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Fraction of particles getting through?

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data

    An unstable ion of mass ##M##, energy ##E## emits a massless particle of energy ##E_\nu## at angle ##\theta##. In the rest frame of the ion, find ##E_\nu^*## and ##cos \theta^*##.

    Ions are now accelerated to ##\gamma=100## and a detector with radius ##r=20m## is placed ##D=200 km## away coaxially. Show that ##\cos \theta^* = \frac{1-\gamma^2 \theta^2}{1+\gamma^2 \theta^2 - \frac{\theta^2}{2}}## where ##\theta \simeq \frac{r}{D}##. Assuming in rest frame of ion, neutrinos are emitted equally in all directions, find the fraction of neutrinos that get through the detector.



    2. Relevant equations


    3. The attempt at a solution

    Part (a)
    By boosting to CM frame, we have
    [tex]E_\nu^* = \gamma E_\nu \left( 1 - \beta \cos \theta\right) [/tex]
    [tex]\cos \theta^* = \frac{\cos \theta - \beta}{1-\beta \cos \theta}[/tex]

    Part (b)
    Expanding small angles, we have
    [tex]\cos \theta^* = \frac{1-\gamma^2 \theta^2}{1+\gamma^2 \theta^2 - \frac{\theta^2}{2}}[/tex]

    How do I find the fraction of particles getting through? I did this in the rest frame:
    [tex]f = \frac{4\int_0^{\theta^*} \sin \theta d\theta \int_0^{\phi^*} d\phi}{4\pi} [/tex]
    [tex]f = \frac{1}{\pi} \left(1 - cos \theta^*\right) \theta^*[/tex]

    Do I divide this by ##\gamma## to account for time dilation to find rate of particles getting through?
     
  2. jcsd
  3. Jun 11, 2015 #2

    mfb

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    In (a), I think everything is supposed to be expressed with the given quantities.

    How did you get the expansion for small angles?

    Compare the solid angle (ion rest frame) where the neutrino hits the detector to the full solid angle.
    There is no rate to calculate.
     
  4. Jun 11, 2015 #3
    I expanded ##\beta(\gamma)## and ##\theta \simeq \frac{r}{D}##.

    So I can do everything in the ion's rest frame? Then the fraction is simply ##f = \frac{4\int_0^{\theta^*} \sin \theta d\theta \int_0^{\phi^*} d\phi}{4\pi}##.
     
  5. Jun 11, 2015 #4

    mfb

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    The ϕ integral is trivial, but the prefactor of 4 looks odd.
    Sure, that's where you have the isotropic emission.
     
  6. Jun 11, 2015 #5
    I used ##\int_{-\theta^*}^{\theta^*} = 2 \int_0^{\theta^*}##.
     
  7. Jun 11, 2015 #6

    mfb

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    θ does not get negative at all, or I misunderstood your coordinate system.
    (Even then, where does the other 2 come from?)
     
  8. Jun 11, 2015 #7
    The cylindrical detector is on axis with the beam. ##\theta \simeq \frac{r}{D}##. So I think it should be ##\int_{-\phi^*}^{\phi^*} d\phi \int_0^{\theta^*} d\theta##
     
  9. Jun 12, 2015 #8

    mfb

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    You need a really weird coordinate system to get negative θ. "The other side" should differ by pi in the angle ϕ, but not in θ.
     
  10. Jun 12, 2015 #9
    Here's how I understand it: ##\theta## is the angle from the centre of the 'square' to the edge, so the area of the path is ##(r \sin \theta d\phi) \times (r d\theta)##. If I do ##\int_0^\theta d\theta \int_0^\phi d\phi##, it only covers 1/4 of the patch?

    Untitled.png
     
  11. Jun 12, 2015 #10

    mfb

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    There is no square, the detector is a disk. With usual conventions for spherical coordinates, the angle goes from 0 to θmax or from -pi/2 to -pi/2+θmax, where the lower value is exactly in the center of the disk.
    What do you mean with "area of the path"?
     
  12. Jun 12, 2015 #11
    I meant the area of the shaded patch. My worry is that the angle ##\theta## is defined as the angle from the centre of the patch to the edge of the patch..and using limits (0,theta) and (0,phi) only covers 1/4 of the patch.
     
  13. Jun 12, 2015 #12

    mfb

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    That is the only reasonable definition.
    It covers 1/2, and if you integrate phi from 0 to 2 pi you cover everything.
     
  14. Jun 12, 2015 #13
    Surely the limits of ##\phi## cant be from to 2pi. Ok, so the limits for the fraction are: (-phi,phi) and (0,theta).
     
  15. Jun 13, 2015 #14

    mfb

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    -pi to pi is possible as well, it depends on the choice of your coordinate system which you still did not reveal. The result of the integral has to be 2 pi, the definition of the angle is not so important.
     
  16. Jun 13, 2015 #15
    I suppose that the origin is where the particles fly out, and the detector is situated ##200km## away.
     
  17. Jun 13, 2015 #16

    mfb

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    Yes that is obvious, but how do you define the angles phi and theta?
     
  18. Jun 13, 2015 #17
    Theta is the angle subtended from point of emission to edge of detector and phi is the same, but sideways.
    proxy.php?image=http%3A%2F%2Fs13.postimg.org%2Fyd3uqmyg7%2FUntitled.png
     
  19. Jun 13, 2015 #18

    mfb

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    That is a very unconventional and problematic coordinate system that leads to ugly integrals.
    Spherical coordinates are so much easier.
     
  20. Jun 13, 2015 #19
    Ok, how would the integrals with limits look like in the spherical coordinate system?
     
  21. Jun 13, 2015 #20

    mfb

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    Theta is the angle to the central axis (polar angle), phi is the azimuthal angle. Phi runs from 0 to 2 pi, theta from 0 to the maximal theta.
     
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