Can a Homogeneous Universe Avoid a Big Crunch?

[GoToLoop]

TL;DR Summary

I do not see how gravity could predict that the universe should contract into a Big Crunch.

[Mentor note: This post has been edited to not fall afoul of the Physics Forums rule prohibiting the posting of personal theories]Simplified Thought Experiment: Imagine that the universe is homogenously filled with particles. We can calculate the force on a given particle as the force due to gravity of all other particles within a sphere of the visible universe around it. Since this particle is at the center of this visibility sphere, the net force will be 0. Apply this same logic to all other particles in the universe, and you continue to get a net gravitational force of 0. Therefore, all matter will behave as a steady state with no expansion or contraction within this simple model.

This seems to have a similar effect for a non-homogenous universe as well. As the sphere of visibility expands, you would expect to get local clumping, with perhaps larger and larger clumps as the visibility increases, but there would be no Big Crunch.

Thoughts? This simple claim has been bothering me for several years. Can someone point out where the argument breaks down, if at all?

 

[Orodruin]

Gravity does not work like that on cosmological scales. To understand this fully you will need to learn general relativity.

It is not about forces and objects falling under their influence. It is about the behavior of spacetime itself.

 

[Nugatory]

This simple claim has been bothering me for several years. Can someone point out where the argument breaks down, if at all?

The history here is interesting... I started to write a summary then realized that it would be more helpful to suggest that you Google for “cosmological constant” - the Wikipedia article has a good summary of that history.

 

[PeroK]

Thoughts? This simple claim has been bothering me for several years. Can someone point out where the argument breaks down, if at all?

Fundamentally, it breaks down as you are using Newton's law of gravity - which on a cosmological scale was superseded by General Relativity in 1915.

 

[Ibix]

The problem is that you are using a Newtonian model of gravity, and Newtonian gravity is only an approximation to general relativity. It's a pretty good approximation when you are studying systems that don't change much over the time it takes light to cross them (and where gravitational fields are weak, although that's not relevant here). That approximation is fine for the solar system and even for clusters of galaxies, whose proper motions are well under 1% of light speed. But IIRC light can circumnavigate a closed universe (one with a Big Crunch) once in the entire lifetime of the universe - in which time it's changed just a bit.

So the problem, as others have pointed out, is that you are using Newtonian gravity. That means that you are using a model that assumes that the universe doesn't change to predict how the universe changes. That's never going to work.

 

[etotheipi]

Please examine the equation of motions $3\left(\dfrac{\dot{a}}{a} \right)^2 = 8 \pi \rho - \dfrac{3k}{a^2}$ and $3 \left( \dfrac{\ddot{a}}{a}\right) = -4(\rho + 3P)$ for homogenous, isotropic universe with stress energy tensor of a perfect fluid. Given that $\rho \propto a^{-n}$ where $P=0 \longrightarrow n=3$ and $P = \rho/3 \longrightarrow n=4$, what is the qualitative behaviour of this system of equations for $k \in \{-1,0,1 \}$?

 

[Bandersnatch]

Uh, oh. The thread might be marked 'I', but I think ^that's above its level.

I don't think the issue here is as much the application of Newton's gravity, as it is its faulty application. The Newtonian approximation is, after all, used pedagogically to provide a heuristic for Friedmann equations as an expression of conservation of energy (e.g. in Liddle).
The misapplication comes from changing frames of reference. I believe this has been discussed at least a few times on this forum.
In the frame of reference of the observer, all massive points at any radius are attracted towards the observer, who, alone, has 0 forces acting on them. If we change the FoR to another observer, we get 0 forces on that observer while every other point (including the previous observer) is attracted to the centre. But it's not unexpected - we did change the frame of reference. It's no different to noticing how in a system of orbiting bodies, in the frame of reference of each body there are no forces acting on it (it's in free fall), yet the system as a whole is not static.
What we can't do, is treat both results as inhabiting the same frame of reference, since that gives us this apparent paradox.
Whereas every observer seeing every other point as attracted to them only is consisted with the global picture of expansion/contraction, since that's what we'd expect every observer to see.

