Differential Cross-Section - Revision history

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:36:05Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:31:20Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:29:24Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:28:31Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:28:08Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:26:53Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:14:45Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:12:24Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T18:07:40Z

Differential Cross-Section

Vanwdani: /* Differential Cross-Section */

2019-01-01T17:57:08Z

Differential Cross-Section

← Older revision		Revision as of 18:36, 1 January 2019
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−	<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{sp^{4}\sin^4{\theta}}\left( p^{4}\cos^4{\theta}+2p^{4}\cos^2{\theta}+p^{4}+4E^{2}p^{2}+4E^{2}p^{2}\cos^2{\theta}+4E^{*4}\right) </math></center>	+	<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{2}p^{4}\sin^4{\theta}}\left( p^{4}\cos^4{\theta}+2p^{4}\cos^2{\theta}+p^{4}+4E^{2}p^{2}+4E^{2}p^{2}\cos^2{\theta}+4E^{4}\right) </math></center>

@@ Line 124: / Line 124: @@
 In the Ultra-relativistic limit as <math> E \approx p</math>
 <center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2\left(3+\cos^2{\theta}\right)^2}{4E^{*2}\sin^4{\theta}} </math></center>
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@@ Line 125: / Line 125: @@
 In the Ultra-relativistic limit as <math> E \approx p</math>
-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left(3+\cos{\theta}\right)^2 </math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2\left(3+\cos^2{\theta}\right)^2}{4E^{*2}\sin^4{\theta}} </math></center>
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@@ Line 125: / Line 125: @@
 In the Ultra-relativistic limit as <math> E \approx p</math>
-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left(3+\cos{\theta}\right^2 </math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left(3+\cos{\theta}\right)^2 </math></center>
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@@ Line 125: / Line 125: @@
 In the Ultra-relativistic limit as <math> E \approx p</math>
-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+2\cos^2{\theta}+1+4+4\cos^2{\theta}+4\right) </math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left(3+\cos{\theta}\right^2 </math></center>
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@@ Line 110: / Line 110: @@
-<center><math>\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{82}p^{*2}\cos^2{\theta}</math></center>
+<center><math>\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}</math></center>
@@ Line 116: / Line 116: @@
 Expressing this as the differential cross-section
-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4sp^{*4}\sin^4{\theta}}\left( 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}+16E^{*2}p^{*2}+16E^{82}p^{*2}\cos^2{\theta}+16E^{*4}\right) </math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4sp^{*4}\sin^4{\theta}}\left( 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}+16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}+16E^{*4}\right) </math></center>
-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{sp^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4E^{*2}p^{*2}+4E^{82}p^{*2}\cos^2{\theta}+4E^{*4}\right) </math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{sp^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4E^{*2}p^{*2}+4E^{*2}p^{*2}\cos^2{\theta}+4E^{*4}\right) </math></center>
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-<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{2s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )</math></center>
+<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )</math></center>
@@ Line 112: / Line 112: @@
 <center><math>\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{82}p^{*2}\cos^2{\theta}</math></center>
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← Older revision		Revision as of 18:07, 1 January 2019
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	<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{2}\left(1+\cos{\theta}\right)4E^{2}}{4p^{2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{2}}=\frac{4E^{2}p^{2}\left(1+\cos{\theta}\right)}{p^{4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{2}p^{2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{2}p^{2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{4}\sin^4{\theta}}</math></center>		<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{2}\left(1+\cos{\theta}\right)4E^{2}}{4p^{2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{2}}=\frac{4E^{2}p^{2}\left(1+\cos{\theta}\right)}{p^{4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{2}p^{2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{2}p^{2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{4}\sin^4{\theta}}</math></center>
		+
		+
		+	Combing like terms further,
		+
		+
		+	<center><math>\left(1,3/4\right)\rightarrow \qquad \qquad 4p^{4}\cos^4{\theta}+8p^{4}\cos^2{\theta}+4p^{*4}</math></center>
		+
		+
		+	<center><math>\left(2,6/5\right) \rightarrow \qquad \qquad 16E^{*4}</math></center>
		+
		+
		+	<center><math>\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{2}p^{2}+16E^{82}p^{*2}\cos^2{\theta}</math></center>
		+
		+
	----		----

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-<center><math>\left(9\right)\qquad \qquad \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}}=\frac{E^{*2}p^{*2}\left(1-\cos{\theta}\right)}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(1-\cos{\theta}\right)\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(-\cos^3{\theta}+3\cos^2{\theta}-3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
+<center><math>\left(9\right)\qquad \qquad \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(-\cos^3{\theta}+3\cos^2{\theta}-3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
-<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
+<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
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