# Gravitational attraction on the cosmic scale?

1. Jun 10, 2014

### Vincentius

Hi, the following two views appear inconsistent to me:

In the infinite perfectly homogeneous universe:

a) the net force of gravity is zero everywhere, so no energy is being exchanged and no particle is pulled in any direction whatsoever.

b) the net force of gravity within a spherical cavity is zero, therefore particles put into the cavity will (only) attract each other, therefore force of gravity on these particles is non-zero and directed to the center of the sphere on average. Applied to arbitrary spheres it implies that the net force of gravity on any particle in the universe must be non-zero.

I like a) better than b) but the latter is mainstream. So what's wrong with a)?

Is this subject being treated anywhere?

2. Jun 10, 2014

### DennisN

The Universe is only homogeneous on very large scales, not on smaller length-scales.

3. Jun 10, 2014

### Vincentius

Of course, but assume a perfectly homogeneous universe to keep things simple.

4. Jun 10, 2014

### DennisN

But I actually don't understand the point of assuming it at all. The assumption is not compatible with the Universe. It might be better if you are more specific about what you are trying to understand, so people here can help.

5. Jun 10, 2014

### marcus

I don't understand why you say "b" is mainstream, Vincent. I never heard such an argument. Do you have a link to where some professional cosmologist argues for "b"?

I think "a" (which you say YOU like but is NOT mainstream) is in fact the mainstream view: In a homog universe (of the sort cosmologists normally work with, namely center-less and boundary-less) there is no pull in any direction.

Indeed "a" is based on a correct argument (from a mainstream point of view) namely the approximate isotropy---the distribution of mass is approximately the same in all directions---so one does not expect net pull in any direction. One only gets a net pull to the extent that the distribution is NOT homogeneous and isotropic.

and as far as we can see the conclusion of "a" is born out by observation. Some people claim to have detected a massive lopsided drift in one direction or the other, which they want to be explained by some massive lopsided inhomogeneous distribution of matter, but AFAIK it has not been confirmed. Aside from the minor effect of some obvious inhomogeneities (which you are neglecting) it seems there is nothing wrong with "a".

I am wondering why you do not ask a different question, namely what is wrong with "b"?

Last edited: Jun 10, 2014
6. Jun 10, 2014

### DennisN

Marcus is more able to extract the issue and help out here than me (thanks for jumping in, Marcus). I thought post#1 was some kind of argument for a static universe...

7. Jun 10, 2014

### marcus

Dennis thanks for the kind encouragement!
But actually I am a little puzzled about Vincent's "b".

I would welcome help in thinking of what is wrong with "b". From you or any of the others---Brian Powell, George Jones, one of the other mods.

It seems to me that it could be argued that there is NO force of gravity on any galaxy that is in free fall. In its own inertial frame the galaxy feels no force. You only have a force defined if you have some preferred external frame of reference.

It seems to me that Vincent, by assuming homogeneity is providing no basis for defining a preferred frame. He wants to argue, in "b" that there is some nonzero force of gravity on every galaxy. To do that it seems to me he has to make explicit some preferred frame. I'm a bit vague on this. Hope someone else will clarify.

8. Jun 10, 2014

### Bill_K

Don't you recognize it? It's a standard paradox. Nothing to do with cosmology or general relativity, it results from attempting to apply theorems relating to Laplace's Equation to a source with an infinite uniform distribution. It can be the Newtonian gravitational potential of a mass distribution, the electrostatic potential of a charge distribution, etc. (a) of course is the correct answer.

9. Jun 10, 2014

### Vincentius

Ok. Perhaps I should rephrase the question: in the Friedmann equation

H2=H02m/a3λ)​

the dust term Ωm/a3 is commonly associated with the attractive force of gravity. So if at the same time no particle is pulled in any direction, i.e. net force is zero, this seems a contradiction. What is the resolution?

Last edited: Jun 10, 2014
10. Jun 10, 2014

### Mordred

a3 is the scale factor .

the dust term is just Ωm

I'll need to think about the rest of it

11. Jun 10, 2014

### marcus

Ahhh! Thanks. I must not have gotten enough sleep last night. Staring stupidly at the problem the thought did cross my mind several times that the outside was infinite. Anyway (b) does not work.

