Difference between revisions of "Relativistic Differential Cross-section"

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As shown earlier
+
<center><math>F=4|E_1 E_2\vec {v}_1-E_1 E_2\vec {v}_2|=4|E_1\vec{p}_2- E_2\vec {p}_1|</math></center>
 
 
The s quantity is known as the square of the center of mass energy (invariant mass)
 
<center><math>s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2</math></center>
 
 
 
 
 
 
 
 
 
 
 
<center><math>s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2} \equiv \mathbf P_1^{'*2}+2 \mathbf P_1^{'* }\mathbf P_2^{'*}+ \mathbf P_2^{'*2}</math></center>
 
 
 
 
 
As shown earlier, the square of a 4-momentum is
 
 
 
 
 
<center><math>\mathbf P^{2} \equiv m^2</math></center>
 
 
 
This gives,
 
<center><math>s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+  m_2^{2} \equiv m_1^{'2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}+  m_2^{'2}</math></center>
 
 
 
 
 
For the case <math>m_1=m_2=m</math>
 
 
 
 
 
 
 
<center><math>s \equiv  2m^{2}+2 \mathbf P_1^* \mathbf P_2^* \equiv  2m^{2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}</math></center>
 
 
 
Using the relationship
 
 
 
 
 
<center><math>\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)</math></center>
 
 
 
 
 
 
 
<center><math>s \equiv 2m^2+2(E_1^*E_2^*-\vec p \ _1^* \vec p \ _2^*)</math></center>
 
 
 
 
 
In the center of mass frame of reference,
 
 
 
<center><math>E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*</math></center>
 
 
 
 
 
<center><math>s_{CM} \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} </math></center>
 
 
 
 
 
Using the relativistic energy equation
 
 
 
<center><math>E^2 \equiv \vec p_1 \ ^2+m^2</math></center>
 
 
 
 
 
<center><math>s_{CM} \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})</math></center>
 
 
 
 
 
<center><math>s_{CM}=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}=4E_1^*E_2^*</math></center>
 
 
 
 
 
<center><math>F=4E_1 E_2|\vec {v}_1-\vec {v}_2|=4(m^2+\vec p_1 \ ^{*2})|\vec {v}_1-\vec {v}_2|</math></center>
 
  
  

Revision as of 21:27, 3 July 2017

Relativistic Differential Cross-section

[math]d\sigma=\frac{1}{F}|\mathcal{M}|^2 dQ[/math]

dQ is the invariant Lorentz phase space factor


[math]dQ=(2\pi)^4\delta^4(\vec p_1 +\vec p_2 - \vec p_1^' -\vec p_2^')\frac{d^3 \vec p_1^'}{(2\pi)^3 2E_1^'}\frac{d^3 \vec p_2^'}{(2\pi)^3 2E_2^'}[/math]


and F is the flux of incoming particles


[math]F=2E_1 2E_2|\vec {v}_1-\vec {v}_2|=4|E_1\vec{p}_2- E_2\vec {p}_1|[/math]


[math]F=4|E_1 E_2\vec {v}_1-E_1 E_2\vec {v}_2|=4|E_1\vec{p}_2- E_2\vec {p}_1|[/math]


The invariant form of F is


[math]F=4\sqrt{(\vec {p}_1 \cdot \vec {p}_2)^2-(m_1m_2)^2}[/math]


[math]F_{cms}=4 \vec p_i\sqrt {s}[/math]


[math]d\sigma=\frac{1}{4 \vec p_i\sqrt {s}}|\mathcal{M}|^2 dQ[/math]


[math]d^3 \vec p_1^'=\vec p^{'3}_1 d \vec p^' d\Omega[/math]


[math](E_1^')^2=(\vec p_1^')^2+(m_1)^2[/math]


[math]E_1^' d E_1^'= \vec p_1^' d \vec p_1^'[/math]


[math]dQ=\frac{1}{(4\pi)^2}\delta (E_1+E_2-E_1^'-E_2^')\frac{\vec p_1^'dE_1^'}{E_2^'}d\Omega[/math]<\center>


[math]W_i \equiv E_1+E_2 \qquad \qquad W_f \equiv E_1^'+E_2^'[/math]


[math]dW_f=dE_1^'+dE_2^'=\frac{\vec p_1^' d \vec p_1^'}{E_1^'}+\frac{p_2^' dp_2^'}{E_2^'}[/math]


In the center of mass frame

[math]|\vec p_1^'|=|\vec p_2^'|=|\vec p_f^'| \rightarrow |\vec p_1^' d \vec p_1^'|=|\vec p_2^' d \vec p_2^'|=|\vec p_f^' d \vec p_f^'|[/math]


[math]dW_f=\frac{W_f}{E_2^'}dE_1^'[/math]


[math]dQ_{cms}=\frac{1}{(4\pi)^2}\delta (W_i-W_f)\frac{\vec p_f dW_f}{W_f}d\Omega[/math]


[math]dQ_{cms}=\frac{1}{(4\pi)^2}\frac{\vec p_f}{\sqrt s}d\Omega[/math]


[math]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s} \frac{\mathbf p_f}{\mathbf p_i}|\mathcal {M}|^2[/math]