## Calculation of Knockout reaction

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 $S_p$ is

$\displaystyle S_p = m_B + m_2 - m_A$

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

$\displaystyle P_A + P_a = P_1 + P_2 + P_B$

$\displaystyle \begin{pmatrix} m_A \\ 0 \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix} + \begin{pmatrix} \sqrt{m_B^2 + k^2} \\ -\vec{k} \end{pmatrix}$

Rearrange and set $P_k = P_A - P_B$

$\displaystyle (P_A - P_B) + P_a = P_k + P_a = P_1 + P_2$

$\displaystyle \begin{pmatrix} m_A - \sqrt{m_B^2 + k^2} \\ \vec{k} \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix}$

set

$\displaystyle P_k = \begin{pmatrix} m_A - \sqrt{m_B^2 + k^2} \\ \vec{k} \end{pmatrix} = \begin{pmatrix} \sqrt{m_k^2 + k^2} \\ \vec{k} \end{pmatrix}$

where $m_A - \sqrt{m_B^2 + k^2} = \sqrt{m_k^2 + k^2}$, simplify and get the effective orbital nucleon mass, that depends on the separation energy and momentum.

$\displaystyle m_k(S, k) = \sqrt{m_A^2 + m_B^2 - 2m_A\sqrt{m_B^2 + k^2}}$

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

$\displaystyle \begin{pmatrix} \sqrt{m_k^2 + k^2} \\ \vec{k} \end{pmatrix} + \begin{pmatrix} m_a + T \\ k_a \hat{z} \end{pmatrix} = \begin{pmatrix} \sqrt{m_1^2 + k_1^2} \\ \vec{k}_1 \end{pmatrix} + \begin{pmatrix} \sqrt{m_2^2 + k_2^2} \\ \vec{k}_2 \end{pmatrix}$

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

$\displaystyle P_c = P_k + P_a = \begin{pmatrix} \sqrt{m_k^2 + k^2} + m_a + T \\ \vec{k} + k_a \hat{z} \end{pmatrix}$

The Lorentz boost to the CM frame is

$\displaystyle \vec{\beta} = \frac{\vec{k} + k_a \hat{z}}{ \sqrt{m_k^2 + k^2} + m_a + T}$

and

$\displaystyle \gamma = \frac{1}{\sqrt{1-\beta^2}}$

A general Lorentz transformation in vector form is

$\displaystyle \begin{pmatrix} E' \\ \vec{k}' \end{pmatrix} = \begin{pmatrix} \gamma E \pm \gamma \vec{\beta}\cdot \vec{k} \\ \pm \gamma E \vec{\beta} + \vec{k} + (\gamma -1) (\hat{\beta}\cdot \vec{k}) \hat{\beta} \end{pmatrix}$

In the CM frame, the 4-momenta are

$\displaystyle P_k' = \begin{pmatrix} \gamma \sqrt{m_k^2 + k^2} - \gamma \vec{\beta} \cdot \vec{k} \\ - \gamma \sqrt{m_k^2+k^2} \vec{\beta} + \vec{k} + (\gamma -1 ) (\hat{\beta}\cdot \vec{k}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_a' = \begin{pmatrix} \gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z} \\ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta} \end{pmatrix}$

The total momentum should be zero,

$\displaystyle \vec{k}_c = - \gamma (m_a + T + \sqrt{m_k^2+k^2} )\vec{\beta} + \vec{k} + k_a\hat{z} + (\gamma -1 ) (\hat{\beta}\cdot ( \vec{k} + k_a \hat{z})) \hat{\beta}$

divide both side by $m_a + T + \sqrt{m_k^2+k^2} = E$,

$\displaystyle \frac{1}{E}\vec{k}_c = - \gamma \vec{\beta} + \vec{\beta} + (\gamma -1 ) (\hat{\beta}\cdot \vec{\beta} ) \hat{\beta} = 0$

The center of mass energy or total rest mass is

$\displaystyle E_c = M_c = \gamma (m_a + T + \sqrt{m_k^2 + k^2} ) - \gamma \vec{\beta} \cdot (\vec{k} + k_a \hat{z} ) \\ = \gamma E ( 1 - \beta^2) = \frac{E}{\gamma}$

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

$\displaystyle p = \frac{1}{2E_c} \sqrt{(E_c^2 - (m_1+m_2)^2)(E_c^2-(m_1-m_2)^2)}$

The scattered 4-momenta are

$\displaystyle P_1' = (\sqrt{m_1^2+p^2}, \vec{p})$
$\displaystyle P_2' = (\sqrt{m_2^2+p^2}, -\vec{p})$

In here, the direction of the vector $\vec{p}$ could be tricky, as the momenta are not parallel to z-axis.

Transform it back to A’s rest frame.

