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

Sure, take another 9 days of stumbling and bumbling

Please Register or Log in to view the hidden image!

 

Log in or Sign up to hide all adverts.

Presuming what you wrote is correct, the final antiderivative is

\( \theta+2 \tan^{-1}{{(r+R)}{\tan{{{\theta}\over{2}}}\over{r-R}}\over{2 r}\)

Which yields an evaluation of

\( {\pi}\over{r} \)

On the condition that the integration above is correct, the observation highlighted in red is FALSE.

The observations marked in green are TRUE, however, and I'm still interested in pursuing the veracity of the generalization marked in blue. It is premature to proclaim that the blue highlighted generalization is false; it is possible that the n-sphere / (n+1) ball is simply the wrong topology to apply as the dimensions grow.

This is not a dodge, Tach, so let me be clear: you were right to proclaim that the math showed the conjecture broke down for the 2-ball, and I was wrong.

QUICK EDIT: that evaluation looks off. The net force at r=0 should be zero. I guess I'll have to study it more...

 

Log in or Sign up to hide all adverts.

Nope : \( -\frac{\theta}{2r}-\frac{1}{r} \tanh^{-1}{{(r+R)}{\tan{{{\theta}\over{2}}}\over{r-R}}\over{2 r}\)

There might be some other substitution mistakes you've made but I don't have the time to track them all.

So, the bottom line is that your conjecture is FALSE.

It is, just make \(r=0\) and the integrand changes, I already pointed this out to you 9 days ago.

Yeah, this is the closest you'll come to admitting that you were wrong.

This part of the conjecture is false as well. There is no way of extending the algorithm I outlined beyond 3 dimensions. I left this for last while I had to wait for you to admit that the 2-ball part of the conjecture is false. One error at a time.

 

Log in or Sign up to hide all adverts.

What's this argument about? If the question is whether Newton's shell theorem generalises for a \(- \frac{1}{r^{n-1}}\) type gravitational field in \(n\) dimensions, then I'd have thought that followed easily from Gauss's theorem and a couple of symmetry arguments.

Regarding the integral around a circle (\(G\): gravitational constant; \(\rho\): linear mass density; minus sign for an attractive force):

\( F_{z} \,=\, -\, G \rho \int_{0}^{2\pi} \mathrm{d}\theta \, \frac{r \,-\, R \cos(\theta)}{s^{2}} \)​

with \(s^{2} \,=\, r^{2} \,+\, R^{2} \,-\, 2 r R \cos(\theta)\), a simple way to approach this is by variable substitution. Specifically, instead of integrating over \(\theta\), integrate the angle \(\varphi\) around the test point at \((z,\, x) \,=\, (r,\, 0)\). From the cosine rule,

\( R^{2} \,=\, r^{2} \,+\, s^{2} \,+\, 2 r s \cos(\varphi) \,. \)​

(The reason for the + sign is that I'm defining \(\varphi\) in such away that \(\varphi \,=\, 0\) coincides with \(\theta \,=\, 0\). In terms of the angle \(\phi\) appearing in the diagram RJBeery provides in [POST=3055461]post #122[/POST], \(\varphi \,=\, \pi \,-\, \phi\).) Combining with the expression for \(s^{2}\) above, this gives a relation between \(\theta\) and \(\varphi\):

\(R \cos(\theta) \,-\, r \,=\, s \cos(\varphi) \,.\)​

From the sine rule we also get

\(R \sin(\theta) \,=\, s \sin(\varphi) \,.\)​

Differentiating the square of either of these and cleaning things up:

\( \begin{eqnarray} R \sin(\theta) \bigl[ R \cos(\theta) \,-\, r \sin(\varphi)^{2} \bigr] \mathrm{d}\theta &=& s^{2} \sin(\varphi) \cos(\varphi) \mathrm{d}\varphi \\ \bigl[ R \cos(\theta) \,-\, r \sin(\varphi)^{2} \bigr] \mathrm{d}\theta &=& s \cos(\varphi) \mathrm{d}\varphi \\ \bigl[ r \cos(\varphi)^{2} \,+\, s \cos(\varphi) \bigr] \mathrm{d}\theta &=& s \cos(\varphi) \mathrm{d}\varphi \\ \bigl( r \cos(\varphi) \,+\, s \bigr) \mathrm{d}\theta &=& s \mathrm{d}\varphi \,. \end{eqnarray} \)​

Substituting all of this into the integral gets us:

\( F_{z} \,=\, G \rho \int_{0}^{2\pi} \mathrm{d}{\varphi} \, \frac{\cos(\varphi)}{r \cos(\varphi) \,+\, s} \,, \)​

and we only need an expression for \(s\) in terms of \(\varphi\). From the expression for \(R^{2}\) above and the quadratic formula,

\(s \,=\, -\, r \cos(\varphi) \,+\, \sqrt{R^{2} \,-\, r^{2} sin(\varphi)^{2}} \,,\)​

and the integral becomes

\( F_{z} \,=\, G \rho \int_{0}^{2\pi} \mathrm{d}{\varphi} \, \frac{\cos(\varphi)}{\sqrt{R^{2} \,-\, r^{2} sin(\varphi)^{2}}} \,. \)​

