Gravitational attraction on the cosmic scale?

[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?

 

[DennisN]

So what's wrong with a)?

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

 

[Vincentius]

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

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

 

[DennisN]

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

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.

 

[marcus]

...

In the infinite perfectly homogeneous universe:

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)?

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"?

 

[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...

 

[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.

 

[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.

 

[Vincentius]

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.

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

H²=H₀²(Ω_m/a³+Ω_λ)​

the dust term Ω_m/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?

 

[Mordred]

a³ is the scale factor .

the dust term is just Ω_m

I'll need to think about the rest of it

 

[marcus]

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.

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.

 

[Mordred]

Yeah looking at the way the question is worded I agree

 

[Vincentius]

If we all agree a) is right and b) is not, then this still leaves the question why Ω_m/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.

 

[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.

 

[Chalnoth]

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.

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.

 

[Vincentius]

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.

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

 

[Chalnoth]

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

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 $p = -\rho$,

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

 

[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.

 

[Chalnoth]

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.

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.

 

[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.

 

[Vincentius]

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.

Maybe we are getting somewhere now. I agree the Newtonian interpretation of gravitational attraction is troublesome, so GR is the proper way to go. However, the Friedmann equation from a GR point of view lacks clear physical interpretation. The equation (for all thinkable forms of mass-energy) just follows from the math. So the key thing here, in my view, is why we necessarily link Ω_m/a³ to gravitating matter? Does this necessarily follow from GR? I doubt. Is perhaps the sole reason for this the Newtonian interpretation, just because it looks the same and it is something we can imagin? In other words, I believe the term is right, but I distrust the interpretation of Ω_m as being matter (density).

 

[Matterwave]

Maybe we are getting somewhere now. I agree the Newtonian interpretation of gravitational attraction is troublesome, so GR is the proper way to go. However, the Friedmann equation from a GR point of view lacks clear physical interpretation. The equation (for all thinkable forms of mass-energy) just follows from the math. So the key thing here, in my view, is why we necessarily link Ω_m/a³ to gravitating matter? Does this necessarily follow from GR? I doubt. Is perhaps the sole reason for this the Newtonian interpretation, just because it looks the same and it is something we can imagin? In other words, I believe the term is right, but I distrust the interpretation of Ω_m as being matter (density).

This term is true for any term that has the stress energy tensor of a perfect fluid. The Friedmann equations are the physical result of applying the Einstein Field Equations to the case of a perfectly homogenous and isotropic distribution of a perfect fluid.

See here: http://en.wikipedia.org/wiki/Perfect_fluid

 

[Chalnoth]

So the key thing here, in my view, is why we necessarily link Ω_m/a³ to gravitating matter? Does this necessarily follow from GR?

Yes.

The way that the energy density of the stuff changes as the universe expands is a function of its pressure, and is taken from the conservation of the stress-energy tensor. This conservation law, which is integral to GR, identically relates $\Omega_m / a^3$ to $p = 0$. And $p = 0$, as I showed above, results in $\ddot{a} < 0$.

 

[Vincentius]

Yes.

The way that the energy density of the stuff changes as the universe expands is a function of its pressure, and is taken from the conservation of the stress-energy tensor. This conservation law, which is integral to GR, identically relates $\Omega_m / a^3$ to $p = 0$. And $p = 0$, as I showed above, results in $\ddot{a} < 0$.

I understand the math. But where it tells it must be matter (dust). A form of energy with the same dependency on the scale factor would work identically. This may no match expectations, but can not be excluded on theoretical grounds. Dark energy by the cosmological constant is not obvious either.

 

[Chalnoth]

I understand the math. But where it tells it must be matter (dust). A form of energy with the same dependency on the scale factor would work identically. This may no match expectations, but can not be excluded on theoretical grounds. Dark energy by the cosmological constant is not obvious either.

The physics of the material that is relevant in this context is contained in the relationship between pressure and energy density for that material. It's not hard to show, using the conservation of stress-energy in General Relativity, that if the ratio of energy density to pressure is a constant, then:

$$w = {\rho \over p}$$
$$\rho(a) \propto a^{-3(1 + w)}$$

If this ratio of energy density to pressure is not a constant, then it can get a bit more complicated. But the only way for the energy density to be proportional to $a^{-3}$ is if $w = 0$, and hence $p = 0$.

 

[Vincentius]

Again, I understand the math. It does not tell that for p=0 ρ must be matter. In fact, matter is somewhat doubtful as per our discussion above. On the other hand, since net force of gravity is zero, particles will conserve kinetic energy, which conservation also points at p=0 (no pressure, so no energy exchange). So perhaps the Ω_x/a³ term is related to kinetic energy, i.e. to the inertial motion of the receding cosmic masses.

