Edited version.

A body $A$ is receding at acceleration $\vec {a}$ with respect to a point $P$ because of the expansion of the universe.

Another body $B$ is accelerating at the rate of $\vec{a}$ with respect to a point $P$ through spacetime. Its movement is not because of the expansion of the universe.

For $P$, body $A$ and $B$ are accelerating in the same way but the cause is different in each case.

So, is there any way of differentiating between the states of motion of the two bodies for $P$ within the laws of physics?

  • $\begingroup$ Energy and momentum conservation. The energy and momentum of a freely moving body remain unchanged. Cosmic expansion, however, changes both. $\endgroup$ May 7 at 17:31
  • $\begingroup$ A freely moving body has a constant momentum and kinetic energy in an observer's rest system. Comic expansion seems to accelerate a body relative to us. The acceleration is very slow, but it is still there. $\endgroup$ May 7 at 18:05
  • $\begingroup$ $d\vec v$ is the differential element of velocity, not velocity. An object with "velocity $d\vec v$" is stationary. You just want $\vec v$. $\endgroup$
    – g s
    May 7 at 18:27
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    $\begingroup$ Since the expansion velocity needs the large cosmological scales to be detected , I expect that the effect on a moving body would be so tiny as not to be detectable. $\endgroup$
    – anna v
    May 7 at 19:23
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1 Answer 1


In general relativity acceleration is more complicated than it is in Newtonian mechanics. We normally think of acceleration as $d^2x/dt^2$ but in GR it is possible to have accelerating rest frames. For example if you are in an accelerating car then in your rest frame $d^2x/dt^2 = 0$ because, well, it's your rest frame so your $x$ coordinate is constant at zero.

But even if all the car windows were blacked out, so could not check your surroundings, you would be able to tell you were accelerating because you could feel the g force pushing you back into your seat. And if you dropped an object you would see it accelerate away from you - it would be you accelerating, not the object, but it would look to you as if it was accelerating away from you.

And in fact this is key to defining a measure of acceleration that all observers will agree on regardless of their rest frame. Suppose you drop an object and it accelerates away from you with some acceleration $a$ that must mean your acceleration is $-a$, and we call the magnitude of this your proper acceleration. This proper acceleration is a Lorentz scalar meaning that all observers will agree on it.

The proper acceleration can be a bit counter intuitive. For example if I drop my pen it will accelerate away from me at $-g$, and that must mean my proper acceleration is $g$ even though in my coordinates I am stationary in my chair. Conversely if you jump out of an airplane above me (ignoring wind resistance) your proper acceleration is zero until you open your parachute, even though I would measure you to be accelerating towards me. In GR terms I would be the one accelerating upwards towards you. For more on this see How can you accelerate without moving? and If gravity isn't a force, then how are forces balanced in the real world?

Anyhow, the objects that we see accelerating away from us due to the expansion of the universe have a proper acceleration of zero. Their acceleration is like your acceleration when you've just jumped out of the plane. Even though we see the objects accelerating those objects would feel themselves to be weightless just as you feel weightless when you're free falling.

By contrast one of Elon Musk's rockets accelerating away from the surface of the Earth has a non-zero proper acceleration and if you were in that rocket you'd feel high g forces not weightlessness. So this difference in the proper acceleration is how we tell the difference between the two types of motion.

  • $\begingroup$ if the body is moving through geodesics it's proper acceleration is zero. If it is not moving through the geodesic path proper acceleration is non-zero. $\endgroup$ May 11 at 7:10
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    $\begingroup$ @AgnibhoDutta Yes, exactly! So in your example objects accelerating due to the expansion of the universe are following geodesics while the rockets are not. $\endgroup$ May 11 at 7:25

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