## 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.

## on Diagonalization (reminder)

since i don’t have algebra book on my hand, so, it is just a reminder, very basic thing.

for any matrix $M$ , it can be diagonalized by it eigenvalue $\lambda_i$  and eigen vector $v_i$, given that it eigenvectors span all the space. thus, the transform represented by the matrix not contractive, which is to say, the dimension of the transform space is equal to the dimension of the origin space.

Let denote, D before Diagonal matrix, with it elements are eigenvalues.

$D_{ij} = \lambda_i \delta_{ij}$

P be the matrix that collect the eigenvectors:

$P_{i j} = \left( v_i \right)_j = \begin {pmatrix} v_1 & v_2 & ... & v_i \end {pmatrix}$

Thus, the matrix $M$ is :

$M = P \cdot D \cdot P^{-1}$

there are some special case. since any matrix can be rewritten by symmetric matrix $S$ and anti-symmetric matrix $A$. so we turn our focus on these 2 matrices.

For symmetric matrix $S$, the transpose of $P$ also work

$S =P \cdot D \cdot P^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

which indicated that $P^T = P^{-1}$. it is because, for a symmetric matrix, $M = M^T$ ,  the eigenvalues are all different, then all eigenvector are all orthogonal, thus $P^T \cdot P = 1$.

For anti-symmetric matrix $A$

$A = P \cdot D \cdot P^{-1}$

since the interchange of row or column with corresponding exchange of eigenvalues in D still keep the formula working. Thus, the case $P = P^T$ never consider.

___________________________________________________________

For example, the Lorentz Transform

$L = \gamma \begin {pmatrix} \beta & 1 \\ 1 & \beta \end {pmatrix}$

which has eigenvalues:

$D = \gamma \begin {pmatrix} \beta-1 & 0 \\ 0 & \beta+1 \end {pmatrix}$

$P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end {pmatrix}$

the eigenvector are the light cone. because only light is preserved in the Lorentz Transform.

and it is interesting that

$L = P \cdot D \cdot P^{-1} = P^{-1} \cdot D \cdot P = P^T \cdot D \cdot (P^T)^{-1} = (P^T)^{-1} \cdot D \cdot P^T$

another example is the Rotation Matrix

$R = \begin {pmatrix} cos(\theta) & - sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}$

$D = \begin {pmatrix} Exp( - i \theta) & 0 \\ 0 & Exp(i \theta) \end {pmatrix}$

$P = \begin {pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$

the last example to give is the $J_x$ of the spin-½ angular momentum

$J_x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$D = \begin {pmatrix} -1 & 0 \\ 0 & 1 \end {pmatrix}$

$P = \begin {pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}$

## detail treatment on Larmor Precession and Rabi Resonance

a treatment on Larmor Precession and Rabi resonance

the pdf is a work on this topic. it goes through Larmor Precession and give example on spin-½ and spin-1 system.

then it introduce Density matrix and gives some example.

The Rabi resonance was treated by rotating frame method and using density matrix on discussion.

the last topic is on the relaxation.

the purpose of study it extensively, is the understanding on NMR.

the NMR signal is the transverse component of the magnetization.

## Natural unit

on Size, Energy and Unit, we know that the speed of light better be equal to 1, that simplify the equation of relativity.

$c = 1$

now, we impose 1 more things, the Reduced Planck constant,  $\hbar$ also set to 1. that simplify all equations with angular momentum or spin.

$\hbar = 1$

the Angular momentum:

$J^2 \left| l,m \right > = j(j+1) \hbar ^2 \left| l,m\right >$

$J_z \left|l,m \right> = m \hbar \left|l,m\right>$

now becomes :

$J^2 \left| l,m \right > = j(j+1) \left| l,m\right >$

$J_z \left|l,m \right> = m \left|l,m\right>$

when we want to calculate the real value, we can recover the $\hbar$ by considering the dimension. the reduced Planck constant has dimension

$[\hbar] = [kg][m^2][s^{-1}] = [energy][second]$

for example,

$E = \omega \hbar$

## Differential Cross Section

In nuclear physics, cross section is a raw data from experiment. Or more precisely differential cross section, which is some angle of the cross section, coz we cannot measure every scatter angle and the differential cross section gives us more detail on how the scattering going on.

The differential cross section (d.s.c.) is the square of the scattering amplitude of the scatter spherical wave, which is the Fourier transform of the density.

$d.s.c = |f(\theta)|^2 = Fourier ( \rho (r), \Delta p , r )$

Where the angle θ come from the momentum change. So, sometime we will see the graph is plotted against momentum change instead of angle.

By measuring the yield of different angle. Yield is the intensity of scattered particle. We can plot a graph of the Form factor, and then find out the density of the nuclear or particle.

However, the density is not in usual meaning, it depends on what kind of particle we are using as detector. For example, if we use electron, which is carry elected charge, than it can feel the coulomb potential by the proton and it reflected on the “density”, so we can think it is kind of charge density.

Another cross section is the total cross section, which is sum over the d.s.c. in all angle. Thus, the plot always is against energy. This plot give us the spectrum of the particle, like excitation energy, different energy levels.

## Special Relativity I

i just state the formula and the usage of it.

the basic equation is

$E^2 = (p c)^2 + (m c^2)^2$

where E is total energy, p is momentum

here we can see the advantage of using MeV as unit of mass. the equation is now further simplified by using MeV/c as momentum unit.

$E^2 = p^2 + m^2$

which is Pythagorean theorem!

the speed of the particle is from the formula

$\beta = \frac {v}{c} = \frac {p}{E}$

For example, proton mass is 940 MeV/c2, if we say an proton is moving at 94MeV, or a 94 MeV proton. we mean, the KINETIC Energy (K.E.) of proton is 94MeV. The total energy is

$E = m + K.E.$

Thus, a 94 MeV proton is moving at 41.7% of light speed. by using a right-angle triangle of base 10, side 11, and the hight is Square-Root 21.

another way around is, a proton at 90% speed of light, how much K.E. it has? which is around 3000 MeV or 3 GeV [Giga eV]