Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Volume element in Relativity

  1. Apr 27, 2017 #1
    In a change of coordinate system we have ##dx^\mu = (\partial x^\mu / \partial \xi^{\kappa})d \xi^{\kappa}##, where the term in round brackets is the Jacobian. That notation implies a sum over all values that ##\kappa## can take. This don't tell us that it's an alternating sum for the case of volume element in the new coord. system, i.e. a sum where the terms have alternating sign, which is what we obtain if we resolve for the Jacobian determinant. So is there a problem with the notation on how the coordinates transform?
     
  2. jcsd
  3. Apr 27, 2017 #2
    I'm wondering whether or not the correct would be to define the volume element by the wedge product, like ##dv = dx \wedge dy \wedge dz##, because then it gives us the correct result when we make the coordinate transformation.
     
  4. Apr 27, 2017 #3

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What does ##\int f\left(x,y,x\right) dxdydx## mean?

    What happens if you change from Cartesian coordinates ##\left\{x,y,z\right\}## to another coordinate system?
     
  5. Apr 27, 2017 #4
    It means the integral of a function ##f## with respect to the ##{x, y, z}## system. This is a scalar and must be the same in all coordinate systems.
    So...
    We need the Jacobian so that the volume is invariant.
     
  6. Apr 27, 2017 #5

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  7. Apr 27, 2017 #6
    Thanks. I understand what is said in these materials. What I don't understand is why the Jacobian determinant is just ##det[ \partial x^\mu / \partial \xi^\mu]## instead of ##det[ (\partial x^\mu / \partial \xi^\kappa)(\partial x^\nu / \partial \xi^\sigma)]## for the transformation ##dy^\mu dy^\nu \rightarrow dx^\sigma dx^\kappa##, as I said in post #1. As an example of what I mean, consider the transform from ##dxdy## to ##drd\theta##.
    mFR4Qay.png
     
  8. Apr 28, 2017 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Start from usual 3D Euclidean space. It's clear that ##\mathrm{d}^3 \vec{x}=\mathrm{d} x \mathrm{d} y \mathrm{d} z## only if you have Cartesian coordinates ##(x,y,z)##.

    Now take any other kind of generalized coordinates ##q^k## (##k \in \{1,2,3\}##). I use the usual notation with upper and lower indices also here in Euclidean space. Now to get the volume elements you can think of some region of space time and divide it in "infinitesimal" volume elements spanned by coordinate lines. These coordinate lines in general will not define local Cartesian basis vectors through there tangents at a point but some general basis spanning an infinitesimal parallelepiped,

    https://en.wikipedia.org/wiki/Parallelepiped

    As is nicely explained on the Wikipedia page, the volume is given by the determined of the spanning vectors (i.e., components taken wrt. Carstesian coordinates). This leads to
    $$\mathrm{d}^3 \vec{x} = \mathrm{det} \frac{\partial(x^1,x^2,x^3)}{\partial (q^1,q^2,q^3)} \mathrm{d}^3 q=\epsilon_{ijk} \frac{\partial x^i}{\partial q^1}\frac{\partial x^j}{\partial q^2} \frac{\partial x^k}{\partial q^3} \mathrm{d}^3 q.$$
    The trouble now is that the Levi-Civita symbol provides only tensor components with respect to Cartesian coordinates.

    But now we have the advantage that there is a metric in our Euclidean space. To that end let's check, how the Levi-Civita symbol behaves under a general coordinate transformation, if we apply the rules as if it would provide covariant tensor components:
    $$\epsilon_{abc}'=\epsilon_{ijk} \frac{\partial x^i}{\partial q^a}\frac{\partial x^j}{\partial q^b}\frac{\partial x^k}{\partial q^c} = J \epsilon_{abc},$$
    i.e., there's the Jacobian of the transformation as an additional factor.

    To see, how the metric components help, let's calculate how their determinant behaves. To transform from Cartesian components ##g_{ij}=\delta_{ij}## to the components with respect to the basis given by the coordinate lines, we have
    $$g_{ab}' = \frac{\partial x^i}{\partial q^a} \frac{\partial x^j}{\partial q^b} \delta_{ij}.$$
    end thus
    $$g'=\mathrm{det}(g_{ab}')=J^2.$$
    Now we assume that ##J>0## (this we can always get by choosing an appropriate order of the ##q^j##; for ##J>0## one says the orientation of the coordinate basis is the same as for the Cartesian basis; i.e., we consider orientation-conserving coordinate transformations only). Then we can write
    $$\epsilon_{abc}'=J \epsilon_{abc}=\sqrt{g'} \epsilon_{abc},$$
    where ##\epsilon_{abc}## is the usual Levi-Civita symbol. So defining
    $$\Delta_{abc}=\sqrt{g} \epsilon_{abc}$$
    we have defined the components ##\Delta_{abc}## of a tensor, the Levi-Civita tensor (under general orientation-preserving coordinate transformations), and thus we have a covariant way to define volume elements in the usual sense of Euclidean geometry:
    $$\mathrm{d} V= \Delta_{abc} \mathrm{d} q^a \mathrm{d} q^b \mathrm{d} q^c.$$
    The same holds true in Riemannian spaces of any dimension since we haven't made use of ##d=3## anywhere in the above considerations.

    In relativity (SR and GR), however, you don't have a Euclidean or Riemannian space but a Lorentzian space or Lorentzian manifold, respectively. Because the signature of the pseudo-metric is (1,3) (west coast) or (3,1) (east coast) the only qualification we must make is that we have to write
    $$\Delta_{\mu \nu \rho \sigma} = -\sqrt{-g} \epsilon_{\mu \nu \rho \sigma},$$
    where ##\epsilon_{\mu \nu \rho \sigma}## is the usual Levi-Civita symbol (the minus sign is convention).

    It is also important to note that there's another sign flip when going to contravariant components, because raising the indices leads to
    $$\Delta^{\alpha \beta \gamma \delta}=g^{\alpha \mu} g^{\beta \nu} g^{\rho \gamma} g^{\sigma \delta} \Delta_{\mu \nu \rho \sigma}=-\frac{1}{g} \sqrt{-g} \epsilon_{\alpha \beta \gamma \delta}=+\frac{1}{\sqrt{-g}} \epsilon_{\alpha \beta \gamma \delta},$$
    because ##\mathrm{det}(g^{\mu \nu})=1/\mathrm{det} (g_{\mu \nu})=1/g##. As we see the signs are opposite then for the Levi-Civita tensors with lower indices! The minus sign in the definition of the covariant components of the Levi-Civita tensor comes simply from the fact that usually in textbooks ##\Delta^{\alpha \beta \gamma \delta}=+\epsilon_{\alpha \beta \gamma \delta}## in Minkowski space in Minkowski coordinates, i.e., the Levi-Civita tensor with upper indices is defined by the Levi-Civita symbol, while the one with upper indices by the negative of the Levi-Civita symbol.
     
  9. Apr 28, 2017 #8
    Thanks vanhees71 for your detailed reply. So the Levi-Civita tensor plays the role of the wedge product that I was thinking of before on this thread?
     
  10. Apr 29, 2017 #9

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Yes. You can also do without a metric, i.e., with a "bare" vector space. Then covariant integration is restricted to the integration of alternating forms.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volume element in Relativity
Loading...