Limit of Energy in Lab Frame - Revision history

Vanwdani at 19:08, 1 January 2019

2019-01-01T19:08:12Z

2018-12-29T17:41:40Z

2018-12-29T17:40:57Z

2018-03-16T16:47:40Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:44:30Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:43:13Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:40:17Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:36:13Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:30:34Z

Minimum Moller Energy in Lab Frame

2018-03-15T18:26:12Z

\theta=0^{\circ}

@@ Line 3: / Line 3: @@
 [[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
 [[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
-[[Relativistic_Units|<math>\vartriangleright </math>]]
+[[Limit_of_Scattering_Angle_Theta_in_Lab_Frame|<math>\vartriangleright </math>]]
 </center>
@@ Line 165: / Line 165: @@
 [[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
 [[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
-[[Relativistic_Units|<math>\vartriangleright </math>]]
+[[Limit_of_Scattering_Angle_Theta_in_Lab_Frame|<math>\vartriangleright </math>]]
 </center>

@@ Line 161: / Line 161: @@
-<center><math>\textbf{\underline{Navigation}}</math>
+<center><math> \underline{\textbf{Navigation} }</math>
-[[S-Channel|<math>\vartriangleleft </math>]]
+[[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
 [[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
-[[U-Channel|<math>\vartriangleright </math>]]
+[[Relativistic_Units|<math>\vartriangleright </math>]]
 </center>

← Older revision		Revision as of 17:40, 29 December 2018
Line 1:		Line 1:
		+	<center><math> \underline{\textbf{Navigation} }</math>
		+
		+	[[Uniform_distribution_in_Energy_and_Theta_LUND_files\|<math>\vartriangleleft </math>]]
		+	[[VanWasshenova_Thesis#Moller_Scattering\|<math>\triangle </math>]]
		+	[[Relativistic_Units\|<math>\vartriangleright </math>]]
		+
		+	</center>
		+
	The t quantity is known as the square of the 4-momentum transfer		The t quantity is known as the square of the 4-momentum transfer

← Older revision		Revision as of 16:47, 16 March 2018
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−	<center><math>5496553579.26(1-\cos \theta_{2^\ 2^{'}})=(E_2^{'}-(.511\ MeV))</math></center>	+	<center><math>5496553579.26\ eV(1-\cos \theta_{2^\ 2^{'}})=(E_2^{'}-(.511\ MeV))</math></center>
		+
		+
		+
		+	At <math>\theta_{2^\ 2^{'}}=0^{\circ}</math>
		+
		+
		+	<center><math>5496553579.26\ eV+(.511\ MeV)=E_2^{'}</math></center>
		+
		+
		+	<center><math>.511\ MeV=E_2^{'}</math></center>

← Older revision		Revision as of 18:44, 15 March 2018
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	<center><math>((.511\ MeV)^2-(53\ MeV)^{2})(1-\cos \theta_{2^\ 2^{'*}})=-(.511\ MeV)(E_2^{'}-(.511\ MeV))</math></center>		<center><math>((.511\ MeV)^2-(53\ MeV)^{2})(1-\cos \theta_{2^\ 2^{'*}})=-(.511\ MeV)(E_2^{'}-(.511\ MeV))</math></center>
		+
		+
		+	<center><math>5496553579.26(1-\cos \theta_{2^\ 2^{'}})=(E_2^{'}-(.511\ MeV))</math></center>

← Older revision		Revision as of 18:43, 15 March 2018
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−	<center><math>(m^2-E_2^{2})(1-\cos \theta_{2^\ 2^{'*}})=(~~m^2-~~E_2^{'}m)</math></center>	+	<center><math>(m^2-E_2^{2})(1-\cos \theta_{2^\ 2^{'*}})=-m(E_2^{'}-m)</math></center>


−	<center><math>((.511\ MeV)^2-(53\ MeV)^{2})(1-\cos \theta_{2^\ 2^{'*}})=((.511\ MeV)~~^2-~~E_2^{'}(.511\ MeV))</math></center>	+	<center><math>((.511\ MeV)^2-(53\ MeV)^{2})(1-\cos \theta_{2^\ 2^{'*}})=-(.511\ MeV)(E_2^{'}-(.511\ MeV))</math></center>

← Older revision		Revision as of 18:40, 15 March 2018
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−	<center><math>2(m^2-E_2^{2})(1-\cos \theta_{2^\ 2^{'*}})=2(m^2-E_2^{'}m)</math></center>	+	<center><math>(m^2-E_2^{2})(1-\cos \theta_{2^\ 2^{'*}})=(m^2-E_2^{'}m)</math></center>
		+
		+
		+	<center><math>((.511\ MeV)^2-(53\ MeV)^{2})(1-\cos \theta_{2^\ 2^{'*}})=((.511\ MeV)^2-E_2^{'}(.511\ MeV))</math></center>
		+

	----		----

← Older revision		Revision as of 18:36, 15 March 2018
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−	<center><math>t=-2p_1^{2}(1- \cos \theta_{1^\ 1^{'}})=-2p_2^{2}(1-\cos \theta_{2^\ 2^{'}})</math></center>	+	<center><math>t=-2p_1^{2}(1- \cos \theta_{1^\ 1^{'}})=-2p_2^{2}(1-\cos \theta_{2^\ 2^{'}})=2(m^2-E_2^{'}m)</math></center>

		+
		+
		+	<center><math>-2p_2^{2}(1-\cos \theta_{2^\ 2^{'*}})=2(m^2-E_2^{'}m)</math></center>
		+
		+
		+	<center><math>2(m^2-E_2^{2})(1-\cos \theta_{2^\ 2^{'*}})=2(m^2-E_2^{'}m)</math></center>

	----		----

← Older revision		Revision as of 18:30, 15 March 2018
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	This implies that the Moller electron has a non-zero momentum, hence it's total energy is more than it's rest mass energy. The momentum that the Moller electron would have would have to be transfered from the incident electron to the "stationary" electron bound to the detector. The binding energy of an electron bound to a hydrogen atom is 13.6 eV		This implies that the Moller electron has a non-zero momentum, hence it's total energy is more than it's rest mass energy. The momentum that the Moller electron would have would have to be transfered from the incident electron to the "stationary" electron bound to the detector. The binding energy of an electron bound to a hydrogen atom is 13.6 eV
		+
		+
		+	<center><math>t=-2p_1^{2}(1- \cos \theta_{1^\ 1^{'}})=-2p_2^{2}(1-\cos \theta_{2^\ 2^{'}})</math></center>
		+
		+
	----		----

← Older revision		Revision as of 18:26, 15 March 2018
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−	==<math>\theta=0^{\circ}</math>==	+	==<math>\theta \approx 0^{\circ}</math>==

	There is no scattering, or no momentum transfer at 0 degrees since the incident momentum direction is the same as the scattered momentum direction. However, at a certain angle enough momentum must be transferred to provide the ionization energy to create a Moller electron.		There is no scattering, or no momentum transfer at 0 degrees since the incident momentum direction is the same as the scattered momentum direction. However, at a certain angle enough momentum must be transferred to provide the ionization energy to create a Moller electron.