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Rutherford scattering - electromagnetic form factor

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi there. This is not really a problem, I am only trying to understand something but I simply can't.
    So Rutherford scattering says that $$ \frac{d\sigma }{d\Omega}=(\frac{Ze^2m}{8\pi \varepsilon _0 p^2})^2\frac{1}{\sin ^4(\Theta/2)}|F(q)|^2$$ where $$F(q)=\int \rho (\vec x)e^{i\vec q \cdot \vec x}d^3 \vec x$$ is the electromagnetic form function. Now for spherical symmetry: $$\rho (\vec x) \rightarrow \rho(x)$$ and $$\vec q \cdot \vec x=qrcos\Theta.$$ Meaning that my form function can be written as $$F(q)=2\pi\int _0 ^1 d(cos\Theta)\int _0 ^\infty r^2dr\rho(r)e^{iqrcos\Theta}=$$ $$=\frac{4\pi }{q}\int _0 ^\infty \rho(r)sin(qr)rdr.$$

    I don't understand that last step. How the hell did we get ##sin(qr)##...?

    2. Relevant equations


    3. The attempt at a solution
    Ok, tried to do it step by step in order to answer this question but this is as close as I could get:
    Starting with
    $$F(q)=2\pi\int _0 ^1 d(cos\Theta)\int _0 ^\infty r^2dr\rho(r)e^{iqrcos\Theta}=$$ $$=2\pi \int \rho (r)r^2dr\int e^{iqrcos\Theta}d(cos\Theta)=$$ $$=2\pi \int \rho (r)r^2dr \frac{1}{iqr}[e^{iqr}-1]=$$ $$=2\pi \int \rho (r)rdr \frac{1}{iq}e^{\frac{iqr}{2}}[e^{\frac{iqr}{2}}-e^{-\frac{iqr}{2}}]=$$ $$=\frac{4\pi }{q}\int \rho (r)re^{\frac{iqr}{2}}sin(\frac{qr}{2})dr$$

    Now I wonder what am I doing wrong...?
     
    Last edited: Apr 6, 2015
  2. jcsd
  3. Apr 6, 2015 #2

    TSny

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    Gold Member

    Make sure you are using the correct limits for the integral ##d(\cos\Theta)##.
     
  4. Apr 6, 2015 #3
    Aaaaaaaamm... Ah, of course, I have to integrate over the entire nucleus (sphere)...

    So the ##d(\cos)## integral is actually from 1 to -1. And this brings me to ##sin(qr) / qr##.
     
    Last edited: Apr 6, 2015
  5. Apr 6, 2015 #4
    Mind = blown
     
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