 

[timmdeeg]

Summary:: I do not see how gravity could predict that the universe should contract into a Big Crunch.

If you through a stone upwards (i) it will reach certain height (ii) and thereafter fall down (iii).

A universe which is supposed to end up into a Big Crunch is first expanding decelerated (era i), will reach a maximal size (era ii) and will thereafter contract (era iii).

The particles in the universe feel a gravitational attraction all the time but being in free fall they don't feel a force.

Please see this analogy with care because the spacetimes are different.

 

[etotheipi]

The "big crunch" occurs for $k=1$. Since $8\pi \rho- \dfrac{3}{a^2} \geq 0$, and $\rho = \gamma a^{-n}$ for some $\gamma$ and with $n \geq 3$, then $a$ is bounded from above by $a_c = \left( \dfrac{8\pi \gamma}{3} \right)^\frac{1}{n-2}$. Further, $|\ddot{a}| = \dfrac{4a}{3}\left( \rho + 3P \right)$ is bounded from below so $a$ does not asymptotically approach $a_c$.

 

[GoToLoop]

Wow! Thanks for all the responses!

I don't think the issue here is as much the application of Newton's gravity, as it is its faulty application. The Newtonian approximation is, after all, used pedagogically to provide a heuristic for Friedmann equations as an expression of conservation of energy (e.g. in Liddle).

So far, I think this answer goes the furthest in providing a counter-point. (Other answers which say that I should use General Relativity instead of Newton's gravity might be correct, in the end. But after all, Newtonian Physics ought to be an excellent approximation for this simple thought experiment, so I've never tried to view it in the more complicated terms of General Relativity. Regardless, I'd like to reach a firm answer under Newtonian physics before considering how the answer might change due to General Relativity.)

The misapplication comes from changing frames of reference. I believe this has been discussed at least a few times on this forum.
In the frame of reference of the observer, all massive points at any radius are attracted towards the observer, who, alone, has 0 forces acting on them. If we change the FoR to another observer, we get 0 forces on that observer while every other point (including the previous observer) is attracted to the centre. But it's not unexpected - we did change the frame of reference. It's no different to noticing how in a system of orbiting bodies, in the frame of reference of each body there are no forces acting on it (it's in free fall), yet the system as a whole is not static.

I'm not sure that all of this is accurate under a Newtonian model. The reason that the net force is 0 in the first case is because the subject is at the center of a homogenous sphere, and therefore all forces due to gravity cancel out. In the case of orbiting bodies where we consider one subject at a time, there are forces acting on them based on the distribution of the surrounding massive objects (e.g. even in the moon's own reference frame, it can easily calculate its own acceleration due to the mass of the earth). This is still the Newtonian model, so the rationalization is basically that a body cannot "feel" the force of gravity in the human sense because the force is acting on every part of it at the same time (whereas you can "feel" a force while in a jet-plane because you get compressed into your seat). But the entire object still undergoes an acceleration.

To re-iterate, my simple thought experiment calculates a force for each object by summing up all the gravitational forces exhibited on it within its own visible universe. In a solar system configuration under Newtonian physics I believe that this produces the expected dynamic behavior that we would normally attribute to the Newtonian model. In a homogenous universe under Newtonian physics, I believe that this produces an effectively static universe, as opposed to a Big Crunch prediction.

Whereas every observer seeing every other point as attracted to them only is consisted with the global picture of expansion/contraction, since that's what we'd expect every observer to see.

Just to clarify, in this setup, each particle in the universe only predicts its own acceleration - not anybody else's. The reason to do this it because the visible universe is different for every particle in question and you cannot accurately predict a particle's acceleration if you use the wrong visible universe.

I'm happy to make further clarifications if people still aren't following me. :)

 

[PeterDonis]

Newtonian Physics ought to be an excellent approximation for this simple thought experiment

No, it isn't. That's the point that multiple responses to you have been making. If it seems to your intuition that it should, that means your intuition is wrong and needs to be retrained.

