# Hubble flow and inertial reference frames

Looks like different questions, but they are basically the same:

1) Let us imagine a mostly empty expanding universe with constant $H$. Two observers (Ginger and Fred) that are separated by a cosmological distance but follow the Hubble flow will see each other accelerating (if measuring proper distances). Will each consider the other as non-inertial? What if Ginger has an electric charge: Will Fred see that this electric charge is accelerated but not radiating?

2) And what about the reverse? Put two opposite charges at the distance at which they are at rest with respect to each other (I assume this distance must exist because if the charges are too close, they will strongly attract and move towards each other; but If they are far enough from each other, their attraction force will be weak and the charges will see each other moving away due to the Hubble law). Do this mean that the static force no longer follows the inverse square law if described in terms of the proper distance? What if we describe it in terms of the comoving distance?

3) If the charges are "bounded" that way, it looks like neither particle will be in an inertial frame: Presumably the center of mass will move with the local hubble flow, and from its point of view the particles are at a fixed distance, at rest relative to the CM, but accelerated relative to nearby particles that follow the hubble flow. Will the nearby particles see that the charges radiate? Will an observer at the CM see these charges at rest but radiating?

• I think that this is an important question. Commented Feb 12, 2020 at 23:13

So the inverse square law still applies. The force exerted on the particles is still proportional to $\frac{1}{r^2}$, you simply have a velocity offset that is proportional to separation that also needs to be taken into consideration.
Say you have a time step $\delta t$. The total velocity, $v$, is a combination of the recessional velocity for the separation, $R$, and peculiar velocity, $v_p$: $$v(t)\propto HR(t)-(v_p(t-\delta t)+\frac{\delta t}{R(t-\delta t)^2})$$ Let's remember this relation just shows proportionality; I didn't include all the constants in the inverse square term. But basically, if you find a place where $v=0$, then the first term won't change in the next moment of time, but the second term will, guaranteed. This means the particles can never remain at rest w.r.t. each other for more than a moment, but you can clearly see the inverse square law affecting the only acceleration term in the expression.