Thanks for your answer. I understand that it's not a real velocity. What I'm trying to do here is applying v = z x c to a scenario in which the formula IS accurate enough (in special limited cases as @marcus said) so that I'm sure I understand how and WHEN the formula should be used. I now indeed understand from Marcus that the formula couldn't be applied in the case of cosmological time.This formula V = H0 D actually uses comoving distances and then it works for any distance, not only for short ones. The comoving distance of any galaxy remains constant forever, because it is defined as the proper distance of that galaxy today. Hence it is D2 in your example. The formula z c = V is an approximation for short distances, because cosmological redshift is not directly proportional to recession velocity.
As Marcus said before, recession velocity does not really mean a velocity - it is really only a recession rate that we can actually do without. All we need is the redshift and we can do most of the cosmology we require.
I have thought of an example I'd like to give in which I think the formula v = z x c is accurate. Please correct me if I'm still using this formula the wrong way in this example:
Since I'm thinking the redshift as a doppler shift which is wrong, let's assume we're now measuring the change in a sound wave from a sound emitting device that is moving away from an observer in an expandable fasion with a constant velocity. What I mean in an expandable fasion is that there are other devices that are also moving away from the observer whom their speeds depend on their distances (such as in a stretching scenario) so that there's a H that can be calculated (and which is changing over time).
The formula combined is: λObs / λEmit = D2 / D1 = z + 1 = (v = H x D) / s (= speed of sound which is 340 m/s)
The observer knows the initial distance D1 being 750m. The timepoint at which the device is at D1 is t0
At t1, the observer measures λObs of the sound wave the device is emitting, already knowing what the initial λEmit was.
At the time he measures this, the device has moved to D2. Thus, the device reaching D2 occurs at t1.
The measured random values the observer gets are λObs and λEmit being 500 and 300 respectively.
Thus, the formula is now: 500 / 300 = D2 / 750 = z + 1 = (H x D) / s
Calculating D2 gives a value of 1250m and z a value of 0,6667, thus 0,6667 = (H x D) / 340
Therefore, 0,6667 x 340 gives the speed of the object, being 226,667 m/s. Thus, 226,667 = H x D.
My question: When the device was at D1 = 750m (t0), what value did H have? Is it 226,667 / 750 = 0,302 m/s/m or is that at t1 when the device is at D2? If this H value is at t0 that means that calculating D gives 226,667 / 0,302 = 750m, thus D is D1. v being 226,667 m/s, can be taken at any time point (t0 and t1) since it's constant. It is therefore the timepoint at which H is measured (t0 or t1) that decides if D is D1 or D2! Even if z is calculated at t1. So if the calculated H is the H value at t0, then I'd say in general that z(t) would give an H at (t - Δt)
Am I making any sense here?