
#1
Mar2411, 05:05 PM

Sci Advisor
P: 5,307

I have a problem with the following scenario:
A friend of mine claims that it's possible to start with equations for a compressible fluid or dust and to derive an expanding solution in Newtonian mechanics (!) He claims that this expansion is homogeneous and isotropic (I understand his reasoning) and that all particles in the cloud are in free fall i.e. feel no force (this what I do not believe). His equations for test particles do indeed look homogeneous and isotropic, but I doubt that he can correctly derive free fall. He claims to be able to derive the motion of a test particle that moves together with the expanding sphere of dust in free fall due to the gravitation of the dust. He does not present the equation for this sphere of dust, but instead he uses a kind of effective equation for a test particle from the very beginning. It's not clear to me how he can bypass the equation for the dust itself. Instead of presenting his derivation I would like to use a different approach. Start with the e.o.m for the dust, i.e. a kind of Euler equations for a compressible fluid taking into account the gravitation of the dust. If this could be done one finds directly how the flow of dust may look like. In that case the motion of the test particle along with the dust could indeed result in free fall. First question: is there a proof whether such a scenario as described above is possible or impossible? is there an analogy with an expanding universe (from GR) in the sense of Newtonian mechanics, i.e. an explosion of a matter distribution in space (instead of expansion of space as in GR)? Second question: is there a simple derivation of the dynamics of a gravitating sphere of dust in Newtonian fluid dynamics? Of course the radius of the sphere shall be infinite in order to achieve an homogeneous and isotropic solution. Third question: does anybody know how to combine the Euler equations for a fluid with the gravitational energy [tex]V = \int d^3x \int d^3y \, \frac{\rho(\vec{x},t)\,\rho(\vec{y},t)}{\vec{x}\vec{y}}[/tex] 



#2
Mar2411, 10:16 PM

P: 688

I think I can answer one of the questions.
You can derive the Friedmann equation for an expanding universe with Newtionion mechanics as long as the universe is flat and no cosmological constant. Just look at the escape velocity of a particle trying to escape the mass/energy of the universe. Actually, you can derive the Friedmann equations with curvature from Newtionion mechanics. It takes some handwaving, however. See "Introduction to Modern Cosmology" by Liddle. 



#3
Mar2511, 01:17 AM

Sci Advisor
P: 5,307

Thanks.
Starting with a scalar density, how can one satisfy both decreasing density and homogenity? Sitting at r I see the density [tex]\rho(\vec{r}, t) = \rho(r, t)[/tex] Sitting at a different point I see [tex]\rho(\vec{r}^\prime, t) = \rho(\vec{r} + \vec{dr}, t) = \rho(r, t) + dr_i\,\partial_i\,\rho(r, t)[/tex] But if the solution shall be homogeneous the second term must vanish and therefore the density must be constant. Where's my mistake? 



#4
Mar2511, 07:21 AM

Sci Advisor
P: 5,307

dynamics of selfgravitating dust
OK, I checked "Introduction to Modern Cosmology" by Liddle.
It seems to be rather clear. The equations are well defined from the very beginning and it seems that Newtonian gravity does more or less exactly the same as GR does. As the derivation is more or less identical to GR (differential equations do not care about their meaning :) the problem must be in the very beginning. I guess it's the step where Liddle introduces the total energy of a mass (volume element) m. In an infinite universe with Newtonian gravity this potential energy is illdefined due to the instantaneous 1/r law. So I would say that if one skips the technical detail that the whole model is not defined, the results of the model agree with GR :) 


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