Absolute polarization measurement by elastic scattering

The magnitude of proton polarization can be measured by NMR technique with a reference. Because the NMR gives the free-induction decay signal, which is a voltage or current. For Boltzmann polarization using strong magnetic field and low temperature, the polarization can be calculated. However, when a reference point is not available, the absolute magnitude of proton polarization can be measured using proton-proton elastic scattering. The principle is the nuclear spin-orbital coupling. That creates left-right asymmetry on the scattering cross section.

Because of spin-orbital interaction:

$V_{ls}(r) = f(r) \vec{l} \cdot \vec{s} ,$

where $f(r)$ is the distance function, $\vec{l}$ is the relative angular momentum, $\vec{s}$ is the spin of the incident proton. In the following picture, the spin of the incident proton can be either out of the plane ($\uparrow$ ) or into the plan ($\downarrow$). When the proton coming above, the angular momentum is into the plane ($\downarrow$). The 4 possible sign of the spin-orbital interaction is shown. We can see, when the spin is up, the spin-orbital force repulses the proton above and attracts the proton below. That creates an asymmetry in the scattering cross section.

The cross section is distorted and characterized using analysing power $A_y$. Analyzing power is proportional to the difference between left-right cross-section. By symmetry (parity, time-reversal) consideration, $A_y = 1 + P sin(2\theta)$ (why?), in center of mass frame. In past post, the transformation between difference Lorentz frame. The angle in the $A_y$ has to be expressed in lab angle. The cross section and $A_y$ can be obtained from http://gwdac.phys.gwu.edu/ .

In scattering experiment, the number of proton (yield) is counted in left and right detectors. The yield should be difference when either proton is polarized. The yield is

$Y(\theta, \phi) = L \epsilon \sigma_0 (1 + cos(\phi)A_y(\theta) P) ,$

where $L$ is the luminosity, $\epsilon$ is the detector efficiency, $\sigma_0$ is the integrated cross-section of un-polarized beam and target of the detector, $P$ is the polarization of either the target or beam. When both target and the beam are polarized, the cross section is

$\sigma = \sigma_0 (1 + (P + P_T)A_y + P P_T C_yy),$

where $C_yy$ is spin-spin correlation due to spin-spin interaction of nuclear force.

Using the left-right yield difference, the absolute polarization of the target or the beam can be found using,

$\displaystyle A_y P = \frac{Y_L - Y_R}{Y_L + Y_R} ,$

where $Y_L = Y(\phi =0)$ and $Y_R = Y(\phi=\pi)$.

Cross section IV

A more microscopic view of the cross section is

$\displaystyle \frac{dN}{dS} \sigma \rho dt dA= dn$,

where $N$ is the number of incident particle, $S$ is the area of the beam, $\sigma$ is the total cross section, $\rho$ is the particle density in the target, $A$ is the area of the target, $t$ is the thickness of the target, and $n$ is the particle detected.

The cross section is a constant, so, after integration on the target area and thickness,

$\displaystyle \sigma = \frac{n}{N \rho t}$.

If the beam is not uniform, $dN/dS = f(S)$ is a function of $S$. The integration has to be careful.

Cross seciton of Coulomb Scattering

The coulomb scattering looks very easy, the formula of the differential cross section in CM frame is,

$\frac{d\sigma}{d\Omega} = (\frac{Z1 Z2 e}{4 E_{cm}})^2 \frac{1}{sin^4(\theta_{cm}/2)}$,

where $e = 1.44 MeV fm$ and $1 fm^2 = 10 mb$. The tricky point is, in most experiment, we are working in Laboratory frame that require frame transformation.

The relationship of the energy in CM frame and the energy in the Lab frame can be found by Lorentz transform, and use the total kinematic energy (both particle 1 and particle 2). In the CM frame, we can image we have a fixed virtual target on the center of mass, and there is only 1 object moving at energy of the total kinematic energy.

