# General Relativity - Curvature of two superimposed perfect pressureless fluids

If we imagine a hypothetical perfect fluid with no pressure, then in its inertial rest frame its stress-energy tensor would only have one component:

$$\quad T_{00} = \rho$$

Solving the field equation would give us a diagonal Ricci tensor with equal components:

$$\quad R{\mu\nu} = \frac12R$$ when $$\mu = \nu$$ and $$0$$ otherwise

So the Ricci curvature is the same in all directions, ie. the curvature is 'spherical' at that point.

Now we could add pressure terms $$T_{ii}$$ to the stress-energy tensor, and then the Ricci tensor would still be diagonal but not uniform, hence elliptical curvature instead.

However, we could also model 'pressure' (or at least normal stress) by having two pressureless fluids with equal densities and opposite velocities, superimposed on each other. Say they are moving in the x-direction. Then, since $$T_{\mu\nu} = \rho u_\mu u_\nu$$, we have

$$\quad T_{\mu\nu} = T_{\mu\nu_1} + T_{\mu\nu_2} = \rho_1u_{\mu_1}u_{\nu_1} + \rho_2u_{\mu_2}u_{\nu_2}$$

But since $$\rho_1 = \rho_2$$, $$u_{x_1} = -u_{x_2}$$, and $$u_{t_1} = u_{t_2}$$, the $$xt$$ components cancel. And since the $$y$$ and $$z$$ velocities are all zero, we only have

$$\quad T_{tt} = 2\rho u_t^2$$ and $$T_{xx} = 2\rho u_x^2$$

In other words we've effectively added pressure. As before, the resulting Ricci curvature will not be uniform. But if you think about the field equation,

$$\quad R_{\mu\nu} - \frac12Rg_{\mu\nu} = T_{\mu\nu}$$

it is linear in the sense that if $$R_{\mu\nu_1}$$ solves $$T_{\mu\nu_1}$$ and $$R_{\mu\nu_2}$$ solves $$T_{\mu\nu_2}$$, then $$R_{\mu\nu_1} + R_{\mu\nu_2}$$ should solve $$T_{\mu\nu_1} + T_{\mu\nu_2}$$, because the Ricci scalar is a linear contraction of the Ricci tensor. So the superposition of these two fluids should correspond to adding their Ricci tensors. But if you add two tensors that are each uniform along every direction of spacetime, the result should still have that property. Put another way, each separate Ricci tensor has the whole Minkowski space as a single eigenspace, so their sum should as well, just with the eigenvalues added. The only difference is that we are not working in the rest frame anymore, but the mentioned properties are frame-independent. So how could the curvature become elliptical when it seems it should still be spherical? I've been thinking about this and I just don't know how to reconcile it.

$$\quad R(U, U) = {U^t}^2 + {U^x}^2 + {U^y}^2 + {U^z}^2$$
Here U is a unit vector. But because the tangent space is hyperbolic, it's not the sum but the difference of squares that's the same for all unit vectors, ie. the $${U^t}^2$$ would need a minus sign. In other words the tensor would have to be a multiple of the metric. Which makes perfect sense because the metric gives the same value for any unit vector. With the exception that time-like and space-like vectors give opposite signs - so the curvature would be positive along any time-like direction and negative along space-like ones, or vice versa. I guess that's the closest thing to spherical curvature in relativity.