# Would an inverse-cube law obey Newton's shell theorem?

Discussion in 'Physics & Math' started by eram, Mar 15, 2013.

1. ### RJBeeryNatural PhilosopherValued Senior Member

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Agreed on the "extreme analysis" but you are neglecting to consider 50 miles of area-constant mass counteracting the closer of the cylinder bases.

3. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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No, my cylinder of 60 miles length makes the one inch diameter ends have negligible effect at point A which is 50 miles from one end and 10 miles from the other, but, if you don´t believe that I can be both more mathematical and more extreme:

Make cylinder 100 light years long and only 0.001 inch in diameter. Put point A on the axis 1 light years away from the mass center. Then there is a section of the cylinder 49 ly long of pulling one way but the first 49 ly on the other side EXACTLY cancells out that pull, leaving the net force due to the distant 2 ly long section pulling (weakly, I admit but not 0 pull) on point A the other way, I.e. trying to move a test mass placed at point A, towards the mass center, as I said before.

Yes, I admit that in the original 60 mile long cylinder there is a much smaller net force due to the tiny end mass which is only 10 miles away and the end mass 50 miles away does not fully cancel it out, but:

Note even if I did not make 60 mile cylinder longer, but only by made diameter be 0.001 inch I reduced the pull of the flat end caps by a factor of a million compared to the original 1 inch diameter cylinder but that reduced the pull of the cylinder walls only by a factor of a thousand.

I.e. even on my original 60 mile long cylinder I can make the effect of the end masses as small as I like wrt to the effect of the cylinder walls. For example instead of the original 1 inch diameter, I change the diameter to only 100 Angstroms.

Ready to say "uncle" yet? or do I need to make the diameter only 100 Bohr radii?

5. ### eramSciengineerValued Senior Member

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What does that mean?
Anyway I agree with your analysis.

RJ: A non-spherical object of finite size will always produce such results.

7. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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In college wrestling at least when neither can pin both shoulders of the other to the mat but one is stronger and dominating, and that one can inflict pain, then to stop the pain, the other says "uncle." I.e. he is admitting defeat even if not officially "pined."

BTW there is nothing you can do IMHO, that will exhaust you as quickly as college wrestling against an equal opponent. As I recall, after 3 minutes with no victor the match is called a drawl - the school doesn´t want either so tired that they die on campus. I seldom (perhaps never?) lasted the full 3 minutes - I either won or was defeated. After a couple of months I quit - my special 5-year program was taking all my time.

8. ### eramSciengineerValued Senior Member

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So saying "uncle" is to admit that the victor is more "senior" than you?

By the way, a running joke among my friends is that MMA stands for "Man-on-Man Action"

9. ### RJBeeryNatural PhilosopherValued Senior Member

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Hah! Late Friday night posting, I was completely hammered. :cheers: This is trivially false; start with a sphere and deform one side towards the probe point such that the sphere's curvature is concave; same mass, but now closer to the probe point. (sorry for the wild goose chase Billy)

Makes me curious though; are there deformations to the massive shell which would maintain the net zero force on the interior (if we allow the mass to grow with the area as needed)?

10. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I take that as saying "uncle."

11. ### RJBeeryNatural PhilosopherValued Senior Member

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Yes, certainly, "uncle"

Take Newton's shell and deform it from the inside such that half of it becomes a cone and the other half remains a semisphere; remembering to allow for mass growth as needed to keep it area-constant, what would you predict the net force to be from the original sphere's center?

12. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I´m not sure what you are suggesting for altitude of the cone. I assume its base still is the same circle as that of the hemisphere and joined to it. To avoid having pi, lets agree to a unit* of length that makes the radius of the hemisphere, R, equal to 1/(square root of pi.) I.e. then the surface area of the hemisphere is 2. Is the altitude of the cone such that its surface area remains 2 also? I.e. no change in the mass of the cone or its surface density. I assume that for rest of this post.

Obviously all parts of the cone now less than R from the original center will pull harder on test point there than they did when shape was simle sphere an those part of the cone more distant than R will pull less. I´m not sure I can easily tell if the net force the cone makes has increased or decreased. I warn you I am lazy** so may not have more to say but certainly that net force (of the cone part) is along the cone axis and directed towards the apex of the cone. The net force of the hemisphere part is also on that axis, but away from the cone apex.