One can go further and get an analytic expression (e.g. substitute \(x \,=\, \frac{r}{R} \sin(\varphi)\)), but it's already clear that this integral will be zero from the fact it's periodic (so it doesn't matter which interval of \(2\pi\) we integrate over) and antisymmetric around \(\varphi \,=\, \pi/2\). Explicitly setting \(\varphi' \,=\, \varphi \,-\, \pi/2\) and integrating from \(-\pi\) to \(\pi\), the expression becomes:

\( F_{z} \,=\, -\, G \rho \int_{-\pi}^{\pi} \mathrm{d}{\varphi'} \, \frac{\sin(\varphi')}{\sqrt{R^{2} \,-\, r^{2} cos(\varphi')^{2}}} \,. \)​

So the end result is \(F_{z} \,=\, 0\), in case Gauss's theorem isn't enough for anyone.

 

Err, you don't need any "variable substitution" since you can integrate directly:

\(F(r) = \frac{Gm}{2 \pi} \int_{0}^{2 \pi} { \frac{(r-Rcos \theta) d \theta}{R^2+r^2-2Rr cos \theta}\)

This is surely a mistake since:

\(s^2+2rscos\varphi+r^2-R^2=0\)

From this elementary error your proof goes downhill.

 

If your idea of "integrate directly" is "stick it in a symbolic solver" then that doesn't always work and you shouldn't just naively copy the result. In this case the integral in question actually depends on whether \(r \,<\, R\) or \(r \,>\, R\), and a more intelligent symbolic solver would actually ask you to specify such things. Example Maxima session:

Code:

Maxima 5.27.0 http://maxima.sourceforge.net
using Lisp SBCL 1.0.57-1.fc17
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) integrate( (r - R * cos(theta)) / (r^2 + R^2 - 2 * r * R * cos(theta)), theta, 0, 2 * %pi );
Is  abs(R) - abs(r)  positive, negative, or zero?

negative;
                                     2 %pi
(%o1)                                -----
                                       r
(%i2) integrate( (r - R * cos(theta)) / (r^2 + R^2 - 2 * r * R * cos(theta)), theta, 0, 2 * %pi );
Is  abs(R) - abs(r)  positive, negative, or zero?

positive;
(%o2)                                  0
(%i3) 

So according to Maxima, by your definition, \(F(r) \,=\, \frac{Gm}{r}\) if \(r \,>\, R\) and \(F(r) \,=\, 0\) if \(r \,<\, R\).

 

I think Wolfram is quite intelligent for a symbolic integrator. So are others. Both agree with Demidovich, the ultimate reference.

Err, in your haste you glossed over the elementary error in your derivation. You need to pay attention to solving quadratic equations.

 

Yes, I say as much earlier in the same post (or rather, \(R^{2} \,=\, r^{2} \,+\, s^{2} \,+\, 2 r s \cos(\varphi)\), which is the same thing). Where's the "error"?

 

Go back and read your "solution" to the above equation. There are two solutions, right? What gives you the right to choose one and discard the other? Both of them are equally valid, so you can't have a unique substitution.

 

Er, you've noticed that \(-\, r \cos(\theta) \,-\, \sqrt{R^{2} \,-\, r^{2} \sin(\theta)^{2}\) is always negative, right? :bugeye:

 

You mean \(-\, r \cos(\varphi) \,-\, \sqrt{R^{2} \,-\, r^{2} \sin(\varphi)^{2}\), right? Yes, it is negative despite the fact that \(-r cos(\varphi)\) is positive.

 

Starting again...

2-Ball Analysis:
Using the picture from Wiki, but treating it as a 2-dimensional massive ring; the probe point lies within this "circle shell". I'm only concerned with lateral forces, since I presume we can all agree that the net force in the y dimension is zero regardless of form of the force function regarding its dependence upon distance.

Please Register or Log in to view the hidden image!

