1

An argument I hear repeatedly is the light-speed limit only applies where spacetime is flat, so faster-than-light speed is possible where spacetime is curved. Thus special relativity does not apply to our expanding universe which is theoretically expanding faster than light beyond the cosmological horizon. Here's the problem: the universe's spacetime is flat. The cosmological constant is a very small number in curvature units. The WMAP has empirically confirmed this flatness. So why wouldn't special relativity apply--especially the light-speed limit?

Qmechanic
  • 201,751
garmichaels
  • 93
  • 5
    You seem to be confusing several things. Faster-than-light speed is not possible in either a flat or curved spacetime. This limit applies to objects moving in spacetime and not to the expansion of spacetime itself. – user341440 Feb 01 '22 at 19:50
  • My answer below addressed a different misconception in your question, but I second the above comment. – Michael Seifert Feb 01 '22 at 19:54
  • 1
    Does this answer your question? Will acceleration rate of expansion of space become faster than speed of light? – rfl Feb 01 '22 at 20:27
  • user341440, granted, but the question is why shouldn't special relativity apply given that the universe's spacetime is flat? – garmichaels Feb 03 '22 at 19:40
  • "This limit applies to objects moving in spacetime and not to the expansion of spacetime itself."

    Right, user341440, but given that the cosmological constant is small and spacetime is flat on a large scale, why wouldn't special relativity apply? And how can you be so sure that the limit does not apply beyond the horizon?

     – garmichaels Feb 03 '22 at 20:00
  • At https://arxiv.org/abs/astro-ph/0310808 , you'll find the preprint of a paper that's very widely-used in explanations of the differences between SR and GR, especially in cosmology. – Edouard Feb 11 '22 at 05:28

2 Answers2

4

"Spatially flat" is not the same thing as "flat spacetime". Even if the geometry at any moment of cosmological time $t$ is plain old Euclidean space, the expansion and contraction of this space as time goes on leads to non-zero spacetime curvature.

By way of analogy: the surface of the Earth can be thought of as the union of a bunch of circular lines of latitude, from 90°N to 90°S. Mathematically, a circle has no intrinsic curvature.* Yet the individual circles (with no curvature) form a two-dimensional surface with curvature.

*It has extrinsic curvature when you draw it on a plane, but extrinsic curvature isn't what we're talking about when we talk about spacetime curvature.

Michael Seifert
  • 48,772
  • "Spatially flat" is not the same thing as "flat spacetime".

    Michael Seifert, no one is claiming flat space is the same as flat spacetime. I mentioned that the cosmological constant is a very small number in curvature units, so spacetime is essentially flat in any case.

     – garmichaels Feb 03 '22 at 19:51
  • @garmichaels: Sorry I misinterpreted you! You mentioned the WMAP results, which famously showed that $\Omega_k = -0.0027 ± 0.0039$ (consistent with zero). But $\Omega_k = 0$ only refers to spatial flatness, rather than the full spacetime curvature, so that's what I assumed you were referring to. – Michael Seifert Feb 03 '22 at 20:32
1

Special relativity do of course also apply to our expanding universe. The statement of special relativity is not that there can be no speed faster than the speed of light, but that no information can be transported at speeds faster than the speed of light. The expansion of space does not transport information, so there is no contradiction.

rfl
  • 6,455
  • rfi, thank you. – garmichaels Feb 03 '22 at 19:53
  • If it can be observed (and it can, by the redshift of light from stars), the expansion of space conveys the fact that space is expanding, and the conveyance of any fact to an observer conveys information to that observer. I'd appreciate hearing any explanation as to why this would not be the case. – Edouard Feb 11 '22 at 04:45
  • @Edouard, that information is not transported at faster than light speed – rfl Feb 11 '22 at 11:11
  • @rfl-From the reference I'd made in my comment on the question, here's the argument for obs of superluminal expansion: "Although the photons are in the superluminal region and therefore recede from us (in proper distance), the Hubble sphere also recedes. In decelerating universes H decreases as ˙a decreases (causing the Hubble sphere to recede). In accelerating universes H also tends to decrease since ˙a increases more slowly than a. As long as the Hubble sphere recedes faster than the photons immediately outside it, D˙ H > vrec −c, the photons end up in a subluminal region and approach us." – Edouard Feb 11 '22 at 14:49
  • It really does take a close look at dashed & dotted lines in Lineweaver & Davis's diagrams, together with a realization that what we can see corresponds only to what the star whose photons are in question looked like when they left it, to "see" what they're talking about. (Also, the dots for the time derivatives copy off-center.) – Edouard Feb 11 '22 at 15:02
  • As near as I can tell, the top ("Proper Distance") diagram in the L&D article is intended to demonstrate the fact that we can (just barely) see the oldest known star (14 billion yrs. old, and nicknamed "Methuselah"), thru the processes they've described. In past- and future-eternal cosmological models by Penrose, Poplawski, and others, it seems unlikely that 1905's SR would suffice, whereas 1915's GR lives a life of its own, and surprised even Einstein. So did 1929's Einstein-Cartan Theory, later used by Poplawski. – Edouard Feb 11 '22 at 19:10
  • This is not to say that SR "doesn't apply": It's fine for preventing train collisions, avoiding ultraviolet catastrophes, and other quite serious terrestrial and galactic matters, and even its explanation of gravity (very well-described, I think, in "Science Asylum" on Youtube) suffices for such relatively math-free beings as myself. – Edouard Feb 11 '22 at 19:32