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Beam dynamics

  1. Dec 12, 2014 #1

    I have this problem - From a generator, I get a compton scattering with the electrons theta and phi angles.
    where I having the following equations for a particle

    px = E_particle * sin (theta) * cos (phi);
    py = E_particle * sin (theta) * sin (phi);
    pz = E_particle * cos (theta);

    where polar angle θ (theta), and azimuthal angle φ (phi). So, I am trying to build some kind of spherical-Cartesian transformation, as I need to have y'=dx/dy and x'=dz/dx in order to build the transfer matrix for beam dynamics (ie transport of the beam inside a magnet following thin lens approximation)

    Would that be possible to have such transformation ?


    Last edited: Dec 12, 2014
  2. jcsd
  3. Dec 12, 2014 #2
    Are you trying to express spherical coordinates in terms of the Cartesian coordinates? This is found in many sources.

    For example http://mathworld.wolfram.com/SphericalCoordinates.html
  4. Dec 12, 2014 #3
    Well, maybe I am missing something - The movement of the particles is along the z-axis (ie the reference obrit) and I need to calculate the y' = ds/dy and x'=ds/dx - So, the y' is actually the theta angle, but the phi angle accounts for the angle in the X-Y plane, while I would need the x' = the angle on the X-Z plane meaning I need the roll angle about the local s-axis...
  5. Dec 12, 2014 #4
    This is entirely derivable in terms of the angles. You are trying to calculate the ratio between the change in y (or x ) to the change in z. Since you are keeping the direction constant, the angles don't change and the change in the total distance ([itex]\Delta r[/itex]) cancels out. Use the relation between the coordinates given in the link to take the ratio and get the results.
  6. Dec 12, 2014 #5
    I guess you mean relation (95) ? And what happens to the dr ? for me I am starting with some energy / px, py,pz instead...
  7. Dec 12, 2014 #6
    The energy doesn't matter here. You have particles moving in some direction [itex] (\theta,\phi) [/itex] and want to calculate [itex] \frac{dx}{dz},\frac{dy}{dz} [/itex](or you can use s instead of z). Relation 95 is much more complicated than what you need since in your case [itex] (\Delta \theta,\Delta \phi) [/itex] are zero. From relations 4,5,6 you can get [itex] (\Delta x, \Delta y, \Delta z [/itex]) for [itex] (\Delta \theta = 0,\Delta \phi = 0) [/itex]. [itex] \Delta r [/itex] cancels out in the ratio and you get want your looking for. Notice that in the formulas in this link the roles of [itex] \theta [/itex] and [itex] \phi [/itex] are switched.
  8. Dec 12, 2014 #7


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    I would expect that to be y'=dy/ds and x'=dx/ds.
    With s=z (to a very good approximation outside magnets if I remember correctly), this is y'=dy/ds=py/pz.
    The energy cancels in this ratio.
  9. Dec 12, 2014 #8
    So, then I guess I need to calculate the actual derivative *not* just the ratio, right ?
  10. Dec 12, 2014 #9
    Yes, that is correct ie y'=dy/ds and x'=dx/ds as you care about the change of y/x on the direction of the movement ;-) So, when you say "outside of magnets" what do you mean ? and what about x' ? is there a similar approximation (and any reference would be great!)

    Thanks bunch!
  11. Dec 12, 2014 #10
    Since the ratio is constant it is the same.
  12. Dec 12, 2014 #11


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    (Dipole) magnets give a changing z-direction with s which is ugly, but should not matter as your compton scattering happens at a single point anyway.

    Same as for y', of course.
    Simple algebra, nothing you would find in a reference I guess. Books of beam dynamics should cover that somewhere.
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