Converting to barns - Revision history

Vanwdani at 17:54, 29 December 2018

2018-12-29T17:54:27Z

Vanwdani at 17:53, 29 December 2018

2018-12-29T17:53:51Z

Vanwdani at 06:34, 9 August 2018

2018-08-09T06:34:25Z

Vanwdani at 01:13, 5 April 2018

2018-04-05T01:13:14Z

Vanwdani at 19:53, 4 April 2018

2018-04-04T19:53:38Z

Vanwdani at 19:31, 4 April 2018

2018-04-04T19:31:55Z

Vanwdani at 19:25, 4 April 2018

2018-04-04T19:25:21Z

Vanwdani at 19:16, 4 April 2018

2018-04-04T19:16:02Z

Vanwdani at 18:56, 4 April 2018

2018-04-04T18:56:46Z

Vanwdani at 18:44, 4 April 2018

2018-04-04T18:44:28Z

← Older revision		Revision as of 17:54, 29 December 2018
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−	~~This~~ can be simplified to the form	+	For ultra-relativistic particles, this can be simplified to the form

@@ Line 1: / Line 1: @@
 Using the equation from Landau and Lifshitz(Quantum Electrodynamics Volume 4)
@@ Line 112: / Line 120: @@
 ----
-=Links=
+<center><math> \underline{\textbf{Navigation} }</math>
-[[DV_RunGroupC_Moller#Comparing_experimental_vs._theoretical_for_M.C3.B8ller_differential_cross_section_11GeV|Back]]
+</center>

← Older revision		Revision as of 06:34, 9 August 2018
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−	Using the equation from ~~[1]~~	+	Using the equation from Landau and Lifshitz(Quantum Electrodynamics Volume 4)

	<center><math>\frac{d\sigma}{d\Omega}=\frac{ e^4 }{8E^2}\left \{\frac{1+cos^4\frac{\theta}{2}}{sin^4\frac{\theta}{2}}+\frac{1+sin^4\frac{\theta}{2}}{cos^4\frac{\theta}{2}}+\frac{2}{sin^2\frac{\theta}{2}cos^2\frac{\theta}{2}} \right \}</math></center>		<center><math>\frac{d\sigma}{d\Omega}=\frac{ e^4 }{8E^2}\left \{\frac{1+cos^4\frac{\theta}{2}}{sin^4\frac{\theta}{2}}+\frac{1+sin^4\frac{\theta}{2}}{cos^4\frac{\theta}{2}}+\frac{2}{sin^2\frac{\theta}{2}cos^2\frac{\theta}{2}} \right \}</math></center>

@@ Line 82: / Line 82: @@
-Where <center><math>E^2=p^2+m^2=(53\ MeV)^2+(0.511\ MeV)^2 \rightarrow E=53.0025\ MeV</math></center>
+Where <center><math>E^2=p^2+m^2=(11000\ MeV)^2+(0.511\ MeV)^2 \rightarrow E=11000\ MeV</math></center>
-<math>\gamma=\frac{E}{m}=\frac{53.0025\ MeV}{0.511\ MeV}=103.72</math>
+<math>\gamma=\frac{E}{m}=\frac{11000\ MeV}{0.511\ MeV}=21526</math>
-<math>z'=\frac{1}{\gamma}z=\frac{1}{103.72}5\ cm=0.048\ cm</math>
+<math>z'=\frac{1}{\gamma}z=\frac{1}{21526}5\ cm=2.32\times 10^{-4}\ cm</math>

@@ Line 82: / Line 82: @@
-Where <center><math>E^2=p^2+m^2=(11000\ MeV)^2+(0.511\ MeV)^2 \rightarrow E=11000.0000119\ MeV</math></center>
+Where <center><math>E^2=p^2+m^2=(53\ MeV)^2+(0.511\ MeV)^2 \rightarrow E=53.0025\ MeV</math></center>
-<math>\gamma=\frac{E}{m}=\frac{11000.0000119\ MeV}{0.511\ MeV}=21526.41</math>
+<math>\gamma=\frac{E}{m}=\frac{53.0025\ MeV}{0.511\ MeV}=103.72</math>
-<math>z'=\frac{1}{\gamma}z=\frac{1}{21526.4188099}5\ cm=2.32e-4\ cm</math>
+<math>z'=\frac{1}{\gamma}z=\frac{1}{103.72}5\ cm=0.048\ cm</math>
-<center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =1.05\times 10^{-1} barns^{-1}</math></center>
+<center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{0.05 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =1.05\times 10^{-3} barns^{-1}</math></center>
 Using the number of incident electrons, for 1 Moller electron:
-<center><math>\frac{1}{\rho_{target}\times l_{target} \times 6\times 10^7}=1.58\times 10^{-7} barns</math></center>
+<center><math>\frac{1}{\rho_{target}\times l_{target} \times 6\times 10^7}=1.58\times 10^{-5} barns</math></center>
 We can use this number to scale the number of electrons per angle to a differential cross-section in barns.  Using the plot of the Moller electron scattering angle theta in the Center of Mass frame,

← Older revision		Revision as of 19:31, 4 April 2018
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	with <math>c=1</math>		with <math>c=1</math>

−	and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}=\frac{E}{\sqrt{E}\sqrt{1-\frac{p^2}{E^2}}}=\frac{E}{m}</math>	+	and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}=\frac{E}{\sqrt{E^2}\sqrt{1-\frac{p^2}{E^2}}}=\frac{E}{m}</math>
		+
		+
		+	Where <center><math>E^2=p^2+m^2=(11000\ MeV)^2+(0.511\ MeV)^2 \rightarrow E=11000.0000119\ MeV</math></center>
		+
		+	<math>\gamma=\frac{E}{m}=\frac{11000.0000119\ MeV}{0.511\ MeV}=21526.41</math>
		+
		+
		+	<math>z'=\frac{1}{\gamma}z=\frac{1}{21526.4188099}5\ cm=2.32e-4\ cm</math>

← Older revision		Revision as of 19:25, 4 April 2018
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	with <math>c=1</math>		with <math>c=1</math>

−	and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}=\frac{E}{m}</math>	+	and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}=\frac{E}{\sqrt{E}\sqrt{1-\frac{p^2}{E^2}}}=\frac{E}{m}</math>

@@ Line 70: / Line 70: @@
-Where <math>v \equiv \frac{p}{m}</math>
@@ Line 77: / Line 79: @@
 with <math>c=1</math>
-and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}</math>
+and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}=\frac{E}{m}</math>

← Older revision		Revision as of 18:56, 4 April 2018
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	This gives, for LH2 in a 5cm long target in the Lab Frame, or in the CM frame:		This gives, for LH2 in a 5cm long target in the Lab Frame, or in the CM frame:
		+
		+
		+	<center><math>z'=\gamma (z-vt)</math></center>
		+
		+
		+	Where <math>v \equiv \frac{p}{m}</math>
		+
		+
		+	and <math>\beta^2 \equiv \frac{v^2}{c^2}=\frac{p^2c^2}{m^2c^2c^2}=\frac{p^2c^2}{E^2}=\frac{p^2}{E^2}</math>
		+
		+	with <math>c=1</math>
		+
		+	and <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}</math>

@@ Line 64: / Line 64: @@
 where ρ<sub>target</sub> is the density of the target material, l<sub>target</sub> is the length of the target, and I<sub>scattered</sub> is the number of incident particles scattered per time.
-This gives, for LH2 in a 5cm long target:
+This gives, for LH2 in a 5cm long target in the Lab Frame, or in the CM frame:
 <center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =1.05\times 10^{-1} barns^{-1}</math></center>