By Headrick M
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Additional resources for A solution manual for Polchinski's String Theory
The terms we are looking for arise from the contraction of the q µ with each other. Specifically, the singular part of the OPE of q µ (y) with q ν (y ′ ) is − 2α′ (y − y ′ )−2 η µν , (42) so the combination e1 · e2 e3 · e4 arises from the contractions of q µ1 (y1 ) with q µ2 (y2 ) and q µ3 (y3 ) with q µ4 (y4 ): igo2 α′−1 Tr(λa3 λa4 λa1 λa2 )e1 · e2 e3 · e4 (2π)26 δ26 ( ′ ki ) i 2α′ k2 ·k3 +1 ′ ×|y12 |2α k1 ·k2 −1 |y13 |2α k1 ·k3 +1 |y23 | y1 × y3 ′ ′ ′ dy4 |y14 |2α k1 ·k4 |y24 |2α k2 ·k4 |y34 |2α k3 ·k4 −2 .
Neglecting the “. . ” and approximating the exponents 2α′ k1 · k2 = 2α′ k3 · k4 = −α′ s − 2 by 2, (58) becomes 32 − sin2 (πx12 ) sin2 (πx34 ) . (63) ′ π(α s + 4) After integrating over the positions xi , the amplitude (53) is S ′ (k1 , . . , k4 ) = −i(2π)26 δ26 ( ki )go4 2−9 (2π)−23 α′−14 i 1 . 12, we calculated the three-point vertex for two open-string tachyons to go to a closed-string tachyon: 4πigc (2π)26 δ26 (k1 + k2 + k3 ). (65) α′ We can reproduce the s-channel pole of (64) by simply writing down the Feynman diagram using this vertex and the closed-string tachyon propagator, i/(s + 4/α′ ): 1 − i(2π)26 δ26 ( ki )gc2 4(2π)2 α′−2 .
The path integral is now ... σ = e−Si . . = ... f f − Si . . f + 1 2 S ... 2 i f + ··· , (72) where f is the path integral calculated using only the free action (70). 13b). It is the Weyl variation of the third term, gives rise to the term in βµν quadratic in H, that we are interested in, and in particular the part proportional to ¯ ν : . . f, d2z : ∂X µ ∂X (73) 31 3 CHAPTER 3 G . This third term is whose coefficient gives the H 2 term in βµν 1 2 S ... 2 i × f = 1 Hωµν Hω′ µ′ ν ′ 2(6πα′ )2 (74) ′ ′ ′ ¯ ν (¯ d2zd2z ′ : X ω (z, z¯)∂X µ (z)∂X z ) :: X ω (z ′ , z¯′ )∂ ′ X µ (z ′ )∂¯′ X ν (¯ z′ ) : .
A solution manual for Polchinski's String Theory by Headrick M