1
$\begingroup$

Suppose that we put Bob and Alice into intergalactic space. If they look around they will see the light from distant galaxies shifted according to the Hubble law. More importantly, the light is (on average) isotropic.

Now suppose we accelerate Bob to e.g. $\beta = 0.999$. He should see the light in forward angles shifted blue and in backward angles shifted more towards red. In other words, the universe isn't isotropic to Bob anymore.

Then again, if we accelerate both of them to $\beta$, the relative situation is identical, but now both 'should' see anisotropic universe. According to this reasoning, there exists a special reference frame where the universe is isotropic. This, of course, isn't what we measure.

What is the solution to this issue?

My take: Suppose the motion is along a common $x$ axis. At $t=0$ (according to Alice) two distant galaxies at $x = a$ and $x = -a$ shoot out a signal. Then Bob sees these events occuring at $t_{\pm} = \pm \gamma \, \frac{\beta}{c} a$. In other words, the backward signal was emitted when this galaxy was much younger and therefore (Hubble law) moving more slowly (giving a small redshift). But as Bob is moving away from it, it redshifts further. Also the forward galaxy is much older and is originally redshifted a lot. But as Bob is moving towards it, it blueshifts.

It would be great is there was some explanation without using the whole machinery of GTR. Thanks!

$\endgroup$
  • $\begingroup$ According to this reasoning, there exists a special reference frame where the universe is isotropic I failed to see the reasoning. Ability to choose a reference frame does not guarantee that a random cosmological solution of GR could become isotropic in any frame. You can do it only if the solution was isotropic in some special frame to begin with. $\endgroup$ – A.V.S. Nov 9 '18 at 3:05
3
$\begingroup$

According to this reasoning, there exists a special reference frame where the universe is isotropic. This, of course, isn't what we measure.

There is such a special frame, and that is what we measure. The special frame is the frame moving with the Hubble flow, and in that frame, the CMB is observed to be uniform in all directions, with no differences in Doppler shifts. The Hubble flow can also be characterized approximately as the frame in which the galaxies are at rest. That's why the universe looks nearly isotropic in the earth's frame. In the earth's frame, the CMB does have a difference between one side (slightly blueshifted) and the opposite one (slightly redshifted). However, because our galaxy is nearly at rest relative to the Hubble flow, these shifts are rather small.

$\endgroup$
  • $\begingroup$ Oh, didn't know that! Thanks for the reply. Also, if an observer moves with relativistic velocity w.r.t. this special frame, then the Hubble law is no longer valid (in the linear, isotropic form) for him? $\endgroup$ – DrLRX Nov 8 '18 at 21:59
  • 1
    $\begingroup$ Two tenets of SR: "there are no preferred frames", and "nothing can travel faster than light", just don't apply on a cosmic scale. $\endgroup$ – JEB Nov 9 '18 at 0:19
  • $\begingroup$ @JEB: "No preferred frame" still holds. The interpretation is the same as in SR: if we have some nearby matter that's in a certain state of motion, then we are free to use that matter to define a frame of reference. $\endgroup$ – Ben Crowell Nov 9 '18 at 2:30
  • $\begingroup$ @BenCrowell I agree, except everyone in the (visible) universe--even outside our light cone--will agree on the same preferred system (in co-moving coordinates, so maybe they're different frames?). $\endgroup$ – JEB Nov 9 '18 at 16:20

Your Answer

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

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