@Bandersnatch in his post referred to Newtonian reasoning being used as a heuristic, i.e., as a sort of plausibility argument for the Friedmann Equation. Note that even in this case, the Newtonian argument reaches a conclusion different from yours. But in any case, such an argument is still just heuristic. It does not take into account the fact that Newtonian physics is simply a wrong model for the universe as a whole. Newtonian physics cannot even comprehend the conceptual scheme we use to model the universe as a whole.

I'm happy to make further clarifications if people still aren't following me. :)

The issue is not that people are not following you. The issue is that you are using the wrong theory. See above.

 

[PeterDonis]

even in the moon's own reference frame, it can easily calculate its own acceleration due to the mass of the earth

Quite apart from the fact that Newtonian physics is the wrong theory for modeling the universe as a whole, your understanding of it even in cases where it is a good approximate model is wrong. This is an example. In the moon's rest frame, the moon has no acceleration, by definition; it is at rest. What you can calculate in the moon's reference frame is the acceleration of the earth due to the mutual gravitational force between the Earth and the moon.

Note that similar reasoning applied to the case of a homogeneous distribution of particles gives the outcome that @Bandersnatch describes in his post.

 

[DrStupid]

Simplified Thought Experiment: Imagine that the universe is homogenously filled with particles. We can calculate the force on a given particle as the force due to gravity of all other particles within a sphere of the visible universe around it.

You seem to refer to Newtonean gravity. That's a long-range interaction without theoretical limit. That means that the force on a given particle is the force due to gravity of all other particles within the entire universe - not only in the visible universe. But in an infinite universe your argumentation doesn't work anymore. That's what Newton wrote about it in a letter to Richard Bentley:

'But you argue in the next paragraph of your letter that every particle of matter in an infinite space has an infinite quantity of matter on all sides & by consequence an infinite attraction every way & therefore must rest in equilibrio because all infinites are equal. Yet you suspect a parallogism in this argument, & I conceive the parallogism lies in the position that all infinites are Equal.
[...]
And so a Mathematician will tell you that if a body stood in equilibrio between any two equal and contrary attracting infinite forces, & if to either of those forces you add any new finite attracting force: that new force how little so ever will destr{oy} the equilibrium & put the body into the same motion into which it would pu{t} it were those two contrary equal forces but finite or even none at all: so tha{t} in this case two equal infinites by the addition of a finite to either of them become unequal in our ways of recconning.'

Your argumentation is correct for the center (and only for the center) of a finite homogeneous spherical mass distribution. But it fails in an infinite homogeneous universe. You can set the particle in the center of concentric homogeneous shells and the resulting total force is zero. But if you set the center of the shells in another position you will get a non-zero net force. You get different results even though there are always infinite forces acting in all directions. The resulting force depends on the arbitrary choice of the center and is therefore as undefined as the center of an infinite space.

All you can do is caclulating the tidal forces between two particles. That's quite easy with Gauss's law:

$\nabla g = - 4 \cdot \pi \cdot G \cdot \rho$

results in the tidal acceleration

$\Delta g\left( {\Delta r} \right) = - {\textstyle{4 \over 3}}\pi \cdot G \cdot \rho \cdot \Delta r$

between each two particles with the distance $\Delta r$. All particles are accelerated towards each other. It is the same tidal acceleration as for random particles in a finite homogeneous spherical mass distribution that can be calculated using the shell theorem.

 

[kimbyd]

Summary:: I do not see how gravity could predict that the universe should contract into a Big Crunch.

[Mentor note: This post has been edited to not fall afoul of the Physics Forums rule prohibiting the posting of personal theories]Simplified Thought Experiment: Imagine that the universe is homogenously filled with particles. We can calculate the force on a given particle as the force due to gravity of all other particles within a sphere of the visible universe around it. Since this particle is at the center of this visibility sphere, the net force will be 0. Apply this same logic to all other particles in the universe, and you continue to get a net gravitational force of 0. Therefore, all matter will behave as a steady state with no expansion or contraction within this simple model.

This seems to have a similar effect for a non-homogenous universe as well. As the sphere of visibility expands, you would expect to get local clumping, with perhaps larger and larger clumps as the visibility increases, but there would be no Big Crunch.

Thoughts? This simple claim has been bothering me for several years. Can someone point out where the argument breaks down, if at all?

As others have noted, you really should use GR for this.

If you want to try to use Newtonian gravity, you can get the same answer, but it requires a bit of fudging.

The Newtonian model of an expanding universe must be finite: there's simply no way to know how to do the calculations for an infinite universe.

If it's finite, what should the shape be? To answer this, consider the symmetry of the system: each observer sees things moving towards or away from them in a uniformly expanding (or contracting) universe. This radial symmetry means that our Newtonian model should be a sphere of matter centered on the observer.

So, if you've got a uniform sphere of matter around you that is currently experiencing uniform expansion (v=Hd), how will that expansion change over time? I won't go through the derivation, but it's not terribly difficult. Because the sphere is finite, we can use Gauss's Law to easily calculate the acceleration due to gravity of every bit of matter in the sphere (it experiences an inward acceleration resulting from the mass contained within that radius, and no acceleration from the stuff at a larger radius).

If you run through the numbers, the equation you get out is exactly the same as the first Friedmann equation for a matter-only universe (and you can even add in the cosmological constant if you do it properly).

Furthermore, if you change coordinates to ask how an observer elsewhere in the sphere would interpret things if they thought themselves to be unaccelerated, they'd predict the exact same behavior of the matter around them as you just calculated. That is, they'd see matter behaving as if it was composed of a finite sphere of matter centered around them instead of around you.

So yes, the right way of doing things is to use General Relativity. But you can, if you use the right construction, show Newtonian physics also predicts the same exact thing. And yes, the Newtonian view will absolutely predict that the sphere will begin to contract if the rate of expansion is too slow relative to the density, just as General Relativity does.

 

[Ibix]

And yes, the Newtonian view will absolutely predict that the sphere will begin to contract if the rate of expansion is too slow relative to the density, just as General Relativity does.

...but that's because it's just a finite spherical cloud of dust, and "what goes up must come down" applies in a way that it doesn't for an infinite FLRW universe, right?

 

[DrStupid]

The Newtonian model of an expanding universe must be finite: there's simply no way to know how to do the calculations for an infinite universe.

Interesting. Where is the problem with an infinite universe? Why is it for example not possible to integrate tidal accelerations over an infinite mass distribution?

 

[kimbyd]

Interesting. Where is the problem with an infinite universe? Why is it for example not possible to integrate tidal accelerations over an infinite mass distribution?

Basically the integrals diverge.

We had a longish discussion on this on this forum not long ago where I refined my understanding.

If you were to take bare Newton's Law of gravitation for an infinite distribution of mass, you'd have to sum up an infinite amount of matter, and that sum diverges. You'd get infinities in your answer that you can't eliminate.

If you get a bit more sophisticated and try to use a way of looking at a finite part of it via Gauss's Law, you have a problem: Gauss's Law is invalid for cases where there is matter at infinity (this is a bit of an oversimplification, but that's the gist of it). Essentially, the derivation of Gauss's Law says that there are extra infinities that you have to account for but have no way to calculate.

So what I'm saying instead is you can pretend that Gauss's Law is valid anyway by selecting a finite sphere of matter and calculating its behavior. Then you can show that that result is the same for any co-moving observer who does the same calculation using any size sphere centered anywhere (provided that center is co-moving).

This isn't the kind of thing that I'd trust in any sort of formal proof. We know that it works because it matches exactly the result from General Relativity, a result which simply does not have the same mathematical problems. But you could do the construction above in another way and it'd fall apart very quickly. Specifically, if Gauss's Law holds, then you don't need to restrict yourself to just looking at a spherical shape: you could use a cylinder instead. And Newtonian gravity doesn't care about motion (the force only depends upon masses and relative position, not velocity), so an infinitely-smooth universe obeys cylindrical symmetry just as well as it obeys spherical symmetry. But the answer you get would be completely different and very incorrect.

