Generally, the best starting point of any kinematics calculation for any reaction is from the CM frame. In the exit channel, the four-momentum in the CM frame is

the boost from CM frame to Lab frame is . Lets define the perpendicular vector on the plane of be , thus, the four-momentum in the Lab frame is

where

We can see, the particle-A is moving with kinematic energy of , hit particle-a, result in particle-b and particle-B.

Suppose the magnetic field is parallel to z-axis

The rotation radius is

where is the charge state of the particle, and is the perpendicular unit vector on xy-plane.

The time for 1 cycle is

And the length for 1 cycle is

When the Lorentz boost is parallel to the B-field direction,

………………. (1)

and the total energy is

…………….. (2)

By eliminating the ,

…………….. (3)

We can see that, the energy and have a linear relation. That simplified a alot of thing. For instance, the intercept is related to , which is further related to the Q-value of the reaction. If we can plot the E-z plot and determine the intercept, we can extract the Q-value, which is related to the excitation energy or some sort.

Without the magnetic field, to extract the Q-value, we really need to know the scattering angle (the azimuthal angle can be skipped due to symmetry in some cases). And usually, the relation is not linear.

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