I want to derive the inhomogeneous Maxwell's equations on curved space time from the principle of equivalence. I assume that for a specific space-time point there is a coordinate system $\xi^\alpha$ such that in that specific coordinate system the Maxwell's equations take their flat space time form:

$$ \frac{\partial F^{\mu\nu}(\xi)}{\partial \xi^\mu}=J^\nu(\xi) $$

Now I do a coordinate transformation to an arbitrary coordinate system $x^\mu$, where the $x^\mu$'s are a function of the $\xi$'s: $x^\mu(\xi^\alpha)$. This relation is also invertible: $\xi^\alpha(x^\mu)$.

To derive the Maxwell's equations in the new coordinate frame I first look at $F_{\mu\nu}(\xi)$. It is define around the space time point that we consider trough the charges that a local observer sees:

$$F^{\mu\nu}(\xi)=\frac{\partial A^\nu(\xi)}{\partial \xi_{\mu}}-\frac{\partial A^\mu(\xi)}{\partial \xi_{\nu}}$$.

Although I don't understand why (?) I now assume that the vector potential $A_\mu$ transforms as:

$$A^\mu(\xi)=\frac{\partial \xi^\mu}{\partial x^\alpha}\tilde{A}^\alpha(x)$$

I get then:

$$F^{\mu\nu}(\xi)=\frac{\partial}{\partial \xi_\mu}\left(\frac{\partial \xi^\nu}{\partial x^\alpha}\tilde{A}^\alpha(x)\right)-\frac{\partial}{\partial \xi_\nu}\left(\frac{\partial \xi^\mu}{\partial x^\alpha}\tilde{A}^\alpha(x)\right)\\= \frac{\partial^2 \xi^\nu}{\partial x^\alpha \partial \xi_\mu}\tilde{A}^\alpha(x) + \frac{\partial \xi^\nu}{\partial x^\alpha}\frac{\partial \tilde{A}^\alpha(x)}{\partial x^\beta }\frac{\partial x^\beta}{\partial\xi_\mu}-\frac{\partial^2 \xi^\mu }{\partial x^\alpha\partial \xi_\nu}\tilde{A}^\alpha(x)-\frac{\partial \xi^\mu}{\partial x^\beta}\frac{\partial \tilde{A}^\beta(x)}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial\xi_\nu}$$

This doesn't show that $F^{\mu\nu}$ is a vector? Ideally I want to have:

$$F^{\mu\nu}(\xi)=\frac{\partial \xi^\mu}{\partial x^\alpha}\frac{\partial \xi^\nu}{\partial x^\beta}\left(\frac{\partial \tilde{A}^\alpha}{\partial x_\beta}-\frac{\partial \tilde{A}^\beta}{\partial x_\alpha}\right)$$

But I cannot see how I should get this from what I have shown above. So right now the problem is to show that $F^{\mu\nu}$ is a tensor in general relativity.

When I have this I can plug this into Maxwell's equations and get the form in a general coordinate system. I should get:


where $;$ denotes the covariant derivative. Written out this is:

$$\frac{\partial F^{\mu\nu}}{\partial x^\mu}+\Gamma^\mu_{\mu \sigma} F^{\sigma\nu}+\Gamma ^{\nu}_{\mu\sigma}F^{\mu\sigma}=-J^\nu$$

  • 1
    $\begingroup$ Just note there are quite a few errors in your equations with indices. Similarly, you have terms like $x/\xi$ which should be $\partial x / \partial \xi$. With these corrections you'll note your transformation for $F^{\mu \nu}$ does indeed give your desired result. $\endgroup$
    – Eletie
    Commented Mar 31, 2023 at 10:24
  • $\begingroup$ Thanks! I corrected a mistake in the transformation of the vector potential. But I still don't get the desired result. What am I doing wrong? $\endgroup$
    – jojo123456
    Commented Mar 31, 2023 at 18:37
  • $\begingroup$ Also do you have an argument why the vector potential transforms like a vector in general relativity? $\endgroup$
    – jojo123456
    Commented Mar 31, 2023 at 18:38

3 Answers 3


The components of $F^{\mu \nu}$ (and $J^\mu$) will also transform when you go to the new coordinate system. Specifically, you will have $$ \tilde{F}^{\mu \nu}(x) = \left[ \frac{\partial x^\mu}{\partial \xi^\alpha} \frac{\partial x^\nu}{\partial \xi^\beta} F^{\alpha \beta}\right]_{\xi(x)} \neq F^{\mu \nu}(\xi(x)) $$ (even at a fixed point in space.)

