# Why can't we fix the metric and its derivatives at boundary, with the variational method?

In general relativity and for its Einstein-Hilbert action, we usually ask that the metric variations $\delta g_{\mu \nu}$ cancel on the boundary $\partial \, \Omega$ of some region $\Omega$ of the spacetime manifold ($\delta g_{\mu \nu} = 0$). But if the variations are of compact support, why can't we also ask that the derivatives cancel at the boundary ; $\partial_{\lambda} \, \delta g_{\mu \nu} = 0$ on $\partial \, \Omega$ ?

Why is this too restrictive, or too abusive ? Why would this be too easy ? (I know about the surface term problem in General Relativity)

Under the variational principle, the variations are arbitrary and don't have to obey the equations of motion (the variations are not on shell). Under the variational method, only $g_{\mu \nu}$ and $\partial_{\lambda} \, g_{\mu \nu}$ are supposed to obey some equation of motion (the Einstein equation in this case).

There's something important that I do not understand here, and I need to clear this up !

EDIT 1 :

Maybe the question is badly formulated and something is missing. The variational method is usually (?) used to find the classical equations of motion (no quantum mechanics here). In this case, the variational method is just a recipe to find the equations. The variation $\delta g_{\mu \nu}$ are then fully arbitrary but are of compact support, and all its derivatives cancels too on the boundary. Is that right ?

If that is the case, then we don't need to care much about the surface terms during some arbitrary variations, and we don't need to add any surface counter-term to the action (aka Gibbons-Hawking-York surface integral). This is what I need to clear up.

Maybe I'm simply confusing this classical recipe with some other uses of the variational principle ?

EDIT 2 :

Under the action integral variation, is the perturbed field $\phi'(x) = \phi(x) + \delta \phi(x)$ "on-shell" or "off-shell" ?

As far as I know, the perturbed field is "off-shell" ; it doesn't necessarily satisfy the field equation, contrary to the unperturbed $\phi(x)$ (which is "on-shell", by definition). Is that right ?

It is possible that there's another interpretation of the stationary action principle (to be confirmed) ; the variation $\delta \phi(x)$ may correspond to a modification of the boundary conditions, that the real field should satisfy. In this case, I guess that the perturbed field $\phi'(x)$ should also satisfy the field equation (the perturbation stay "on-shell"). Someone confirm this interpretation ?

• Just a comment : Suppose there's that unknown field in a region of spacetime. The variational principle states that the action of that field over the region should be stationnary for ANY small variations of the field in the bulk. The variations could be of compact support (they vanish on the boundary of the region), and be restricted to be smooth (not necessarily analytic). Their derivatives may also vanish on the boundary. If so, then we don't need to care about the surface terms that comes out of the variations, since these terms are trivially 0. Is that right ? – Cham Nov 13 '15 at 18:19
• Also take note that the region $\Omega$ is arbitrary here, around any point in spacetime. It is not necessarily the whole spacetime manifold $\mathcal{M}$. – Cham Nov 13 '15 at 18:34
• Hi Cham. Welcome to Phys.SE. Here are some writing tips: 1. Write for the readers who see your post for the first time. No need to keep obsolete material in the main post (only exception: unless some answers point to it). Interested readers can instead browse the edit history. 2. Ask only one question. 3. Try to limit total number of edits if possible. – Qmechanic Nov 13 '15 at 21:36
• According to this paper : arxiv.org/abs/1203.2736, there are two versions of the stationary action principle for classical particles. This suggest that there should be two versions of the same principle for fields. This may (or may not) explain the issue. – Cham Nov 14 '15 at 20:10