What is the Average Density of the Universe and Its Scale Over Time?

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SUMMARY

The average density of the universe at a redshift of 11.1 is calculated to be 5.32 × 10^-24 kg/m³, which corresponds to approximately 1.92 × 10²⁴ hydrogen atoms per cubic meter. The calculation involves scaling the distance between galaxies by a factor of 12.1, resulting in a volume that is 1772 times smaller than today. A critical error identified in the discussion is the incorrect application of Avogadro's number, which should not be multiplied when converting density from kg/m³ to hydrogen atoms/m³. Proper unit tracking is essential to avoid such mistakes in cosmological calculations.

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Thomas Smith
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Homework Statement
The current average density of the universe is roughly 3 x 10^-27 km m^-3. What was the average density of the universe at the time the light was emitted by a galaxy with the red shift of z=11.1? Express your answer in terms of a number of H atoms per cubic meter.
Relevant Equations
R(t)/R(t0) = 1/1+z

Na x p/mH Where Na is Avogardo's number in terms of atoms, mH is the mass of the hydrogen atom in kg and p is the average density.
Firstly i worked out the scale factor of the universe
R(t)/R(t0) = 1/1+z = 1/1+11.1 = 1/12.1 = 12.1^3 = 1/1772

The distance between the galaxies were 12.1 times less than today and the volume was 1772 times smaller than today.

Then I think the average density in the universe at that time is (3×10^-27 )×1772 = 5.32 × 10^-24kg m^-3

then the average density in terms of the hydrogen atom
= 6.023×10^23 × (5.32×10^-24/1.67×10^-24) = 1.92×10^24 hydrogen atoms per cubic meter.

This does not seem right to me at all!
 
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If you have the density in kg/m^3, and you want the density in hydrogen atoms/m^3, then you need to divide by kg/hydrogen atom, which you correctly have as 1.6x10^-24. Why did you then multiply by Avogadro's number?
 
phyzguy said:
If you have the density in kg/m^3, and you want the density in hydrogen atoms/m^3, then you need to divide by kg/hydrogen atom, which you correctly have as 1.6x10^-24. Why did you then multiply by Avogadro's number?
I thought that in regards to the number of hydrogen atoms you needed to include it
 
Thomas Smith said:
I thought that in regards to the number of hydrogen atoms you needed to include it
Including it is fine. Including it twice is not. If you had tracked units carefully, the error would have made made itself more obvious.

There is another error in your calculation. One that tracking units will not catch. If you scale down distance by a factor of 1772, what does that do to volume?
 
jbriggs444 said:
There is another error in your calculation. One that tracking units will not catch. If you scale down distance by a factor of 1772, what does that do to volume?
Maybe I'm wrong, but I think the OP did this correctly. The distance scales by a factor of 12.1, which scales the volume by 12.1^3 = 1772.
 
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Thomas Smith said:
I thought that in regards to the number of hydrogen atoms you needed to include it

No. including it gives you the mass of a mole of hydrogen atoms. You want the mass of a single hydrogen atom.
 

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