I am a little confused over integration using proper volume element. When do we use $\sqrt{-g}$ in calculations?

For example, in many calculations involving stars, say when using TOV equation, this never appears even when doing integral over density $\rho(r)$ to obtain $M(r)$, where $M(r)$ is enclosed mass. Conceptually I know it is merely change of coordinates since $\sqrt{-g}$ is a Jacobian of sorts. Could anyone provide an explicit instance when this calculation is necessary?


The coordinate invariant volume element/measure on a manifold with metric $g$ is $$ d^d x \sqrt{|g|} $$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$ d^d x' \sqrt{g'} = d^d x \sqrt{g} $$ In general relativity, every physical quantity should be coordinate invariant. It is therefore convenient to work with coordinate invariant quantities, even in intermediate steps so that we can be certain our final result is diffeomorphism invariant.

The factor of $\sqrt{|g|}$ comes from the fact that the volume form is (up to a normalization) $$ \varepsilon_{\mu_1 \cdots \mu_d} = \sqrt{|g|} {\tilde \varepsilon}_{\mu_1 \cdots \mu_d} ~. $$ where $ {\tilde \varepsilon}_{\mu_1 \cdots \mu_d} $ is a completely anti-symmetric form with $$ {\tilde \varepsilon}_{012\cdots(d-1)} = +1 $$ It can be clearly shown that only with the factor of $\sqrt{|g|}$ above is the quantity $\varepsilon_{\mu_1 \cdots \mu_d}$ a tensor.

  • $\begingroup$ thanks, I get the physical motivation how it works mathematically but most of physical calculations I encounter actually do not end up needing $sqrt{-g}$, except in the trivial case when I, say, change coordinates (since it is Jacobian). Do you have a situation (e.g. solving some equations of motion) that uses this? $\endgroup$ – Everiana May 15 '16 at 21:00

Prahar is correct, but here are two more things to note.

If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable (pick an orientation) and the frame is consistently oriented, there is a unique $n$-form $\omega$ such that $\omega(E_1,\dotsc, E_n)=1$ wherever the frame is defined. If $\{x^i\}$ are coordinates in a chart consistently oriented with the frame, then $\omega=\sqrt{\rvert g\lvert}\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ in the domain of the chart. For a proof, see Lee's Introduction to Smooth Manifolds. We can then use this form as a measure on $M$, and can define the integral of a function by $\int_Uf:=\int_Uf\omega$ where $U\subset M$ is $n$-dimensional.

We can also define an $C^\infty(M)$-linear isomorphism between the modules of $p$-forms and $(n-p)$-forms called the Hodge dual ($\star$), with some special properties. In particular, $\star:C^\infty (M)\to \Omega^n(M)$ defines a volume measure if $M$ is orientable by $\star (1)$. One can show that $\star(1)=\omega$ (from above). For a proof, see Jost's Riemannian Geometry and Geometric Analysis. Thus $\int_Uf:=\int_U\star f=\int_Uf\star(1)=\int_Uf\omega$ is also a viable integral.


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