Possible Mandelstam $s$-values Assume $\phi\phi\rightarrow \phi\phi$ scattering.
Mandelstam variable $s=(p_1+p_2)^2$. In center of mass frame $s=(E_1+E_2)^2=4E^2$.
In non-relativistic limit $s \approx 4m^2$. In relativistic limit $s = 4m^2+4p^2$.
So, it seems that $4m^2<s<4m^2+4p^2$ are possible $s$ values.
At the same time we know that $s+t+u=4m^2$. From this I can assume that $s < 4m^2$.
And now I am getting confused. It can be both true at the same time. Where am I making a mistake? What are actual values of s that are allowed? 
 A: The Mandelstam variables are $s=(p_1+p_2)^2, t=(p_1-p_3)^2$ and $u=(p_1-p_4)^2$. Let $\mathbf p, \mathbf p'$ and $\theta$ be C.O.M momenta and angle between incoming and outgoing constituents in C.O.M frame. Then we have the following relations $$s = 4(m^2 + |\mathbf p|^2), \,\,\,\, t = -2|\mathbf p|^2(1-\cos \theta) \,\,\,\,\text{and}\,\,\,\ u = -2|\mathbf p|^2(1+\cos \theta)$$
In particular if $\theta = \pi$, $s >0, t<0$ and $u=0$ with $s+t+u=4m^2$ preserved.  More generally, the kinematically allowed region for the process is constrained to lie in a plane orthogonal to $(1,1,1)$ in the space spanned by $s,t,u$ at a distance $4m^2/\sqrt{3}$ from the origin, lying in the region $s \geq 4m^2, t \leq 0, u \leq 0$.
A: $s$, $t$ and $u$ can take any complex values.  The physical region, hwoever, is the range that is accessible in experiments. If particles 1 and 2 are incoming (the "s-channel") then $s=(p_1+p_2)^2> (m_1+m_2)^2$ can be arbitrary large, while $t$ and $u$ are both negative.  In the $t$-channel with particles 2 and 4 incoming, it is $t$ that is greater than $(m_2+m_4)^2$ etc.
