Re: Gravitational field of a homogeneous universe
From: Theo Wollenleben (alpha0x89_at_yahoo.de)
Date: 09/10/04
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Date: 10 Sep 2004 01:23:20 -0700
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.
> 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. It's not
a big surprise that you can get another result. I suppose you can get
more different results when choosing other coordinates. Compare to the
similar situation when calculating a series:
http://mathworld.wolfram.com/RiemannSeriesTheorem.html
> But I simply don't see where the flaw in the derivation of this result
> lies!
Mathematically both results are flawed.
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