Difference between revisions of "Simulations of Particle Interactions with Matter"

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== Overview ==
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==Class Admin==
  
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=== Particle Detection ===
 
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A device detects a particle only after the particle transfers energy to the device.
 
  
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Energy intrinsic to a device depends on the material used in a device
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[[TF_SPIM_ClassAdmin]]
  
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Some device of material with an average atomic number (<math>Z</math>) is at some temperature (<math>T</math>).  The materials atoms are in constant thermal motion (unless T = zero degrees Klevin).
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== Homework Problems==
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[[HomeWork_Simulations_of_Particle_Interactions_with_Matter]]
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Statistical Thermodynamics tells us that the canonical energy distribution of the atoms is given by the Maxwell-Boltzmann statistics such that
 
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<math>P(E) = \frac{1}{kT} e^{-\frac{E}{kT}}</math>
 
−
 
 
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<math>P(E)</math> represents the probability of any atom in the system having an energy <math>E</math> where
 
−
 
 
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<math>k= 1.38 \times 10^{-23} \frac{J}{mole \cdot K}</math>
 
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Note:  You may be more familiar with the Maxwell-Boltzmann distribution in the form
 
−
 
 
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<math>N(\nu) = 4 \pi N \left ( \frac{m}{2\pi k T} \right ) ^{3/2} v^2  e^{-mv^2/2kT}</math>
 
−
 
 
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where <math>N(v) \Delta v</math> would represent the molesules in the gas sample with speeds between <math>v</math> and <math>v + \Delta v</math>
 
−
 
 
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==== Example 1: P(E=5 eV) ====
 
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;What is the probability that an atom in a 12.011 gram block of carbon would have and energy of 5 eV?
 
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First lets check that the probability distribution is Normailized; ie: does <math>\int_0^{\infty} P(E) dE =1</math>?
 
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<math>\int_0^{\infty} P(E) dE = \int_0^{\infty} \frac{1}{kT} e^{-\frac{E}{kT}} dE = \frac{1}{kT} \frac{1}{\frac{1}{-kT}} e^{-\frac{E}{kT}} \mid_0^{\infty} = - [e^{-\infty} - e^0]= 1</math>
 
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<math>P(E=5eV)</math> is calculated by integrating P(E) over some energy interval ( ie:<math> N(v) dv</math>).  I will arbitrarily choose 4.9 eV to 5.1 eV as a starting point.
 
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<math>\int_{4.9 eV}^{5.1 eV} P(E) dE =  - [e^{-5.1 eV/kT} - e^{4.9 eV/kT}]</math>
 
−
 
 
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<math>k= (1.38 \times 10^{-23} \frac{J}{mole \cdot K} )  =  (1.38 \times 10^{-23} \frac{J}{mole \cdot K} )(6.42 \times 10^{18} \frac{eV}{J})= 8.614 \times 10^{-5} \frac {eV}{mole \cdot K}</math>
 
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assuming a room empterature of <math>T=300 K</math>
 
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then<math>kT = 0.0258 \frac{eV}{mole}</math>
 
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and
 
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<math>\int_{4.9 eV}^{5.1 eV} P(E) dE =  - [e^{-5.1/0.0258} - e^{4.9/0.0258}] = 4.48 \times 10^{-83} - 1.9 \times 10^{-86} \approx 4.48 \times 10^{-83}</math>
 
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or in other words the precise mathematical calculation of the probability may be approximated by just using the distribution function alone
 
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<math>P(E=5eV) = e^{-5/0.0258} \approx 10^{-85}</math>
 
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This approximation breaks down as <math>E \rightarrow 0.0258 eV</math>
 
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Since we have 12.011 grams of carbon and 1 mole of carbon = 12.011 g = <math>6 \times 10^{23} </math>carbon atoms
 
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We do not expect to see a 5 eV carbon atom in a sample size of <math>6 \times 10^{23} </math> carbon atoms when the probability of observing such an atom is <math>\approx 10^{-85}</math>
 
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The energy we expect to see would be calculated by
 
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<math><E> = \int_{0}^{\infty} E \cdot P(E) dE</math>
 
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If you used this block of carbon as a detector you would easily notice an event in which a carbon atom absorbed 5 eV of energy as compared to the energy of a typical atom in the carbon block.
 
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----
 
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;Silicon detectors and Ionization chambers are two commonly used devices for detecting radiation.
 
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approximately 1 eV of energy is all that you need to create an electron-ion pair in Silicon
 
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<math>P(E=1 eV) = e^{-1/0.0258} \approx 10^{-17}</math>
 
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approximately 10 eV of energy is needed to ionize an atom in a gas chamber
 
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<math>P(E=10 eV) = e^{-10/0.0258} \approx 10^{-169}</math>
 
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−
 
 
−
 
 
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The low probability of having an atom with 10 eV of energy means that an ionization chamber would have a better Signal to Noise ratio (SNR) for detecting 10 eV radiation than a silicon detector
 
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But if you cool the silicon detector to 200 degrees Kelvin (200 K) then
 
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<math>P(E=1 eV) = e^{-1/0.0172} \approx 10^{-26} << 10^{-17}</math>
 
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So cooling your detector will slow the atoms down making it more noticable when one of the atoms absorbs energy.
 
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also, if the radiation flux is large, more electron-hole pairs are created and you get a more noticeable signal.
 
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Unfortunately, with some detectore, like silicon, you can cause radiation damage that diminishes it's quantum efficiency for absorbing energy.
 
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=== The Monte Carlo method ===
 
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; Stochastic
 
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: from the greek word "stachos"
 
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: a means of, relating to, or characterized by conjecture and randomness.
 
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A stochastic process is one whose behavior is non-deterministic in that the next state of the process is partially determined.
 
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Quantun Mechanics is perhaps the best example such a non-deterministic systems.  The canonical systems in Thermodynamics is another example.
 
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Basically the monte-carlo method uses a random number generator (RNG) to generate a distribution (gaussian, uniform, Poission,...) which is used to solve a stochastic process based on an astochastic description.
 
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==== Example 2 Calculation of <math>\pi</math>====
 
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;Astochastic description:
 
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: <math>\pi</math> may be measured as the ratio of the area of a circle of radius <math>r</math> divided by the area of a square of length <math>2r</math>
 
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[[Image:PI_from_AreaRatio.jpg]]<math>\frac{A_{circle}}{A_{square}} = \frac{\pi r^2}{4r^2} = \frac{\pi}{4}</math>
 
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You can measure the value of <math>\pi</math> if you physically measure the above ratios.
 
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; Stochastic description:
 
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: Construct a dart board representing the above geometry, throw several darts at it, and look at a ratio of the number of darts in the circle to the total number of darts thrown (assuming you always hit the dart board).
 
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; Monte-Carlo Method
 
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:Here is an outline of a program to calulate <math>\pi</math> using the Monte-Carlo method with the above Stochastic description
 
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[[Image:MC_PI_fromAreaRatio.jpg]]
 
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begin loop
 
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  x=rnd
 
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  y=rnd
 
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  dist=sqrt(x*x+y*y)
 
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  if dist <= 1.0 then numbCircHits+=1.0
 
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  numbSquareHist += 1.0
 
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end loop
 
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  print PI = 4*numbCircHits/numbSquareHits
 
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=== A Unix Primer ===
 
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To get our feet wet using the UNIX operating system, we will try to solve example 2 above using a RNG under UNIX
 
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==== List of important Commands====
 
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# ls
 
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# pwd
 
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# cd
 
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# df
 
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# ssh
 
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# scp
 
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# mkdir
 
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# printenv
 
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# emacs, vi, vim
 
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# make, gcc
 
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# man
 
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# less
 
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# rm
 
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−
----
 
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Most of the commands executed within a shell under UNIX have command line arguments (switches) which tell the command to print information about using the command to the screen.  The common forms of these switches are "-h", "--h", or "--help"
 
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ls --help
 
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ssh -h
 
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'' the switch deponds on your flavor of UNIX''
 
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if using the switch doesn;t help you can try the "man" (sort for manual) pages (if they were installed). 
 
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Try
 
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man -k pwd
 
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the above command will search the manual for the key word "pwd"
 
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==== Example 3: using UNIX ====
 
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Step
 
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# login to inca.<br> [http://physics.isu.edu/~tforest/Classes/NucSim/Day1/XwindowsOnWindows.html click here for a description of logging in if using windows]
 
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# mkdir src
 
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# cd src
 
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# cp -R ~tforest/NucSim/Day1 ./
 
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# ls
 
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# cd Day1
 
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# make
 
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#./rndtest
 
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[http://physics.isu.edu/~tforest/Classes/NucSim/Day1/RNG/Marsaglia/noviceExample/ Here is a web link to the source files you can copy in case the above doesn't work]
 
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=== A Root Primer ===
 
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==== Example 1: Create Ntuple  and Draw Histogram====
 
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=== Cross Sections ===
 
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==== Definitions ====
 
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;Total cross section
 
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:<math>\sigma</math> = <math>\equiv \frac{\# particles\; scattered} {\frac{ \# incident \; particles}{Area}}</math>
 
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;Differential cross section
 
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:<math>\sigma(\theta)</math> = <math> \frac{d \sigma}{d \Omega} \equiv \frac{\frac{\# particles\; scattered}{solid \; angle}} {\frac{ \# incident \; particles}{Area}}</math>
 
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; Solid Angle
 
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:[[Image:SolidAngleDefinition.jpg]]
 
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: <math>\Omega</math>= surface area of a sphere covered by the detector
 
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: ie;the detectors area projected onto the surface of a sphere
 
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:A= surface area of detector
 
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:r=distance from interaction point to detector
 
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:<math>\Omega = \frac{A}{r^2} </math>sterradians
 
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: <math>A_{sphere} = 4 \pi r^2</math> if your detector was a hollow ball
 
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:<math>\Omega_{max} = \frac{4 \pi r^2}{r^2} = 4\pi</math>sterradians
 
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;Units
 
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:Cross-sections have the units of Area
 
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:1 barn = <math>10^{-28} m^2</math>
 
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; [units of <math>\sigma(\theta)</math>] =<math>\frac{\frac{[particles]}{[sterradian]}} {\frac{ [ particles]}{[m^2]}} = m^2</math>
 
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[[Image:FixedTargetScatteringCrossSection.jpg]]
 
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; Fixed target scattering
 
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: <math>N_{in}</math>= # of particles in = <math>I \cdot A_{in}</math>
 
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:: <math>A_{in}</math> is the area of the ring of incident particles
 
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:<math>dN_{in} = I \cdot dA = I (2\pi b) db</math>= # particles in a ring of radius <math>b</math> and thickness <math>db</math>
 
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You can measure <math>\sigma(\theta)</math> if you measure the # of particles detected <math>d N</math> in a known detector solid angle <math>d \Omega</math> from a known incident particle Flux (<math>I</math>)  as
 
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<math>\sigma(\theta) = \frac{\frac{d N}{ d \Omega}}{I}</math>
 
  
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Alternatively if you have a theory which tells you <math>\sigma(\theta)</math> which you want to test experimentally with a beam of flux <math>I</math> then you would measure counts (particles)
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=Introduction=
  
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<math>dN = I \sigma(\theta) d \Omega = I \sigma(\theta)  \frac{d A}{r^2} = I \sigma(\theta) \frac{r^2 \sin(\theta) d \theta d \phi}{r^2}</math>
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[[TF_SPIM_Intro]]
  
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;Units
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= Energy Loss =
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: <math>[d N] = [\frac {particles}{m^2}][m^2] [sterradian] </math> = # of particles
 
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: or for a count rate divide both sides by time and you get beam current on the RHS
 
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: integrate and you have the total number of counts
 
  
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;Classical Scattering
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[[TF_SPIM_StoppingPower]]
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: In classical scattering you get the same number of particle out that you put in (no capture, conversion,..)
 
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: <math>d N_{in} = dN</math>
 
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:<math>d N_{in} = I dA = I (2\pi b) db</math>
 
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: <math>d N = I \sigma(\theta) d \Omega =  I \sigma(\theta) \sin(\theta) d \theta d \phi = I \sigma(\theta) \sin(\theta) d \theta (2 \pi )</math>
 
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:<math>  I (2\pi b) db =  I \sigma(\theta) \sin(\theta) d \theta (2 \pi )</math>
 
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:<math>    b  db =  \sigma(\theta) \sin(\theta) d \theta </math>
 
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:<math>\sigma(\theta) =  \frac{b}{\sin(\theta)}\frac{db}{d \theta}</math>
 
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:<math>\frac{db}{d \theta}</math> tells you how the impact parameter <math>b</math> changes with scattering angle <math>\theta</math>
 
  
−
==== Example 4: Elastic Scattering ====
 
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This example is an example of classical scattering.
 
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Our goal is to find <math>\sigma(\theta)</math> for an elastic collision of 2 impenetrable spheres of diameter <math>a</math>.  To solve this elastic scattering problem we will describe the collision using the Center of Mas (C.M.) coordinate system in terms of the reduced mass.  As we shall see, by using C.M. coordinate system the 2-body collision becomes a 1-body problem.  Then we will describe the motion of the reduced mass in the C.M. Frame.
 
