I am considering a point R inside an sphere of radius 1 in dimension d. The point is at a distance x from the center of the sphere, and the axis 0x is thus defined.

The distance between the point and the sphere surface is $L(x,\mathbf{\omega})$, where $\mathbf{\omega}$ is the angle between R and the point considered on the sphere. In 2D, $\mathbf{\omega}$ is an angle $\theta$, in 3D, it is $\theta,\phi$, etc...

I am interested in the mean distance (to a power $p$) between R and the sphere surface, projected on the axis Ox. The average is done over all possible angles (rather than over all points of the sphere surface). Physically, this value can be seen as the net force $f$ on R along Ox, when for all angles, the point measures $L$ and gets pulled to the surface with a force $L^p$.

For exemple, in d=1 :

$f^1_p (x)=(1+x)^p-(1-x)^p$

In d=2 :

$f^2_p (x)= \int_0^{2 \pi} d\theta \sin(\theta)L(x,\theta)^p$

with $L(x,\theta)=\sqrt{1-x^2 + x^2 \sin^2(\theta)}$

Interestingly enough, I found numerically that this measure has very funky properties. Namely, $f_p^d$ is linear if p=1 or p=d+1, and is sublinear in between.

My goal was to prove this linearity for p=d+1 recursively, but I am now stuck. Any chance this is a known problem, or that there is a simple solution I overlooked?

Note : If I'm correct, one can show that :

$f_p^d = 2 \int_0^1 \frac{u^{d-1}}{\sqrt{1-u^2}} f_p^{d-1} \left( \frac{x}{R_1(x,u)} \right) \left(u R_1(x,u) \right)^p du \qquad $ With $R_1(x,u)= \sqrt{ \frac{R_0^2 - x^2}{u^2} + x^2 } $

Edit : numerically, it seems that the mean p-distance $<l_d^p>$from R to the sphere, averaged over all angles $\omega$ is a constant if $p=d$. I'll call this constant $C_d$.

Let us take that as an assumtion to compute $f_{d+1}^d$ :

$f_{d+1}^d = \int_\Omega d\omega \mathbf{u}_x \mathbf{.} \mathbf{u}_\omega L(x,\omega)^{d+1} $

I thought I could integrate by parts, using $f_1^d = \int_\Omega d\omega \mathbf{u}_x \mathbf{.} \mathbf{u}_\omega L(x,\omega)^1 $ and using $f_{1}(x)=x$. However, this does not really achieve anything.

  • $\begingroup$ Numerically, it also looks like, under this definition of the mean, the mean p-distance (L^p) from a point R to the sphere surface in dimension d is constant if p=d. This is most likely corollary to what I want to show. $\endgroup$ – SergeD May 25 '18 at 12:04
  • $\begingroup$ When attempted in cartesian coordinates, you need to integrate $$\int \cosh^{p+1} u\, du$$ $\endgroup$ – Paul Sinclair May 25 '18 at 18:34
  • $\begingroup$ Mmm I'm not sure I see how and why... $\endgroup$ – SergeD May 26 '18 at 14:54
  • $\begingroup$ I really left that comment to help me find this post again when I had more time. The idea is that you define the axes so that $R$ is on the $x_n$ axis, and then do iterated integration with that axis first. This integral will have the form $$\int \sqrt{a^2 + (x_n - r)^2}dx_n$$ where $a^2$ is the contribution from all the other $x_i$. The substitution $x_n - r = a\sinh u$ then converts the integral into the form above. Unfortunately Wolfram Alpha gives a result that will be incredibly hard to integrate for the remaining $x_i$, so I don't think it is that helpful. $\endgroup$ – Paul Sinclair May 26 '18 at 17:43
  • $\begingroup$ Thank you ! I did find an easier way to prove it recursively, but it's going take me a bit of time to write it down. Will do it very soon, though. $\endgroup$ – SergeD May 27 '18 at 14:17

I wrote the solution in a PDF eventually. It's not very complicated algebra but it is a bit long.


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