# I What does "expansion of the universe" mean?

1. Jun 23, 2016

### MeJennifer

[Moderator's note: thread spun off from post in a different thread quoted below.]

That seems to me a rather extraordinary claim.

Could you demonstrate how for instance cosmological redshift can be explained in a coordinate independent fashion?

Last edited by a moderator: Jun 27, 2016
2. Jun 23, 2016

### Staff: Mentor

It is pretty straightforward. The comoving congruence is coordinate independent, as is the fact that the resulting expansion tensor is expanding.

That is just calculating a null geodesic between two lines in the comoving congruence.

3. Jun 23, 2016

### MeJennifer

The expansion tensor is not a spacetime tensor but a tensor on some 3+1 foliation.

Let's stay scientific!

4. Jun 23, 2016

### Staff: Mentor

The expansion tensor is a perfectly valid tensor, both theoretically and operationally.

5. Jun 23, 2016

### MeJennifer

It is misleading to use this tensor as an argument!

6. Jun 23, 2016

### Staff: Mentor

Why? It is coordinate independent and it shows that the universe expanding. It therefore directly addresses what you considered to be an extraordinary claim.

Simply calling it misleading or unscientific is hardly a well reasoned argument. Your aggressive tone to pervect is not well considered, nor is your current tone with me.

7. Jun 23, 2016

### Staff: Mentor

Incorrect. The definition of the expansion tensor uses the projection tensor $h_{ab} = g_{ab} + X_a X_b$ which projects out the part of an arbitrary vector or tensor field that is orthogonal to the vector field $X$. But such projected vectors, tensors, etc. are still 4-vectors, 4-tensors, etc.

This is incorrect, not only because of the above, but because it is not always possible to even find such a 3+1 foliation in a spacetime containing a given timelike congruence, such that every 3-surface in the foliation is orthogonal to every timelike curve in the congruence. It happens to be possible for the congruence of comoving observers in FRW spacetime, but you can't depend on it as a general property.

8. Jun 23, 2016

### MeJennifer

Let me ask you this would you use the expansion tensor as an argument to prove coordinate independence?

9. Jun 23, 2016

### Staff: Mentor

You don't need an argument to prove coordinate independence; coordinate independence is an obvious fact about any observable quantity in GR, since any such quantity must be expressible as a scalar invariant. So I don't understand what your issue is here.

A photon emitted by a particular comoving observer has a 4-momentum vector determined at the event of emission by the photon's frequency as measured by the emitter--this emitted frequency is just the contraction of the photon's 4-momentum with the emitter's 4-velocity at the event of emission. The photon's 4-momentum vector determines a unique null geodesic from the event of emission to the event of reception; the photon's 4-momentum is parallel transported along this null geodesic to the event of reception. At the event of reception, the comoving observer who measures the photon's frequency has a 4-velocity, which is contracted with the photon's 4-momentum to give the measured frequency (modulo a factor of Planck's constant if you are not using natural units). This entire process is manifestly coordinate independent. The fact that the process results in a redshift is easily derived from the fact that the congruence of comoving observers has a positive expansion scalar, i.e., that its expansion tensor has a positive trace. This derivation is also manifestly coordinate independent.

10. Jun 23, 2016

### MeJennifer

Cosmological redshift is not a coordinate dependent quality, ok it seems I have a lot to learn!

11. Jun 24, 2016

### Staff: Mentor

This is true, but as @PeterDonis mentions this isn't what is usually meant by the term pseudo tensor in GR.

To your point however, if the trace of the expansion tensor is positive then the congruence is separating. If you time reverse everything (manifold and congruence) then you get a negative trace indicating that the congruence is contracting.

So the one additional thing that we have to do to say that the universe is expanding in a frame invariant manner is to invariantly identify which direction along the timelike congruence is the positive direction. For that we can simply appeal to features of the universe like entropy or the weak force interaction.

12. Jun 24, 2016

### RockyMarciano

Sure, but that appeal in no way can affect how a tensor and a pseudotensor are defined mathematically, and yet you claimed it was a tensor, wich is an object defined in mathematical terms only.

13. Jun 24, 2016

### MeJennifer

How about observers who are not comoving?
Or are those perhaps discounted as non relevant to general relativity?

14. Jun 24, 2016

### Staff: Mentor

Words are defined however a group of people chooses to define them, and often different groups of people use the same word to mean different things.

My comment that it is a valid tensor is correct, as is your comment that it is a pseudotensor. We were just using the definitions of different groups of people.

15. Jun 24, 2016

### Staff: Mentor

What about them? Why would they be discounted in any way?