 

[Chalnoth]

Again, I understand the math. It does not tell that for p=0 ρ must be matter. In fact, matter is somewhat doubtful as per our discussion above. On the other hand, since net force of gravity is zero, particles will conserve kinetic energy, which conservation also points at p=0 (no pressure, so no energy exchange). So perhaps the Ω_x/a³ term is related to kinetic energy, i.e. to the inertial motion of the receding cosmic masses.

For $p=0$, it must be made of something which has non-zero rest mass and non-relativistic velocity. Generally we call this kind of stuff matter (including both dark matter and baryonic matter).

Yes, in the early universe normal matter behaved as radiation.

 

[Vincentius]

For $p=0$, it must be made of something which has non-zero rest mass and non-relativistic velocity. Generally we call this kind of stuff matter (including both dark matter and baryonic matter).

Ok, thanks! If that's right, it doesn't look as energy. Why is it necessary for p=0 that there is non-zero rest mass? Can you give a reference?

However, won't you agree that p= 0 also in case of the kinetic energy density of inertial motion of receding masses?

If it is matter though, then my original problem is still there: zero net force vs. (something that acts as) attracting matter.

 

[Chalnoth]

Ok, thanks! If that's right, it doesn't look as energy. Why is it necessary for p=0 that there is non-zero rest mass? Can you give a reference?

It requires both a non-zero rest mass and non-relativistic velocity (though technically I suppose you could just use the second condition, as you can't have a non-relativistic velocity if it has zero rest mass).

Anyway, I hope you'll agree that the pressure of a fluid is a function of the kinetic energy of the individual components of that fluid. If it has no kinetic energy, and no rest mass, then it has no energy density either (In fact, in a very real sense I think you could state that it doesn't exist at all).

However, won't you agree that p= 0 also in case of the kinetic energy density of inertial motion of receding masses?

Ahh, but here the kinetic energy is with respect to the local matter. The recession velocity plays no part in the kinetic energy as it relates to the pressure of the fluid.

If it is matter though, then my original problem is still there: zero net force vs. (something that acts as) attracting matter.

Again, consider a finite sphere of matter, instead of the force a single particle feels.

 

[Vincentius]

Anyway, I hope you'll agree that the pressure of a fluid is a function of the kinetic energy of the individual components of that fluid. If it has no kinetic energy, and no rest mass, then it has no energy density either (In fact, in a very real sense I think you could state that it doesn't exist at all).

I am afraid I don't agree. No pressure means no (ex)change of energy. i.e. pressureless particles can still carry constant kinetic energy>0. With no forces acting on them this energy is indeed constant.

 

[Chalnoth]

I am afraid I don't agree. No pressure means no (ex)change of energy. i.e. pressureless particles can still carry constant kinetic energy>0. With no forces acting on them this energy is indeed constant.

What do you mean by a "pressureless particle"?

Pressure is not a function of an individual particle, but instead of a fluid of particles.

 

[Vincentius]

Ok. I mean radially moving particles (Hubble flow) which do not collide. I suppose this is a pressureless fluid.

 

[Chalnoth]

Ok. I mean radially moving particles (Hubble flow) which do not collide. I suppose this is a pressureless fluid.

If your particles' only motion is the Hubble flow, then those particles are not moving with respect to the Hubble flow, and thus have no kinetic energy compared to said flow and produce no pressure.

 

[Vincentius]

Indeed, no pressure. But no kinetic energy? That depends on the definition of kinetic energy, but I don't know the answer. There must be kinetic energy in the receding masses. Imagine what would happen if everything moves at the same speed but inward. Kinetic energy is a relational property between masses, hence this favors a Machian definition of kinetic energy between every pair of masses (Schrödinger!). Kinetic energy relative to an imaginary frame is physically meaningless.

Again I don't know the answer. The point though is that if there is kinetic energy stored in the cosmic recession of matter (by whatever definition), then it is conserved since net force is zero.

 

[Chalnoth]

Indeed, no pressure. But no kinetic energy? That depends on the definition of kinetic energy, but I don't know the answer. There must be kinetic energy in the receding masses. Imagine what would happen if everything moves at the same speed but inward. Kinetic energy is a relational property between masses, hence this favors a Machian definition of kinetic energy between every pair of masses (Schrödinger!). Kinetic energy relative to an imaginary frame is physically meaningless.

Again I don't know the answer. The point though is that if there is kinetic energy stored in the cosmic recession of matter (by whatever definition), then it is conserved since net force is zero.

You have to use the definition that's relevant to the situation at hand. In this case, the relevant frame of reference is the co-moving frame of reference.

It's easier if you think of it as an expanding gas than as a set of individual particles.