Euler's equations in differential forms

[spaghetti3451]

Homework Statement

Euler's equations can be written using vector calculus as

$\displaystyle{\frac{\partial v_{i}}{\partial t}+v^{j}\left(\frac{\partial v_{i}}{\partial x^{j}}\right) = -\left(\frac{1}{\rho}\right)\frac{\partial p}{\partial x^{i}}+f_{i}}.$
Euler's equations can also be written using differential forms as

$\displaystyle{\mathcal{L}_{{\bf{v}}+\partial / \partial t}(\nu) = \textbf{d} \left\{\frac{1}{2}||{\bf{v}}||^{2}+\phi-\int\frac{dp}{\rho}\right\}}$
Under the assumption that $f_{i} = \text{grad}_{i}\phi$, that $p=p(\rho)$ and that $\nu$ is the $1$-form with components $v_i$, how can you prove that the two formulations above are equivalent?

Homework Equations

The Attempt at a Solution

$\displaystyle{\frac{\partial v_{i}}{\partial t}+v^{j}\left(\frac{\partial v_{i}}{\partial x^{j}}\right) = -\left(\frac{1}{\rho}\right)\frac{\partial p}{\partial x^{i}}+f_{i}}$

$\displaystyle{\frac{\partial v_{i}}{\partial t}+\frac{\partial}{\partial x^{j}}\left(v^{j}v_{i}\right)-v_{i}\frac{\partial v^{j}}{\partial x^{j}} = \text{grad}_{i}\ \left(-\int\frac{dp}{\rho}\right)+\text{grad}_{i}\ \phi}$

$\displaystyle{\frac{\partial v_{i}}{\partial t}+\frac{\partial}{\partial x^{j}}\left(v^{j}v_{i}\right) = v_{i}\frac{\partial v^{j}}{\partial x^{j}}+\text{grad}_{i}\ \left(-\int\frac{dp}{\rho}\right)+\text{grad}_{i}\ \phi}$

$\text{I have to fill up this missing line}$

$\displaystyle{\frac{\partial}{\partial x^{j}}\left(v^{j}v_{i}\right) + \frac{\partial v_{i}}{\partial t} = \text{grad}_{i}\left(\frac{1}{2}||{\bf{v}}||^{2}\right)+\text{grad}_{i}\ \left(-\int\frac{dp}{\rho}\right)+\text{grad}_{i}\ \phi}$

$\displaystyle{\frac{\partial\nu}{\partial t}+\mathcal{L}_{\bf{v}}(\nu) = \textbf{d}\left\{\frac{1}{2}||{\bf{v}}||^{2}+\phi-\int\frac{dp}{\rho}\right\}}$

$\displaystyle{\mathcal{L}_{{\bf{v}}+\partial / \partial t}(\nu) = \textbf{d} \left\{\frac{1}{2}||{\bf{v}}||^{2}+\phi-\int\frac{dp}{\rho}\right\}}$

 

[mighty2000]

To prove that the two formulations are equivalent, we can start by expanding the terms in the first equation using the given assumptions:

$\displaystyle{v^{j}\left(\frac{\partial v_{i}}{\partial x^{j}}\right) = v_{i}\frac{\partial v^{j}}{\partial x^{j}}}$

$\displaystyle{-\left(\frac{1}{\rho}\right)\frac{\partial p}{\partial x^{i}} = -\text{grad}_{i}\left(\frac{dp}{\rho}\right)}$

Substituting these into the first equation, we get:

$\displaystyle{\frac{\partial v_{i}}{\partial t}+\frac{\partial}{\partial x^{j}}\left(v_{i}v^{j}\right) = v_{i}\frac{\partial v^{j}}{\partial x^{j}}-\text{grad}_{i}\left(\frac{dp}{\rho}\right)+f_{i}}$

Using the definition of the Lie derivative, we can rewrite this as:

$\displaystyle{\frac{\partial v_{i}}{\partial t}+\mathcal{L}_{\bf{v}}(\nu) = \textbf{d}\left\{\frac{1}{2}||{\bf{v}}||^{2}+\phi-\int\frac{dp}{\rho}\right\}}$

Which is equivalent to the second equation given. Therefore, we have proven that the two formulations are equivalent.