# Why the square-root of the metric

#### stevendaryl

Staff Emeritus
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 $\sqrt{-g}$?

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#### pervect

Staff Emeritus
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 [Broken] (Ben Crowell's online book) talks a little about tensor densities in an informal way.

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#### PeterDonis

Mentor
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...dxn. 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.

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