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The form factor

  1. Dec 7, 2009 #1
    Here is the problem:

    Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge
    distribution [tex]\rho[/tex](r), which in the spherically symmetric case gives

    [tex] F(q) = \int \frac{4\pi \hbar r}{q}\rho(r) sin(qr/\hbar))dr [/tex]

    to find an expression for F(q) for a simple model of the proton considered as a uniform
    spherical charge distribution of radius R.

    Using your calculated expression for F(q), demonstrate that in the limit
    [tex]\frac{qR}{\hbar} << 1[/tex]
    the form factor reduces to 1.

    So, what I have tried so far:

    I said that for 0 < r < R, the charge density is constant, and could be taken outside the integral along with the other constants, leaving me with

    [tex] F(q) = \frac{4\pi \hbar \rho}{q}\int^R_0 r.sin(qr/\hbar)dr[/tex]

    which when I integrate by parts leads to

    [tex] F(q) = \frac{4\pi \hbar^3 \rho}{q^3}sin(qR/\hbar) - \frac{4\pi \hbar^2 \rho R}{q^2}cos(qR/\hbar) [/tex]

    I am certain that I have done the integration correctly.

    Then I come to the part where I make the approximation that [tex]\frac{qR}{\hbar} << 1[/tex] and this happens:
    [tex] sin(qR/\hbar) \approx qR/\hbar [/tex]

    and

    [tex] cos(qR/\hbar) \approx 1 [/tex]

    which gives me

    [tex] F(q) = \frac{4\pi \hbar^2 \rho R}{q^2} - \frac{4\pi \hbar^2 \rho R}{q^2} = 0[/tex]

    but I am supposed to be getting F(q) = 1 when I make this approximation.

    What am I doing wrong here? My only thought was that the question mentions something about normalization, and I thought that may have something to do with something, but I couln't think what. Any suggestions?

    Thanks.
     
  2. jcsd
  3. Dec 7, 2009 #2
    I used a charge density of e/volume (3 e /4 pi R^3) integrated over all space in spherical polars & normalised it then put it in the equation. It basically just got rid of the e

    But i still get 0 for the 2nd bit!



    edit: Hooray, i get 1 using 2nd order approximations for sin & cos
     
    Last edited: Dec 7, 2009
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