# A Variational derivative and Euler-Poincare equations

#### eipiplusone

Hi,

I'm trying to understand the Euler-Poincare equations, which reduce the Euler-Lagrange equations for certain Lagrangians on a Lie group. I'm reading Darryl Holm's "Geometric mechanics and symmetry", where he suddenly uses what seems to be a variational derivative, which I'm having a hard time understanding. The Euler-Poincare reduction theorem (and equation) goes as follows: It is the $\frac{\delta l}{\delta \xi}$ - derivative which I'm guessing is a variational derivative. It must be a covector, and in computations it seems to reduce to partial derivatives of $l$ wrt. $\xi$ (in suitable coordinates) - I have also seen other sources where the EP-equation is stated in terms of $\frac{\partial l}{\partial \xi}$ instead of the abovementioned.

He defines it in two different ways. The first definition is found in one of the later chapters of the book (and it seems to be in a more general setting): And in another book ("Geometric mechanics - part 2") he defines it as: My questions are:

- which definition should I focus on?
- Is the pairing in the latter definition the usual covector-vector pairing?

Any explanations, hints or references would be greatly appreciated! Thanks.

<mentor: fix latex>

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#### fresh_42

Mentor
2018 Award
My questions are:

- which definition should I focus on?
$11.13$ The other definition $(2.1.1)$ is basically just the definition of a derivative. The crucial point is what is variated, namely the direction $v:=\delta q$, resp. $v:=\delta u$.
You could also combine them and get both in one:
$$\delta l[ u ]= \lim_{t \to 0}\dfrac{l[ u+t\delta u ] - l[ u ]}{t}=\left. \dfrac{d}{dt}\right|_{t=0}l[ u+t\delta u ]=\left\langle \dfrac{\delta l}{\delta u},\delta u \right\rangle = \int \dfrac{\delta l}{\delta u} \cdot \delta u \,dV$$
- Is the pairing in the latter definition the usual covector-vector pairing?
Yes. It's the view of the derivative as a linear function, similar to the gradient in $\mathbb{R}^n$: $d l(u)(v)=\langle \nabla l(u) , v \rangle$, except that we do not take the total differential here but only the partial along $\delta u$.

#### eipiplusone

Thanks for your answer. I am still confused though :/
• I don't understand the last equality; $\langle \frac{\delta l}{\delta u} , \delta u \rangle = \int \frac{\delta l}{\delta u} \cdot \delta u \hspace{1mm} d V$. I would think that $\frac{\delta l}{\delta u} \cdot \delta u$ is the euclidean inner product representation of the evaluation $\frac{\delta l}{\delta u}[\delta u] \in \mathbb{R}$, where $\frac{\delta l}{\delta u}, \delta u$ are euclidean vectors. But in that case the integral doesn't make sense, since $\langle \frac{\delta l}{\delta u} , \delta u \rangle = \frac{\delta l}{\delta u} \cdot \delta u$ . So... why the need for the integral? I'm probably missing something fundamental.
• Is it correct to say that if $l$ is an ordinary function of a vector $u$, then $\frac{\delta l}{\delta u} = \frac{\partial l}{\partial u}$?

#### fresh_42

Mentor
2018 Award
Thanks for your answer. I am still confused though :/
• I don't understand the last equality; $\langle \frac{\delta l}{\delta u} , \delta u \rangle = \int \frac{\delta l}{\delta u} \cdot \delta u \hspace{1mm} d V$. I would think that $\frac{\delta l}{\delta u} \cdot \delta u$ is the euclidean inner product representation of the evaluation $\frac{\delta l}{\delta u}[\delta u] \in \mathbb{R}$, where $\frac{\delta l}{\delta u}, \delta u$ are euclidean vectors. But in that case the integral doesn't make sense, since $\langle \frac{\delta l}{\delta u} , \delta u \rangle = \frac{\delta l}{\delta u} \cdot \delta u$ . So... why the need for the integral? I'm probably missing something fundamental.
The integral comes from the definition of the inner product in $L^2(I)\, : \,\langle f,g \rangle = \int_I f(x)g^*(x)dx$ only that $f,g$ in our case aren't functions but differential forms, and $I$ not an interval but a vector field $V$.
• Is it correct to say that if $l$ is an ordinary function of a vector $u$, then $\frac{\delta l}{\delta u} = \frac{\partial l}{\partial u}$?
Yes. However, partials are usually meant as coordinate directions, and $u$ doesn't have to be one.

The main difficulty, at least mine, is to distinguish the roles a derivative is playing in a certain context. I've made the fun and listed a couple (10) of them here: https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ but this has meant to be more as a hint on how easy it is to get confused rather than an explanation. Fun fact: slope wasn't even on the list.

My attempt to sort things out was this one: https://www.physicsforums.com/insights/pantheon-derivatives-part-ii/#toggle-id-0
If you look at the pdf, then it's probably better to read, resp. easier to search for keywords, e.g. Gâteaux or Noether as in our case here.

#### eipiplusone

Thank you. The documents you link to looks very nice, I will see if they can make things clear to me.

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