−
 
−
[[Image:SPIM_ElasCollis_Lab_CM_Frame.jpg | 500 px]]
 
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[[Media:SPIM_ElasCollis_Lab_CM_Frame.xfig.txt]]
 
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; Variable definitions
 
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:<math>b</math>= impact parameter ; distance of closest approach
 
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:<math>m_1</math>= mass of incoming ball
 
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:<math>m_2</math>= mass of target ball
 
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:<math>u_1</math>= iniital velocity of  incoming ball in Lab Frame
 
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:<math>v_1</math>= final velocity of  <math>m_1</math> in Lab Frame
 
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:<math>\psi</math>= scattering angle of <math>m_1</math> in Lab frame after collision
 
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:<math>u_1^{\prime}</math>= iniital velocity of  <math>m_1</math> in C.M.  Frame
 
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:<math>v_1^{\prime}</math>= final velocity of  <math>m_1</math> in C.M. Frame
 
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:<math>u_2^{\prime}</math>= iniital velocity of  <math>m_2</math> in C.M.  Frame
 
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:<math>v_2^{\prime}</math>= final velocity of  <math>m_2</math> in C.M. Frame
 
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:<math>\theta</math>= scattering angle of <math>m_1</math> in C.M. frame after collision
 
−
 
−
 
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;Determining the reduced mass:
 
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−
 
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[[Image:SPIM_2Body-1BodyCoordSystem.jpg | 500 px]]
 
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; vector definitions
 
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:<math>\vec{r}_1</math> = a position vector pointing to the location of <math>m_1</math>
 
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:<math>\vec{r}_2</math> = a position vector pointing to the location of <math>m_2</math>
 
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:<math>\vec{R}</math> = a position vector pointing to the center of mass of the two ball system
 
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:<math>\vec{r} \equiv \vec{r}_1 - \vec{r}_2</math> = the magnitude of this vector is the distance between the two masses
 
−
 
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In the C.M. reference frame the above vectors have the following relationships
 
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# <math>\vec{R} = 0 = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \Rightarrow m_2 \vec{r}_1 = -m_2 \vec{r}_2</math>
 
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# <math>\vec{r}_1 - \vec{r}_2 = \vec{r}</math>
 
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solving the above equations for <math>\vec{r_1}</math> and <math>\vec{r_2}</math> and defining the reduced mass <math>\mu</math> as
 
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:<math>\mu = \frac{m_1 \cdot m_2}{m_1 + m_2} \equiv</math> reduced mass
 
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leads to
 
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: <math>\vec{r}_1 = \frac{\mu}{m_1} \vec{r}</math>
 
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: <math>\vec{r}_2 = \frac{\mu}{m_2} \vec{r}</math>
 
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We can use the above reduced mass relationships to construct the Lagrangian in terms of <math>\vec{r}</math> instead of <math>\vec{r}_1</math> and <math> \vec{r}_2</math> thereby reducing the problem from a 2-body problem to a 1-body problem.
 
−
 
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; Construct the Lagrangian
 
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The Lagrangian is defined as:
 
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<math>\mathcal{L} = T - U</math>
 
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where
 
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<math>T \equiv</math> kinetic energy of the system
 
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<math>U \equiv</math> Potential energy of the system which describes the interaction
 
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<math>\mathcal{L} = \frac{1}{2} |\vec{\dot{r}}_1|^2 + \frac{1}{2} |\vec{\dot{r}}_2|^2 - U</math>
 
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:= <math>\frac{1}{2} m_1 \left (\frac{m_2}{m1+m_2} \right )^2  |\vec{\dot{r}}|^2 + \frac{1}{2} m_2 \left (\frac{m_1}{m1+m_2} \right )^2  |\vec{\dot{r}}|^2 -U(\vec{r})</math>
 
−
 
−
after substituting derivative of the expressions for <math>\vec{r_1}</math> and <math>\vec{r}_2</math>
 
−
 
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: = <math>\frac{1}{2} \mu |\vec{\dot{r}}|^2 -U(\vec{r})</math>  The 2-body problem is now described by a 1-body Lagrangian
 
−
 
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Lagranges equations of motion are given by
 
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: <math>\frac{\partial \mathcal{L}}{\partial q} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\dot{q}}</math>
 
−
where <math>q</math> represents on of the coordinate (cannonical variables).
 
−
 
−
To get the classical scattering cross section we are interested in finding an expression for the dependence of the impact parameter on the scattering angle,<math>\frac{d b}{d \theta}</math>. 
 
−
 
−
Now lets redraw the collision in terms of a reference frame fixed on <math>m_2</math> (before collision its the Lab Frame but not after collision).
 
−
 
−
[[Image:SPIM_ElasCollis_CMFrame.jpg]] [[Media:SPIM_ElasColls_CMFrame_xfig.txt]]
 
−
 
−
The C.M. Frame rides along the center of mass, the above coordinate system though has its origin on <math>m_2</math> and only overlaps in space with the CM frame at the collision point sufficiently to illustrate <math>\theta</math>.  If <math>b > a</math> then there is no collision (<math>\theta=0</math>),  otherwise a collision happens when r=a (the distance between the balls is equal to their diameter).  A head on collision is defined as <math>b=0</math> (<math>\theta=\pi</math>).
 
−
 
−
;Observation
 
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: as <math>\theta</math> gets smaller, <math>b</math> gets bigger
 
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: <math>\frac{d b}{d \theta} < 0</math>
 
−
 
−
Using plane polar coordinates (<math>r, \phi</math>) we can describe the problem in the lab frame as:
 
−
 
−
<math>\vec{v} = \dot{R} \hat{e}_r + r \dot{\phi} \hat{e}_{\phi}</math>
 
−
 
−
<math>T = \frac{1}{2} \mu ( \dot{r)}^2 + r^2 \dot{\phi}^2)</math>
 
−
 
−
 
−
<math>U(r) = \left \{  {0 \; r > a \atop \infty \; r \le a} \right .</math>
 
−
 
−
<math>\mathcal{L} = T -U = \frac{1}{2} \mu ( \dot{r}^2 + r^2 \dot{\phi}^2) - U(r)</math>
 
−
 
−
Lagranges Equation of Motion:
 
−
 
−
<math>\frac{\partial \mathcal {L}}{\partial \phi} = \frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}</math>
 
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<math>0 = \frac{d}{d t} [ \mu r^2 \dot{\phi}] \Rightarrow</math>  there is a constant of motion ( Constant angular momentum)
 
−
 
−
<math>\ell \equiv \mu r^2 \dot{\phi} = \vec{r} \times \vec{p} = \vec{r} \times \mu \vec{v} = r^2 \mu \dot{\phi}</math>
 
−
 
−
substitute <math>\ell</math> into <math>\mathcal{L}</math>
 
−
 
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<math>\mathcal{L} = \frac{1}{2} ( \mu  \dot{r}^2 + \frac{\ell}{\mu r^2} ) - U(r)</math>
 
−
 
−
The two equations above are in terms of <math>r</math> and <math>\phi</math> whereas our goal is to find an expression for <math>\frac{ d b}{ d \theta}</math>.  Since <math>r</math> is related to <math>b</math> and <math>\phi</math> is related to<math> \theta</math> (<math>\theta = \pi - 2\phi</math>; see figure above) we should try and find expressions for <math>d \phi</math> in terms of <math>r(b)</math>
 
−
 
−
;Trick
 
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: <math>\dot{\phi} = \frac{d \phi}{d t} = \frac{d \phi}{d r} \frac{d r}{d t}</math>
 
−
:<math>\Rightarrow \ell = \mu r^2 \frac{d \phi}{d r} \dot{r}</math>
 
−
:or
 
−
: <math>d \phi = \frac{\ell}{\mu r^2 \dot{r}} dr</math>
 
−
 
−
We now need an expression for <math>\dot{r}</math> in order to integrate the above equation to determine the functional dependence of <math>\phi</math> and hence<math> \theta</math>.
 
−
 
−
Since Energy is conserved (Elastic Scattering), we may define the Hamiltonian as
 
−
 
−
<math>H = T + U = \frac{1}{2} (mu \dot{r}^2 + \frac{\ell}{\mu r^2}) + U(r) = constant \equiv E</math>
 
−
 
−
solving for <math>\dot{r}</math>
 
−
 
−
<math>\dot{r} = \pm \sqrt{\frac{2(E-U(r))}{\mu} - \frac{\ell^2}{\mu^2 r^2}}</math>
 
−
 
−
substituting the above into the equation for <math>d \phi</math> and integrating:
 
−
 
−
<math>\phi = \int d \phi = \int_{r_{min}}^{r_{max}} \frac{\ell}{\mu r^2 \dot{R}} dr</math>
 
−
 
−
<math>r_{min} = a  \; \; \;  r_{max}= \infty  \; \; \;  U(r) = 0 : a \le r \le \infty</math>
 
−
 
−
<math>\phi = \int_a^{\infty} \frac{\ell} {r^2 \sqrt{2 \mu E - \frac{\ell^2}{r^2}} }dr</math>
 
−
 
−
For <math>a \le r \le \infty</math> : <math>E = \frac{1}{2} \mu v^2_{cm} \Rightarrow v_{cm} = \sqrt{\frac{2E}{\mu}}</math>
 
−
 
−
<math>\vec{\ell} = \vec{r} \times \vec{p} \Rightarrow |\vec{\ell}| = |\vec{r}| |\vec{p}| \sin(\phi) = r \mu v_{cm} \sin(\phi) = r \mu \left ( \sqrt{\frac{2E}{\mu}} \right) \sin(\phi) = \sqrt{2 \mu E} r\sin(\phi) =\sqrt{2 \mu E} b</math>
 
−
 
−
substituting this expression for <math>\ell</math> into the last expression for <math>\phi</math> above :
 
−
 
−
<math>\phi =\int_a^{\infty}  \frac{b dr}{r\sqrt{(r^2-b^2)}}</math>
 
−
 
−
;Integral Table
 
−
: <math>\int  \frac{dx}{x\sqrt{(\alpha x^2+\beta x+\gamma)}} = \frac{-1}{\sqrt{-\gamma}} \sin^{-1} \left (\frac{\beta x+2\gamma}{|x|\sqrt{\beta^2-4\alpha \gamma}} \right )</math>
 
−
 
−
let <math>x=r \;\; \alpha=1 \;\; \beta=0 \;\; \gamma=-b^2</math>
 
−
 
−
then
 
−
 
−
<math>\phi = \left . b \frac{1}{\sqrt{-(-b^2)}} \sin^{-1} \left (\frac{-2b^2}{r\sqrt{0-4(1)(-b^2) } }\right ) \right |_a^{\infty} = \sin^{-1} (0)- \sin^{-1}(-\frac{b}{a})</math>
 
−
 
−
or
 
−
 
−
<math>\sin(\phi) = \frac{b}{a} = \sin \left ( \frac{\pi}{2} - \frac{\theta}{2} \right ) = \cos \left ( \frac{\theta}{2} \right )</math>
 
−
 
−
:<math>\Rightarrow b = a \cos \left( \frac{\theta}{2} \right)</math>
 
−
 
−
 
−
; Now substitue the above into the expression for <math>\sigma(\theta)</math>
 
−
 
−
<math>\sigma(\theta) = \frac{b}{\sin(\theta)} \frac{d b}{d \theta} = \frac{a \cos(\theta/2)}{sin(\theta)} a[-\sin(\theta/2)]\frac{1}{2}
 
−
= \frac{a^2}{2} \frac{\cos(\theta/2) \sin(\theta/2)}{\sin(\theta)}</math>
 
−
 
−
drop the negative sign, sqrt in denominator allows this, and use the trig identity
 
−
 
−
:<math>\sin \left (\frac{\theta}{2} + \frac{\theta}{2} \right ) = \cos \left (\frac{\theta}{2} \right) \sin \left (\frac{\theta}{2} \right ) + \cos \left ( \frac{\theta}{2} \right ) \sin \left (\frac{\theta}{2} \right )</math>
 
−
:<math>\sin(\theta) = 2 \cos \left (\frac{\theta}{2} \right ) \sin \left (\frac{\theta}{2} \right )</math>
 
−
 
−
<math>\sigma(\theta) = \frac{a^2}{2} \frac{1}{2} = \frac{a^2}{4}</math>
 
−
 
−
<math>\sigma = \int \sigma(\theta) d \Omega = \frac{a^2}{2} \frac{1}{2} 4 \pi  = \pi a^2</math>
 
−
 
−
 
−
;compare with result from definition
 
−
 
−
:<math>\sigma</math> = scattering cross-section <math>\equiv \frac{\# particles\; scattered} {\frac{ \# incident \; particles}{Area}}</math>
 
−
:number of particles scattered = number of incident particles
 
−
: Area = <math> \pi a^2</math> = The area profile in which a collision occurs[[Image:ClassicalEffectiveScatteringArea.jpg | 200 px]]
 
−
 
−
<math>\sigma = \frac{{N}}{\frac{ N}{\pi a^2}} = \pi a^2</math>
 
−
 
−
==== Lab Frame Cross Sections ====
 
−
 
−
The C.M. frame is often chosen to theoretically calculate cross-sections even though experiments are conducted in the Lab frame.  In such cases you will need to transform cross-sections between two frames.
 
−
 
−
The total cross-section should be frame independent
 
−
 
−
:<math>\sigma_{C.M.} = \sigma_{Lab}</math>
 
−
 
−
or
 
−
 
−
: <math>\sigma(\theta) d \Omega = \sigma(\psi) d \Omega^{\prime}</math>
 
−
 
−
where
 
−
 
−
<math>\theta</math> is in the CM frame and <math>\psi</math> is in the Lab frame.
 