* All units of length should have a name. This one is not meters or feet but the BTRJ unit. So more correctly stated the area of the hemisphere is 2 BTRJ square.

** I like to think but not to calculate any more. My quick, intutive guess, is that the total net force is towards the cone apex. This is because each tiny differential part*** of the part of the cone less than R from the center is not only pulling with with stronger force now (due to being only r^(-2) diminished (r < R) but also the component along the cone axis is a greater fraction of the force it makes).

I.e. consider differential rings of the sphere´s surface as having "migrated" to the surface of the cone. Of course the net force of each such ring is along the cone axis only (by symetry).

*** Say 0.0001 BTRJ square in size.

Last edited by a moderator: Mar 30, 2013
13. ### RJBeeryNatural PhilosopherValued Senior Member

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haha I was really just thinking out loud, not necessarily asking for an answer. I might give the math a shot on this...

14. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Integrating up a set of differental ring forces from cone base to apex is probably not very hard as only a 1-D integration. Chose your force unit such that the differental ring that is still R from the center has force 1 and make all rings have the same area (& same mass). (Ones near the Apex are much wider than those near the near base.)

I bet it is best to break the integration into two parts - considering the rings with less than R from the center separately from those more than R from the center. I.e. let the cancellation between reduced and increased forces be done after the two integrations are done.

Before you know the answer do as I did. Make a WAG as to whether or not the net force is towards the cone apex.

OTHERS PLEASE DO THE SAME - GIVE US YOU WAGs as to which direction the net force pulls.

15. ### eramSciengineerValued Senior Member

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Does WAG stand for Wild Ass Guess?

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yes

17. ### eramSciengineerValued Senior Member

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Originally it's a thin shell, with area density (mass per unit area).

So a hemisphere has a certain amount of mass. One hemisphere is deformed into a cone, which presumably still mass the same amount of mass.

We must specify if the cone still has the same area density, otherwise the height of the cone may vary. This will most probably produce variable results.

18. ### RJBeeryNatural PhilosopherValued Senior Member

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If the hemisphere is deformed into a cone of the same mass then there would be only a single value for the height of the cone such that the net internal force from the center remains zero. The question I posed was, "independent of height (> R), what do we predict if the mass increases/decreases as needed on the cone such that the area density remains the same constant as it was before being deformed?"

19. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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First note the unit of lenth was defined in post 207 and called the BTRJ unit. It is not same as feet or meters but needed a a name.
These two, (1) & (2) are independent conditions. You can not impose both on the same cone. Because, if the cone mass and surface mass density remains uchanged, that alone sets the cone altitude. I.e. the altitude of the cone is such that the cone mass and surface area remans the same I.e. 2 BTRJ square units, I´ll call that cone C1.

Now if the altitude of cone is very small, a quasi-flat cone, called C2, as it is almost a flat sheet, and if the mass remains the same then the surface density is increased, approximately doubled, as a nearly flat cone has only slightly more than 1 BTRJ square units of surface area. Obviously the net force of both hemisphere and nearly falt cone C2 is directed towards the apex of the "flat cone"

In contrast, if the altitude of the cone (a cone called C3) is very large so near the hemisphere it is nearly a cylinder, then the cone area is greatly increased and the mass surface density is greatly reduced, if the total mass of the cone remains unchanged. In this case, equally obviously the net force of both cone and henisphere is still along the cone axis but now directed away for the cone apex.

Thus there is some cone (cone 4) with altitude between cones C2 & C3 where this net force changes direction - I.e. net force reverses by becoming zero net force as it changes sign. A "zero net force" cone (called cone 4) with unchaged mass (form the original hemisphere it was deformed from).

There is no reasons to expect that the altitude of C1 and C4 are the same as you supposed they are in your "double condition" (1) & (2) statement of your post´s first sentence.

Last edited by a moderator: Mar 31, 2013
20. ### Guest254Valued Senior Member

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Hi Billy,

No, unfortunately what you've said here does not suddenly transform your argument into a proof. And in addition, I will point out that you certainly did not prove "as t approaches 0 that the ratio X/Y approaches unity", but rather you accepted this fact because I told you it was true. Let the gravitational force due to a patch Δ on our surface be denoted by F(Δ). Whenever we say "patch" we will mean the type of patch obtained by intersecting a cone with the surface, as in your post. At the moment, we have*

For each sufficiently small patch Δ on our surface, there is another disjoint patch Δ' = Δ'(Δ) (i.e. the set Δ' depends on the set Δ) such that

$\lim_{|\Delta|\rightarrow 0}\left| \frac{F(\Delta)}{F(\Delta')}\right| =1, \qquad (**)$

where |Δ| denotes the area (or "measure") of the set Δ.

