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I read this formula in Sean Carroll's book of GR:


where n is the 4-vector orthogonal to the hypersurface $\partial\Sigma$ and $\gamma$ the induced metric defined by

$$\gamma_{ij} = \partial_i X^{\mu}\partial_jX^{\nu}g_{\mu\nu}.$$

The only trouble I have is with the appearance of this induced metric, because I have the formula


which then should give me straightforwardly


by the classical Stokes' theorem. I was thinking that it may be something with the differential element: going from $d^nx$ to $d^{n-1}x$ forced the apparition of this indeuced metric but I didn't manage to figure out how so far.

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up vote 7 down vote accepted

There are a couple of problems you're running into: first, $\sqrt{g}$ is part of the integration measure. When you're relating the integral over a manifold to the integral over its boundary, you need to use the appropriate integration measure for each region. And also, Stokes' theorem is defined in terms of a covariant derivative $\nabla_\mu$, not a coordinate derivative $\partial_\mu$.

$$\int_{\Sigma}\partial_{\mu}(\cdots)\mathrm{d}^nx \neq \int_{\partial\Sigma}(\cdots)\mathrm{d}^{n-1}x$$

Basically, the "classical" Stokes' theorem, as you're thinking about it, just doesn't work for arbitrary manifolds, at least not with arbitrary coordinate systems. (You can get away with it by using Gaussian normal coordinates, which are specially constructed so that the integration measure on the boundary is a simple "dimensional reduction" of the measure on the manifold, but in that case $\sqrt{\gamma} = \sqrt{g}$.)

What Stokes' theorem really does is relate the integral of an $n$-form over a boundary to the integral of its exterior derivative over the enclosed submanifold.

$$\int_{\partial\Sigma}\omega = \int_{\Sigma}\mathbf{d}\omega$$

When you go to apply this, if $\omega$ is the dual of a vector field you get

$$\begin{align}\omega &= \color{blue}{n_\mu V^\mu}\color{red}{\sqrt{\gamma}\mathrm{d}^{n-1} y} \\ \mathbf{d}\omega &= \color{blue}{\nabla_\nu V^\nu}\color{red}{\sqrt{g}\mathrm{d}^{n} x}\end{align}$$

The details of how to work this out are in appendix E of Sean Carroll's book, but basically, note that the exterior derivative does two things: it effectively replaces the normal vector with a covariant derivative, and it replaces the integration measure of the hypersurface (including the factor of $\sqrt{\gamma}$) with that of the manifold.

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