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  4. Which bit of the Shell theorem is not working?
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Which bit of the Shell theorem is not working?

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Offline Eternal Student (OP)

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Re: Which bit of the Shell theorem is not working?
« Reply #20 on: 16/10/2023 07:30:37 »
Hi again,
Quote from: Halc on 14/10/2023 22:22:28
That   (how Newton considered the problem) would be quite relevant to this discussion, especially given the way the question was framed.

   If you are interested here's some details:

    The University of Pittsburgh seem to have taken a copy of an article writen by John Norton.   I was going to assume that website is safe enough.   This does seem to have all the people that were mentioned by the Historian on the Stack Exchange site.
https://sites.pitt.edu/~jdnorton/papers/cosmological-woes-HGR4.pdf

    I've only spent 20 minutes glancing over it so far - but it has been interesting reading.

1.   It does relate to the issues discussed in this forum post.   The Newtonian cosmological model under consideration is precisely that of an infinite space with a uniform mass distribution.

2.    The first person credited with challenging Newton's laws on this issue  is Seeliger   circa. 1895.   Issac Newton was dead by then, of course.   It seems that others had noticed the problem before then and had directly challenged Newton but the response from Newton is somewhat dissapointing (see later).

3.    There are some very easy to understand methods of showing that Newton's statements about gravity will not be useable in this sort of space.    Their page numbers are 272 ~ 274.    The author divides a sphere up in such a nice way that they can reduce everything to just an infinite sum       A - A + A - A + A - .... (an infinite alternating sum of +/- a finite number A).    As I'm sure you know, such a sum is non convergent, you can re-arrange the order in which you add and subtract the terms to end up with anything     (infinity, non-zero but finite, zero ... whatever you want).    Seeliger was extremely thorough and apparently spent many pages challenging Newton's law of gravity in this sort of universe   For example, he also demonstrated that Newton's formulation of gravity would put undefined or infinite tidal forces on every part of space.    It was only this level of rigour that was thought to have finally made it impossible for the pro-Newton camp to continue ignoring the problem any longer.

4.   The most immediate obstacle to Newton's laws for gravity is the very notion that the force of gravity on an object could or should be the resultant sum of forces from all masses in the universe.    In an infinite space with uniform mass distribution, that sum does not converge.   

5.   So what was Newton's own response when these sorts of problems were brought to his attenntion while he was alive?    It seems that for the most part, he just denied that the problem(s) existed.    There are a few letters between  Newton and Bentley (a bishop who was interested in Cosmology and how consistent with Christianity it might be) and in those letters there seems to be some recognition of problems handling infinities.   By and large it was considered as a mathematical oddity and not something that should really need to influence the physics (or Christianity for that matter).   Newton was a sufficiently recognised force that when he denied the problem existed, the problem just did not exist.

Best Wishes.
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Offline Eternal Student (OP)

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Re: Which bit of the Shell theorem is not working?
« Reply #21 on: 16/10/2023 07:58:30 »
Hi.

Quote from: Petrochemicals on 16/10/2023 07:06:41
Has my post been deleted, and Paul's?
   I don't know.

Quote from: Petrochemicals on 16/10/2023 07:06:41
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.
Wikipedia gives a short definition or explanation of the shell theorem:

