I was very busy last month that I went to CERN to join two experiments. The experiments were using ISOLDE Solenoid Spectrometer, which is a HELIOS-type equipment. In fact, I planned to post a series about HELIOS spectrometer, mainly focus on the kinematics, to address different detail, like offset-beam effect, tilted-beam effect, the constant-thetaCM line, etc., but I was very busy, probably until Dec. Anyway~

The pervious post lay the foundation for HELIOS spectrometer. The energy and z-position is linear, that simplified a lot of things. And the Q-value or excitation energy is in the intercept and in most cases, can be linearly approximated, i.e. . We can derive the exact relation later.

In the above plot, 28Mg(d,p) at 9.43 MeV/u. The dashed line is for infinite detector, the red line is for finite detector, and the blue line is 0 deg center of mass angle. We can see, for smaller energy, the change of the hit-position is larger and non-linear. It is because at smaller energy, the scattered angle is small, and the particle hits the detector at very small angle ( with respect to the detector plane, not detector normal), so that the particle will be stopped much earlier.

Today, I will show the algebraic solution for finite detector effect.

Finite detector effect, from it name, because the detector has size, so the hit position is difference from the on-axis position. Below is the x-y plane view, the detector plane is the orange line, with distance from the z-axis of a and make an angle to the x-z plane. The blue line is the helix curve of the charged particle in the magnetic field. The radius of the helix curve is , and the emittance angle of the particle is .

If the particle hit back the z-axis, the , but due to finite detector size, it will not complete a cycle.

The equation of the detector plane is

The equation of the locus of the particle is

where, is the particle emittance with respect to the z-axis. Solving these two equation for , we have

where is the number of time that the particle crossed the detector plan. Note that, the length of a cycle is

In the case when , , and we want the particle hit from the outside, i.e. , the solution becomes

When ,

In transfer reaction, when the beam energy is fixed, the degree of freedom is 2: the excitation energy () and center of mass angle (. After the HELIOS, the observables are energy () and z-position (). With the finite detector effect, the mapping look like this

We can see, there are some regions that the manifold of is folded. So, there is no way to have an inverse transform to get back the and $latex_{cm}$ from the observables. Unless, we can, somehow, extrapolating the z-axis position.