( {\varepsilon}_{abcd}\varepsilon^{demn} (\ast F)_{mn} \right) $$ Then, we use the property $$ {\varepsilon}_{abcd}\varepsilon^{demn} = 6 \delta^e_{[a} \delta^m_b \delta^n_{c]} $$ Finally, $$ (\ast J)_{abc} = 3 \delta^e_{[a} \delta^m_b \delta^n_{c]} \partial_e (\ast F)_{mn} = 3 \partial_{[a} (\ast F)_{bc]} = ( d \ast F )_{abc} $$ where again, the last equility is the definition of $d$. Thus, we see that $\nabla_a F^{ab} = J^b \implies d \ast F = \ast J$.

QED.

PS - To answer your last question. The differential form notation is aware of the connection through the Hodge dual in which $\sqrt{\det g}$ enters. Note also that in the divergence of any $p$-form, only the following component of the connection appears - $\Gamma^a_{ab}$ which depends entirely on $\sqrt{\det g}$. Other components never appear, i.e. in full generality $$ \nabla_a T^{[abc\cdots]} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} T^{[abc\cdots]} \right)~. $$

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  • $\begingroup$ @Prabar Thank you for your comment in my post physics.stackexchange.com/questions/358050/… I now see why I obtained a result which differs from Andi's expression by a sigh. The reason is that you used a non-standard definition of Hodge star in Eq.(1). $\endgroup$ – Wein Eld Sep 19 '17 at 15:35
  • $\begingroup$ @Prabar Usually the definition is $(*J)_{abc}=J^d \varepsilon_{dabc}$ which differs from Eq.(1) by a sign. (For this definition, see Nakahara or John Baez etc.) $\endgroup$ – Wein Eld Sep 19 '17 at 15:37
  • $\begingroup$ @WeinEld - That's right, but this is the convention used by Strominger in the notes. $\endgroup$ – Prahar Sep 19 '17 at 15:42
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The two formulations (1) & (2) are equivalent, mainly because:

  1. the covariant and the partial derivative of an antisymmetric $(0,2)$ tensor $F_{\mu\nu}$ is equivalent for a torsion-free connection.

  2. the Levi-Civita connection preserves the metric $\nabla_{\lambda} g_{\mu\nu}=0$.

  3. OP's eq. (2a) reads in local coordinates $$ \pm J^{\nu}~=~\nabla_{\mu} F^{\mu\nu}~\equiv~\partial_{\mu} +\Gamma_{\mu\lambda}^{\mu}F^{\lambda\nu} +\Gamma_{\mu\lambda}^{\nu}F^{\mu\lambda}~=~\frac{1}{\sqrt{|g|}} \partial_{\mu}(\sqrt{|g|} F^{\mu\nu})~=~(\delta F)^{\nu} \tag{2a}$$ in Minkowski signature $(\pm, \mp,\mp,\mp)$. Here $\delta$ is the Hodge co-differential, up to sign conventions.

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    $\begingroup$ Thanks! Do you know of any resources where the equivalence is made explicit? $\endgroup$ – MrRobot Jul 18 '17 at 11:06
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    $\begingroup$ @MrRobot Zee, Einstein Gravity in a Nutshell. $\endgroup$ – Ryan Unger Jul 18 '17 at 12:20
  • $\begingroup$ @MrRobot: It's also an exercise in Chapter 4 of Wald's General Relativity. $\endgroup$ – Michael Seifert Jul 18 '17 at 17:26
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If $\nabla$ is a torsion-free connection (not necessarily Levi-Civita), then for an arbitrary $k$-form $\omega_{a_1...a_k}$ we have $(d\omega)_{a_1...a_{k+1}}=(k+1)\partial_{[a_1}\omega_{a_2...a_{k+1}]}=(k+1)\nabla_{[a_1}\omega_{a_2...a_{k+1}]}$. If you expand in terms of connection coefficients, the symmetric connection coefficients will be killed by the antisymmetrization.

So $\nabla_{[a} F_{bc]}=0\Leftrightarrow dF=0$.

For the first relation $\nabla_a F^{ab}=J^b$, look up the codifferential. It is defined as $\delta\omega=(-1)^k\star^{-1}d\star\omega$, so it is basically $\delta=\pm\star d\star$ with the usual sign clusterf*ck one has to deal with when using the Hodge star. The codifferential is in some sense the "adjoint" operator of $d$, and it decreases the degree of a differential form by one, and also knows $\delta\delta=0$ (and I think its own version of the Poincaré-lemma as well).

It can be shown (see for example General Relativity by Norbert Straumann), that the codifferential acts on a differential form by taking its divergence via the Levi-Civita connection (remember that the Hodge star and so the codifferential requires a metric, so it also "sees" the Levi-Civita connection), so $(\delta\omega)^{a_1...a_{k-1}}=\pm\nabla_a\omega^{aa_1...a_{k-1}}$ (once again annoying signs).

Now, the Maxwell equation in question is given by $d\star F=\mathcal{J}$, but then $\mathcal{J}$ is a 3-form here. So let us instead have $\mathcal{J}=\star J$, then $$ d\star F=\star J, \\ \star^{-1} d\star F=J=\delta F. $$ But by the previous discussion $(\delta F)^b=\pm\nabla_aF^{ab}$, so your formula is given.

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The calculation goes as follows 1) dF =0:
$F_{[\mu,\nu;\lambda]}=\frac{1}{3}(F_{\mu\nu;\lambda}+F_{\nu\lambda;\mu}+ F_{\lambda\mu;\nu})= \frac{1}{3}(F_{\mu\nu,\lambda}-\Gamma^\tau_{\mu\lambda}F_{\tau\nu}- \Gamma^\tau_{\nu\lambda}F_{\mu\tau} + F_{\nu\lambda,\mu}-\Gamma^{\tau}_{\nu\mu}F_{\tau\lambda}-\Gamma^{\tau}_{\lambda\mu}F_{\nu\tau} + F_{\lambda\mu,\nu} - \Gamma^{\tau}_{\lambda\nu}F_{\tau\mu} - \Gamma^{\tau}_{\mu\nu}F_{\lambda\tau})=\frac{1}{3}(F_{\mu\nu,\lambda}+F_{\nu\lambda,\mu}+ F_{\lambda\mu,\nu})$ making use of the antisymmetry of the 2 indices of the electromagnetic field tensor and the symmetry of the 2 lower indices of the Christoffel-symbols (Torsion assumed to be zero) the terms with the Christoffel-symbols cancel out.

2) d*F=J Probably similar, however, at the moment I cannot work it out.

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