Another surprise that no post on the kinematics of transfer reaction!

Suppose the reaction is , where is the incident particle, is the target nucleus that is at rest, are the outgoing particle.

The Q-value for the reaction is the difference between initial mass and final mass

When the incident energy is too low, lower then the Q-value (we will see the relationship later) , the reaction is impossible, because the nuclei-1 and 2 cannot be created.

The equation for the reaction is

where are 4-momenta. The number of freedom is 2. The total unknown is 8 from , the masses of particle-1, and -2 give 2 constrains, the equations give 4 constrains.

where is the incident energy. We first transform the system into center of momentum (CM) frame. Since the total momentum is zero in the CM frame, we construct a CM 4-momentum

The Lorentz beta is .

The invariance mass of the CM 4-momentum is unchanged, which is equal to the total energy of the particles in the CM frame. We can check

The total energy in CM frame is

After the reaction, the momenta of the particle-1 and -2 are the same an opposite direction in the CM frame.Balancing the energy and momentum, the momentum is

In the equation, if either of the bracket is negative, the momentum cannot be formed. The smaller term is

If we set the line for this quantity be zero,

$latex 2m_BT_A + 2(m_A + m_B)Q – Q^2 = 0 \\ T_A = \frac{Q^2 – 2(m_A+m_B)}{2m_B} \\ = \frac{1}{2m_B} ( (Q-m_A-m_B)^2 – (m_A+m_B)^2) \\ = \frac{1}{2m_B} ( (m_1+m_2)^2 – (m_A+m_B)^2)$

Thus, if the is smaller then that value, the reaction is impossible.

The 4-momenta of the outgoing particles can be constructed.

Back to Lab frame,

For simplicity, set ,

We can see, the locus in the momentum space is

This is an ellipse with length , and centered at . If the relativistic effect is small, it is a circle.

Here is an example,