## Posters for Postdoc symposium

I made two posters, one for the experiment on 207Hg and the other is about real-time beam diagnostic.

## Argonne HELIOS spectrometer & its scientific success

I have a chance to give a small seminar in Hong Kong University. Here I attached the ppt as pdf file.

argonne helios spectrometer &amp; its scientific success

The talk is an overview of the HELIOS spectrometer in Argonne National Laboratory ( which I am mainly working on). This talk also help me what is done and what is not explored. Hope you will find it interesting.

## Updated android program – Rel.Cal.

A page for HELIOS kinematics was added in the previous program.

Click HERE to get the apl file

Here is the screenshot

The plot has a lot to improve, say

• auto-adjustment of the origin.
• Added a dot to indicate the current thetaCM
• able to change the line for constant thetaCM

## HELIOS – an inverse problem

When give incident energy, the transfer reaction (or 2-body reaction in general) in a magnetic field is a transform:

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

We already give the constant $E_x$ line and constant $\theta_{cm}$ line is previous post. Now, we study an inverse problem that, given the energy and z-position, or a point on the EZ plot, what is the excitation energy and center of mass angle?

The solution is

$\displaystyle \begin{pmatrix} E_x \\ \theta_{cm} \end{pmatrix} = \begin{pmatrix} - m_B + \sqrt{E_t^2 + m_b^2 - 2\gamma E_t (E - \alpha \beta z)} \\ \frac{\gamma \beta E - \alpha \gamma z}{ \sqrt{\gamma^2 (E-\alpha \beta z)^2 - m_b^2}} \end{pmatrix}$

The above solution is for infinite detector, which is not really practical.

Assume the rotation radius is much larger than the detector size $a$. The coupled equations are:

$\displaystyle E = \gamma E_{cm} - \gamma \beta p \cos(\theta_{cm})$

$\displaystyle \alpha \beta \gamma z = (\gamma E - E_{cm})\left( 1- \frac{1}{2\pi} \frac{a}{\rho} \right)$

$\displaystyle \rho = \frac{ p \sin(\theta_{cm}} {2\pi a}$

eliminate $\theta_{cm}$,

$\displaystyle \alpha \beta \gamma z = (\gamma E - E_{cm})\left( 1- \frac{\alpha \gamma \beta a}{\sqrt{2\gamma E E_{cm} - E^2 - m^2 \gamma^2 - p^2}} \right)$

Since $E_{cm}^2 = m^2 + k^2$

Change of variable

$\displaystyle k \rightarrow m \tan(x) , 0 < x < \pi/2$

$\displaystyle z = \frac{\gamma E - m \sec(x)}{\alpha \beta \gamma }\left( 1- \frac{\alpha \gamma \beta a}{\sqrt{2\gamma E m \sec(x) - E^2 - m^2 \gamma^2 - m^2 \tan^2(x)}} \right)$

The square root can be rearranged as,

$\displaystyle 2\gamma E m \sec(x) - E^2 - m^2 \gamma^2 - m^2 \tan^2(x) \\ = (E^2-m^2) \gamma^2 \beta^2 - (\gamma E - m \sec(x))^2 = H^2 - K^2$

Use

$\alpha \gamma \beta z \rightarrow Z \\ \beta \gamma \alpha a \rightarrow G$

We have,

$\displaystyle Z = K \left( 1 - \frac{G}{\sqrt{H^2 - K^2}} \right)$

Next, change or replace,

$\displaystyle K \rightarrow H \sin(\phi), -\frac{\pi}{2} < \phi < \frac{\pi}{2}$

$\displaystyle Z = H \sin(\phi) - G \tan(\phi)$

Since $H , G > 0$, and $G < H$, as

$\displaystyle \frac{G}{\sqrt{H^2 - K^2 }} = \frac{a}{2\pi\rho} < 1$

The function

$f(\phi) = H \sin(\phi) - G \tan(\phi)$

looks like the orange line below

The blue line is $f(\phi) = Z$. It seems that there are multi-solution for $\phi$. When $\theta_{cm} >> 0$,

$f'(\phi) = H \cos(\phi) - G \sec^2(\phi) > 0$

Thus, there is only one solution. However, when $\theta_{cm}$ is small, there is multi-solution. That need to be careful.

