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

$F=2E_1 2E_2|\vec {v}_1-\vec {v}_2|=4|E_1E_2\vec v_{21}|$

where $v_{21}$ is the relative velocity between the particles in the frame where particle 1 is at rest

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

Using the relativistic definition of energy

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

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

Letting $E_{21}\equiv E_2$ be the energy of particle 2 wiith respect to particle 1, the relativistic energy equation can be rewritten such that

$|p_{21}^2| =E_{21}^2-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2}{m^2}-m^2=\frac{(\mathbf P_1 \cdot \mathbf P_2)^2-m^4}{m^2}$

where similarly $p_{21}$ is defined as the momentum of particle 2 with respect to particle 1.

The relative velocity can be expressed as

$v_{21}=\frac{|\vec p_{21}|}{E_{21}}$

$F=2E_1 2E_2|\vec {v}_1-\vec {v}_2|=4|mE_{21}\vec v_{12}|=4|mE_{21}\frac{|\vec p_{21}|}{E_{21}}|=4m|\vec p_{21}|$

The invariant form of F is

$F=4\sqrt{(\mathbf P_1 \cdot \mathbf P_2)^2-m^4}$

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

where in the center of mass frame $E_1=E_2$ and $\vec p_1^*=-\vec p_2^*$

$\mathbf P_1 \cdot \mathbf P_2 = E_1^2+\vec p_1\,^{*2}$

$F=4\sqrt{(E_1^2+\vec p_1\,^{*2})^2-m^4}=4\sqrt{(\vec p_1 \,^{*2}+m^2+\vec p_1\,^{*2})^2-m^4}$

$F=4\sqrt{(2\vec p_1\,^{*2}+m^2)^2-m^4}=4\sqrt{4\vec p_1\,^{*4}+m^4+4\vec p_1\,^{*2}m-m^4}$

$F_{cms}=4 \sqrt {4m^2\vec p_1\,^{*2}+4\vec p_1 \,^{*4}}$

$F_{cms}=4 \sqrt {\vec p_1\,^{*2}4(m^2+\vec p_1 \,^{*2})}$

As shown earlier

$s_{CM}=4(m^2+\vec p_1 \,^{*2})$

$F_{cms}=4 \sqrt {\vec p_1^*\,^2s}$

$F_{cms}=4 \vec p_1^*\sqrt {s}$

$d\sigma=\frac{1}{4 \vec p_1\,^*\sqrt {s}}|\mathcal{M}|^2 dQ$

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

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

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

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

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

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

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

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

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