QFT: differential cross section from center of mass to lab frame

Question

I have the following process: two ingoing particles, a photon hitting a nucleus, and two outgoing particles, the nucleus and a pion. I have computed $|M|^2$ and the differential cross section in the center of mass frame $frac{d sigma}{d Omega_{CM}}$; I now have to go into the lab frame, where the nucleus is initially at rest, and consider the limit of a infinite massive nucleus $M_N to infty$, and compute $frac{d sigma}{d Omega_{lab}}$.

Is there a general procedure to go from the first to the second?
I first wrote $frac{d sigma}{dt}$ and then multiplied it for a rather complicated expression that I found on a book to obtain $frac{d sigma}{d Omega_{lab}}$. However, taking the infinite massive nucleus limit, the result I get is not what I'm supposed to.

guest · Answer

A Lorentz boost can be used to go from the center of mass frame to the lab frame. Mandelstam variables are invariant under a boost. Here is the boost procedure for Compton scattering:

In the center of mass frame, let $p_1$ be the inbound photon, $p_2$ the inbound electron, $p_3$ the scattered photon, $p_4$ the scattered electron.

begin{equation*}
p_1=begin{pmatrix}omega omegaend{pmatrix}
qquad
p_2=begin{pmatrix}E-omegaend{pmatrix}
qquad
p_3=begin{pmatrix}
omega
omegasinthetacosphi
omegasinthetasinphi
omegacostheta
end{pmatrix}
qquad
p_4=begin{pmatrix}
E
-omegasinthetacosphi
-omegasinthetasinphi
-omegacostheta
end{pmatrix}
end{equation*}

where $E=sqrt{omega^2+m^2}$.

It is easy to show that

begin{equation}
langle|mathcal{M}|^2rangle
=
frac{e^4}{4}
left(
frac{f_{11}}{(s-m^2)^2}
+frac{f_{12}}{(s-m^2)(u-m^2)}
+frac{f_{12}^*}{(s-m^2)(u-m^2)}
+frac{f_{22}}{(u-m^2)^2}
right)
end{equation}

where

begin{equation}
begin{aligned}
f_{11}&=-8 s u + 24 s m^2 + 8 u m^2 + 8 m^4

f_{12}&=8 s m^2 + 8 u m^2 + 16 m^4

f_{22}&=-8 s u + 8 s m^2 + 24 u m^2 + 8 m^4
end{aligned}
end{equation}

for the Mandelstam variables $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$, $u=(p_1-p_4)^2$.

Next, apply a Lorentz boost to go from the center of mass frame to the lab frame in which the electron is at rest.

begin{equation*}
Lambda=
begin{pmatrix}
E/m & 0 & 0 & omega/m
0 & 1 & 0 & 0
0 & 0 & 1 & 0
omega/m & 0 & 0 & E/m
end{pmatrix},
qquad
Lambda p_2=begin{pmatrix}m  0  0  0end{pmatrix}
end{equation*}

The Mandelstam variables are invariant under a boost.
begin{equation}
begin{aligned}
s&=(p_1+p_2)^2=(Lambda p_1+Lambda p_2)^2

t&=(p_1-p_3)^2=(Lambda p_1-Lambda p_3)^2

u&=(p_1-p_4)^2=(Lambda p_1-Lambda p_4)^2
end{aligned}
end{equation}

In the lab frame, let $omega_L$ be the angular frequency of the incident photon
and let $omega_L'$ be the angular frequency of the scattered photon.
begin{equation}
begin{aligned}
omega_L&=Lambda p_1cdot(1,0,0,0)=frac{omega^2}{m}+frac{omega E}{m}

omega_L'&=Lambda p_3cdot(1,0,0,0)=frac{omega^2costheta}{m}+frac{omega E}{m}
end{aligned}
end{equation}

It follows that
begin{equation}
begin{aligned}
s&=(p_1+p_2)^2=2momega_L+m^2

t&=(p_1-p_3)^2=2m(omega_L' - omega_L)

u&=(p_1-p_4)^2=-2 m omega_L' + m^2
end{aligned}
end{equation}

Compute $langle|mathcal{M}|^2rangle$ from $s$, $t$, and $u$ that involve $omega_L$ and $omega_L'$.
begin{equation*}
langle|mathcal{M}|^2rangle=
2e^4left(
frac{omega_L}{omega_L'}+frac{omega_L'}{omega_L}
+left(frac{m}{omega_L}-frac{m}{omega_L'}+1right)^2-1
right)
end{equation*}

From the Compton formula
begin{equation*}
frac{1}{omega_L'}-frac{1}{omega_L}=frac{1-costheta_L}{m}
end{equation*}

we have
begin{equation*}
costheta_L=frac{m}{omega_L}-frac{m}{omega_L'}+1
end{equation*}

Hence
begin{equation*}
langle|mathcal{M}|^2rangle=
2e^4left(
frac{omega_L}{omega_L'}+frac{omega_L'}{omega_L}+cos^2theta_L-1
right)
end{equation*}

The differential cross section for Compton scattering is
begin{equation*}
frac{dsigma}{dOmega}propto
left(frac{omega_L'}{omega_L}right)^2langle|mathcal{M}|^2rangle
end{equation*}

QFT: differential cross section from center of mass to lab frame

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