−
 
−
 
−
;A non-relativistic transformation:
 
−
 
−
: <math>\sigma(\theta) d \Omega = \sigma(\psi) d \Omega^{\prime}</math>
 
−
: <math>\sigma(\theta) 2 \pi \sin(\theta) d \theta = \sigma(\psi) 2 \pi \sin (\psi) d \psi</math>
 
−
: <math>\Rightarrow \sigma(\psi) = \frac{\sin(\theta)}{\sin(\psi)} \frac{d \theta}{d \psi} \sigma(\theta)</math>
 
−
 
−
The transformation is governed by the dependence of <math>\theta</math> on <math> \psi</math> <math> \left( \frac{d \theta}{d \psi} \right )</math>
 
−
 
−
Lets return back to our picture of the scattering Process
 
−
 
−
[[Image:SPIM ElasCollis Lab CM Frame.jpg | 500 px]]
 
−
 
−
if we superimpose the vectors <math>\vec{v}_1</math> and <math>\vec{v}_1^{\prime}</math> we have
 
−
 
−
[[Image:SPIM ElasCollis Lab CM Frame_Velocities.jpg]]
 
−
 
−
Trig identities (non-relativistic Gallilean transformation) tell us
 
−
 
−
<math>v_1 \sin(\psi) = v_1^{\prime} \sin(\theta)</math>
 
−
 
−
 
−
<math>v_1 cos(\psi) = v_{cm} + v_1^{\prime} \cos(\theta)</math>
 
−
 
−
solving for <math>\psi</math>
 
−
 
−
<math>\tan(\psi) = \frac{\sin(\psi)}{\cos(\psi)} = \frac{v_1^{\prime} \sin(\theta)/v_1}{\frac{v_{CM}}{v_1} + \frac{v_1^{\prime} \cos(\theta)}{v_1} }
 
−
= \frac{\sin(\theta)}{\cos(\theta) + \frac{v_{CM}}{v_1^{\prime}}}</math>
 
−
 
−
For an elastic collision only the directions change in the CM Frame: <math>u_1^{\prime}= v_1^{\prime}</math>  & <math>u_1^{\prime}= v_2^{\prime}</math>
 
−
 
−
;From the definition of the C.M.
 
−
;<math>\vec{v}_{CM} = \frac{m_1 \vec{u}_1 + m_2 \vec{u}_2}{m_1+m_2} = \frac{m_1}{m_1+m_2} \vec{u}_1</math>
 
−
 
−
;conservation of momentum in CM Frame <math>\Rightarrow</math> :
 
−
:<math>m_1 u_1^{\prime} = - m_2 u_2{\prime}</math>
 
−
 
−
:<math> \Rightarrow v_1^{\prime} = u_1^{\prime} = \frac{-m_2}{m_1} u_2^{\prime}</math>
 
−
 
−
; Gallilean Coordinate transformation:
 
−
;<math>\vec{u}_1 = \vec{u}_1^{\prime} + \vec{v}_{CM} = \vec{u}_1^{\prime} + \frac{m_1}{m_1+m_2} \vec{u}_1</math>
 
−
:<math>\Rightarrow u_1{\prime} = \left [ 1 - \frac{m_1}{m_1 + m_2} \right ] u_1 = \frac{m_2}{m_1+m_2}u_1</math>
 
−
:<math>\Rightarrow v_1^{\prime} = u_1^{\prime}  =\frac{m_2}{m_1+m_2} u_1</math>
 
−
 
−
; another experission for <math>\psi</math>
 
−
 
−
using the above gallilean transformation we can do the following
 
−
 
−
:<math>\frac{v_{CM}}{v_1^{\prime}}= \frac{\frac{m_1}{m_1+m_2} u_1}{\frac{m_2}{m_1+m_2} u_1} = \frac{m_1}{m_2}</math>
 
−
 
−
or
 
−
 
−
: <math>\tan(\psi) = \frac{\sin(\theta)}{\cos(\theta) + \frac{m_1}{m_2}}</math>
 
−
 
−
after a little trig substitution
 
−
 
−
<math>\Rightarrow \frac{m_1}{m_2} = \frac{sin(\theta - \psi)}{\sin(\psi)} =</math> constant
 
−
 
−
now use the chain rule to find <math>\frac{d \theta}{d \psi}</math>
 
−
 
−
: <math>f \equiv  \frac{sin(\theta - \psi)}{\sin(\psi)} =</math> constant
 
−
:<math>df = 0 = \frac{ \partial f}{\partial \psi} d \psi  + \frac{ \partial f}{\partial \theta} d \theta </math>
 
−
: <math>\Rightarrow \frac{d \theta}{d \psi} = \frac{-\frac{ \partial f}{\partial \psi} }{\frac{ \partial f}{\partial \theta} }</math>
 
−
 
−
:<math>-\frac{ \partial f}{\partial \psi} = \frac{\cos(\theta - \psi)}{\sin(\psi)} + \frac{\sin(\theta - \psi)}{\sin(\psi)}</math>
 
−
:<math>\frac{ \partial f}{\partial \theta }= 1 + \frac{\sin(\theta - \psi) \cos(\psi)}{\cos(\theta - \psi) \sin(\psi)}</math>
 
−
 
−
after substitution:
 
−
: <math>\sigma(\psi) = \frac{\sin(\theta)}{\sin(\psi)} \frac{d \theta}{d \psi} \sigma(\theta)</math>
 
−
: <math>=\frac{\sin(\theta)}{\sin(\psi)} \left [ 1 + \frac{\sin(\theta - \psi) \cos(\psi)}{\cos(\theta - \psi) \sin(\psi)} \right ] \sigma(\theta)</math>
 
−
 
−
For the above equation to be more useful one would prefer to recast it in terms of only <math>\psi</math> and masses.
 
−
 
−
:<math>\sigma(\psi) = \frac{\left [ \frac{m_1}{m_2}\cos(\psi) + \sqrt{1-\left ( \frac{m_1 \sin(\psi) }{m_2} \right )^2 }\right ]}{\sqrt{1 - \left ( \frac{m_1 \sin(\psi)}{m_2}\right )^2 }}</math>
 
−
 
−
== Stopping Power ==
 
 
Ann. Phys. vol. 5, 325, (1930)
 
Ann. Phys. vol. 5, 325, (1930)
−
=== Bethe Equation ===
 
−
====Classical Energy Loss ====
 
−
 
−
Consider the energy lost when a particle of charge (<math>ze</math>) traveling at speed <math>v</math> is scattered by a target of charge (<math>Ze</math>).  Assume only the coulomb force causes the particle to scatter from the target as shown below.
 
−
 
−
[[Image:SPIM_Bethe_ClassCoulScat.jpg]]
 
−
 
−
; Notice:
 
−
: as <math>ze</math> is scattered the horizontal component of the coulomb force (<math>F</math>) flips direction; ie no horizontal force
 
−
 
−
:<math>F_{vertical} = k \frac{zZe^2}{r^2} \sin(\theta) = k \frac{zZe^2}{r^2}  \frac{b}{r}</math>
 
−
 
−
where
 
−
: k =<math>\frac{1}{4 \pi \epsilon_0}</math>
 
−
: r = distance between incident projectile and target atom
 
−
: b= impact parameter of collision
 
−
 
−
 
−
Using the definition of Impulse one can determine the momentum change of <math>ze</math> as
 
−
 
−
: <math>\Delta p = \int F dt</math>
 
−
 
−
Let's assume that the energy lost by the incident particle <math>ze</math> is absorbed by an electron in the target atom.  This energy may be cast in terms of the incident particles momentum change as
 
−
 
−
:<math>\frac{(\Delta p)^2}{2m_e}</math>
 
−
 
−
By calculating the change in momentum (<math>\Delta p</math>) of the incident particle we can infer the amount of energy lost by the incident particle and absorbed one of the target materials atomic electrons.
 
−
 
−
:<math>\Delta P = \int F dt = \int k \frac{zZe^2b}{r^3}  dt</math>
 
−
 
−
using  <math>dt = \frac{dx}{v} = \frac{d x}{\beta c}</math> we have
 
−
 
−
: <math>= k \frac{zZe^2b}{\beta c} \int_{-\infty}^{+\infty} \frac{ dx}{(x^2+b^2)^{3/2}}</math>
 
−
: <math>=\frac{kzZe^2b}{\beta c b^2} \int_{-\infty}^{+\infty} \frac{ dx/b}{(1+\frac{x}{b^2})^{3/2}}</math>
 
−
 
−
:<math>\int_{-\infty}^{+\infty} \frac{ dx/b}{(1+\frac{x}{b^2})^{3/2}}=2</math>
 
−
 
−
:<math> \Delta p = \frac{2kzZe^2b}{\beta c b^2}</math>
 
−
 
−
casting this in terms of the classical atomic electron radius <math>r_e</math>
 
−
 
−
: <math>r_e = \frac{k e^2}{m_e v^2} \sim \frac{k e^2}{m_e c^2}</math>  just equate <math>F = \frac{ke^2}{r_e^2} = m \frac{v^2}{r_e}</math>
 
−
 
−
Then
 
−
 
−
:<math> \Delta p = \frac{2zZr_e m_e c}{\beta  b}</math>
 
−
 
−
and
 
−
 
−
: <math>\Delta E = \frac{(\Delta p)^2}{2m_e} = 2 \left ( \frac{r_e m_e}{\beta b}\right )^2 \frac {z^2 Z^2 c^2}{m_e}</math> : <math>Z</math> = 1 here because I shall assume the energy is lost to just the electron and the Atom is a spectator
 
−
 
−
Now let's calculate an expression representing the average energy lost for an incident particle traversing a material of some thickness.
 
−
 
−
Let
 
−
 
−
: <math>P(\Delta E)</math> = Probability of an interaction taking place which results in an energy loss <math>\Delta E</math>
 
−
 
−
If we let
 
−
 
−
Z  = Atomic Number = # electrons in target Atom = number of protons in an Atom
 
−
 
−
N = Avagadros number  = <math>6.022 \times 10^{23}  \frac{Atoms}{mol}</math>
 
−
 
−
A = Atomic mass =  <math>\frac{g}{mole}</math>
 
−
 
−
<math>dP(\Delta E)</math>  = probability of hitting an atomic electron in the area of an annulus of radius (<math>b + db</math>) with an energy transfer between <math>\Delta E</math> and <math>\Delta E + d(\Delta E)</math>
 
−
 
−
Then
 
−
 
−
:  <math>\frac{-dE }{dx}= \int_0^{\infty} dP(\Delta E) \Delta E</math> = energy lost by the incident particle per distance traversed through the material
 
−
 
−
I am just adding up all the energy losses weighted by the probability of the energy loss to find the total energy loss.
 
−
 
−
:<math>dP(\Delta E) = \frac{N}{A} d \sigma =\frac{N}{A} (2 \pi b db) Z</math>  : classically <math>\sigma = \pi b^2 ; d \sigma = 2\pi b db
 
−
</math>
 
−
 
−
<math>\Rightarrow \frac{-dE}{dx} =  \int_0^{\infty} \frac{N}{A} (2 \pi b db) Z \Delta E = \frac{2 \pi N Z}{A} \int_0^{\infty} \Delta E b db</math>
 
−
:= <math>\frac{2 \pi N Z}{A} \int_0^{\infty} \left [ \frac{2 r_e^2 m_e c^2 z^2}{\beta^2 b^2}\right ] b db</math>
 
−
:= <math>4 \pi N r_e^2 m_e c^2 \frac{z^2 Z}{A \beta^2} \int_0^{\infty} \frac{db}{b}</math>
 
−
:=<math>\frac{\mathcal{K} }{A} \frac{z^2 Z}{\beta^2} \int_0^{\infty} \frac{db}{b}</math>
 
−
 
−
where
 
−
<math>\frac{\mathcal{K}}{A} = \frac{4 \pi N r_e^2 m_e c^2}{A} = 0.307 \frac{MeV cm^2}{g}</math>''' if A=1'''
 
−
 
−
The limits of the above integral should be more physical in order to reflect the limits of the physics interaction.  Let b_{min}  and b_{max} represent the minimum and maximum possible impact parameter where the physics is discribed,  as shown above,  by the coulomb force.
 
−
 
−
;What is <math>b_{min}</math>?
 
−
 
−
if <math>b \rightarrow 0</math> then <math>\frac{d E}{dx}</math> diverges and the energy transfer <math>\rightarrow \infty : \Delta E \sim \frac{1}{b}</math>.  Physically there is a maximum energy that may be transferred before the physics of the problem changes (ie: nuclear excitation, jet production, ...).  The de Borglie wavelength of the atom is used to estimate a value for <math>b_{min}</math> such that
 
−
 
−
: <math>b_{min} \sim  \frac{1}{2} \lambda_{de Broglie} = \frac{h}{2p} = \frac{h}{2 \gamma m_e \beta c}</math>
 
−
 
−
;What is <math>b_{max}</math>?
 
−
 
−
As <math>b</math> gets bigger the interaction is "softer" and longer.  If the interaction time (<math>\tau_i</math>) is so long that it is equivalent to an electron orbit (<math>\tau_R</math>) then the atom looks more like it is neutrally charged.  You move from an interaction in which the electron orbit is perturbed adiabatically such that there is no orbit change and the minimum amount of energy is transferred to no interaction taking place because the atom is neutral.
 
