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This post explains what the units of the Hubble Constant are, but not why they are what they are. The units are clear from the fact the constant relates distance and observed speed of galaxies.

But the Hubble constant is supposed to measure the rate at which space is expanding, right?

So I would expect more appropriate units to be $\frac{L^3}{T}$ rather than $T^{-1}$ .

Perhaps there is another metric that measures in these units?

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    $\begingroup$ You already have answers but you might like this way of putting it: a larger volume expands faster than a smaller volume, in the sense that more units of volume are added per second. A Hubble parameter like you propose wouldn't be useful because it would depend on the volume you're measuring. $H$ measures the relative expansion rate: how fast your volume is expanding, divided by the original volume. $\endgroup$ – Javier Apr 22 '17 at 14:39
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Typically, one thinks of Hubble's constant in terms of astronomical observation. This begins with Hubble's empirical law:

$$v = H d$$

Where $v$ is the velocity derived from the redshift of a distant galaxy, and $d$ is the distance to it. The classic observational units for $H$ are therefore $\frac{\rm m/s}{\rm Mpc}$. Relativists hate carrying around more than one unit, though, so we convert the Megaparsecs to meters, and end up with inverse seconds (or meters, since $c$ gives us a way to convert back and forth).

More simply, we can also think of Hubble's constant in terms of the Robertson-Walker metric:

$$ds^{2} = -dt^{2} + a^{2}(t)d^{3}{\vec x}$$

Where $d^{3}{\vec x}$ is the 3-metric of a homogenous space. Then, $a(t)$ tells us how "big" space is at the current time, and we can think of the rate of expansion as $H(t) = \frac{\dot a}{a}$${}^{1}$, which obviously has units of inverse time.

Note that $H$ is actually a function of $t$ and is not constant in time, except for some special cases.

EDIT:

Note, if what you care about is the time evolution of 3-volumes, you can see that these are proportional to $a^{3}V_{0}$. A time derivative of this then gives you $\frac{dV}{dt} = 3{\dot a}a^{2}V_{0} = 3 H a^{3}V_{0}$, which has the units you want, but is still logically anterior to the ordinary Hubble constant.

${}^{1}$We divide by $a$ so that $H$ doesn't depend on our arbitrary choice of the scale of $a$, and simply reflects the relative rate of expansion.

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  • $\begingroup$ $v$ is a velocity not a "redshift". You probably meant that "$v$" is the radial velocity derived from the light frequency shift (redshift) assuming that it is caused by the Doppler effect. $\endgroup$ – freecharly Mar 7 '17 at 18:03
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Yes in fact there is another metric. That is the frequency: $$f=\frac{1}{T}$$ That is what $T^{-1}$ is supposed to mean. Also the equation for the expansion velocity is $$v = H\cdot d$$, so in terms of units $s^{-1}$ makes perfect sense. Though from my knowledge the unit of Hubbles constant is usually represented with: $\frac{\text{km}}{\text{Mpc} \cdot \text{s}}$.

However I do understand your point to deal with it in terms of volume expansion rate $V=\frac{L^3}{T}$, still let us think about the meaning of this equation. The further two objects are, the faster they will be moving away from each other. The rate of expansion is constant, it is $H$, but due to the expansion of the space the phenomenon of distance comes in. The easiest way to imagine this is using the balloon analogy. Two dots on a ballon will have the distance of 10mm, two other dots a distance of 1000mm. If we know double the size of the balloon in a given time $t$ then the first pair will have a distance of 20mm, the second a distance of 2000mm. So one pair expanded with 10m/s, the other with a 1000m/s. The rate of expansion remained constant, the expansion speed depends on two given points due to the effects of geometry.

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  • $\begingroup$ I agree units make sense with regards to what they relate to. What does not make sense to me is relation of the constant to the expansion of space. Units of space seem to be missing from the 'constant'. $\endgroup$ – docscience Mar 7 '17 at 18:05
  • $\begingroup$ But doesn't the equation represent the speed at which two distant points in space would move away from each other? So the dimension of space is included in this regard. $\endgroup$ – Magnetar Mar 7 '17 at 18:06
  • $\begingroup$ Yes but if you interpret it that way you would say the further away from earth you are, space is expanding more rapidly. But that's not true. The rate of expansion should be the same everywhere. $\endgroup$ – docscience Mar 7 '17 at 18:09
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    $\begingroup$ That is exactly what the law is saying. The further away two objects are, the more rapidly space is expanding. A galaxy that is further away than another will be moving away from us faster than the closer galaxy. This is one of the key aspects of Hubbles Law. $\endgroup$ – Magnetar Mar 7 '17 at 18:13
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    $\begingroup$ Think of this as an expanding earth. If your friend lives 10km away from you and earth is doubling its volume in that given time, than after this time he will be away 20km, with an expansion velocity of 20km/t. If somebody is living away 1000km, after the time t he will be away 2000km, having expanded with 2000km/t. So yes, the rate of expansion is constant, in this case it is doubling, in the universes case, well take Hubbles constant. And I hope this analogy has cleared the problem with further objects moving away from us at faster speeds. $\endgroup$ – Magnetar Mar 7 '17 at 18:18
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If $H$ would be defined as $H=x\frac{\frac {m} {sec}} m=x\frac 1 {sec}$, the value would, of course, be much smaller, but then we have to express $d$ not in $(Mpc)$ but in $(m)$ for the recess velocity (expressed in $\frac m {sec}$) to remain equal at distance d.

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