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

P_{cm} = \begin{pmatrix} \sqrt{m^2 + p^2} = E_{cm} \\  \vec{p} \end{pmatrix}

the boost from CM frame to Lab frame is \vec{\beta} . Lets define the perpendicular vector on the plane of \vec{p} be \hat{n} , thus, the four-momentum in the Lab frame is

P = \begin{pmatrix} E \\ \vec{k} \end{pmatrix} =\begin{pmatrix} \gamma E_{cm} + \gamma \beta \hat{\beta}\cdot \vec{p} \\ \left(  \gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p}) \right) \hat{\beta} + (\hat{n} \cdot \vec{p} ) \hat{n} \end{pmatrix}


\displaystyle E_{cm} = \frac{1}{2E_t}\left( E_t^2 + m_b^2 -m_B^2 \right)

\displaystyle p^2 = \frac{1}{4E_t^2} \left(E_t^2 - (m_b + m_B)^2\right) \left(E_t^2 - (m_b - m_B)^2\right)

\displaystyle E_t^2 = (m_a+m_A)^2 + 2m_aT

\displaystyle \beta = \frac{\sqrt{(m_A+T)^2 - m_A^2}}{m_a + m_A + T}, \gamma = \frac{1}{\sqrt{1-\beta^2}}

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

Suppose the magnetic field is parallel to z-axis

\vec{B} = B \hat{z}

The rotation radius is

\displaystyle \rho = \frac{\vec{k} \cdot \hat{xy} }{cZB}

where Z is the charge state of the particle, and \hat{xy} is the perpendicular unit vector on xy-plane.


The time for 1 cycle is

\displaystyle T_{cyc} = \frac{2\pi \rho}{v_{xy}} = \frac{2\pi}{cZB} \frac{\vec{k}\cdot\hat{xy}}{v_{xy}} = \frac{2\pi}{c^2 ZB} E

And the length for 1 cycle is

\displaystyle z_{cyc} = v_z T_{cyc} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{xy} \frac{v_z}{v_{xy}} = \frac{2\pi}{cZB} \vec{k}\cdot \hat{z}

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

\vec{\beta} \cdot \hat{z} = 1, \hat{n} \cdot \hat{z} = 0

\displaystyle z_{cyc} = \frac{2\pi}{cZB} \left(\gamma \beta E_{cm} + \gamma (\hat{\beta} \cdot \vec{p})\right) ………………. (1)

and the total energy is

\displaystyle E = \gamma E_{cm} + \gamma \beta (\hat{\beta}\cdot \vec{p}) …………….. (2)

By eliminating the \hat{\beta} \cdot \vec{p} ,

\displaystyle E = \frac{1}{\gamma}E_{cm} + \frac{cZB}{2\pi} \beta z_{cyc} …………….. (3)

We can see that, the energy and z_{cyc} have a linear relation. That simplified a alot of thing. For instance, the intercept is related to E_{cm}, 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.