johne1618
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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?
<br /> M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}<br />
where R(t) is the radius of the Universe at cosmological time t.
By making the substitution r=\sin \chi one finds that the above integral gives:
<br /> M = \pi^2 R(t)^3 \rho.<br />
According to wikipedia the hyperarea of a 3-sphere is 2\pi^2 R^3 so I'm out by a factor of two.
Is this the correct integral?
<br /> M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}<br />
where R(t) is the radius of the Universe at cosmological time t.
By making the substitution r=\sin \chi one finds that the above integral gives:
<br /> M = \pi^2 R(t)^3 \rho.<br />
According to wikipedia the hyperarea of a 3-sphere is 2\pi^2 R^3 so I'm out by a factor of two.
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