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#### etotheipi

I wanted to ask about a step I couldn't understand in Tong's notes$$\int_M d^n x \partial_{\mu}(\sqrt{g} X^{\mu}) = \int_{\partial M} d^{n-1}x \sqrt{\gamma N^2} X^n = \int_{\partial M} d^{n-1}x \sqrt{\gamma} n_{\mu} X^{\mu}$$we're told that in these coordinates ##\partial M## is a surface of constant ##x^n##, and further that ##g = \mathrm{det}(g_{\mu \nu}) = \gamma N^2## where ##\gamma_{ij}## is the pull-back of ##g_{\mu \nu}## to ##\partial M##. Also, the normal vector is ##n^{\mu} = (0, \dots, 1/N)##, or with downstairs components ##n_{\mu} = (0, \dots, N)##.

But, I thought the usual divergence theorem was stated$$\int_M d^n x \partial_{\mu} V^{\mu} = \int_{\partial M} d^{n-1}x V^{\mu} n_{\mu}$$in which case I'd get

$$\int_M d^n x \partial_{\mu}(\sqrt{g} X^{\mu}) = \int_{\partial M} d^{n-1}x \sqrt{\gamma N^2} X^{\mu} n_{\mu} = N \int_{\partial M} d^{n-1}x \sqrt{\gamma} n_{\mu} X^{\mu} $$i.e. I'd still have the extra factor of ##N##. What am I forgetting...? Thanks!

But, I thought the usual divergence theorem was stated$$\int_M d^n x \partial_{\mu} V^{\mu} = \int_{\partial M} d^{n-1}x V^{\mu} n_{\mu}$$in which case I'd get

$$\int_M d^n x \partial_{\mu}(\sqrt{g} X^{\mu}) = \int_{\partial M} d^{n-1}x \sqrt{\gamma N^2} X^{\mu} n_{\mu} = N \int_{\partial M} d^{n-1}x \sqrt{\gamma} n_{\mu} X^{\mu} $$i.e. I'd still have the extra factor of ##N##. What am I forgetting...? Thanks!

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