Csdrhrt

I'm new in variations of metric. Let be $g$ a metric with signature (1,-1,-1,-1) on a manifold.
If I consider a family of variations $g+varepsilon h$ (depending on $h$), used to derive the Eulero-Lagrange equations, I think that $g+varepsilon h$ must be metrics with signature (1,-1,-1,-1). So I think that $h$ can't be a generic function but it must be a function in a way that $g+varepsilon h$ is a metric with signature (1,-1,-1,-1) and $h$ cannot be merely the zero function otherwise we cannot derive the Euler-Lagrange equations using Fundamental lemma of calculus of variations (or something similar). What is the suitable set where $h$ are in?

My question is strictly mathematical and it would be interesting solve this question also in general in calculus of variations.

To answer your question let's go schematically through the derivation of the Euler-Lagrange equation and see explicitly where the set from which we pick the $h$'s enters and if the signature really matters.

When deriving the Euler-Lagrange equation we start with a functional (action) $S[g] = int mathcal{L}[g]{rm d}^nx$ where $mathcal{L}[g]$ is the Lagrangian and consider a variation $S[g+epsilon h] - S[g]$. Expanding this it a series in $epsilon$ we end up with

$$frac{S[g+epsilon h] - S[g]}{epsilon} = int[F_{munu}h^{munu} + mathcal{O}(epsilon)]{rm d}^nx$$

In such a derivation we often have to do integration by parts giving us boundary terms and for these to vanish we have to impose some conditions on the $h$ we allow (just like for scalars where we typically only allow variations that vanish on the boundary).

Now we demand that the directional derivative $lim_{epsilonto 0}frac{S[g+hepsilon]-S[g]}{epsilon} = 0$ for all the allowed $h$'s giving us the Euler-Lagrange equation $F_{munu} = 0$ whose solution gives us the metric $g$ for which the action is stationary. This requires us to consider a large enough set of $h$'s such that $int F_{munu}h^{munu}{rm d}^nx = 0 text{ for all $h$}implies F_{munu} = 0$.

The exact choice of this set, apart from restrictions coming from boundary conditions, plays no big role in the derivation apart from indirectly: it should be such that $int F_{munu}h^{munu} = 0$ for all $h$ implies $F_{munu} = 0$ and also be such that the integrals exist. Trying to exactly specify this set is just tedious, which is why nobody really specify it very formally, and comes with no real advantage other than "being rigorous for rigor's sake".

Nowhere does the signature of $g$ or the variation $g+hepsilon$ really enter the derivation. We are anyway taking the limit $epsilonto 0$ so we only care about "infinitesimal" $epsilon$'s and the signature of $g$ will be the same as $g+hepsilon$ for small enough $epsilon$ if $g$ is non-degenerate. And even if it was not, in general the metric signature can change so I don't see why one would demand $g+hepsilon$ to have the same signature as $g$ (but if the metric turns out to be non-degenerate and regular everywhere which is the usual case then the metric signature doesn't change). You are of course free to consider only $h$'s with the same signature as you want and $epsilon > 0$ if that makes you sleep better at night, but note that this doesn't really change anything in the derivation.

Let us sketch what's at stake:

1. Assuming that the Lagrangian density ${cal L}(x,g(x),partial g(x))$ is a differentiable function of the metric $g(x)$ (and its derivative), then in accordance with the variational principle, in an interior point $x$, the first variation $h(x)$ of the metric is (and must be) an arbitrary symmetric matrix (at least up to a multiplicative constant, cf. pt. 2 & 3).




2. Note in particular that the metric $g(x)+varepsilon h(x)$ must have the same signature as $g(x)$ for sufficiently small $varepsilon$.




3. If the region is compact, we can use uniform continuity to argue that such sufficiently small $varepsilon$ exists globally.




4. To make the variation problem well-posed it is usually necessary to impose pertinent boundary conditions on $h$.