# Static universe, spacetime and the stress-energy tensor

1. Mar 22, 2015

### timmdeeg

Einstein's static universe obeys $\rho = 2\lambda$. So, attractive and repelling gravity cancel each other.

I'm curious about the spacetime in this universe. Because the scale factor is constant, it seems that neighboring co-moving test particles don't show relative acceleration, thus no geodesic deviation. So, from this I would expect the spacetime to be flat. On the other side, as this universe contains energy, the spacetime can't be flat like Minkowski-spacetime.

Obviously I am missing something and I will appreciate any help.

Perhaps it is important to look at the 4x4 matrix of the stress-energy tensor describing the static universe. Unfortunately I couldn't find any reference. It seems correct to say that all matter particles are at rest to each other. If so, does it mean that the non-diagonal elements, representing shear stress and momentum flux would vanish?

2. Mar 22, 2015

### Staff: Mentor

Correct.

No, because the geodesic deviation of comoving test particles is not the only possible kind of geodesic deviation. For the spacetime to be flat, all possible kinds of geodesic deviation must be zero. They're not in the Einstein static universe.

3. Mar 22, 2015

### timmdeeg

So there is geodesic deviation of non-comoving particles, though all matter particles are at rest (hard to grasp).
Does this mean, that the mere possibility of geodesic deviation creates sources of gravity? Then I would expect them to be components of the stress-energy tensor. If true, could you kindly give me a little more background regarding shear stress and momentum flux resp. in this special case?

4. Mar 22, 2015

### Staff: Mentor

No, I didn't say that. Remember that there can be geodesic deviation along spacelike geodesics as well as timelike geodesics. In the particular case of the Einstein static universe, there is no timelike geodesic deviation, but there is spacelike geodesic deviation. In other words, freely falling objects don't experience relative acceleration, but if, for example, you try to make triangles from spacelike geodesics, the sum of their angles won't be 180 degrees; it will be larger, because of the spacelike geodesic deviation. (It's basically a 3-d version of what happens with great circles on the surface of the Earth.)

Yes, the SET of the Einstein static universe is not zero. It is the SET of a perfect fluid with $p = - \frac{1}{3} \rho$, i.e., the pressure is negative and its magnitude is 1/3 of the magnitude of the energy density. (If you work out the details, this is because you have $p_{\Lambda} = - \rho_{\Lambda}$ for the cosmological constant and $p_M = 0$ for the matter. Combining those with the fact that $\rho_M = 2 \rho_{\Lambda}$ gives the net balance of $p = - \frac{1}{3} \rho$ for the SET as a whole.) This means that $\rho + 3 p = 0$ (note that I'm using units where $G = c = 1$), so the Friedmann equations give $a$ constant.

5. Mar 23, 2015

### timmdeeg

Would it be fair to say that the spacetime of the Einstein static universe - though being spatially curved - is flat? I don't mean flat in the Minkowskian sense, but flat with regard to the absence of timelike geodesic deviation. Or is the strict requirement to have 'flat spacetime' the vanishing of the Riemann tensor, or perhaps something else?

Thanks for explaining the SET of the static universe. One more question. As you mentioned the perfect fluid, is it the general consequence that in this case the SET is described by its trace only?

6. Mar 23, 2015

### julcab12

.... Einstein original FE solution has no spatial curvature, hence flat in general; or mathematically approximately flat. Since the observable universe is very close to homogeneous and isotropic.

"Einstein was interested in finding static ( = 0) solutions, both due to his hope that general relativity would embody Mach's principle that matter determines inertia, and simply to account for the astronomical data as they were understood at the (1) A static universe with a positive energy density is compatible with (5) if the spatial curvature is positive (k = +1) and the density is appropriately tuned; however, (6) implies that will never vanish in such a spacetime if the pressure p is also nonnegative (which is true for most forms of matter, and certainly for ordinary sources such as stars and gas). Einstein therefore proposed a modification of his equations, to

where is a new free parameter, the cosmological constant. Indeed, the left-hand side of is the most general local, coordinate-invariant, divergenceless, symmetric, two-index tensor we can construct solely from the metric and its first and second derivatives. With this modification, the Friedmann equations become

....and

These equations admit a static solution with positive spatial curvature and all the parameters [PLAIN]http://ned.ipac.caltech.edu/level5/GIFS/rho2.gif, [Broken] p, and http://ned.ipac.caltech.edu/level5/GIFS/Lambda.gif nonnegative..

http://ned.ipac.caltech.edu/level5/Carroll2/Carroll1_2.html

Last edited by a moderator: May 7, 2017
7. Mar 23, 2015

### Staff: Mentor

No. See below.

That isn't what "flat" means. AFAIK there isn't a brief term for a spacetime that has no timelike geodesic deviation.

Yes.

No. A perfect fluid has an SET of the form $diag(\rho, p, p, p)$ in the rest frame of the fluid. Knowing the trace alone does not contain all of the information in the tensor.

8. Mar 23, 2015

### Staff: Mentor

Yes, it does. Spatial slices of the Einstein static universe, in the standard FRW chart, are positively curved.

9. Mar 23, 2015

### julcab12

... He once tried the old flat Minkowski spacetime assuming no gravitational effects -- zero spatial curvature for simplicity until Ed and tensor came. or am i misunderstanding something.

Yes, positively curved (closed).

10. Mar 23, 2015

### timmdeeg

Thanks for clarifying flat spacetime.
Thanks.

May I ask a little more regarding timelike geodesics in the Einstein static universe?

Freely falling comoving particles: as their proper distance is constant over time, their geodesics should be parallel.

Freely falling particles, initially moving parallel with respect to the fluid:
Does your statement "freely falling objects don't experience relative acceleration" also include this case?
Would their geodesics converge linearly like great circles on earth towards the northern pole?

To the second case: I wonder If I misunderstand John C. Baez in http://[URL [Broken]']The Meaning of Einstein's Equation[/URL]. He says on page 13/14 that in the static universe the volume of a ball of freely falling particles shrinks accelerated, if the ball is moving relative to the fluid, because of increasing energy in the rest frame of the ball.
Now, provided this is correct and means relative acceleration, would then freely falling objects, which are not comoving experience tidal forces in contrast to comoving objects?
And further, is it this kind of information, which isn't contained in the trace of the SET?

Last edited by a moderator: May 7, 2017
11. Mar 23, 2015

### Staff: Mentor

That gives a static universe, yes--one completely empty of matter and energy. AFAIK Einstein never considered this as a valid solution for our actual universe.

12. Mar 23, 2015

### Staff: Mentor

Yes.

Baez' calculation, which you link to, appears to say that the answer is no; and that calculation looks correct to me, as far as it goes. I think I was too hasty in saying that there is no "timelike geodesic deviation" in the Einstein static universe. What is true is that, for the Einstein static universe, there are no nonzero components of the Riemann curvature tensor with $t$ as one of the indexes. So any geodesic deviation along timelike geodesics can only happen if those geodesics have components in directions other than the $t$ direction, i.e., if the freely falling particles are moving in comoving coordinates.

That's what Baez' calculation shows, yes.

Yes, that's what Baez' calculation indicates. The shrinkage of the non-comoving ball of test particles is due to tidal gravity.

Yes.

13. Mar 24, 2015

### timmdeeg

PeterDonis, you have helped me a lot to improve my understanding. Thank you very much.

14. Mar 24, 2015

### Staff: Mentor

You're welcome! Glad I could help.