Re: Gravitational field of a homogeneous universe

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 09/10/04 

Date: Fri, 10 Sep 2004 12:42:33 +0200

Theo Wollenleben wrote:
> Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<chpojm$jl6$1@news.urz.uni-heidelberg.de>...
>
>>Well, after these preliminaries ;-), on to the actual problem. Let's
>>imagine we have an infinite, flat threedimensional space filled with
>>homogeneous matter, i.e. the density rho does not depend on the position.
>>
>>Now, one approach to calculate the gravitational field in this space
>>would be to use a formula analogous to the one used in electrostatics
>>to calculate the electric field:
>>
>>vec{g}(vec{r}) = G int dV' rho(vec{r}')
>> (vec{r}-vec{r'})/|vec{r}-vec{r'}|^3
>>
>>(using a LaTeX-like notation), where vec{g} is the gravitational
>>acceleration at the point vec{r}, G is the gravitational constant, and
>>the integral runs over all the volume. Now, since the density is
>>constant, we can take it out of the integral:
>>
>>vec{g}(vec{r}) = G rho int dV' (vec{r}-vec{r'})/|vec{r}-vec{r'}|^3
>>
>>Since the remaining integrand is an odd function, the integral gives
>>zero:
>>
>> vec{g}(vec{r}) = 0
>>
>
>
> The integral does not converge. You computed sort of a principal value
> of the integral.

Good point... But why does the Robertson-Walker metric apparently give
the same result then?

>>But there is another approach (which is used e.g. by the cranks Marcel
>>Luttgens and Jim Jastrzebski), which gives a different result - and I am
>>not sure where the flaw with that approach lies.
>>
>>Let's "take out" a sphere from this homogeneous, infinite universe.
>>The matter outside the sphere is spherically symmetric around the center
>>vec{r}_0 of the sphere and therefore does not contribute to the
>>gravitational field in its inside. So the gravitational field on the
>>surface of the sphere is determined only by the matter contained
>>therein; and therefore we get that the gravitational field at a point
>>vec{r} is given by
>> vec{g}(vec{r}) = G M_enclosed (vec{r}-vec{r}_0)/|vec{r}-vec{r}_0|^3
>> = G (4/3) pi (vec{r}-vec{r}_0).
>
>
> Now you are computing infinity minus infinity in another way.

Sorry, I don't see how I am doing that here.

> It's not
> a big surprise that you can get another result. I suppose you can get
> more different results when choosing other coordinates.

Yes, indeed. See this post of mine in sci.astro:
Google Message ID <ch9rq4$odb$1@news.urz.uni-heidelberg.de>

[snip]

>>But I simply don't see where the flaw in the derivation of this result
>>lies!
>
>
> Mathematically both results are flawed.

Then what *is* the gravitational field of a homogeneous universe, in
your opinion? And how would one go about calculating it?

Bye,
Bjoern

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