# Solutions to Einstein Field Equations where $T_{\mu \nu} = 0$

My Level/Background:

I have just completed my first year of undergrad. In high school, I completed AP Physics C Mechanics and Electricity and Magnetism. In my first year of undergrad, I completed a course on Newtonian Mechanics and a course on Special Relativity and Electromagnetism which both approximately followed the sections on those topics in the Feynman Lectures on Physics.

The Question

I am starting to dive into tensor analysis and general relativity in my free time and I'm having some confusion about the Einstein Field Equation.

The Einstein Field Equation (without cosmological constant) states that $$G_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$ where $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$$ is the Einstein curvature tensor.

In most pop-science explanations of GR they say that matter and energy (or their density and flow I guess), which are represented by $$T_{\mu\nu}$$, cause spacetime to curve, which I assume is represented by the curvature tensor $$G_{\mu\nu}$$. Objects then move along the shortest path proper time path (geodesic) in this distorted spacetime.

They often do this by giving the pretty misleading picture of placing a large mass on a trampoline, where the fabric of the trampoline is spacetime, and showing how the large mass causes the fabric to bend and how this affects the motion of smaller objects thrown on the trampoline.

In the case of a spherical non-rotating planet, I'm assuming $$T_{\mu\nu}$$ is $$0$$ everywhere except for where the planet is. So that means $$G_{\mu\nu} = 0$$ everywhere not inside the planet.

My question is does that mean there is no curvature outside the planet (or is Einstein curvature a different thing than regular curvature)? Since this seems to imply that there would be no curvature in spacetime outside the planet which is clearly wrong since objects do orbit the Sun.

Or does the value of $$T_{\mu\nu}$$ inside the planet (where it is nonzero) affect the curvature of spacetime outside the planet (where it is zero) in a large radius around it?

In summary, what is the best way to think about how mass and energy affect the curvature of spacetime around them?

• weird caltech flex but okay – PhysMath Jul 21 '20 at 2:57

## 2 Answers

There are four different curvature tensors at play here. The complete information about the curvature is encoded into the Riemann tensor $$R^{\sigma}_{\;\mu \tau \nu}$$, and the other three tensors are all derived from it.

The Ricci tensor is a contraction $$R_{\mu \nu} = R^{\sigma}_{\;\mu \sigma \nu} = g^{\sigma \tau} R_{\sigma \mu \tau \nu}.$$

The Ricci scalar is a contraction $$R = g^{\mu \nu} R_{\mu \nu}.$$

The Einstein tensor is $$G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}.$$

The vanishing of $$G_{\mu \nu}$$ implies the vanishing of $$R_{\mu \nu}$$. It is easy to show: contract the definition of $$G_{\mu \nu}$$ with the inverse metric $$g^{\mu \nu}$$, you will obtain $$0 = G_{\mu \nu} g^{\mu \nu} = \left( 1 - \frac{d}{2} \right) R.$$

Here $$d = g^{\mu \nu} g_{\mu \nu}$$ is the dimensionality of space-time. Unless $$d = 2$$, we must have $$R = 0$$. Now plug this result into the definition of $$G_{\mu \nu}$$ to obtain $$0 = G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} \cdot 0 \cdot g_{\mu \nu} = R_{\mu \nu}.$$

Hence, in the vacuum, the Ricci tensor vanishes. In fact, Einstein came to this conclusion even before the final form of his equations for gravity was finalized. He tried generalizing it as $$R_{\mu \nu} = \kappa T_{\mu \nu}$$ first, and that didn't work out, which lead him to the definition of $$G_{\mu \nu}$$.

However, $$R_{\mu \nu} = 0$$ does not imply $$R^{\mu}_{\;\nu \sigma \tau} = 0$$. The space-time outside the region where the planet is located is still curved, even though the Ricci tensor vanishes. Your intuition is also correct: if the full Riemann tensor was to vanish outside the inner region occupied by the planet, test bodies in its vicinity wouldn't feel its gravity, which is not at all what we observe in nature.

• @mihirb it is a differential equation, a lot like other differential equations like Maxwell. Remember how in electrodynamics, the Coulomb's inverse-square law can be derived from a local differential equation $\nabla \vec{E} = \rho / \varepsilon_0$? A similar thing is going on here, only more mathematically involved. Look at solutions of Einstein equations for examples, like e.g. the one for the collapsing star. Vacuum solutions like Schwarzschild are also insightful, but they are obtained for the limiting case where the planet collapses to a point (a singularity). – Prof. Legolasov Jul 21 '20 at 0:51
• I see. I looked it up for the Schwarzschild metric and it seems like you can first solve the equations for the metric taking into account the symmetries that you are assuming, and then you can use the metric to find the value of the Riemann curvature at any point in spacetime. So mathematically, the value of $T_{\mu \nu}$ in the equations is affecting the Riemann curvature at any point in spacetime through the metric. – mihirb Jul 21 '20 at 1:00
• @mihirb yeah, the derivation of the Schwarzschild metric takes $T_{\mu \nu}$ into account, though in a somewhat strange way due to the source of that metric being an infinitely dense singularity. – Prof. Legolasov Jul 21 '20 at 1:01
• @Prof.Legolasov Schwarschild solution is exterior solution, when spacetime is spherically symmetric. It does describe spacetime outside any perfectly spherically symmetrical nonrotating object, not just collapsed object (albeit, only in this case it describes spacetime everywhere). – Umaxo Jul 21 '20 at 4:25
• @mihirb that mental model is essentially correct, but with a caveat. The situation you described is physically impossible, because $T_{\mu \nu}$ must satisfy $T^{\mu \nu}_{\;\;\;\;;\nu} = 0$. However, it is possible to make a precise statement along your line of thinking, which would make you conclude that perturbations in the matter distribution cause the ripples (gravitational waves) in the space-time geometry to propagate outwards with local velocity $c$. I'm not going to make this precise statement here as it is a long story, and comments are not meant for extended discussion. – Prof. Legolasov Jul 21 '20 at 21:17

It is true:

$$G_{\mu\nu} = 0$$

at, say, the space station...yet it doesn't just sit there, does it?

Look at Maxwell's equation:

$${\bf \nabla \cdot E} = \rho/\epsilon_0$$

we could just as well say "charge tell the electric field how to diverge, and the electric field tells charge how to move" (to paraphrase J. A. Wheeler), but a zero divergence near a charge doesn't mean zero electric field.

Likewise, $$G_{\mu\nu}=0$$ doesn't mean $$g_{\mu\nu} = \eta_{\mu\nu}$$.

• Thanks! The comparison to electromagnetism and $\nabla \cdot \vec{E} = \rho / \epsilon_0$ was very helpful. – mihirb Jul 21 '20 at 1:05
• One question though. What would be the field in GR that corresponds to the electric field in electromagnetism? spacetime? i.e. any mass affects curvature in a region of spacetime around it similar to how a charge produces an electric field in a large region around it? I guess the local Einstein equations at each point can tell you how the mass's (region where $T_{\mu\nu} \neq 0$) effect on spacetime can "ripple out" and affect the curvature of spacetime at the location of the space station. – mihirb Jul 21 '20 at 19:26