Divergence theorem step

[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!

 

[martinbn]

The theorem uses a unit normal vector, yours is not. In the first step it is not $n$.

 

[etotheipi]

Ah, so did I need to use a normal vector which is unit with respect to the Riemannian metric $\delta_{\mu \nu}$, e.g. $m^{\mu} = (0,\dots, 1)$, and not $n^{\mu}$ which is unit with respect to the $g_{\mu \nu}$ metric? These would be related by $n_{\mu} = N m_{\mu}$ and I could write$$\int_M d^n x \partial_{\mu}(\sqrt{g} X^{\mu}) = \int_{\partial M} d^{n-1}x \sqrt{\gamma N^2} X^{\mu} m_{\mu} = \int_{\partial M} d^{n-1}x \sqrt{\gamma} n_{\mu} X^{\mu}$$which is what we want...