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Calculating cross sections

  1. Dec 2, 2004 #1
    The textbook says, "The cross section, [tex] \sigma [/tex], is the transition or reaction rate per scatterer in the target, per unit incident flux". It gives the formula

    [tex] \sigma = transition \ rate \cdot \frac{1}{\# \ of \ scatterers \ in \ target} \cdot \frac{1}{unit \ incident \ flux} [/tex]

    [tex] = \frac{transition \ prob.}{unit \ time} \cdot \frac{1}{\# \ of \ scatterers} \cdot \frac{1}{unit \ incident \ flux} [/tex]
    (7.132) ​

    It states, "Flux is defined as the number of particles crossing an area, A, in a certain amount of time, T". It gives the formula

    [tex] flux = \frac{\# \ of \ particles}{AT} \cdot \frac{| \vec{v_1} |}{| \vec{v_1} |} = \frac{\# \ of \ particles \ | \vec{v_1} |}{V} [/tex]

    It states that V is the volume containing the particles. It gives

    [tex] flux = \frac{| \vec{v_1} |}{V} [/tex]
    (7.133) ​

    because in this case, they are considering only one particle. It then gives the cross section as

    [tex] \sigma = \frac{transition\ probability}{unit\ time \times unit\ volume} \cdot 1 \cdot \frac{1}{| \vec{v_1} |} [/tex]
    (7.134) ​

    And I'm wondering, how all the sudden did volume get into the denominator? Something doesn't sound right.

    BTW, this IS quantum mechanics, as the textbook is about to plug in the formulas from previously computed Feynman diagrams, which it has done a B- job of explaining so far. It's "Quantum Field Theory of Point Particles and Strings" by Hatfield.
     
    Last edited: Dec 2, 2004
  2. jcsd
  3. Dec 3, 2004 #2
    Now, let's calculate some probability amplitudes.

    Let's consider a simple Fock space which satisfies

    [tex] a^\dagger(p_1) | 0 \rangle = | p_1 \rangle [/tex]

    [tex] a(p_1) | 0 \rangle = 0 [/tex]

    [tex] [a(p_1),a^\dagger(p_2)] = \delta(p_1 - p_2) [/tex]

    from which follows

    [tex] \langle p_1 | p_2 \rangle = \delta(p_1 - p_2) [/tex]

    [tex] \langle p_1 p_2 | p_3 p_4 \rangle = \delta(p_1 - p_3) \delta(p_2 - p_4) + \delta(p_1 - p_4) \delta(p_2 - p_3) [/tex]

    My textbook multiplies the delta functions by scaling factors, but we're keeping things simple.

    The probability for two particles is given by

    [tex] \frac{(\langle p_1 | p_2 \rangle)^2}{\langle p_1 | p_1 \rangle \langle p_2 | p_2 \rangle} = \frac{(\delta(p_1 - p_2))^2}{(\delta(0))^2} [/tex]

    The probability for four particles is

    [tex] \frac{(\langle p_1 p_2 | p_3 p_4 \rangle)^2}{\langle p_1 p_2 | p_1 p_2 \rangle \langle p_3 p_4 | p_3 p_4 \rangle} = \frac{(\delta(p_1 - p_3) \delta(p_2 - p_4) + \delta(p_1 - p_4) \delta(p_2 - p_3))^2}{((\delta(0))^2 + (\delta(p_1 - p_2))^2)((\delta(0))^2 + (\delta(p_3 - p_4))^2} [/tex]
     
  4. Dec 4, 2004 #3

    Janitor

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    I interpret your question to mean ‘Why does [tex] AT v_1 = V [/tex]?’ The slug of beam particles in question is a prism (or a cylinder in a loose sense of the word, with arbitrary cross section shape, but with a pair of opposite ends that lie in planes parallel to one another). The ends of the prism each have area [tex] A [/tex], while the length of the prism is the distance the particles travel in time [tex] T [/tex], which is of course just their common speed [tex] v_1 [/tex] multiplied by the time lapse [tex] T [/tex]. The volume of the prism is the product of these terms.

    Or am I missing the point of your question?
     
  5. Dec 4, 2004 #4

    dextercioby

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    Yes,that was the only question he posted.But i'm having difficulties in undersranding the reason for his second post,as it had little to nothing to do with the question itself. :confused:
     
  6. Dec 4, 2004 #5
    That's right. I understand that. But what I don't understand is that when you plug (7.133) into (7.132), the quotient gets flipped, so [tex] V [/tex] should end up in the numerator. But (7.134) has it in the denominator as "unit volume". Notice that [tex] v_1 [/tex] got flipped just as it should. (7.134) should read

    [tex] \sigma = \frac{transition\ probability \times unit\ volume}{unit\ time} \cdot 1 \cdot \frac{1}{| \vec{v_1} |} [/tex]
     
  7. Dec 5, 2004 #6

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    Okay, I see that you were putting emphasis in your question on 'denominator,' not on 'volume.'

    If 'unit time' just has units T and 'unit volume' just has units L^3, then (7.134) is claiming that cross section has units T^-1 T^-1 L^-3 (L/T)^-1, which works out to T^-1 L^-4 instead of the L^2 that it ought to. Now I understand your concern! Not sure what to make of this.
     
  8. Dec 5, 2004 #7

    dextercioby

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    So far,everything is correct.

    You're right.It doesn't sound right,as the substitution for the unit flux of incident particles is wrongly made.The volume unit should not be in the denominator,but in the numerator,just as you posted.The equation would be then dimentionally correct.
    Maybe it was a typo in the book.Does the book have "errata"??
     
  9. Dec 6, 2004 #8
    I don't know.
     
    Last edited: Dec 6, 2004
  10. Dec 6, 2004 #9
    * number of scatterers= density_scatterers*unit volume
    * density of scatterers in the unit volume=1 => nbr of scatterers=unit volume
    * unit incident flux= v1
    => your final relation sigma relation (one scatterer in the unit volume for a unit incident flux)

    Seratend.
     
  11. Dec 7, 2004 #10
    It's probably a typo. The volume should definitely be in the numerator.
     
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