Determining H value from scalefactor

[JohnnyGui]

I’m probably missing something obvious here but I can’t figure out what.

Consider a galaxy, starting at an initial distance of ##D_0##, recessing with a constant velocity. After a time Δt, we’d measure a larger distance ##D_{t}##. From this scenario I’d conclude that the Hubble value based on this object is calculated by:
$$H = \frac{D_t – D_0}{Δt} \cdot \frac{1}{D_0}$$
That is, the velocity, which is obtained by the first fraction, divided by the initial distance ##D_0## of the galaxy gives the Hubble value when the galaxy was at distance ##D_0##

Now, from what I understand, the scale factor ##a## is how large the reached distance ##D_t## is relative to the initial distance ##D_0##. Thus; ##a = \frac{D_t}{D_0}##
Furthermore, the rate of change of the scale factor ##\dot a## is how much the distance has increased per unit of time, relative to the initial distance ##D_0##. So ##\dot a## could be calculated by adding the velocity (a distance per unit time) to the initial distance ##D_0## and dividing that by ##D_0##:
$$\dot a = (\frac{D_t – D_0}{Δt} + D_0) \cdot \frac{1}{D_0}$$
When combining these formulations for ##H##, ##a## and ##\dot a## I’d get: ##H+1 = \dot a##.
This is obviously not true since it should be ##H = \frac{\dot a}{a}##.

What am I missing?

 

[PeterDonis]

from what I understand, the scale factor aa is how large the reached distance ##D_t## is relative to the initial distance ##D_0##.

Not with the definition of ##H## you are using. If ##H = \dot{a} / a##, then ##a## has units of distance, which means ##a## at time ##t## is just ##D_t##, and ##a## at time ##t_0## is just ##D_0##.

So ##\dot a## could be calculated by adding the velocity (a distance per unit time) to the initial distance ##D_0##

No. The derivative ##\dot{a}## is just the change in distance over the change in time (in the limit as the change in time goes to zero), so it would be

$$
\dot{a} = \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

So ##H = \dot{a} / a## at time ##t_0## would be

$$
\frac{\dot{a}}{a} = \frac{1}{D_0} \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

 

[John Park]

So ˙ a a could be calculated by adding the velocity (a distance per unit time) to the initial distance D 0 D 0 and dividing that by D 0 D 0 :

Your resulting equation for a-dot makes no sense dimensionally.

 

[JohnnyGui]

Not with the definition of HHH you are using. If H=˙a/aH=a˙/aH = \dot{a} / a, then aaa has units of distance, which means aaa at time ttt is just DtDtD_t, and aaa at time t0t0t_0 is just D0D0D_0.

Thanks for the reply. Could you please explain a bit more why my defintion of ##H## says that ##a = D_0## or ##a = D_t## at ##t_0## and ##t## respectively? I thought a scale factor is the ratio between the new ##D_t## and the original distance ##D_0##.

 

[PeterDonis]

I thought a scale factor is the ratio between the new ##D_t## and the original distance ##D_0##.

It isn't. It's just the distance. More precisely, it's just the distance if you are defining ##H## as ##\dot{a} / a##. That should be evident from the fact that ##H## is the fractional rate of change of distances, i.e., the ratio of recession speed to distance, which is the ratio of distance per unit time to distance. So if ##H = \dot{a} / a##, then ##\dot{a}## must be recession speed--distance per unit time--and ##a## must be distance.

 

[JohnnyGui]

It isn't. It's just the distance. More precisely, it's just the distance if you are defining ##H## as ##\dot{a} / a##. That should be evident from the fact that ##H## is the fractional rate of change of distances, i.e., the ratio of recession speed to distance, which is the ratio of distance per unit time to distance. So if ##H = \dot{a} / a##, then ##\dot{a}## must be recession speed--distance per unit time--and ##a## must be distance.

I'm surprised, because this wiki does use ##D_t = a \cdot D_0## and ##H = \frac{\dot a}{a}## simultaneously. From the article it does look like they distinguish the distance from the scale factor.

 

[PeterDonis]

From the article it does look like they distinguish the distance from the scale factor.

What they are doing amounts to measuring distances in units of ##D_0## instead of in units of, say, light-years. So ##a = 1## just means the distance is ##D_0## in units of light-years (or whatever ordinary units you want to use).

Mathematically, if you really want to interpret ##D_0## this way, as a distance unit, then you would write

$$
H = \frac{\dot{a}}{a} = \frac{D_0}{D_t} \frac{d}{dt} \frac{D_t}{D_0}
$$

Since ##D_0## doesn't change with time, the ##D_0## factors just cancel out and you have the same thing I wrote before. In terms of the limits, you would have

$$
\dot{a} = \lim_{\Delta t \rightarrow 0} \frac{a_t - a_0}{\Delta t} = \frac{1}{D_0} \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

And the ##D_0## will just cancel out, once again, when you evaluate ##\dot{a} / a##.

 

[JohnnyGui]

Explanation

Thanks a lot for the detailed explanation. It makes sense now.

I just noticed my error myself as well. The Wiki is considering ##D_t## as a distance in the past while ##D_0## being the current distance at this moment. I was considering them to be the other way around whch led me to erroneously think that ##a## should be >1.

After correcting for this, when one writes the distance in units of ##D_0##, the ##\dot{a}(t)## is the speed at distance ##D_t## relative to the distance ##D_0## just like you wrote it.
If the speed has been constant all throughout history, then ##\dot{a}(t)## would equal the current Hubble value ##H_0##. If the speed hasn't been constant, then that means one would have determine ##\dot{a}(t)## at time ##t## and then use...
$$\dot{a}(t) \cdot D_0 = v(t)$$
...to calculate the speed at time ##t##. This calculated speed divided by the distance ##D_t## at that time would give the Hubble value at that time, ##H(t)##.
$$\frac{\dot{a}(t) \cdot D_0}{D_t} = H(t)$$
Since ##D_t = a(t) \cdot D_0## (a(t) being <1) this means that:
$$\frac{\dot{a}(t) \cdot D_0}{a(t) \cdot D_0} = \frac{\dot{a}(t)}{a(t)} = H(t)$$