Do Einstein field equations only relate local spacetime curvature to local energy-momentum of matter?
If so, can we extend Einstein field equations globally relating global spacetime curvature to global energy-momentum of matter?
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$\begingroup$ The Einstein field equations are defined globally, its just that physicists tend to work with them in component form. The global equation is $$\operatorname {Ric} +\frac 12 Rg = \kappa T.$$ $\endgroup$– Daniel WatersCommented May 3, 2022 at 13:38
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3$\begingroup$ @DanielWaters That still works with the local objects (at one point of spacetime) $\endgroup$– OONCommented May 3, 2022 at 13:55
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1$\begingroup$ What, exactly, do you mean by "global curvature" (or "global" energy-momentum) as opposed to "local curvature"? $\endgroup$– ACuriousMind ♦Commented May 3, 2022 at 15:09
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$\begingroup$ @ACuriousMind I’m not really sure either but in certain spacetimes like minkowski space, the curvature is zero everywhere (globally) and is a vacuum solution of the EFE. Unless I’ve misunderstood OP’s question or got something wrong I think the answer is in general no, the equations cannot be extended. $\endgroup$– ThatpotatoisaspyCommented May 3, 2022 at 15:16
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1$\begingroup$ @OON, yes, Thatpotatoisaspy's answer clarifies the physical reason why you can't really define the EFE globally, but the Einstein curvature itself is a global object, so one can (in principle) solve the vacuum EFE globally, such solutions are known as Einstein manifolds. $\endgroup$– Daniel WatersCommented May 4, 2022 at 1:10
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2 Answers
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Yes, they cannot be extended to relate global curvature to global energy-momentum, not in general at least. You can see this by noting that the Einstein Field Equations can be derived by demanding that the stress energy tensor is locally conserved:
$${{\nabla}^{μ}T_{μν}} = 0$$
and noting that
$$\nabla^μ(R_{μν}-\frac{1}{2}g_{μν}R+Λg_{μν})= 0$$
is an identity in (pseudo) Riemannian geometry.
In general, the Einstein Field Equations cannot be extended to relate global spacetime curvature to global energy- momentum because global energy-momentum is not always well-defined. Noether’s theorem relates conservation of energy to time translation symmetry. In spacetimes like the friedmann expanding universe for example, time translation symmetry is not obeyed so no such globally conserved quantity can be defined.
In specific spacetimes you may be able to define a globally conserved energy. However the fundamental idea of general relativity is local, not global.
answered May 3, 2022 at 15:12
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$\begingroup$ So what i understand is that for the universe we are like a bug which try to solve Einstein's field equation to understand the universe locally or in other words we are trying to understand physical phenomenon of universe locally by solving EFE. Please correct me if i am wrong. $\endgroup$ Commented May 4, 2022 at 10:12
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1$\begingroup$ @KeshavShrestha The solutions of the EFE are called metric tensors which describe the geometry of spacetime at points, that is locally. So i suppose you could say we’re like bugs that are solving for local descriptions. The 1 to 1 equivalence of curvature to energy-momentum is a local statement. The EFE say “the curvature at this point is proportional to the stress-energy at the same point”. However since we haven’t defined what that point is we can use this local description to analyse any region of spacetime we choose. It allows you to solve for example, the path of a test particle. $\endgroup$ Commented May 4, 2022 at 11:18
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$\begingroup$ as you said "we can use this local description to analyse any region of spacetime we choose" does that mean we are analysing spacetime globally through pointwise local description? $\endgroup$ Commented May 4, 2022 at 16:31
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In the limit for small velocities and masses the Einstein solutions must match the Newtonian ones.
Suppose a point inside a tunnel from pole to pole of a spherical planet without atmosphere. The point is vacuum locally because a small neighborhood of the point is in the vacuum, but the Poisson equation $\nabla^2 \phi = 4\pi G \rho$, where $\rho$ is the density of the planet, is valid, which can be verified by substituting the known solution $$\mathbf a = -\nabla \phi = -GM\mathbf r$$ Where $M$ is the mass of the planet.
The same must be valid for the Einstein field equations. In this case, $T^{00}$ is related to the density of the planet as a whole, and not locally to a small neighborhood of the point (which would be zero). The Ricci tensor is also not zero as in vacuum solutions.
answered May 4, 2022 at 0:31