# Why does the factor $\sqrt{-g}$ make the volume element invariant?

My question is an extension on this and this question. The question is, how or "in what sense" does the factor $$\sqrt{-g}$$ make the measure invariant?

Suppose, I do not add this factor to the measure. This would cause the absolute value of the Jacobi determinate to appear in the integral. This seems fine to me. I can still calculate the integral. Why is the factor necessary?

• Yes, you can calculate things. But then you're calculating physics of something else. In GR, the relevant symmetry is general coordinate invariance; changing coordinates should not affect physics. Not adding $\sqrt{-g}$ changes the action, and therefore physics. Commented Jun 16, 2019 at 18:06
• $\sqrt{-g}$ is the absolute value of the determinant of the Jacobian matrix! Commented Jun 16, 2019 at 18:08
• Possible duplicates: physics.stackexchange.com/q/63950/2451 and links therein. Commented Jun 16, 2019 at 18:11
• No. See page 2 of reed.edu/physics/courses/Physics411/html/page2/files/…. Commented Jun 16, 2019 at 18:58
• $\det{g_{\mu\nu}}=(\det{J^\mu_\nu})^2$ Commented Jun 16, 2019 at 19:08

You include this factor $$\sqrt{-g}$$ in order to make the action invariant under diffeomorphisms. If you did not include it, it would be the integral of a scalar tensor density of weight different from zero (I think +1 or -1, I don't really remember how the convention works), meaning that as you said the measure would change of a determinant factor.
By including $$\sqrt{-g}$$ you remove this change, since $$g=det(g_{\mu\nu})\rightarrow det\left(g_{\tau\lambda}\frac{\partial x^{\tau}}{\partial x'^{\mu}}\frac{\partial x^{\lambda}}{\partial x'^{\nu}}\right)=g\,\cdot\,(|J|^{-1})^2=g'$$ and $$d^4x\rightarrow det\left(\frac{\partial x'}{\partial x}\right)d^4x=|J|\,d^4x=d^4x'$$
So in conclusion $$d^4x\sqrt{-g}\quad\rightarrow\quad d^4x\sqrt{-g}|J|\cdot|J|^{-1}=d^4x\sqrt{-g}$$ and the action is invariant, provided the Lagrangian density is a scalar.