So, what do you need to do now? Well, first you want to cover your surface with a collection (call it A) of disjoint patches, such that if Δ is in A, then Δ' is also in A (so you can use some sort of cancellation argument). This is not particularly easy and I imagine one would need to invoke some variant of Vitali's covering lemma. Alternatively, you can choose to cover all the the surface less some remainder, and show that the gravitational force from this remainder can be made arbitrarily small with finer and finer finite coverings. The second option is probably easier to implement, but I'll use the first option because the language is clearer. The given our covering A the total force is

$F = \sum_{\Delta \in A} F(\Delta) = \sum_{(\Delta,\Delta') \in A\times A} F(\Delta) + F(\Delta')$

where in the second sum we ran over all ordered pairs (Δ,Δ') in A. Note that this will be an infinite sum. Now you want to some how use the cancellation argument. We can certainly show that R(Δ)=F(Δ)+F(Δ') tends to zero if |Δ| goes to zero, but that means we have to work with an even cleverer collection of sets {A(n)} for which the area of all its members tends to zero as n gets larger, so you can use (**). Let's pretend we've managed to construct this collection of sets (by means of a similar argument used to construct A). Then

$F = \sum_{(\Delta,\Delta') \in A_n\times A_n} F(\Delta) + F(\Delta').$

We know that as n tends to infinity, the area |Δ| for each Δ in A(n) tends to zero. This is good, because now we can try to invoke (**). We can show that for each Δ in A(n), the limit of R(Δ) is zero as n tends to infinity. But this is pretty unexciting anyway, because this is necessary for the sum to converge! What we need to show is that R(Δ) decays fast enough so that the sum is actually zero.

As you can see, there is a lot of non-trivial mathematics needed to make your argument into any sort of proof! This can certainly be done, but your post doesn't include any of it! If you'd like to have a crack at it, I'd recommend the alternative argument, i.e. covering your surface with a finite collection of disjoint patches plus the remainder, and show that the gravitational force due to the remainder can be made arbitrarily small. After arguing as above, your main task will be to show that R(Δ) is better than just o(1) as Δ goes to zero (which is obvious), but something like o(Δ) [here o() is the little-oh-symbol].

*Note: in no way have you actually proved this, but I'm willing to let that go because the argument is simple in comparison to others needed to turn what you've written into a proof!

21. ### Guest254Valued Senior Member

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You mean the post in which Billy certainly did *not* prove anything? And from the non-proof you, with your well developed vision of gravitational fields inside higher dimensional spheres, you "saw" what the answer was? Or, do you mean guess? There's nothing wrong with the word "guess", but let's not try to elevate it beyond what it is. I'm assuming you're forced to elevate the status of guesses because they are all you can offer.

But I don't think I need to say much more here, you're willing to make the point for me...

I can see that the thread has moved on a bit since I was here last. If anyone wants to know the gravitational potential at a point x inside a closed surface S with uniform density, it will be proportional to

$\int_S \frac{1}{|\mathbf{x}-\mathbf{y}|}\, \mathrm{d}S(\mathbf{y}).$

This comes from solving Poisson's equation $\Delta \phi = \delta_S$, where $\delta_S$ denotes the Dirac measure supported on S.

22. ### RJBeeryNatural PhilosopherValued Senior Member

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Yes, relying on intuition is pretty close to "guessing", but it's really more like a "guestimation". You're implying that it's completely random.
Yes, but only with beer goggles. You conveniently left out my follow-up post the next morning.
It might do you some good to let your mind wander, Guest. Of Tach's many character flaws at least he puts himself out there and allows himself to make mistakes; that's how progress and learning happens. I know I learned something with the whole relativistic mirrored/matte wheel debacle and I suspect many others did too.

23. ### TachBannedBanned

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Based on your posts I doubt that you learned anything from your debacle, you are just as incapable of formulating problems in a mathematical formalism now as then.