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Charge distribution with exponential form. Halzem and Martin. exercise 8.4

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data

    In the exercise 8.4 from Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin we can see:


    if the charge distribution [itex]\rho(r)[/itex] has an exponential form [itex]e^{-mr}[/itex], then:


    [tex] F(q) \propto (1 - \frac{q^2}{m^2})^{-2} [/tex]


    where F(q) is:

    [tex] F(q) = \int\rho(x) e^{iq.x} d^3x [/tex]


    3. The attempt at a solution


    The book says that first we integrate the angular part and obtain:


    [tex] F(q) = 2\pi \int\rho(r) (\frac{e^{iqr}-e^{-iqr}}{iqr}) r^2 dr [/tex]


    Please, can anyone say me how can I obtain [itex] (\frac{e^{iqr}-e^{-iqr}}{iqr}) [/itex]
     
  2. jcsd
  3. May 28, 2012 #2

    vela

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    As the problem says, you integrate the angular part. Show us your work at doing that.
     
  4. May 29, 2012 #3
    Hello Vela. Thank you for your answer.

    In spherical coordinates, we have

    [tex] F(q) = \int\rho(r) e ^{i q r}r^2 dr \int_{0}^{\pi} sen \theta d\theta \int_{0}^{2\pi} d\phi = 2 * 2\pi \int\rho(r) e ^{i q r} r^2 dr [/tex]

    But then, from where come:

    [tex] (\frac {e ^{i q r}- e ^{-i q r}}{iqr} )[/tex]

    I think that it comes from a property of Fourier transform. It looks like a [itex] sen (iqr) [/itex] but from where?

    And moreover, I am not sure if [itex] e^{iqx}[/itex] in spherical coordinates were [itex] e^{iqr}[/itex]

    Thank you.
     
  5. May 29, 2012 #4

    vela

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    ##\vec{q}\cdot\vec{x}## isn't equal to qr. Remember the dot product depends on the angle between the two vectors, not only on their magnitudes.
     
  6. May 29, 2012 #5
    Ok. Then I must write:

    [tex] F(q) = \int\rho(r)\ e ^{i \ q \ r \cos \alpha}r^2 dr \int_{0}^{\pi} sen \theta \ d
    \theta \int_{0}^{2\pi} d\phi = 2 * 2\pi \int\rho(r) e ^{i \ q \ r \ cos \alpha} r^2 dr
    [/tex]

    And [itex]\alpha [/itex] is the angle between q an r. Sorry, but now I don´t kown how to continue. I can´t integrate by parts... I don´t know which way to take.
     
  7. May 29, 2012 #6

    vela

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    When evaluating the integral, orient the coordinate system so that q lies along the z-axis.
     
  8. May 29, 2012 #7
    I think that I have it already. I must make a little change:

    [tex] F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr[/tex]

    [itex] \theta [/itex] is the angle between z-axe and r, and q is along z-axe, so that [itex] \theta [/itex] is the angle between q and r.

    [tex] \frac {d}{d\theta} e^{i \ q \ r \ cos \theta} = -i \ q \ r \ sen \theta \ e^{i \ q \ r \ cos \theta} [/tex]

    so that:

    [tex] \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta = [\frac{e^{i \ q \ r \ cos \theta}}{- \ i \ q \ r}]_{0}^{\pi} = [\frac{e^{i \ q \ r \ cos \theta}}{i \ q \ r}]^{0}_{\pi} = \frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}[/tex]

    And finally we have:

    [tex] F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr = 2\pi \int \rho (r) \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr[/tex]


    It´s incredible. I didn´t think never that I would be able to solve this problem. I have an exam of "Elementary particles" in the University in a few weeks. I work and I can´t visit the classes. I don´t know anybody that can help me becuase I don´t know the other students. When I am not able to solve a problem, then nobody can help me and it is frustrating.

    I am very grateful, vela. Thank you very much.
     
