3
$\begingroup$

This is a question I've been mulling over for a while and I'm hoping someone here can point me in the right direction. Sorry if it's a bit of a novice question. For the record, I don't fully know GR, but don't let that stop you from using it in the answer.

Since the universe is expanding - that is, the spacetime metric is expanding by way of a near-exponentially increasing scale factor - we can say that the distance between any two non-bound objects is increasing over time. Herein lays my dilemma; if there were two objects separated by a large distance that had no relative velocities initially, after a long time, the effects of expansion would cause them to have large apparent velocities away from each other. Given that there hasn't been any acceleration to cause these velocities, are there still relativistic effects in play? That is, is there time dilation between the two frames?

Furthermore, given long enough time, the rate of increasing distance between the two objects could place them outside of their visible horizon (ie they are travelling away from each other at superluminal velocities). Since there was still no acceleration to achieve this feat, what can one say about the relativistic effects in this case?

At first I thought this was an easy question. I thought of course there would be relativistic effects and when the objects go superluminal, the visible horizon is there to ensure there can never be causal contact and thus preserve physics. But then I thought what if spacetime stopped expanding abruptly (seems crazy but as far as I know, nothing makes this completely impossible)? Since there was no initial relative velocities, wouldn't the two objects return to being in the same inertial frame? And seeing as none of them experienced any sort of acceleration, how then could we describe their two final states? By which I mean, if we were to assume there were relativistic effects during transit, how would we overcome such simple paradoxes like the twin paradox, or other relevant ones?

At this point, I'm stumped. I even attended a lecture by Miguel Alcubierre since he would have had to consider these types of effects in his design... No help. Equations are great to illustrate a point, but I'm really going to need a conceptual answer as well to fully understand this.

$\endgroup$
2
  • 1
    $\begingroup$ So what exactly is the question? The difference between acceleration-induced velocities and expansion-induced velocities, relativistically speaking? $\endgroup$
    – Jim
    Sep 3, 2013 at 14:41
  • $\begingroup$ @Jim In essence, yes but it doesn't sit well with me to simplify it that much. The question came about while reading about the Alcubierre warp drive. So in essence, I'm asking about the effects of superluminal travel due to expansion when causal contact is re-established afterward. But now I fear I'm maddening the unhelpful $\endgroup$
    – Jack Dozer
    Sep 3, 2013 at 14:50

2 Answers 2

3
$\begingroup$

Given that there hasn't been any acceleration to cause these velocities, [...]

As a side issue, even in Newtonian mechanics, accelerations don't cause velocities. Accelerations are just a measure of how rapidly velocities are changing.

What you're running into here is the fact that general relativity doesn't have any notion of how to measure the motion of object A relative to a distant object B. It is neither true nor false that A and B gain relative velocity due to cosmological expansion. It is neither true nor false that A and B have nonzero accelerations relative to one another. Frames of reference in GR are local, not global. It's valid to say that distant galaxies are moving away from us at some velocity. It's also valid to say that everything is standing still, but the space between us and the distant galaxy is expanding.

[...] are there still relativistic effects in play? That is, is there time dilation between the two frames?

Kinematic time dilation is well defined in SR, which means that in GR it's only defined locally. Gravitational time dilation is only well defined in GR in the case of a static spacetime, but cosmological spacetimes aren't static. So it is neither true nor false that there is time dilation between us and a distant galaxy. Concretely, you could measure Doppler shifts. If you feel like interpreting these shifts in purely kinematic terms, you can assign a velocity to the distant galaxy relative to us. But this is not mandatory and actually doesn't really work very well, in the sense that the velocity you get is usually several times smaller than the rate at which the proper distance between the galaxies is increasing. (Proper distance is defined as the distance you would measure with a chain of rulers, each at rest relative to the CMB, at a moment in time defined according to a notion of simultaneity defined by cosmological conditions such as the temperature of the CMB.) In particular, there are galaxies that we can observe that are now and always have been receding from us at $v>c$, if you define $v$ as the rate of change of proper distance. The fact that we can observe them tells us that their Doppler shifts are finite and correspond to $v<c$.

Here is a nice popular-level article that explains a lot of this kind of stuff:

Davis and Lineweaver, "Misconceptions about the Big Bang," http://www.scientificamerican.com/article.cfm?id=misconceptions-about-the-2005-03

It's paywalled, but there are lots of copyright-violating copies floating around on the web.

The following is a presentation of the same material at a higher level:

Davis and Lineweaver, "Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe," http://arxiv.org/abs/astro-ph/0310808

$\endgroup$
6
  • $\begingroup$ so what you're saying is no one knows for sure yet. $\endgroup$
    – Jack Dozer
    Sep 3, 2013 at 21:21
  • 1
    $\begingroup$ @DanII: We do know for sure. We know for sure that the question doesn't make sense. It's like asking how many pounds a thought weighs, or which tree in the forest is the most honest. $\endgroup$
    – user4552
    Sep 3, 2013 at 21:47
  • 1
    $\begingroup$ @BenCrowell my thoughts weigh about 20i pounds (note that thoughts are imaginary :-P) but they could probably stand to lose a couple and the most honest tree in the forest is obviously a birch and that's just common knowledge; I shouldn't have to explain it (not like those lying willows all shifty and whatnot). But as for DanII's question, I just don't know the answer. You? $\endgroup$
    – Jim
    Sep 3, 2013 at 23:15
  • $\begingroup$ @Jim: Thanks for making me laugh this morning. But as for DanII's question, I just don't know the answer. You? My answer explains the reasons why there is no literal answer to the question as posed, because the question makes assumptions that are false. It's not an unsolved mystery. It's been well understood since 1915 that this question is a type of question that is ill-posed, for the reasons I explained above. $\endgroup$
    – user4552
    Sep 4, 2013 at 15:49
  • 1
    $\begingroup$ It's not really true that there isn't time dilation between us and a distant galaxy. GR predicts that time runs slower in a galaxy seen at redshift $z$ by a factor $(1+z)$, which is indeed measured — e.g. in supernova lightcurves — at all redshift from 0 to ~10. $\endgroup$
    – pela
    Nov 20, 2017 at 21:29
0
$\begingroup$

The way I think about GR is that we simply set the absolute frame of the Universe with respect to G and Z, the gravitational constant and the impedance of space. Then find all the lines of symmetry in distance, mass and time that keep these two variables constant. We do that because space does not have the trig functions, space can only use Taylor series expansions about these two variables.

One might ask if these two are constant, and that is a better question. Einstein himself pointed out that the vacuum has another property, inherent noise, or energy. Why not conjecture about the fundamental lines of symmetry for vacuum noise, then just put all the equations of physics in those units?

$\endgroup$
1
  • 1
    $\begingroup$ Is there an answer here? $\endgroup$
    – Jim
    Apr 23, 2014 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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