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.


Well, I think I get it now, so I'll post an answer in case anyone's curious.

A diagonal Ricci tensor with equal components does NOT have the same value along all directions of spacetime. It is a multiple of the identity matrix, so it takes the sum of squares of a vector's components:

$\quad R(U, U) = {U^t}^2 + {U^x}^2 + {U^y}^2 + {U^z}^2$

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.

So what does this sum-of-squares tensor do then? Well, if we draw all the unit vectors on a standard space-time diagram, it would take the square of their length on that diagram. In Minkowski space, all unit vectors lie along a hyperbola, so the one along the time axis has the shortest length, and they get longer as you approach the asymptote (the speed of light). So the faster you're moving relative to the fluid, the greater the Ricci curvature, ie. convergence of free-fall paths, you experience. Which makes sense because, in that observer's frame, the moving fluid is Lorentz-contracted, hence has higher density, and also has pressure by virtue of its motion - so it's stress-energy tensor components are larger, meaning its curvature should be as well.

This is alluded to in the exercises of the "Spatial Curvature" section of John Baez's awesome GR Tutorial which someone on SO had kindly referred me to, but I didn't see the connection until now!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.