1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mean square radius of sphere

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    why is the mean square radius of a sphere given as <r2> = 3/5 R2

    3. The attempt at a solution

    i saw this thread


    but i don't understand the helper (malawi_glenn)'s post

    he put in a p(r), but what is that? density?


  2. jcsd
  3. Aug 16, 2011 #2


    User Avatar
    Homework Helper

    What malawi_glenn was describing is the solution for a general density function dependent on radial distance from the center of the sphere. The problem you're asking about is just using a constant density [tex]\rho .[/tex] In that case, we have spherical symmetry and can just divide the sphere of radius R into spherical shells. What we are calculating is the (volume) average of the function [tex]r^{2}[/tex] for the sphere, which comes out as

    [tex] \frac{\rho \int_{0}^{R} r^{2} \cdot 4\pi r^{2} dr}{\rho \int_{0}^{R} 4\pi r^{2} dr}[/tex]

    [we're integrating over concentric spherical shells]

    [tex] = \frac{4\pi\rho \int_{0}^{R} r^{4} dr}{\frac{4\pi}{3} \rho R^{3} }[/tex]

    [the denominator is just the volume of a sphere, hence we will get a "volume average"]

    [tex] = \frac{4\pi\rho \cdot \frac{1}{5} R^{5} }{\frac{4\pi}{3} \rho R^{3} } = \frac{3}{5} R^{2} .[/tex]

    (Ahh... I'd forgot the "pleasures" of editing TeX expressions...}

    In practice, the density function is used to describe, say, the electron distribution about a nucleus, so the numerical factor can be something other than 3/5 . (This is akin to the moment integral used to find a moment of rotational inertia, but there we use cylindrical coordinates around the rotation axis.)

    Ironically, malawi_glenn had asked about this topic himself a couple years before the thread you cited, but received no reply... [ https://www.physicsforums.com/showthread.php?t=159471 ]
  4. Aug 17, 2011 #3
    wow thank you very much

    it suddenly dawned on me that the expression looks a lot like the formula to calculate center of mass.

    are they by chance the same formula? a sort of average formula?

    also, what is the significance of

    why is r2 linked to volume? shouldn't it be r3 instead? also, what does 3/5 R2 tell us about the volume average? is it just an averaged value? but how does this value describe th electron distribution about a nucleus?
  5. Aug 17, 2011 #4


    User Avatar
    Homework Helper

    Integrating the function r2 in the radial direction as we did is called a "second moment integral". If we used just r , we would have a "first moment integral".

    When you take an average [tex] \frac{x_{1} f(x_{1}) + ... + x_{n} f(x_{n})}{f(x_{1}) + ... + f(x_{n}) } ,[/tex] whether it is finding class average score for an exam or the center of mass of a finite set of masses, you are calculating the average of the "first moment" of the distribution. As an integral, the first moment becomes [tex]\int x f(x) dx .[/tex]

    When we are looking at something like variance of a set of data or the rotational inertia about an axis of a set of masses, we take an average with higher powers of x , in this case, x2 : [tex] \frac{x_{1}^{2} f(x_{1}) + ... + x_{n}^{2} f(x_{n})}{f(x_{1}) + ... + f(x_{n}) } . [/tex] The numerator gives the "second moment". The integral form is [tex]\int x^{2} f(x) dx .[/tex]

    I made the remark I did earlier because the mean square radius integral is effectively finding a second moment of the mass of the sphere about its center for uniform density.

    It isn't: that's the function we need in the integrand to obtain "mean squared radius". The volume came in when we integrated spherical shells of infinitesimal volume [tex] 4\pi r^{2} dr . [/tex] Dividing this integral by the volume of a sphere, which is the result of the integral in the denominator, gives us the "volume-averaged" result that makes this the "mean squared radius". (If we'd written a first-moment integral [tex]\int r \cdot 4\pi r^{2} dr [/tex] in the numerator, we would have found a "volume-averaged" mean radius, [tex]\frac{3}{4} R . [/tex]

    The integral we found here doesn't do that: it provides a "mean squared radius" for a uniform density distribution. If we'd wanted to find a mean squared radius for an electron distribution, we would set, say, [tex]\rho(r) = K e^{-\alpha r} [/tex] and used that in the ratio

