Is the Force on a Particle in a Homogeneous Static Universe Always Zero?

[Higgsono]

Consider an infinite homogeneous static universe with a constant mass density $$\rho$$. If we were to calculate the force on a test particle located at a certain point accoring to Newtons law of gravity. It would be logical to conclude from a symmetry argument that the force on the particle should be zero. But is this true? For as we know, the force on that particle depends on how we are adding up the contributions of mass in the universe. So one way to calculate the force will give a 0 net force on the particle, and another way of doing it will give a net force in a certain direction. In fact, depending on how we do the sum, we can come to the conclusion that the particle can be subject to any force, even an infinite force.

So how is this problem resolved? From a mathematical standpoint it should have a solution.

 

[Buzz Bloom]

So one way to calculate the force will give a 0 net force on the particle, and another way of doing it will give a net force in a certain direction.

Hi Higgsorno:

There is a third option. The effect of the mass is on space (rather than on a test particle) causing space to expand or contract.

Regards,
Buzz

 

[Higgsono]

Hi Higgsorno:

There is a third option. The effect of the mass is on space (rather than on a test particle) causing space to expand or contract.

Regards,
Buzz

Yes, but if we look at the problem from a pure mathematical perspective, assuming Newtons law of gravity. What is the answer then?

 

[Buzz Bloom]

assuming Newtons law of gravity

Hi Higgsono:

What has been learned in the past century is that Newton's laws do no apply to cosmology. General Relativity has superseded Newton's laws. However, if you want to ignore that, Newton's laws would leave your hypothetical universe static.

Regards,
Buzz

 

[Higgsono]

Hi Higgsono:

What has been learned in the past century is that Newton's laws do no apply to cosmology. General Relativity has superseded Newton's laws. However, if you want to ignore that, Newton's laws would leave your hypothetical universe static.

Regards,
Buzz

But why do I get different answers for the force on the particle depending on how I add up the mass in this infinite universe?

 

[Bandersnatch]

It would be logical to conclude from a symmetry argument that the force on the particle should be zero.

Should it?
Can you apply shell theorem to the scenario? Choose an arbitrary point at distance R from the point mass and calculate the force of gravity acting along the vector R.

What has been learned in the past century is that Newton's laws do no apply to cosmology. General Relativity has superseded Newton's laws. However, if you want to ignore that, Newton's laws would leave your hypothetical universe static.

You can get Friedmann equations from Newton's laws, though. I'm willing to bet that every single cosmology textbook does it before moving on to more advanced stuff.

 

[Buzz Bloom]

You can get Friedmann equations from Newton's laws, though. I'm willing to bet that every single cosmology textbook does it before moving on to more advanced stuff.

Hi Bandersnatch:

That is quite fascinating. In An Introduction to the Science of Cosmology, Raine & Thomas, the derivation from GR equations is presented, (pg 74, 79) but there is no derivation from Newton's Equations. I would much appreciate seeing such a derivation. Can you post a citation including a page number?

Regards,
Buzz

 

[Buzz Bloom]

But why do I get different answers for the force on the particle depending on how I add up the mass in this infinite universe?

Hi Higgsono:

I am not sure what you meas by "how you add up the mass". Are you asking about different way to calculate the force using Newton's laws, or asking about the difference between Newton and GR? In both the same symmetry rules holds. With a uniform mass distribution in a symmetric space the force on a test particle is zero.

Regards,
Buzz

 

[Orodruin]

You can get Friedmann equations from Newton's laws, though. I'm willing to bet that every single cosmology textbook does it before moving on to more advanced stuff.

This is an instance of lying to children. It essentially works because given the dimensions of the quantities involved, it is the only form that the resulting differential equation can take. That you get the correct numerical constants in front is to a large extent a coincidence.

 

[Bandersnatch]

Can you post a citation including a page number?

Well, I would have lost that bet pretty fast, as after skimming through Raine's book I can't find any mention of it. There's a chance it's been relegated to one of the problem sets, but having learned from my past mistakes, I wouldn't bet on it.
You can find it e.g. in:
'Cosmology. The origin and evolution of cosmic structure' by Coles & Lucchin (p.24)
'An introduction to modern cosmology' by Liddle (p.17)
'Introduction to cosmology' by Roos (p.19)
In any case here's an example:
http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf

This is an instance of lying to children. It essentially works because given the dimensions of the quantities involved, it is the only form that the resulting differential equation can take. That you get the correct numerical constants in front is to a large extent a coincidence.

Thanks for the deeper insight.
But does it count as lies to children, if it's not meant as a substitute for the real thing? Whenever I see it in texts, it's always labelled with warnings about fudging involved, and accompanied by GR-based derivation.

 

[Higgsono]

Hi Higgsono:

I am not sure what you meas by "how you add up the mass". Are you asking about different way to calculate the force using Newton's laws, or asking about the difference between Newton and GR? In both the same symmetry rules holds. With a uniform mass distribution in a symmetric space the force on a test particle is zero.

