5
$\begingroup$

I came up with this query after @Rob Jeffries's answer to a previous question of mine.

So, is there any evidence that distant galaxies are moving away from us with speeds greater than $c$, due to the expansion of space, or is it just an artifact of Hubble's equation, $v=H_0D$?

If indeed this is a fact, does it determine the shape/geometry of our universe?

$\endgroup$
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/26549/2451 and links therein. $\endgroup$ – Qmechanic Oct 18 '16 at 20:52
  • $\begingroup$ Please do not make trivial edits (such as "space expansion" -> "expansion of space" or "Is there any" -> "What is the") as each such edit bumps the question as "active". $\endgroup$ – ACuriousMind Nov 13 '16 at 17:45
4
+50
$\begingroup$

There is evidence that galaxies at red shifts greater than about $\sim1.5$ are indeed moving away from us faster than light. We can measure the red shifts, and observe their light.

The General Relativity (GR) equation for redshift is not the same as it is in special relativity, which maxes at c for an infinite redhift. The relation of redshift to velocity is a linear approximation to both, and as you get to about z=0.5 and higher it starts diverging from the GR equation. V=HD which is Hubbles relation for velocity as function of distance and the Hubble constant, remains valid, even after v>c. Both GR and SR (special relativity) are approximately linear, v = cz, for velocity and redshift, at lower speeds. SR tops out at c, GR moves further up.

This is all pretty well explained in books like Dodelson, and you can also see it explained and graphed very nicely (all three curves, GR, SR, and the linear relation) in the following Arxiv article. Figure 1 particularly graphs all 3 curves, but the article has the equations. See,
https://arxiv.org/pdf/astro-ph/0011070v2.pdf

This effect does not determine the shape or geometry, it is rather fully determined by them. Specifically, the Lamda Friedman Robertson Walker solution of Einstein's Field Equations (with a cosmological constant), with isotropy and homogeneity assumed, leads exactly to the observed values for z as function of distance, for cosmological distances.

Edit: answer to @Yogi DMT's comment below.

No, the other way around. The crests are the same, except space stretches, and the distance between crests also. Say it emitted N cycles. It has to be the same number of complete cycles received, but the space covered by the same number of complete cycles is bigger, so the wavelength is larger. See a couple sites.

This one gives a basic interpretation of the gravitational redshift, and including the cosmological redshift: http://curious.astro.cornell.edu/physics/104-the-universe/cosmology-and-the-big-bang/expansion-of-the-universe/610-what-is-the-difference-between-the-doppler-redshift-and-the-gravitational-or-cosmological-redshift-advanced. It doesn't answer your question exactly but gives you a sense of the gravitational and cosmological redshift.

This next one gives more detail and detailed differences, a lot more specific, and also in the section on Expansion of Space derives the equation for the cosmological redshift in the standard cosmology model. It shows that 1+z = $a_{now}$/$a_{then}$, where the a's are the scale factor of the universe, i.e. The ratio in the equation is the ratio of the universe sizes (or say the distance to a distant specific Galaxy) now and then. Now is when the light is received, then is when emitted, in the comoving time coordinate. This next one gives more detail and detailed differences, also in the section on Expansion of Space derives the equation for the cosmological redshift in the standard cosmology model. It shows that 1+z = $a_{now}$/$a_{then}$, where the a's are the scale factor of the universe, i.e. the ratio in the equation is the ratio of the universe (or say the distance to a distant specific Galaxy) now and then. Now is when the light is received, then is when emitted, in the comoving time coordinate which is that used for the cosmological metric. It is at https://en.m.wikipedia.org/wiki/Redshift

You always have to be careful in these calculations to not shift coordinate systems in the same equation or concept. Easy to get confused, but the math is sort of easy.

$\endgroup$
  • $\begingroup$ If it's space itself that expanding wouldn't that mean the distance between crests doesn't change? Why then would the frequency of a wave be affected? $\endgroup$ – Yogi DMT Oct 20 '16 at 14:19
  • $\begingroup$ @Yogi DMT. See my answer to this question in my expansion edited in at the end, in my posted answer. $\endgroup$ – Bob Bee Oct 21 '16 at 1:59
  • 1
    $\begingroup$ This is not evidence, but a further description of what is expected to be the case. Your edit should mention that $c$ is invariant, that's why the frequency of a wave is affected. $\endgroup$ – Rob Jeffries Oct 21 '16 at 6:39
  • 1
    $\begingroup$ Isn't it that galaxies with $z>1.5$ are currently receding at $>c$? $\endgroup$ – Rob Jeffries Oct 21 '16 at 6:44
  • $\begingroup$ @BobBee What then is the difference between space stretching and objects just moving within a single "space density"? $\endgroup$ – Yogi DMT Oct 21 '16 at 13:30
-3
$\begingroup$

If the galaxy was traveling away faster than the speed of light, then we wouldn't be able to see it. (obviously) By very definition, it exist outside of the "observable universe". Not only can we not "observe" it with our eyes, but no information can reach us at all. (similar to the inside of a black hole). It cannot affect us in any way.

Asking wether things outside of the observable universe "exist" is somewhat of a "Zen riddle". It opens an very deep philosophical or meta-physical debate about the definition of the word "exist"

$\endgroup$
  • 5
    $\begingroup$ Quite untrue. Objects receding at $>c$ now can be seen now. They include all galaxies at redshifts $>1.5$. $\endgroup$ – Rob Jeffries Oct 21 '16 at 6:41
  • $\begingroup$ But you are seeing them as they were in the past. At that point in time they were traveling < c relative to us. You can't observe something traveling away at > c. Therefore you can't prove that such a thing exists, you can only extrapolate. That's all I meant. $\endgroup$ – Lorry Laurence mcLarry Oct 21 '16 at 10:23
  • 3
    $\begingroup$ I cannot see anything now if that's the definition you are using, since there is a finite light travel time for everything. The objects currently receding at $>c$ are part of the observable universe, since we observe them . $\endgroup$ – Rob Jeffries Oct 21 '16 at 11:17
  • $\begingroup$ I think we are degrading into semantics here. All observed galaxies travel slower than light. Therefore it is a matter of extrapolation that they might exist in some other state at some instant other than the one in which they were observed; not a matter of empirical evidence. Therefore the answer to the question is "Yes... but technically no." $\endgroup$ – Lorry Laurence mcLarry Oct 22 '16 at 17:28
  • 2
    $\begingroup$ No, this is simply not true. Not all observed galaxies travel slower than. You can indeed observe galaxies that traveled faster than $c$ when it emitted the light, and still travel faster than $c$ today when we observe it. That's a (non-trivial, admittedly) feature of the expanding Universe, which is outside the scope of special relativity. $\endgroup$ – pela Nov 13 '16 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy