Riemann Zeta Function & The Four-Point Veneziano Amplitude Proposition of Freund-Witten Johar M. Ashfaque The four-point Veneziano amplitude is A(s, t) = B(−α(s), −α(t)) where α(s) is the Regge trajectory usually taken as a linear function in terms of the squared four momenta (the Mandelstam variables) s = −(k1 + k2 )2 and t = −(k2 + k3 )2 for external momenta ki=1,2,3,4 . The four-point amplitude with s, t and u = −(k1 + k3 )2 symmetrized, is the crossing symmetrized amplitude A4 (s, t, u) = A(s, t) + A(t, u) + A(s, u). Traditionally, one takes the linear Regge trajectory to be s 2 and works in units where the mass squared of the tachyon is −2. This implies s + t + u = −8 and α(s) = 1 +
α(s) + α(t) + α(u) = −1. Proposition 0.1 The crossing symmetrized amplitude A4 (s, t, u) defined as A4 (s, t, u) = A(s, t) + A(t, u) + A(s, u) subject to α(s) + α(t) + α(u) = −1 can be written entirely in terms of the Riemann zeta function as A4 (s, t, u) = B(−α(s), −α(t)) + B(−α(t), −α(u)) + B(−α(s), −α(u)) =
1
Y ζ(1 + α(x)) . ζ(−α(x)) x=s,t,u
The Proof
We will need to employ the well-known Euler reflection formula π Γ(z)Γ(1 − z) = sin πz which follows from the Weierstrass product expansion for Gamma and sine so that Γ(−α(s) − α(t))−1
1 sin(π(−α(s) − α(t)))Γ(1 + α(s) + α(t)) π 1 = − sin(πα(u))Γ(−α(u)) π
=
since α(s) + α(t) + α(u) = −1. We will also need the half circle trigonometric identity X Y πα(x) sin(πα(x)) = −4 cos . 2 x=s,t,u x=s,t,u Now
1 A4 (s, t, u) = 2π
Y
2Γ(−α(x)) cos
x=s,t,u
1
πα(x) . 2
Recall 1.1 The duplication formula for Euler Gamma function √ 1 Γ(z)Γ z + = 21−2z πΓ(2z) 2 which can be expressed as 1 z 1 Γ(z) = π − 2 2z−1 Γ Γ (z + 1) . 2 2 Note. For z→ we have cos
πz 2
1 (z + 1) 2
−1 1 1 . = π Γ (z + 1) Γ (1 − z) 2 2
The factor
2Γ(−α(x)) cos
becomes
πα(x) 2
α(x) Γ − 2 √ 1 1 α(x) 2π − 2 2−α(x)−1 Γ − Γ (1 − α(x)) = π2−α(x) 2 2 1 Γ 2 (1 + α(x))
so that upon using α(s) + α(t) + α(u) = −1 again A4 (s, t, u)
α(x) 2
Γ − π2−α(x) 1 x=s,t,u Γ 2 (1 + α(x)) α(x) Γ − 2 √ Y . π = x=s,t,u Γ 1 (1 + α(x)) 2 1 2π
=
Y √
Now using the famous functional equation for the Riemann zeta function πz z z−1 ζ(z) = 2 π sin Γ(1 − z)ζ(1 − z) 2 which is valid for complex z. Multiplying both sides by z Γ 2 and making use of the identity Γ(z)Γ(1 − z) =
π sin πz
to exchange sin with
πz 2
−1 z z π Γ Γ 1− 2 2
and finally Γ(1 − z) for 1
π − 2 2−z Γ
1 z (1 − z) Γ 1 − 2 2 2
by the duplication formula we have 1 z z− 12 ζ(z) = π Γ (1 − z) ζ(1 − z) Γ 2 2 which can be rewritten as an equality of ratios for −z z Γ −2 ζ(1 + z) − 1 −z = π 2 ζ(−z) 1 Γ 2 (1 + z) Upon substitution into α(x) Γ − 2 √ Y A4 (s, t, u) = π x=s,t,u Γ 1 (1 + α(x)) 2 and using α(s) + α(t) + α(u) = −1 and we arrive at the required result.
3