Regards,
Buzz

No, the force on the test charge depends on where you choose the origin. If you choose the origin to be at the test cherge and adding the contribution to the force from spherical shells, then the net force is zero. But if you choose the origin to be at som other point, and add upp the mass the same way using spherical shells, you will get a net force on the particle which is not zero.

 

[newjerseyrunner]

I think the first thing to do would be to reduce the complexity of the problem to a one dimensional vector, with the origin being the point in question. Then determine the force applied by an infinite vector of homogeneous density, which I would expect would be convergent. If you get a finite value, then you can assume basic symmetries about the universe and say that the same force applied in all directions result in 0.

 

[Chalnoth]

Hi Higgsono:

What has been learned in the past century is that Newton's laws do no apply to cosmology. General Relativity has superseded Newton's laws. However, if you want to ignore that, Newton's laws would leave your hypothetical universe static.

That's not true. Newton's laws recreate the first Friedmann equation exactly for the case of a uniform, matter-only fluid. If you modify Newton's laws to account for the cosmological constant, then that is included correctly as well. Radiation isn't included correctly, however.

 

[Chalnoth]

One way you can do the Newtonian derivation, by the way, is to use Gauss's law. Since the universe is spherically-symmetric about every point, it's possible to simply reduce the problem to a single sphere: pick an origin and a radius, and you only need to consider the mass inside that sphere for calculating the acceleration due to gravity of the particles on the surface of the sphere. The result is the first Friedmann equation.

 

[PeterDonis]

Consider an infinite homogeneous static universe with a constant mass density

This is not a possible model according to GR. A homogeneous static universe must be finite.

 

[Higgsono]

ok, I'm not getting the answer I'm looking for.

This is not a possible model according to GR. A homogeneous static universe must be finite.

My concern was not with general relativity. It is a mathematical problem. And I have already solved the problem in two different ways, but have got different answers. Read my original post and my answer to another post, and you will know what I mean.

 

[Higgsono]

I think the first thing to do would be to reduce the complexity of the problem to a one dimensional vector, with the origin being the point in question. Then determine the force applied by an infinite vector of homogeneous density, which I would expect would be convergent. If you get a finite value, then you can assume basic symmetries about the universe and say that the same force applied in all directions result in 0.

Yes, then you will get a unique answer i assume. But if you add up the mass using spherical shells in 3 dimensions you will not get a unique answer, because the answer depends on the choice of origin. But I don't understand, because the problem should have a solution I think, in mathematical terms. Because the problem is well defined.

 

[Orodruin]

Because the problem is well defined.

No it isn't. Your problem is essentially to solve Poisson's equation with a constant inhomogeneity in an infinite domain. In order to do this, you will have to specify the behaviour of the gravitational potential at infinity. This essentially boils down to selecting one point to be "the center of the Universe".

 

[Chalnoth]

ok, I'm not getting the answer I'm looking for.

If you want to see the derivation, see here, for example:
http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf

The short version is that doing the calculation the way you did was incorrectly taking account of the symmetry of the system. When you use the symmetry of the system to make use of Gauss's Law, that result already takes into account the entire mass distribution of the universe. You don't then balance it by doing Gauss's law from a different point of origin and add the result.

When done correctly, the result you get is basically the same as the full General Relativity result, which is that there can be no static, homogeneous universe: either it is expanding or collapsing (or, if you have a cosmological constant, unstable).

 

[Higgsono]

If you want to see the derivation, see here, for example:
http://www.astronomy.ohio-state.edu/~dhw/A5682/notes4.pdf

The short version is that doing the calculation the way you did was incorrectly taking account of the symmetry of the system. When you use the symmetry of the system to make use of Gauss's Law, that result already takes into account the entire mass distribution of the universe. You don't then balance it by doing Gauss's law from a different point of origin and add the result.

When done correctly, the result you get is basically the same as the full General Relativity result, which is that there can be no static, homogeneous universe: either it is expanding or collapsing (or, if you have a cosmological constant, unstable).

Thanks!

 

[Chronos]

These issues are not entirely unlike those Einstein cited for preferring a finite universe. Our mathematical tools are highly effective at predicting physical interactions, but, become questionable when we attempt to apply them over unbounded parameter spaces.

 

[Khashishi]

Infinity isn't a valid number in mathematics. Even in the hyperreal number system, there are an infinite number of infinities, so you need to be more specific in how you define an infinite universe. For instance, you should probably definite a finite universe of size R and define a limit as R goes to infinity. If you define it this way, then it's clear that the Newtonian universe has a center of mass. Objects will be pulled toward the center of mass. It wouldn't be possible to have a static solution unless you included something analogous to the cosmological constant.

 

[l0st]

So how is this problem resolved? From a mathematical standpoint it should have a solution.

Well, did you try to calculate it? You can just pick any axis, and calculate force along that line. That should give you a simple 3-dim integral, which will either converge, or not, no?