# Do objects outside the Hubble sphere violate special relativity?

According to special relativity, objects can move with speed at most c.

However, objects outside the Hubble sphere recede from us faster than the speed of light.

How can these be reconciled?

• The first fact concerns local velocity; the second concerns global velocity.
– J.G.
Jun 22, 2022 at 13:00
• I really don’t know enough about general relativity to know which answer I should accept Jun 23, 2022 at 15:05

Dale's answer is wrong, inasmuch as it suggests that this is related to parallel transport in curved spacetime.

If you parallel transport two sublight velocities to the same spacetime location and compare them in the special-relativistic fashion, the relative speed you get will always be less than $$c$$. "Sublight" here means that the velocity four-vector is timelike. All galaxies, no matter how distant from us, have sublight velocities in that sense, so any relative speed defined by parallel transport will be less than $$c$$.

In the Milne cosmology, which is the zero-energy-density limit of standard cosmology, the universe is spatially infinite and Hubble's law applies, so cosmological recession speeds larger than $$c$$ are possible. But the Milne cosmology is just special relativity in disguise: it's a patch of Minkowski space in nonstandard coordinates. A relative speed measured by parallel transport, regardless of path, will always be the special-relativistic relative speed, which is not equal to the cosmological recession speed.

The cosmological recession speed is just a number. It has "speed" in its name, and has units of speed, but it has no useful physical interpretation as a speed. A cosmological recession speed of $$c$$ isn't the speed of light; a light beam can always recede faster than a galaxy regardless of the galaxy's cosmological recession speed. In the Milne cosmology, a cosmological recession speed of $$c$$ corresponds to a special-relativistic relative speed of $$c\tanh 1 \approx 0.76c$$, which has no special significance as a speed.

To summarize: objects outside the Hubble sphere do not recede faster than light. They do recede faster than $$c$$ in a certain sense, but not in a sense that has any physical significance.

• The physical significance is that a photon flying in your direction while at the Hubble radius is keeping a constant distance, that's why we say space recedes at the speed of light there Jun 22, 2022 at 18:56
• @Yukterez The distance you're talking about has no physical significance. Again, look at the Milne cosmology. A light beam at $x=t\tanh 1$ in inertial coordinates, moving toward $x=0$, is keeping a constant distance from $x=0$ by the FLRW definition of distance. Jun 22, 2022 at 20:15
• $\rm x$ is just a $\rm t$-dependend coordinate. In relativity the distance along $\rm x$ at time $\rm t$ is not $\rm x_2-x_1$ but $\int_{\rm x_1}^{\rm x_2} \sqrt{g_{\rm x x}}$, especially since Milne has a $\rm t$ in his $g_{\rm x x}$ Jun 22, 2022 at 21:11
• @Yukterez The integral depends not just on the endpoints but the whole path. The path looks straight in FLRW coordinates, but is actually curved (i.e. not a spacetime geodesic) unless $a'=0$. In the Milne case the integral is over curves of constant $t^2-x^2$, which is why you get the odd behavior of the light momentarily standing still with respect to that distance. Jun 22, 2022 at 22:17
• SR is locally valid. How do you demonstrate that SR is not globally valid without parallel transporting quantities between different points in spacetime and showing that there's "more to it" than just flat space?
– HTNW
Jun 23, 2022 at 11:42

Yes, this violates special relativity. Special relativity is only valid in a region of spacetime which is small enough that curvature effects are negligible. The Hubble sphere is not a small region of spacetime in this sense. Be aware that the size of this region is not one fixed size but it depends on both the curvature and the sensitivity of the experiment.

That said, the violation here is a subtle one. The violation is not specifically that $$v>c$$. The guarantee that $$v only applies for massive objects in inertial frames. In non-inertial frames, like rotating frames, it is not problematic to have $$v>c$$. What would be problematic is for a massive object to be moving locally faster than light, which is not the same as $$v>c$$ in non-inertial frames. In the case of the Hubble sphere this condition (massive object moving locally faster than light) is not violated.

What does violate SR is that the relative velocity of a distant object is not unique. In flat spacetime, when you parallel transport a distant velocity in order to compare it to a local velocity, there is one unique answer regardless of the path chosen for the parallel transport. In curved spacetime, you will get different relative velocities by choosing different paths for parallel transport. This violates SR, but it is a more subtle violation than simply $$v>c$$, and it applies to all distant objects not merely those sufficiently distant to have $$v>c$$.

• Can you really say it violates SR when GR says the velocity between us and a faraway object is not even well defined?
– HTNW
Jun 22, 2022 at 13:03
• Special relativity makes experimental predictions which are wrong in the presence of curvature
– Dale
Jun 22, 2022 at 13:05
• That's well and good, but "special relativity doesn't accurately describe what's going on with the expanding universe" is not the same as "the superluminal velocity (due to the expansion of the universe) between us and faraway objects violates SR". The question asks about the second statement and you seem to affirm it. It would be nice if you would also explain that there is no real "superluminal velocity" in the first place, too.
– HTNW
Jun 22, 2022 at 13:10
• Hmm, actually, I am reconsidering. In general curvature violates SR, but I am not sure this specific effect is a violation. I am deleting this while I think
– Dale
Jun 22, 2022 at 13:18
• @JohnDavis the fact that coordinate velocity can exceed c in SR is explicitly discussed in the answer. Downvote if you feel you must, but in the presence of curvature the experimental predictions of SR are wrong. Calling that a “violation” is perfectly reasonable and not misleading. “Applies only locally” is also fine, but I see no reason to change it.
– Dale
Jun 23, 2022 at 11:50

The speed of light is the absolute speed that massless particles like photons can travel at. Such particles are also called luxons.

A photon travels in spacetime. However, spacetime in Einstein's GR is dynamical. This means that the metric can change. And this means spatial distances can become larger and there is no limit to this speed. It can be higher than c.

Objects outside the Hubble sphere are receding faster than light because the place at where they are at is expanding faster than light from the centre. They, themselves, need not be moving.

The super-luminal recession velocities of objects beyond the Hubble sphere of course do not violate SR.

The limiting value c only applies to the velocity of an object as measured locally, i.e. relative to the local comoving frame. This means that the peculiar velocities of galaxies (i.e. their velocities relative to the Hubble flow) must always be less than c. But this does not impose any restrictions on the possible values of the recession velocities due to the expansion of the universe (the metric can expand faster than $$c$$).

In fact, super-cluster of galaxies are almost comoving objects, which means that they are almost locally at rest, i.e. their peculiar velocities $$v_{pec} \simeq 0$$ (they simply move along with the Hubble flow). However, distant super-clusters of galaxies are moving away from us with recession velocities $$v_{rec}>c$$ due to the cosmic expansion of space. But, I insist, the velocity of these super-clusters with respect to the local comoving (inertial) frame is practically zero, that is, $$v_{pec}<, and no holy law is violated.