  
−
Let
+
=Interactions of Electrons and Photons with Matter=
  
−
: <math>\tau_i = \frac{b_{max}}{v} (\sqrt{1-\beta^2})</math>  : fields at high velocities get Lorentz contracted
+
[[TF_SPIM_e-gamma]]
−
: <math>\tau_R \equiv \frac{h}{I}</math> : I <math>\equiv</math> mean excitation energy of target material ( <math>E = h \nu = h/ \tau</math>)
 
  
−
Condition for <math>b_{max}</math> :
+
= Hadronic Interactions =
−
:<math>\tau_i = \tau_R</math>
 
−
<math>\Rightarrow b_{max} = \frac{h \gamma \beta c}{I}</math>
 
  
−
<math>-\frac{dE}{dx} = \frac{\mathcal{K} }{A} \frac{z^2 Z}{\beta^2} \int_0^{\infty} \frac{db}{b}</math>
+
[[TF_SPIM_HadronicInteractions]]
−
: <math>= \frac{\mathcal{K} }{A} \frac{z^2 Z}{\beta^2} \ln \frac{b_{max}}{b_{min}}</math>
 
−
: <math>= \frac{\mathcal{K} }{A} \frac{z^2 Z}{\beta^2} \ln \frac{2 \gamma^2 m_e \beta^2 c^2}{I}</math>
 
  
−
=====Example 5: Find <math>\frac{dE}{dx}</math> for a 10 MeV proton hitting a liquid hydrogen (<math>LH_2</math>) target=====
+
= Final Project=
−
A = Z=z=1<br>
 
−
<math>m_e c^2</math> = 0.511 MeV <br>
 
−
I = 21.6 eV : see solid data point From  Figure 27.5 on pg 6 of [http://pdg.lbl.gov/2007/reviews/passagerpp.pdf PDG] below.<br>
 
−
[[Image:PDG_IonizationPotential.jpg | 500 px]]
 
  
−
Just need to know <math>\gamma</math> and <math>\beta</math>
+
A final project will be submitted that will be graded with the following metrics:
  
−
"a 10 MeV proton" <math>\Rightarrow</math> Kinetic Energy (K.E.) = 10 MeV = <math>(\gamma - 1) mc^2</math>
+
1.) The document must be less than 15 pages.
  
−
:<math>\Rightarrow \gamma = \frac{K.E.}{mc^2} + 1 = \frac{10 MeV}{938 MeV} + 1 \sim 1 = \frac{1}{\sqrt{1-\beta^2}}</math>
+
2.) The document must contain references in a bibliography (5 points) .
  
−
Proton is not relativistic
+
3.) A comparison must be made between GEANT4's prediction and either the prediction of someone else or an experimental result(30 points).
  
−
: <math>v^2 = \frac{2 K.E.}{m} = \frac{2 \cdot 10 MeV}{938 MeV/c^2} = 2 \times 10^{-2} c^2 \Rightarrow \beta^2 = \frac{v^2}{c^2} = 2\times 10^{-2}</math>
+
4.) The graphs must be of publication quality with font sizes similar or larger than the 12 point font (10 points).
  
−
Plugging in the numbers:
+
5.) The document must be grammatically correct (5 points).
−
: <math>\frac{dE}{dx} = \left ( 0.307 \frac{MeV \cdot cm^2}{g}\right ) (1)^2 (1) \frac{1}{2 \times10^{-2}} \ln \left( \frac{2 (1) (0.511 MeV) (2 \times10^{-2})}{21.6 eV} \frac{10^6 eV}{MeV}\right)</math>
 
−
: <math>= 105 \frac{MeV cm^2}{g}</math>
 
  
−
;How much energy is lost after 0.3 cm?
+
6.) The document format must contain the following sections: An abstract of 5 sentences (5 points) , an Introduction(10 points), a Theory section (20 points) , if applicable a section describing the experiment that was simulated, a section delineating the comparisons that were made, and a conclusion( 15 points).
  
−
'''Notice that the units for energy loss are normalized by the density of the material'''
+
=Resources=
−
<math>\rho_{LH_2}</math> = 0.07 <math>\frac{g}{cm^3}</math><br>
 
  
−
To get the actual energy lost I need to multiply by the density.   So for any given atom the energy loss will depend on the state (solid, gas, liqid) of the atom as this effects the density of the material.   
+
[http://geant4.web.cern.ch/geant4/ GEANT4 Home Page]
  
−
:<math>\Delta E = (105 \frac{MeV cm^2}{g}) (0.07 \frac{g}{cm^3}) (0.3 cm)</math> = 2.2 MeV
+
[http://root.cern.ch ROOT Home page]
  
−
[[Image:SPIM_HydrogenStoppingPower.pdf]] Compare with Triumf Kinematics Handbook, 2nd edition, September 1987, L.G. Greeniaus
+
[http://conferences.fnal.gov/g4tutorial/g4cd/Documentation/WorkshopExercises/  Fermi Lab Example]
  
−
====Bethe-Bloch Equation ====
 
  
−
While the classical equation above works in a limited kinematic regime, the Bethe-Bloch equation includes the corrections needed to cover most kinematic regimes for heavy particle energy loss.
+
[http://physics.nist.gov/PhysRefData/Star/Text/ESTAR.html NIST Range Tables]
  
−
:<math>-\frac{dE}{dx} = \mathcal{K} z^2 \frac{Z}{A} \frac{1}{\beta^2} \left [ \frac{1}{2} \ln \left (\frac{2 m_2 c^2 \beta^2 \gamma^2 T_{max}}{I^2} \right) - \beta^2 - \frac{\delta}{2}\right ]</math>[http://pdg.lbl.gov/2007/reviews/passagerpp.pdf PDG reference Eq 27.1 pg 1]
+
[http://ie.lbl.gov/xray/ X-ray specturm]
  
−
where
+
[[Installing_GEANT4.9.3_Fsim]]
  
−
; <math>T_{max} = \frac{2 m_e c^2 \beta^2 \gamma^2}{1+ \frac{2 \gamma m_e}{M} + \frac{m_e}{M}}</math>
+
== Saving/restoring Random number seed==
−
:= Max K.E. transferable to the Target of mass <math>M</math> in a single collision.
 
  
−
;<math>-\beta^2</math>
+
You save the current state of the random number generator with the command
−
: =  correction for electron spin and very distant collisions which deform the electron atomic orbits each process reducing dE/dx by <math>\frac{\beta}{2}</math>
 
  
−
; <math>\frac{\delta}{2}</math>
+
/random/setSavingFlag 1
−
:= density correction term: in the classical derivation the material is treated as just a system of <math>N</math> atoms uniformly distributed in space.  These Atoms, however, give the material polarizability which can reduce the electric field (dielectric).
 
  
−
==== GEANT 4 implementation ====
+
/run/beamOn 100
  
−
The GEANT4 file  (version 4.8.p01)
+
/random/saveThisRun
  
−
source/processes/electromagnetic/standard/src/G4BetheBlockModel.cc
+
A file is created called
  
−
is used to calculate hadron energy loss.
+
currentEvent.rndm
  
−
line 132 <math>\Rightarrow</math>
+
/control/shell mv currentEvent.rndm currentEvent10.rndm
  
−
:<math>-\frac{dE}{dx} =  \log \left ( \frac{2 m_e c^2 \tau (\tau +2) E_{min}}{I^2}\right) - \left (1 - \frac{E_{min}}{E_{max}} \right ) \beta^2</math>
 
  
−
where
+
You can restore the random number generator and begin generating random number from the last save time
  
−
:<math> \tau = \frac{K.E.}{M}</math>
+
/random/resetEngineFrom currentEvent.rndm
  
−
line 143 <math>\Rightarrow</math>
+
==Building GEANT4.11==
  
−
:<math>\frac{dE}{dx} -= \log  ( \tau (\tau + 2) ) -cden</math> = density corection = <math>\frac{\delta}{2}</math>
+
===4.11.2===
 +
[[TF_GEANT4.11]]
  
−
line 148 <math>\Rightarrow</math>
+
==Building GEANT4.10==
  
−
:<math>\frac{dE}{dx} -= \frac{2c}{Z_{target}}</math> = shell correction, corrects for the classical asumption that the atomic electron velicity is initially zero; or the the incident particles velocity is far greater than the atomic electron's velocity.
 
  
−
line 154 <math>\Rightarrow</math>
+
===4.10.02===
  
−
: <math>\frac{dE}{dx} *= \frac{2 \pi m_e c^2 r_e^2 z^2}{\beta^2} \rho_e \;\;\;\; \rho_e \propto \frac{NZ}{A}</math>
+
[[TF_GEANT4.10.2]]
 +
===4.10.01===
  
−
==== Energy Dependence ====
+
[[TF_GEANT4.10.1]]
  
−
[[Image:SPIM_EnergyLoss_EnergyDependence.jpg | 500 px]]
+
==Building GEANT4.9.6==
  
−
The above curve shows the energy loss per disntace traveled (<math>\frac{dE}{dx}</math>) as a function of the incident particles energy. There are three basic regions.  At low incident energies ( < 10^5 eV) the incident particle tends to excite or even ionize the atoms in the material it is penetrating.  The maximum amount of energy loss per distance traveled is defined at as the Bragg peak.  The region after the Bragg peak in which the energy loss per distance traveled reaches its smalest value is reffered to as the point of minimum ionizing.  Minimimum ionizing particles will have incident energies corresponding to this value or larger.  The characteristic of the minimum ionizing particles is that their energy loss per distance traveled is essentially constant making simulations easier until the drop below the minimum ionizing energy level as they are passing through the material.
+
[[TF_GEANT4.9.6]]
  
−
In general the Bethe-Bloch equation breaks down at low energies (below the Bragg peak)  and is a good description (to within 10%) for
+
==Building GEANT4.9.5==
  
−
:<math>10 \frac{MeV}{a.m.u.} < E < 2 \frac{GeV}{a.m.u.}</math>  and <math>Z</math> < 26 (Iron)
+
[[TF_GEANT4.9.5]]
  
−
the <math>\frac{1}{\beta^2}</math> term in the Bethe-Bloch equation dominates between the Bragg peak and the minimum ionization region.
+
An old version of Installation notes for versions prior to 9.5
  
−
the <math>\ln</math> term and its corrections influence the dependence of  <math>\frac{dE}{dx}</math> as you move up in energy beyond the minimum ionization point.
+
[http://brems.iac.isu.edu/~tforest/NucSim/Day3/ Old Install Notes]
  
−
=== Energy Straggling ===
 
  
−
While the Bethe-Bloch formula gives you a way to quantify the amount of energy a heavy charged particle  looses as a function of the distance traveled, you should realize that when you calculate the total energy lost via
+
Visualization Libraries:
  
−
:<math> \Delta E = \int_{E_i}^{E_f} \left ( \frac{dE}{dx} \right ) dx</math>
+
[http://www.opengl.org/ OpenGL]
  
−
you are only determining the AVERAGE energy loss.  In other words, Bethe-Bloch is the Astochastic process describing energy loss.
+
[http://geant4.kek.jp/~tanaka/DAWN/About_DAWN.html DAWN]
  
−
In reality the energy loss process is a stochastic process because of the statistical fluctuations which occur in the actual number of collisions which take place.
 
  
 +
[http://doc.coin3d.org/Coin/  Coin3D]
  
−
==== Thick Absorber ====
+
==Compiling G4 with ROOT==
  
−
A thick absorber is one in which a large number of collisions takes place.  In this situation the central limit theorem from statistics tells you that the larger number of random variables <math>N</math> involved will result in observables which are distributed in a Gaussian manner.
+
These instruction describe how you can create a tree within ExN02SteppingVerbose to store tracking info in an array (max number of steps in a track is set to 100 for the desired particle)
  
−
The gaussian probability function is defined as
+
[[G4CompileWRootforTracks]]
  
−
: <math>P(x,\Delta) \propto e^{\frac{(\Delta - \bar{\Delta})^2}{ \sigma^2}}</math>
+
==Using SLURM==
  
−
where the Full Width at Half Max (FWHM) of the distribution = <math>2 \sqrt{2 \ln 2} \sigma</math>
+
http://slurm.schedmd.com/quickstart.html
  
−
In the case of energy loss, the variance using the Bethe-Bloch equation should be
 
  
−
:<math>\sigma_0^2 = 4 \pi N r_e^2 (m_e c^2)^2 \rho \frac{Z}{a} x</math>
+
https://rc.fas.harvard.edu/resources/documentation/convenient-slurm-commands/
  
−
the realitivistic variance is
+
===simple batch script for one process job===
  
−
: <math>\sigma^2 = [\frac{1-\beta^2/2}{1-\beta^2} ]\sigma_0^2</math>
+
create the file submit.sbatch below
  
−
for very thick absorbers see
+
<pre>
 +
#!/bin/sh
 +
#SBATCH --time=1
 +
cd src/PI
 +
./PI_MC 100000000000000
 +
</pre>
  
−
C. Tschaler, NIM '''64''', (1968) 237 ; ''ibid'', '''61''', (1968) 141
+
the execute
  
−
When simulating energy loss of heavy charged particles the Bethe-Bloch equation may be used to calculate a <math>\frac{dE}{dx}</math> which can determine the average energy loss at the given kinetic energy of the particle. This average is then smeared according to a gaussian distribution of variance
+
:sbatch submit.sbatch
  
−
: <math>\sigma^2 =4 \pi N r_e^2 (m_e c^2)^2 \rho \frac{Z}{a} x [\frac{1-\beta^2/2}{1-\beta^2} ]</math>
+
check if its running with
  
−
====Thin Absorbers====
+
:squeue
  
−
In thin absorbers the number of collisions is small preventing the use of the central limit theorem to describe the stochastic process of energy loss in terms of a Gaussian distribution.  The Large energy transfers that are possible cause the energy loss distribution to look like a Gaussian one with a high energy tail (or foot).
+
to kill a batch job
  
−
The skewness of the resulting energy loss distribution is quantified as
+
:scancel JOBID
  
−
:<math>\kappa = \frac{\bar{\Delta}}{W_{max}}</math>
+
===On minerve===
  
−
: <math>\Delta  \equiv 2 \pi N_a r_e^2 m_e c^2 \rho \frac{Z}{A} \left ( \frac{z}{\beta}\right)^2 x </math> = lead term in Bethe Bloch equation
+
Sample script to submit 10 batch jobs.
  
