Which bit of the shell theorem broke down and why?The assumption of homogeneity
We'll have a perfectly uniform distribution of Hydrogen gas spread out over space AND we'll also assume that space is infinite.[NOTE: It is not necessary or guaranteed that the universe is or was infinite, that doesn't matter, for this hypothetical situation we are having an infinite space]. For a star to form, some hydrogen has to clump together under gravity and form an overdense region of Hydgrogen.Which kind of departs from your premise of a perfectly uniform distribution. Yes, anomalies form early on and escalate into more dense regions. It is a stretch to suppose that they are spherically symmetrical anomolies.
We can choose to put an origin somewhere and draw a sphere of radius r around that. Now, consider a particle of Hydrogen on the boundary (surface) of that sphere. By the shell theorem, we can see that the Hydrogen particle is attracted to the origin by gravity.Sure, but it is alto attracted in all other directions equally since there is a similar sphere at which it is on the edge, in any direction. The particle doesn't accelerate since the net force is zero. Your application of the theorem is obviously invalid, and your logic stands up, so the theorem cannot apply to an unbounded uniform distribution of mass since it doesn't satisfy the definition of a spherical distribution.
Now, consider a particle of Hydrogen
The space is homogeneous for the purpose of this connundrumPick one.
The particle is the test mass being attracted to the homogeneous everything-else by a net force of zero. It doesn't even need a specific mass since it is the coordinate acceleration at its location that we're after, the gradient of the gravitational field there.Now, consider a particle of HydrogenThe space is homogeneous for the purpose of this connundrumPick one.
The space is homogeneous for the purpose of this connundrum.Thank you @Halc , that is the essence of it. This is a hypothetical situation and there is no need to tarnish it by suggesting the real world isn't as perfect as the model.
Yes, you could try and explain why a real world situation wouldn't be like the situation in the model.Well, at some point you have to explain how matter started to clump rather than remain perfectly homogeneous. For that, you have to go back to the earliest epochs (long before there were protons and such) when there was fantastic energy density, and then apply quantum fluctuations to the situation. Lacking a unified field theory, this is all still pretty speculative. One also has to go about explaining the annoying matter/antimatter imbalance.
We do need a 1/r2 relationship for the shell theorem to hold.that's right, and 1/r2 doesn't hold up for extreme cases (like near neutron stars and such). But the shell theorem is totally inapplicable here, and by the simplest symmetry, there should be zero net forces on anything except due to those local variances which come only out of quantum theory, so GR isn't going to be a huge help over just doing this the Newtonian way.
We only need to consider the initial action on a test particle given the assumed uniform initial distribution of mass across an infinite space.This suggests the former of the two.
Even if you want to assign every particle some random initial velocity, there's no force on any particle at initial time, so it's velocity does not change and the density remains uniform if particles do move randomly.Except in an expanding metric, any initial velocity drops off almost instantly, so in pretty much no time, nothing has any velocity except due to local accelerations from local interactions. Kinetic energy is not conserved under non-inertial metrics, which includes expanding metrics.
So the picture at time = initial time + δt is identical to the picture at time = initial time).Heck no. The density went way down over that time, but sans local fluctuations, the place is still nice and uniform energy density and there should be no net 'forces' acting on anything.
This is a hypothetical situation and there is no need to tarnish it by suggesting the real world isn't as perfect as the model.Well, it is those imperfections that eventually brought about galaxies and such. Without them, the universe would be a permanent gray paste. That's why I asked if this topic was about the actual homogeneous case, or if it was about how the clumping starts.
We can choose to put an origin somewhere and draw a sphere of radius r around that. Now, consider a particle of Hydrogen on the boundary (surface) of that sphere. By the shell theorem, we can see that the Hydrogen particle is attracted to the origin by gravity. We also see that we can ignore the attraction to anything outside that sphere of radius r.Back to the OP argument, since one can 'frame' my rebuttal in a way that refutes my rebuttal.
We'll have a perfectly uniform distribution of Hydrogen gas spread out over space AND we'll also assume that space is infinite. [NOTE: It is not necessary or guaranteed that the universe is or was infinite, that doesn't matter, for this hypothetical situation we are having an infinite space]. For a star to form, some hydrogen has to clump together under gravity and form an overdense region of Hydgrogen.Eppur si muove (Galileo)
I'm not sure if this topic is about the applicability of the shell theorem to an infinite homogeneous mass distribution (answer: not applicable), or if it is about how matter started to clump (need a different model to guide us) early in the game.
...the theorem cannot apply to an unbounded uniform distribution of mass since it doesn't satisfy the definition of a spherical distribution.For an infinite space with uniform density everything is spherically symmetric about every point in that space.
