Metric Volume, Divergence, and Integration

A nondegenerate metric converts an orientation into a canonical volume form. In coordinates its coefficient is $\sqrt{\lvert\det g\rvert}$, whose Jacobian factor precisely compensates that of the coordinate wedge. The same density defines divergence as local volume change; for the Levi-Civita connection it equals $\nabla_\mu X^\mu=\lvert g\rvert^{-1/2}\partial_\mu(\lvert g\rvert^{1/2}X^\mu)$. Its flux identity is an immediate instance of Stokes’ theorem.

Required background. Tensor Fields and Natural Operations supplies tensor components, determinants under coordinate change, and vector fields.

Helpful background. Calculus and Linear Algebra for Relativity supplies Jacobians; Exterior Algebra and Differential Forms supplies orientation and Stokes’ theorem; Events, Intervals, and Minkowski Geometry supplies the flat Lorentzian normalization.

The metric volume form

Let $(M,g)$ be an oriented pseudo-Riemannian $n$-manifold and write $g=\det(g_{\mu\nu})$ in a positively oriented chart. The metric volume form is

$$\mathrm{vol}_g =\sqrt{\lvert g\rvert}\, dx^1\wedge\cdots\wedge dx^n.$$

For a four-dimensional Lorentzian metric of signature $(-,+,+,+)$, $g\lt0$ and the site orientation convention becomes

$$\mathrm{vol}_g =\sqrt{-g}\,dx^0\wedge dx^1\wedge dx^2\wedge dx^3, \qquad \epsilon_{0123}=+\sqrt{-g}.$$

Why is this coordinate independent? Under $x=x(x')$, the metric transforms as $g'=J^{\mathsf T}gJ$, so $\sqrt{\lvert g'\rvert}=\lvert\det J\rvert\sqrt{\lvert g\rvert}$. Meanwhile the ordered coordinate wedge transforms by the inverse Jacobian. In orientation-preserving overlap maps the two factors cancel. An orientation-reversing map changes the sign of the ordered wedge, which is why an orientation is needed for a volume form. Without one, $\sqrt{\lvert g\rvert}\,d^nx$ is still a density that can be integrated.

The relation between metric determinant, density, and orientation is treated in Tong 2019, §3.1.3, pp. 95–96, PDF.

Divergence from volume change

The divergence of a vector field $X$ can be defined without first choosing coordinates by the infinitesimal change of volume:

$$\mathcal L_X\mathrm{vol}_g =(\operatorname{div}_gX)\,\mathrm{vol}_g.$$

In a positively oriented coordinate chart, the Lie derivative of a top form gives

$$\operatorname{div}_gX =\frac{1}{\sqrt{\lvert g\rvert}} \partial_\mu\!\left(\sqrt{\lvert g\rvert}\,X^\mu\right).$$

For the Levi-Civita connection this equals the trace $\nabla_\mu X^\mu$. Indeed,

$$\nabla_\mu X^\mu =\partial_\mu X^\mu+\Gamma^\mu{}_{\mu\nu}X^\nu, \qquad \Gamma^\mu{}_{\mu\nu} =\partial_\nu\log\sqrt{\lvert g\rvert}.$$

The last identity follows by inserting the Christoffel formula and using $\partial_\nu\log\lvert\det g\rvert =g^{\alpha\beta}\partial_\nu g_{\alpha\beta}$. Tong derives these identities in Tong 2019, §3.2.4, pp. 111–113, PDF. For an arbitrary affine connection with torsion or nonmetricity, its traced covariant derivative need not agree with the volume-change definition; the Levi-Civita specialization is consequential.

Polar-coordinate check

Flat Euclidean two-space provides a sharp coordinate test. In polar coordinates,

$$ds^2=dr^2+r^2d\phi^2, \qquad \sqrt{g}=r.$$

For a radial field $X=f(r)\partial_r$,

$$\operatorname{div}X =\frac1r\frac{d}{dr}\bigl(rf(r)\bigr) =f'(r)+\frac{f(r)}r.$$

In particular, $\operatorname{div}\partial_r=1/r$ on the punctured plane, even though the displayed component $X^r=1$ is constant. The missing term records the widening of radial rays. The qualification matters: $\partial_r=(x\partial_x+y\partial_y)/r$ is undefined at the origin, so it is not a smooth vector field on a full disk. On the annulus $A_{\varepsilon,R}$,

$$\int_{A_{\varepsilon,R}}\operatorname{div}\partial_r\,\mathrm{vol} =2\pi(R-\varepsilon).$$

The outer-circle flux is $2\pi R$, while the annulus’s outward normal on the inner circle is $-\partial_r$, giving flux $-2\pi\varepsilon$. Their sum matches the bulk integral. For a smooth full-disk check, take $X=r\partial_r=x\partial_x+y\partial_y$. Then $\operatorname{div}X=2$, so both the bulk integral and the boundary flux equal $2\pi R^2$. These two calculations independently check the density formula, its normalization, and the regularity hypothesis in the divergence theorem.