1.         A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.

2.   If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

 - - - - - - - - - -

   So the shell is made of some material and that material does gravitate - it attracts other stuff to it.    When you're outside that shell it behaves exactly as you would expect .   You would have a force of attraction identical to the force of attraction you would have if you imagine all of the shell was just squahed down into one really small ball and placed at the centre position of that shell.
    It's only when you're inside the hollow shell where it may seem you get an unexpected result.   You are still attracted to each part of that shell.   If you happen to be at the centre of the shell then you are equally attracted in all directions and there is no net force, as you may expect.    However, even when you move away from that centre and toward one piece of the shell,  you still feel no net force of attraction in any direction.   Provided you stay inside the shell, you find no net force acts on you.
    There are many articles and YT videos that provide explanation in varying levels of detail and sophistication.   The simplest explanation is merely suggestive or hand-wavy:     When you move toward one piece of the shell,   the distance between you and that piece decreases,   so by Newton's law the force of attraction to that piece (the bit infront of you) should increase.   That does happen.   However,   since you have moved off-centre, there is now a bit more of the shell behind you rather than infront of you.   That tends to increase the attraction to the direction behind you.   It just so happens that both things happen in the right proportion and the net (or total) force stays at 0.
      The acceleration can be determined in various ways,  you could use any inertial reference frame you want   OR   you can just identify that acceleration more directly by giving the person insde the shell an accellerometer.    To get the right result with an accelerometer in the real world you'd need a really, really big shell and a person who is actually floating in outer space so that they weren't experincing any forces other than the gravitational attraction of the shell.

Best Wishes.
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Re: Which bit of the Shell theorem is not working?
« Reply #22 on: 16/10/2023 08:53:24 »
I see, so it is gravitational potential within the shell, the "net gravitational force?? The spherically symmetrical shell being uniform in density.
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Offline alancalverd

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Re: Which bit of the Shell theorem is not working?
« Reply #23 on: 16/10/2023 09:02:09 »
Quote from: Eternal Student on 16/10/2023 01:12:41
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.

If you insert an arbitrary point mass in an infinite homogeneous medium, the barycenter is the point mass itself! 
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Offline Halc

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Re: Which bit of the Shell theorem is not working?
« Reply #24 on: 16/10/2023 15:41:39 »
Quote from: Eternal Student on 16/10/2023 01:12:41
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'.

Quote
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.

Quote from: Eternal Student on 14/10/2023 00:44:17
   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.
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.
The acceleration from the hole is real, seen in action, and is a good illustration of one's inability to treat the rest of uniform mass space using the shell theorem.
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Offline Eternal Student (OP)

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Re: Which bit of the Shell theorem is not working?
« Reply #25 on: 16/10/2023 16:11:54 »
Hi.

yes @alancalverd I can see what you're saying in the first part.
The second part is just optimism or blind faith where the mathematics fails.
Quote from: alancalverd on 16/10/2023 09:02:09
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.

The R.H.S.  is a non-convergent sum.   The L.H.S. is an undefined product since M = the sum of masses = ∑mi is also a non-convergent sum.

Best Wishes.
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Re: Which bit of the Shell theorem is not working?
« Reply #26 on: 16/10/2023 16:14:15 »
Another thought, if by a homogenous shell gravity is nil and void within it, does this give rise to ponderances of gravitational control and even anti gravittational understanding? If by some thought one could equate the exterior of the earth to  a sphere interior through some mechanism or device should I float at my level?
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Re: Which bit of the Shell theorem is not working?
« Reply #27 on: 16/10/2023 16:14:58 »
Quote from: Petrochemicals on 16/10/2023 08:53:24
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.


Quote from: alancalverd on 16/10/2023 09:02:09
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.
« Last Edit: 16/10/2023 16:19:39 by Halc »
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Re: Which bit of the Shell theorem is not working?
« Reply #28 on: 16/10/2023 18:27:17 »
Depends on how you define the system! There is always an earthwards vector component and a sunwards component acting on the moon. I suspect the sum points more or less (depending on the position of the other planets) towards the barycenter of the 3 bodies.

In a simple 2-body orbiting system the acceleration is always towards the barycenter, otherwise you wouldn't have an orbit.
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Offline Eternal Student (OP)

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Re: Which bit of the Shell theorem is not working?
« Reply #29 on: 16/10/2023 20:47:09 »
Hi.

Quote from: Halc on 16/10/2023 15:41:39
Force per unit mass is acceleration, so it would be an acceleration field.
    Yes....   people do blur the distinction between force and acceleration.
The template or "poster boy" for a force field is the Electric field.    We understand that   E(at position x)  is the force per unit charge at position x and call that the field strength at x.

