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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?

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migrated from skeptics.stackexchange.com Jul 16 '15 at 14:42

This question came from our site for scientific skepticism.

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    $\begingroup$ 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. $\endgroup$ – DavePhD Jul 16 '15 at 13:45
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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.

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  • $\begingroup$ 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. $\endgroup$ – Kostas Aug 13 at 19:47
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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.

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  • $\begingroup$ Going by standard usage, mpc is actually a milliparsec. You want Mpc instead. $\endgroup$ – Kyle Kanos Jan 15 '17 at 22:01
  • $\begingroup$ Also where did that $c$ come from in the last line? $\endgroup$ – Kyle Kanos Jan 15 '17 at 22:04
  • $\begingroup$ @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. $\endgroup$ – Gary Godfrey Jan 16 '17 at 8:42
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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!

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