Lie derivative of volume form The volume from on an oriented 4-dimensional (pseudo-) Riemannian manifold $(M,g)$ is given by $$\Omega :=\sqrt{\lvert g\rvert}\,dx^{0}dx^{1}dx^{2}dx^{3}:=\sqrt{\lvert g\rvert}\,d^{4}x$$ where $g=\text{det}(g_{\mu\nu})$ is the determinant of the metric tensor $g_{\mu\nu}$ (apologies for my sloppy physicist notation).
Given a diffeomorphism $\phi:M\rightarrow M$, the volume form $\Omega$ should change by a Lie derivative, i.e. $$\delta\Omega =\mathcal{L}_{X}\Omega$$ where $X$ is the vector field generating the diffeomorphism. Now I know that $$\delta\left(\sqrt{\lvert g\rvert}\right)=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right) =\frac{1}{2}\sqrt{\lvert g\rvert}g^{\mu\nu}\delta g_{\mu\nu}=\sqrt{\lvert g\rvert}g^{\mu\nu}\nabla_{\mu}X_{\nu}$$ where I have used that $\delta g_{\mu\nu}=\mathcal{L}_{X}g_{\mu\nu}=\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}$. And so, so far I have $$\mathcal{L}_{X}\Omega=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right)d^{4}x+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)=g^{\mu\nu}\nabla_{\mu}X_{\nu}\,\Omega+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)$$
However, I'm not sure how $d^{4}x$ transforms, so my question is: what is the Lie derivative of $d^{4}x$? i.e. what is $$\mathcal{L}_{X}\left(d^{4}x\right)\;?$$ I'm a physicist and so my knowledge of differential geometry is not that extensive, any help would be much appreciated.
 A: Let $X = X^i \partial_i$ be a vector field generating a flow $\phi_t$.
Then you can use three methods to compute $\mathcal L_X \Omega$ :
$$
\mathcal L_X \Omega = \left.\frac{d}{d t}\right|_{t=0} \phi_t^*\Omega
$$
or Cartan's magic formula :
$$
\mathcal L_X \Omega = \iota_X d \Omega + d \iota_X \Omega \\
= d \iota_X \Omega \\
= d(\sqrt{|\det g_{ij}|} \iota_X d^4 x)
$$
or even use Leibniz rule then Cartan's formula :
$$
\mathcal L_X \Omega
= \mathcal L_X (\sqrt{|\det g_{ij}|}d^4 x) \\
= \mathcal L_X (\sqrt{|\det g_{ij}|})d^4 x + \sqrt{|\det g_{ij}|}\mathcal L_X (d^4 x) \\
= \left(\iota_X d \sqrt{|\det g_{ij}|}\right)d^4 x + \sqrt{|\det g_{ij}|} d \iota_X \left(d^4 x\right)
$$
Those two last methods are equivalent and are direct to compute using $X=X^i \partial_i$.
Lets consider the last method.
Lets fix $f:=\sqrt{|\det g_{ij}|}$ for clarity.
Then :
$$
\mathcal L_X \Omega = \left(\iota_X d \sqrt{|\det g_{ij}|}\right)d^4 x + \sqrt{|\det g_{ij}|} d \iota_X \left(d^4 x\right) \\
= \left(\iota_X d f\right)d^4 x + f d \iota_X \left(d^4 x\right) \\
= (X^i \partial_i f)d^4 x + f d(X^1 dx^2 \wedge dx^3\wedge dx^4 - ...+... - X^4 dx^1 \wedge dx^2 \wedge dx^3) \\
= (X^i \partial_i f)d^4 x + f (\partial_1 X^1 dx^1 \wedge dx^2 \wedge dx^3\wedge dx^4 + ... + \partial_4 X^4 dx^1 \wedge dx^2 \wedge dx^3\wedge dx^4) \\
= (X^i \partial_i f)d^4 x + f (\partial_i X^i)d^4 x \\
= (X^i \partial_i f)d^4 x +  (\partial_i X^i) \Omega \\
$$
Now, using derivative formulas on $f$ you can continue from here.
