# I Is time absolute?

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1. Nov 2, 2016

### Doctor Strange

If I take a spaceship and park it near the event horizon of a black hole and then measure the age of the universe by observing SNe Ia, then travel back out to normal space (no gravitational forces, at rest with respect to CMB), will the dates agree? That is, if the measured age of the universe is 13.8 billion years near the event horizon, and it takes me 100 million years (proper time) to travel back out to normal space, will my new measurement of SNe Ia agree with a date of 13.8 + 0.1 = 13.9 Billion years? If that is true, can we say that time is absolute (i.e. all observers will agree on the age of the universe when using SNe Ia)?

Last edited: Nov 2, 2016
2. Nov 2, 2016

### Staff: Mentor

Supernova data does not tell you the age of the universe. It can be used to find other things, such as the distance to the host galaxy, and with enough data you can determine how the expansion of the universe is behaving (accelerating vs not).

In any case, the essence of your question appears to be about how different observers measure the age of the universe. There is no absolute answer, as time passes differently for observers. However, they can, if they want to, agree that the time in a specific frame of reference be used as the de-facto age of the universe. The most convenient frame is probably a co-moving frame, which, in cosmology, is a frame that is at rest with respect to the CMB. Barring small differences in the gravitational forces in the local area, all observers in these co-moving frames will measure the age of the universe to be the same.

3. Nov 3, 2016

### GeorgeDishman

There seem to be two obvious problems. First, to "park" a ship by which I assume you mean it remains at a constant radius, you would need to fire its engine constantly so you are talking about an accelerating frame. Second, the ship observer would see a universe that would appear far from "homogeneous and isotropic" so it's not clear to me that you could even define a single age that would apply to all he could observe. I think he would have to apply some correction to remove the observational distortions which would effectively give him the view "as if the black hole wasn't there" which could then be analysed, but that would give the conventional age.

4. Nov 3, 2016

### Chronos

Absolute time requires an absolute reference frame, which, as far as we know does not exist. Cosmologists use an arbitrarily chosen reference frame called the CMB reference frame, but, it is no more absolute than the clock in your kitchen. Einstein dispelled the notion of absolute reference frames over a century ago.

5. Nov 3, 2016

### substitute materials

Chronos isn't it a bit misleading to say that the CMB reference frame is arbitrary? It's not an absolute reference frame, but it does make sense to choose this reference frame to define cosmic time, since we live in a homogenous and isotropic universe where co-moving coordinates can be conveniently defined, in which most particles are generally at rest in regards to the CMB, right? We see a dipole anistropy in the CMB that corresponds to a roughly 600 km/sec motion of Earth relative to this radiation, but these motions are randomly distributed on the largest scales. Hence we can meaningfully talk about the age of the universe.

6. Nov 3, 2016

### Doctor Strange

I've been working through this mental experiment and I keep on coming back to the same place. Roughly speaking, the age of the universe is the inverse of the Hubble constant. You can tweak it with FLRW, but for all intents and purposes, your clock is the inverse of the Hubble constant. If you begin moving relativistically with respect to the Hubble flow, the galaxies in front of you will become blue-shifted and the ones behind you will be more redshifted. It seems to me that any observer in any reference frame can correct the asymmetries and arrive at a proper calculation of the Hubble constant. So why isn't the age of the universe an objective and absolute value?

Last edited: Nov 3, 2016
7. Nov 3, 2016

### Bandersnatch

Because there's nothing absolute about the CMBR-rest frame. Of course you can convert what you observe to any frame you like - otherwise the reality of relativity would not be consistent. But in the end, the choice of the CMBR-rest frame is arbitrary, if convenient.

8. Nov 3, 2016

### Doctor Strange

I confused the issue by bringing the CMB rest frame into it. I corrected my comment to use the Hubble flow. If you begin moving relativistically with respect to the Hubble flow, you will measure an asymmetry in the red-shift of the galaxies around you. Everyone will be able to easily correct this asymmetry and arrive at the same measurement of the Hubble constant. That's the part I think you're missing. You can have any number of observers moving at any number of different velocities, they can all use the Hubble constant to figure out the objective age of the universe.

Last edited: Nov 3, 2016
9. Nov 3, 2016

### Bandersnatch

The issue here, I think, is that when we say 'arbitrary choice of reference frame', we mean that in the relativistic sense - the physics is the same in all frames, and there's nothing special about the CMBR-rest frame, apart from the fact that it makes description of the universe easier.