−
<math>\rho</math> = density of absorbing material.
+
the filename is minervesubmit and you run like
 +
source minervesubmit
  
−
:<math>W_{max} = \frac{(pc)^2}{\frac{1}{2} \left [  m_e c^2 + \left ( \frac{M^2 c^2}{m_e} \right ) \right ] + \sqrt{(pc)^2 + (Mc^2)^2}}</math> = max energy transfered in 1 collision (headon / knock out collision)
+
<pre>
 +
cd /home/foretony/src/GEANT4/geant4.9.5/Simulations/N02wROOT/batch
 +
qsub submit10mil
 +
qsub submit20mil
 +
qsub submit30mil
 +
qsub submit40mil
 +
qsub submit50mil
 +
qsub submit60mil
 +
qsub submit70mil
 +
qsub submit80mil
 +
qsub submit90mil
 +
qsub submit100mil
 +
</pre>
  
−
This comes from the relativistic kinematics of an Elastic Collisions.<br>
+
The file submit10mil looks like this
−
[[Image:SPIM_ThinAbsorbers_Scatering.jpg | 400 px]]
+
<pre>
 +
#!/bin/sh
 +
#PBS -l nodes=1
 +
#PBS -A FIAC
 +
#PBS -M foretony@isu.edu
 +
#PBS -m abe
 +
#
 +
source /home/foretony/src/GEANT4/geant4.9.5/geant4.9.6-install/bin/geant4.sh
 +
cd /home/foretony/src/GEANT4/geant4.9.5/Simulations/N02wROOT/batch/10mil
 +
../../exampleN02 run1.mac > /dev/null
 +
</pre>
  
−
Conservation of Momentum <math>\Rightarrow</math> :
 
  
−
: <math>\vec{p} = \vec{p}^{\; \prime \prime} + \vec{p}^{\; \prime}</math>
+
use
  
−
Conservation of Energy <math>\Rightarrow</math> :
+
qstat
  
−
: <math>E_{tot} + m_ec^2 = E_{tot}^{\prime \prime} + E_{tot}^{\prime}</math>
+
to check that the process is still running
−
: <math>\sqrt{(pc)^2 + (Mc^2)^2} + m_ec^2 = \sqrt{(p^{\; \prime \prime} c)^2 + (Mc^2)^2}  + E_k +  m_e c^2</math>
 
−
:<math> \Rightarrow  (p^{\prime}c)^2 = E_k^2 + 2E_km_ec^2</math>
 
  
−
: <math>\gamma = \frac{E_{tot}}{mc^2} = \frac{ \sqrt{(pc)^2 + (mc^2)^2}}{mc^2}</math>
+
use
−
:<math>\beta= \frac{p}{\gamma mc} = \frac{pc}{E_{tot}}</math>
 
  
−
using conservation of E & P as well as substituting for <math>p^{\prime}</math> you can show
+
qdel jobID#
  
−
:<math>(p^{\; \prime \prime}c)^2 = (pc)^2 - 2E_k\sqrt{(pc)^2 +(Mc^2)^2} + E_k^2</math> : cons of E
+
if you want to kill the batch job, the jobID number shows up when you do stat.
−
:<math>= (pc)^2 + E_k^2 + 2E_km_ec^2 -2pc\sqrt{E_k^2+2E_km_ec^2} \cos(\theta)</math> : cons of P
 
  
−
<math>\Rightarrow</math>
+
for example
 +
<pre>
 +
[foretony@minerve HW10]$ qstat
 +
Job id                    Name            User            Time Use S Queue
 +
------------------------- ---------------- --------------- -------- - -----
 +
27033.minerve            submit          foretony        00:41:55 R default       
 +
[foretony@minerve HW10]$ qdel 27033
 +
[foretony@minerve HW10]$ qstat
 +
</pre>
  
−
: <math>pc \cos(\theta) \sqrt{1+\frac{2m_ec^2}{E_k}} = \sqrt{(pc)^2+(Mc^2)^2} + m_ec^2</math>
+
==Definitions of Materials==
  
−
solving for <math>E_k</math>
+
[[File:MCNP_Compendium_of_Material_Composition.pdf]]
  
−
:<math>E_k = \frac{2m_ec^2(pc)^2\cos^2 (\theta)}{[\sqrt{(pc)^2 + (Mc^2)^2} +m_ec^2]^2 - (pc)^2 \cos^2 (\theta)}</math>
+
==Minerve2 GEANT 4.10.1 Xterm error==
  
−
===== (Landau THeory) =====
 
−
<math>\kappa \leq 0.01</math>
 
  
−
Landau assumed
+
On OS X El Capitan V 10.11.4 using XQuartz
−
:# <math>W_{max} = \infty</math> is max energy transfer
 
−
:# electrons are free (energy fransfer is so large you can neglect binding)
 
−
:# incident particle maintains velocity (large momentum transfer from big mass to small mass) (bowling ball hits ping pong ball)
 
  
−
L. Landau, "On the Energy Loss of Fast Particles by Ionization", J. Phys., vol 8 (1944), pg 201
+
~/src/GEANT4/geant4.10.1/Simulations/B2/B2a/exsmpleB2a
  
−
instead of a gaussian distribution Landau used
+
<pre>
  
−
: <math>P(x,\Delta) \propto \frac{1}{\bar{\Delta}\pi} \int_0^{\infty} e^{-u \ln u - u \lambda} \sin(\pi u) du</math>
+
# Use this open statement to create an OpenGL view:
 +
/vis/open OGL 600x600-0+0
 +
/vis/sceneHandler/create OGL
 +
/vis/viewer/create ! ! 600x600-0+0
 +
libGL error: failed to load driver: swrast
 +
X Error of failed request:  BadValue (integer parameter out of range for operation)
 +
  Major opcode of failed request:  150 (GLX)
 +
  Minor opcode of failed request:  3 (X_GLXCreateContext)
 +
  Value in failed request:  0x0
 +
  Serial number of failed request:  25
 +
  Current serial number in output stream:  26
 +
</pre>
  
−
where
 
  
−
: <math>\lambda = \frac{1}{\bar{\Delta}} \left [ \Delta - \bar{\Delta} \ln \bar{\Delta} - \ln \epsilon + 1 -C \right ]</math>
 
−
: <math>\bar{\Delta} = 2\pi N_a r_e^2 m_e c^2 \rho \frac{Zz^2}{A \beta^2}x</math>
 
−
:<math>\ln \epsilon = \ln \left [ \frac{(1-\beta^2)I^2}{2m_ec^2 \beta^2} \right ]</math>
 
−
: <math>C = 0.577</math>
 
  
−
[[Image:SPIM_Landau_ThinkAbsorberDist.jpg]]
 
  
−
===== (Vavilou's Theory) =====
 
  
−
Vavilous paper
+
[[TF_SPIM_OLD]]
−
 
 
−
P.V. Vavilou, "Ionization losses of High Energy Heavy Particles", Soviet Physics JETP, vol 5 (1950? )pg 749
 
−
 
 
−
describe the physics for the case
 
−
 
 
−
;<math>0.01 < \kappa < \infty </math>
 
−
 
 
−
The distribution function derived is shown below as well as a conceptual overlay of Vavilou's and Landau's distributions.  (The <math>\zeta f(x,\Delta)</math> in the picture should be a <math>\bar{\Delta}P(x,\Delta)</math> )
 
−
 
 
−
 
 
−
: <math>P(x,\Delta) = \frac{1}{\bar{\Delta}\pi} x e^{x(1+\beta^2C)} \int_0^{\infty} e^{xf_1} \cos(y \lambda_1 + xf_2) dy</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>f_1 = \beta^2 \left [ \ln(y) - C_i(y)\right ] - \cos(y) - y S_i(y)</math>
 
−
: <math>f_2 = y\left [ \ln(y) - C_i(y)\right ] + \sin(y) + \beta^2  S_i(y)</math>
 
−
: <math>C_i(y) \equiv - \int_y^{\infty} \frac{\cos(t)}{t} dt</math>
 
−
: <math>S_i(y) \equiv \int_0^{y} \frac{\sin(t)}{t} dt</math>
 
−
: <math>C = 0.577</math>
 
−
 
 
−
[[Image:SPIM_Vavilou_Landau_ThinAbsorber.jpg | 400 px]]
 
−
 
 
−
==== GEANT4's implementation ====
 
−
 
 
−
GEANT 4 uses the skewness parameter <math>\kappa</math> to determine if it will use a "fluctuations model" to calculate energy straggling or the gaussian model described in section 2.2.1.
 
−
 
 
−
===== kappa > 10 =====
 
−
If
 
−
: <math>\kappa \equiv \frac{ \bar{\Delta}}{W_{max}}</math> > 10
 
−
and we have a thick absorber ( large step size) then the Gausian function in 2.2.1 is used to calculate energy straggling.
 
−
 
 
−
What happens is <math>\Delta E</math> is calculated via <math>\int_{E_i}^{E_f} \frac{dE}{dx} dx</math> then the actual energy loss predicted by the simulation is chosen from a Gaussian distribution to account for energy straggling such that the <math>\sigma</math> of this Gaussian distribution is given by:
 
−
 
 
−
:<math>\sigma^2 = 2 \pi r_2^2m_ec^2N_{el} \frac{Z_h}{\beta^2} T_C s (1 - \frac{\beta^2}{2})</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>N_{el}</math> = electron density of the medium
 
−
:<math>Z_h</math> = charge of the incident particle
 
−
:<math>s</math> = step size
 
−
:<math> T_C</math> = cutoff kinetic energy for <math>\delta </math>-electrons
 
−
 
 
−
<math>T_C</math> tells GEANT where to put the cutoff for using the Gaussian distribution for energy straggling.  This tells the simulation the low energy cutoff where Bethe-Bloch starts to fail due to ionization.
 
−
 
 
−
=====Delta-electrons =====
 
−
What is a <math>\delta</math> - electron?
 
−
 
 
−
<math>\delta</math> - electrons are also known as "knock -on" electrons and delta rays.
 
−
 
 
−
As heavy particles traverse a medium they can ionize electrons from atoms.
 
−
 
 
−
In a cloud chamber (a supercooled volume of super saturated water vapor which ionizes as charged particles pass through)  such and event would look like:
 
−
 
 
−
[[Image:SPIM_DeltaRay_CloudChamber.jpg | 400 px]]
 
−
 
 
−
The physics of ionization is different from the physics used to calculate Bethe-Bloch energy loss.  Remember Bethe-Bloch  starts to break down at low energies below the Bragg peak. 
 
−
 
 
−
Because of this GEANT 4 sets the cutoff for this process to be
 
−
 
 
−
: <math>T_{cut}</math> > 1 keV
 
−
 
 
−
 
 
−
Note:  The BE energies of an electron in Hydrogen is 13.6 ev and the electrons in Argon have binding energies between 15.7 eV and 3.2 keV.
 
−
 
 
−
===== Fluctuations Model: kappa < 10=====
 
−
 
 
−
If <math>\kappa \equiv \frac{ \bar{\Delta}}{W_{max}} < \frac{\Delta E}{T_C}</math>
 
−
 
 
−
Then GEANT 4 uses a "Fluctuations Model" to determine energy loss instead of Bethe-Bloch.
 
−
 
 
−
; Fluctuations Model
 
−
:# the atom is assumed to have on 2 energy levels <math>E_1</math> and <math>E_2</math>
 
−
:# you can excite the atom and lose either <math>E_1</math> or <math>E_2</math> or you can ionize the atom and lose energy according to a <math>\frac{1}{E^2}</math> function <math>u_j</math>.
 
