# Noether's Theorem For Functionals of Several Variables

1. May 8, 2014

### bolbteppa

My question is on using a form of the single variable Noether's theorem to remember the multiple variable version.

Noether's theorem, for functionals of a single independent variable, can be translated into saying that, because $\mathcal{L}$ is invariant, we have

$$\mathcal{L}(x,y_i,y_i')dx = \sum_{j=1}^n p_i d y_j - \mathcal{H}dx = \mathcal{L}(x^*,y_i^*,y_i'^*)dx^* = \sum_{i=1}^n p_i d y_i^* - \mathcal{H}dx^* = C$$

It is usually stated by saying that

$$\sum_{i=1}^n \frac{\partial \mathcal{L}}{\partial ( \tfrac{d y_i}{dx})} \frac{\partial y_i^*}{\partial \varepsilon} - \left[\sum_{j=1}^n \frac{\partial \mathcal{L}}{\partial ( \tfrac{d y_j}{dx})} \tfrac{\partial y_j }{\partial x} - \mathcal{L}\right]\frac{\partial x^*}{\partial \varepsilon}$$

is conserved, but this seems to be equivalent to what I've written above.

(I've offered a hopefully unnecessary explanation of the details of the equivalence, posed as a question, http://math.stackexchange.com/questions/787011/noethers-theorem-for-functionals-of-several-variables [Broken]).

I like the above expression, it's great for remembering Noether's theorem.

Can we generalize it to functionals of several variables?

The statement of the multivariable Noether I know is that, for

$$\mathcal{L} = \mathcal{L}(x_i,u_j,\frac{\partial u_j}{\partial x_i})$$

we have that

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}\left(\frac{\partial u_j^*}{\partial \varepsilon_k} - \sum_{i=1}^n\frac{\partial u_j}{\partial x_i}\frac{\partial x_i^*}{\partial \varepsilon _k}\right) + \mathcal{L}\frac{\partial x_i^*}{\partial \varepsilon _k}\right] = 0$$

I can hardly remember this, and as I've indexed it I can't turn it into anything involving what I *think* is the Hamiltonian for a functional of several independent variables

$$\mathcal{H} = \sum_{j=1}^np_{ij}\frac{\partial u_j}{\partial x_i} - \mathcal{L} = \sum_{j=1}^n \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}\frac{\partial u_j}{\partial x_i} - \mathcal{L}$$

Can this be turned into something similar to my main equation, perhaps using $\delta_{ij}$'s or $g_{\mu \nu}$'s or something?

An attempt:

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}\left(\frac{\partial u_j^*}{\partial \varepsilon_k} - \sum_{i=1}^n\frac{\partial u_j}{\partial x_i}\frac{\partial x_i^*}{\partial \varepsilon _k}\right) + \mathcal{L}\frac{\partial x_i^*}{\partial \varepsilon _k}\right] = 0$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}\left(d u_j^* - \sum_{i=1}^n\frac{\partial u_j}{\partial x_i}d x_i^* \right) + \mathcal{L}d x_i^* \right] = 0$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}d u_j^* - \sum_{k=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_k}{\partial x_i})}\sum_{l=1}^n\frac{\partial u_k}{\partial x_l}d x_l^* + \mathcal{L}d x_i^* \right] = 0$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}d u_j^* - \sum_{k=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_k}{\partial x_i})}\sum_{l=1}^n\frac{\partial u_k}{\partial x_l}d x_l^* + \sum_{l=1}^n\delta^l_i \mathcal{L}d x_l^* \right] = 0$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}d u_j^* - \sum_{l=1}^n (\sum_{k=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_k}{\partial x_i})}\frac{\partial u_k}{\partial x_l} - \delta^l_i \mathcal{L})d x_l^* \right] = 0$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m \frac{\partial \mathcal{L}}{\partial (\tfrac{\partial u_j}{\partial x_i})}d u_j^* - \sum_{l=1}^n \mathcal{H}d x_l^* \right] = 0.$$

$$\sum_{i=1}^n\frac{\partial}{\partial x_i}\left[\sum_{j=1}^m p_{ij}d u_j^* - \sum_{l=1}^n \mathcal{H}d x_l^* \right] = 0.$$

I don't know if that's right.

Last edited by a moderator: May 6, 2017
2. May 9, 2014

### Orodruin

Staff Emeritus
3. May 9, 2014

### bolbteppa

Brilliant, thank you.