# Interpretation of Einstein field equations

I have a beginners question concerning the Einstein field equations.

Although I've read some basic texts about G.R. and understand some of the formula's and derivations I still have the feeling there's something I don't properly understand. Hope my question is not too vague.

The Einstein field equations state: $$G_{\mu \nu } =\kappa T_{\mu \nu }$$.

I am told that these equations can be justified on the basis of two fundamental properties:

1. They are coordinate independent, meaning that these are tensors therefore they can be defined without reference to a specific coordinate system. For example 'Gravitation' by Misner, Thorne and Wheeler devotes a paragraph on the importance of 'coordinate independent thinking'.

2. The beauty of above equations is that they generate a curvature of spacetime on the left side in which the stress energy tensor on the right is automatically conserved: $${T^{\alpha\nu}}_{;\nu}=0$$.

Now how does this compare to classical mechanics?

Newtons laws imply that we are describing our system with respect to an inertial frame of reference. Hamiltonial mechanics gives us a way to use more general coordinates. But still the assumption is that spacetime is flat and we always know how to transform back to our 'normal' inertial coordinates.

A typical approach in classical mechanics would be to specify some distribution of mass at $$t=0$$ as a boundary condition, then use Newton or Euler-Lagrange to calculate the evolution of this distribution in time thereby arriving at a (conserved) value for the stress-energy tensor at every time and at every point in space.

In general relativity the equation $$G_{\mu \nu } =\kappa T_{\mu \nu }$$ somehow suggests to me that our 'boundary condition' now requires us to specify some $$T_{\mu \nu }$$ for all of spacetime at once (meaning at every point and at every time). With that we can now calculate the curvature of our spacetime, which in turn guarantees us that stress-energy is automatically conserved.

I know I'm probably wrong, but it seems to me that in this way we can specify just any stress energy tensor on the right (since it's automatically conserved anyway). After solving the field equations we are done, only to conclude that (since we've calculated that spacetime must now be curved in some way) we don't exactly know in what coordinates we are working anymore.

Also having to specify $$T_{\mu \nu }$$ over all of spacetime suggests to me that we have to know something about the time evolution of the system before we can even begin to calculate the metric (which in turn should then be able to tell us about the time evolution of the system??).

Is it perhaps that we have to start with postulating some form of $$T_{\mu \nu }$$ with certain properties, and then solve for $$g_{\alpha\beta}$$ and work our way back to interpret how we would transform our spacetime coordinates to arrive at acceptable coordinates (local Lorentzian coordinates perhaps)? So that we have now rediscovered our preferred local Lorentzian coordinates after which we can look at our $$T_{\mu \nu }$$ again to draw some conclusions about it's behaviour?

So my question is:

Is there a straightforward way to explain in what way the Einstein field equations can have the same predictive value as in the classical mechanical case where we impose a boundary condition in a coordinate system we understand and then solve to find the time evolution of that system in that same coordinate system?

I know the question is very vague, I wouldn't ask if I wasn't confused. But I hope that maybe someone recognizes their own initial struggle with the subject and point me in the right direction.

• A few general remarks: Einstein's field equations are not unique. There are modifications (cosmological constant, torsion, scalar terms etc.) that are perfectly "physical" and that have been explored to some extent. The reason why you may think that the coordinate system problem doesn't exist in Newtonian mechanics is because you were never taught Newtonian mechanics in arbitrary coordinates and in Hamiltonian mechanics the textbooks usually pick from pretty trivial "well suited" coordinate systems for you. That, however, is not the whole story there either. Oct 31, 2022 at 20:05
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Oct 31, 2022 at 20:08
• @Community I know. I'll see if I can elaborate and come up with a more specific example tomorrow. Oct 31, 2022 at 21:13
• It sounds to me like you are looking for the ADM formalism, a Hamiltonian approach to GR which evolves a 3-metric in time. Oct 31, 2022 at 22:05
• @Ghoster Thanks, I think that might help me gain a better understanding. I'll look into it. Nov 1, 2022 at 8:45

There are no Lorenzian coordinates in General Relativity. The properties of space-time, such as geodesic lines, conservation laws and singularities, can be described in any suitable coordinate system.

