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

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

1. ### eramSciengineerValued Senior Member

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Let's say an inverse-cube law of force applies to point masses.

Will the force exerted by a spherically symmetric body be equivalent to a mass in its center?

3. ### rpennerFully WiredRegistered Senior Member

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No. Neither will the force exerted by a spherical shell on its interior equal zero.

5. ### eramSciengineerValued Senior Member

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Hmm, I thought that being spherically symmetric was the only condition necessary for it to be equivalent to a point mass in the geometric center.

The inverse square law has some interesting properties indeed.

7. ### AlphaNumericFully ionizedRegistered Senior Member

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The inverse square law and the shell theorem are intimately linked because the fact anywhere inside the shell experiences zero force is because there is a very particular relationship between how the material is distributed and how much each little piece pulls on you. When you are located somewhere not in the middle of the sphere you are closer to some of the shell but there's now more shell on the other side of you, if you see what I mean. The additional amount is precisely balanced by the different strength of attraction due to closer proximity of the closer piece of the shell.

Spherical symmetry tells you that at the centre of the shell you would feel no resultant force, regardless of what the $F(r)$ expression was, since you're pulled in all directions equally. If you allow the force to depend on angle, so some direction causes a stronger pull than others, then you would obviously break this. A spherically symmetric shell under a force which depends only on distance will always have zero force at the centre. Off the centre and it becomes dependent on the force's precise form.

8. ### eramSciengineerValued Senior Member

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Yup. I was reading on how Newton conceived an inverse-cube law, but since a sphere becomes distinguishable from a point mass, it makes it more implausible I guess.

9. ### RJBeeryNatural PhilosopherValued Senior Member

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It's an interesting question. I should probably think before posting, but I wonder if Newton's Shell would work with an inverse cube law and a higher dimensional sphere...

10. ### TachBannedBanned

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Spherical symmetry proves that it does, independent of the function form $F(r)$ and independent of the number of dimensions.

11. ### AlphaNumericFully ionizedRegistered Senior Member

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Tach, if the Shell Theorem includes the result that the force experienced inside a shell is zero then the form F(r) is important, as it would become dependent upon the number of dimensions. It is not immediate that spherical symmetry is enough to enforce the other result from the Shell Theorem, that a spherically symmetric shell has the same gravitational field as a point mass of the same mass for all locations outside the shell. Given a gravitational profile F(r,m) for a point mass of mass m is it immediate that the force produced by a spherical shell will look like F(r,M) where M is the total shell mass, for any F? No. Or were you referring to something else in the theorem?

12. ### TachBannedBanned

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This is not what I am saying, what I am saying that the resulting force inside the shell is zero. I should have been more precise.

13. ### RJBeeryNatural PhilosopherValued Senior Member

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Wait, you believe this is true for any force function?

14. ### TachBannedBanned

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Yes, the spherical symmetry dictates that as I already mentioned. Doing the complete calculations for a force $F = k /r^3$ shows that to be true only in the center of the shell since $F = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \left( 1/s + \frac{r^2 - R^2}{s^3} \right) \, ds.$, as opposed to the case $F =k / r^2$ where the resultant force is identically zero everywhere inside the shell since $F = \frac{GMm}{4r^2 R} \int_{R-r}^{R+r} \left( 1 + \frac{r^2 - R^2}{s^2} \right) \, ds.$.

Last edited: Mar 18, 2013
15. ### eramSciengineerValued Senior Member

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Nice, that's even more interesting.

Would an inverse-cube point mass be indistinguishable from a 3-sphere?

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

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Except for the central point, the force at a point inside a uniform density mass shell is zero only for inverse r^2 gravity. Proof:

Consider a point not the center inside the shell and imagine a very narrow "double cone" with apex, A, at that point. From A inside these cones are two thin disk circles of mass, one R and other closer one to the shell is r from the shell. (r+R is the diameter of the shell.) The mass at R distant from A is ~R^2 and its force on point A towards it self is ~R^2/(R^p) where p is some power, 2 for real gravity or electric charge force. Force from mass at R is ~R^(2-p).

Likewise the attraction at A from the smaller mass at r is ~r^(2-p). The ratio of these two force is (R/r)^(2-p) and unity (equal forces) ONLY if p = 2.

For example if p = 1, then the force ratio (from mass at R compared to that from mass at r) is R/r - I.e. a net pull to the more distant side where R´s mass is, dominates as (R>r); however, if p = 3 then the ratio of the force from R to that of the force from r is r/R. I.e. the force from R mass is less than from mass at r.

If you don´t like math, consider r/R very small compared to 1 and p = 3 (case of the question) I.e. point a is close to an inside wall and the larger by (R/r)^2 mass at R has it gravity greatly decrease by the cube law. (Rapidly falls off.) The smaller mass at the near wall insde the cone has its force also decreased by the cube but of a much smaller number than R.

17. ### Q-reeusValued Senior Member

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How is it then that in GR, it is a given that an equipotential region exactly applies interior to a spherical mass shell? Given that in GR the Newtonian inverse square law does NOT apply exactly!

Last edited: Mar 18, 2013
18. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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To observe any deviaton from inverse square very large spaces are envolved. In the famous test of GR, the light from distant star was barely observed to bend even by the mass of the sun while traveling about a solar radius of space.

As some instrument inside the hollow sphere trying to detect the deviation from inverse square, would have mass and thus a small gratitonal field, there might be some variation of gravitational potential inside the hollow sphere it could oberve, but but it would be at least 100 billion times larger than that caused by gravity in truth not being exactly inverse square. I.e. your chance of noticing the GR effect on gravity would be same as that of seeing tiny unicorn running around the inside surface of the hollow sphere.

19. ### Q-reeusValued Senior Member

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Sure under 'normal' conditions any departure from Newtonian gravity is minute. But when approaching the other, 'black hole' extreme......
Nice try but I'm not by that impressed. Notice the word exactly was deliberately used twice - on purpose. Currently I'm not sure there is full consistency in the GR position of exact equipotential everywhere interior to a spherical mass shell. Or whether there is some subtle issue that goes like: "Hypothetical departure from inverse square law in a flat background spacetime is not equivalent to departure from inverse square law in a nonlinear GR spacetime arena". If that is the 'explanation', it needs quite a bit of elaboration imo.

20. ### eramSciengineerValued Senior Member

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Ah... , I was hoping no one would bring this up so soon. High chance of derailment

You mean a 100 billion times smaller?

I think the equipotential region is just an approximation.

If not then you can blame it on spacetime.

21. ### Q-reeusValued Senior Member

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Sure you're not becoming derailment sensitive here? So is the content of #13 already a derailment - since the response there to your rather specialized inverse cube query addresses it via a more generalized setting of inverse-anything-other-than-r^2.
You think or know? It is not an approximation in GR. Do you need specific references to prove that?
Let's see if expert opinion can clear that one up.

22. ### eramSciengineerValued Senior Member

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Fair enough but his post wouldn't carry on very much.

I said I think. Anyway, I also think BillyT was saying the same thing in #15

yup.

23. ### RJBeeryNatural PhilosopherValued Senior Member

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Tach, if you weren't so hypercritical when others make mistakes I would let this go, but the above clearly shows you proved yourself wrong in the very same post where you arrogantly proclaim "Yes, the spherical symmetry dictates that as I already mentioned". This is trivially false for a 1-sphere anyway: a point within the shell will feel a greater force from whichever side it is closer to regardless of the F function (unless F does not vary with distance of course).