Beam dynamics

  • Thread starter Alkass
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  • #1
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Hello

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 ?

Thanks

Alex
 
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Answers and Replies

  • #2
129
12
Hello

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'=dx/d 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 ?

Thanks

Alex
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
 
  • #3
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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...
 
  • #4
129
12
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...
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.
 
  • #5
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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 (Δr\Delta r) cancels out. Use the relation between the coordinates given in the link to take the ratio and get the results.
I guess you mean relation (95) ? And what happens to the dr ? for me I am starting with some energy / px, py,pz instead...
 
  • #6
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I guess you mean relation (95) ? And what happens to the dr ? for me I am starting with some energy / px, py,pz instead...
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.
 
  • #7
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y' = ds/dy and x'=ds/dx
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.
 
  • #8
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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.
So, then I guess I need to calculate the actual derivative *not* just the ratio, right ?
 
  • #9
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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.
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!
 
  • #10
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So, then I guess I need to calculate the actual derivative *not* just the ratio, right ?
Since the ratio is constant it is the same.
 
  • #11
34,039
9,881
So, when you say "outside of magnets" what do you mean ?
(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.

and what about x' ?
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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