Obviously this is too longI tend to do that as well, but I'll try to break up my reply, which is slow in coming since I've been kinda busy.
Suppose we have an infinite space with an initially uniform density of material, ρ, everywhere. Consider a test particle located somewhere in that space, at position x . What force would that test particle experience?First of all, the sorts of forces you describe cannot be 'experienced', which is why an accelerometer reads zero despite its coordinate acceleration in orbit, and it is very much coordinate acceleration you're after here, not proper, which happens due to something usually contacting you on one side and deforming you. So the acceleration of the particle seems coordinate system dependent, thus the question of if it is really accelerating towards that arbitrary CoM you defined, or the stuff at that origin is accelerating towards the particle, is an abstract question, not a physical one.
If the Hydrogen was at rest initially (which as you correctly point out for an expanding universe is actually a very reasonable assumption as the scale factor increases) then it just sits there and stays at rest for ever more.Well, it's a hot universe at first under great pressure, meaning particles are not stationary, and they do experience force as they bump into each other. It should all cancel out, but not perfectly, and the anomalies, especially early on in such close quarters, introduce variations in local mass density, which, via gravity, tends to amplify the variations as gravity wells begin to form. In short, the homogeneous state is in sort of an unstable equilibrium.
However, applying the shell theorem (as it was done in the OP) can get a different answer. The Hydrogen can be shown to have a net force in the direction of the origin.That's a coordinate effect, as stated above. Coordinate acceleration is coordinate system dependent, and can be zero in some, such as the coordinate system putting the origin at the particle in question. Yes, stuff will be attracted to each other. The expansion (if there is any) should slow measurably. If all is stationary, I think collapse occurs, even in Newtonian physics.
Sure, but it is alto attracted in all other directions equally since there is a similar sphere at which it is on the edge, in any direction. The particle doesn't accelerate since the net force is zero.More and more, I think my reply there doesn't hold water. I'm willing to say it is spherically symmetric. The inability to find a center of mass, or to identify a gravitational gradient (not a frame dependent thing) seems a better way to disqualify the applicability of the shell theorem in this Newtonian analysis. The particle doesn't accelerate since there's no downhill (a region of lower potential) available to it.
You seem to agree that the shell theorem would show a non-zero net force exists in some direction. You also seem to agree that direction is arbitrary. The last sentence doesn't need to follow though. You don't apply the Shell theorem with all possible origins and then sum all predicted net forces to obtain the actual real and very final net force.Selecting an origin is a coordinate thing. You can't select more than one at a time, so adding up the forces from different frames doesn't work.
Worse that that, even it's magnitude is arbitrary (just move the origin further away but in the same direction, the inner sphere grows ~ r3 in volume and hence in mass, while the gravitational force from it falls off only as ~1/r2, making the total force vary ~ r = the distance from our test particle to the origin).Just so, which is why distant star are receding (and have greater acceleration) than the nearby ones.
You sometimes make arguments that the space isn't suitably spherically symmetricI'm backing away from that.
All that exists in the space is the stuff we have stipulated will have a constant density, ρ, everywhere. There are no points in space where a lack of spherical symmetry is exhibited.Backing away because that's a pretty good counter to my comment.
Example, trajectories of stars in a galaxy can be estimated by considering only that mass which is in a spherical region interior to that star etc.This is incorrect since a galaxy isn't spherically symmetric. With a disk, a particle is attracted to rings of material outside its radius as well as the material inside. So it all needs to be considered. Kudos to the guys that work out where all the mass needs to be to get the velocity curves we see. It's a lot of calculus to do.
Anyway, just to be clear, what you're suggesting is the following:I cannot say I suggest this, at least not anymore. Like I said, there are reason the shell theorem doesn't apply, but this logic seems faulty.
A particle in the interior of a shell experiences no force due to gravity from the material of that shell. That holds for shells of some finite thickness (assuming the usual things like spherical symmetry). However, when that shell is so thick that it's of unbounded extent, the result may not hold, the particle might experience a force.
Much the same question was asked [on stack exchange]Cheat! The fun was in working it out ourselves.
n the negative side, I don't especially agree with the highest ranked answer. Their response pivots around the inability to identify a unique gravitational potential function, Φ, in such a space.That sounds really strong to me. Gravitational potential is not frame dependent, and thus seems to be flat everywhere in this universe, hence no vector (direction) to any physical force.
The other answers vary in quality and relevance. There is one about some history and how major figures like Newton may also have noticed this problem that the shell theorem produces in this situation.That would be quite relevant to this discussion, especially given the way the question was framed.
there is no reason why a well defined gravitational field g can't exist throughout space
There is no reason why fairies should not exist,..... (etc.)....
