Old work - Revision history

Vanwdani at 19:29, 9 March 2016

2016-03-09T19:29:18Z

Vanwdani: /* Moller Differential Cross Section */

2016-03-09T19:28:25Z

Moller Differential Cross Section

Vanwdani: /* Moller Differential Cross Section */

2016-03-09T19:27:50Z

Moller Differential Cross Section

Vanwdani: Created page with "===Moller Differential Cross Section=== Using the equation from [1] $\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{2}}\left \{\frac{1+cos^4(\frac{\theta^}{2})}{sin…"$

2016-03-09T19:26:58Z

Created page with "===Moller Differential Cross Section=== Using the equation from [1] <center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{*2}}\left \{\frac{1+cos^4(\frac{\theta^*}{2})}{sin…"

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===Moller Differential Cross Section===
Using the equation from [1]

<center><math>\frac{d\sigma}{d\Omega '_1}=\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>where\ \alpha=\frac{e^2}{\hbar c}\quad with\quad \hbar = c =1\ and\ \theta^*=\theta^*_1=\theta^*_2</math></center>

This can be simplified to the form

<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}</math></center>

Plugging in the values expected for 2 scattering electrons:

<center><math>\alpha ^2=5.3279\times 10^{-5}</math></center>

<center><math>E^*\approx 106.031 MeV</math></center>

Using unit analysis on the term outside the parantheses, we find that the differential cross section for an electron at this momentum should be around

<center><math>\frac{5.3279\times 10^{-5}}{4\times 1.124\times 10^{16}eV^2}=1.18\times 10^{-21} eV^{-2}=\frac{1.18\times 10^{-21}}{1eV^2}\times \frac{1\times 10^{18} }{1\times 10^{18}}=\frac{.0012}{GeV^2}</math></center>

Using the conversion of

<center><math>\frac{1}{1GeV^2}=.3894 mb</math></center>

<center><math>\frac{.0012}{1GeV^2}=\frac{.0012}{1}\frac{1}{1GeV^2}=.0012\times .3894 mb=.467\times 10^{-3}mb</math></center>

We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>

← Older revision		Revision as of 19:29, 9 March 2016
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	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>		<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+	===Substituting for Moller range and energies===
		+	Converting the number of electrons to barns,
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega_{2}'}=\frac{dN}{\mathcal L d\Omega}</math></center>
		+
		+
		+
		+
		+	<center><math>\Longrightarrow dN=\frac{d \sigma}{d \Omega} \mathcal L</math></center>
		+
		+
		+	<center><math>where \mathcal L=i\, \rho I</math></center>
		+
		+
		+	where ρ<sub>target</sub> is the density of the target material, l<sub>target</sub> is the length of the target, and N is the number of incident particles scattered.
		+
		+
		+	<center><math>\mathcal L=\frac{.86g}{1 cm^3}\times \frac{(100cm)^3}{1m^3} \times \frac{1 kg}{1000 g}\times \left[ \left(.75 \frac{1 mole}{1.01 g} \times \frac{1000g}{1 kg} \right)+\left(.25 \frac{1 mole}{14.01 g} \times \frac{1000g}{1 kg} \right)\right] \times \frac{6.022\times10^{23}particles}{1 mole} \times \frac{1cm}{100 cm} \times \frac{1 m}{ } \times \frac{10^{-28} m^2}{barn} </math></center>
		+
		+
		+
		+	<center><math>\mathcal L=\frac{860kg}{1 m^3}\times \left[ \left(\frac{742.574 mole}{kg} \right)+\left(\frac{17.844 mole}{1 kg} \right)\right] \times \frac{6.022\times 10^{-7}\ particles\cdot m^3}{1 mole\cdot barn} </math></center>
		+
		+
		+
		+
		+	<center><math>\mathcal L=\frac{5.170\times 10^{-4}kg\cdot particles}{1 mole\cdot barn} \left(\frac{760.418 mole}{kg} \right)</math></center>
		+
		+
		+
		+	<center><math>\mathcal L=\frac{.394\ particles}{1 barn} </math></center>
		+
		+
		+
		+	<center><math>\Longrightarrow N=\sigma \frac{.394\ particles}{1 barn}</math></center>
		+
		+
		+
		+	{\| class="wikitable" align="center" border=1
		+	\|+ '''Number of electrons from Moller electron Momentum'''
		+	\|-
		+	! <math>p_2^{'}(MeV/c)</math>
		+	! <math>\sigma(b)</math>
		+	! <math>Number of electrons</math>
		+	\|-
		+	\| <math>\equiv 10000</math>
		+	\| <math>3.751\times 10^{5}</math>
		+	\|<math>\approx 1.478\times 10^5</math>
		+	\|-
		+	\| <math>\equiv 5000 </math>
		+	\| <math>1.501\times 10^{6}</math>
		+	\|<math>\approx5.914\times 10^5</math>
		+	\|-
		+	\| <math>\equiv 1000 </math>
		+	\|<math>3.751\times 10^{7}</math>
		+	\|<math>\approx 1.478\times 10^7</math>
		+	\|-
		+	\| <math>\equiv 500</math>
		+	\| <math>1.501\times 10^{8}</math>
		+	\|<math>\approx 5.914\times 10^7</math>
		+	\|}
		+
		+
		+	<center>[[File:Lab_Frame_Moller_DiffX.png]]</center>

← Older revision		Revision as of 19:28, 9 March 2016
Line 42:		Line 42:

