Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have often seen $\sqrt{-g}$ in integrals, especially actions, where $g=\mathrm{det}(g_{\mu \nu})$. Does anyone know of a derivation that shows that this is indeed the volume element which must be utilized? I have searched online, but all the literature I have encountered assumes you accept it as the volume element.

share|cite|improve this question
Towards the end of Chapter 2 of Sean Carroll's lecture notes: – DJBunk May 9 '13 at 12:34
Doesn't the wikipedia article explain it sufficiently? It seems to have a nice simple example too. – twistor59 May 9 '13 at 12:45

I) Pragmatically speaking, the most important property of $\sqrt{-g}$ for model building purposes, is not per se the fact that $\sqrt{-g}d^{4}x$ measures the volume element of a 4-dimensional Parallelepiped with infinitesimal edges $dx^0, \ldots, dx^3$.

II) A more important property is that $\sqrt{-g}d^{4}x$ transforms as a scalar (i.e. is invariant) under general coordinate transformations $$x^{\prime \mu} ~=~ f^{\mu}(x^0 , \ldots, x^3).$$ In more detail, the factor $\sqrt{-g}$ transforms as a density under general coordinate transformations, cf. also this Phys.SE answer.

III) Example: The Einstein-Hilbert action

$$S_{EH}[g]~:=~\int R \sqrt{-g}d^{4}x$$

is invariant under change of coordinates, because the scalar curvature $R$ and volume element $\sqrt{-g}d^{4}x$ are both invariant under general coordinate transformations. Thus the action $S_{EH}[g]$ is a geometric quantity that doesn't depend on the choice or coordinates.

IV) The factor $\sqrt{-g}$ is the so-called canonical density on a Lorentzian manifold. If the Lorentzian manifold is endowed with another density $\rho$, one could in principle use that to build actions and get interesting theories.

V) Finally let us mention that on a symplectic manifold $(M,\omega)$, the canonical density is instead given by the Pfaffian

$$ {\rm pf}(\omega_{ij}) $$

of the symplectic matrix $\omega_{ij}$.

share|cite|improve this answer
Another thing one could mention is that given an (ordered) basis of orthonormal 1-forms $\theta_i$ this volume form can be written as $\theta_1 \wedge \theta_2 \wedge \dots \wedge \theta_n$ up to sign (which depends on choice of orientation and the signature of the metric) – alexarvanitakis May 9 '13 at 20:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.