<p>The Hubble constant effectively has a unit of inverse time only: <span class="math-container">${\rm s}^{-1}$</span>:</p>
<p>It is usually given as <span class="math-container">${\rm km}\cdot{\rm s}^{-1}{\rm Mpc}^{-1}$</span>. Yet looking at it, km and Mpc are both units of length, so they cancel eachother out when converted to the same unit. Thus add the appropriate unit conversion from km to Mpc and it remains an inverse time only:</p>
<p><span class="math-container">$$t = \frac{1}{H_0} = \frac{\rm Mpc}{60 {\rm km}\cdot{\rm s}^{-1}} = \frac{3\cdot 10^{22} {\rm m}{\rm s}}{60\cdot 1000 {\rm m}} = 5\cdot 10^{17}{\rm s} = 15{\rm Ga}$$</span></p>
<p>Add dimensionless factors to the equation to account for different cosmological models and inflation theories. This is anyway only a ballpark number - but an easy one to get for the age of the universe.</p>

CC BY-SA 4.0
2021-02-12T21:50:41.873Z
I fixed the LaTeX-typesetting for you, hope you don't mind.
B--rian
CC BY-SA 4.0
2021-02-13T07:57:55.770Z
On the contrary, much appreciated
planetmaker
CC BY-SA 4.0
2021-02-12T07:07:49.390Z
2021-02-12T21:49:25.503Z
2021-02-12T21:49:25.503Z
planetmaker
<p>I'm a grad student and I feel like this is an extremely elementary question that I <em>should</em> know, but I admit that I don't. Occasionally in science I find myself solving an equation for some answer where it feels like the units don't add up.</p>
<p>For instance, in a spatially flat matter-dominated universe with nonrelativistic matter, the accepted age of such a universe is:</p>
<p><span class="math-container">$$t_0 = \frac{2}{3}H_0$$</span></p>
<p>where <span class="math-container">$H_0$</span> has units of <span class="math-container">${\rm km} \cdot {\rm s}^{-1} \cdot {\rm Mpc}^{-1}$</span>.</p>
<p>Why am I getting a time from this? I know it works, I know this is correct. This isn't the only example of this I've ever seen, but it is the freshest in my memory. It conceptually makes sense, but usually I'm in the habit of doing meticulous dimensional analysis and just wondered if there's some deep metaphysics or philosophy behind when that applies and when it doesn't.</p>

CC BY-SA 4.0
2021-02-12T21:47:49.127Z
Welcome to astronomy SE! I pretty-formated your formulas and units for you.
B--rian
<units>
Units and some equations
ObiDonKenobi
# Units and some equations

(Asked by: username_0 on 2021-02-12)

I'm a grad student and I feel like this is an extremely elementary question that I *should* know, but I admit that I don't. Occasionally in science I find myself solving an equation for some answer where it feels like the units don't add up.

For instance, in a spatially flat matter-dominated universe with nonrelativistic matter, the accepted age of such a universe is:

$$t\_0 = \frac{2}{3}H\_0$$

where $H\_0$ has units of ${\rm km} \cdot {\rm s}^{-1} \cdot {\rm Mpc}^{-1}$.

Why am I getting a time from this? I know it works, I know this is correct. This isn't the only example of this I've ever seen, but it is the freshest in my memory. It conceptually makes sense, but usually I'm in the habit of doing meticulous dimensional analysis and just wondered if there's some deep metaphysics or philosophy behind when that applies and when it doesn't.

**Tags:** units
> **Comments:**
> **username_1**: Welcome to astronomy SE! I pretty-formated your formulas and units for you.  

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## 1 answers

### Answer by username_2 on 2021-02-12. Score: 6 | Accepted answer

The Hubble constant effectively has a unit of inverse time only: ${\rm s}^{-1}$:

It is usually given as ${\rm km}\cdot{\rm s}^{-1}{\rm Mpc}^{-1}$. Yet looking at it, km and Mpc are both units of length, so they cancel eachother out when converted to the same unit. Thus add the appropriate unit conversion from km to Mpc and it remains an inverse time only:

$$t = \frac{1}{H\_0} = \frac{\rm Mpc}{60 {\rm km}\cdot{\rm s}^{-1}} = \frac{3\cdot 10^{22} {\rm m}{\rm s}}{60\cdot 1000 {\rm m}} = 5\cdot 10^{17}{\rm s} = 15{\rm Ga}$$

Add dimensionless factors to the equation to account for different cosmological models and inflation theories. This is anyway only a ballpark number - but an easy one to get for the age of the universe.

> **Comments:**
> **username_1**: I fixed the LaTeX-typesetting for you, hope you don't mind.  
> **username_2**: On the contrary, much appreciated  

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