What Happens to Euler-Lagrange in Field Theories (ADM)?

[nonequilibrium]

Hello,

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

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

However, it seems that if we now choose a time-splitting, like for example in ADM where the essence is to rewrite [itex]S = \int \mathrm d^4 x \; \mathcal L(g_{\mu \nu}, \partial_\rho g_{\mu \nu})[/itex] as [itex]\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)}[/itex]

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

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

 

[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!

 

[Orodruin]

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.

 

[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.

 

[haushofer]

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