We take a similar approach with gravity.    "The charge" is now "the mass", so it's a bit unfortunate that the force per unit charge just turns out to be the force per unit mass which is an acceleration.   I suppose it's fortunate rather than unfortunate - you choose.

   As I'm sure you know, the entire concept of gravity as a force and therefore as something which exists as a field of force throughout space is a Newtonian (not GR) concept.

Quote from: Halc on 16/10/2023 15:41:39
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.
     It's the spatial spread of the mass that is most of the problem in our example (an ininfite homegenous space).    You could imagine that you have a finite total mass of  100 Kg in the entire universe and just spread that out over an infinite space.  That gives you a density of  ?   0 ?   Not sensible or sensibly different from a universe which just had 0 mass in it, unless we put all the 100 Kg  in some finite radius,  say 100 Km of the origin and have all the rest of space empty. There are other distributions we could have (like density falling of as 1/r3) but no sensible way of having a uniform density throughout the infinite space with a total mass that would still be finite.

    I think it's actually easier to see why you always would obtain a suitable potential function (in sensible situations) and then just look at what goes wrong in unusual situations.   I'm also at risk of repeating much of what has been said before but just in different ways.  I don't think anyone wants to read pages of repetitious stuff and I've already had several goes at writing it and I can't do it in less than a page.   So we'll only outline the thing.
     You can find a potential function for just one mass    (it's    -GM/r,  see any textbook).     So you can find a suitable potential for two masses just by adding up two individual potentials.    ∇(A+B) = ∇A + ∇B    so the function which is the sum of the individual potentials will be suitable as the potential for the gravity arising from the two masses.   (This is simply because Newton did state the gravitational force on our test object is equal to the vector sum of the force caused by each individual mass,   the sum on the RHS  ∇A + ∇B  is precisely that vector sum and so the potential function A+B is precisely what you want, it's a function whose gradient is precisely that force).
   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.    The fun stops when you have an infinite collection of masses.   You desperately want   ∇(infinite sum of functions) =   infinite sum of  ∇(each function) but you just don't have it.   Summing an infinite collection of terms is not something you can do but you can do something good enough and re-write the expression you wanted where you understand that each infinite sum was just shorthand for the limit of an infinite sum.    It turns out that the expression you want to hold, will hold provided those limits do actually exist. 
   So, you will have the relationship you want when the mass distribution through space falls off fast enough as you move away from the test particle.   So in those situations, all is good, you would always be able to find a suitable potential function.
    So the only examples where there may NOT be a potential function would have to be when the density does not fall fast enough.   In those situations the total mass in the universe ~   ∫over space    ρ  d(Volume)   ---> ∞.
   So we will find no examples for you involving only finite mass in the universe  (or indeed only a finite number of masses), sorry.   Well, not just using Newtonian mechanics anyway.   Using GR the entire concept of gravitational potential energy is ill defined anyway but still of some interest to think about.

Best Wishes.
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Re: Which bit of the Shell theorem is not working?
« Reply #30 on: 17/10/2023 00:18:02 »
Hi.

Quote from: Halc on 16/10/2023 15:41:39
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.

- - - - - - - - -

Quote from: Petrochemicals on 16/10/2023 16:14:15
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.
   Indeed you can use the hollow shell theorem and planet earth in some fashion similar to that you asked about.   Drill a tunnel in toward the centre of the earth.   The further in you go, the lower the force of gravity will be on you.   If you could also scoop the centre bit of the earth out then you would have a nice little region where you can float about and  experience Zero-G.

Best Wishes.

