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When we have Hubble constant and it's inverse Hubble Time (1/H) what units are they measured in? I know Hubble constant is in "km/s per Mpc" but is there any other units which are popular used with it, and if so is there any conversion between the two? And would Hubble Time unit still be in billion years, seeing it is just displaying a time?

Hatmix5
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2 Answers2

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Although in theory we should all be using SI units, for things that are very large or very small these units are an inconvenient size and it's common to invent new units that are more convenient. So, for example particle physicists measure mass in GeV (strictly speaking GeV/$c^2$) and cosmologists measure distance in light years and/or parsecs.

In the case of the Hubble constant it has dimensions of $T^{-1}$ so the SI unit would be $s^{-1}$. The Hubble time has units of $T$ so the SI unit would be the second. However if we take the value for $H$ measured by Planck, $67 \text{km} \space \text{s}^{-1}/\text{Mpc}$, and convert to units of per second the value is about $2.2 \times 10^{-18} \text{s}^{-1}$, which is a lot harder to remember than the number $67$. That's why cosmologists use those rather strange units. As long as all cosmologists use the same units it doesn't really matter what the units are.

The Hubble constant isn't actually constant and will change in the future. Exactly how it changes depends on the behaviour of dark energy, which is somewhat uncertain at the moment.

Response to comment:

To convert the value of $H$ to $s^{-1}$:

The units of $H$ are (rearranging slightly) $s^{-1}\text{km}/\text{MPc}$. Kilometers and Megaparsecs are both units of length, so their ratio is a dimensionless number. If $N$ is the number of kilometers in a megaparsec then the ratio is just $1/N$. Google helpfully tells us that the number of km in an Mpc is $3.09 \times 10^{19}$, so to convert the Hubble constant to units of per second just divide it by $3.09 \times 10^{19}$.

John Rennie
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  • What is the conversion you used to go from kms/Mpc to s^-1? Also i'm aware of it's 'non constant' behavior, ahaha. But i like to call it a constant. It kind of reminds me of why people call String Theory a theory.. – Hatmix5 Sep 10 '14 at 11:13
  • @NickI: I've edited my answer to respond to your comment – John Rennie Sep 10 '14 at 11:22
  • @NickI: you've lost me. What does H per billion years mean? Do you mean the change in the value of H with time? – John Rennie Sep 10 '14 at 15:17
  • Just look at the type of units: km/s/MPc is length per time per length... which when rearranged, gives you length / length / time, which is just 1/time (T^-1) – Dois May 31 '16 at 00:19
  • I suggest an edit to make this post consistent with cosmology conventions; the "Hubble constant" $H_0$, by definition, is constant. The "Hubble parameter", $H(t)$, is not. $H_0$ is simply defined to be $H(t)$ where $t = t_0$, or when $a(t) = a(t_0) =1$. So I suppose it is a question of whether you consider $a = 1$ always for any observer (even if one is observing a billion years from now), or if $a = 1$ refers to the current time, and $a > 1$ a billion years from now. Adopting the latter convention would ensure $H_0$ is always constant. – pretzlstyle Jan 03 '18 at 20:06
  • Also, it is ignoring history to say that the units of kms$^{-1}$Mpc$^{-1}$ are used as a convenience of affording a more memorable numerical value. I would sooner say that those units are used because they are the original units of Hubble's axes. – pretzlstyle Jan 03 '18 at 20:08
  • Sorry for the basic question, but why is Hubble's Constant defined in units of "distance per time per distance" when most acceleration values are "distance per time per time"? – Jonathan Nov 22 '18 at 08:57
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    @Jonathan Hubble's constant tells us how the recession speed increases with distance away from us i.e. it is the constant in the equation $v = Hd$, where $v$ is the recession speed and $d$ is the distance away from us. So it has units of velocity per meter i.e. (m/s)/m. – John Rennie Nov 22 '18 at 09:04
  • I see so it's a function of distance, thus distance is a required unit. Makes sense – Jonathan Nov 27 '18 at 21:17
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The Hubble constant is, technically, a reciprocal time. If you note, two of the units in the combination (km/s)/Mpc are actually distance units, thus they cancel to give reciprocal time, i.e. $\mathrm{s}^{-1}$, but with a prefactor. The Hubble time is then just the reciprocal thereof.

In fact, I'd say to think of it as reciprocal time is more natural than to use speed per distance, because the real "gist" of the constant, at least with a steady Hubbleian expansion, is that distances are scaling up exponentially in the same fashion as "magnifying" a bitmap continuously with steadily increasing zoom power. Effectively, the ultrastructure of the Universe is blowing up like just such a bitmap over a very long time.

The other comment that SI units cannot provide a reasonable scale is wrong: that's what SI prefixes are for, to form larger/smaller multiple units of the base units in an easily-convertible way. Instead of $\mathrm{s}^{-1}$, a better unit is $\mathrm{Es}^{-1}$, that is, reciprocal exaseconds, where one exasecond is $10^{18}\ \mathrm{s}$. To get a feel for this timescale, the current age of the Universe is around 0.44 Es, and the Earth about 0.14 Es. One Es is thus a bit over twice the elapsed age of the Universe so far.

Thus, in the fully-utilized SI unit system, the value is

$$H_0 \approx 2.3\ \mathrm{Es}^{-1}$$

with broad uncertainty, as there seems to be some interesting (not yet resolved!) conflict between various ways of assessing this constant. What this value means is that in each exasecond, the scale of the ultrastructure undergoes 2.3 "$e$-foldings", or zoomings each by a factor of $e\times$ magnification, i.e. $\approx 2.718\times$ each.

The reason that speed per distance is useful, though, is that oftentimes we want to estimate the recessional speed of a distant object. If distance to an object is going up exponentially with time, i.e.

$$D(t) := D_0 e^{H_0 t}$$

then the apparent speed it is moving away with is given by

$$\frac{dD}{dt} = H_0 D_0 e^{H_0 t}$$

where $D_0$ is the distance at present time, hence the recession speed at present time ($t = 0$) is $H_0 D_0$, Hubble constant times present distance, so it can be interpreted as speed per distance and we can fold this into the unit for simplicity.

The_Sympathizer
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