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S. 12). To this end we first remark that A2 (u, t) coincides (modulo sign) with A1 (s, t) (see the proof of the Pomeranchuk theorem in Lecture 1). 12). For large z ∝ s the dominant contribution to the integral over z1 and √ 1, since according to our assumption the z2 arises from z1 ∼ z2 ∼ z integrand grows linearly with z1 and z2 (A1 (zi ) zi f (t)). 7b) for K(z1 , z2 , z): K(z1 , z2 , z) z(z − 2z1 z2 ). 12) and omitting irrelevant s-independent factors we obtain an estimate ρ(s, t) ∝ z1 dz1 z2 dz2 z(z − 2z1 z2 ) , z1 > 1, z2 > 1, z > 2z1 z2 .

This explains why the physics of the t-channel is important for large s. 2 Physics of the t-channel and complex angular momenta In the previous lecture we have discussed analytic properties of the invariant amplitude A(s, t, u) describing the scattering of neutral spinless particles (π 0 mesons). As was already stressed, it is convenient to depict the kinematics of the reactions and the location of the singularities of A(s, t) on the Mandelstam plane; see Fig. 1. u=0 u = 4µ s=0 2 s = 4µ2 t-channel t = 4µ2 u s t t=0 u-channel Fig.

What can be said about them? It turns out, quite a lot. Let us assume that the amplitude at large s is polynomially bounded 0 uniformly in t, that is A± 1 (s, t) < s , 0 = const for any t. 1) is convergent and f (t) has no singularities for any t. Notice that up to this point we did not discuss the convergence properties of the integrals but rather only local properties of the Mandelstam representation (the location of the singularities). Since a moving singularity in , = (t) leads in turn to a singularity in t = t( ), then for > 0 , f (t) has no moving singularities in t.