12. Jun 10, 2014

### Mordred

Yeah looking at the way the question is worded I agree

13. Jun 10, 2014

### Vincentius

If we all agree a) is right and b) is not, then this still leaves the question why Ωm/a3 is commonly associated with gravitational attraction (matter decelerates the expansion). This term indeed behaves by its form as if there were gravitational attraction and indeed the argument b) coincides with the Newtonian derivation of the Friedmann equation. However, given correctness of a) one would refute this common interpretation.

14. Jun 10, 2014

### Matterwave

First, a correction, in the equation you gave, you really only have $\dot{a}^2$ which applies for both $\dot{a}>0$ and $\dot{a}<0$. By that equation, you really can't tell if the matter term is attractive or not since a higher density makes contractions contract faster and expansions expand faster (the contracting sphere is the time reversal of the expanding sphere).

It's actually the other Friedmann equation, involving $\ddot{a}$ which shows that a density will lead to a slowing of expansion and a speeding up of contraction ($\ddot{a}<0$). This is really a general relativistic result, and can be derived from either applying Einstein's field equations directly to the FLRW metric, or by (covariant) conservation of the stress-energy tensor.

15. Jun 11, 2014

### Chalnoth

Actually, you can get the same result in Newtonian gravity. See here, for example:
http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf

What you gain from General Relativity is the effect that pressure has on the expansion. As normal matter experiences no pressure on cosmological scales, the result is the same for normal matter.

16. Jun 11, 2014

### Vincentius

For general density, yes. For pressureless dust Ωm/a3 implies attraction, which is the common interpretation of the dust term, right?

17. Jun 11, 2014

### Chalnoth

Technically yes, though it takes a few steps from there to get to that conclusion. The conclusion can be drawn most directly from the second Friedmann equation, as Matterwave pointed out. That equation is (with most of the constants eliminated for clarity):

$${\ddot{a} \over a} = -{1 \over 2}(\rho + 3p)$$

Here $\rho$ is the energy density and $p$ is the pressure. For a cosmological constant, for instance, where $p = -\rho$,

$${\ddot{a} \over a} = -\rho$$

18. Jun 11, 2014

### Vincentius

Thanks for pointing this out Chalnoth and Matterwave. We agree that according to the Friedmann equations pressureless dust implies attraction. So the question how to reconcile this with zero net gravity in the homogeneous universe still stands.

19. Jun 11, 2014

### Chalnoth

Take a look at the Newtonian derivation I linked earlier.

The way to estimate this is not to consider one dust particle's effect on another, but to separate out a finite sphere of dust particles. Due to the symmetry of the rest of the universe, the gravitational effect of everything else on this sphere will be identically zero and can thus be ignored. But the sphere itself will have a tendency to slow its expansion (or collapse inward on itself). As this argument must be true everywhere, it implies a universal change in expansion rate brought about by the matter density.

20. Jun 11, 2014

### Matterwave

The Newtonian way to think about it would be as Chalnoth pointed out. Let's just start out with initial condition of no contraction or expansion. All spheres that you draw are forced into contracting by symmetry. This leads to the conclusion that all of space contracts (again, not to a point, but just everything gets closer to each other, just as the reverse of expansion, you have to remember that the universe is infinite in this scenario).

I think the confusion is really that usually one always has in mind contraction to a point. Like, e.g. a sphere contracts to a point. The difference with this argument is that the universe will stay homogenous and isotropic in its contraction. It's not like the different spheres are contracting to points. Every point is getting closer to every other point.

I am of the opinion that such a Newtonian derivation is more of a trick than anything else. There's no real spatial contraction of course in Newtonian mechanics, space-time is flat. I would always appeal to the general relativistic result from using the Einstein field equations. The Einstein field equations give us a different way of thinking of this problem than the Newtonian way. The Einstein field equations tells us that really matter gives curvature to space-time (let's again apply the same initial condition as before) and this curvature in space-time is such that for space-like slices of space-time, the 3-D metric on them contracts as time moves forward. Here there's no appeal to any forces. Again, I want to say that the result can be obtained by covariant conservation of the stress-energy tensor. It is really THIS physics, that in my mind, makes the matter appear to slow down expansion and speed up contraction.