    Last edited: May 29, 2012
  9. Jun 23, 2012 #8
    I need help with this integral: charge distribution

    1. The problem statement, all variables and given/known data


    In the exercise 8.4 from Quarks and Leptons. An Introductory Course in Modern Particle Physics - F.Halzem,A.Martin we can see:



    if the charge distribution [itex]\rho(r)[/itex] has an exponential form [itex]e^{-mr}[/itex], then:


    [tex] F(q) \propto (1 - \frac{q^2}{m^2})^{-2} [/tex]


    where F(q) is:

    [tex] F(q) = \int\rho(x) e^{iq.x} d^3x [/tex]



    2. Relevant equations

    In spherical coordinates, we have

    [tex] F(q) = \int _{0}^{2\pi} d\phi \int_{0}^{\pi} e^{i \ q \ r \ cos \theta} sen \theta \ d\theta \int \rho (r) \ r^2 \ dr = 2\pi \int \rho (r) \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr[/tex]



    3. The attempt at a solution

    I am not able to solve this integral:

    [tex] F(q) = 2\pi \int e^{-mr} \ (\frac{e^{i \ q \ r} - \ e^{- \ i \ q \ r }}{i \ q \ r}) \ r^2 \ dr[/tex]

    I have tried:

    [tex] F(q) = \frac{2\pi}{iq} \int (e^{Ar} - e^{-Br}) r \ dr = \frac{2\pi}{iq} ( \int (e^{Ar} \ r \ dr - \int e^{-Br} r \ dr)[/tex]

    with [itex] A = iq-m[/itex] and [itex] B = iq+m[/itex]

    Finally I have this solution :

    [tex] F(q) = \frac{2\pi}{iq} ( [...] + \frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (2(m^2-q^2) - B^2 e^{Ar} - A^2 e^{-Br})) [/tex]

    with [...] igual to the initial conditions when I integrate by parts. I suppose that it is possible to choose initials conditions so that [...] -> 0. But I think that there are too many coefficients. Why do I obtain [itex] 4q^2m^2[/itex] ? Is it a mistake or is the result good?

    Thank you very much.
     
  10. Jun 23, 2012 #9
    Re: I need help with this integral: charge distribution

    I made an error with a sign. The result is:

    [tex] F(q) = \frac{2\pi}{iq} ( [...] + \frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (4i q m + A^2 e^{-Br} - B^2 e^{Ar})) [/tex]

    And if [...] -> 0, then:

    [tex] F(q) = 2\pi \ (\frac{1}{(1-\frac{q^2}{m^2})^2 m^4 + 4q^2m^2} (4m + \frac{A^2 e^{-Br} - B^2 e^{Ar}}{iq})) [/tex]

    But, what about [itex] 4q^2m^2[/itex]?
     
  11. Jun 23, 2012 #10

    vela

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    It's kind of hard to follow what you did because you left out a lot of steps. Show how you integrated
    $$\int_0^\infty re^{-r(m\pm iq)}\,dr$$ If you're familiar with Laplace transforms, you could also combine the exponentials to get
    $$\int_0^\infty r\sin(qr)e^{-mr}\,dr$$ which is the Laplace transform of ##r\sin qr##.
     
  12. Jun 23, 2012 #11
    I thought also this integral that you say:
    $$\int_0^\infty r\sin(qr)e^{-mr} dr $$
    but, how can I integrate it?, by parts? When I integrate by parts I use:
    $$ \int u \ dv = u \ v - \int v \ du $$
    In your integral, who is ##u## and who is ##v##?


    Now I show you the steps in the calculation of my integral:


    $$ F(q) = \frac{2\pi}{iq} \int (e^{Ar} - e^{-Br}) r \ dr = \frac{2\pi}{iq} ( \int (e^{Ar} \ r \ dr - \int e^{-Br} r \ dr) $$

    with ##A = iq-m## and ## B = iq+m##

    $$ F(q) = \frac{2\pi}{iq} \left\{ ( \left[ r´ \frac{e^{Ar´}}{A} \right]_0 ^r - \int \frac{e^{Ar}}{A} dr ) - (\left[ r´ \frac{e^{-Br´}}{-B} \right]_0 ^r - \int \frac{e^{-Br}}{-B} dr) \right\}$$

