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The Question

What I want to know is how to calculate the mean square distance between a fixed point which is any random point in the sphere and rest of the points.

My Understanding

I'm clueless on how to calculate it. I'm a beginner. First I started from the straight line equation between the two points then equate it to radius of the sphere. then we can take another point. but this is not gonna help right? I think I should use integration here..

enter image description here Why it is an useful quantity to calculate?

For example, in calculating magnetic moment of an electron around the nucleus the mean squared radius is used.

Venus8588
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  • I realized that after posting. My bad – Venus8588 Feb 15 '21 at 18:35
  • The problem here is that the electron is not distributed on the surface of a sphere but in the whole space around the nucleus. You would have some sort of density of the electron $\rho(x,y,z)$ such that $\iiint\rho(x,y,z)dxdydz=1$, probably provided as a solution to the Schrodinger's equation, and the question is to calculate $\iiint(x^2+y^2+z^2)\rho(x,y,z)dxdydz$. I am not a physicist so I don't know what the physical interpretation of the mean squared distance is in this case. –  Feb 15 '21 at 18:40
  • For the electron case, we can calculate mean squared distance by multiplying radial wavefunction by r^2 and then integrating it. So can we do something like that? – Venus8588 Feb 15 '21 at 18:42
  • @StinkingBishop Yes. That's why we calculate the probability density which gives some distance where electron lies from the nucleus. – Venus8588 Feb 15 '21 at 18:49
  • If one point is the center of the sphere and the other is anywhere on its surface the mean squared distance between them is $R^2$, where $R$ is the radius of the sphere. – user Feb 15 '21 at 19:30
  • @user I had the same remark, but in fact the second point is anywhere in the ball with radius $R$ not on the sphere, as remarked by StinkingBishop. – Jean Marie Feb 15 '21 at 19:58
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    @JeanMarie OP was changed after my comment. Before this the point was explicitly assumed to reside on surface of the sphere. :) – user Feb 15 '21 at 20:02
  • What do you mean by the phrase "fixed point"? It does not seem to mean a fixed position. What about the rest of the points: are their positions fixed? – user Feb 16 '21 at 10:24
  • @user just mark a point P on the surface of the sphere. Now we can try finding the straight line distance between the point P and other points on the surface or inside the surface. We will definitely get some average distance. This is actually what I want to find out. – Venus8588 Feb 16 '21 at 14:10
  • What is known about the distribution of the other points? – user Feb 16 '21 at 14:27
  • @user I don't understand. Should we have a distribution? – Venus8588 Feb 16 '21 at 14:35
  • @user I have a question from a test which is similar. Let me edit the question and add it there. – Venus8588 Feb 16 '21 at 14:40
  • Of course. Actually it depends only on distribution. – user Feb 16 '21 at 14:54
  • @user In case of the mean distance square of the electrons, we have a distribution.. The following link has the distributions https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html . We can try for R(1,0) state – Venus8588 Feb 16 '21 at 15:00
  • In this case you also have a distribution of a special form: $\delta(x^2+y^2+(z+R)^2-R^2)$ where $\delta(x)$ is the Dirac's $\delta$-function. – user Feb 16 '21 at 16:29
  • @user I'm don't know what to do. Anyways Thank you so much for trying this question. It means so much. – Venus8588 Feb 16 '21 at 19:11

1 Answers1

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Hint:

To compute a mean square distance of a point distribution from a given point $(x_0,y_0,z_0)$ you need to compute the integral: $$ \langle r^2\rangle=\frac{\int_V[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]\rho(x,y,z)\,dV}{\int_V\rho(x,y,z)\,dV}.\tag1 $$ where $\rho(x,y,z)$ is the distribution density. In your example: $$ \rho(x,y,z)\sim\delta(x^2+y^2+z^2-R^2) $$ and $$x_0=(0,0,R), $$ where the origin of coordinates is assumed to be at the center of the sphere.

It remains only to substitute the expressions into (1) and evaluate the integral in spherical coordinates ($x=R\sin\theta\cos\phi,\,y=R\sin\theta\sin\phi, z=R\cos\theta)$.

Can you take it from here?

user
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