# EFE and Local Minkowski

Suppose we view the Einstein Field Equations (EFE) in the context of a boundary value problem with a given stress-energy tensor and boundary conditions. The problem is solved by finding a pseudo-metric.

Is there an unspoken condition that the metric to be found is locally Minkowski or is this implied by the EFE?

Specifically, do you get a different resulting pseudo-metric if you choose a different signature such as (+---) rather than (-+++)?

• The manifold on which the metric is placed is taken to be pseudo-Riemannian, i.e. in any local patch one can always set up a flat coordinate system. May 30 '16 at 16:50
• By "locally Minkowski", do you mean of 1+3 signature? May 30 '16 at 16:59
• Thanks Robin, my question is -Is a 1+3 signature ( or other standard) an unspoken condition? May 30 '16 at 21:37
• Note from my answer: the metric signature is invariant for the whole manifold. But a $-2$ signature is the same as a $+2$: the only distinct signatures in four dimensions are $\pm 4$, $\pm 2$, $\pm 0$. The rest is just sign conventions.
– user107153
May 31 '16 at 14:14
• @Prahar What do you mean by "in any local patch one can always set up a flat coordinate system"? A flat coordinate system is usually taken to be one in which the coordinate vector fields are parallel (Jost 2011). The existence of such a coordinate system is equivalent to the vanishing of the curvature. Jun 1 '16 at 2:53

Given mild differentiability conditions on the metric (I thought this might be $C^\infty$ which is not very mild, but see comments to this answer by 0celo7 below) then, for any point $p$, you can always pick a coordinate system $\left\{x^i\right\}$ which is locally flat, which means that

• $g_{ij}(p) = \pm\delta_{ij}$ -- tangent vectors along coordinate curves are orthonormal at $p$;
• $\partial g_{ij} / \partial x_k \rvert_p = 0$ -- it's a good approximation;
• $\partial^2 g_{ij} / \partial x_k \partial x_l \rvert_p \ne 0$ in general -- but not that good an approximation.

In addition it is a theorem that the metric signature is invariant (this is because you can pick a basis for the whole manifold where the metric components are $\pm\delta_{ij}$ although this is not a coordinate basis in general of course, and it follows from continuity of the metric that its signature can therefore not vary).

Between them these two results are sufficient to show that, on sufficiently small neighbourhoods, things look like Minkowski space in GR. Note that this result just depends on differentiability conditions, not on the particular form of the field equations.

• I think the metric only needs to be $C^2$ for such coordinates to exist, not $C^\infty$. Also, it is not possible to pick coordinates (in general) in which the metric components are $\delta_{ij}$ globally because the existence of a global chart is only available to spacetimes which are homeomorphic to $\mathbb{R}^4$. May 30 '16 at 20:45
• @0celo7 I thought that too, but I think the result relies on Taylor series (which would imply $C^\omega$ though, not just $C^\infty$, so ...). The signature thing just relies on there being a basis for $TM$ where the cpts of the metric are $\pm\delta_{ij}$ everywhere: that basis does not need to be a coordinate basis (and never will be except in flat spacetime as you say).
– user107153
May 30 '16 at 21:18
• Oh I misread that (and made a dumb error), of course such a basis exists. I'm assuming the "locally flat" coordinates are what mathematicians call "normal coordinates"? I've never seen them defined by a power series. The metric needs to be $C^1$ so that we have a connection. Then there is a small neighborhood $N$ containing $p$ such that the exponential map is a diffeomorphism of a neighborhood of $0$ in $T_pM$ onto $N$. Then, let $\{e_i\}$ be a basis of $T_pM$, and the normal coordinates are defined by $N\ni r=\exp_p(x^ie_i)$. May 30 '16 at 22:55
• The first condition is nontrivial, but the second one is easy: In normal coordinates geodesics are linear, i.e. the geodesic with initial velocity $v$ has components $x^i(t)=v^it$, so the geodesic equation in these coordinates at $p$ is $\Gamma^i{}_{jk}(p)v^jv^k=0$. By symmetry we have $\Gamma^i{}_{jk}(p)=0$. Some algebra should yield the second condition from that. The third follows because the Riemann tensor does not generally vanish at $p$. But the curvature is defined for a $C^1$ connection, so the metric has to be $C^2$. No Taylor expansion needed. May 30 '16 at 23:00
• @0celo7: thanks! I have added a note to point to your comments. I think but am not at all sure without thinking further that the original $C^\infty$ requirement might have been because the reference I got this from was trying to show this before introducing a connection. But, anyway, the main thing is that such coordinates exist in all reasonable cases I guess.
– user107153
May 31 '16 at 14:11