For example, we have a target of mass $m_2$, a projectile with mass $m_1$ and energy $T_1$, a classical energy in CM frame is

$E_{cm} = \frac{m_2}{m_1+m_2} T_1$

In fact the $E_{cm}$ has only 5% difference between relativistic and non-relativistic even up to 500 MeV

When calculating the integrated cross section, we can do it in the CM frame, but it is more intuitive to do it in the Lab frame. In this case, we need to transform the differential cross section from the CM frame to the Lab frame, mathematically, the transformation is done by a factor called Jacobian.

We can compare the result using the kinematic calculator in LISE++.

In the above plot, the blue line is the d.s.c. in CM frame, and the red line is d.s.c. of the 9Be in Lab frame. Jacobian was added, therefore, the zero degree d.s.c. of 9Be is larger than the 180 degree d.s.c. in the CM frame.

The grazing angle of the scattering, can be calculated by the shorted distance between the target and the projectile. In the Lab frame, the target is not fixed, so it is not easy to know the shortest distance. But in the CM frame, the virtual target is fixed, and we can calculate the distance using the $E_{cm}$ and $\theta_{cm}$.

Very short introduction to Partial-wave expansion of scattering wave function

In a scattering problem, the main objective is solving the Schrödinger equation

$H\psi=(K+V)\psi=E\psi$

where H is the total Hamiltonian of the scattering system in the center of momentum, K is the kinetic energy and V is the potential energy. We seek for a solution $\psi$,

$\displaystyle \psi_{k}^{+}(r)=e^{i\vec{k}\cdot \vec{r}}+f(\theta)\frac{e^{ikr}}{kr}$

The solution can be decomposed

$\displaystyle \psi_{k}^{+}(r)=R_{l}(k,r)Y_{lm}(\theta,\phi)=\frac{u_{l}(k,r)}{kr}Y_{lm}(\theta,\phi)$

The solution of $u_{l}(k,r)$ can be solve by Runge-Kutta method on the pdf

$\displaystyle \left(\frac{d^2}{d\rho^2} + 1 - \frac{l(l+1)}{\rho^2} \right)u_{l}(k,\rho)=U(\rho)u_{l}(k,\rho)$

where $\rho=kr, k=\sqrt{2\mu E}/\hbar, \mu=(m_1+m_2)/(m_1 m_2)$ and $U=V/E$.

For $U = 0$, the solution of $u_l$ is

$\displaystyle u_{l}(k,r)=\hat{j}_l(\rho) \xrightarrow{r\rightarrow \infty} \sin(r') = \frac{e^{ir'}-e^{-ir'}}{2i}$

where $r' = kr-l\pi/2$ and $\hat{j}_l$ is the Riccati-Bessel function. The free wave function is

$\displaystyle \phi_k(r)=e^{i\vec{k}\cdot\vec{r}}=\sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ikr}i^l (e^{ir'}-e^{-ir'})$

where $P_l(x)$ is the Legendre polynomial.

Note that, if we have Coulomb potential, we need to use the Coulomb wave instead of free wave, because the range of coulomb force is infinity.

For $U\neq 0$, the solution of $u_l(r can be found by Runge-Kutta method, where R is a sufficiency large that the potential $V$ is effectively equal to 0.  The solution of $u_l(r>R)$ is shifted

$\displaystyle u_{l}(k,r>R)=\hat{j}_l(\rho)+\beta_l \hat{n}_l(\rho) \xrightarrow{r\rightarrow \infty} \frac{1}{2i}(S_l e^{ir'}-e^{-ir'})$

where $S_l$ is the scattering matrix element, it is obtained by solving the boundary condition at $r = R$. The scattered wave function is

$\displaystyle \psi_k(r)=\sum\limits_{l=0} P_l(\cos(\theta)) (2l+1) i^l \frac{u_l(r)}{kr}$

put the scattered wave function and the free wave function back to the seeking solution, we have the $f(\theta)$

$\displaystyle f(\theta) = \sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ik} (S_l - 1)$

and the differential cross section

$\displaystyle \frac{d\sigma}{d\Omega}=|f(\theta)|^2$.

