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In E&M in Minkowski space, the Lorenz and Coulomb gauges are typically used since they make things vastly simpler. On a curved background, Maxwell's equations (without sources) can be written as:

\begin{align} \nabla_a F^{a b} &= 0 \\ F_{a b} &= \nabla_a A_b - \nabla_b A_a = \partial_a A_b - \partial_b A_a. \end{align}

Assuming the Lorenz condition, this can be written in the visually pleasing form:

\begin{align} \Box A^a = R^a_bA^b. \end{align}

Unfortunately, the coordinate expression of this equation in general coordinates is not very pretty. Conversely, using the fact that $F_{ab}$ is antisymmetric, we can write Maxwell's equations (in any gauge) as,

\begin{align} \partial_a\left(\sqrt{-g}g^{a c}g^{b d}\left(\partial_c A_d - \partial_d A_c\right)\right) = 0 \end{align}

which, upon identifying the variables,

\begin{align} A_{ab} &= \partial_b A_a \\ \Pi^i &= \sqrt{-g}g^{t c}g^{i d}\left(A_{d c} - A_{c d}\right) \\ \end{align}

can be nicely decomposed into first order form,

\begin{align} \partial_t \Pi^i &= -\partial_j \left(\sqrt{-g}g^{j c}g^{i d}\left(A_{d c} - A_{c d}\right)\right) \\ \partial_t A_{a b} &= \partial_b A_{at} \\ \partial_i \Pi^i &= 0. \end{align}

I may have gotten these slightly wrong as I am going mostly from memory and it is pretty late, but the general idea still stands (You invert the definition of the $\Pi^i$'s to get the $A_{it}$'s). If one evolves these equations as is, you are forced to solve the elliptic equation to get the $A_t$ component (which is computationally expensive). If you instead use the Lorenz gauge ($\nabla_a A^a = 0$), you can recover a hyperbolic form of the equations which is relatively simple in general coordinates.

The thing is, adopting the Weyl/temporal gauge ($A_t=0$) makes the equations much simpler than the Lorenz gauge and is still hyperbolic. Is there some property of the Weyl/temporal gauge that makes it unsuitable for this kind of calculation which would explain why it is not more commonly used?

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  • $\begingroup$ The Weyl (or "temporal") gauge is sometimes used, but people are reluctant to use it more generally because it's not a relativistically covariant gauge condition. However, as with all gauge choices, none is "better" than any other, certain ones are just more convenient for certain situations where "convenient" is a matter of personal opinion. Therefore, I think this question is primarily opinion-based. $\endgroup$
    – ACuriousMind
    Sep 8, 2016 at 13:56
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    $\begingroup$ I am just wondering if I am missing something obvious. The Coulomb gauge isn't relativistic covariant either, but it and it's generalisations are commonly used. I am just looking for a reason; perhaps when implemented computationally it is in practice difficult to maintain the elliptic constraint? The point is that I don't know and I am just trying to find others thoughts on the matter. $\endgroup$
    – Base
    Sep 8, 2016 at 15:31
  • $\begingroup$ In general, I am looking for a gauge, such that Maxwell's equations in an arbitrary background can be written as a system of hyperbolic PDEs with an elliptic constraint on the hypersurface where the initial data is defined. I would also like this system to be as simple as possible, so I would like suggestions for good gauge choices. $\endgroup$
    – Base
    Sep 8, 2016 at 15:35

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Since OP mentions canonical momenta $\pi$, it seems relevant to recall the Hamiltonian formulation of E&M in a curved spacetime. The Dirac-Bergmann analysis yields a primary constraint $\pi^0\approx 0$ and a secondary constraint (Gauss law), which both become first class constraints, and which generate$^1$ Hamiltonian gauge symmetry.

From a Hamiltonian perspective, there are now at least 2 ways to proceed:

  1. We immediately discard the $\pi^0$ variable, i.e. we work with 7 variables $(A_{\mu}; \pi^1,\pi^2,\pi^3)$. Then the Weyl/temporal gauge $A_0\approx 0$ is a bad gauge choice because it is not transversal to the Hamiltonian gauge symmetry, i.e. it does not intersect the gauge orbits.

  2. We keep the $\pi^0$ variable, i.e. we work with 9 variables $(A_{\mu}; \pi^{\nu};\lambda)$, where $\lambda$ is a Lagrange multiplier to impose the first class constraint $\pi^0\approx 0$. In this case we have 2 gauge symmetries, so we need 2 gauge-fixing conditions, i.e. the Weyl/temporal gauge $A_0\approx 0$ is an incomplete gauge choice.

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$^1$ For a secondary opinion, check out this Phys.SE post. Note that the Lagrangian gauge symmetry $\delta A_{\mu}=d_{\mu}\Lambda$ and the Hamiltonian gauge symmetry are not identical.

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