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I Momentum density in curvilinear coordinates

  1. Sep 4, 2016 #1
    Hi,

    In an article on theoretical fluid dynamics I recently came across the following equation:

    $$M_i = \sqrt{g} \rho v_i$$

    where ##M_i## denotes momentum density, ##v_i## velocity, ##\rho## the mass density and g is the determinant of the metric tensor. It is probably quite obvious, but I do not get why you have to put the ##\sqrt{g}## in there. Aren't both densities related to the same volume form? I would appreciate any help!
     
  2. jcsd
  3. Sep 5, 2016 #2

    vanhees71

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    Let ##\vec{b}_i## (##i \in \{1,2,3 \}##) denote any basis of ##\mathbb{R}^3## (it can also be a basis defined locally by generalized coordinates) and ##\vec{e}_i## arbitrary Cartesian coordinates. Further define the coordinates of the position vector as
    $$\vec{x}=\tilde{x}^i \vec{b}_i=x^j \vec{e}_j.$$
    Now we need a manifestly covariant description of the volume element. In Cartesian coordinates we simply have
    $$\mathrm{d}V=\mathrm{d}^3 \vec{x}=\mathrm{d} x^1 \mathrm{d} x^2 \mathrm{d} x^3.$$
    The trick is to rewrite this as
    $$\mathrm{d}^3 \vec{x}=\Delta_{jkl} \mathrm{d} x^j \mathrm{d} x^k \mathrm{d} x^l,$$
    where ##\Delta_{jkl}## is the Levi-Civita symbol, which is totally antisymmetric under interchange of its indices and ##\Delta_{123}=1##.

    Unfortunately the Levi-Civita symbol is not a general tensor, because defining the transformation matrix between the general and the Cartesial coordinates as
    $${T^j}_k=\frac{\partial x^j}{\partial \tilde{x}_k}$$
    we have
    $$\mathrm{d}^3 \vec{x}=\Delta_{jkl} {T^{j}}_{j'} {T^{k}}_{k'} {T^{l}}_{l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'}=\det \hat{T} \mathrm{d} \Delta_{j'k'l'} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'},$$
    i.e., you get the Jacobian of the transformation in addition, as it must be for the proper transformation of the volume element. I assume that this determinant is positive (i.e., that the orientation of the basis ##\vec{b}_j## is the same as that of the Cartesian basis ##\vec{e}_j##, which you can always get by choosing the right order of the basis vectors).

    Now we have a Euclidean space, i.e., there is the scalar product. In Cartesian components it's simply represented by ##\delta_{jk}##, i.e., you have for any two vectors ##\vec{x}## and ##\vec{y}##
    $$\vec{x} \cdot \vec{y} = \delta_{jk} x^j y^k=\delta_{jk} {T^{j}}_{j'} {T^{k}}_{k'} \tilde{x}^j \tilde{y}^k := g_{j'k'} \tilde{x}^j \tilde{y}^k.$$
    In matrix-vector notation you thus have
    $$\hat{g}=\hat{T}^t \hat{T} \; \Rightarrow g=\det{\hat{g}}=(\det \hat{T})^2.$$
    So you can write the volume element as
    $$\mathrm{d}^3 \vec{x}=\det T \Delta_{j'k'l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'} = \sqrt{g} \Delta_{j'k'l'} \mathrm{d} \tilde{x}^{j'} \mathrm{d} \tilde{x}^{k'} \mathrm{d} \tilde{x}^{l'}.$$
    Since for Cartesian coordinates ##g_{jk}=\delta_{jk}## we can define the Levi-Civita tensor components as an invariant tensor via
    $$\epsilon_{jkl}=\sqrt{g} \Delta_{jkl}.$$
    Then the general covariant definition of the volume element in terms of the position-vector components reads
    $$\mathrm{d}^3 \vec{x} =\epsilon_{jkl} \mathrm{d} \tilde{x}^j \mathrm{d} \tilde{x}^k \mathrm{d} \tilde{x}^l.$$
    Now if you have a scalar density (like mass density ##\rho(\vec{x})## in non-relativistic fluid dynamics), a mass element in a small volume around ##\vec{x}## is given by [corrected according to #3]
    $$\mathrm{d} m=\mathrm{d}^3 \vec{x} \rho(\vec{x})=\sqrt{g} \rho \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3.$$
    For the mass-current-density then you have [corrected according to #3]
    $$\mathrm{d} m v^i= \sqrt{g} \rho v^i \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3.$$
     
    Last edited: Sep 6, 2016
  4. Sep 5, 2016 #3
    Thank you a lot for this elaborate answer! It helped me very much to understand these calculations. In the second last equation, you wrote
    I believe this was a typing error, and it should be $$ \mathrm{d} m=\mathrm{d}^3 \vec{x} \rho(\vec{x})=\sqrt{g} \rho \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3$$Therefore, the last equation should say $$\mathrm{d} m v^i= \sqrt{g} \rho v^i \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3$$ Since ##\mathrm{d} m v^i## is the momentum of a fluid parcel and momentum density (or mass-current-density) ##M_i## is defined via ##\mathrm{d} m v^i = M_i \mathrm{d} \tilde{x}^1 \mathrm{d} \tilde{x}^2 \mathrm{d} \tilde{x}^3##, by comparing the equations we get $$M_i = \sqrt{g} \rho v^i$$ Please correct me if I'm wrong!
     
  5. Sep 6, 2016 #4

    vanhees71

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    Sure, I'll correct the typos in the original posting too!
     
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