\(F = {GMm\over{s^1}} dM = {{M\over{2\pi}}d\theta} cos \phi=\frac{r^2+s^2-R^2}{2rs} s^2 = R^2 + r^2 - 2Rrcos(\theta) s^2 = {R^2 + r^2 - 2Rr\cos{\theta}} dF = {{GmdM}\over{s}} = {{GmM}\over{2\pi s}} d\theta dF_x = -dFcos(\phi) = -{{GmM}({r^2+s^2-R^2})\over{2\pi s^2 r}} d\theta = -{{GmM}({r^2+(R^2 + r^2 - 2Rr\cos{\theta})-R^2})\over{2\pi r(R^2 + r^2 - 2Rr\cos{\theta})}} d\theta = -{{GmM}({r - R\cos{\theta})}\over{\pi(R^2 + r^2 - 2Rr\cos{\theta})}} d\theta F_x = \int{dF_x} = -{{GmM}\over{\pi}}\int_{\theta=0}^{2\pi}{({r - R\cos{\theta})}\over{(R^2 + r^2 - 2Rr\cos{\theta})}} d\theta \int{{a+bcos(x)}\over{c+dcos(x)}} = {{bx}\over{d}} + {{{2(bc - ad)}tanh^{-1}{{(c-d)tan({{x}\over{2}})}\over{\sqrt{d^2-c^2}}}\over{d\sqrt{d^2-c^2}}}} a = r b = -R c = (R^2 + r^2) d = -2Rr x = \theta Therefore F_x = -{{GmM}\over{\pi}} * {{-R\theta}\over{-2Rr}} + {{{2((-R)(R^2 + r^2) - r(-2Rr))}tanh^{-1}{{((R^2 + r^2)-(-2Rr))tan({{\theta}\over{2}})}\over{\sqrt{(-2Rr)^2-(R^2 + r^2)^2}}}\over{(-2Rr)\sqrt{(-2Rr)^2-(R^2 + r^2)^2}}}} F_x = -{{GmM}\over{\pi}} * {{{\theta}}\over{2r}}+arctanh{{(r+R)}{\tan{{{\theta}\over{2}}}\over{r-R}}\over{r} \)


...if nothing else I'm getting a wonderful crash course on Calc. After all of this effort it's just as important for me to understand the math as it is to know the truth. Alas, the above analysis still has probs I believe...

 

Yes.

So if it's not clear, \(s\) is the distance between the test point at radius \(r\) and a point on the circle. I specifically want the positive solution here.

 

I know exactly what \(s\) is.

 

How does the term in \(sin^2(\varphi)\) show up in the LHS? The LHS is the differential of \(2R^2sin^2(\theta)\) so, it should look exactly like the RHS, i.e. \(R^2 sin(\theta) \cos(\theta) d\theta\). Your solution is so complicated that the odds that you made a mistake are quite high. The presence of \(sin^2(\varphi)\) in LHS s highly suspect, how did you make it appear?

 

Everyone's making terribly hard work of this integral. Set \(\epsilon = r/R\). Then the integral is

\( \frac{1}{R} \int_0^{2\pi} \frac{ \epsilon - \cos \theta}{\epsilon^2 - 2\epsilon \cos\theta +1}\, \mathrm{d}\theta = \frac{1}{R} \oint \frac{ \epsilon - \frac{1}{2}(z+\frac{1}{z})}{\epsilon^2 - \epsilon(z+\frac{1}{z}) + 1}\, \frac{\mathrm{d}z}{\mathrm{i}z} = \frac{1}{\mathrm{i}R} \oint \frac{ \epsilon z - \frac{1}{2} z^2 - \frac{1}{2}}{(\epsilon z -1)(\epsilon -z)}\, \frac{\mathrm{d} z}{z}\)

where we've set \(z=e^{\mathrm{i}\theta}\) and z-integral is taken round the unit circle. We pick up a residue at \(z=0\) and another at \(z=\epsilon\) or \(z=1/\epsilon\), depending on how big \(\epsilon\) is. If we're inside the shell \(\epsilon<1\) so we pick up a residue from \(z=\epsilon\), hence

\( \frac{1}{R} \int_0^{2\pi} \frac{ \epsilon - \cos \theta}{\epsilon^2 - 2\epsilon \cos\theta +1}\, \mathrm{d}\theta = \frac{2\pi \mathrm{i}}{\mathrm{i} R} \left[ \frac{1}{2\epsilon} - \frac{ \frac{1}{2} \epsilon^2 - \frac{1}{2}}{\epsilon (\epsilon^2 -1)} \right] = 0. \)

If we're outside the shell, \(\epsilon>1\) so we pick up a residue from \(z=1/\epsilon\) rather than from \(z=\epsilon\). This again gives the expected result.

 

Good. So that's settled then.

 

From differentiating \(s^{2} \sin(\varphi)^{2}\) on the RHS and moving the term in \(\mathrm{d}\theta\) to the LHS. You get such a term because \(s^{2} \,=\, r^{2} \,+\, R^{2} \,-\, 2 r R \cos(\theta)\) depends on \(\theta\).

The rest is just substituting identities derived from the sine and cosine rules.

No, it doesn't work like that. Either I made an error or I didn't, and if you want to suggest I have it is up to you to point it out.

In any case, I've now got a number of independent confirmations of the end result: 1) I can derive it from Gauss's theorem (this would be the first instinct of most physicists), 2) my own evaluation of the integral in [POST=3055711]post #144[/POST], 3) the Maxima CAS system finds the integral is zero, and 4) Guest254's approach using residue calculus.

 

Yes, this is clearly correct. Residue theorem produces the correct answer for both the inside and the outside of the circle. I managed to get Wolfram to produce the correct result as well.

 

OK Guest I'll give props when they're due, nice job.

And, Tach, I'm not going to gloat nor am I going to ask for the admission that you promised, we all know what went down here.