a GEANT4 Simulation


The GEANT4 program structure was borrow from the example B5. I found that the most confusing part is the Action. Before that, let me start with the main().

GEANT4 is a set of library and toolkit, so, to use it, basically, you add a alot GEANT4 header files on your c++ programs. And, every c++ program started with main(). I write the tree diagram for my simplified exampleB5,

 +- DetectorConstruction.hh
    +- Construct()
       +- HodoscopeSD.hh     // SD = sensitive detector
          +- HodoscopeHit.hh //information for a hit
          +- ProcessHits()   //save the hit information
       +- ConstructSDandField() //define which solid is SD
       +- ConstructMaterials()  //define material
+- PhysicsList.hh  // use FTFP_BERT, a default physics for high energy physics
+- ActionInitialization.hh
   +- PrimaryGeneratorAction.hh // define the particle gun
   +- RunAction.hh // define what to do when a Run start, like define root tree
   +- Analysis.h  // call for g4root.h
   +- EventAction.hh //fill the tree and show some information during an event


A GEANT4 program contains 3 parts: Detector Construction, Physics, and Action. The detector construction is very straight forward. GEANT4 manual is very good place to start. The physics is a kind of mystery for me. The Action part can be complicated, because there could be many things to do, like getting the 2ndary beam, the particle type, the reaction channel, energy deposit, etc…

Anyway, I managed to do a simple scattering simulation, 6Li(2mm) + 22Ne(60A MeV) scattering in vacuum. A 100 events were drawn. The detector is a 2 layers plastic hodoscope, 1 mm for dE detector, 5 mm for E detector. I generated 1 million events. The default color code is Blue for positive charge, Green for neutral, Red for negative charge. So, the green rays could be gamma or neutron. The red rays could be positron, because it passed through the dE detector.

Screenshot from 2016-01-30 00:34:34.png

The histogram for the dE-TOF isScreenshot from 2016-01-29 23:26:34.png

We can see a tiny spot on (3.15,140), this is the elastics scattered beam, which is 20Ne. We can see 11 loci, started from Na on the top, to H at the very bottom.

The histogram of dE-E

Screenshot from 2016-01-29 23:30:38.png

For Mass identification, I follow the paper T. Shimoda et al., Nucl. Instrum. Methods 165, 261 (1979).

Screenshot from 2016-01-30 00:06:02.png

I counted the 20Na from 0.1 billion beam, the cross section is 2.15 barn.


β-delayed proton emission branches in 43Cr

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DOI: 10.1103/PhysRevC.83.014306

β-delayed proton emission means, a positron emitted firstly, then proton emit. so, a proton converted to a neutron and a positron, then  a proton emit. as a result, the mass number reduced by 1 from lost 2 proton and gain 1 neutron. therefore, this process should be only happened in neutron-deficient nuclei.

Chromium-43 has atomic number 24, the neutron number is 19.

it was reported that a first time observation of β-delayed 3 proton decay in 45Fe. and the same decay was discovered in 43Cr in this paper. both process was recorded by an Imaging Time Projection Chamber.

the feature of this chamber is, it can capture the photos for decay process. in the paper, they shows a clear picture of 3-protons decay. and also, the measurement is very accurate. they can found 12 events of 3 protons decay among the total 12524 events of proton decay, which is about 1044 to 1. base on this precision, they deduced the relative branching ratio to be 91.8% for 1-proton case, 8.1% for 2 protons case, and 0.096% of 3 protons case.

by the chamber, they also recorded non-proton decay events, which may come from non-decay or β-deacy. the energy of β-decay is too small to detected.

However, they use the Maximum likelihood method to deduce the decay probability of β-decay. they found that it was 12%. since 2 β-decay has not been observed or either not possible, they deduced the absolute branching ratio.

a discussion on the extreme small branching ration of 3-protons decay was presented at the end of paper.













Method II (decay)

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there is another way to study nuclear physics, which is by observe the decay process.

there are 3 major decays, the alpha, beta and gamma. the alpha decay is an excited nucleus go to a lower energy state by emitting an Helium ( 2 protons and 2 neutrons) nucleus. This process change the nucleus constitution and make it lighter. the beta decay is an excited nucleus go to a lower energy state by emitting an electron ( or positron, or capture an electron). This process also change the nucleus constitution but the change in mass is very small. the gamma decay is an excited nucleus go to a lower energy state by emitting a photon ( or the light particle). this process does not change the nucleus constitution.

there are many other decays, like neutron, proton, and even fission can regard as decay. in general, decay is a process that a nucleus go to a lower energy state and become more stable. But other decays has too short for lifetime, so that after the earth was formed, they are almost gone. and only alpha, beta and gamma has long left time to survive.

