# Noether's Theorem For Functionals of Several Variables

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

Does Noether's theorem, for functionals of a single independent variable, just say that, because $\mathcal{L}$ is invariant, we have

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

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, so is this equivalent to what I've written above (I've hopefully justified all of this below)?.

(If there is any ambiguity in this question, I've offered a hopefully unnecessary explanation of the details below)

The only problem with it I see is that it works fine for conservation of linear momentum, but it seems to show that conservation of energy and angular momentum are always zero, since $dx^* = 0$, unless I add a constant to the above, i.e.

$$\mathcal{L}(x,y_i,y_i')dx = \mathcal{L}(x^*,y_i^*,y_i'^*)dx^* + C$$

which seems fine as adding constants to Lagrangians do not affect the EOM, so that's the only problem I see with it.

I like the above expression, it's great for remembering Noether's theorem, it's more general than what almost every physics book presents as Noether's theorem (it's not just conservation of energy!), explains what Landau's additivity of integrals of motion comment means mathematically etc...

Can we generalize it to Lagrangian densities?

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

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

(I have no idea why Jackson adds a $g^{\alpha \beta}$)

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 this is right.

References:

1. Gelfand, Calculus of Variations - Ch. 3, 4 & 7
2. Jackson, Classical Electrodynamics - Ch. 12 Sec 10

Noether's theorem for functionals of a single independent variables states that, given a continuous one-parameter family of infinitesimal transformations of the form

$$T(x,y_i,\varepsilon) = (x^*(x,y_i,\varepsilon),y_i^* (x,y_i,\varepsilon))= (x^*,y_i^*)= (x+\frac{\partial x^*}{\partial \varepsilon}\varepsilon,y_i+\frac{\partial y_i^*}{\partial \varepsilon}\varepsilon),$$

if a Lagrangian

$$\mathcal{L} = \mathcal{L}(x,y_i,\frac{dy_i}{dx})$$

is invariant

$$\mathcal{L}(x,y_i,\frac{dy_i}{dx}) = \mathcal{L}(x^*,y_i^*,\frac{dy_i^*}{dx})$$

under $T$ then the scary-looking quantity

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

is conserved (more precisely, you do it in terms of the functional but that is irrelevant for this discussion).

Examining this quantity for two seconds let's us notice we can write this conserved quantity as

$$\sum_{i=1}^n \frac{\partial \mathcal{L}}{\partial ( \tfrac{d y_i}{dx})} \frac{\partial y_i^*}{\partial x} - \mathcal{H}\frac{\partial x^*}{\partial \varepsilon}$$

in terms of the Hamiltonian

$$\mathcal{H} = \sum_{j=1}^n \frac{\partial \mathcal{L}}{\partial ( \tfrac{d y_j}{dx})} \tfrac{\partial y_j }{\partial x} - \mathcal{L} = \sum_{j=1}^n p_i \tfrac{\partial y_j }{\partial x} - \mathcal{L}$$

or, taking away the parameter, as

$$\sum_{i=1}^n p_i d y_i^* - \mathcal{H}dx^*$$

Notice the similarity between the quantity in Noether's theorem and this:

$$\mathcal{H} = \sum_{j=1}^n p_i \tfrac{\partial y_j }{\partial x} - \mathcal{L} \rightarrow \mathcal{L} dx = \sum_{j=1}^n p_i d y_j - \mathcal{H}dx$$

Does Noether's theorem just say that, because $\mathcal{L}$ is invariant we have, solving for $\mathcal{L}$, that

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