## HELIOS constant thetaCM line

somehow I confused the term “line”, because sometimes it means a “curve”, anyway~

There is a plot of the energy – z-pos plot (EZ-plot) in the last post. In the EZ plot, the constant excitation energy line is already given in the equation (3) in this post. And the finite-detector correction is presented in the last post. The constant $\theta_{CM}$ line is

$\displaystyle E = \frac{-\sin^2(\theta_{CM}) \alpha \beta \gamma^2 z + \cos(\theta_{CM})\sqrt{\alpha^2 z^2 + m_b^2 (1-\sin^2(\theta_{CM})\gamma^2)}}{1-\sin^2(\theta_{CM})\gamma^2}$

where $\alpha = \frac{cZB}{2\pi}$

you can guess/see that, this is a solution of a quadratic equation. In fact, from the equation (1) and (2) in this post, by eliminating the $\cos(\theta_{CM})$ in $\vec{\beta} \cdot \vec{p}$, we got the constant $E_x$ line. By eliminating the $m_B$ (the mass of the heavier particle, or the excited energy) in $E_{cm}$ (see $E_{tot}$ in this post ), we got the above equation.

When $\theta_{CM} = 0$, the equation reduces to

$\displaystyle E = \sqrt{\alpha^2 z^2 + m_b^2}$

OK, I know I skipped a lot of step, which is not my style. From the equation (1) and (2), eliminate $E_{cm}$ gives you,

$\displaystyle E = \frac{\alpha}{\beta}z - \frac{1}{\gamma}(\vec{\beta}\cdot\vec{p})$

it is a linear! But if you look closely, the center of mass momentum has explicit $E_x$ dependence, it is a constant of motion only when $E_x$ is constant. The higher the excitation energy, the less energy of the recoil energy $E$ of the proton! Thus, it is like $p = p(E_x) = p(E_x(E))$. A more easy way to eliminate $m_B$ or $E_x$, is expand $E_{cm}$ into

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

Thus, eliminating $m_B = (m_B)_{g.s.} + E_x$ by isolate $m_B$.

## Finite detector effect in HELIOS

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. $E_x \propto E_{cm}$. 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 $\phi_p$ 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 $\rho$, and the emittance angle of the particle is $\phi$.

If the particle hit back the z-axis, the $\Delta \phi = 2\pi$, but due to finite detector size, it will not complete a cycle.

The equation of the detector plane is

$\displaystyle x\cos(\phi_p)+ y\sin(\phi_p) = a$

The equation of the locus of the particle is

$\displaystyle \begin{pmatrix} x \\ y \end{pmatrix} = \rho \begin{pmatrix} \sin(\tan(\theta) \frac{z}{\rho} - \sin(\phi)) \\ \cos(\phi) - \cos(\tan(\theta) \frac{z}{\rho} \end{pmatrix}$

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

$\displaystyle z = \frac{\rho}{\tan(\theta)}\left( \phi_p - \phi + n\pi + (-1)^{n} \sin^{-1}\left( \frac{a}{\rho} + \sin(\phi-\phi_p) \right) \right), n=0,1,2,...$

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

$\displaystyle z_0 = 2\pi \frac{\rho}{\tan(\theta)}$

In the case when $\phi = 0$, $\phi_p = \pi$, and we want the particle hit from the outside, i.e. $n = 2m+1, m = 0, 1,2,..$, the solution becomes

$\displaystyle z = z_0 \left( 1 - \frac{1}{2\pi}\sin^{-1}\left( \frac{a}{\rho} \right) \right)$

When $\rho >> a$,

$\displaystyle z \approx z_0 \left( 1 - \frac{1}{2\pi} \frac{a}{\rho}\right)$

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

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

## 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.