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Re: Which bit of the Shell theorem is not working?
« Reply #31 on: 18/10/2023 01:44:15 »
Quote from: Eternal Student on 16/10/2023 20:47:09
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.
Due to Earth, 6400 km at 1g, Maybe 100k km due to sun. Maybe 15 million km due to the galaxy. OK, so we expend enough energy to climb 15 billion meters at 1g, putting us in intergalactic space. Are we at zero yet?  Hell no. Now we start playing games.
Define a shell around us, a million light years thick, starting nearby and working outward. Each shell has mass in it contributing to the depth of our gravitational potential.
A shell nearby contributes X (km). A shell of similar thickness, twice as far away (so 4x the area) masses 4x as much, but being twice as far away, contributes only 2x the potential of the nearby shell. In other words, the amount each shell contributes to our gravitational depth goes up in proportion to the distance, so more distant galaxies contribute more than do the nearby ones because there's so many more of them. So the depth here is -X -2X -3X -4X ... which certainly doesn't converge on any number. Even if they were all just -1X it would still put us at unbounded depth.

The exercise is apparently invalid, but it is also utterly meaningless to suggest express our potential in absolute terms where zero is potential at the place that is infinitely far away from any mass, and clocks there run at full speed. If time is something that flows, the actual time (undilated by gravity) runs infinitely faster than does what any physical clock measures (all presuming that clocks measure the flow of time, which of course they don't).
The absurdity of trying to express this is one of my pieces of evidence against models with flowing time.
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Offline Eternal Student (OP)

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Re: Which bit of the Shell theorem is not working?
« Reply #32 on: 19/10/2023 21:46:06 »
Hi.

   Yes I can see what you're saying @Halc .
I'm deeply sympathetic.   I don't suppose that will help.

Let's try and be a bit more useful.    You seem to be trying to do the impossible, so don't.
You discuss time dilation and gravitational potential.   Most of the results we have for that would tend to come from using the Schwarzschild metric in GR.    Around a massive spherical body, the Schwarzschild metric is a fair approximation of what you have and how spacetime behaves.    For that solution of the EFE (Einstein Field equations), the co-ordinate time is exactly as you have stated.   It is the rate at which clocks progress when you are infinitely far from the mass.
    After that you are generalising the result and imagining that time would always run more slowly in a region of lower potential than some other region you are comparing against.   It might, it does seem reasonable but we do not have the Schwarzschild solution applying across all of space.   We know the universe is not a vacuum around one spherical mass.

   Indeed there is a completely different metric that is often used when discussing the Universe,  the Robertson-Walker metric.    That is obtained as a solution to the EFE under a very different set of assumptions.   In particular the universe is not a vacuum, it's actually too much unlike any vacuum.  Nowhere is less dense than anywhere else,  the universe is filled homogeneously with whatever cosmological fluids you have chosen to have (matter, radiation, dark energy  etc.)    With the Robertson-Walker metric, time dilation does not happen when you move from one gravitational potential to another because   (i)  gravitational potential is ill defined anyway,   (ii) nowhere is any different to anywhere else, there is no place of different potential.    That is not to say that time-dilation isn't an effect you can observe at all, you certainly can.   Special relativity is automatic in the machinery of GR, so movement will cause time dilation.
    The time co-ordinate in the Robertson-Walker metric is not and does not need to be related to how time may flow when you are infinitely far away from all mass.   Indeed it turns out that there are lots of places in a FRW universe where time flows at the same rate as the time co-ordinate.   Provided a clock is co-moving with the expansion of the universe then it will progress at a rate identical to the time co-ordinate.
    The passage of co-ordinate time is not abstract, it is something that a real clock can measure.  There is no requirement to be in some special place, for example to be in a place with the lowest gravitational potential, or infinitely far from all mass.   For the Robertson-Walker metric, the co-ordinate time is not that.   

     There is a natural tendancy to blend the two metrics together and assume that our real universe will have properties of both,  with more of one in some places (close to a body the Schwarzschild metric prevails) than other places.
None-the-less it is an error to assume that the co-ordinate time used in the Schwarzschild metric is precisely the same as the co-ordinate time in the FRW metric,   or that either of these is precisely the same as the underlying co-ordinate time that our real universe with its real metric would use.   
   
Quote from: Halc on 18/10/2023 01:44:15
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.

Best Wishes.
« Last Edit: 19/10/2023 22:01:49 by Eternal Student »
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