## Momentum Matching in Transfer Reaction

In transfer reaction, the kinematics determines the scattering angles and momenta, but it does not tell the probability of the reaction. Not all angle have same cross section, because when a particle is being transfer to or pickup from a nucleus, the momentum has to be matched. It is like a spaceship need a proper angle and speed in order to orbit around the earth, it the incident angle or speed are not good, the spaceship will just go out or crash into the earth.

Suppose our reaction is $A(B,1)2$, $A$ is incident particle, $B$ is the target, $1, 2$ are the outgoing particles. The momentum transfer from $A$ via $2$ to $B$ or $1$ is

$\vec{q} = \vec{p}_2 - \vec{p}_A = \vec{p}_f - \vec{p}_i$

where $\vec{p}_i$ is initial momentum and $\vec{p}_f$ is final momentum.

Assume the reaction take place at the surface of nucleus $B$. The maximum angular momentum it can create is

$\vec{L} = \vec{r} \times \vec{q} \rightarrow L = r q$

In QM, the angular momentum must be $L = \sqrt{l(l+1)} \hbar$. Here we pause for unit conversion. In this blog, we usually using nuclear unit, that momentum in [MeV/c] and $\hbar = 197.327 \textrm{MeV fm/c}$, thus, to calculate the angular momentum, simple multiple the momentum in MeV/c and nuclear radius in fm. The radius of a nucleus is roughly $r = 1.25 A^{1/3} \textrm{fm}$.

So, if the momentum $\sqrt{(l+1)(l+2)} \hbar > r|q| > \sqrt{l(l+1)} \hbar$, the possible angular momentum is $\sqrt{l(l+1)} \hbar$ or the l-orbit.

In the transfer reaction, the equation for $\vec{p}_k$ and $\vec{p}_i$ are known. It is straight forward to calculate $r q / \hbar$, plot against scattering angle, and we look for the value for $\sqrt{l(l+1)}$.

Or, we can scale the y-axis by

$\displaystyle y' = \sqrt{\left(q\frac{r}{\hbar}\right)^2 + \frac{1}{4}} - \frac{1}{2}$

The new y-axis is in unit of $\sqrt{l(l+1)}$.

We can also plot the contour of momentum matching on the $p_k$ versus scatering angle. The momentum matching is

$\displaystyle q^2 = \frac{l(l+1)\hbar^2}{r^2} = p_f^2 + p_i^2 - 2 p_f p_i \cos(\theta)$

where $\theta$ is the scattering angle of particle $1$. Solve for $p_f$

$\displaystyle p_f = p_i \cos(\theta) \pm \sqrt{ \left( \frac{l(l+1) \hbar^2}{r^2} \right) - k_i^2 \sin^2(\theta) }$

We can plot the contour plot for difference $l$. and then overlap with the momentum and the scattering angle of the outgoing particle $2$.

a correction, the y-axis should be $p_2$, momentum of particle 2.

## Kinematics of Transfer Reaction

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

Suppose the reaction is $A(B, 1)2$, where $A$ is the incident particle, $B$ is the target nucleus that is at rest, $1, 2$ are the outgoing particle.

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

$Q = m_A+ m_B - m_1-m_2$

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

$P_A + P_B = P_1 + P_2$

where $P_i$ are 4-momenta. The number of freedom is 2. The total unknown is 8 from $P_1, P_2$, the masses of particle-1, and -2 give 2 constrains, the equations give 4 constrains.

$P_A = ( E_A , 0, 0, k_A) = ( m_A + T_A, 0, 0, \sqrt{2 m_A T_A + T_A^2})\\ P_B = (m_B, 0, 0, 0)$

where $T_A$ 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

$P_{cm} = (P_A+P_B)/2 = ( m_A+ m_B + T_A, 0, 0, k_A)$

The Lorentz beta is $\beta = k_A/(m_A+m_B+T_A)$.

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

$P_A' = ( \gamma E_A - \gamma \beta k_A, 0, 0, -\gamma \beta E_A + \gamma k_A) \\ P_B' = ( \gamma m_B , 0, 0, -\gamma \beta m_B )$

The total energy in CM frame is

$E_{tot} = \gamma( m_A+m_B + T_A) - \gamma \beta k_A = \sqrt{(m_A+m_B+T_A)^2 - k_A^2}$

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

$\displaystyle k = \frac{1}{2 E_{tot}} \sqrt{(E_{tot}^2 - (m_1+m_2)^2)(E_{tot}^2 - (m_1-m_2)^2)}$

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

$E_{tot}^2 - (m_1+m_2)^2 = (m_A+m_B+T_A)^2 - k_A^2 - (m_1+m_2)^2 \\ =(m_A+m_B+T_A)^2 - k_A^2 - (m_A+m_B-Q)^2 \\ = 2m_BT_A + 2(m_A + m_B)Q - Q^2$

If we set the line for this quantity be zero,

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

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

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

$P_1' = ( \sqrt{m_1^2 - k^2} , k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , k \cos(\theta)) \\ P_2' = ( \sqrt{m_2^2 - k^2} , -k \cos(\phi) \sin(\theta) ,- k \sin(\phi) \sin(\theta) , -k \cos(\theta))$

Back to Lab frame,

$P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta) \\ k \cos(\phi) \sin(\theta) \\ k \sin(\phi) \sin(\theta) \\ \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta)\end{pmatrix}$

For simplicity, set $\phi = 0$,

$P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta) \\ k \cos(\phi) \sin(\theta) \\ 0 \\ \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta) \end{pmatrix}$

We can see, the locus in the momentum space is

$( k cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta))$

This is an ellipse with length $k (1, \gamma)$, and centered at $(0, \gamma \beta \sqrt{m_1^2 + k^2})$. If the relativistic effect is small, it is a circle.

When the ellipse long axis $\gamma k$ is longer then the center position $\gamma \beta \sqrt{m_1^2 + k^2}$, the scattering angle can be to whole circle, i.e.

$\displaystyle \gamma \beta \sqrt{m_1^2 + k^2} < \gamma k \rightarrow \gamma \beta m_1 < k$

in here, we can treat $\gamma \beta m_1 = k_\beta(1)$, this is the momentum “gain” from shifting from CM frame to Lab frame. By compare $k_\beta(1)$ or $k_beta(2)$ to $k$ we can know the span of scattering angle.

Here is an example,