In the last post, we outline the calculation of knockout reaction. Here, we will give a detail calculation of the knockout reaction and compare with transfer reaction.

Suppose the reaction is A(a,12)B, where a = 1, and A = B + 2. Notice that the number 1, 2 are symbols and not represent any value.

The separation energy is

In A’s rest frame, the 4-momenta of the 5 particles are

Rearrange and set

set

where , simplify and get the effective orbital nucleon mass, that depends on the separation energy and momentum.

Thus, we transformed the knockout reaction into quasi 2-body reaction.

For 2-body reaction, we have to transform into the center of momentum frame. The 4-momentum of the center of momentum frame is

The Lorentz boost to the CM frame is

and

A general Lorentz transformation in vector form is

In the CM frame, the 4-momenta are

The total momentum should be zero,

divide both side by ,

The center of mass energy or total rest mass is

Starting from here, thing will be the same as 2-body transfer reaction. The momentum after scattering is

The scattered 4-momenta are

In here, the direction of the vector could be tricky, as the momenta are not parallel to z-axis.

Transform it back to A’s rest frame.

Or, we can use another Lorentz boost to the a’s rest frame by

but the form for and will keep in the same.

The difference from transfer reaction is that, the CM frame energy is a function of . So, it is not a reaction constant anymore. In summary,

Unwrapping,

The Lorentz boost can be either to A’s rest frame (normal kinematics)

or to a’s rest frame (inverse kinematics)