Anyway, just to be clear, what you're suggesting is the following:
A particle in the interior of a shell experiences no force due to gravity from the material of that shell. That holds for shells of some finite thickness (assuming the usual things like spherical symmetry). However, when that shell is so thick that it's of unbounded extent, the result may not hold.
That (how Newton considered the problem) would be quite relevant to this discussion, especially given the way the question was framed.
Has my post been deleted, and Paul's?I don't know.
If an object within a shell has no gravitational effect on any object within it, surely that is from the point of view of the outside Observer? One would assume that the shell would indeed have gravitational attraction, if the she'll where 100, 000km thick and 10 million in diameter?There may be some minor errors in the English in that question. I think I know what you're asking.
I don't recall Newton ever saying that gravity was the gradient of a potential. It is just the force that a unit mass would experience when it is put somewhere.No. It is (and always has been) an attractive force between masses. The vector has magnitude GMm/r2 and direction toward the barycenter of the masses. In the absence of a barycenter it has no direction.
I don't recall Newton ever saying that gravity was the gradient of a potential.Being mostly unfamiliar with what Newton actually says, I cannot contest that. Can you think of a case (with finite mass) where it isn't so? I mean, with infinite mass, potential is pretty much undefined at all. There are perhaps relative differences in potential, but no meaningful displacement from 'zero'.
It is just the force that a unit mass would experience when it is put somewhere.Force per unit mass is acceleration, so it would be an acceleration field.
Anyway, just to be clear, what you're suggesting is the following:Didn't see this before, but it makes sense. Using the bad rubber sheet analogy, an infinite sheet covered with a unform layer of marbles is in unstable equilibrium. Take one marble away (the equivalent of our hollow infinite-thick shell), then all the marbles near the vacancy move away from it, being 'repelled' by gravity as it were. This is seen in cosmology: there is the Dipole repeller and the lesser know Cold Spot repeller ,both regios of low mass density from which all nearby objects accelerate away. This acceleration from a hole presumably counters the acceleration towards the CoM of the arbitrary sphere you defined in the OP.
A particle in the interior of a shell experiences no force due to gravity from the material of that shell. That holds for shells of some finite thickness (assuming the usual things like spherical symmetry). However, when that shell is so thick that it's of unbounded extent, the result may not hold.
If you insert an arbitrary point mass in an infinite homogeneous medium, the barycenter is the point mass itself!The Barycentre is the position R so that the product MR = the sum ∑miri.
I see, so it is gravitational potential within the shell, the "net gravitational force?? The spherically symmetrical shell being uniform in density.Force and potential are not the same thing, having different units. Inside the shell, the scalar potential is the same everywhere, but lower than it is outside the shell. The force vector inside the shell is zero, and greater than zero (pointing towards the shell) outside of it.
The [force] vector has magnitude GMm/r2 and direction toward the barycenter of the masses. In the absence of a barycenter it has no direction.This only works for simple two-body systems. I can easily set up some orbiting masses where the acceleration of the test particle is away from the barycenter of the system, or in any other direction I like. The moon for instance (in say an otherwise empty Sun/Earth/moon system) rarely accelerates directly towards either Earth or Sun, and on occasion it accelerates directly away from Earth. If the moon was closer to Earth (as it was long ago), it would be capable of at some point accelerating away from the Barycenter of the system.
Force per unit mass is acceleration, so it would be an acceleration field.Yes.... people do blur the distinction between force and acceleration.
Can you think of a case (with finite mass) where it (gravity being the gradient of a potential function) isn't so?No, not without spreading the mass out so thinly that there is none of it in any finite volume.
This is seen in cosmology: there is the Dipole repeller and the lesser know Cold Spot repeller...... (etc.)That bit will be on my mind for quite a while. Yes, there are some stars (and things) in a region that is mostly like a hollow empty shell and they aren't doing as the hollow shell theorem would suggest, they seemingly are experiencing an acceleration. Thanks for mentioning it, it's a nice concrete example where the criteria of the hollow shell theorem are almost met but the usual result is not.
Another thought ... does this give rise to ponderances of gravitational control and even anti gravittational understanding?Not usually while I'm awake. You do already have some control over gravity, move two things apart and you have reduced the force of gravity between them.
You can keep doing this (just adding up individual potentials), so in the general case, you can always identify a suitable potential function for any finite number of masses.I sort of did something like this when I wondered about gravitational potential being negative. How negative? I liked to measure potential in km. It can also be expressed as a speed.
The absurdity of trying to express this is one of my pieces of evidence against models with flowing time.Maybe a clock infintely far from all mass would progress infinitely faster than all real clocks in the universe. That doesn't need to matter and it doesn't demonstrate the absurdity of time as something that flows. The underlying co-ordinate time of the real universe and its real metric is probably not the thing that such a hypothetical clock in this hypothetical place is presumed to show.