Thanks, I think this has put me on the right track. My current attempt:
Using the product rule, we get
$$\mathcal{L}_v(v_i dx^i) = \mathcal{L}_v (v_i) dx^i + v_i \mathcal{L}_v (dx^i)$$
Interpreting \mathcal{L}_v as a directional derivative the first term equals v^j \partial_j v_i dx^i . For...
Hello,
I have a maybe unusual question. In a paper, I recently found the equation $$\mathcal{L}_v(v_i dx^i) = (v^j \partial_j v_i + v_j \partial_i v^j) dx^i$$
Where v denotes velocity, x spatial coordinates and \mathcal{L}_v the Lie derivative with respect to v. Now I'm an undergraduate who...
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 $$ \mathrm{d} m=\mathrm{d}^3 \vec{x} \rho(\vec{x})=\sqrt{g} \rho \mathrm{d} \tilde{x}^1 \mathrm{d}...
Hi,
In an article on theoretical fluid dynamics I recently came across the following equation:
$$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...