To put this all together, there's no way using only Newtonian Gravitation to calculate how an infinite distribution of matter behaves. You can hack it, but the derivation is dicey. Using General Relativity, the result is simple and unambiguous. But fortunately the hacked result and the GR result are identical, so we know it's the right one.

 

[Orodruin]

Gauss's Law is invalid for cases where there is matter at infinity (this is a bit of an oversimplification, but that's the gist of it). Essentially, the derivation of Gauss's Law says that there are extra infinities that you have to account for but have no way to calculate.

This is not really the case though. Gauss law is local and valid for any finite volume and you typically derive it for just a finite volume without any reference to behavior at infinity. Instead, the problem is one of the behavior of the solution at the boundary. While the matter distribution itself can be assumed isotropic and homogeneous (ie, constant density), this is not compatible with boundary conditions that are. In other words, the solution cannot be isotropic and homogeneous.

There is a way around this though. If you consider just the case of a universe with spatially constant matter density and a “center” where the gravitational field is zero and around which you have isotropic boundary conditions, then you will end up in the same situation as you described in the finite case. If you make a transformation to the accelerated frame of any other particle P, the inertial force resulting from the transformation will transform the gravitational field in such a way that the situation around P looks exactly the same as around your original center.

 

[kimbyd]

This is not really the case though. Gauss law is local and valid for any finite volume and you typically derive it for just a finite volume without any reference to behavior at infinity. Instead, the problem is one of the behavior of the solution at the boundary. While the matter distribution itself can be assumed isotropic and homogeneous (ie, constant density), this is not compatible with boundary conditions that are. In other words, the solution cannot be isotropic and homogeneous.

There is a way around this though. If you consider just the case of a universe with spatially constant matter density and a “center” where the gravitational field is zero and around which you have isotropic boundary conditions, then you will end up in the same situation as you described in the finite case. If you make a transformation to the accelerated frame of any other particle P, the inertial force resulting from the transformation will transform the gravitational field in such a way that the situation around P looks exactly the same as around your original center.

Yes, I did oversimplify. I believe the actual limitation is that the potential vanishes at infinity. Which is the case for any finite distribution of matter, and I believe even remains the case for some infinite distributions of matter (e.g., an infinite rod, which is a common shape to use for electricty and magnetism calculations, as it's used to represent a wire).

A more in-depth discussion is here:
https://web.mit.edu/8.286/www/lecn18/ln03-euf18.pdf

They go into a lot of detail about how dealing with infinity mucks things up in the Newtonian case. And they also point out a historical fact I did not know: Previous to General Relativity, people genuinely thought (and Newton indeed argued) the OP here was correct: that a uniform, non-expanding universe would experience no net gravitational forces and simply stay static.

Edit: I think you can consider cases with the potential not vanishing at infinity as well, but it has to be finite. Here it is not.

 

[kimbyd]

To elaborate a little more, I believe you can ignore the behavior at infinity if, and only if, you can precisely specify the potential across a finite surface surrounding the system you're describing. If you can do that, you're golden. I don't think you can do that in this case because you've got an infinite amount of matter sitting outside any boundary you draw that you don't know how to deal with.

 

[Orodruin]

I believe the actual limitation is that the potential vanishes at infinity.

Again, this is not the case. Gauss law holds, it is a mathematical fact. The issue is that the solution to Poisson’s equation with a constant source term is incompatible with the assumption of zero potential at infinity. This is an issue for Poisson’s equation, not for Gauss law itself.

To elaborate a little more, I believe you can ignore the behavior at infinity if, and only if, you can precisely specify the potential across a finite surface surrounding the system you're describing. If you can do that, you're golden. I don't think you can do that in this case because you've got an infinite amount of matter sitting outside any boundary you draw that you don't know how to deal with.

That is true also in the case of a line charge or surface charge. Yet in both of those cases we know the solutions (although we have to make certain assumptions about the potential’s behavior at infinity). In both those cases we can also impose the full symmetry of the system on the boundary conditions, something we cannot do in the constant density case.

If you can specify the behavior of the potential on a finite surface you will always be able to find a solution as this just sets up a regular Poisson’s equation problem in a compact region.

