TF DerivationOfCoulombForce - Revision history

Foretony at 11:54, 24 September 2014

2014-09-24T11:54:39Z

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2009-03-03T03:17:17Z

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2009-03-03T03:05:43Z

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2009-03-03T02:54:45Z

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2009-03-02T22:43:16Z

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← Older revision		Revision as of 11:54, 24 September 2014
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	:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{\hbar}{2m i} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>		:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{\hbar}{2m i} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>
		+
		+	[[Forest_NucPhys_I#Electro_Magnetic_Interactions]]

← Older revision		Revision as of 03:17, 3 March 2009
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	Check: <math>\frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^}{\partial X}) = \frac{\partial \psi^}{\partial X} \frac{\partial \psi}{\partial X} + \psi^\frac{\partial^2 \psi}{\partial X^2} - \frac{\partial \psi}{\partial X} \frac{\partial \psi^}{\partial X} - \psi \frac{\partial^2 \psi^*}{\partial X^2}</math>		Check: <math>\frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^}{\partial X}) = \frac{\partial \psi^}{\partial X} \frac{\partial \psi}{\partial X} + \psi^\frac{\partial^2 \psi}{\partial X^2} - \frac{\partial \psi}{\partial X} \frac{\partial \psi^}{\partial X} - \psi \frac{\partial^2 \psi^*}{\partial X^2}</math>
		+
		+	:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{\hbar}{2m i} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>

← Older revision		Revision as of 03:05, 3 March 2009
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	:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* (- \frac{\hbar^2}{2m}) \frac{\partial^2 \psi}{\partial X^2} - \psi (-\frac{\hbar^2}{2m})\frac{\partial^2 \psi}{\partial X^2} ] = 0</math>		:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* (- \frac{\hbar^2}{2m}) \frac{\partial^2 \psi}{\partial X^2} - \psi (-\frac{\hbar^2}{2m})\frac{\partial^2 \psi}{\partial X^2} ] = 0</math>

−	:::~~<math>~~<math>\frac{\partial \psi^* \psi}{\partial t} - \frac{i\hbar}{2m} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>	+	:::<math>\frac{\partial \psi^* \psi}{\partial t} - \frac{i\hbar}{2m} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>
		+
		+	Check: <math>\frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^}{\partial X}) = \frac{\partial \psi^}{\partial X} \frac{\partial \psi}{\partial X} + \psi^\frac{\partial^2 \psi}{\partial X^2} - \frac{\partial \psi}{\partial X} \frac{\partial \psi^}{\partial X} - \psi \frac{\partial^2 \psi^*}{\partial X^2}</math>

← Older revision		Revision as of 03:01, 3 March 2009
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	If free particle <math>H = \frac{p^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial X^2}</math>		If free particle <math>H = \frac{p^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial X^2}</math>
		+
		+	;Cross-section
		+
		+	:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* (- \frac{\hbar^2}{2m}) \frac{\partial^2 \psi}{\partial X^2} - \psi (-\frac{\hbar^2}{2m})\frac{\partial^2 \psi}{\partial X^2} ] = 0</math>
		+
		+	:::<math><math>\frac{\partial \psi^* \psi}{\partial t} - \frac{i\hbar}{2m} [ \frac{\partial}{\partial X} (\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) ] = 0</math>

← Older revision		Revision as of 02:54, 3 March 2009
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	:::<math>\sigma = \int (\frac{d\sigma}{d\Omega}) d\Omega =</math> probability of scattering in any direction.		:::<math>\sigma = \int (\frac{d\sigma}{d\Omega}) d\Omega =</math> probability of scattering in any direction.

−	Let j \equiv \frac{#particles}{area}	+	Let<math> j \equiv \frac{#particles}{area}</math>
		+
		+	<math>j_{scattered} = \frac{#Nucleons scattered}{area}</math>
		+
		+	From Q.M. <math>j = \frac{\hbar}{2mi}(\psi^* \frac{\partial \psi}{\partial X} - \psi \frac{\partial \psi^*}{\partial X}) =</math> particle current density
		+
		+	This comes from the continuity equation
		+
		+
		+	:::<math>\frac{\partial \ro}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0</math>
		+
		+
		+	Time dependent Shrodinger equation \Longrightarrow
		+
		+	:::<math>\frac{\partial \psi}{\partial t} = -\frac{i}{\hbar} H \psi</math>
		+
		+	:::<math>\frac{\partial \psi^}{\partial t} = \frac{i}{\hbar} H \psi^</math>
		+
		+	<math>\frac{\partial}{\partial t}\psi^* \psi =</math> time derivative of the particle density
		+
		+	:::<math>\psi^* \frac{\partial \psi}{\partial t} + \psi \frac{\partial \psi^*}{\partial t}</math>
		+
		+	:::<math>\psi^* (-\frac{iH}{\hbar}\psi) + \psi (\frac{iH}{\hbar} \psi^*)</math>
		+
		+	:::<math>\frac{\partial \psi^* \psi}{\partial t} + \frac{i}{\hbar} [ \psi^* H \psi - \psi H \psi^*] = 0</math>
		+
		+	If free particle <math>H = \frac{p^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial X^2}</math>

← Older revision		Revision as of 22:43, 2 March 2009
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	If you measure the phase shift you can determine "sign" of V.		If you measure the phase shift you can determine "sign" of V.

−	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math> \frac{#scattered/ solid-angle}{#in / Area} </math>	+	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math> \frac{#scattered/ solid-angle}{#in / Area} [m^2]</math>
		+
		+	:::=probability of scattering into the solid angle <math>( d\Omega )</math>
		+
		+	:::<math>\sigma = \int (\frac{d\sigma}{d\Omega}) d\Omega =</math> probability of scattering in any direction.
		+
		+	Let j \equiv \frac{#particles}{area}

← Older revision		Revision as of 22:39, 2 March 2009
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	If you measure the phase shift you can determine "sign" of V.		If you measure the phase shift you can determine "sign" of V.

−	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math>\frac{# scattered / solid-angle}{#in / Area}</math>	+	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math> \frac{#scattered/ solid-angle}{#in / Area} </math>

← Older revision		Revision as of 22:38, 2 March 2009
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	If you measure the phase shift you can determine "sign" of V.		If you measure the phase shift you can determine "sign" of V.

−	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math>~~\frac{~~\frac{# scattered}{solid-angle}~~}{\frac~~{#in}{Area}}</math>	+	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math>\frac{# scattered / solid-angle}{#in / Area}</math>

← Older revision		Revision as of 22:37, 2 March 2009
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	If you measure the phase shift you can determine "sign" of V.		If you measure the phase shift you can determine "sign" of V.

−	Cross-section:<math> \frac{d \sigma}{d \~~omega~~} =</math> differential X-section = <math>\frac{\frac{# scattered}{solid-angle}}{\frac{#in}{Area}}</math>	+	Cross-section:<math> \frac{d \sigma}{d \Omega} =</math> differential X-section = <math>\frac{\frac{# scattered}{solid-angle}}{\frac{#in}{Area}}</math>

@@ Line 428: / Line 428: @@
 :::::<math>k_{II}^2 = \frac{2m E}{\hbar^2}</math>
-If V=0 <math>\Longrightarrow</math> <math>k_I = k_{II}</math> <math>\Longrightarrow</math> <math>\delta = 0</math>.
+If V=0 <math>\Longrightarrow</math> <math>k_I = k_{II}</math> <math>\Longrightarrow</math> <math>\delta = 0</math>:
 Free neutron; No target; No phase shift.
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