Relativistic Differential Cross-section

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_1E_2\vec v|[/math]

where v is the relative velocity between the particles. In the frame where one of the particles (particle 1) is at rest , the relative velocity can be expressed as

[math] v_2=\frac{|\vec p_2|}{E_2}[/math]

[math]\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)= E_{1}E_{2}[/math]

Using the relativistic definition of energy

[math]E \equiv p^2+m^2=m^2[/math]

[math]\mathbf P_1 \cdot \mathbf P_2 = mE_{2}[/math]

[math]\rightarrow E_2=\frac{\mathbf P_1 \cdot \mathbf P_2}{m}[/math]

[math]|p_2^2| =E_2^2-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2}{m^2}-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2-m^4}{m^2}[/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]