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The Role of Riemann's Zeta Function in Mathematics and Physics

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The Role of Riemann's Zeta Function in Mathematics and Physics universe Article The Role of Riemann’s Zeta Function in Mathematics and Physics †,‡ Walter Dittrich Institut für Theoretische Physik, Universität Tübingen, Auf der Morgenstelle 14, D-72076 Tübingen, Germany; [email protected] † This article is dedicated to Martin Reuter on the occasion of his 60th birthday, as well as to a mutual hero of ours: Bernhard Riemann. ‡ This paper is based on a talk given at the conference “Quantum Fields—From Fundamental Concepts to Phenomenological Questions”, Mainz, Germany, 26–28 September 2018. Received: 28 January 2019; Accepted: 11 March 2019; Published: 14 March 2019 Abstract: In particular, Riemann’s impact on mathematics and physics alike is demonstrated using methods originating from the theory of numbers and from quantum electrodynamics, i.e., from the behavior of an electron in a prescribed external electromagnetic field. More specifically, we employ Riemann’s zeta function to regularize the otherwise infinite results of the so-called Heisenberg–Euler Lagrangian. As a spin-off, we also calculate some integrals that are useful in mathematics and physics. Keywords: Riemann; zeta function; functional equation; Heisenberg-Euler Lagrangian 1. Usefulness of Riemann’s Functional Equation for the Zeta Function in Physics Riemann introduced in his talk in Berlin in 1859 [1] (p. 147) Z ¥ s − s s −1 − s − 1 1 G p 2 z(s) = dxy(x) x 2 + x 2 2 − . (1) 2 1 s(1 − s) where y(x) is related to one of Jacobi’s q functions. Notice that there is no change of the right-hand side − s 2 s under s ! (1 − s). p G 2 z(s) has simple poles at s = 0 (from G) and s = 1 (from z). To remove 1 these poles, we multiply by 2 s(s − 1). This is the reason why Riemann defines 1 − s s x(s) = s(s − 1)p 2 G z(s), (2) 2 2 which is an entire function (z is a meromorphic function). Obviously, we have x(s) = x(1 − s) together with the symmetrical form of the functional equation, which was proved by Riemann for all complex s: s − s 1 − s − 1 (1−s) G p 2 z(s) = G p 2 z(1 − s). (3) 2 2 Notice that the right-hand side is obtained from the left-hand side by replacing s by 1 − s. Before we continue with Equation (3), we make use of two important formulae due to Euler and Legendre. Universe 2019, 5, 79; doi:10.3390/universe5030079 www.mdpi.com/journal/universe Universe 2019, 5, 79 2 of 10 Legendre’s duplication formula: G functions of argument 2s can be expressed in terms of G functions of smaller arguments. − 1 2s− 1 1 G(2s) = (2p) 2 2 2 G(s)G s + 2 1 1 = p 22s−1G(s)G s + . (4) p 2 s When we replace s ! 2 , we obtain: 1 s s + 1 G(s) = p 2s−1G G p 2 2 p p s s + 1 or G(s) = G G . (5) 2s−1 2 2 Euler’s reflection formula for the G function: p G(s)G(1 − s) = . (6) sin ps s+1 s 1 Replace s ! 2 = 2 + 2 : s + 1 s + 1 p G G 1 − = ps p 2 2 sin 2 + 2 s + 1 1 − s p or G G = ps . (7) 2 2 cos 2 s+1 Now multiply Equation (3) by G 2 to obtain − s s s + 1 − 1−s s + 1 1 − s p 2 G G z(s) = p 2 G G z(1 − s). (8) 2 2 2 2 Then, we get from Equations (5) and (7) the following equations: p − s p − 1−s p 2 ( ) ( ) = 2 ( − ) p s−1 G s z s p ps z 1 s , 2 cos 2 2 ps z(1 − s) = cos G(s)z(s). (9) (2p)s 2 Here, we replace s ! 1 − s thereby obtaining 2 p(1 − s) z(s) = cos G(1 − s)z(1 − s) (10) (2p)1−s 2 ps or z(s) = 2sps−1 sin G(1 − s)z(1 − s). (11) 2 The latter equation is of great importance in the following sections. 2. Correction of the Classical Electromagnetic Lagrangian by Vacuum Electrons In 1936, Werner Heisenberg and Hans Euler [2] wrote down the first effective Lagrangian in quantum field theory, which incorporates a quantum correction to the classical Lagrangian of a constant electromagnetic field; this correction is due to the polarization of the quantum vacuum (Dirac’s idea), i.e., the effect of an external constant electromagnetic field on the motion of the vacuum electrons. Universe 2019, 5, 79 3 of 10 To simplify matters, we only consider a constant magnetic field in z direction. For this special case, the modified Lagrangian takes the form—in Schwinger’s representation: 