6
$\begingroup$

Consider an object traveling across the universe and consider two observers billions of light years away from each other at rest with respect to their local universes. The object will pass by the first observer at time $t=0$ and the object will pass by the second observer at time $t=T$. Would the expansion of space cause any difference in the relative velocity between the object and the first observer at t=0 and the relative velocity between the object and the second observer at t=T?

$\endgroup$
2
  • 2
    $\begingroup$ I would suggest "who measure isotropic cosmic microwave background" instead of "at rest with respect to their local universes. I'm pretty sure it's what you mean to mean. $\endgroup$
    – g s
    Commented Dec 20, 2023 at 6:48
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Dec 20, 2023 at 7:16

2 Answers 2

11
$\begingroup$

Yes, the object would move slower with respect to the second observer. The more general way of framing this is to define the object's peculiar velocity as its velocity with respect to its local universe, wherever it is. Peculiar velocities decay proportionally to $1/a$, where $a$ is the expansion factor.

This is essentially the cosmological redshift for massive particles. It can be understood as a self-sorting effect. All of the observers "at rest with respect to their local universe" are moving away from you, with more distant observers receding more quickly. If you start to move in any direction, you will gradually overtake slower-moving observers in that direction, asymptotically approaching the observer moving at the same velocity as you. So your velocity with respect to the instantaneously local observer decreases over time, asymptotically approaching zero.

$\endgroup$
5
  • $\begingroup$ So kinetic energy is not conserved? Assuming the universe was closed, small and the expansion slow enough to not have horizons, then the object will eventually return back to the first observer with a slower speed? $\endgroup$ Commented Dec 21, 2023 at 5:21
  • $\begingroup$ @PatoGalmarini Kinetic energy is frame dependent, and every fundamental observer is in a different frame. Note also that a closed universe always collapses too soon for anything, even light, to circumnavigate it! However, indeed there can be no global energy conservation in cosmology. Essentially due to the reference frame issue, it's not possible to define a global conserved energy. $\endgroup$
    – Sten
    Commented Dec 21, 2023 at 9:46
  • 1
    $\begingroup$ (An exception to a closed universe collapsing too soon for circumnavigation is if it's supported against collapse by dark energy, so for example you could circumnavigate Einstein's original static universe, although it is an unstable configuration.) $\endgroup$
    – Sten
    Commented Dec 21, 2023 at 9:55
  • $\begingroup$ Thanks! So in such case of circumnavigation, once the object returns back the speed will be indeed less than what it was the first time it passed? $\endgroup$ Commented Dec 21, 2023 at 20:56
  • 1
    $\begingroup$ @PatoGalmarini well to be clear, Einstein's static universe is static, so in that case the speed would be the same. But if you found a configuration that expands while still allowing circumnavigation, then yes, the speed would be less. For a collapsing configuration, the speed would be greater. $\endgroup$
    – Sten
    Commented Dec 21, 2023 at 23:26
-2
$\begingroup$

The expansion of the universe happens for everyone in the universe at the same rate.No location is special.If that was the case it would indicate a point of origin of the universe which has been disproven.So no they would measure the same time interval(assuming spacetime curvature locally is the same).

$\endgroup$
3
  • 3
    $\begingroup$ How is this relevant to the OP's question about the velocities measured by the two observers? $\endgroup$
    – PM 2Ring
    Commented Dec 20, 2023 at 6:22
  • $\begingroup$ It is important because $u = \frac{\Delta x}{\Delta \tau}$.If 2 regions of spacetime with the same local curvature have the same $\Delta \tau$ then u cannot possibly be different. $\endgroup$ Commented Dec 20, 2023 at 7:07
  • 3
    $\begingroup$ What is $\Delta x$ supposed to be here? How do you even define that in a curved spacetime? $\endgroup$
    – Sandejo
    Commented Dec 20, 2023 at 18:46

Your Answer

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

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