The thing is, however, that you can do this in any case where you have specified enough for your problem to have a unique solution. In the case of constant density, specifying boundary conditions at infinity will also uniquely determine the field. The issue here is that those boundary conditions cannot be made isotropic and homogeneous. As I mentioned already in the previous post, there are ways around this.

 

[Orodruin]

A more in-depth discussion is here:
https://web.mit.edu/8.286/www/lecn18/ln03-euf18.pdf

They go into a lot of detail about how dealing with infinity mucks things up in the Newtonian case.

Regarding this: I do generally agree with the results. However, it also omits a few key details in the dealing with infinity. Whether this is done to spare the student or just overlooked I cannot say. It is perfectly possible to deal with the infinite situation as such, but it needs to be made sure that the imposed boundary conditions at infinity are compatible with Poisson’s equation. The requirement imposed on the shape used for the matter distribution of having zero quadrupole moment effectively translates to an assumption on the behavior of the potential at infinity and its symmetry.

 

[DrStupid]

Basically the integrals diverge.

The integral of the 1/r potential diverges. It would require renormation to get a useful result.

The integral of the 1/r² force does not diverge but there is a constant remaining that cannot be determined. As mentioned above the result depends on the arbitrary choice of the frame of reference.

But I am talking about tidal accelerations which scale with 1/r³. That means they behave like short-range-interactions. The integral shouldn't diverge.

 

[stevendaryl]

Again, this is not the case. Gauss law holds, it is a mathematical fact.

What are you saying is a mathematical fact? Gauss’ law for gravity is mathematically equivalent to $\nabla \cdot \vec{g}= 4\pi G\rho$ where $\vec{g}$ is the gravitational acceleration and $\rho$ is the mass density. But the issue is whether the differential form of Newton’s law of gravity is equivalent to his inverse square law, in the case of an infinite mass distribution that doesn’t go to zero at infinity.

Strictly speaking, the inverse square law doesn’t give a consistent answer, because the integral over all space depends on how you do the integral. Or summation, if you consider point masses instead of a continuous distribution.

 

[DrStupid]

Strictly speaking, the inverse square law doesn’t give a consistent answer, because the integral over all space depends on how you do the integral.

The integral of the inverse square law results in the gravitational acceleration of a single test particle in the infinite mass distribution, which is indeed not well defined. But in order to answer the initial question we just need the relative acceleration between each two test particles (aka the tidal acceleration) or the gradient of the gravitational acceleration. The unkown constant that is included in individual accelerations is not an issue in this case.

 

[Orodruin]

What are you saying is a mathematical fact? Gauss’ law for gravity is mathematically equivalent to $\nabla \cdot \vec{g}= 4\pi G\rho$ where $\vec{g}$ is the gravitational acceleration and $\rho$ is the mass density. But the issue is whether the differential form of Newton’s law of gravity is equivalent to his inverse square law, in the case of an infinite mass distribution that doesn’t go to zero at infinity.

Strictly speaking, the inverse square law doesn’t give a consistent answer, because the integral over all space depends on how you do the integral. Or summation, if you consider point masses instead of a continuous distribution.

The problem with the inverse square law is one of the assumed boundary conditions being that the field is zero at infinity. Solving the differential equation correctly, the field should diverge at infinity and if you do solve the differential equation (preferably of the potential) it indeed does so. The inverse square law only appears as the Green's function of the problem with homogeneous boundary conditions, which is incompatible with the solution for a constant mass distribution (except the trivial one) and therefore cannot be used consistently for that purpose. The solution is therefore to solve the differential equation with appropriate boundary conditions.

The differential equation is Poisson's equation with a constant source term
$$
-\nabla^2 \phi = 4\pi G \rho.
$$
Assuming spherical symmetry around the origin (this imposes a particular boundary condition), we find that indeed
$$
\phi(r) = \frac{2\pi G \rho r^2}{3}
$$
and hence
$$
\vec g = - \frac{4\pi G \rho \vec x}{3},
$$
which is compatible with the only possible spherically symmetric (i.e., isotropic around the origin) boundary condition that is compatible with the differential equation.

