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Mandelstam variablesThe variabless, t, uare calledMandelstam variables. They are a great shorthand, usedalmost exclusively in2→2scattering and in1→3decays, although there are generalizations formore momenta. For2→2scattering, with initial momentap1andp2and final momentap3andp4, they are defined bys≡(p1+p2)2= (p3+p4)2(104)t≡(p1−p3)2= (p2−p4)2(105)u≡(p1−p4)2= (p2−p3)2(106)These satisfys+t+u=summationdisplaymj2(107)wheremjare the invariant masses of the particles.16Section 5
As we saw in the previous example,s, tanducorrespond to particular diagrams wheremomentum in the propagator has invariantpμ2=s,toru. We says-channel for annihilationdiagrams. In these the intermediate state haspμ2=s >0. Thet- andu- channels are scatteringdiagramss−channelt−channelu−channels, tanduare great because they are Lorentz invariant. So we computeM(s, t, u)in the center-of-mass frame, and then we can easily find out what it is in any other frame, for example theframe of the lab in which we are doing the experiment. We will uses, tandua lot.5.2Derivative couplingsSuppose we have an interaction with derivatives in it, likeLint=λφ1(∂μφ2)(∂μφ3)(108)where I have included 3 different scalar fields for clarity. In momentum space, these∂μ’s givefactors of momenta. But now remember thatφ(x) =integraldisplayd3p(2π)312ωpradicalbig(ape−ipx+ap†eipx)(109)So if the particle is being created (emerging from a vertex) it gets a factor ofipμ, and if it’sbeing destroyed (entering a vertex) it gets a factor of−ipμ. So a−for incoming momentum anda+for outgoing momentum.In this case, it’s quite important to keep track of whethermomentum is flowing into or out of the vertex.For example, take the diagramφ2φ1φ3φ2φ1(110)Label the initial momentap1μandp2μand the final momentap1μ′andp2μ′.The exchangedmomentum iskμ=p1μ+p2μ=p1μ′+p2μ′. Then this diagram givesiM= (iλ)2(−ip2μ)(ikμ)ik2(ip2ν′)(−ikν) =−iλ2[p2·p1+ (p2)2][p2′·p1′+ (p2′)2](p1+p2)2(111)As a cross check, we should get the same answer if we use a different Lagrangian related tothe one we used by integration by parts:Lint=−λφ3[(∂μφ1)(∂μφ2) +φ1squareφ2](112)Now our one diagram becomes four diagrams, from the two types of vertices on the two sides, allof which look like Eq. (110). It’s easiest to add up the contributions to the vertices before multi-plying, which givesM= (iλ)2[(−ip2μ)(−ip1μ) + (−ip2)2]ik2[(ip2ν′)(ip1′ν) + (ip2′)2](113)=−iλ2[p2·p1+ (p2)2][p2′·p1′+ (p2′)2](p1+p2)2(114)Examples17
which is exactly what we had above. So integrating by parts does not affect the matrix ele-ments, as expected. Thus the Feynman rules passed our cross check.