The force of gravity does not exist

I understand(-ish) that, following general relativity, an apple falls onto earth, not because there is a force pulling earth and the apple toward each other but because, both the earth and the apple are travelling through time in the same direction and that their mass curve the space-time fabric causing the geodesic of the movement through time to "fall" down the same location in space.

Why does mass and energy curve space-time?

I don't understand why mass (and energy) curve space-time. I am "familiar" (I never saw any explanation about it) with Einstein equation putting $T_{\mu\nu}$ (Mass and energy) and $G_{\mu\nu}$ (space-time geometry) in a linear relationship but I fail to get an intuitive explanation for why such relationship between $T_{\mu\nu}$ and $G_{\mu\nu}$ exist and why it has to be linear. Note that I don't really understand the true meaning of $T_{\mu\nu}$ and $G_{\mu\nu}$ but only the vague description that I made of them.


In short, my question is

Can you give me an intuition for why mass and energy curve the space-time fabric and for why this relationship is linear?

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    $\begingroup$ Although it's true that the relationship between stress-energy and Einstein curvature is linear in form, GR is a nonlinear theory. For instance, if I can write down a solution to the field equations for a gravitational wave, and the field equations for some other gravitational wave, I cannot add them together the way I could for electromagnetic waves. $\endgroup$ – Ben Crowell Dec 31 '17 at 21:39
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    $\begingroup$ In the weak field limit, GR reduces to Newtonian gravity, which is linear, so to me that seems like a sensible reason to expect the form of the Einstein field equations to be linear. If you try to add a nonlinearity to the EFE, it's pretty difficult to do it in a way that (1) is tensorial, (2) still gives the correct Newtonian limit, and (3) doesn't introduce any new dimensionful constant. For instance, I could multiply one side of the EFE by a factor of $1+kR$, where $R$ is the trace of the Ricci tensor, but then I would have a new dimensionful constant $k$. How would we guess its value? $\endgroup$ – Ben Crowell Dec 31 '17 at 21:44

Imagine you don't know of any kind of gravity at all, just all the other interactions matter can have. When you write down an action in Lagrangian field theory, you can use it to derive the zero-gravity equations of motion. Now pretend instead that gravity due to spacetime curvature is the only thing you do know about: again, what happens follows from a stationary action principle. So how do we make a fuller theory that uses both pieces of information? Well, we just add two actions together, with a proportionality constant that governs the strength of gravity. (In theory there could be an extra interaction term in the action, but in general relativity there isn't.) In short, the reason $T_{\mu\nu}-\kappa^{-1}G_{\mu\nu}=0$ comes from varying the full action is because the full action is of the form $S_{\text{Gives}T}-\kappa^{-1}S_{\text{Gives}G}$.


There are several reasons:

  1. What we know is that matter interacts. Gravity, whatever it is, is a certain interaction which is different from all the other interaction we know. So there must exist an equation to describe the dynamics given a matter distribution and ignoring all other interaction (we can limit ourselves to EM interaction since the other two are "quantum-effects").
  2. The fact that the space-time is curved is a natural consequence of searching for a theory which DOES NOT privilege a certain system of coordinates. Think about what happens if you change coordinates in the plane to the surface of a sphere, (by for instance a stereographic projection), you "curved" your plane, but just because you changed point of view. But, since Einstein pointed out that gravity is an expression of a change of coordinates, then what is caused by matter is something pretty similar to what is done by changing coordinates. Ergo, the space-time must be curved by matter!
  3. The linearity in the equation is because the Ricci tensor is 2nd order in the metric and, as in every fundamental equation of physics, we want to stay up to the second order in the solution; putting higher orders into our equation just makes weaker our system, since we need more and more fixed initial conditions, which is a pretty hard problem even though we stopped just at the 2nd order.
  4. Finally, we set $T_{\mu\nu}$ and $R_{\mu\nu}$ because we need the "right symmetries" in our field equation and also because we know that where "there is no matter" there is gravity instead (e.g. Schwarzschild solution), so for $T_{\mu\nu}=0$ we still want gravity.

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