Fix a closed hypersurface $\Sigma$ in a Riemannian $3$-manifold $(M,g)$. Let $\Sigma_t$ be a family of closed hypersurfaces evolving from $\Sigma$ by inverse mean curvature flow (IMCF) in $M$. That is, given $x\in\Sigma$, they obey $$\frac{d}{dt}\Phi_t(x)=-\frac{\mathbf{H}}{H^2}=\frac{1}{H}\nu,\tag{D}$$ where $\mathbf{H}$ is the mean curvature vector of $\Sigma_t$ and $\Phi_{t\in(-\epsilon,\epsilon)}:M\to M$ is a one-parameter family of diffeomorphisms such that $\Phi_0$ is the identity map and $\Phi_t(\Sigma)=\Sigma_t$. If $d\mu_{\Sigma_t}$ denotes the Riemannian volume form on $\Sigma_t$, I would like to show that $$\frac{\partial}{\partial t}d\mu_{\Sigma_t}=d\mu_{\Sigma_t}.\tag{1}$$ This formula is included in Geometric Relativity by Dan A. Lee, and he says that it can be inferred from Proposition 2.10, which reads:
The $X$ here is a vector field on $M$ defined by $$X(p)=\frac{d}{dt}|_{t=0}\Phi_t(p),$$ and that $\hat{X}$ is the tangential part of $X$ obtained from the decomposition relative to the unit normal vector field $\nu$.
Idea: Since we are talking about the hypersurface case with $X$ being purely normal, I would opt for (P2) with vanishing $\hat{X}$, which gives me $$\frac{d}{dt}|_{t=0}\mu(\Sigma_t)=\int_\Sigma H\varphi d\mu_\Sigma=\int_\Sigma d\mu_\Sigma=(\text{the area of $\Sigma$}).\tag{2}$$ This is still far from the desired equation (1). As a matter of fact, I don't even know how the author ditched that particular time instant $t=0$. Could somebody tell me what else he might have done to translate (2) into (1)? Thank you.
Added: Maybe I missed something important. The author said Proposition 2.10 tells us that for any $t$, $$\frac{d}{dt}(\Phi_t^*d\mu_{\Sigma_t})=\Phi_t^*(\mathrm{div}_{\Sigma_t}X_t)(\Phi_t^*d\mu_{\Sigma_t}),$$ where $X_t$ is defined by $$X_t(\Phi_t(p))=\frac{d}{dt}\Phi_t(p),$$ but I don't see how this works.
The following formula is well known and probably found in most expositions of minimal (hyper)surfaces: Let $M$ be a manifold with Riemannian metric $g_M$ and $\Sigma$ be an $(n-1)$-dimensional manifold. Let $\Phi_t: \Sigma \rightarrow M$ be a smooth family of embeddings and $g_t = \Phi_t^*g_M$ be the family of induced Riemannian metrics such that at $t=0$, $$ \partial_t\Phi = \phi \nu, $$ where $\phi$ is a smooth function on $\Sigma$ and $\nu$ is the unit normal to $\Sigma_0$ in the direction of $\partial_t\Phi$. Then the volume measure $d\mu_t$ of $g_t$ satisfies $$ \partial_td\mu_t = \phi H d\mu_t. $$
Here is a calculation using local coordinates $(x^1, \dots, x^{n-1})$ on $\Sigma$ that shows this: For $t \ge 0$, let \begin{align*} g_{ij} &= g_M(\partial_i\Phi_t,\partial_j\Phi_t). \end{align*} It follows that \begin{align*} \partial_tg_{ij} &= g_M(\nabla_t\partial_i\Phi,\partial_j\Phi) + g_M(\partial_i\Phi,\nabla_t\partial_j\Phi)\\ &= g_M(\nabla_i\partial_t\Phi,\partial_j\Phi) + g_M(\partial_i\Phi,\nabla_j\partial_t\Phi)\\ &= g_M(\nabla_i(\phi\nu),\partial_j\Phi) + g_M(\partial_i\Phi,\nabla_j(\phi\nu))\\ &= \phi(g_M(\nabla_i\nu,\partial_j\Phi)+g_M(\partial_i\Phi,\nabla_j\nu))\\ &= 2\phi A_{ij}, \end{align*} where $A_{ij}\,dx^i\,dx^j$ is the second fundamental form. On the other hand, \begin{align*} \partial_td\mu_t &= \partial_t(\sqrt{\det g})\,dx. \end{align*} The formula now follows from the standard formula for the derivative of the determinant (when it is positive): \begin{align*} \partial_t(\log\det M)&= \operatorname{trace} M^{-1}\partial_tM. \end{align*}
A key trick here is to pull everything back to the fixed manifold $\Sigma$ and view everything as time-dependent functions on $\Sigma$. Viewing them as functions on $\Sigma_t$ is more confusing for me, so I avoid it.
You'll have to verify carefully that each step is valid. If you haven't done calculations like this in the past, you might want to use local coordinates on $M$ as well. There are slicker ways to justify the calculation above, but I always recommend working it out in local coordinates first.