In this very brief introduction, we can see

• How the scattering matrix $S_l$ is obtained
• How the scattering amplitude $f(\theta)$ relates to the scattering matrix

But what is scattering matrix? Although the page did not explained very well, especially how to use it.

Angular distribution of Neutrons from the Photo-Disintegration of the Deuteron

this paper was written on 1949. at that time, deuteron just discovered 20 years. this paper presents a method on detecting the diffraction cross section of the neutron from a disintegrated deuteron by gamma ray of energy 2.76MeV. and by this, they found the photo-magnetic to photo-electric cross section ration. the ratio is 0.295 ± 0.036.

the photo-electric dipole transition and photo-magnetic dipole transition can both be induced by the gamma ray. Photo carry 1 angular momentum, the absorption of photon will excited the spherical ground state $^1S$ into $^3P$. the 2 mechanisms of the disintegrations results 2 angular distributions of the neutrons. by examine the angular distribution, they find out the ratio.

the photo-magnetic cross section is isotropic and the photo-electric cross section is follow a of a $sin^2$ distribution. the average intensity of neutron detected on a angle is:

$I(\gamma ) = \int_{\gamma_1}^{\gamma_2} {(a + b sin^2(\gamma)) sin(\gamma) d\gamma } / \int_{\gamma_1}^{\gamma_2} {sin(\gamma) d\gamma }$

where a is the contribution from the photo-magnetic interaction and b is from photo-electric interaction. and $\gamma_1$ and $\gamma_2$ are the angle span by the finite size of the target and detector. the integration is straight forward and result is:

$I(\gamma) = a+b( 1 - 1/3 ( cos^2(\gamma_1) + cos(\gamma_1) cos(\gamma_2) + cos^2 ( \gamma_2) )$

and the author guided us to use the ration of 2 angle to find the ration of a and b. and the ration of a and b is related to the probability of the magnetic to the electric effect by

$a/b = 2/3 \tau$

. and the photo-magnetic to photo-electric cross section ratio is:

$\tau/(\tau+1)$

the detector was described in detail on 4 paragraphs. basically, it is a cylindrical linear detector base on the reaction $B^{10} ( n,\alpha)Li^7$. it was surrounded by paraffin to slow down fast nuetrons.

on the target, which is heavy water, $D_2 O$, they use an extraordinary copper toriod or donut shape container. it is based on 3 principles:

• The internal scattering of neutron
• Departure from point source
• The angular opening of the γ – ray source

they place the γ – ray source along the axis of the toriod, move it along to create different scattering angle.

they tested the internal scattering of the inside the toriod and found that it is nothing, the toriod shape does not have significant internal scattering.

they test the reflection of neutron form surrounding, base on the deviation from the inverse-square law. and finally, they hang up there equipment about 27meters from the ground and 30 meters from buildings walls. (their apparatus’s size is around 2 meters. They measured 45, 60, 75 and 90 degree intensity with 5 degree angular opening for each.

Differential Cross Section III

a single diagram can illustrate everything — the relation between the yield ( the number of particle detected per second in the detector ) and the differential cross section.

the upper drawing is a big view, and the lower one is a zoom-in. the red-circles on the upper drawing are same as the lower one.

From the upper one, the number of particles that scatted by the target and go to direction (θ,φ) is :

$\frac {N_0}{S} \times ( \Delta \sigma \times N) = n$

the left hand side can be interpolated at follow:

$\frac{N_0}{S}$ is the number of particle per second per unit area, or the flux.

$\Delta \sigma \times N$ is the total area of  the cross section that deflect or scatter particle to direction (θ,φ).

The left hand side can also be viewed as :

$N_0 \times \frac { \Delta \sigma \times N } {S}$

where the fraction after multiplication is the chance of getting scattered to the direction (θ,φ). ( the requirement of the flux $N_0$ is in HERE. )

and the right hand side is the number of particle detected at direction (θ,φ).

since both $n$ and $\Delta \sigma$ depend on (θ,φ). thus we can differential it and get the angle dependent of these 2.