−
 
 
−
The total energy loss in a step will be
 
−
 
 
−
: <math>\Delta E = \Delta E_{exc} + \Delta E_{ion}</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>\Delta E_{exc} = \eta_1 E_1 + \eta_2 E_2</math>
 
−
 
 
−
: <math>\Delta E_{ion} = \sum_{j=1}^{\eta_3} \frac{I}{1 - u_j \frac{T_{up}-I}{T_{up}}}</math>
 
−
 
 
−
:<math>\eta_1</math>, <math>\eta_2</math>, and <math>\eta_3</math> are the number of collisions which are sampled from a poison distribution
 
−
 
 
−
:<math>u_j = \int_{I}^{E_j} \frac{I T_{up}}{T_{up} - I} \frac{dx}{x^2}</math>
 
−
 
 
−
: <math>E_j = \frac{I}{1- rand  \frac{T_{up}-1}{T_{up}}}</math> : rand = random number between 0 and 1
 
−
 
 
−
:<math>T_{up} = \left \{  {~ 1 keV \;  threshold \;energy \;for \; \delta- ray \; production \atop T_{max} \; \;\;\; if \; T_{max} < 1 keV} \right .</math>
 
−
 
 
−
: <math>I</math> = mean ionization energy
 
−
 
 
−
:<math>E_2 \approx (10 eV) Z^2</math>
 
−
 
 
−
:<math>\ln E_1 = \frac{\ln (I) - f_2 \ln (E_2)}{f_1}</math>
 
−
 
 
−
:<math>f_1 + f_2 =1</math>
 
−
 
 
−
:<math>f_2 =\left \{  {0 \; z=1 \atop \frac{2}{z} \; z \ge 2} \right .</math>
 
−
 
 
−
The fluctuation model was comparted with data in
 
−
 
 
−
K. Lassila-Perini and L. Urban, NIM, A362 (1995) pg 416
 
−
 
 
−
The cross sections used for excitation and ionization may be found in
 
−
 
 
−
H. Bichel, Rev. Mod. Phys., vol 60 (1988) pg 663
 
−
 
 
−
=== Range Straggling===
 
−
 
 
−
;Def of Range (R):
 
−
: The distance traveled before all the particles energy is lost.
 
−
 
 
−
:<math>R \equiv \int_0^T \frac{dE}{\frac{dE}{dx}}</math>
 
−
:  = theoretical calucation of the path length traveled by a particle of incident energy <math>T</math>
 
−
 
 
−
: Note units: <math>\left [ R \right ] =  \frac{g}{cm^2} ; \left [ \frac{dE}{dx} \right ] = \frac{MeV \cdot cm^2}{g}</math>
 
−
 
 
−
the Energy Straggling introduced in the previous section results is particles penetrating material to different depths.  The energy straggling results in Range straggling.
 
−
 
 
−
If we do a shielding experiment where we have a source of incident particles of energy E and we count how many "punch" through a material of thickness (x) we would see a trnsmission coeffient <math>\left ( \frac{N_{out}}{N_{in}} \right) </math> which would look like
 
−
 
 
−
[[Image:SPIM_RangeStraggling.jpg | 400 px]]
 
−
 
 
−
====Fractional Range Straggling ====
 
−
 
 
−
<math>\frac{\sigma_R}{R} \equiv</math> fractional range stragling
 
−
 
 
−
Assuming the energy loss of a non-relativistic heavy ion through matter follows a Gaussian (thick absorber)
 
−
 
 
−
Then it can be shown that
 
−
 
 
−
<math>\frac{\sigma_R}{R} \approx \frac{1}{2} \sqrt{\frac{M}{A}}</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>M</math> = mass of the target electrons
 
−
 
 
−
: <math>A</math> = atomic mass of the Projectile
 
−
 
 
−
since
 
−
 
 
−
:<math>m_e = 9.11 \times 10^{-31}</math> kg
 
−
 
 
−
and
 
−
 
 
−
: 1 a.m.u. = <math>1.66 \times 10^{-27}</math> kg
 
−
 
 
−
then
 
−
 
 
−
: <math>\frac{\sigma_R}{R} \approx \frac{1}{2} \sqrt{\frac{9.11 \times 10^{-31}}{1.66 \times 10^{-27}A}}</math>
 
−
: = 1.17 % if using a proton (A=1)
 
−
 
 
−
The above is a "back of the envelope" estimate.  The experimentaly measured values for Cu, Al, and Be target using a proton projectile are
 
−
 
 
−
[[Image:SPIM_RangeStrag_SigmaR_overR.jpg | 400 px]]
 
−
 
 
−
If the incident projectile is an electron then <math>\frac{\sigma_R}{R}  \approx \frac{1}{2}</math> making electron range straggling a vague concept.
 
−
 
 
−
Therre are several definitions of electron range
 
−
 
 
−
;1.) Maximum Range (<math>R_0</math>):
 
−
:This range is defined using the continuous slowing down approximation (CSDA) in the the electrons are assumed to have many collisions over very small dimensions making it appear to be continuous energy loss instead of discrete.  The range is then calculated by integrating over these average energy losses <math>\frac{dE}{dx} \cdot s</math>.
 
−
 
 
−
;2.) Practical Range (<math>R_P</math>):
 
−
: This stopping distance is defined by extrapolating the electron transmission cure to zero (see below).
 
−
 
 
−
[[Image:SPIM_PracticalRangStraggline_4Electrons.jpg | 400 px]]
 
−
 
 
−
=== Electron Capture and Loss ===
 
−
====Bohr Criterion====
 
−
:"A rapidly moving nucleus is fully ionized if its velocity exceeds that of its most tightly bound electron"
 
−
 
 
−
The Bohr Model:
 
−
:<math>\Rightarrow E = \frac{mz^2e^4}{8 \epsilon_0^2 h^2 n^2}</math>
 
−
 
 
−
for the inner most electron (<math>n=1</math>)
 
−
 
 
−
:<math>\frac{1}{2} mv^2 = \frac{mz^2e^4}{2(4\pi \epsilon_0)^2 \hbar^2} \Rightarrow v = \frac{z e^2}{4 \pi \epsilon_0 \hbar}</math>
 
−
 
 
−
 
 
−
:the fine structure constant <math>\alpha \equiv \frac{e^2}{4 \pi \epsilon_0 \hbar c} = \frac{1}{137}</math>
 
−
 
 
−
:<math> v = zc \alpha</math>
 
−
 
 
−
If <math>v > zc \alpha</math> the the nucleus is fully ionized
 
−
 
 
−
or
 
−
 
 
−
if <math>\frac{z}{v/c} = \frac{z}{\beta} < \frac{1}{\alpha} = 137</math>
 
−
 
 
−
alternatively if the ion is moving through a material with a speed such that
 
−
 
 
−
:<math>\frac{z}{\beta} > \frac{1}{\alpha} =137</math>
 
−
 
 
−
 
 
−
Then electrons may be captured by the projectile and lost by the target.
 
−
 
 
−
==== Z-effective====
 
−
Describing the charge state of your heavy ion traveling through matter at a velocity below the Bohr criterion is very complicated.  There is a competition between electron capture and loss.  Accurate cross sections are needed to simulate the process reliably.
 
−
 
 
−
Some insight into this process can be found using the Thomas-Fermi model to describe an atom moving slow enough so it has captured many electrons but fast enough so its not neutral.  In the Thomas-Fermi model the distribution of electrons in an atomic is described as being uniformly distributed such that there are 2 electrons iin each discrete volume of phase space( the space in which all possible states of a system are represented)  defined using planks constant as <math>h^3</math>. 
 
−
 
 
−
For the purpose of simulations you would like a relationship for <math>Z_{eff}</math> in terms of <math>\beta</math> and <math>Z</math>. 
 
−
 
 
−
It is usually adequate to use fits for empirical data as long as we know that we are in the kinematic range in which those fits are valid.
 
−
 
 
−
 
 
−
when <math>E < 10</math> MeV the data indicates that
 
−
 
 
−
: <math>Z_{eff} = Z(1 - e^{-\beta\frac{B}{Z^{2/3}}})</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>B = 130 \pm 5</math>
 
−
:<math>Z_{eff} \equiv</math> effective charge f the projectile = <math>Z - \bar{q}_c</math>
 
−
: <math>Z</math> = number of protons
 
−
:<math>\bar{q}_c</math> = average number of captured electrons
 
−
 
 
−
 
 
−
 
 
−
'''When calculating stopping power for E < 10 MeV you use Zeff in the Bethe=Bloch equation.'''
 
−
 
 
−
Note:  As the ions charge state fluctuates while it slows down (or if accelerated through materials) you will need to recalcualte the energy loss, and as a result you will get larger energy loss fluctuations in this energy range.
 
−
 
 
−
For thin aboserber you will look for stripping and loss cross sections.
 
−
 
 
−
: Here a thin absorber is one whose thickness is less than the charge equilibrium distance defined as the distance traveled until the projectile's velocity is <math> v \ll zc\alpha</math>
 
−
 
 
−
A rule of thumb is that a thin absorber for low energy ions has a thickness <math>\le \frac{5 \frac{\mu g}{cm^2}}{\rho}</math>
 
−
 
 
−
For thick absorbers:  The experimentally determined expression for the change in <math>Z_{eff}</math> from <math>Z</math> is
 
−
 
 
−
:<math>\Delta Z_{eff} = \frac{1}{2} \sqrt{ \left [  Z_{eff} (1 - \frac{Z_{eff}}{Z}^{1.67}\right ] }</math>
 
−
 
 
−
=== Multiple Scattering ===
 
−
 
 
−
The Bethe-Bloch equation tells us how much energy is lost and the work of Moliere describes the angular deflection of the particle which lost the energy thereby leading to a prediction of the Cross-section.  GEANT4 though uses the more complete Lewis theory to describe Multiple Couloumb Scattering (MCS) sometimes generically referred to as multiple scattering.
 
−
 
 
−
There are 3 regions in which coulomb scattering is calculated
 
−
 
 
−
; 1.) Single Scattering:
 
−
: For thin materials.
 
−
: If the probability of more than 1 coulomb scattering is small
 
−
:The use the Rutherford formula for <math>\frac{d \sigma}{d \Omega}</math>
 
−
 
 
−
;2.)Multiple Scattering:
 
−
: In this case the number of independent scatterings is large (N > 20) and the energy loss is small such that the problem can be treated statisticaly to obtain a probability distribution for the net deflection angle <math> [P(\theta)]</math> as a function of the material thickness that is traversed.
 
−
 
 
−
 
 
−
;3.) Plural Scattering:
 
−
: If N <math>\le</math> 20 then you can't use Rutherford to describe the scattering nor use a normal random statistical description.
 
−
 
 
−
see E. Keil, Z. Naturforsch, vol 15 (1960), pg 1031
 
−
 
 
−
 
 
−
Reviews of rigorous multiple scattering calculations may be found in
 
−
: P.C. Hemmer, et. al., Phys. Rev, vol 168 (1968), pg 294
 
−
 
 
−
==== GEANT4's implementation of MSC (N>20) ====
 
−
 
 
−
GEANT4 models MSC when N>20 using model functions to determine the angular and spatial distributions chosen to give the same moments of these distributions as the Lewis theory.
 
−
 
 
−
:H.W. Lewis, Phys. Rev., vol 78 (1950), pg 526
 
−
 
 
−
modern versions of the above are at
 
−
 
 
−
: J.M. Fernandez-Varea, et. al., NIM, B73 (1993), pg 447
 
−
: I. Kawrakow, et. al., NIM, B142 (1998) pg 253
 
−
 
 
−
When N>20 multiple scattering can be described as a statistical process using a modified version of the Boltzman transport equation from statistical mechanics. 
 
−
 
 
−
;Note: The simulation step size is chosen such that (N>20),  If you have materials so thin that N < 20 then GEANT4 will likely skip the material.  (one way around this is to increase the thickness and change the density).  If the material thickness can't be increased because its sandwhiched between two other materials then you will need to write a special step algorithm for the volume and have GEANT4 use it for the step.
 
−
 
 
−
 
 
−
Let <math>f(s,\vec{x},\hat{v}) \equiv</math> the distribution function for a system of incident particles traveling through a material.
 
−
 
 
−
where
 
−
 
 
−
:<math>s =</math> arc length of the particle's path through the material
 
−
:<math>\vec{x} =</math> position of a charged particle
 
−
: <math>\hat{v} =</math> direction of motion of the particle <math>\frac{\vec{v}}{|\vec{v}|}</math>
 
−
 
 
−
The multiple scattering experienced by a single charged particle traveling through the material is then simulated by sampling from the distribution <math>f(s,\vec{x},\hat{v} )</math>
 
−
 
 
−
The governing transport/diffusion equation is based on the continuity equation but with a "sink" term representing the possibility of collisions ejecting particles out of the volume.
 