Via a decay process, we can know the energy level, lifetime, parity of each level of a nucleus. These informations can help us to build nuclear models and theories to understand and predict nuclear properties, like the strong and weak force.

i will present a series about decay in future posts. watch out!

on angular momentum adding & rotation operator


the angular momentum has 2 kinds – orbital angular momentum L , which is caused by a charged particle executing orbital motion, since there are 3 dimension space. and spin S , which is an internal degree of freedom to let particle “orbiting” at there.

thus, a general quantum state for a particle should not just for the spatial part and the time part. but also the spin, since a complete state should contains all degree of freedom.

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

when we “add” the orbital angular momentum and the spin together, actually, we are doing:

J = L \bigotimes 1 + 1 \bigotimes S

where the 1 with L is the identity of the spin-space and the 1 with S is the identity of the 3-D space.

the above was discussed on J.J. Sakurai’s book.

the mathematics of L and S are completely the same at rotation operator.

R_J (\theta) = Exp( - \frac {i}{\hbar} \theta J)

where J can be either L or S.

the L can only have effect on spatial state while S can only have effect on the spin-state. i.e:

R_L(\theta) \left| s \right> = \left| s\right>

R_S(\theta) \left| x \right> = \left| x\right>

the L_z can only have integral value but S_z can be both half-integral and integral. the half-integral value of Sz makes the spin-state have to rotate 2 cycles in order to be the same again.

thus, if the different of L and S is just man-made. The degree of freedom in the spin-space is actually by some real geometry on higher dimension. and actually, the orbital angular momentum can change the spin state:

L \left| s \right> = \left | s' \right > = c \left| s \right>

but the effect is so small and

R_L (\theta) \left| s\right > = Exp( - \frac {i}{\hbar} \theta c )\left| s \right>

but the c is very small, but if we can rotate the state for a very large angle, the effect of it can be seen by compare to the rotation by spin.

\left < R_L(\omega t) + R_S(\omega t) \right> = 2 ( 1+ cos ( \omega ( c -1 ) t)

the experiment can be done as follow. we apply a rotating magnetic field at the same frequency as the Larmor frequency. at a very low temperature, the spin was isolated and T_1 and T_2 is equal to \infty . the different in the c will come up at very long time measurement and it exhibit a interference pattern.

if c is a complex number, it will cause a decay, and it will be reflected in the interference pattern.

if we find out this c, then we can reveal the other spacial dimension!


the problem is. How can we act the orbital angular momentum on the spin with out the effect of spin angular momentum? since L and S always coupled.

one possibility is make the S zero. in the system of electron and positron. the total spin is zero.

another possibility is act the S on the spatial part. and this will change the energy level.


an more fundamental problem is, why L and S commute? the possible of writing this

\left| \Psi \right> = \left| x,t \right> \bigotimes \left| s \right>

is due to the operators are commute to each other. by why?

if we break down the L in to position operator x and momentum operator p, the question becomes, why x and S commute or p and S commute?

[x,S]=0 ?

[p,S]=0 ?

[p_x, S_y] \ne 0 ?

i will prove it later.


another problem is, how to evaluate the Poisson bracket? since L and S is not same dimension. may be we can write the eigenket in vector form:

\begin {pmatrix} \left|x, t \right> \\ \left|s\right> \end {pmatrix}

i am not sure.



For any vector operator, it must satisfy following equation, due to rotation symmetry.

[V_i, J_j] = i \hbar V_k   run in cyclic


where J is rotation operator. but i am not sure is it restricted to real space rotation. any way, spin is a vector operator, thus

$latex [S_x, L_y] = i \hbar S_z = – [S_y, L_x] $

so, L, S is not commute.

On mass deficit

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The mass deficit is due to the potential energy loss.

But, why there are potential?

Coz there is a force associate with a potential.

Or in more correct way, the potential energy is due to the force. When 2 bodies in space has a attractive force, then it naturally go together. And the force accelerate the body and increase their kinetic energy.

Thus, we think the force created a field, which stored potential energy. When the force accelerates body, the potential energy converted into kinetic energy.

