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Why the square-root of the metric

  1. Dec 13, 2012 #1

    stevendaryl

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

    [itex]L = \sqrt{-g} ((\nabla_\mu \Phi)(\nabla^\mu \Phi) - m^2 \Phi^2)[/itex]

    Is there a simple explanation for why this is scaled by [itex]\sqrt{-g}[/itex]?
     
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  3. Dec 13, 2012 #2

    pervect

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    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.
     
    Last edited by a moderator: May 6, 2017
  4. Dec 13, 2012 #3

    PeterDonis

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    Because the Lagrangian density is integrated over spacetime to obtain the action, and without the factor of [itex]\sqrt{-g}[/itex], the integral isn't invariant under a change of coordinate charts. [Edit: This is another way of saying what pervect said.]
     
  5. Dec 13, 2012 #4
    Because [itex]\sqrt{-g}=det(J)[/itex], 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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