Boundary flux and its hypotheses

Let $D$ be a compact oriented $n$-dimensional region with piecewise smooth boundary. Stokes’ theorem applied to the $(n-1)$-form $\iota_X\mathrm{vol}_g$ gives the signature-independent statement

$$\int_D(\operatorname{div}_gX)\,\mathrm{vol}_g =\int_{\partial D}\iota_X\mathrm{vol}_g.$$

The boundary orientation is outward-normal-first. If a boundary component is non-null and has outward unit normal $N$, set $\epsilon=N^2=\pm1$ and define $\mathrm{vol}_{\partial D}=\iota_N\mathrm{vol}_g$. Since the normal component of $X$ is $\epsilon g(X,N)N$, the right side is

$$\int_{\partial D}\epsilon\,g(X,N)\,\mathrm{vol}_{\partial D}.$$

The factor is $+1$ on a timelike boundary and $-1$ on a spacelike boundary. Some conventions absorb it into a directed surface element; the displayed formula keeps it explicit. A null boundary has no unit normal and a degenerate induced metric; the differential-form flux remains valid, but this unit-normal formula cannot be copied unchanged.

Neither compact support nor a compact domain may be silently dropped: noncompact integrals require convergence and control of flux at infinity. Coordinate volume $d^nx$ alone is not invariant. The expression $\sqrt{-g}$ is real when the determinant is negative, including Lorentzian index one; $\sqrt{\lvert g\rvert}$ covers every nondegenerate signature.

Cartan’s formula makes the relation to Stokes especially transparent. Since an $n$-form on an $n$-manifold has zero exterior derivative,

$$\mathcal L_X\mathrm{vol}_g =d(\iota_X\mathrm{vol}_g).$$

Comparing this with the definition of divergence and integrating immediately produces the flux identity. This route uses the metric only to construct $\mathrm{vol}_g$; after that, the theorem is an orientation statement.

For a scalar field $\phi$, the same density formula gives the Laplace–Beltrami operator in a Riemannian metric or the wave operator in the Lorentzian convention:

$$\Box\phi =\nabla_\mu\nabla^\mu\phi =\frac1{\sqrt{\lvert g\rvert}} \partial_\mu\!\left( \sqrt{\lvert g\rvert}\,g^{\mu\nu}\partial_\nu\phi \right).$$

This expression is valid for the Levi-Civita connection. The absolute value inside the square root keeps it meaningful for any fixed nondegenerate signature; in four-dimensional Lorentzian applications it is $\sqrt{-g}$. A coordinate singularity can make the density vanish, as $r^2\sin\theta$ does on a polar axis, without making the metric geometrically degenerate there—the chart itself has failed at that locus. By contrast, a zero determinant in a regular chart signals a genuinely degenerate metric.

Exercises

In three-dimensional Euclidean space with spherical coordinates, $ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta\,d\phi^2$, let $X=f(r)\partial_r$ extend smoothly through $r=0$. Compute its divergence and verify the flux identity on a ball of radius $R$.

Solution

Here $\sqrt{g}=r^2\sin\theta$, so

$$\operatorname{div}X =\frac{1}{r^2\sin\theta} \partial_r(r^2\sin\theta\,f) =\frac1{r^2}\frac{d}{dr}(r^2f).$$

Smooth extension of a rotationally invariant radial field requires $f(r)=r h(r^2)$ for a smooth function $h$ near the origin; in particular, $r^2f(r)\to0$. Integrating over the ball gives $4\pi\int_0^R d(r^2f)/dr\,dr=4\pi R^2f(R)$. This equals the outward boundary flux $f(R)\operatorname{Area}(S_R^2)$.

References

• Tong, David. General Relativity. University of Cambridge lecture notes, 2019, §3.1.3, pp. 95–96, and §3.2.4, pp. 111–113. Official course PDF.