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Lecture 6

Chasse_neige

Four Particle Test

From 3 pt amplitude, using consistency to guess the result of 4 pt.

Tension: non-zero spin will not be compatible with locality.

We assume perturbativity in the 4 pt test and amplitudes are doinated by the on-shell pole terms. (+ analytic terms required by consistency), this means that Ms are rational functions of all kinetic variables. We call this “tree-level approximation”.

DoF Counting

4×4410=2s(p1+p2)2t(p1+p4)2u(p1+p3)2

And we have

s+u+t=0

The 4 pt amplitude can be expressed as

M4=M4(s,t,u)

We assume that M is a rational function.

Most General Form of M4

Self-Interact Massless Scalar

On-shell Factorization Theorem gives that in the s channel

lims0iM4(1,2,3,4)=iM3(p1,p2,ps)×iM3(ps,p3,p4)×isiϵ

Doing the same thing for t and u channels, we can derive that

M4=μ2(1s+1t+1u)+regular terms

The regular term will not contain any 1-order terms of stu

regular term=λ0+λ1(s2+t2+u2)+λ2stu+

From the field theory we can claim that

1s+1t+1uμϕ3×μϕ3Constλ0ϕ4s2+t2+u2λ1(μϕ4)stuλ2(μϕ)2(νϕ)2

Yang-Mills Theory

Massless spin-1 particles (n3), we call these particles gluons, and the species colors. The amplitude

M4(1a+,2b+,3c,4d)=[12]2342f(s,t,u)

s-channel (s0)

psp1+p2=|s[s|

Under this representation

lims0iM4=iseiM3(1a+,2b+,se)×iM3(se+,3c,4d)=iseκabeκcde[12]3[1s][s2]3433ss4

Simplify

[1s]s4=[11]14+[12]24=[12]243s[s2]=34[42]

Define

α=eκabeκecd

Then the amplitude can be expressed as

lims0M4=[12]2342αst

t-channel (t0)

pt=|t[t|

Amplitude is

limt0iM4=iteiM3(1a+,4d,te+)×iM3(te,2b+,3c)+iM3(1a+,4d,te)×iM3(te+,2b+,3c)=iteκdaeκbce[[1t]3[14][4t]3t3322t+[2t]3[23][3t]4t3411t]

while t0, we have

[14]41O(t)[23]32O(t)

We suppose that 14O(t),[14]O(1), so we can derive that

1t4tO(t)

[23]O(t),23O(1)

[23][2t][3t]O(t)

Under this assumption, the amplitude has the form

limt0M4=[12]2342βtu

Similarly, in the u-channel, we have

limu0M4=[12]2342γus

From these limit forms, we can guess the rational function f(s,t,u)

f(s,t,u)=Ast+Btu+Cus

where

AC=αBA=βCB=γ

So this form is only possible if

eκabeκcde+κadeκbce+κaceκdbe=0

called Jacobi’s Identity.

Consistant interaction among n (n3) massless spin-1 particles (by minimal coupling) must have a Lie-algebra structure.

Self-Interaction of Spin-j (j > 1)

lims0M4=αs([12]2342t)j

u and t channels are similar.

Gravity (j = 2)

M4(1a+2,2b+2,3c2,4c2)=[12]4344Astu{Ast2(s0)Atu2(t0)Aus2(u0)

This gives out

α=β=γeκabeκcde=eκadeκbce=eκaceκbde

With a proper choice of basis, we can give out a simple form of κ

κabc=κaδabδac

There can not be more than one massless spin-2 particles interaction by each other through minimal coupling.

Amplitude:

M4=[12]4344κ2stu

Consider spin-j (j>2)

M4(1a+j,2b+j,3cj,4dj)=[12]2j[34]2jf(s,t,u)[M4]=0[f]=4j

Do consistent factorization similarly, which requires fκ2,[κ]=1j

f=κ2f~(s,t,u)

where f~ is a rational function of s,t,u

[f~]=22j

If j>2, we can say that [f~]<6, which is impossible because this will cause double-pole singularity in any channel.

So massless spin-j particles cannot have minimal self-interaction.

Remarks

  1. No underlying theory assumed in our derivation. Tension: spin locality
  2. j=1,2 4pt amplitudes’s form is nearly “uniquely” fixed.
  3. “constructable”: 3pt n-pt (n>3) “BCFW”

Gravitational Couplings of all Massless Particles

Simplify: we only consider how 1 species of spin-j particles coulple with gravitons.

M(1+j,2+j,3j,4j)

is impossible if j>2 (same reason as analyzed above, they can’t minimal couple with gravitons), so we only consider j=0,1,2

Gravitional Compton Scattering

Gravitional coupling κj of a spin-j particles equals the graviton cubic self-coupling κ.

If we consider more than 1 massless spin-2 particles, then any other massless can only couple gravitationally to one of them.

Remarks

  1. Minimal couplings are assumed in all our derivation.
  2. Massive particles can also be derived using the same trick, but the spinor-helicity formalism will be more complicated (need to spinors to construct one momentum).

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