<p>So to estimate the answer, you need to convert the mass density to number density by $n_H = \rho_H / m_H$. After getting the current number density you need to redshift it to finite $z$, here $z=6$. Keep in mind the density of hydrogen evolves $\sim (1+z)^3$ due to the expansion of the universe. </p>

<p>That said, the assumption 'let universe be completely made from hydrogen' renders this question highly unrealistic, hence I agree with @James K that this is not real world astronomy. In the real world, only about 5% of the total energy is made of baryons, among which Hydrogen abundance is $\sim 75\%$. Also, at redshift $z\sim 6$, star formation etc. lowers the neutral hydrogen gas density, c.f. recent observations <a href="https://arxiv.org/pdf/1506.02037.pdf" rel="nofollow noreferrer">here</a> and <a href="https://arxiv.org/pdf/1503.05553.pdf" rel="nofollow noreferrer">here</a>. </p>

CC BY-SA 4.0
2018-06-25T18:34:12.503Z
2018-06-25T18:34:12.503Z
Boson Bear
<p>Let universe be completely made from hydrogen. And also we have redshift $z= 6$. with Hubble constant $H_{0} = 2.1941747572815535\times 10^{-18}\:\mathrm{s}^{-1}$. We also know that density of the universe $\rho =\frac{3H^2}{8\pi G}$ ($G= 6.67\times 10^{-11}\:\mathrm{m^3\ kg^{-1}\ s^{-2}}$).</p>

<p><strong>Is the following approach right to find the number density of hydrogen in universe?</strong></p>

<p>First we calculate density $\rho =\frac{3H^2}{8\pi G}$, then we have
$1+z = \sqrt{\frac{1+v/c}{1-v/c}}$. From that we get $v = H\times R$ and from this we calculate $R$, the radius of observable universe. With that, we calculate volume of universe $V = (4\pi/3)R^3$ and we get Mass $M=\rho V$. Then we calculate number of Hydrogen atoms with $N = M/m_{\mathrm{hydrogen}}$. Now we have number density of hydrogen $\rho = N/V$.</p>

<p>Is this solution right? Or we must solve it another way?</p>

<p><sub>The source of question is National Olympiad of Iran which is not available in English so the above is my translation for the question.</sub></p>

CC BY-SA 3.0
2017-04-16T15:52:39.003Z
H to 16 significant figures? LOL.
ProfRob
CC BY-SA 3.0
2017-04-16T16:11:19.457Z
@RobJeffries It is calculated by changing the non-SI usual unit KM/sMpc to SI units. I know we don't know the hubble constant with this accuracy and the number is made by changing H= 67.8 KM/sMpc. Parsec in the denominator caused this!
titansarus
CC BY-SA 3.0
2017-04-16T17:11:22.973Z
@RobJeffries Sorry, It was H0 in the main question. I forgot the index. Also, I Know we don't gain precision by doing a unit conversion, We use H and in the formula and at the last part, we substitute it by an appropriate approx, like 2.2.
titansarus
CC BY-SA 3.0
2017-04-16T21:12:19.447Z
I think there some rules about check-my-homework questions, but here are some hints: Your definition of redshift is the Doppler shift, while you need the cosmological definition 1+z = 1/a; your definition of the Hubble law confuses radius with distance; and you don't need to care about the volume if you're only interested in the density.
pela
CC BY-SA 3.0
2017-04-16T21:34:16.887Z
I'm voting to close this question as off-topic because It seems to be a maths question, not about real world astronomy
James K
CC BY-SA 4.0
2018-06-26T12:13:00.060Z
There are Doppler's shift, Hubble's law, universe density. I think that is astro?
Kornpob Bhirombhakdi
<cosmology><redshift><density><hubble-constant>
Approximating density of hydrogen in [observable] universe
titansarus
# Approximating density of hydrogen in [observable] universe

(Asked by: username_0 on 2017-04-16)

Let universe be completely made from hydrogen. And also we have redshift $z= 6$. with Hubble constant $H\_{0} = 2.1941747572815535\times 10^{-18}\:\mathrm{s}^{-1}$. We also know that density of the universe $\rho =\frac{3H^2}{8\pi G}$ ($G= 6.67\times 10^{-11}\:\mathrm{m^3\ kg^{-1}\ s^{-2}}$).

**Is the following approach right to find the number density of hydrogen in universe?**

First we calculate density $\rho =\frac{3H^2}{8\pi G}$, then we have
$1+z = \sqrt{\frac{1+v/c}{1-v/c}}$. From that we get $v = H\times R$ and from this we calculate $R$, the radius of observable universe. With that, we calculate volume of universe $V = (4\pi/3)R^3$ and we get Mass $M=\rho V$. Then we calculate number of Hydrogen atoms with $N = M/m\_{\mathrm{hydrogen}}$. Now we have number density of hydrogen $\rho = N/V$.

Is this solution right? Or we must solve it another way?

The source of question is National Olympiad of Iran which is not available in English so the above is my translation for the question.

**Tags:** cosmology, redshift, density, hubble-constant

> **Comments:**
> **username_1**: H to 16 significant figures? LOL.  
> **username_0**: @RobJeffries It is calculated by changing the non-SI usual unit KM/sMpc to SI units. I know we don't know the hubble constant with this accuracy and the number is made by changing H= 67.8 KM/sMpc. Parsec in the denominator caused this!  
> **username_0**: @RobJeffries Sorry, It was H0 in the main question. I forgot the index. Also, I Know we don't gain precision by doing a unit conversion, We use H and in the formula and at the last part, we substitute it by an appropriate approx, like 2.2.  
> **username_2**: I think there some rules about check-my-homework questions, but here are some hints: Your definition of redshift is the Doppler shift, while you need the cosmological definition 1+z = 1/a; your definition of the Hubble law confuses radius with distance; and you don't need to care about the volume if you're only interested in the density.  
> **username_3**: I'm voting to close this question as off-topic because It seems to be a maths question, not about real world astronomy  
> **username_4**: There are Doppler's shift, Hubble's law, universe density. I think that is astro?  

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## 1 answers

### Answer by username_5 on 2018-06-25. Score: 1

So to estimate the answer, you need to convert the mass density to number density by $n\_H = \rho\_H / m\_H$. After getting the current number density you need to redshift it to finite $z$, here $z=6$. Keep in mind the density of hydrogen evolves $\sim (1+z)^3$ due to the expansion of the universe.

That said, the assumption 'let universe be completely made from hydrogen' renders this question highly unrealistic, hence I agree with @username_3 that this is not real world astronomy. In the real world, only about 5% of the total energy is made of baryons, among which Hydrogen abundance is $\sim 75\%$. Also, at redshift $z\sim 6$, star formation etc. lowers the neutral hydrogen gas density, c.f. recent observations [here](https://arxiv.org/pdf/1506.02037.pdf) and [here](https://arxiv.org/pdf/1503.05553.pdf).


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