Usually Einstein's equations are solved for a specific type of matter, for example electromagnetic field, for which: $$T_{\mu\nu}=\frac{1}{4\pi}\left(F_{\mu\alpha}F_{\nu}{}^{\alpha}-\frac{1}{4}g_{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right)$$ Then the equation $$T^{\mu\nu}{}_{;\nu}=0$$ leads to Maxwell's equations: $$F^{\mu\nu}{}_{;\nu}=0$$. Einstein's equations are solved for both metric $$g_{\mu\nu}$$ and electromagnetic field tensor $$F_{\mu\nu}$$. Both $$g_{\mu\nu}$$ and $$F_{\mu\nu}$$ must be specified on a spacelike hypersurface (such as $$t=0$$ in Minkowski space; these are initial conditions) and at the spatial infinity (boundary conditions); Einstein equations allow to find the evolution of the system in the future.

Also, as a side remark, equation $$T^{\mu\nu}{}_{;\nu}=0$$ isn't a conservation law in the common meaning. If a space-time has a symmetry with respect to transformation $$x'^{\mu}=x^\mu+\xi^\mu$$, then vector $$\xi^\mu$$ satisfies the Killing equation: $$\xi_{\mu;\nu}+\xi_{\nu;\mu}=0$$ Then $$(T^{\mu\nu}\xi_{\nu})_{;\mu}=0$$, and $$J=\int T^{0\nu}\xi_{\nu}\sqrt{-g}d^3x$$ will be a conserved quantity. If there are no Killing vectors, then there are no conservation laws.

• Thanks! I think your remark that BOTH $g_{\mu\nu}$ and $F_{\mu\nu}$ must be specified TOGETHER on a spacelike hypersurface answers my question! Somehow I missed that until now. I allways viewed the right hand side as the 'source to be specified' and the left hand side as the 'thing to be calculated'. Solving for $g_{\mu\nu}$ and $F_{\mu\nu}$ together makes much more sense to me now.. Thanks again. Nov 1, 2022 at 8:41

The situation is far more similar in Newtonian gravity than you might expect.

these can be justified on the basis of two properties:

1. coordinate independent
2. stress energy automatically conserved Now how does this compare to classical mechanics?

The EFEs state how gravity relates to matter distribution. The Newtonian counterpart is $$\nabla_i\nabla^i\phi=-4\pi G\rho$$. Like the EFEs, this is coordinate-independent, with a tensor on both sides (although this time it's of rank $$0$$, not $$2$$; note also Newtonian tensors' indices run over space only, which affects how $$\nabla$$ is defined). The conservation law is $$\partial_t\rho+\nabla_ij^i=0$$, with $$j^i$$ the mass flux density.

Let's see how your other concerns would apply to Newtonian mechanics:

the assumption is that spacetime is flat and we always know how to transform back to our 'normal' inertial coordinates

Actually, Newtonian mechanics also works in curved space, but for historical reasons we didn't need to use it there. I used $$\nabla$$ above instead of $$\partial$$, partly because of that, partly because if you work in e.g. spherical polar coordinates rather than Cartesian ones you don't want to conflate $$\partial^i$$ with $$\partial_i$$.

specify some distribution of mass at $$t=0$$ as a boundary condition, then use Newton or Euler-Lagrange to calculate the evolution of this distribution in time

Yes, pair the two equations above.

our 'boundary condition' now requires us to specify some $$T_{\mu \nu }$$ for all of spacetime at once

Again, this applies in Newtonian physics (for each value of $$t$$ we specify what happens over all of space), but it's not upsetting because most problems find either negligible effects at long distances or, if we attempt a Newtonian equivalent of the Friedmann equations, a tractable characterization of large-scale effects that neglects small-scale details.

we can specify just any stress energy tensor on the right (since it's automatically conserved anyway)

No; if you choose $$T_{\mu\nu}$$ or $$j^i$$ that isn't conserved, it won't come up in a solution of the two equations, be it EFE+conserved $$T$$ or Newtonian gravity+conserved $$j$$.

(since we've calculated that spacetime must now be curved in some way) we don't exactly know in what coordinates we are working anymore

No; you choose a coordinate system in which to solve the equations up front.

we have to know something about the time evolution of the system before we can even begin to calculate the metric (which in turn should then be able to tell us about the time evolution of the system??)

If Newton calculated how a projectile moves over Earth's surface, at first he'd neglect the projectile's gravity, time-dependence in Earth's gravity etc. Once one knows the solution this obtains, perturbative calculations can make it more precise. So, in at least some problems - indeed, many problems of interest - we can do a pretty good job of this, if only numerically. Of course, many idealizations don't even need that.