Relativistic Differential Cross-section

Relativistic Differential Cross-section

$d\sigma=\frac{1}{F}|\mathcal{M}|^2 dQ$

dQ is the invariant Lorentz phase space factor

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

and F is the flux of incoming particles

$F=2E_1 2E_2|\vec {v}_1-\vec {v}_2|=4|E_1E_2\vec v|$

where v is the relative velocity between the particles.

$\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)= E_{1}E_{2}$

Using the relativistic definition of energy

$E \equiv p^2+m^2=m^2$

$\mathbf P_1 \cdot \mathbf P_2 = mE_{2}$

$\rightarrow E_2=\frac{\mathbf P_1 \cdot \mathbf P_2}{m}$

$|p_2^2| =E_2^2-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2}{m^2}-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2-m^4}{m^2}$

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

$v_2=\frac{|\vec p_2|}{E_2}$

$v_2^2=\frac{|\vec p_2|^2}{E_2^2}$

The invariant form of F is

$F=4\sqrt{(\vec {p}_1 \cdot \vec {p}_2)^2-(m_1m_2)^2}$

$F_{cms}=4 \vec p_i\sqrt {s}$

$d\sigma=\frac{1}{4 \vec p_i\sqrt {s}}|\mathcal{M}|^2 dQ$

$d^3 \vec p_1^'=\vec p^{'3}_1 d \vec p^' d\Omega$

$(E_1^')^2=(\vec p_1^')^2+(m_1)^2$

$E_1^' d E_1^'= \vec p_1^' d \vec p_1^'$

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

$W_i \equiv E_1+E_2 \qquad \qquad W_f \equiv E_1^'+E_2^'$

$dW_f=dE_1^'+dE_2^'=\frac{\vec p_1^' d \vec p_1^'}{E_1^'}+\frac{p_2^' dp_2^'}{E_2^'}$

In the center of mass frame

$|\vec p_1^'|=|\vec p_2^'|=|\vec p_f^'| \rightarrow |\vec p_1^' d \vec p_1^'|=|\vec p_2^' d \vec p_2^'|=|\vec p_f^' d \vec p_f^'|$

$dW_f=\frac{W_f}{E_2^'}dE_1^'$

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

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

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