Now, the interesting thing occurs when you change coordinates to the accelerating frame of a particle located a distance $\vec d$ away. We find the new coordinates $\vec y = \vec x - \vec d$ and the gravitational field
$$
\vec g' = - \frac{4\pi G \rho (\vec y + \vec d)}{3} - \vec g(\vec d) = -\frac{4\pi G\rho \vec y}{3}
$$
so the situation is indeed spherically symmetric about $\vec y = 0$ as well as about $\vec x = 0$ if we pick the correct coordinate system. This is equivalent to taking the spherical solution with a finite radius $R$ and letting $R \to \infty$.

 

[kimbyd]

This is not really the case though. Gauss law is local and valid for any finite volume and you typically derive it for just a finite volume without any reference to behavior at infinity. Instead, the problem is one of the behavior of the solution at the boundary. While the matter distribution itself can be assumed isotropic and homogeneous (ie, constant density), this is not compatible with boundary conditions that are. In other words, the solution cannot be isotropic and homogeneous.

There is a way around this though. If you consider just the case of a universe with spatially constant matter density and a “center” where the gravitational field is zero and around which you have isotropic boundary conditions, then you will end up in the same situation as you described in the finite case. If you make a transformation to the accelerated frame of any other particle P, the inertial force resulting from the transformation will transform the gravitational field in such a way that the situation around P looks exactly the same as around your original center.

This is pretty much exactly what I've been saying. So I don't know why you're quibbling here.

As for Gauss's Law, my point is that actually using the law to determine field strength requires an assumption about the symmetry of the field. And you can't know the symmetry of the field in a homogeneous, isotropic distribution from first principles using Newtonian mechanics alone. I think you can get there if you assume that a homogeneous, isotropic distribution must stay homogeneous and isotropic (pretty sure this is equivalent to the assumption of spherical symmetry). But I don't think you can get there without that assumption tacked on.

As for the comment that the matter distribution being isotropic and homogeneous is not compatible with boundary conditions that are, I just don't think that's a useful statement to make. By using a spherical boundary around the origin, you produce a universe which is locally homogeneous and isotropic for every co-moving observer within the boundary, as derived from their measurements of the accelerations of the matter around them. Thus we can interpret that boundary as being a fictitious structure created to make the math work out, rather than something that actually breaks the symmetry in any real sense.

 

[Orodruin]

This is pretty much exactly what I've been saying. So I don't know why you're quibbling here.

The point is: there is no need to ever refer to an ever increasing sphere. You can solve the Poisson equation right off the bat if you assume appropriate boundary conditions. I do not see why this seems to be controversial to some.

locally homogeneous and isotropic

Locally homogeneous? The entire point of honogeneity is being able to translate between points, ie, non-locality. You cannot just translate between points and retain isotropy. You need to also include the acceleration.

As a solution to and property of the Poisson equation, it is very important. I sm talking about the mathematical structure of the problem. This should not be controversial.

And you can't know the symmetry of the field in a homogeneous, isotropic distribution from first principles using Newtonian mechanics alone. I think you can get there if you assume that a homogeneous, isotropic distribution must stay homogeneous and isotropic (pretty sure this is equivalent to the assumption of spherical symmetry). But I don't think you can get there without that assumption tacked on.

Which, again, is equivalent to the assumption of the appropriate boundary conditions. You could technically choose to also introduce other multipoles. A dipole can be transformed away by acceleration, but quadrupoles and higher cannot.

I am also not sure what you mean by ”Newtonian mechanics”. The inverse square law? To me, Newtonisn gravity is the Poisson equation. Newton may not have formulated it this way, but the inverse square law is the inverse square law because that is how the gradient of the Green’s function of the Laplace operator behaves.

 

[kimbyd]

Locally homogeneous? The entire point of honogeneity is being able to translate between points, ie, non-locality. You cannot just translate between points and retain isotropy. You need to also include the acceleration.

Yes, obviously I'm talking about considering only co-moving observers.