What happens to Euler-Lagrange in field theories? (ADM?)

1. Jul 28, 2015

nonequilibrium

Hello,

So in the familiar case of non-relativistic particle Lagrangians/actions, we know the equations of motions are given by $$\frac{\partial \mathcal L}{\partial x^i} = \frac{\mathrm d }{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot x^i} \right)$$

In the familiar case of relativistic field theory Lagrangians/actions, we have
$$\frac{\delta \mathcal L}{\delta \phi} = \partial_\mu \left( \frac{\delta \mathcal L}{\delta ( \partial_\mu \phi )} \right)$$

However, it seems that if we now choose a time-splitting, like for example in ADM where the essence is to rewrite $S = \int \mathrm d^4 x \; \mathcal L(g_{\mu \nu}, \partial_\rho g_{\mu \nu})$ as $\boxed{ S = \int \mathrm d t \; \mathrm d^3 x \; \mathcal L(g_{i j}, \partial_k g_{i j}, \dot g_{ij}, N, N^i)}$

In this case it seems the equation of motion is given by
$$\frac{\delta \mathcal L}{\delta g_{ij}} = \frac{\mathrm d }{\mathrm dt} \left( \frac{\delta \mathcal L}{\delta ( \dot g_{ij} )} \right)$$

This seems a bit weird. Is it obvious the latter two equations of motions are compatible?

2. Jul 28, 2015

nonequilibrium

In fact, I might be completely wrong about that last equation of motion. I suppose that would resolve my confusion. Can anyone confirm/disconfirm my last equation of motion? Thanks!

3. Jul 29, 2015

Orodruin

Staff Emeritus
The equations of motion do not change just because you rewrite the integral. The action is still a four-dimensional integral and the action is what you extremise.

4. Jul 29, 2015

Ben Niehoff

I find it more straightforward to write the equations of motion as

$$\frac{\delta S}{\delta \varphi} = 0$$
for any field $\varphi$. The variation operator $\delta$ behaves very much like a differentiation operator, e.g.

$$\delta (A_\mu A^\mu) = \delta (g^{\mu\nu} A_\mu A_\nu) = A_\mu A_\nu \, \delta g^{\mu\nu} + 2 g^{\mu\nu} A_\mu \, \delta A_\nu.$$
If your action has derivatives in it (as it must to give dynamical equations of motion), then you will have to integrate by parts to move derivatives off of $\delta$ terms and onto the usual fields:

$$\delta (\partial_\mu \varphi \partial^\mu \varphi) = 2 \partial_\mu \varphi \, \delta (\partial^\mu \varphi) \overset{\text{i.b.p.}}{\longrightarrow} - 2 \partial^\mu \partial_\mu \varphi \, \delta \varphi.$$
In such cases, strictly speaking you must take into account boundary terms in your action. Usually the boundary is at infinity and you assume sufficiently fast fall-off of your fields that the boundary terms are zero. But if you do anything that violates these assumptions, take care.

5. Jul 30, 2015

haushofer

You should view the particle case as a 1-dimensional field theory, whereas GR is in general a D-dim. field theory.