Euler's equation for a incompressible inviscid fluid is

$\displaystyle \frac{\partial \textbf{v}_t}{\partial t}+(\textbf{v}_t \cdot \nabla)\textbf{v}_t=-\nabla p_t$

where $\textbf{v}_t=\textbf{v}_t(r)$ is the fluid velocity and $p_t=p_t(r)$ is the pressure. Let $\nu_t(r)$ be the $1$-form associated to $\textbf{v}_t(r)$, ie,

$\displaystyle \nu_t=\sum_{i=1}^{3}v_t^idr^i$

Show that

$\displaystyle \frac{\partial \nu_t}{\partial t}+L_{\textbf{v}_t}\nu_t=-d(p_t-\frac{1}{2}v_t^2)$

where $v_t^2=\textbf{v}_t \cdot \textbf{v}_t$.

HINT: If $\omega=\omega_i dx^i$ is a $1$-form and $\mathbb{X}=X^ie_{(i)}$ a vector field on $\mathbb{R}^n$ then $\displaystyle L_{\mathbb{X}}\omega=(\frac{\partial X^j}{\partial x^i}\omega_j + X^j \frac{\partial \omega^i}{\partial x^j})dx^i$.

I am having serious problems attempting this question but I believe once going this question shouldnt be too bad. I believe it is ideal to start with the LHS and try and work to the RHS.

I am unsure on how to express the first equation in a more differential form like form and so be able to use it in the question.

UPDATE 2: So I get so far

$$\begin{align} \frac{\partial \nu_t}{\partial t}+L_{\textbf{v}_t}\nu_t &= \frac{\partial \nu_t}{\partial t}+\left(\frac{\partial v^j}{\partial r^i}v_j + v^j \frac{\partial v_i}{\partial r^j}\right)dr^i. \end{align} $$

working backwards,

$$ \begin{align} \left(-\frac{\partial p}{\partial r^i}+v^j\frac{\partial v^j}{\partial r^i} \right)dr^i &= \frac{\partial}{\partial r^i}(-p_t+\frac{1}{2}v^2)dr^i \\ &= -d(p_t-\frac{1}{2}v^2) \end{align} $$

Somehow I need to connect the two parts.

Please answer if possible using basic differential calculus I am not familiar with using tensors, and Riemmannian manifolds, tangent spaces and bundles... I am only covering a basic course on differential calculus. Sorry for the inconvenience to those who looked at the question earlier. I believe my question should indicate the standard I am working at.

  • $\begingroup$ is this not the bernoulli's equation? $\endgroup$
    – abel
    Dec 29 '14 at 17:46
  • $\begingroup$ Possibly. It wasnt stated in the question. $\endgroup$
    – Trajan
    Dec 29 '14 at 18:04
  • $\begingroup$ @abel Bernoulli's equation is $\partial v/\partial t + \omega \times v = -\nabla (p + \frac12 v^2)$, where $\omega = \text{curl} \,v$. I might be wrong on the first $+$, but the second $+$ is definitely a $+$ and not a $-$. $\endgroup$ Dec 30 '14 at 2:48

Sketched derivation: Given a (time-independent) Riemannian manifold $(M,g)$. Let the connection $\nabla$ be the Levi-Civita connection. Let there be given a scalar function $p\in C^{\infty}(M)$; and a vector field $V\in\Gamma(TM)$. Define one-form/co-vector field $\nu:=V^{\flat}\in\Gamma(T^{\ast}M)$ via the musical isomorphism.

The Euler's equation for a incompressible inviscid fluid reads in components

$$\tag{1} \partial_t V^k +\nabla^k p ~=~-(\nabla_VV)^k ~\equiv -V[V^k] -V^i \Gamma^k_{ij}V^j .$$

If we lower the vector index in eq. (1), we get

$$\tag{2} \partial_t \nu_k +\partial_k p ~=~\frac{1}{2}V^iV^j\partial_k g_{ij}-V[\nu_k]. $$

Equation (2) leads to the sought-for equation $$\tag{3} \partial_t \nu + \mathrm{d}p ~=~ \frac{1}{2}V^iV^j\mathrm{d}g_{ij}-V[\nu_k] \mathrm{d}x^k ~\stackrel{(4)}{\equiv}~\frac{1}{2}\mathrm{d}\left(\nu[V] \right)-{\cal L}_V\nu. $$

In the last step, we used the hint $$\tag{4} {\cal L}_V\nu ~\equiv~V[\nu_k] \mathrm{d}x^k +\nu_k \mathrm{d}V^k.$$

  • $\begingroup$ Thanks for the answer, unfortunately it is of too higher level for me to verify. I have edited my question to include more information. $\endgroup$
    – Trajan
    Dec 28 '14 at 15:03

Use the hint with $\mathbb X = \mathbf v$ and $\omega = \nu$ (suppressing the $t$ subscript). Then you get $$\displaystyle L_{\mathbf{v}}\nu=\left(\frac{\partial v^j}{\partial r^i}v_j + v^j \frac{\partial v_i}{\partial r^j}\right)dr^i.$$

Now see that $$\frac{\partial v^j}{\partial r^i}v_j = \frac12 \frac{\partial }{\partial r^i}(v^j v_j) = \frac12 \frac{\partial }{\partial r^i}(v^2) ,$$ and $$v^j \frac{\partial v_i}{\partial r^j} = \mathbf v \cdot \nabla v_i.$$
It is mostly unwrapping the notation, and using implicit differentiation of $v^2$.

Remember $dr^i$ are simply the standard basis vectors expressed as $1$-forms. And $dp$ is just a fancy way to write $\nabla p$.


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