What does general relativity's metric tensor have to do with quantum electrodynamics? Sabine Hossenfelder recently posted a YouTube video titled, The Closest we have to a Theory of Everything.
At 9:15, she shows the action $S$ for electrodynamics and, immediately after, the Einstein-Hilbert action for general relativity (with also the matter Lagrangian).
Both equations show the square root of negative $g$, $g$ being the determinant of the metric tensor.
I can't find the equation for the $S$ of electrodynamics anywhere else... It looks very similar to the regular $S$ used in QED, the glaring addition being the square root of $-g$...
Is this $S$ that is shown in the video the same $S$ used in QED?
Does this equation have a special name, as the Einstein–Hilbert one does?
 A: The action for QED in flat spacetime is
$$
S = \int d^4 x \left( - \frac{1}{4e^2} F_{\mu\nu} F^{\mu\nu} + i {\bar \psi} \gamma^\mu D_\mu \psi + m {\bar \psi} \psi\right) , \quad D_\mu = \partial_\mu + i A_\mu , \quad  F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu . 
$$
Here $\gamma_\mu$ is a $4\times 4$ Dirac matrix satisfying $\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu = - 2 \eta_{\mu\nu}$ and $\psi$ is the fermion field (it is a column vector) and ${\bar \psi} = \psi^\dagger \gamma^0$ is its so-called Dirac conjugate (it is a row vector). The action of electrodynamics is just the first term above.
In curved spacetimes with a general metric $g_{\mu\nu}$ it is a bit more complicated because we have a fermion field $\psi$. To construct such an action, we introduce a vielbein $e^\mu_a$ which satisfies $e^\mu_a e^\nu_b \eta^{ab} = g^{\mu\nu}$. We then have to introduce a spin connection $\omega$ which satisfies
$$
\partial_\mu e_\nu^a - \Gamma^\lambda_{\mu\nu} e^a_\lambda + \omega_\mu{}^a{}_b e^b_\nu = 0 .
$$
The action is then given by
$$
S = \int d^4 x \sqrt{-g} \left( - \frac{1}{4e^2} F_{\mu\nu} F^{\mu\nu} + {\bar \psi} e_a^\mu \gamma^a D_\mu \psi + m {\bar \psi} \psi\right) , \qquad D_\mu = \partial_\mu + \frac{1}{4} \omega_\mu^{ab} \gamma_a \gamma_b + i A_\mu . 
$$
A: As hinted in the comments and in the other answer, it is not the same action, but it is related. Let me be more specific. The action you're used to is probably
$$S = \int \left(- \frac{1}{4} F_{\mu\nu}F^{\mu\nu} + j_\mu A^{\mu}\right) \mathrm{d}^4x. \tag{1}$$
Sabine's version is
$$S = \int \left(- \frac{1}{4} F_{\mu\nu}F^{\mu\nu} + j_\mu A^{\mu}\right) \sqrt{-g} \mathrm{d}^4x. \tag{2}$$

Is this  that is shown in the video the same  used in QED?

The difference is that Eq. (1) is a special case of Eq. (2). Namely, Eq. (1) is Eq. (2) in the case where the metric $g_{\mu\nu}$ is the Minkowski metric $\eta_{\mu\nu}$ and the coordinates being used are Cartesian coordinates, a case in which $\sqrt{- \eta} = 1$ (in other coordinate systems, $\sqrt{-g}$ would be just the Jacobian). The $\sqrt{-g}$ factor is a modification necessary for formulating the action in arbitrary coordinates or in more general spacetimes (the latter probably being Sabine's interest in the video).

Does this equation have a special name, as the Einstein–Hilbert one does?

Not that I'm aware of.
As mentioned in Prahar's excellent answer and in DanielC's comment, it is a bit more complicated to correctly write down the action for QED specifically in curved spacetime, due to the need of correctly dealing with fermionic fields.
At last, I think it is worth mentioning that the actions in Eqs. (1) and (2) are not necessarily those of QED. They describe the electromagnetic field $A_\mu$ as coupled to a source $j_\mu$, which need not be a fermionic field. If one considers $j_\mu$ as an external source and extremize the action by varying $A_\mu$, one will arrive at the classical form of Maxwell's equations (either in flat or curved spacetime, depending on the action you employ).
