Mandelstam variables The variables s t u are called Mandelstam variables They

Mandelstam variables the variables s t u are called

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Mandelstam variables The variables s, t, u are called Mandelstam variables . They are a great shorthand, used almost exclusively in 2 2 scattering and in 1 3 decays, although there are generalizations for more momenta. For 2 2 scattering, with initial momenta p 1 and p 2 and final momenta p 3 and p 4 , they are defined by s ( p 1 + p 2 ) 2 = ( p 3 + p 4 ) 2 (104) t ( p 1 p 3 ) 2 = ( p 2 p 4 ) 2 (105) u ( p 1 p 4 ) 2 = ( p 2 p 3 ) 2 (106) These satisfy s + t + u = summationdisplay m j 2 (107) where m j are the invariant masses of the particles. 16 Section 5
As we saw in the previous example, s, t and u correspond to particular diagrams where momentum in the propagator has invariant p μ 2 = s , t or u . We say s -channel for annihilation diagrams. In these the intermediate state has p μ 2 = s > 0 . The t - and u - channels are scattering diagrams s channel t channel u channel s, t and u are great because they are Lorentz invariant. So we compute M ( 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 the frame of the lab in which we are doing the experiment. We will use s, t and u a lot. 5.2 Derivative couplings Suppose we have an interaction with derivatives in it, like L int = λφ 1 ( μ φ 2 )( μ φ 3 ) (108) where I have included 3 different scalar fields for clarity. In momentum space, these μ ’s give factors of momenta. But now remember that φ ( x ) = integraldisplay d 3 p (2 π ) 3 1 2 ω p radicalbig ( a p e ipx + a p e ipx ) (109) So if the particle is being created (emerging from a vertex) it gets a factor of ip μ , and if it’s being destroyed (entering a vertex) it gets a factor of ip μ . So a for incoming momentum and a + for outgoing momentum. In this case, it’s quite important to keep track of whether momentum is flowing into or out of the vertex. For example, take the diagram φ 2 φ 1 φ 3 φ 2 φ 1 (110) Label the initial momenta p 1 μ and p 2 μ and the final momenta p 1 μ and p 2 μ . The exchanged momentum is k μ = p 1 μ + p 2 μ = p 1 μ + p 2 μ . Then this diagram gives i M = ( ) 2 ( ip 2 μ )( ik μ ) i k 2 ( ip 2 ν )( ik ν ) = 2 [ p 2 · p 1 + ( p 2 ) 2 ][ p 2 · p 1 + ( p 2 ) 2 ] ( p 1 + p 2 ) 2 (111) As a cross check, we should get the same answer if we use a different Lagrangian related to the one we used by integration by parts: L int = λφ 3 [( μ φ 1 )( μ φ 2 ) + φ 1 square φ 2 ] (112) Now our one diagram becomes four diagrams, from the two types of vertices on the two sides, all of which look like Eq. (110). It’s easiest to add up the contributions to the vertices before multi- plying, which gives M = ( ) 2 [( ip 2 μ )( ip 1 μ ) + ( ip 2 ) 2 ] i k 2 [( ip 2 ν )( ip 1 ν ) + ( ip 2 ) 2 ] (113) = 2 [ p 2 · p 1 + ( p 2 ) 2 ][ p 2 · p 1 + ( p 2 ) 2 ] ( p 1 + p 2 ) 2 (114) Examples 17
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.

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