# How do you combine the expansion rates of two galaxies moving away from each other?

How do you combine the expansion rates of two galaxies moving away from each other? According to special relativity, you use the addition formula to sum the velocities of two particles moving through space, but it is believed that if two galaxies are far enough apart, a superluminal expansion rate is possible. This seems to me to be easily observable. Let's say you observe galaxy A moving away from you at .75c. You then turn 180 degrees and observe galaxy B moving away from you at .75c. Is the total rate 1.5c? Has this experiment ever been done to confirm a superluminal expansion rate?

• Do you mean recession as opposed to expansion? Feb 3, 2022 at 21:15
• If you wanted to experimentally measure Galaxy A's speed as seen from Galaxy B, wouldn't you have to travel to Galaxy B to do so? That might take a while. Feb 3, 2022 at 21:25
• Michael Seifert, Do I have to travel to galaxy B or can I be an observer in between and combine two speeds? Feb 3, 2022 at 21:36
• Joseph, if you like recession better, go with recession. Feb 3, 2022 at 21:37
• You do realise that when something expands it literally means “get larger” whereas recede means “moving away from”? I think I will stick with the latter because it’s correct grammatically. Cheers Feb 4, 2022 at 0:35

Yes, it is as simple as that. But…

#### Homologous expansion

Hubble's law tells us that the recession velocity $$v$$ between two galaxies separated by a distance $$d$$ is equal to $$v=H_0 d$$, where $$H_0$$ is the Hubble constant. In other words, velocity is proportional to distance. We also say that the expansion os "homologous".

Hence, if two galaxies A and B are exactly opposite from you at distances $$d_\mathrm{A}$$ and $$d_\mathrm{B}$$, respectively, then because their mutual separation $$d_\mathrm{AB}$$ is the sum of those two distances, we have that the velocity of A with respect to B is $$\begin{array}{rcl} v_\mathrm{AB} & = & H_0 \, d_\mathrm{AB}\\ & = & H_0 \,(d_\mathrm{A} + d_\mathrm{B})\\ & = & H_0 d_\mathrm{A} + H_0 d_\mathrm{B}\\ & = & v_\mathrm{A} + v_\mathrm{B}. \end{array}$$

If A and B are not exactly opposite, it gets a little trickier, but the principle is the same.

#### The "but"…

However, we measure neither velocities, nor distances. We measure redshifts $$z$$ and/or luminosities, and convert these into velocities and distances, assuming an underlying cosmological model that tells us the connection between these quantities. The accepted model that seems to describe our Universe well is the so-called ΛCDM model, based on the FLRW metric and the Friedmann equations.

For nearby galaxies, the velocity is close to the usual Doppler velocity $$v=cz$$, where $$c$$ is the speed of light. But this approximation quickly becomes invalid for more distant galaxies. In fact, in our model, any galaxy with a redshift larger than roughly 1.48 recedes faster than the speed of light. Our model tells us that this redshift corresponds to a (current) distance of roughly 14.5 billion lightyears (called "Glyr").

We can't travel to galaxy A and measure the redshift/velocity of galaxy B, but we don't have to! We routinely observe galaxies with redshifts much larger than 1.48. The current record-holder, GN-z11 has a redshift of $$z=11.1$$, hence lying 32 Glyr from us and currently receding at $$v=2.2c$$.

In our model, a galaxy would be deduced to be receding at $$0.75c$$ if we observed its redshift to be $$z=0.97$$. The distance to such a galaxy would be roughly 11 Glyr, so A and B would be some 22 Glyr apart. Our model tells us that a hypothetical observer in A would measure B's redshift to be $$z_\mathrm{AB}\simeq3.1$$. Assuming that they use the same model, that redshift can then be used to deduce the distance of $$22\,\mathrm{Glyr}$$, and a recession velocity of $$v=H_0\times(22\,\mathrm{Glyr})$$.