    [tex]< r^{2}> = \frac{\int_0^{R} K e^{-\alpha r} \cdot r^{2} \cdot 4\pi r^{2} dr }{\int_0^{R} K e^{-\alpha r} \cdot 4\pi r^{2} dr } . [/tex]

    If we find that the mean-squared radius is smaller than 3/5 R2, that tells us that the electron distribution is concentrated more closely to the center of the sphere (the nucleus) than a uniform distribution, for example.
  6. Aug 18, 2011 #5
    i see, i check up wiki
    http://en.wikipedia.org/wiki/Moment_(mathematics [Broken])

    and it says the first moment is usually the average ? expectation value?

    the 2nd moment is the variance?

    so from what you said, in the electron case, if i were to use the 2nd moment 3/5 R2, that would give the the variation of the electron position in the sphere right?

    so now if i use the 1st moment, 3/4 R, what does it tell me? does it tell me the average number of electrons? but if i use a non-uniform density, (like in the electron variation case where smaller value than 3/5R2 means closer to center of sphere), what does a smaller value than 3/4R tell me now?

    just for interest, what does the 3rd moment tell us? if it exist.
    Last edited by a moderator: May 5, 2017
  7. Aug 18, 2011 #6


    User Avatar
    Homework Helper

    More precisely, the first moment is the numerator in calculating an average. When we multiply exam scores by the number of students who received each score and then add up those terms [tex][ N_{1}x_{1} + ... + N_{n}x_{n} ] ,[/tex] we are calculating the first moment of the score distribution. We then divide that by the number of students (sometimes called the "zeroth moment" because we can be said to be adding [tex] N_{1}x_{1}^{0} + ... + N_{n}x_{n}^{0} )[/tex] to obtain the "class average", or the mean [tex]\mu.[/tex]

    Here, we have to be a little more careful. In a statistical distribution, the "variance" is more completely called the "variance about the mean". The expression for that is [tex] \sigma^{2} = \frac {N_{1}(x_{1}-\mu)^{2} + ... + N_{n}(x_{n}-\mu)^{2} }{N_{1} + ... + N_{n} } .[/tex] The square root of this is the standard deviation, [tex]\sigma.[/tex]

    When we calculate the second moment of some physical distribution of, say, mass, we are generally finding that about some reference point. For rotational inertia, this is the radius at the rotation axis, r = 0 , hence, [tex]\int_{0}^{R} r^{2} dm ; [/tex] for the mean-squared radius, we measured outward from the center of the sphere.

    To obtain the mean-squared radius, we then divided the second moment of mass by the mass of the sphere. Since this involved a volume integral also, we would properly refer to our result as a "volume-averaged" mean-squared radius. In other sorts of calculations, different results can be obtained, depending upon what quantity the average of the moment was taken over.

    Not really. It is simply a measure of how the distribution is arranged. To cover the remainder of your questions, the various moments and their averages can be used as a way to provide more information about the nature of a distribution.

    As an example, if I tell you that the average annual salary of employees in a company is $80,000 , does that mean that everyone is receiving just about $80,000 a year or that a few are receiving much more than that and most of the employees rather less than $80,000? The average alone can't answer that for you. (This is how averages are sometimes used deceptively.) I could get more information about the distribution if I knew the variance about that average. Let's say the square root of the variance, the standard deviation, is $40,000 . Now I see that there is a lot of spread around the average, so many people are making much more or much less than $80,000 year. But is that symmetrical around the average, skewed toward the high end, toward the low end? I still don't know. I could then ask for a third moment, etc. in order to build up a better understanding of the "shape" of the distribution.

    This kind of moment analysis has its uses, but it does have a limitation. Notice that each moment (or averaged moment) only gives us one number, when what might give us a clearer picture of the distribution would be a histogram. But for some kinds of physical measurements, moments or averaged moments are all an experiment can give you. In the example of the electron distribution, we can't actually compile information on the location of electrons (probing will put momentum and energy into the electron, altering what we're trying to measure). But some experiments can indirectly measure quantities such as the mean-squared radius of the distribution (because we are looking at the collective behavior of a lot of atoms at once), so moment measurement sometimes is what we have to be satisfied with.
    Last edited by a moderator: May 5, 2017
  8. Aug 19, 2011 #7
    i see thank you very much. very indepth :D
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Mean square radius of sphere
  1. Expected Mean Squares (Replies: 0)