We know, the the vector form of Maxwell equations \begin{align} \vec\nabla\cdot\vec{E} &= 4\pi\rho \label{Diff I}\\ \vec\nabla\times\vec{B} &= \dfrac{4\pi}{c} \vec{j}+\dfrac{1}{c}\dfrac{\partial\vec{E}}{\partial t} \label{Diff IV}\\ \vec\nabla\times\vec{E} &= -\dfrac{1}{c}\dfrac{\partial\vec{B}}{\partial t} \label{Diff III}\\ \vec\nabla\cdot\vec{B} &= 0. \label{Diff II} \end{align}

The last two of them allow us to introduce the potentials: \begin{align} \vec{E} &= -\frac1c \frac{\partial \vec{A}}{\partial t} - \vec\nabla\phi\\ \vec{B} &= \vec\nabla\times\vec A \end{align} which tells us about gauge invariance of equations.

All four of Maxwell's equations can be written compactly as

\begin{align} \partial_{\mu}F^{\mu\nu} &= \frac{4\pi}{c}j^{\nu} \tag{1}\\ \partial_{[\mu}F_{\alpha\beta]} &= 0\;. \tag{2} \end{align}

And according to the last one equation, the first one we can rewrite (use preferred gauge) in form: \begin{equation} \Box A^{\mu} = -\frac{4\pi}{c} j^{\mu}. \end{equation}

Now we consider the Einstein GR equations: \begin{equation} R_{\mu\nu} = 8\pi G (T_{\mu\nu} - \frac12g_{\mu\nu}T). \end{equation}

Or in "$\Gamma-$field" form (indexes are omitted): \begin{equation} \partial \Gamma - \partial \Gamma + \Gamma\Gamma - \Gamma\Gamma = 8\pi G (T_{\mu\nu} - \frac12g_{\mu\nu}T). \end{equation}

We know, in weak field limit the equations get simply form externally similar to Maxwell ones (so called gravitomagnetism) in 3-vector form, or in covariant form: \begin{equation} \Box h_{\mu\nu} = -16\pi G (T_{\mu\nu} - \frac12\eta_{\mu\nu}T). \end{equation}

Thus, the question: Why for covariant form of Maxwell equations we need two different entities of equations, but for the GR the only one? Or another words, can we write the Einstein GR equations for weak field limit similar to Maxwell equations in field form, not via potentials?

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2 Answers 2


If you want to compare Maxwell's EM with GR, note they're respectively obtainable from extending a global $U(1)$ variance to a local one and Lorentz invariance to invariance under general coordinate transformations. Thus the equivalent of introducing $A_\mu$ in the gauge covariant derivative is introducing Christoffel symbols in the connection, while the commutator $F_{\mu\nu}$ of gauge covariant derivatives is analogous to the commutator $R_{\mu\nu\rho\sigma}$ of connections acting on a vector field. So the equivalent of (1) is Einstein's usual equations, while the equivalent of (2) is $R_{\mu[\nu\rho\sigma]}=0$. In neither case do we need the second equation; a Lagrangian formulation obtains the first, but the second is just a tautology.

  • $\begingroup$ Ok, as I understand, the answer, the Bianchi relation is the GR analog to the (2). $\endgroup$
    – Sergio
    Dec 24, 2018 at 8:24
  • $\begingroup$ I believe it is the other Bianchi identity $R^{\alpha}_{\;\beta[\mu\nu ;\sigma ]}=0$ that is the analog of the EM Bianchi identity. $\endgroup$
    – fewfew4
    Aug 14, 2021 at 0:07
  1. The counterpart to $(\text{metric }g,\text{curvature }R)$ in GR is $(\text{gauge potential }A,\text{field strength }F)$ in E&M.

  2. If you write $F=F(A)$ in E&M, then the Bianchi relation (2) is not needed/a tautology, cf. e.g. this Phys.SE post.

  3. For how GEM appears as a limit of GR, see e.g. this Phys.SE post.


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