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Why the square root of the determinant of the metric?

  1. Jul 31, 2009 #1
    We use as an integration form in Riemannian geometry the covariant
    [tex]\int \sqrt{g}d\Omega[/tex]
    I understand how this is invariant under an arbitrary change of coordinates (both Jacobian and metric square root transformation coefficient will cancel each other), what I don't understand is why don't an integral can be expressed only by
    [tex]\int d\Omega[/tex]
    I mean, it's true that a Jacobian will appear, but its effect will be "canceled" by the change of the integrand under the same change of variables, and it will give the same result! What is the difference between a flat and a curved space in this precise case, why the non flatness will make this integral non well-defined (without the [tex]\sqrt{g}[/tex] )?
    e.g: 1 dimension
    [tex]I = \int dx (x^{2} - 1)[/tex]
    change of variables :
    [tex]y = x^{2}[/tex]
    Will give :
    [tex]I' = \int dy \frac{(y - 1)}{2 \sqrt{y}}[/tex]
    [tex]I' = I[/tex]
    and this works for higher dimensions ....
    So what will change if one introduce a metric ????
  2. jcsd
  3. Jul 31, 2009 #2


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    Nothing changes. It's exactly the same. In your 1d example, (y-1) is the integrand, it's equal to (x^2-1). dy is d(omega). The sqrt(g) part is 1/(2*sqrt(y)). It's the part that comes from the change of measure.
  4. Jul 31, 2009 #3


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  5. Aug 1, 2009 #4
    I know that nothing will change, and that is precisely the problem for me: if nothing will change, why do we introduce a covariant volume element?
    For the book I don't have it. For the link, they simply say that we have to multiply by the square root of the metric determinant to obtain a covariant volume form, but they don't say WHY WE HAVE TO USE THIS COVARIANT VOLUME ELEMENT AS THE INTEGRAL MEASURE IN ORDER THAT THIS INTEGRAL BE INVARIANT UNDER AN ARBITRARY CHANGE OF COORDINATES!
  6. Aug 1, 2009 #5
    The theory is covariant under general coordinate transformations, so it makes sense to demand that the Largrangian density is (manifestly) covariant under general coordinate transformations.
  7. Aug 1, 2009 #6
    Sorry I'm not convinced by simply saying "make sens" :frown:, what I need is a mathematical explanation that force the volume element to be covariant in order that the integral becomes well defined.
    Another example:
    Here is an explicit example:
    [tex]I = \int d^{d}x f(x)[/tex]
    [tex]I' = \int d^{d}x' f(x')[/tex]
    [tex] = \int J[\frac{\partial x'}{\partial x}] d^{d}x f(x'(x))[/tex]
    Will this give the same result in a curved space-time? or the covariant measure is mandatory?
  8. Aug 1, 2009 #7
    It is not that you have to put the det(g) in there to make the integral well defined. It is simply that from a physics point of view, the integrand has a special meaning; it should be of the same form in any arbitrary coordinate system.
  9. Aug 1, 2009 #8
    Why should it be so???
  10. Aug 1, 2009 #9
    Well, Einstein postulated that the laws of physics are covariant under general coordinate transformations and that gravity would appear naturally in this picture. It doesn't ncessarly had to be this way, but it turns out that Nature does work this way.

    This means that you can generalize any tensor equation from special relativity by replacing derivatives by covariant derivatives (as long as you don't ave higher derivatives). The idea is that you can always choose a locally free falling coordinate system where the Christoffel symbols are zero at some spcific point. Then at that point in these coordinates, the usual equations from special relativity are valid, so you can then transform these back to arbitrary coordinates. To do this in a manifestly covariant way, you add the Christoffel symbols to change derivatives to covariant derivatives (they are zero anyway).
  11. Aug 1, 2009 #10

    I do not know if the problem is with me but ... I know all that! but even with that I cannot understand why the integral is not covariant if one does not use the covariant volume form !!!
  12. Aug 1, 2009 #11
    The problem is not the value of the integral but the form of the integrand. If you change coordinates, the integrand will be of a different form due to the Jacobian. If the integrand is not supposed to change then you can arrange for that by including the det(g) in there.

    If you have an integrand with a well defned physical meaning, e.g. a volume, and you write it in Cartesian coordinates, then including a det(g) in there gives you the correct expression for the volume in any arbitrary coordinate system.
  13. Aug 1, 2009 #12
    This helps me a lot, I guess I got it, in GR the problem is not to calculate the "value" of the Einstein-Hilbert action, but our aim is to do a Lagrangian formulation, i.e: write out a functional of the fields such that the field configuration which extremize the later is the solution of the equations of motion of the theory. It's clear now that we need the form of the integrand to be the same in all coordinate systems, because the theory is precisely invariant under diffeomorphisms. am I right?
    Thanks this too helps me a lot :rolleyes:
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