# Can anyone explain the unit for rate of expansion of universe?

If you google for 'what is rate of expansion of universe' you get

Space itself is pulling apart at the seams, expanding at a rate of 74.3 plus or minus 2.1 kilometers (46.2 plus or minus 1.3 miles) per second per megaparsec (a megaparsec is roughly 3 million light-years).

Can anyone please explain what does the unit mean? what is 'per second per megaparsec'? I believe x km/s would suffic the rate of expansion. Not sure what 'per megaparsec' specifies?

## migrated from skeptics.stackexchange.comJul 16 '15 at 14:42

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• It means that the rate of expansion varies with distance. The rate of separation of points zero distance apart is zero. The rate for points separated by 1 megaparsec is 74.3 kilometers per second. The rate for points separated by 2 megaparsec is 148.6 kilometers per second; etc. But definitely off topic here. – DavePhD Jul 16 '15 at 13:45

The hubble relation is:

$$v = H d$$

where $v$ is the velocity of the galaxy relative to the Milky way, and $d$ is the distance of the galaxy relative to the milky way. The velocity is measured using redshift. The distance is measured through a complicated series of standard candles, along with the relationship $I = \frac{I_{0}}{4\pi r^{2}}$.

If you notice, these are related by Hubble's constant $H$. This tells us how fast something is moving apart, based on how far away it is. Namely, it's the rate of expansion of the universe. Since astronomers measure distance in a unit called a parsec, and velocity at these high speeds is naturally measured in km/s, we can see that the "natural" units astronomers use for $H$ are (km/s)/parsec., since that's the only way to get km/s out when you multiply by something measured in parsec.

• Good, standard explanation. To be complete we need to mention that as far as the units go, the km and Mpc are fungible, so actually the SI unit of expansion is 1/s, roughly $H=1.7 \times 10^{-18}/s$ and the inverse of that is more or less the age of the universe. – Kostas Aug 13 at 19:47

As time goes along, everything in the universe expands by a dimensionless scale factor a(t). This $a(t)$ is the scale factor seen in the Robertson-Walker metric. Some galaxy at a distance $x_0$ at time $t_0$, will be at $x(t)$ at time $t$. $$x(t)=a(t) \quad x_0$$ Taking the derivative wrt t gives $$\dot{x}=\dot{a}\quad x_0$$ and dividing by the first equation gives $$\frac{\dot{x(t)}}{x(t)}=\frac{\dot{a(t)}}{a(t)}\equiv H(t)$$ where the Hubble Constant is a function of time and its value now is measured to be $H_0=70 [\frac{km/sec}{Mpc}]$ so we get the familiar "now" Hubble Law $$\dot{x}=H_0 x$$ Astronomers like megaparsecs (Mpc) as a unit of distance, where $$1 [Mpc] = 3*10^6[lyrs] = 3 * 10^{22}[m]$$ The Hubble Constant can be expressed in may different units. For example $$H_0=70[\frac{km/sec}{Mpc}]*10^3[\frac{m}{km}]*\frac{1}{3*10^{22}}[\frac{Mpc}{m}] =2*10^{-18}[\frac{m/sec}{m}]$$ or $$c*H_0=70[\frac{km/sec}{Mpc}]*10^3[\frac{m}{km}]*\frac{1}{3*10^{22}}[\frac{Mpc}{m}]*3*10^8[\frac{m}{sec}] =7*10^{-10}[\frac{m}{sec^2}]$$

which is an acceleration in nice everyday mks units.

• Going by standard usage, mpc is actually a milliparsec. You want Mpc instead. – Kyle Kanos Jan 15 '17 at 22:01
• Also where did that $c$ come from in the last line? – Kyle Kanos Jan 15 '17 at 22:04
• @Kyle: Thank you, I have added c*H to the last equation. This is an example of getting in the habit of leaving c's out of equations. I alsochanged mpc to Mpc. – Gary Godfrey Jan 16 '17 at 8:42

Can anyone please explain what does the unit mean? what is 'per second per megaparsec'? I believe x kms per sec would suffic the rate of expansion. Not sure what 'per megaparsec' specifies?

The mega-parsec unit is required because each mega-parsec of distance is expanding at a rate of 74.3 km/s. So, if two side-by-side mega-parsecs expand in one second, the total expansion is 148.6 km. The "per mega-parsec" forces you to do a summation over vast distances.

Great question BTW!