Non-linearity of Einstein's field equations How to show that, for two Schwarzschild- metrics, the Ricci tensors of two metric tensors do not sum up linearly:
$R_{\mu\nu} (g_1+g_2) \neq R_{\mu\nu} (g_1)+R_{\mu\nu} (g_2)$
while
the Ricci tensor is given by
$$
R_{\mu\nu} = \partial_\lambda \Gamma_{\mu\nu}^\lambda - \partial_\mu \Gamma^\lambda_{\nu\lambda}+ \Gamma^\lambda_{\lambda\tau}\Gamma^\tau_{\mu\nu} - \Gamma^\lambda_{\tau\mu}\Gamma^\tau_{\nu\lambda}
$$
where $\partial_\mu \equiv \frac{\partial}{\partial x^\mu}$ and sum over repeated indices is implied ($a_\mu b^\mu \equiv \sum\limits_{\mu=0}^3 a_\mu b^\mu$).
The Christoffel symbols are further given by
$$
\Gamma^\lambda_{\mu\nu} = \frac{1}{2} g^{\lambda \tau} ( \partial_\mu g_{\nu\tau} + \partial_\nu g_{\mu\tau} - \partial_\tau g_{\mu\nu} ) . 
$$
$g^{\mu\nu}$ is the metric inverse of $g_{\mu\nu}$,
and the Schwarzschild-metric is given by
$$ ds^2 = g_{\mu\nu}dx^\mu dx^\nu= - c^2(1-\frac{r_s}{r})dt^2 + \frac{1}{1-\frac{r_s}{r}} dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2)$$
At first, I'm not sure how to write down $g_1$ and $g_2$ in detail. OK, $g_1$ may stay as-is (the $g_1$ of the Schwarzschild-metric). But then, $g_2$ uses the same coordinates $r$, $\theta$, $\phi$ as $g_1$. There is more then one case:
Case 1 (the simplest): At the beginning, there are two universes with one black hole, respectively. BH1 is happy in its universe (as the only thing in the universe), and BH2  emerges out-of-nothing (from another universe) at the same $r$, $\theta$, $\phi$.
Case 2: At the beginning, again, two universes with one black hole in each. Then, BH2 emerges at an arbitrary point in the universe of BH1. At the end, the two are moving around the joint center of gravity - $R_{\mu\nu}(g_1+g_2)$. Thats quite complicated, I suppose (but more realistic than case 1)
Supposedly it's sufficient, at least to start with, to calculate case 1, because that's easier.
The point is, I do not see why the two Ricci tensors should not just add up linearly. It's a complicated thing and I don't see it. Sorry for that, I'm still learning. If one could write down the calculation (or provide a link to where it is done) - I could hopefully see and understand.
To explain a bit more in detail my confusion: The Ricci tensor can be interpreted as the volume change due to the curvature in comparison to the flat spacetime. By adding an additional black hole (from another universe) to an existing one - shouldn't the volumes simply add up? You see, I am confused. A proper caculation of the difference between $R_{\mu\nu}(g_1 + g_2)$ and $R_{\mu\nu} (g_1)$ + $R_{\mu\nu} (g_2)$ would help tremendously.
 A: The Schwarzschild solution $g$ is a solution to the vacuum Einstein equations
$$
R_{\mu \nu}(g) = 0
$$
on $\mathbb R^{1 + 3}\setminus\{0\}$, and this is the unique spherically symmetric solution, up to scaling the mass parameter and multiplying the metric by a constant (this uniqueness is non-trivial, but is perhaps justified by looking at a derivation of the Schwarzchild metric).
Note that if you have two different Schwarzchild solutions, then their sum is not a multiple of the Schwarzchild metric. For example, let
$$
g_s = -\left(1 - \frac s r\right)dt^2 + \left(1 - \frac s r\right)^{-1}dr^2 + r^2d\Omega^2.
$$
Then
$$
g_s + g_t = -\left(2 - \frac{s + t}{r}\right)dt^2 + \left(\frac{2r^2-r(s + t)}{(r-s)(r-t)}\right)dr^2 + 2r^2 d\Omega^2.
$$
Now for general $s, t$,
$$
\frac{2r^2-r(s + t)}{(r-s)(r-t)} \neq 2 \left(1 - \frac{s + t}{2r}\right)^{-1} 
$$
and hence $g_s + g_t$ is not a solution to the vacuum Einstein equations (by the uniqueness of the Schwarzchild solution), and so
$$
R_{\mu \nu}(g_s + g_t) \neq 0.
$$
A: Here is the simplest explanation of nonlinearity of the Ricci tensor in terms of the metric.  Start with the Riemann curvature tensor $R^{l}_{ijk}$, regarded as a function of the metric $g$. Then for each (positive) constant $a$, $R^{l}_{ijk}(ag)= R^{l}_{ijk}(g)$.
Indeed, it follows from the definition of the Levi-Civita connection that metrics $g$ and $ag$ have exactly the same Levi-Civita connection. Indeed, the compatibility condition of the connection and the metric reads:
$$
X g(Y,Z)= g(\nabla_XY, Z) + g(Y, \nabla_XZ) 
$$
for all vector fields $X, Y, Z$ on the manifold.
Rescaling the metric does not alter this identity.
Now, the Riemann curvature tensor is defined entirely in terms of the connection $\nabla$:
$$
R(X,Y)Z= \nabla_Y \nabla_X Z- \nabla_X \nabla_Y Z + \nabla_{[X,Y]}Z.
$$
Hence, rescaling the metric does not change $R$. (Some sources use the opposite sign convention.)
From this, it follows that the Ricci tensor, which is the contraction $R_{ik}=\sum_{j} R^{j}_{ijk}$ of the Riemann curvature tensor, also does not change if you rescale the metric. Hence, $R_{ik}(g+g)=R_{ik}(2g)= R_{ik}(g)$. Hence, $R_{ik}(g+g)\ne R_{ik}(g) + R_{ik}(g)$, unless the metric is Ricci-flat.