$\displaystyle P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 +p^2} + \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_1^2 +p^2} \vec{\beta} + \vec{p} + (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_2 = \begin{pmatrix} \gamma \sqrt{m_2^2 +p^2} - \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_2^2 +p^2} \vec{\beta} - \vec{p} - (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

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

$\displaystyle \vec{\beta}_{ca} = \frac{ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta}}{\gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z}}$

but the form for $P_1$ and $P_2$ will keep in the same.

The difference from transfer reaction is that, the CM frame energy $E_c$ is a function of $S_p, \vec{k}$. So, it is not a reaction constant anymore. In summary,

$\displaystyle P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 +p^2} + \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_1^2 +p^2} \vec{\beta} + \vec{p} + (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

$\displaystyle P_2 = \begin{pmatrix} \gamma \sqrt{m_2^2 +p^2} - \gamma \vec{\beta}\cdot \vec{p} \\ \gamma \sqrt{m_2^2 +p^2} \vec{\beta} - \vec{p} - (\gamma -1) (\hat{\beta}\cdot \vec{p}) \hat{\beta} \end{pmatrix}$

Unwrapping,

$\displaystyle p = \frac{1}{2E_c} \sqrt{(E_c^2 - (m_1+m_2)^2)(E_c^2-(m_1-m_2)^2)}$

$\displaystyle E_c = M_c = \frac{E}{\gamma}$

$\displaystyle E = m_a + T + \sqrt{m_k^2+k^2} = m_A + m_a + T - \sqrt{m_B^2 + k^2}$

$\displaystyle S_p = m_B + m_2 - m_A$

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

$\displaystyle \vec{\beta} = \frac{\vec{k} + k_a \hat{z}}{ \sqrt{m_k^2 + k^2} + m_a + T} = \frac{\vec{k} + k_a \hat{z}}{E}$

or to a’s rest frame (inverse kinematics)

$\displaystyle \vec{\beta}_{ca} = \frac{ - \gamma (m_a + T)\vec{\beta} + k_a\hat{z} + (\gamma -1 ) k_a (\hat{\beta}\cdot \hat{z}) \hat{\beta}}{\gamma (m_a+T) - \gamma k_a \vec{\beta} \cdot \hat{z}}$

$\displaystyle \gamma = \frac{1}{\sqrt{1-\beta^2}}$

## Kinematics of Knockout reaction

It is super surprise for me that there is no any calculation for knockout kinematics in this blog!!! Part of my PhD is about (p,2p) knockout reaction!

Assume the reaction is A(a,cd)B, so that A = B+b, and b becomes d after the reaction. The separation energy of b is $S_p$,

$S_p + m_A = m_d + m_B$

The master equation for the kinematic calculation is

$P_A+P_a = P_c+P_d +P_B$

where $P_i$ is the 4-momentum of particle i. This equation is true for any reference frame.

We can rearrange the master equation

$(P_A - P_B) +P_a = P_b + P_a = P_c+P_d$

Here $P_b$ is the 4-momentum of the to-be-knockout particle b.  Let expand the master equation in A’s rest frame (or normal kinematic).

$\displaystyle \begin{pmatrix} m_A \\ 0 \end{pmatrix} + \begin{pmatrix} m_a \gamma \\ m_a\gamma \beta \hat{z} \end{pmatrix} = \begin{pmatrix} E_c \\ \vec{k}_c \end{pmatrix} + \begin{pmatrix} E_d \\ \vec{k}_d \end{pmatrix} + \begin{pmatrix} \sqrt{m_B^2+k^2} \\ -\vec{k} \end{pmatrix}$

where $\beta$ is the Lorentz boost.

$\displaystyle P_b = P_A-P_B = \begin{pmatrix} m_a - \sqrt{m_B^2+k^2} \\ \vec{k} \end{pmatrix}$

Since the introduction of the vector $\vec{k}$ for the to-be-knockout particle, 3 extra degree of freedoms : $|\vec{k}|, \theta_k, \phi_k$

To calculate the 4-momenta particle c and d, transform to the NN-frame so that the total momentum is ZERO, by setting

$P_c = P_b + P_a$

extract the Lorentz boost to the NN-frame $\vec{\beta}_c = \vec{k}_c/E_c$. In the NN-frame, the incident channel is particle a and b, the exist channel is particle c and d. In this frame, introduce 2 degree of freedom $\theta_{NN}, \phi_{NN}$ for the scattering angle. The problem becomes a 2-body scattering problem.

