How to calculate mass of closed Universe?

1. Jul 29, 2013

johne1618

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?

$$M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}$$

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:

$$M = \pi^2 R(t)^3 \rho.$$

According to wikipedia the hyperarea of a 3-sphere is $2\pi^2 R^3$ so I'm out by a factor of two.

Last edited: Jul 29, 2013
2. Jul 29, 2013

Simon Bridge

3. Jul 29, 2013

johne1618

4. Jul 29, 2013

dchartier

5. Jul 29, 2013

Simon Bridge

The former applies the cube to the brackets, the latter applies the cube to the function.
$R(t)^3=Rt^3$
$R^3(t)=\big [ R(t)\big ]^3$

compare: $\cos(kx)^3$ with $\cos^3(kx)$
Since it is also valid to write $\cos kx$ the first version does not make it clear what is intended.
Is it the cosine of the cube of a product or the cube of the cosine of a product? OP only had one variable in the brackets, which gives notice to the reader.

... its really just a notation foible.
I don't think anyone would get confused in such a small equation but in big long multiline equations it gets easier to be misread.

Last edited: Jul 29, 2013
6. Jul 29, 2013

dchartier

I understand, thanks!