# Newtonian derivation of Friedmann equation

1. Oct 25, 2006

### Mike2

I'm trying to derive Friedmann's equation for cosmology using Newtonian physics. I've got the force equation F=ma for the case without a cosmological constant. But now I'm trying to incorporate the cosmological constant into this force equation.

But I'm having trouble seeing how the cosmological constant can appear to apply a force on some particle at the edge of a spherical universe of radius R. I thought that I might get somewhere through dimensional analysis if I could understant the units of the cosmological constant from its use in the Einstein Field Equation where it was initially introduced. But I find I don't know the units of any of the other entities in the EFE - the units for the metric gab or the Ricci tensor or the Ricci scalar or the energy-momentum tensor. Do the units change for different values of ab? It seems everyone likes using elaborate mathematics - and I don't see much practical use - not even dimensional units of the things they talk about. Any help out there? Thanks.

2. Oct 25, 2006

### cesiumfrog

It isn't obvious that what you want to do is even possible. Why don't you do a course on general relativity first (or even just say the track one exercises in Misner, Wheeler & Thorne's book)?

3. Oct 27, 2006

### Mike2

In the pdf file, starting with page 9, at:

http://www.astro.caltech.edu/~george/ay21/Ay21_Lecture02.pdf

the author derives the Friedmann equation from newtonian mechanics using the force produced from the mass density of the universe.

And at:

http://en.wikipedia.org/wiki/Cosmological_Constant

the cosmological constant is expressed in terms of an energy density also as:

$$$\Lambda = \frac{{8\pi G}}{{c^4 }}\rho vac$$$

from which we can just as easily derive the Friedmann equation which includes the cosmological constant.

But I don't know how they got the last equation above. I think it may have something to do with the dimensional analysis of the terms in the Einstein field equations shown below. And I'm hoping someone here knows. Thank you.

$$$Rab - \frac{1}{2}Rgab + \Lambda gab = 8\pi Tab$$$