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?
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Sign up to join this communityAccording 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?
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.
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 $v<c$ 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 $v>c$.
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, $v_{pec}<<c$, and no holy law is violated.