This is similar to how we cans ay that the FoR of the Sun is no more special than the FoR of the Earth, even though the former definitely helps with describing how the solar system works - as was proposed by Copernicus.

In other words, are there reference frames that are convenient? Yes. Are they absolute in the relativistic sense? No.

10. Nov 3, 2016

### Staff: Mentor

Because time is relative. An observer moving at relativistic speeds could "correct" for this and perhaps get a value similar to our own, but their is no reason to think that the time measured by their own clock is wrong. Just different.

11. Nov 3, 2016

### Doctor Strange

If I put Edwin Hubble's telescope in a spaceship, then leave in any direction and accelerate to 0.9c, then take his telescope out and measure the red-shift of all the stars in a sphere around me, what value will I get for the Hubble constant?

12. Nov 3, 2016

### Bandersnatch

It would be different depending on direction. Same as we get when measuring it here, on Earth - providing the precision is sufficient. You could then transform the observations into a frame where the $H_0$ is the same in every direction, which would be convenient.

13. Nov 3, 2016

### Staff: Mentor

No, they can all use the Hubble constant to figure out the age of the universe according to comoving observers. But they can also use their observations to figure out the age of the universe according to themselves--that is, according to an observer who has been moving along a worldline with the same observed anisotropy in the Hubble flow as they observe, since the Big Bang--and this age will be different from the age according to comoving observers. So there is no "objective age of the universe" in the sense of one that's the same for all observers.

14. Nov 4, 2016

### Doctor Strange

It would help to have a concrete answer to the question I posed: You take off from Earth, accelerate to 0.9c (in any direction), then measure Hubble's constant. What value will you get?

Last edited: Nov 4, 2016
15. Nov 4, 2016

### phinds

As Bandersnatch already told you in post #12, It would be different depending on direction.

16. Nov 4, 2016

### Doctor Strange

Whatever direction and whatever speed you head off in, you are going to observe an asymmetry in the red-shift. The galaxies in front of you will appear more blue shifted, the galaxies behind will appear more red-shifted (than an observer at rest wrt the Hubble flow). So how do you calculate a constant $H_0$ under those conditions?

17. Nov 4, 2016

### andrew s 1905

Roberto Mangabeira and Lee Smolin in "The Singular Universe and The Reality of Time" take a difference stance. The argue "that there must exist a preferred and global conception of time" in essence one based on a frame defined by the CMB/Hubble flow. They claim it is compatible with GR and SR etc. They reference shape dynamics, Gomes, H et. al. Einstein gravity as a 3D conformally invariant theory arXiv:1010.2481, 2010, and other idea in support of their conclusions.

I don't have the background to judge but it seems to be a possibility.

Regards Andrew
PS They explicitly state their book is not pop science and Smolin gives many technical references in addition to the one I included.

18. Nov 4, 2016

### Staff: Mentor

By using your redshift anisotropy measurements to calculate your velocity relative to a comoving observer at your spatial location. Then you can compute any observation the comoving observer would make, including what Hubble constant he would observe, based on your observations and that known relative velocity.

19. Nov 4, 2016

### Staff: Mentor

The word "preferred" has different possible connotations. A particular frame/conception of time can be "preferred" in the sense of matching up with some particular symmetry of the spacetime (in this case, the fact that there is a set of observers, the comoving observers, who always see the universe as homogeneous and isotropic, and the surfaces of constant time for those observers are therefore picked out by the geometry of the spacetime), without being "preferred" in the sense of being built into the underlying laws of physics. As far as I understand Smolin's claims, he is only saying that a particular frame/conception of time is "preferred" in the former sense, not the latter sense. But I have not read through his work in great detail, so I might be missing something.

20. Nov 4, 2016

### Doctor Strange

So you would calculate the same value for $H_0$ as an observer on Earth: 67.3. The age of the universe is essentially the inverse of the Hubble constant (you can refine it with the FLRW metric, but materially, it's the inverse). So if someone travelling at 0.9c measures 67.3 for $H_0$ and someone at rest (wrt Hubble flow) measures 67.3 for $H_0$, how are we not dealing with an absolute (like the speed of light, which is the same for any observer)?