16. Jun 24, 2016

### MeJennifer

Well how would this tensor work for non-comoving observers?

17. Jun 24, 2016

### Staff: Mentor

The same way it works for comoving observers. There is no restriction or requirement that an observer only use congruences that include their worldline.

For example, both a rotating and an inertial observer may use the same congruence to study a rotating disk.

Last edited: Jun 24, 2016
18. Jun 24, 2016

### Staff: Mentor

Non-comoving observers would be a different congruence of worldlines, so they would have a different expansion tensor. The reason the comoving observers are used to define the expansion "of the universe" is that those observers are picked out by a particular physical property: they all see the universe as homogeneous and isotropic for all time. No other congruence of worldlines has this property.

19. Jun 24, 2016

### MeJennifer

Exactly!

20. Jun 24, 2016

### Staff: Mentor

Exactly what?

They do form different congruences, but nothing prevents a non-comoving observer from identifying the comoving congruence and calculating their expansion tensor.

The identification of the comoving congruence does not depend on the motion of the observer doing the identification , nor does it depend on the choice of coordinates.

21. Jun 25, 2016

### MeJennifer

Expansion of the universe is obvious for comoving observers but not necessarily true for other observers. That is basic general relativity!

A continuation on this argument seems pointless as several people are prepared to backup an earlier statement by a poster who made the claim that expansion of the universe is valid for all observers.

22. Jun 25, 2016

### Ibix

If expansion isn't true for some observers, wouldn't that mean that the CMB was still at hundreds of millions of Kelvin for those observers?

23. Jun 25, 2016

### Staff: Mentor

This is not quite correct. What is correct is that the expansion scalar (the trace of the expansion tensor) of a congruence of worldlines (i.e., a family of observers) is a property of the congruence. The congruence of "comoving" observers in the universe has a positive expansion scalar; but one could construct other congruences that didn't.

However, the term "expansion of the universe" refers specifically to the expansion scalar of the congruence of "comoving" observers, because, as I said before, those observers are picked out by a particular symmetry of the spacetime: they are the ones who see the universe as always homogeneous and isotropic. So although one could find other congruences that did not have a positive expansion, the expansion of those other congruences is not what is meant by "expansion of the universe", because those other congruences don't match up with the symmetry of the spacetime in the same way.

It is also worth emphasizing that the spacetime geometry of the universe itself is the same regardless of what congruence of observers you decide to construct. See further comments below.

No. What it would mean is that those observers (the ones in some hypothetical congruence that did not have a positive expansion scalar) would not see the CMB as isotropic. So the CMB would not have a single temperature to such observers; its temperature would vary with position on the sky. (More precisely, it would vary much more than it would for a "comoving" observer, since the CMB is not perfectly isotropic even to comoving observers, as the WMAP and Planck satellite observations have shown.)

Another thing to keep in mind is that the properties of the CMB, as seen by a given observer at a given event, are constrained by the spacetime geometry of the universe itself, and in particular by how much the CMB has redshifted at a given event since the surface of last scattering. For example, suppose we found some congruence of observers with a negative expansion scalar, and we picked one observer from the congruence who happened to be passing Earth right now. The CMB photons passing that observer at this point are the same ones that are passing Earth, which means they are redshifted by a factor of about 1000 from the surface of last scattering, and that's just as true for the other observer as for us on Earth. The actual spectrum of the CMB that is measured by this other observer might be different because of his motion relative to us; but for him to see a CMB temperature of hundreds of millions of degrees (note that this is already many orders of magnitude higher than the temperature at the last scattering surface, which was only a few thousand degrees), he would have to be moving relative to us at ultra-relativistic speed, and even then he would only see that high CMB temperature in one particular direction (the direction of his motion relative to us--see my comment above about the CMB being anisotropic for such an observer).

24. Jun 25, 2016

### Staff: Mentor

It is a tensor, so it is true for all observers, even if it isn't obvious. (With the mathematical caveats mentioned above)

A congruence depends on the thing being observed, not on the observer. If a rotating observer and an inertial observer are both observing the same disk, then they will use the same congruence. They will calculate the same expansion tensor for the disk, and any related physical predictions such as material failure.

Similarly for a comoving and non comoving observers both observing the universe. They will all use the same congruence to observe the universe and reach the same conclusion about its expansion.

25. Jun 25, 2016

### MeJennifer

This statement is what I disagreed with:

Which was then vigorously defended by, in my view, false and misleading arguments.

Is still everybody but I fully agreeing with the statement that in general relativity the universe is expanding and that this is a coordinate independent fact?

Yes or no?