The total energy, or the total mass of the NN-frame is

$M_c = E_{NN} = \sqrt{E_c^2 - |\vec{k}|^2}$

The momentum of particle c and d is

$\displaystyle |\vec{p}| = p = \frac{1}{2M_c} \sqrt{(M_c^2 - (m_b +m_c)^2)(M_c^2 - (m_b - m_c)^2)}$

In this frame, the normal vectors of a and b are not necessary to be on the z-axis (the beam axis), but at some angle. Thus, the calculation of normal vectors of c and d has to be careful. After the calculation, transform the 4-momenta in to the desire frame.

Experimentally, we measure the energy and momentum of the scattered particle c and d and usually ignored the momentum of the particle B. How to calculate the separation energy ? There are two methods. the first one is calculate the 4-momenta of particle A, a, c, and d. And using

$S_p = m_B + m_d - m_A$

$m_B^2 = (E_A + E_a - E_c -E_d)^2 - |\vec{k}_A+\vec{k}_a-\vec{k}_c-\vec{k}_d|^2$

And 2nd method, particularly for inverse kinematic, in Lab’s frame, the sum of 4-momenta of c and b is

$\displaystyle P_c + P_d = \begin{pmatrix} mc + T_c + m_d + T_d \\ \vec{k}_c + \vec{k}_d \end{pmatrix}$

In A’s rest frame, the Lorentz boost is $-\beta$, the energy part is

$(P_c'+P_d)_E = \gamma(m_c + T_c + m_d + T_d) - \gamma \beta (\vec{k}_c + \vec{k}_d)\cdot \hat{z}$

also

$(P_c'+P_d')_E = (P_A' + P_a' - P_B')_E = m_A + \gamma m_a - E_B$

using

$\displaystyle S_p = m_B + m_d - m_A = \sqrt{E_B^2 - |\vec{k}_b|^2} + m_d - m_A$

to replace $m_A$

$(P_c'+P_d')_E = \sqrt{E_B^2 - |\vec{k}_b|^2} + m_d - S_p + \gamma m_a - E_B$

approximate

$\displaystyle \sqrt{E_B^2 - |\vec{k}_b|^2} - E_B \approx - \frac{|\vec{k}_b|^2}{2E_B}$

put everything together,

$\displaystyle S_p = m_d + \gamma(m_a - m_c -m_d) - \gamma(T_c + T_d) + \gamma \beta (\vec{k}_c + \vec{k}_d)\cdot \hat{z} - \frac{|\vec{k}_b|^2}{2E_B}$

The last term is safe to be neglected as the orbital momentum is usually ~30 MeV/c and the total energy of particle B is close to the mass, which is so much better.

For (p,2p) reaction. $m_a = m_b = m_c = m_p$

$\displaystyle S_p \approx (1-\gamma)m_p - \gamma(T_1 + T_2) + \gamma \beta (\vec{k}_1 + \vec{k}_2)\cdot \hat{z}$

## HELIOS for (p,2p) experiment

The HELIOS concept is kinematics compression using a magnetic field, so that all charged particles spiral back to the z-axis. It is particularly useful in a special 2-body reaction where the degree of freedom is 2, which are center of mass angle $\theta_{cm}$ and the excitation energy of the heavier recoil $E_x$. These 2 degree of freedom are mapped into the experimental space of the light recoil energy $T$ and axial position of the light recoil $z$. i.e.

$\displaystyle \begin{pmatrix} \theta_{cm} \\ E_x \end{pmatrix} \rightarrow \begin{pmatrix} T \\ z \end{pmatrix}$

In (p,2p) reaction, the degree of freedom is 6, which are the momentum of the to-be-knockout proton $(k_b, \theta_b, \phi_b)$, the NN-scattering angle $\theta_{NN}, \phi_{NN}$, and the separation energy of the to-be-knockout proton $S_p$. Because of this, the scattered proton energy and axial position will be washed out. Below is an example of (p,2p) reaction. We can see that because of the orbital motion of the to-be-knockout proton, the scattered energy and angle of the scattered protons can have many value.

Below is the energy vs z plot for a single proton. We can see that the usual straight line becomes diffused.

However, if we measure the two protons at the same time, and sum up the energy and position, a magic appear.

This is because the sum of energy and sum of angle is correlated even without magnetic field. When the energies summed up, the NN scattering angles are gone. This left 4 degree of freedom. (the following has to be checked.) And the axial position is related to the scattering angles, and they related to the angles of the to-be-knockout proton. Thus, only $k_b$ and $S_p$ is left.

In fact, the separation energy can be calculated by

$\displaystyle S_p = (1-\gamma) m_p - \gamma(T_1 + T_2) + \beta \gamma (\vec{k_1} + \vec{k_2})\cdot \hat{z} - \frac{k_b^2}{2 E_B}$

the term $\vec{k}\cdot \hat{z} = \frac{cB}{2\pi} z_{cyc}$. Therefore, the sum of proton energies is linearly related to the sum of axial positions.