$\frac { N_0 N}{S} \frac {d \sigma}{d \Omega} = \frac {d n}{d \Omega}$

if we set the detector moving as radius R. thus the detector area is:

$D_A = R^2 d \Omega$

Therefore, the number of particles will be detected per second on the detector with some area (= Yield ) is:

$Y = \frac {d n}{d \Omega} R^2 d \Omega = \frac { N_o N}{S} R^2 \frac{ d \sigma}{d \Omega} d \Omega$

finish. Oh, the unit of the differential cross section is barn = $10^{-28} m$, recall that a nucleus radius is about $10^{-14} m$

the relation between the differential cross section to the nuclear potential was discussed on HERE.

Differential Cross Section II

Last time, the differential cross section discussion is based on quantum mechanics. This time, i try to explain it will out any math. so, that my mum ask me, i can tell her and make her understand. :)

in a scattering experiment, think about a target, say, a proton fixed in the center, it is positive charged. if another proton coming with some energy. it will get repelled, due to the repulsive nature of Coulomb force of same charge. it should be easy to understand, if the proton coming with high energy, it will get closer to the target, or even enter inside the target.

the repelling angle of the proton is not just depend on the energy it carry, but also on the impact parameter ( we usually call it b , but i like to call it r). the impact parameter is the shortest distance between the target and the line of the moving direction of the proton at long long away.

if the impact parameter is large, the proton miss the target. it almost cannot feel the target affection. thus, it go straight and unaffected. when the r is zero, it will hit the target head on head. and due to the repulsion. it will return back.  so, we can understand. the smaller the impact parameter, the deflection will be larger. since it can feel the force stronger.

For same impact parameter, the higher energy proton will have less deflection, since it travel faster, spend less time by the force, and the deflection get less.

Thus, we have an idea that the angle of deflection is high for small impact parameter and high energy. And most important, it only depends on these 2 factors and the effect from the target.

since our detector can only detect some small angle over some small area. So, we can place out detector on some angle, get the yield, and this is the name –  differential cross section come from.

Now, we have a uniform flow of particle with energy E. they will be deflected by the target and go to some angles. If we detect at the deflection angle, see how many particles ( the yield ) we can get in each angle. we can calculate back the effect of the target. For example, for a small angle, the particle get little defected, and this means the particle is from large impact parameter. for a large angle, the particles are from small impact parameter.

In some cases, the number of particle detected will be very high at some particular angle then others angles and this means, the cross section is large. and this means something interesting.

Moreover, don’t forget we can change the energy of the beam. for some suitable energy, the particle will being absorbed or resonance with the target. that given us low or high cross section on the energy spectrum.

( the graph is an unauthorized from the link: http://www.astm.org/Standards/E496.htm )

The above diagram is the differential cross section obtained from a Deuteron to a Tritium ( an isotope of Hydrogen with 2 neutrons and 1 proton) target, and the reaction change the Tritium into Helium and a neutron get out.

The reaction notation is

$X(a,b)Y$

where a is incident particle, X is target, b is out come particle, and Y is the residual particle.

the horizontal axis is detector angle at lab-frame. and the vertical axis is energy of Tritium. we always neglect the angle 0 degree, because it means no deflection and the particle does not “see” the target. at low energy, the d.c.s. is just cause by Coulomb force. but when the energy gets higher and higher, there is a peak around 60 degree. this peak is interesting, because it penetrated into the Helium and reveal the internal structure of it. it tells us, beside of the Coulomb force, there are another force inside. that force make the particle deflects to angle 60 degree. for more detail analysis, we need mathematic. i wish someday, i can explain those mathematics in a very simple way.

Therefore, we can think that, for higher energy beam, the size we can “see” will be smaller. if we think a particle accelerator is a microscope. higher energy will have larger magnification power. That’s why we keep building large and larger machines.