  • $\begingroup$ Ok thanks! But how do I know that the transformation under a general coordinate transformation looks like that? I know that $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. But then I still need the transformation of the gauge fields to get what you wrote down? $\endgroup$
    – jojo123456
    Commented Mar 31, 2023 at 0:22
  • $\begingroup$ @jojo123456: See here for the general rules. (Most GR textbooks will also have these transformation laws in there somewhere.) $\endgroup$ Commented Mar 31, 2023 at 1:02
  • $\begingroup$ My question is actually a bit more general and I did not find the answer in the standard text books. $F_{\mu\nu}(\xi)= \frac{\partial A_\mu(\xi)}{\partial \xi^\nu}-\frac{\partial A_\nu(\xi)}{\partial \xi^\mu}$. When I transform this to the frame $x$ I get $ \frac{\partial \tilde{A}_\mu(x)}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial\xi^\nu}-\frac{\partial \tilde{A}_\nu(x)}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial \xi^\mu}$. So I still need the transformation of the vector potentials under a general coordinate transformation. But I don't know what that is $\endgroup$
    – jojo123456
    Commented Mar 31, 2023 at 1:15
  • $\begingroup$ @jojo123456: The general transformation for a constant vector is in the link I gave above. Or all I misunderstanding what you're looking for? $\endgroup$ Commented Mar 31, 2023 at 1:36
  • $\begingroup$ I guess my question is why $A_\mu$ is a vector in general relativity? In special relativity this is clear to me but not in general relativity. $\endgroup$
    – jojo123456
    Commented Mar 31, 2023 at 1:38

Locally there exist coordinates where the inhomogeneous Maxwell equations may be represented as they are in Minkowski space:

$$\frac{\partial F^{\alpha\beta}}{\partial x^\beta} = J^\alpha$$

Generally the partial derivatives of the components of a tensor are themselves not components of a tensor. So it's difficult to transform this expression to arbitrary coordinates. However in this local coordinate system, the Christoffel symbols vanish! We may therefore equivalently write this equation as

$$F^{\alpha\beta}_{;\beta} = J^\alpha$$

Now the components of the covariant derivative of a tensor are themselves components of a tensor! Therefore this expression of Maxwell's equations holds in arbitrary coordinates.

  • $\begingroup$ This uses the principle of general covariance (as defined by Weinberg) which I believe the questioner is trying to avoid, based on their previous question. It is quite an easy exercise, mentioned in the book by Weinberg, to derive the general equation in curved space using only the principle of equivalence. The other answer contains the reason the questioner is having trouble. $\endgroup$ Commented Mar 30, 2023 at 22:12
  • 1
    $\begingroup$ This does illustrate however, @jojo, why the principle of covariance is useful! Compare this 2 line argument with the 4 line argument based on the 'raw' principle of equivalence! $\endgroup$ Commented Mar 30, 2023 at 22:13

This is really a large comment on the correct answer above.

For context for others. I believe the questioner is really interested in the following problem:

  • Derive, using the principle of equivalence (in the Weinbergian sense), the general form of Maxwell equations in an arbitrary gravitational field, given that one knows the Maxwell equations in a non-gravitational frame.

This exercise originates from the first paragraph of chapter 4, section 1 in the book on gravitation by Weinberg. In that section, Weinberg is introducing the principle of general covariance, but beforehand he notes that one could in principle derive derive the form of the electromagnetic and gravitational equations purely from the principle of equivalence, rather than use the principle of general covariance.

In essence this is simple, if a little tedious - we write the Maxwell equations in inertial coordinates $\{\xi^\alpha\}$ and then perform a coordinate transform to general coordinates $\{x^\mu\}$, which by the principle of equivalence, must be the correct equation in that frame.

For this to match the correct formula, one must define $F^{\mu\nu}$ to be a tensor, transforming as $F^{\alpha \beta}(\xi) = \frac{\partial \xi^\alpha}{\partial x^\mu} \frac{\partial \xi^\beta}{\partial x^\nu} \tilde{F}^{\mu \nu}(x)$. That is, one should be able to define the (invariant) differential form $F = F_{\mu\nu} dx^\mu dx^\nu$.

Of course, a priori, $F^{\mu \nu}$ does not have to be a tensor. We could have defined $F^{\mu \nu}$ to be a scalar for each $\mu$ and $\nu$. We could then "covariantize" that equation by performing a general coordinate transformation from a non-gravitational frame. That would have given us a different physical equation (which would be experimentally wrong).

One way to argue that $F^{\mu \nu}$ should indeed be a tensor is as follows. Consider the Lorentz force \begin{equation} f^\alpha = e F^\alpha_{\;\;\beta} \frac{d x^\beta}{d\tau}\;. \end{equation} We would like this equation to be covariant, and let us say we know that $f^\alpha$ must be a vector. It then follows that $F^\alpha_{\; \;\beta}$ must be a tensor. (This is the same sort of argument that one uses to show that the magnetic field $B$ is a pseudovector, but kind of generalised.)

I suppose one could question why $f^\alpha$ need be a vector... but I suppose that follows from Newton's second law.

(Note that the charge $e$ is a scalar!)

Ps. Careful about defining $F$ as the derivative of a potential $A$. Usually, one develops the (classical) subject by starting with $F$ and deducing $A$ using Poincaré's lemma and the equation $dF = 0$. (Although this is not terribly important, and in QFT one typically does treat $F$ as derived from some gauge field $A$.)


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