Why the square-root of the metric

[stevendaryl]

In a paper about field theory in curved spacetime, an author says that the Lagrangian density for a free scalar particle is

$L = \sqrt{-g} ((\nabla_\mu \Phi)(\nabla^\mu \Phi) - m^2 \Phi^2)$

Is there a simple explanation for why this is scaled by $\sqrt{-g}$?

 

[pervect]

The simple explanation is the Lagrangian is a tensor density. You want to map a volume to a number (a Lorentz scalar), but representing the volume requires the factor of sqrt(g).

http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html (Ben Crowell's online book) talks a little about tensor densities in an informal way.

 

[PeterDonis]

Because the Lagrangian density is integrated over spacetime to obtain the action, and without the factor of $\sqrt{-g}$, the integral isn't invariant under a change of coordinate charts. [Edit: This is another way of saying what pervect said.]

 

[elfmotat]

Because $\sqrt{-g}=det(J)$, where J is the Jacobian matrix transforming from Cartesian coordinates to the coordinates being used in the particular problem.

The volume element in a set of arbitrary coordinates is given by dV=det(J)dx₁dx₂...dx_n. For example, in Cartesian coordinates in 3D space dV=dxdydz. The determinant of the Jacobian transforming to spherical coordinates is r²sinθ, so the volume element in spherical coordinates is dV=r²sinθdrdϕdθ.

You need the factor in front of dⁿx in order to keep things invariant under a coordinate transformation.