- #1
johne1618
- 371
- 0
How does one integrate the mass density over a closed Universe (a 3-sphere?) to obtain the total mass of that Universe?
Is this the correct integral?
[tex]
M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}
[/tex]
where [itex]R(t)[/itex] is the radius of the Universe at cosmological time [itex]t[/itex].
By making the substitution [itex]r=\sin \chi[/itex] one finds that the above integral gives:
[tex]
M = \pi^2 R(t)^3 \rho.
[/tex]
According to wikipedia the hyperarea of a 3-sphere is [itex]2\pi^2 R^3[/itex] so I'm out by a factor of two.
Is this the correct integral?
[tex]
M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}
[/tex]
where [itex]R(t)[/itex] is the radius of the Universe at cosmological time [itex]t[/itex].
By making the substitution [itex]r=\sin \chi[/itex] one finds that the above integral gives:
[tex]
M = \pi^2 R(t)^3 \rho.
[/tex]
According to wikipedia the hyperarea of a 3-sphere is [itex]2\pi^2 R^3[/itex] so I'm out by a factor of two.
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