    $$ F(q) = \frac{2\pi}{iq} \left\{ (\frac{re^{Ar}}{A}) - \left[\frac{e^{Ar´}}{A^2} \right]_0 ^r - ((-\frac{re^{-Br}}{B}) - \left[\frac{e^{-Br´}}{B^2} \right]_0 ^r) \right\} $$

    $$ F(q) = \frac{2\pi}{iq} \left\{ (\frac{re^{Ar}}{A}) - \left[\frac{e^{Ar}-1}{A^2} \right] + (\frac{re^{-Br}}{B}) + \left[\frac{e^{-Br}-1}{B^2} \right] \right\} $$

    $$ F(q) = \frac{2\pi}{iq} \left\{ r(\frac{e^{Ar}}{A} + \frac{e^{-Br}}{B}) - \frac{e^{Ar}}{A^2} + \frac{e^{-Br}}{B^2} + (\frac{1}{A^2} - \frac{1}{B^2}) \right\} $$

    Here I imagine that we can choose the initial conditions so that ## r(\frac{e^{Ar}}{A} + \frac{e^{-Br}}{B}) = 0 ##

    $$ F(q) = \frac{2\pi}{iq} \left\{ \frac{A^2e^{-Br}-B^2e^{Ar}}{A^2B^2} + \frac{B^2-A^2}{A^2B^2} \right\} $$

    Now:

    $$A^2 = (iq-m)^2 = m^2 - q^2 - 2i \ qm $$
    $$B^2 = (iq+m)^2 = m^2 - q^2 + 2i \ qm $$
    $$B^2 - A^2 = 4i \ qm $$
    $$A^2B^2 = (m^2 - q^2)^2 + 4q^2m^2 $$

    $$ F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(m^2 - q^2)^2 + 4q^2m^2} \right\} $$

    $$ F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2} \right\} $$

    $$ F(q) = 2\pi \left\{ \frac{1}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2} (4m + \frac{A^2e^{-Br}-B^2e^{Ar}}{iq}) \right\} $$

    So, ##F(q) \propto \frac{1}{(1 - \frac{q^2}{m^2})^2m^4 + 4q^2m^2}##. But the problem says that ##F(q) \propto \frac{1}{(1 - \frac{q^2}{m^2})^2}##

    I think that something is wrong in my calculations because the problem don´t says nothing about ##m^4## and ##4q^2m^2##.

    I thank you vela again for your inestimable help.
     
  13. Jun 23, 2012 #12

    vela

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    In the denominator, expand ##(m^2-|\vec{q}|^2)^2##. Also, note that Halzen and Martin are using the convention ##q^2 = -|\vec{q}|^2##.
     
  14. Jun 23, 2012 #13

    vela

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    I'd actually use ##re^{-mr} = \frac{\partial}{\partial m}e^{-mr}## to get
    $$\int_0^\infty r\sin(qr)e^{-mr}\,dr = -\frac{\partial}{\partial m}\int_0^\infty \sin(qr)e^{-mr}\,dr$$
     
  15. Jun 24, 2012 #14
    Ok. Finally I understand. From here:

    $$ F(q) = \frac{2\pi}{iq} \left\{ \frac{4i \ qm + A^2e^{-Br}-B^2e^{Ar}}{(m^2 - q^2)^2 + 4q^2m^2} \right\} $$

    I was working with a vector and finally I transform it into a four-vector using: ##q^2 = -|\vec{p}|^2##

    $$ (m^2 - |\vec{q}|^2)^2 + 4|\vec{q}|^2m^2 = m^4 + |\vec{q}|^4 - 2m^2|\vec{q}|^2 + 4|\vec{q}|^2m^2 = m^4 + |\vec{q}|^4 + 2m^2|\vec{q}|^2 = (m^2 + |\vec{q}|^2)^2 = (m^2 - q^2)^2 $$

    Oh vela!! , you are really great!!

    Thank you very much. When I make my exam next week I´ll keep you in my mind.
     
  16. Jun 24, 2012 #15

    vela

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    The variable r shouldn't be in your final result you're integrating with respect to it.
     
  17. Jun 24, 2012 #16
    Ok. Thanks.
     
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