Mandelstam Variables for 2->2 Scattering with Equal Masses in CoM Frame

[vanhees71]

I'd put $u=4m^2-s-t$. Then the cross section depends only on $s$ and $t$. As shown above through $t$ the expression depends on the scattering angle $\vartheta$. Then you can check, where the expression at fixed $s=E_{\text{cm}}^2$ is maximal as a function of $t$ and thus which angles dominate.

 

[spaghetti3451]

I'd put $u=4m^2-s-t$. Then the cross section depends only on $s$ and $t$. As shown above through $t$ the expression depends on the scattering angle $\vartheta$. Then you can check, where the expression at fixed $s=E_{\text{cm}}^2$ is maximal as a function of $t$ and thus which angles dominate.

Ah! I see!

So, you suggest the following:

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{|\mathcal{M}|^{2}}{64\pi^{2}E_{\text{cm}}^{2}}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}E_{\text{cm}}^{2}}\left|\frac{i}{s-m^{2}} + \frac{i}{t-m^{2}} + \frac{i}{u-m^{2}}\right|^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{u-m^{2}}\right)^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{(4m^{2}-s-t)-m^{2}}\right)^{2}}$

$\displaystyle{= \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{t-m^{2}} + \frac{1}{3m^{2}-s-t}\right)^{2}}$

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s-m^{2}} + \frac{1}{-\sqrt{s-4m^{2}}(1-\cos\theta)-m^{2}} + \frac{1}{3m^{2}-s+\sqrt{s-4m^{2}}(1-\cos\theta)}\right)^{2}}$

For $s\gg \sqrt{m}$,

$\displaystyle{\frac{d\sigma}{d\theta} = \frac{g^{4}}{64\pi^{2}s}\left(\frac{1}{s} + \frac{1}{-\sqrt{s}(1-\cos\theta)-m^{2}} + \frac{1}{-s+\sqrt{s}(1-\cos\theta)}\right)^{2}}$

So, the differential cross-section, for a fixed $s$, is a maximum at angles $\theta = 2\pi n$, where $n=0, \pm 1, \pm 2, \dots$.

What do you think?

 

[Jilang]

I think the square root looks wrong and the expression is dimensionally inconsistent.