Differential cross-section for a 2-particle process in the LAB frame

Question

This should really be a straightforward calculation, but somehow, I keep confusing myself and failing over and over again. I did the calculation so many times that I don't even know what I'm looking at anymore. Some help would be greatly appreciated.

Let me show you what I tried to do, so you can check my understanding (or lack thereof).

(1 + 2 $\rightarrow$ 1' + 2') elastic scattering in the LAB frame
This means: $$ m_1 = m_1'$$ $$m_2 = m_2'$$ $$\mathbf{p_2} = \mathbf{ 0}$$ As you can see, we choose to work in the rest frame of particle 2 before the scattering.

I managed to calculate the differential cross section in the CM (center of mass) frame, which is (in natural units): $$\left( \frac{d\sigma}{d \Omega }\right)_{CM} = \frac{|\mathcal{M}|^2}{(8 \pi)^2 E_{CM}^2}$$ Here, $\mathcal{M}$ is the invariant amplitude and $E_{CM}=E_1 + E_2 = E_1' + E_2'$.

Following the technique used in the CM frame, I use these known relations: $$ d\sigma = \frac{V}{T} \frac{dP}{|\mathbf{v_1} - \mathbf{v_2}|}$$ $$dP = \frac{|\left< f | S|i\right>|^2}{\left< f | f\right> \left<i | i\right>} d\Pi$$ $$\implies d\sigma = \frac{|\mathcal{M}|^2}{(2E_1)(2E_2)|\mathbf{v_1} - \mathbf{v_2}|}d\Pi_{LIPS}$$ This leads me to $$d\sigma = \frac{|\mathcal{M}|^2}{4 E_1 E_2|\mathbf{v_1} - \mathbf{v_2}|} (2\pi)^4 \delta^4(p_1' + p_2' - p_1 - p_2) \frac{d^3 \mathbf{p}_1'}{(2\pi)^3 2 E_1'} \frac{d^3 \mathbf{p}_2'}{(2\pi)^3 2 E_2'} $$ Using $p_2 = m_2$ and integrating over the momenta for particle 2, $$d\sigma = \frac{|\mathcal{M}|^2}{(8 \pi)^2 E_1 m_2|\mathbf{v_1} - \mathbf{v_2}|} \delta(E_1' + E_2' - E_1 - m_2) \frac{d^3 \mathbf{p}_1'}{E_1' E_2'} $$ Next, I switch to spherical coordinates, $d^3 \mathbf{p}_1' = d\Omega \, p^2 dp$ and I factor out the solid angle differential. Finally, I plug in the integral for the remaining degree of freedom $(p)$, to get this: $$\left( \frac{d\sigma}{d \Omega }\right)_{CM} = \frac{1}{(8 \pi)^2 E_1 m_2|\mathbf{v_1} - \mathbf{v_2}|} \int_0^\infty \frac{dp \, p^2 }{E_1' E_2'} |\mathcal{M}|^2 \delta(E_1' + E_2' - E_1 - m_2) $$ This is where I get confused... I tried defining a new variable $x=E_1' + E_2' - E_1 - m_2$ and substituting it in the integral, but everything always gets messy because both $E_1'$ and $E_2'$ depend on $p=|\mathbf{p}_1'|$: $$E_1' = \sqrt{m_1^2 + p^2}$$ $$E_2' = \sqrt{m_2^2 + (\mathbf{p}_1 - \mathbf{p}_1')^2}$$

Messy, messy, messy... so I tried to do it another way:

I recast the CM formula in a Lorentz-covariant form (for known $\mathbf{p}_1$) using Mandelstam variables, i.e. $$ s = (p_1 + p_2)^2 = (p_1' + p_2')^2$$ $$ t = (p_1 - p_1')^2 = (p_2 - p_2')^2$$ $$ dt = 2 | \mathbf{p}_1| \, |\mathbf{p}_1'| \frac{d \Omega_{CM}}{2 \pi}$$ $$\implies \frac{d \sigma}{dt} = \frac{|\mathcal{M}|^2}{64 \pi s |\mathbf{p_1}|^2}$$

Next, I write $s$ and $t$ in the LAB frame in hopes of getting the correct formula, but to no avail... I get something extremely complicated or something simply wrong...

THIS is what I should get at the end, but I can't seem to get it: $$\left( \frac{d\sigma}{d \Omega }\right)_{LAB} = \frac{|\mathcal{M}|^2}{(8 \pi)^2} \frac{|\mathbf{p}_1'|^2}{|\mathbf{p}_1|} \frac{1}{m_2 (E_1 + m_2) \big| |\mathbf{p}_1'| - |\mathbf{p}_1| E_1' \cos{\theta} \big|}$$

So, what do you think?

Answer 1

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{equation} \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} \end{equation}

The take $B$ to be at rest. Thus $p_B^\mu = (m_B,0,0,0)$ and $p_A^\mu = (E_A, \vec{p}_A)$. Plugging this into the equation above, we get \begin{equation} \begin{split} d\sigma &= \frac{1}{(2p_A)(2m_B)} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} \end{equation} where we have used $|\vec{v}_A| = \frac{p_A}{E_A}$. We can integrate over $\vec{p}_2$ to get \begin{equation} \begin{split} d\sigma &= \frac{1}{(2p_A)(2m_B)} \frac{d^3p_1}{(2\pi)^2} \frac{1}{2E_1} \left| {\cal M} \right|^2 \frac{1}{2E_2} \delta(E_A + m_B - E_1 - E_2) \\ \frac{d\sigma}{d\Omega} &= \frac{1}{(2\pi)^2(2p_A)(2m_B)} dp_1 p_1^2 \frac{1}{2E_1} \left| {\cal M} \right|^2 \frac{1}{2E_2} \delta(E_A + m_B - E_1 - E_2) \end{split} \end{equation} where $E_2^2 = (\vec{p}_1 - \vec{p}_A)^2 + m_2^2 = p_1^2 + p_A^2 - 2 p_1 p_A \cos\theta + m_2^2 $ and $E_1^2 = p_1^2 + m_1^2$. To integrate over $p_1$, we see that the $\delta$-function becomes \begin{equation} \begin{split} \delta & \left(E_A + m_B - \sqrt{p_1^2 + m_1^2} - \sqrt{p_1^2 - 2 p_1 p_A \cos\theta + p_A^2 + m_2^2 }\right) \\ &= \delta (p_1 - p_f) \left(\frac{p_f}{E_1} + \frac{p_f - p_A\cos\theta}{E_2} \right)^{-1} \end{split} \end{equation} where $p_1 = p_f$ is the solution to the argument in the delta function. Integrating over $p_1$, we get \begin{equation} \begin{split} \boxed { \frac{d\sigma}{d\Omega} = \frac{1 }{64\pi^2m_B} \left[ E_2 + E_1\left(1 - \frac{p_i}{p_f} \cos\theta\right)\right]^{-1} \frac{p_f}{p_i}\left| {\cal M} \right|^2 } \end{split} \end{equation} where $E_1 = \sqrt{p_f^2 + m_1^2}$, $E_2 = \sqrt{(\vec{p}_f - \vec{p}_i)^2 + m_2^2}$