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Form factor - spherically symmetric

  1. Nov 26, 2006 #1
    1) Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge distribution p(r), which in the spherically symmetric case gives
    ,

    [tex] F(q) = \int \frac{4\pi\hbar r}{q}p(r)sin(\frac{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.

    This just means I can use coulomb's law as an expression for p(r). [tex] p(r) = \frac{q}{4\pi\epsilon_0R} [/tex] ?
     
    Last edited: Nov 26, 2006
  2. jcsd
  3. Nov 27, 2006 #2

    OlderDan

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    I take this to mean

    [tex] \rho(r) = \frac{q}{V} = \frac{3q}{4 \pi R^3} [/tex] for r < R and zero elsewhere.
     
  4. Nov 27, 2006 #3
    Thanks :). That's is what I was thinking after I posted. Don't see many [tex]\epsilon_0 [/tex] floating around in these things.
     
    Last edited: Nov 27, 2006
  5. Nov 27, 2006 #4

    Meir Achuz

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    You know it has nothing to do with Coulomb's law.
     
  6. Dec 5, 2006 #5
    So I evaulated the integral with [tex] \rho(r) = \frac{3e}{4 \pi R^3} [/tex]
    q = momentum transfer, e = proton charge.

    [tex]
    \frac{3e\hbar^2(\hbar\sin[\frac{qr}{\hbar}] - qr\cos[\frac{qr}{\hbar}])}{q^3 R^3} + C[/tex]
    Evaluated between R and 0, (F(q) = 0 between R and infinity since p(r) = 0.)


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

    It asks me to show that for [tex] \fracq{qR}{\hbar} << 1 [/tex] the form factor reduces to 1, i'm not seeing how the charge e can disappear here
     
    Last edited: Dec 5, 2006
  7. Dec 5, 2006 #6

    OlderDan

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    What does normalised charge distribution mean?

    I wonder if that means you should be dividing by a factor something like

    [tex] \int \rho(r) dr [/tex]

    which would remove the quantity of charge from the calculation
     
  8. Dec 5, 2006 #7
  9. Dec 5, 2006 #8

    [tex] 1 = \int \rho(r) d^3r [/tex]
     
  10. Dec 5, 2006 #9

    OlderDan

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    I like that even better. I even wrote my integral that way at first and changed it because yours just had the r integral.
     
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