−
 
 
−
[[Image:SPIM_MultScatDiffEq.jpg | 400 px]]
 
−
 
 
−
:<math>\frac{\partial f(s,\vec{x},\hat{v} ) }{\partial s} + \hat{v} \bullet \vec{\nabla}f(s,\vec{x},\hat{v} ) = N \int \sigma(\hat{v} \bullet\hat{v}^{\prime} )\left [ f(s,\vec{x},\hat{v}^{\prime} )  - f(s,\vec{x},\hat{v} ) \right ] d \hat{v}^{\prime}</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>N</math> = number of atoms per volume
 
−
:<math>\sigma(\hat{v} \bullet\hat{v}^{\prime} )</math> = cross sections for eleastic scattering per Solid angle <math>\left ( \frac{d \sigma}{d \Omega} \right )</math>
 
−
 
 
−
To solve the above diffusion equation the distribution function is expanded in Spherical Harmonics ( <math>Y_{\ell}^m(\theta,\phi)</math> ) and expand <math>\sigma</math> in Legendre Polynomials (<math>P_N(cos \theta)</math>)
 
−
 
 
−
;Note: For Coulomb Scattering in polar coordinates you can write the potential in terms of Legendre Polynomials such that:
 
−
 
 
−
:<math>U=k \frac{q}{r}</math>
 
−
:= <math>k\frac{q}{\sqrt{r^2-a^2-2ar \cos \theta}}</math>  in polar coordinates
 
−
:= <math>k\frac{q}{r} \sum_{n=0}^{\infty} P_n(\cos \theta) \left ( \frac{a}{r}\right )^n</math> (the sqrt term above is expanded using binomial series
 
−
 
 
−
:  <math>f(s,\vec{x},\hat{v} ) = \sum_{\ell,m} f_{\ell,m}(\vec{x},s) Y_{\ell}^m(\hat{v})</math>
 
−
 
 
−
after substituting into the diffusion equation and doing the integral on the righ hand side you get
 
−
 
 
−
:<math>\frac{\partial f_{\ell,m}(\vec{x},s) }{\partial s} + \frac{f_{\ell,m}(s,\vec{x},\hat{v} }{\lambda_{\ell}} = - \sum_{\lambda, i, j} \vec{\nabla} f_{i,j}(\vec{x},s ) \bullet \int Y_{\ell,m}^{\star} \hat{v} Y_{i,j} d \hat{v} \; \; \; \; \; \; \; \;\hat{v} = f(\theta.\phi)</math>
 
−
 
 
−
where
 
−
: <math>\frac{1}{\lambda_{\ell}} = 2 \pi N \int_0^{\pi} \left [ 1-P_{\ell}(\cos \theta)\right ] \sigma(\theta) \sin(\theta) d \theta</math> = <math>\ell^{th}</math> transport mean free path for the <math>f_{\ell}</math> distribution function ( <math>\phi</math> symmetry is assumed making it <math>m</math> independent)
 
−
 
 
−
From the above one can find the average distances traveled and the average deflection andle of the distribution.  Again, see :
 
−
 
 
−
: J.M. Fernandez-Varea, et. al., NIM, B73 (1993), pg 447
 
−
 
 
−
 
 
−
The "moments" of <math>f(s,\vec{x},\hat{v}) </math>  are defined as
 
−
 
 
−
:<math><z> = 2 \pi \int z f(s,\vec{x},\hat{v}) \sin(\theta) d \theta d |\vec{x}| = \lambda_1 \left [ 1-e^{-s/\lambda_1}\right ]</math> = mean geometrical path length
 
−
:<math><\cos(\theta)> = 2 \pi \int_{-1}^1 \sum_{\ell} P_{\ell}(\cos \theta) \int f(s,\vec{x},\hat{v}) \sin(\theta) d \theta d |\vec{x}| = e^{-s/\lambda_1}
 
−
</math>
 
−
:<math>\frac{1}{\lambda_1} = 2 \pi N \int_0^{\pi} \left [ 1-P_1(\cos \theta)\right ] \sigma(\theta) \sin(\theta) d \theta</math>
 
−
 
 
−
Notice there are 3 lengths
 
−
 
 
−
[[Image:SPIM_MultScatDiffEq_PathLength.jpg | 400 px]]
 
−
 
 
−
:<math>s</math> = geometrical path length between endpoints of the step =<math> \left \{  {line \; if \; \vec{B} = 0 \atop arc \; if \; \vec{B} \ne 0 } \right .</math>
 
−
:<math>t</math> = true path length = actual length of the path taken by particle
 
−
:<math><z></math> - mean geometrical path length along the z-axis
 
−
 
 
−
In GEANT4 the <math>\lambda_{\ell}</math>'s are taken from
 
−
 
 
−
If 100 eV < K.E. of electron or positron < 10 MeV
 
−
 
 
−
:D. Liljequist, J. Applied Phys, vol 62 (1987), 342
 
−
:J. Applied Phys, vol 68 (1990), 3061
 
−
 
 
−
If K.E. > 10 MeV
 
−
 
 
−
:R. Mogol, Atomic Data, Nucl, Data tables, vol 65 (1997) pg 55
 
−
 
 
−
 
 
−
 
 
−
with <z> now known GEANT will try to determine "t" for the energy loss and scattering calculations
 
−
 
 
−
as model is used for this where
 
−
 
 
−
:<math>t=\frac{1}{\alpha} \left [ 1 - (1- \alpha \omega z)^{\frac{1}{\omega}})\right ]</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>\omega = 1 + \frac{1}{\alpha \lambda_{10}}</math>
 
−
: <math>\alpha =\left \{  {\frac{\lambda_{10} - \lambda_{11}}{s \lambda_{10}}\;\;\;\; K.E. \ge M_{particle} \atop \frac{1}{R}\;\;\;\; K.E. < M_{particle}} \right .</math>
 
−
: <math>s</math> = stepsize
 
−
: <math>\lambda_{10} - \frac{\lambda_1}{1-\alpha s}</math>
 
−
:<math>\lambda_{11} = \lambda_1</math> at end of strep
 
−
 
 
−
while <math><cos \theta ></math> is calculable, GEANT4 evaluates <math>\cos (\theta)</math> from a probability distribution whose general form is
 
−
 
 
−
:<math>g[\cos(\theta)] - p \left ( qg_1[\cos(\theta)] + (1=q)g_3[\cos(\theta)] \right ) + (1-p)g_2[\cos(\theta)]</math>
 
−
 
 
−
where
 
−
 
 
−
:<math>g_1(x) = C1e^{-a(1-x)}</math>
 
−
:<math>g_2(x) = \frac{C_2}{(b-x)^d}</math>
 
−
:<math>g_3(x) = C_3</math>
 
−
 
 
−
:<math>C_1, C_2, C_3</math> are normalization constants
 
−
:<math>p,q,a,b,d</math> are parameters which follow the work reported in
 
−
 
 
−
:V.L. Highland, NIM, vol 219 (1975) pg497
 
−
 
 
−
The GEANT4 files in version 4.8 were located in
 
−
 
 
−
/source/processes/electromagnetic/utils/src/G4VMultipleScattering.cc
 
−
 
 
−
and
 
−
 
 
−
/source/processes/electromagnetic/standard/src/G4MscModel.cc
 
−
 
 
−
/source/processes/electromagnetic/standard/src/G4MultipleScattering.cc
 
−
 
 
−
== Interactions of Electrons and Photons with Matter==
 
−
=== Bremsstrahlung===
 
−
;Definition:  Radiation produced when a charged particle is deflected by the electric field of nuclei in a material.
 
−
:Note: There is also electron-electron brehmstrahlung but the interaction is with the electric field of the materials atomic electrons.
 
−
 
 
−
The Cross section formula given in Formula 3Cs, pg 928 of reference H.W. Koch & J.W Motz, Rev. Mod. Phys., vol 31 (1959) pg 920 as
 
−
 
 
−
;Note: Bethe & Heitler first calculated this radiation in 1934 which is why you will sometimes hear Bremstrahlung radiation refererd to as Bethe-Heitler.
 
−
 
 
−
:<math>d \sigma = 4 Z^2r_e^2 \alpha \frac{d \nu}{\nu} \left \{ \left (1 + \left( \frac{E}{E_0} \right )^2 \right ) \left [ \frac{\phi_1(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] - \frac{2E}{3E_0} \left [  \frac{\phi_2(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] \right \} </math>
 
−
 
 
−
where
 
−
 
 
−
: <math>E_0</math> = initial total energy of the electron
 
−
:<math>E</math> = final total energy of the electron
 
−
: <math>\nu = \frac{E-E_0}{h}</math> = energy of the emitted photon
 
−
:<math>Z</math> = Atomic number = number of protons in target material
 
−
: <math>\gamma = \frac{100 m_ec^2 h \nu}{E_0 E Z^{1/3}}</math> = charge screening parameter
 
−
:<math>f(Z) = (Z \alpha)^2 \sum_1^{\infty} \frac{1}{ n [ n^2 + (Z \alpha)^2]}</math>
 
−
: <math>\sim (Z \alpha)^2 \left \{ \frac{1}{1+(Z \alpha)^2} +0.20206 - 0.0369(Z \alpha)^2 + 0.0083 (Z \alpha)^4 - 0.002 (Z \alpha)^6\right \}</math>
 
−
:<math>\alpha = \frac{1}{137}</math>
 
−
: <math>\phi_1</math> and <math>\phi_2</math>  = screening functions that depend on Z
 
−
 
 
−
if <math>Z \ge 5</math>
 
−
 
 
−
:<math>\phi_1(\gamma) = 20.863 - 2 \ln[1+(0.55 \gamma)^2] - 4[1-0.6e^{-0.98} - 0.4e^{-3 \gamma/2}]</math>
 
−
:<math>\phi_2(\gamma) = \phi_1(\gamma) - \frac{2}{3}(1+6.5 \gamma + 6 \gamma^2)</math>
 
−
 
 
−
 
 
−
For Z<5 see Tsai, Rev.Mod. Phys., vol 46 (1974) pg 815
 
−
 
 
−
:if <math>3 \ge Z < 5</math> use Equation 3.46 and 3.47
 
−
 
 
−
:if <math> Z < 2</math> use Equation 3.25 and 3.26
 
−
 
 
−
;Note: Energy loss via Bethe-Bloch is due to coulomb deflection and is a '''continuous''' process while Bremstrahlung is a '''discrete''' process (emission of photons)
 
−
 
 
−
;We now know 2 ways charged particles can loose energy when passsing through matter.
 
−
 
 
−
;Energy loss: <math>\left ( \frac{dE}{dx} \right )_{tot} = \left ( \frac{dE}{dx} \right )_{rad} + \left ( \frac{dE}{dx} \right )_{col}</math>
 
−
:<math>{rad} \Rightarrow</math> : Bremstrahlung
 
−
:<math>{col} \Rightarrow</math> : Bethe-Bloch (collision)
 
−
 
 
−
;<math>-\left ( \frac{dE}{dx} \right )_{rad} = N \int_o^{\nu_0} \left ( h \nu \right ) \left ( \frac{d \sigma}{d \nu} \right ) d \nu</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>N = \frac{ number\; atoms}{cm^3} = \frac{\rho N_a}{A} = \frac{density \;\times\; Avagadros\;\;Number}{Atomic number}</math>
 
−
:<math>\left ( h \nu \right )</math> = Energy of emmitted photon
 
−
:<math>\left ( \frac{d \sigma}{d \nu} \right )</math> = Probabitlity of Energy loss
 
−
 
 
−
 
 
−
The quantity<math> \Phi_{rad} </math> is defined such that
 
−
 
 
−
:<math>\Phi_{rad} \equiv \frac{1}{E_0} \int_o^{\nu_0} \left ( h \nu \right ) \left ( \frac{d \sigma (E_0,\nu)}{d \nu} \right ) d \nu</math>
 
−
 
 
−
<math>\Phi_{rad}</math> is a macroscopic function of a given material rather than just the energy <math>\nu</math> which we will use to define a common property of materials known as the radiation length <math>\left ( R_0=\frac{1}{N\Phi_{rad}} \right )</math>
 
−
 
 
−
:<math>\Phi_{rad} = 4Z^2r_e^2 \alpha \left \{ { \left [  \ln(\frac{2E}{m_ec^2}) - \frac{1}{3} - f(z) \right ] \gamma >> 1 \atop  \left [  \ln(183E^{-1/3}) + \frac{1}{18} - f(z) \right ] \gamma \sim 0} \right .</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>\gamma >> 1</math> case is no screening and <math>1 \ll \frac{E_0}{m_e c^2}  < \frac{1}{\alpha Z^{1/3}}</math>
 
−
: <math>\gamma \; \sim \; 0</math> case has <math> \frac{E_0}{m_e c^2}  \gg \frac{1}{\alpha Z^{1/3}}</math>
 
−
 
 
−
The energy loss equation becomes
 
−
 
 
−
: <math>- \frac{dE}{ddx} = N E_0 \Phi_{rad}</math>
 
−
 
 
−
;Note: for intermediate value of <math>\gamma</math> you need to integrate numerically
 
−
:<math>\left ( \frac{dE}{dx} \right )_{rad}\propto Z^2E</math> : Bremstrahlung
 
−
:<math>\left ( \frac{dE}{dx} \right )_{col}\propto Z \ln (E)</math> : Bethe-Bloch
 
−
 
 
−
The illustration below shows the relative contributions of Bethe-Bloch and Bremsstrahlung to the total energy loss according to the above functional dependence.  At low energies the physics of collisions dominates the loss (Bethe-Bloch) and as energy increases the discrete loss by radiation begins to dominate. 
 
−
 
 
−
[[Image:SPIM_Bethe-Brem_Eloss-vs-Energy.jpg | 300 px]]
 
−
 
 
−
 
 
−
;Critical Energy<math> E_C</math>:
 
−
At the critical energy <math>E_C</math> the two energy loss processes contribute equally to the total energy lost by a charged particle interacting with matter.
 
−
:<math>E_C \equiv</math> energy at which <math>\left ( \frac{dE}{dx} \right )_{rad} = \left ( \frac{dE}{dx} \right )_{col}</math>
 
−
 
 
−
In the PDG
 
−
 
 
−
:<math>E_C \sim \frac{800 MeV}{Z+1.2}</math>
 
−
 
 
−
;Examples:
 
−
{| border="3"  cellpadding="20" cellspacing="0"
 
−
|colspan="2"|
 
−
====E_C====
 
−
|-
 
−
|Material
 
−
|<math>E_C</math> (MeV)
 
−
|-
 
−
|Pb
 
−
|9.51
 
−
|-
 
−
|Fe
 
−
|27.4
 
−
|-
 
−
|Cu
 
−
|24.8
 
−
|-
 
−
|Al
 
−
|51
 
−
|}
 
−
 
 
−
==== Electron-Electron Bremstrahlung ====
 
−
; Electron electron bremstrahlung: The radiation produced as 2 electrons pass near eachother
 
−
: <math>d \sigma</math> is essentially the same except you have <math>z</math> instead of <math>Z</math> thereby adding a <math>Z</math> term and not a <math>Z^2</math> term
 
−
 
 
−
refernce:pg 947 from Koch and Motz, Rev. Mod. Phys, vol 31 (1959) pg 920 [[Image:SPIM_Koch andMotz_RevModPhysv31_1959pg920.pdf]]
 
−
 
 
−
as a result
 
−
 
 
−
<math>d \sigma_{tot} = \frac{Z(Z+1)}{Z^2} d \sigma_{Brem}</math>
 
−
: = <math>4 Z(Z+1)r_e^2 \alpha \frac{d \nu}{\nu} \left \{ \left (1 + \left( \frac{E}{E_0} \right )^2 \right ) \left [ \frac{\phi_1(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] - \frac{2E}{3E_0} \left [  \frac{\phi_2(\gamma)}{4} - \frac{1}{3} \ln Z -f(Z)\right ] \right \} </math>
 
−
 
 
−
Most calculations ignore electron-electron Brehmstrahung because its linear in Z and doesn;t become important until low Z where measured atomic form factors are actually used and not Form factors calulated by the Thomas-Fermi-Moliere Model (Z>4).
 
