# I Momentum density in curvilinear coordinates

1. Sep 4, 2016

### daxowax

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. Sep 5, 2016

### vanhees71

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
3. Sep 5, 2016

### daxowax

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!

4. Sep 6, 2016

### vanhees71

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