Imagine there are 2 bodies, A & B, Which at different location and have different force. If A is further away, when it comes at B, it has more kinetic energy than B, coz B is at rest at that position. Thus, further away, higher the potential.

Can thus potential be as high as infinite? In reality, because thing has surface, has size. The minimum distance between the 2 bodies is the sum of their radius. And if we sum up all potential, from the surface to infinity. We found that it is finite. And we like to set the potential at infinity is zero. Thus, it makes all attractive force has negative potential.

But, electric and positron do not has size, so, and electron and a positron can come together and the potential energy they can release is infinite! Since when they are apart infinite and the attractive force accelerate it, when they meet, their speed will be infinite!

Wait! The speed cannot be infinite, the highest speed is the speed of light and for an object has mass, it never move at the speed of light!

So, when electron and positron hit each other from infinity. However, relativity does not limit the energy and the kinetic energy. The result is, it can release infinite energy.

But at first, infinite exist. In reality, we can just approximate the “infinite” by far far away. When the potential change very little. But far far away is still very short compare to infinite. Thus, we don’t have infinite energy source.

In laboratory, we can accelerate electron and position at very high speed and contain several GeV. Thus when those particles annihilate, they release GeV energy due to the kinetic energy and their mass, which is just 0.000511GeV. Does not help much.


Larmor Precession (quick)

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Magnetic moment (\mu ) :

this is a magnet by angular momentum of charge or spin. its value is:

\mu = \gamma J

where J is angular momentum, and \gamma is the gyromagnetic rato

\gamma = g \mu_B

Notice that we are using natural unit.

the g is the g-factor is a dimensionless number, which reflect the environment of the spin, for orbital angular momentum, g = 1.

\mu_B is Bohr magneton, which is equal to

\mu_B = \frac {e} {2 m} for positron

since different particle has different mass, their Bohr magneton value are different. electron is the lightest particle, so, it has largest value on Bohr magneton.

Larmor frequency:

When applied a magnetic field on a magnetic moment, the field will cause the moment precess around the axis of the field. the precession frequency is called Larmor frequency.

the precession can be understood in classical way or QM way.

Classical way:

the change of angular momentum is equal to the applied torque. and the torque is equal to the magnetic moment  cross product with the magnetic field. when in classical frame, the angular momentum, magnetic moment, and magnetic field are ordinary vector.

\vec {\Gamma}= \frac { d \vec{J}}{dt} = \vec{\mu} \times \vec{B} = \gamma \vec {J} \times \vec{B}

solving gives the procession frequency is :

\omega = - \gamma B

the minus sign is very important, it indicated that the J is precessing by right hand rule when \omega >0 .

QM way:

The Tim dependent Schrödinger equation (TDSE) is :

i \frac {d}{d t} \left| \Psi\right> = H \left|\Psi\right>

H is the Hamiltonian, for the magnetic field is pointing along the z-axis.

H = -\mu \cdot B = - \gamma J\cdot B = -gamma B J_z = \omega J_z

the solution is

\left|\Psi(t) \right> = Exp( - i \omega t J_z) \left| \Psi(0) \right>

Thus, in QM point of view, the state does not “rotate” but only a phase change.

However, the rotation operator on z-axis is

R_z ( \theta ) = Exp( - i \frac {\theta}{\hbar} J_z )

Thus, the solution can be rewritten as:

\left|\Psi (t)\right> = R_z( \omega t) \left|\Psi(0)\right>

That makes great analogy on rotation on a real vector.

Hall effect

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It is a short review. for more detail, wiki is a good place.

The hall probe is perpendicular to the B field( pointing up) and have a current I passing through ( going forward ).

Due to the Lorentz force. The positron is moving to right and accumulate. The accumulating charge creates a electric force to the left to against further positron accumulate. The magnetic force will be balanced by the electric force. Due to the electric force, there is associated voltage across the hall probe. This voltage is called hall voltage.

F_B = e v B = V_H d

Where e is positron charge, v is speed of positron, B is the magnetic field, V_H is the Hall voltage and d is the distance across the hall probe.

The current I is

I = A n e v

Where A is the cross section area of the hall probe, n is density of the positron carrier, v is the positron velocity.


V_H = \frac { B}{ V n e } I

Where V is the volume of the hall probe. But the V n is equal to the total number of positron N.

V_H = \left ( \frac { B}{N e} \right ) I

Which is to say, the hall voltage is proportional to the magnetic field.