TF SPIM Intro - Revision history

Foretony: /* Definitions */

2025-01-22T17:41:50Z

Definitions

Foretony: /* Example 1: Create Ntuple and Draw Histogram */

2025-01-17T17:47:59Z

Example 1: Create Ntuple and Draw Histogram

Foretony: /* Example 1: Create Ntuple and Draw Histogram */

2025-01-17T17:47:30Z

Example 1: Create Ntuple and Draw Histogram

Foretony: /* Example 1: Create Ntuple and Draw Histogram */

2025-01-17T17:47:01Z

Example 1: Create Ntuple and Draw Histogram

Foretony: /* A Root Primer */

2025-01-17T17:43:04Z

A Root Primer

Foretony: /* Example 3: using UNIX to compile a RNG */

2025-01-17T17:38:42Z

Example 3: using UNIX to compile a RNG

Foretony: /* Example 4: Elastic Scattering */

2016-09-02T14:54:52Z

Example 4: Elastic Scattering

Foretony: /* The Monte Carlo method */

2016-08-24T17:44:27Z

The Monte Carlo method

Foretony: /* Example 1: P(E=5 eV) */

2016-08-24T17:35:58Z

Example 1: P(E=5 eV)

Foretony: /* Example 1: P(E=5 eV) */

2016-08-24T17:35:18Z

Example 1: P(E=5 eV)

@@ Line 238: / Line 238: @@
 ; Solid Angle
 :[[Image:SolidAngleDefinition.jpg]]
-: <math>\Omega</math>= surface area of a sphere covered by the detector
+: <math>\Omega</math>= solid angle of a sphere covered by the detector
 : ie;the detectors area projected onto the surface of a sphere
 :A= surface area of detector

@@ Line 202: / Line 202: @@
 === Example 1: Create Ntuple  and Draw Histogram===
-Look for the program "ascii2root.C" in Moodle
+Look for the program "asci2root.C" in Moodle
 copy and paste it into you editor on the machine you would like to run root on.

@@ Line 208: / Line 208: @@
 then try the following
-#root
+ root
-your shell promt will change to look like thei : root [0]
 type
-#.x asci2root.C
+ .x asci2root.C
 then exit the root program with
-#.q
+ .q
 and restart it with
-#root -l sim.root
+ root -l sim.root
 and try the command
-Simm->Draw("evt.x");
+ Simm->Draw("evt.x");
 == Cross Sections ==

← Older revision		Revision as of 17:47, 17 January 2025
Line 201:		Line 201:

	=== Example 1: Create Ntuple and Draw Histogram===		=== Example 1: Create Ntuple and Draw Histogram===
		+
		+	Look for the program "ascii2root.C" in Moodle
		+
		+	copy and paste it into you editor on the machine you would like to run root on.
		+
		+	then try the following
		+
		+	#root
		+	your shell promt will change to look like thei : root [0]
		+
		+	type
		+
		+	#.x asci2root.C
		+
		+	then exit the root program with
		+
		+	#.q
		+
		+	and restart it with
		+
		+	#root -l sim.root
		+
		+	and try the command
		+
		+	Simm->Draw("evt.x");

	== Cross Sections ==		== Cross Sections ==

@@ Line 179: / Line 179: @@
 == A Root Primer ==
 In bash shell do
-  export ROOTSYS=~tforest/src/ROOT/root
+  export ROOTSYS=~foretony/src/ROOT/root
 if chsh do
-  setenv ROOTSYS ~tforest/src/ROOT/root
+  setenv ROOTSYS ~foretony/src/ROOT/root
 To start the root program type

@@ Line 168: / Line 168: @@
 Step
-# login to minerve.<br> [http://physics.isu.edu/~tforest/Classes/NucSim/Day1/XwindowsOnWindows.html click here for a description of logging in if using windows] otherwise (ssh username@minerve.cose.isu.edu)
+# login to thorshammer (ssh username@thorshammer.rdc.isu.edu)
 # mkdir src
 # cd src
-# scp -r foretony@minerve.cose.isu.edu:src/PI ./
+# mkdir PI
 # ls
-# cd PI
+# g++ -o PI PI.cc
-# make
+#./PI
 == A Root Primer ==

@@ Line 257: / Line 257: @@
 This example is an example of classical scattering.
-Our goal is to find <math>\sigma(\theta)</math> for an elastic collision of 2 impenetrable spheres of diameter <math>a</math>.  We need to look for a relationship between the impact parameter<math> b</math> and the scattering angle<math> \theta</math>.  To find this relationship, let's solve this elastic scattering problem by  describing the collision using the Center of Mass (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.
+Our goal is to find <math>\sigma(\theta)</math> for an elastic collision of 2 impenetrable spheres of diameter <math>a</math>.  We need to look for a relationship between the impact parameter <math> b</math> and the scattering angle <math> \theta</math>.  To find this relationship, let's solve this elastic scattering problem by  describing the collision using the Center of Mass (C.M.) coordinate system in terms of the reduced mass.  As we shall see, the 2-body collision becomes a 1-body problem  when a C.M. coordinate system is used.  Then we will describe the motion of the reduced mass in the C.M. Frame.
 [[Image:SPIM_ElasCollis_Lab_CM_Frame.jpg | 500 px]]

@@ Line 103: / Line 103: @@
 The above particle detector was an example of describing a stochastic process using a probability distribution to determine the likely hood of finding an atom with a certain energy.
-Quantun Mechanics is perhaps the best example such a non-deterministic systems.  The canonical systems in Thermodynamics is another example.
+Physics at the Quantum Mechanics  scale contains some of the clearest examples of such a non-deterministic systems.  The canonical systems in Thermodynamics is another example.
 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.

@@ Line 58: / Line 58: @@
 Since we have 12.011 grams of carbon and 1 mole of carbon = 12.011 g = <math>6 \times 10^{23} </math>carbon atoms, we would 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>.  Note: The mass of the earth is about <math>10^{27}</math> g <math>\approx 10^{50}</math> atoms, so a carbon atom with an energy of 5 eV would be difficult to observe  in a detector the size of the earth .
-The energy we expect to see would be calculated by
+The average energy we expect to see would be calculated by
 <math><E> = \int_{0}^{\infty} E \cdot P(E) dE</math>

@@ Line 35: / Line 35: @@
 <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.
+Physically, <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.
@@ Line 50: / Line 50: @@
 <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>
-or in other words the precise mathematical calculation of the probability may be approximated by just using the distribution function alone
+or in other words the probability may be approximated by just using the distribution function alone
 <math>P(E=5eV) = e^{-5/0.0258} \approx 10^{-85}</math>
@@ Line 56: / Line 56: @@
 This approximation breaks down as <math>E \rightarrow 0.0258 eV</math>
-Since we have 12.011 grams of carbon and 1 mole of carbon = 12.011 g = <math>6 \times 10^{23} </math>carbon atoms
+Since we have 12.011 grams of carbon and 1 mole of carbon = 12.011 g = <math>6 \times 10^{23} </math>carbon atoms, we would 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>.  Note: The mass of the earth is about <math>10^{27}</math> g <math>\approx 10^{50}</math> atoms, so a carbon atom with an energy of 5 eV would be difficult to observe  in a detector the size of the earth .
 The energy we expect to see would be calculated by