# 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!