	We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>		We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>
		+
		+	===Different p<sub>2</sub><sup>1</sup> Values===
		+
		+	Using the conversion of
		+
		+
		+	<center><math>\frac{1}{1GeV^2}=.3894 mb</math></center>
		+
		+
		+	<center><math>\sigma=\int d\sigma=\int \frac{d\sigma}{d\Omega_2'}d\Omega</math></center>
		+
		+
		+	The range of the detector is considered to be <math> .10 \le \theta \le .87</math>,<math>-\pi \le \phi \le \pi</math>
		+
		+
		+	<center><math>\sigma=\int_{ .611}^{2.531} \int_{-\pi}^{\pi} \frac{d\sigma}{d\Omega_2'}sin\theta \,d\theta \, d\phi </math></center>
		+
		+
		+
		+	<center><math>\sigma=2\pi \int_{.611}^{2.531} \frac{d\sigma}{d\Omega_2'} sin\theta \,d\theta </math></center>
		+
		+
		+
		+	<center><math>\sigma=2\pi (1.638)\frac{d\sigma}{d\Omega_2'} </math></center>
		+
		+
		+
		+	<center><math>\sigma=(10.294) \frac{d\sigma}{d\Omega_2'} </math></center>
		+
		+
		+	{\| class="wikitable" align="center" border=1
		+	\|+ '''Differential Cross Section Scale for Different p<sub>2</sub><sup>1</sup> Values'''
		+	\|-
		+	! <math>p_{2}'(MeV)</math>
		+	! <math>\frac{d\sigma}{d\Omega_{2}^'}(eV^{-2})</math>
		+	! <math>\frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2})</math>
		+	! <math>\frac{d\sigma}{d\Omega_{2}^'}(mb)</math>
		+	! <math>\frac{d\sigma}{d\Omega_{2}^'}(b)</math>
		+	! <math>\sigma(b)</math>
		+	\|-
		+	\| <math>10000</math>
		+	\| <math>9.357\times 10^{-11}</math>
		+	\| <math>9.357\times 10^{7}</math>
		+	\| <math>3.644\times 10^{7}</math>
		+	\| <math>3.644\times 10^{4}</math>
		+	\| <math>3.751\times 10^{5}</math>
		+	\|-
		+	\| <math>5000 </math>
		+	\| <math>3.743\times 10^{-10}</math>
		+	\| <math>3.743\times 10^{8}</math>
		+	\| <math>1.458\times 10^{8}</math>
		+	\| <math>1.458\times 10^{5}</math>
		+	\| <math>1.501\times 10^{6}</math>
		+	\|-
		+	\| <math>1000 </math>
		+	\| <math>9.357\times 10^{-9}</math>
		+	\| <math>9.357\times 10^{9}</math>
		+	\| <math>3.644\times 10^{9}</math>
		+	\| <math>3.644\times 10^{6}</math>
		+	\|<math>3.751\times 10^{7}</math>
		+	\|-
		+	\| <math>500</math>
		+	\| <math>3.743\times 10^{-8}</math>
		+	\| <math>3.743\times 10^{10}</math>
		+	\| <math>1.458\times 10^{10}</math>
		+	\| <math>1.458\times 10^{7}</math>
		+	\| <math>1.501\times 10^{8}</math>
		+	\|}

	===CM to Lab Frame===		===CM to Lab Frame===

← Older revision		Revision as of 19:27, 9 March 2016
Line 42:		Line 42:

	We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>		We find that the differential cross section scale is <math>\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b</math>
		+
		+	===CM to Lab Frame===
		+	We can substitute in for <math>\theta</math>
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{2}}\frac{ (3+cos^2\theta^)^2}{sin^4\theta^*}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{2}}\frac{ (3+cos^2\theta^)^2}{sin(\theta^)sin(\theta^)sin(\theta^)sin(\theta^)}</math></center>
		+
		+
		+	Using,
		+	<center><math>sin(\theta^)=sin(\theta_{2}^)=\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)</math></center>
		+
		+
		+	{\| class="wikitable" align="center"
		+	\| style="background: gray" \| <math>\frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\, \partial \Omega} = \frac{\partial ^2\sigma^(E^,\, \theta^* ,\, \phi^)}{\partial E^\,\partial \Omega^}\frac{p}{p^{}}</math>
		+	\|}
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{2}}\frac{ (3+cos^2\theta^)^2}{\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (3+cos^2\theta^*)^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	Now, using the trigometric identity,
		+	<center><math>sin^2 t+cos^2 t=1\Longrightarrow cos^2(\theta^)=1-sin^2(\theta^)</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (3+1-sin^2(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (4-sin(\theta^)sin(\theta^))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (4-\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^}\ sin \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (4-\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{4}}{4E^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+	Substituting,
		+	<center><math>p_{2}^=\sqrt{E_{2}^{2}-m^2}</math></center>
		+
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (\sqrt{E_{2}^{2}-m^2})^4}{4E^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (E_{2}^{2}-m^2)^2}{4E^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	Substituting in for m, E<sub>2</sub><sup>,</sup>and E<sup></sup>
		+	<math>\alpha^2=5.3279\times 10^{-5}</math>
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=(\frac{ 5.3279\times 10^{-5}( ((53.015MeV)^{2}-(.511MeV)^2)^2}{4\times (106.031MeV)^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>
		+
		+
		+
		+	<center><math>\frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)}</math></center>

Old work - Revision history

Vanwdani at 19:29, 9 March 2016

Vanwdani: /* Moller Differential Cross Section */

Vanwdani: /* Moller Differential Cross Section */

Vanwdani: Created page with "===Moller Differential Cross Section=== Using the equation from [1] \frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{*2}}\left \{\frac{1+cos^4(\frac{\theta^*}{2})}{sin…"

Vanwdani: Created page with "===Moller Differential Cross Section=== Using the equation from [1] $\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{2}}\left \{\frac{1+cos^4(\frac{\theta^}{2})}{sin…"$