Einstein Field Equations Derivation Can someone give a derivation of the Einstein field equations, even a heuristic one? In most of the treatments I've seen, it somewhat falls out of the sky.  Is the basic postulate that curvature is directly proportional to the curvature?  How do we get those multiple Ricci terms?
 A: I am including here an excellent answer by Youtube user Sen Der (@sender1496) to my question posed on a ScienceClic General Relativity video series – an excellent review of the topic.

I'm reading Wald's book on relativity, and it has a somewhat decent derivation I think. First, it defines a generic covariant derivative operator $∇$. After that, it defines a derivative operator associated with a said metric, which intuitively can be thought of as a sort of gradient that takes the curvature of space into account. This leads to the definition of the Riemann tensor in terms of $∇$: $(∇_a ∇_b - ∇_b ∇_a)ω_c = R_{abc}^d ω_d$, which can be shown to describe both the acceleration between geodesics (for example, two initially parallel geodesics fail to remain parallel if there is curvature) and parallel transport deviations in a small closed loop. All of this holds simply because of the axioms of $∇$ along with the fact that it's associated with the metric $(∇_a g_{bc} = 0)$. Furthermore, all geometry is derived from $g$. So for example, inner products are $g_{ab} v^a w^b$ (i.e. $g(v, w)$) and orthogonal vectors fulfill $g_{ab} v^a w^b = 0$. Now, based purely on these definitions, you can show a few interesting facts:

*

*The Riemann tensor is a highly symmetric object, generally speaking. For example, $R_{abcd} = -R_{bacd}, R_{abcd} = R_{cdab}, R_{abcd} = -R_{abdc}$, and so on.

*Because of 1, there is only one meaningful contraction of the Riemann tensor that's non-zero. This is the contraction with respect to the first- and third index (or second and fourth, this is the same thing due to symmetry). Any other contraction is zero. This is why the Ricci tensor is defined in the way it is.

*It can be shown that $∇^a(R_{ab} - \frac1 2 g_{ab} R) = 0$. This is due to the bianchi identity of the Riemann tensor, which can be proven from its definition (this is easy to do).

So what is the derivation then? Well in the book, you start by considering the fact that tidal forces on a body (which isn't an "absolute" measurement of gravity, since those can be problematic in relativity) is given by $-(x \cdot ∇)∇Φ$ in Newtonian dynamics, where $x$ is the deviation vector (here, $∇$ is just the regular gradient). By setting this equal to what the tidal forces should be in general relativity ($-R_{cbd}^a v^c x^b v^d$, might look complicated, but this is simply the acceleration between two geodesics, discussed above), we land at $R_{ab} = 4 π T_{ab}$ (we had to use Poisson's equation $∇^2 Φ = 4 π ρ$ and the stress tensor $T_{ab} v^a v^b = ρ$ to get here). This was first suggested by Einstein but it can quickly be seen to be wrong. The reason for this is that $∇^a T_{ab} = 0$ has to be fulfilled by the stress-energy tensor, but this leads to the contraction of T having to be constant, which isn't true in general. So clearly, this Newtonian substitution is wrong.
The solution is to remember that 3 above holds. As a very good guess, we add the term $-\frac 1 2 g_{ab} R$ to the left hand side of $R_{ab} = 4 π T_{ab}$. Applying $∇^a$ to both sides then yields zero on each side due to statement 3. Moreover, it can be shown that when relativistic effects are negligible (this can be done by writing R in terms of $T = T^a_a$ and setting $T_a^a = -ρ$ in the Newtonian limit), it simplifies to Newtonian gravity, apart from a constant factor. If we change the $4π$ to $8π$ in the equation, this makes it correctly simplify to the Newtonian case when relativistic effects are negligible. Hence the modified equation $R_{ab} - \frac 1 2 g_{ab} R = 8 π T_{ab}$ is both consistent with the conservation equation $∇^a T_{ab}$ and correctly simplifies to Newtonian observations when relativistic effects are small.
(end Sen Der's answer)

A related line of reasoning I've gathered from watching L. Susskind's GR lectures I will summarize here (which somewhat overlaps the previous one):

*

*Begin with the hypothesis that the curvature of space and time is determined by the presence, distribution and motion of matter, energy, and momentum in the region.


*In the process of accounting for the matter, energy and momentum fluxes in a space, it is shown that the stress-energy $T_{\mu\nu}$ must be a Rank 2 (pseudo)tensor


*Accounting for the curvature of space and time in the framework of Riemann gives us the metric tensor $g_{\mu\nu}$ by which intervals and distances in space time are related, viz. $ds^2=g_{\mu\nu}dx^{\mu} dx^{\nu}$ and from it the Riemann tensor $R_{\alpha\beta\gamma\delta}$


*There are only 3 possible Rank 2 tensors that can be formulated from the metric and Riemann tensors: Ricci tensor $R_{\mu\nu}$, Ricci scalar times metric tensor $R g_{\mu\nu}$, and arbitrary scalar times metric sensor $\Lambda g_{\mu\nu}$ (though the third was only added later in the theory's development).  And because conservation of energy & momentum requires covariant derivative of $T_{\mu\nu}$ to be zero, this means the covariant derivative of the Einstein Tensor $G_{\mu\nu}$ must be zero.  Mathematically, because of the Bianchi identity, the simplest form of $G_{\mu\nu}$ whose covariant derivative always vanishes is $G_{\mu\nu} \equiv R_{\mu\nu} – \frac1 2 g_{\mu\nu} R $.  So the EFE becomes $R_{\mu\nu} – \frac1 2 g_{\mu\nu} R = k \, T_{\mu\nu}$ (neglecting cosmological constant term).  Then to find $k$, we take the Newtonian Limit of $T_{00}$ which produces $k = \frac{8\pi G}{c^4}$.