−
 
 
−
==== Radiation Length (Xo)====
 
−
 
 
−
;Radiation Length<math> (X_0)</math>: The distance an electron travels through matter until loosing <math>\frac{1}{e}</math> of its energy due to radiation <math>\left ( \frac{dE}{dx} \right )_{rad}</math>.
 
−
 
 
−
in the high energy limit where <math>\left ( \frac{dE}{dx} \right )_{col}</math> can be ignored <math>( E > E_C )</math>
 
−
 
 
−
:<math>\frac{dE}{dx} = NE_0 \Phi_{rad} \Rightarrow \int_{E_0}^{E} \frac{dE}{E} = -\int_0^X N \Phi_{rad} dx</math>
 
−
: <math>\Rightarrow \ln(\frac{E}{E_0}) =-N \Phi_{rad}X</math>
 
−
 
 
−
or
 
−
 
 
−
: <math>E = E_0 e^{-N \Phi_{rad}X} = E_0 e^{-\frac{X}{X_0}}</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>X_0 \equiv \frac{1}{N \Phi_{rad}}</math> = Radiation Length of a given material
 
−
 
 
−
ie:
 
−
:if <math> X = X_0</math>  Then <math>E=\frac{1}{e} E_0</math> = Energy of electron after traveling a distance of <math>X_0</math> through the material
 
−
 
 
−
 
 
−
{| border="3"  cellpadding="20" cellspacing="0"
 
−
|colspan="2"|
 
−
 
 
−
====Table of Radiation Lengths for several materials====
 
−
|-
 
−
|-
 
−
|Material
 
−
|<math>X_0</math> (cm)
 
−
|-
 
−
|Air
 
−
|30,050
 
−
|-
 
−
|Al
 
−
|8.9
 
−
|-
 
−
|Cu
 
−
|1.43
 
−
|-
 
−
|Fe
 
−
|1.76
 
−
|-
 
−
|H2O
 
−
|36.1
 
−
|-
 
−
|NaI
 
−
|2.59
 
−
|-
 
−
|Pb
 
−
|0.56
 
−
|-
 
−
|Polystyrene
 
−
|42.9
 
−
|-
 
−
|Scintillators
 
−
|42.2
 
−
|-
 
−
|}
 
−
 
 
−
 
 
−
;If we have complete screening <math> (\gamma=0)</math>
 
−
:Then <math>\frac{1}{X_0} = N \Phi_{rad} = 4 \alpha r_e^2 \frac{N_A}{A} \left \{ Z^2 \left [ L_{rad} - f(Z)\right ] + ZL_{rad}^{\prime}\right \}</math>
 
−
: = <math>\frac{Z^2 \left [ L_{rad} - f(Z)\right ] + ZL_{rad}^{\prime}}{716.408 \frac{g}{cm^2}A}</math>
 
−
 
 
−
where
 
−
 
 
−
: <math>L_{rad} \equiv  \frac{1}{4} [\phi_1(\gamma=0) - \frac{4}{3} \ln(Z)] = \left \{ {1 + \int_0^{m_e} [ 1-\frac{F(q)}{Z}]^2 \frac{dq}{q} \;\;\; Z\le4\atop \ln(184.15 Z^{-1/3} \;\;\; Z>4} \right . </math>= radiation logarithm for elastic Atomic scattering
 
−
:<math>L_{rad}^{\prime} \equiv \frac{1}{4} [ \phi_2(\gamma=0) - \frac{8}{3}\ln Z]= \left \{ {1 + \frac{1}{2}\int_0^{m_e} \frac{G_2^{inel}(t)}{Z}\frac{dt}{t} \;\;\; Z\le4\atop \ln(1194 Z^{-2/3} \;\;\; Z>4} \right .</math> = radiation logarithm for inelastic Atomic scattering
 
−
:<math>f(Z) = \alpha^2 Z^2 \left [ \frac{1}{1+\alpha^2 Z^2} + 0.20206 - 0.0369 \alpha^2 Z^2 + 0.0083 \alpha^4 Z^4 -0.002 \alpha^6 Z^6 \right ]</math>  :Z < 92
 
−
 
 
−
 
 
−
;Quick <math>X_0</math> Estimates
 
−
: <math> X_0 = \frac{716.4 \left ( \frac{g}{cm^2} \right ) A }{Z(Z+1) \ln \left ( \frac{287}{\sqrt{Z}} \right )}</math>
 
−
 
 
−
;Examples of Radiation length
 
−
 
 
−
;<math>\frac{1}{e} = \frac{1}{2.72} \sim \frac{1}{3}</math> : an electron has lost 1/3 of its original energy after traveling 1 radiation length (1 <math>X_0</math>) through the material
 
−
 
 
−
;<math>\frac{1}{e^2}  \sim \frac{1}{7}</math> : an electron has lost 1/7 of its original energy after traveling 2 radiation lengths (2 <math>X_0</math>) through the material
 
−
 
 
−
;<math>\frac{1}{e^3}  \sim \frac{1}{20}</math> : an electron has lost 1/20 of its original energy after traveling 3 radiation lengths (3 <math>X_0</math>) through the material
 
−
 
 
−
; After 2.3 radiation lengths the electron energy is down by a factor of 10 from its original value.
 
−
 
 
−
====Bremstrahlung in GEANT 4 ====
 
−
 
 
−
GEANT4 uses an energy cut off  <math>(T_c, k_c)</math> to decide whether to use a continuous energy loss algorithm (msc, Bethe-Bloch) or to generate a secondary particle (photon) and use Bremstrahliung.
 
−
 
 
−
:<math>T_C</math> = incident particle K.E. cutof = secondary particle production threshold
 
−
:<math>k_c</math> = photon energy cutoff below which photons are treated as continuous energy loss.
 
−
 
 
−
;if <math>E_{secondary}<T_C</math> then no photon is created and the effect of the soft reaction is treated as a continuous energy loss via
 
−
:<math>E_{loss}^{Brem} (Z,T,k) = \int_0^{k_c} \left ( \frac{ d \sigma(z,T,k)}{dk}\right ) k dk</math> = continuous energy loss via "soft" photon emission
 
−
 
 
−
:<math>\frac{ d \sigma(z,T,k)}{dk}</math> = cross sections parametrerized by the Evaluated Electrons Data Library (EEDL)
 
−
: reference: J. Tuli, "Evaluated Nuclear Structure Data File", BNL-NCS - 51655 -Rev 87, 1987 from Brookhaven Nat. Lab
 
−
: see  [http://www.nndc.bnl.gov National Nuclear Data Center]:
 
−
 
 
−
To improve simulation speed though, GEANT 4 actually uses a fit to the above cross sections such that
 
−
 
 
−
:<math>E_{loss}^{Brem} (Z,T,k) = (2 - C_{th}Z^{1/4}) \frac{Z(Z+\epsilon_{\ell})(T+M)^2}{T+2M} \left [ \frac{k_C}{T}\right ]^{\beta} \frac{a+b\frac{T}{T_{\ell m}}}{1+c \frac{T}{T_{\ell m}} \frac{f_{\ell}}{N_a}}</math>
 
−
 
 
−
=== Photo-electric effect===
 
−
=== Compton Scattering ===
 
−
=== Pair Production ===
 
−
 
 
−
== Hadronic Interactions ==
 
−
=== Neutron Interactions ===
 
−
==== Elastic scattering====
 
−
 
 
−
==== Inelasstic Scattering====
 
−
 
 
−
== Homework Problems==
 
−
[[HomeWork_Simulations_of_Particle_Interactions_with_Matter]]
 

Latest revision as of 22:31, 15 January 2024

Class Admin

TF_SPIM_ClassAdmin

Homework Problems

HomeWork_Simulations_of_Particle_Interactions_with_Matter

Introduction

TF_SPIM_Intro

Energy Loss

TF_SPIM_StoppingPower

Ann. Phys. vol. 5, 325, (1930)

Interactions of Electrons and Photons with Matter

TF_SPIM_e-gamma

Hadronic Interactions

TF_SPIM_HadronicInteractions

Final Project

A final project will be submitted that will be graded with the following metrics:

1.) The document must be less than 15 pages.

2.) The document must contain references in a bibliography (5 points) .

3.) A comparison must be made between GEANT4's prediction and either the prediction of someone else or an experimental result(30 points).

4.) The graphs must be of publication quality with font sizes similar or larger than the 12 point font (10 points).

5.) The document must be grammatically correct (5 points).

6.) The document format must contain the following sections: An abstract of 5 sentences (5 points) , an Introduction(10 points), a Theory section (20 points) , if applicable a section describing the experiment that was simulated, a section delineating the comparisons that were made, and a conclusion( 15 points).

Resources

GEANT4 Home Page

ROOT Home page

Fermi Lab Example


NIST Range Tables

X-ray specturm

Installing_GEANT4.9.3_Fsim

Saving/restoring Random number seed

You save the current state of the random number generator with the command

/random/setSavingFlag 1

/run/beamOn 100

/random/saveThisRun

A file is created called

currentEvent.rndm

/control/shell mv currentEvent.rndm currentEvent10.rndm


You can restore the random number generator and begin generating random number from the last save time

/random/resetEngineFrom currentEvent.rndm

Building GEANT4.11

4.11.2

TF_GEANT4.11

Building GEANT4.10

4.10.02

TF_GEANT4.10.2

4.10.01

TF_GEANT4.10.1

Building GEANT4.9.6

TF_GEANT4.9.6

Building GEANT4.9.5

TF_GEANT4.9.5

An old version of Installation notes for versions prior to 9.5

Old Install Notes


Visualization Libraries:

OpenGL

DAWN


Coin3D

Compiling G4 with ROOT

These instruction describe how you can create a tree within ExN02SteppingVerbose to store tracking info in an array (max number of steps in a track is set to 100 for the desired particle)

G4CompileWRootforTracks

Using SLURM

http://slurm.schedmd.com/quickstart.html


https://rc.fas.harvard.edu/resources/documentation/convenient-slurm-commands/

simple batch script for one process job

create the file submit.sbatch below 

#!/bin/sh
#SBATCH --time=1
cd src/PI
./PI_MC 100000000000000

the execute

sbatch submit.sbatch

check if its running with

squeue

to kill a batch job

scancel JOBID

On minerve

Sample script to submit 10 batch jobs.

the filename is minervesubmit and you run like 

source minervesubmit

cd /home/foretony/src/GEANT4/geant4.9.5/Simulations/N02wROOT/batch
qsub submit10mil
qsub submit20mil
qsub submit30mil
qsub submit40mil
qsub submit50mil
qsub submit60mil
qsub submit70mil
qsub submit80mil
qsub submit90mil
qsub submit100mil

The file submit10mil looks like this 

#!/bin/sh
#PBS -l nodes=1
#PBS -A FIAC
#PBS -M foretony@isu.edu
#PBS -m abe
#
source /home/foretony/src/GEANT4/geant4.9.5/geant4.9.6-install/bin/geant4.sh
cd /home/foretony/src/GEANT4/geant4.9.5/Simulations/N02wROOT/batch/10mil
../../exampleN02 run1.mac > /dev/null


use 

qstat

to check that the process is still running

use 

qdel jobID#

if you want to kill the batch job, the jobID number shows up when you do stat.

for example 

[foretony@minerve HW10]$ qstat
Job id                    Name             User            Time Use S Queue
------------------------- ---------------- --------------- -------- - -----
27033.minerve             submit           foretony        00:41:55 R default        
[foretony@minerve HW10]$ qdel 27033
[foretony@minerve HW10]$ qstat

Definitions of Materials

File:MCNP Compendium of Material Composition.pdf

Minerve2 GEANT 4.10.1 Xterm error

On OS X El Capitan V 10.11.4 using XQuartz

~/src/GEANT4/geant4.10.1/Simulations/B2/B2a/exsmpleB2a 


# Use this open statement to create an OpenGL view:
/vis/open OGL 600x600-0+0
/vis/sceneHandler/create OGL
/vis/viewer/create ! ! 600x600-0+0
libGL error: failed to load driver: swrast
X Error of failed request:  BadValue (integer parameter out of range for operation)
  Major opcode of failed request:  150 (GLX)
  Minor opcode of failed request:  3 (X_GLXCreateContext)
  Value in failed request:  0x0
  Serial number of failed request:  25
  Current serial number in output stream:  26



TF_SPIM_OLD

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