For the elastic scattering of identical particles, $$A + A → A + A$$, what are the Mandelstam variables?

## Solution ##

The Mandelstam variables are defined, for a process $$1 + 2 → 3 + 4$$, as

\begin{eqnarray} s &=& (p_1 + p_2)^2 \nonumber \\ t &=& (p_1 − p_3)^2 \nonumber \\ u &=& (p_1 − p_4)^2 \nonumber \end{eqnarray}

where the $$p$$s are the four-momenta.

$$s$$ variable

For the $$s$$ variable, the scattering is something like two particles hitting each other head-on with $$\vec p_2=-\vec p_1$$. We can calculate it for the case of identical particles like so: \begin{eqnarray} s = (p_1 + p_2)^2 &=& p_1^2+ 2p_1p_2 + p_2^2 \nonumber \\ &=& m_1^2 + 2p_1p_2 + m_2^2 \nonumber \\ &=& 2m_1^2 + 2(E_1E_2 − \vec p_1 \vec p_2) \nonumber \\ &=& 2m_1^2 + 2(E_1^2 + {\vec p_1}^2) \nonumber \\ &=& 2m_1^2 + 2\big((m_1^2 + {\vec p_1}^2) + {\vec p_1}^2\big) \nonumber \\ &=& 4(m^2 + \vec p^2) \end{eqnarray}

$$t$$ variable

The relevant collision for the $$t$$ variable is shown in the diagram. The momenta of particles 1 and 2 are equal and oppsite. Since they have the same mass, they must have the same energy too. After the collision, the particles again move with equal and opposite momentum, so $$E_1=E_2=E_3=E_4$$ and $$|\vec p_1|=|\vec p_2|=|\vec p_3|=|\vec p_4|$$. The calculation is at the bottom. The last relation \eqref{eq:lte} is true for any t-channel process.

$$u$$ variable

The $$u$$ variable is basically the same process as for the $$t$$ variable, except the angle of scattering is $$\phi$$ in the diagram instead of $$\theta$$. This leads to the argument of the cosine being shifted by $$\pi$$, thus changing its sign. Back to $$t$$:

\begin{eqnarray} t=(p_1-p_3)^2&=& p_1^2 – 2p_1p_3 + p_3^2 \nonumber \\ &=& 2m_1^2 – 2E_1E_3 + 2 \vec p_1 \vec p_3 \nonumber \\ &=& 2m_1 – 2E_1^2 + 2 {\vec p_1}^2 \cos \theta \nonumber \\ \label{eq:lte} &=& -2 {\vec p}^2 (1-\cos \theta) \le 0 \end{eqnarray}

Therefore, for $$u$$:

\begin{eqnarray} u=-2 {\vec p}^2 (1+\cos \theta) \end{eqnarray}