1 L(B) = L(0) + L(1) , L(0) = − B2 , 2 ¥ 1 Z ds 2 1 L(1)(B) = e−im s (eBs) cot(eBs) + (eBs)2 − 1 . (12) 8p2 s3 3 0 The integral was explicitly calculated for the first time in [3] by dimensional regularization and thereafter in [4] by the so-called zeta-function regularization. The findings of these two different methods agree exactly, whereby the result obtained by the zeta-function regularization is finite without the usual subtraction of divergent counterterms. The result turns out to be 1 1 L(1)(B) = − −3m4 + 4(eB)2 − 4z0(−1) + 4m2(eB)(ln 2p − 1) 32p2 3 2eB 2eB 4 2eB − 2m4 ln − 4m2(eB) ln − (eB)2 ln m2 m2 3 m2 2 9 + m 1 2eB > Z => −16(eB)2 dx ln G(x) . (13) > 1 ;> eB For those values of the field strength, i.e., for strong fields m2 1, the integral over the logarithm of the gamma function only yields a constant. = eB With b m2 , we obtain 1+ 1 1+ 1 Z 2b Z 2b 2 2 d b dx ln G(x) ≈ b dx ln G(1) + ln G(x)j = (x − 1) dx x 1 1 1 + 1 1 2b Z 1 1 = b2Y(1) dx(x − 1) = Y(1) = − C. (14) 8 8 1 Here, C is Euler’s number, C = ln g = 0.57721566490, and the digamma function d G0(x) Y(x) = ln G(x) = at x = 1 is given by Y(1) = − ln g. dx G(x) Therefore, by only considering the dominant forms for large magnetic field strength, we obtain for the asymptotic form of the one-loop effective Lagrangian in spinor quantum electrodynamics (QED) 1 1 4 2eB L(1)( ) = − ( )2 − 0(− ) − ( )2 lim B 2 4 eB 4z 1 eB ln 2 eB !¥ 32p 3 3 m m2 aB2 eB e2 = ln + 12z0(−1) − 1 + ln 2 , a = . (15) 6p m2 4p On the other hand, we find in Ritus’ paper [5] under Formula (60) the expression aB2 eB 6 L(1)( ) = + 0( ) lim B ln 2 2 z 2 . (16) eB !¥ 6p gpm p m2 Universe 2019, 5, 79 4 of 10 Since Equations (15) and (16) are just two different representations of the same strong-field Lagrangian L(1)(B), we have the equality 2eB 2eB 6 ln − 1 + 12z0(−1) = ln + z0(2) m2 m22pg p2 or 6 − 1 + 12z0(−1) = − ln(2pg) + z0(2). (17) p2 This important equation is in fact a direct consequence of a variant of the famous functional equation of Riemann’s zeta function [1] ps z(s) = 2sps−1 sin G(1 − s)z(1 − s). (18) 2 Note that the result in Equation (17), which arises from long, complicated field theoretic calculations, follows from solving a physics problem—not from analytical theory of numbers, namely by studying the behavior of vacuum electrons in presence of a constant external strong classical magnetic field. (Later, we meet z0(−1) and z0(2) again in connection with the Glaisher–Kinkelin 1 constant.) Thus far, we consider nonlinear spinor QED where spin 2 particles with mass m are coupled to an external constant magnetic field. The corresponding effective Lagrangian is given by Equation (12). Now, we study charged spinless particles with mass m, associated with a complex scalar field, which interact with a constant magnetic field. Here, the starting point is given by the Heisenberg–Euler effective Lagrangian, which in Schwinger’s proper time representation reads: Z ¥ (1) 1 1 −im2s eBs 1 2 Lscalar(B) = − e − (eBs) − 1 ds. (19) 16p2 0 s3 sin(eBs) 6 Without going into detail, we just repeat the former results for spinor particles obtained for strong magnetic fields, when performing the calculation in both the z function and proper time regularization. Here are the results for scalar QED: aB2 2eB L(1) ( ) = + 0(− ) − + lim scalar B ln 2 12z 1 1 ln 2 (20) eB !¥ 24p m m2 aB2 2eB 6 = ln + − ln gp + z0(2) (21) 24p m2 p2 This brings us back to Equation (17), which is equivalent to Riemann’s functional equation for the zeta function. This functional equation is evidently independent of the masses involved, 1 1 2eB eB be they fermionic or scalar. Were it not for the factors 24p instead of 6p and ln m2 instead of ln m2 in the spinor case, we could have guessed Equations (20) and (21). However, arriving from the proper-time integrals of Equation (12) or Equation (19) at Equations (15), (16), (20), and (21) is a highly challenging undertaking. Finally, let us mention that the results of Equations (15) and (20) can be used in the Callan–Szymanzik renormalization equation to calculate the bz (a) function for spinor and scalar QED to result in 2 a 1 a b (a) = (spinor), b (a) = (scalar).
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