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

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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>
 
<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.
  
 
=== The Monte Carlo method ===
 
=== The Monte Carlo method ===

Revision as of 15:53, 31 August 2007

Overview

Particle Detection

A device detects a particle only after the particle transfers energy to the device.

Energy intrinsic to a device depends on the material used in a device

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).

Statistical Thermodynamics tells us that the canonical energy distribution of the atoms is given by the Maxwell-Boltzmann statistics such that

[math]P(E) = \frac{1}{kT} e^{-\frac{E}{kT}}[/math]

[math]P(E)[/math] represents the probability of any atom in the system having an energy [math]E[/math] where

[math]k= 1.38 \times 10^{-23} \frac{J}{mole \cdot K}[/math]

Note: You may be more familiar with the Maxwell-Boltzmann distribution in the form

[math]N(\nu) = 4 \pi N \left ( \frac{m}{2\pi k T} \right ) ^{3/2} v^2 e^{-mv^2/2kT}[/math]

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]

Example 1

What is the probability that an atom in a 12.011 gram block of carbon would have and energy of 5 eV?

First lets check that the probability distribution is Normailized; ie: does [math]\int_0^{\infty} P(E) dE =1[/math]?


[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]

[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.

The Monte Carlo method

A Unix Primer

A Root Primer

Example 1: Create Ntuple and Draw Histogram

Cross Sections

Deginitions

Example : Elastic Scattering

Lab Frame Cross Sections

Stopping Power

Bethe Equation

Classical Energy Loss

Bethe-Bloch Equation

Energy Straggling

Thick Absorber

Thin Absorbers

Range Straggling

Electron Capture and Loss

Multiple Scattering

Interactions of Electrons and Photons with Matter

Bremsstrahlung

Photo-electric effect

Compton Scattering

Pair Production

Hadronic Interactions

Neutron Interactions

Elastic scattering

Inelasstic Scattering