I am wondering how to get below equation (above eq.(5.104)) on Peskin & Schroeder's QFT book page 167:
$$
y\simeq\frac{1}{2}(1-\cos\theta)\simeq 1-\frac{\chi^{2}}{4}.
$$
Here, the book try to consider an inverse compton scattering (a low-energy photon beam collide with a high energy electron beam), with initial photon energy $\bar{\omega}$, initial electron energy $E$, and label the scattered photon energy $E^{\prime}=yE$. The book use the relation $s=4E\bar{\omega}$ (I thought this already use the approximation $E\sim \bar{\omega}$). Then, to get the expression of $y$, P & S said: "by computing $2k\cdot k^{\prime}$ in the center-of-mass frame and in the lab frame". I am really troubled for this.
Here is my attempt:(initial photon in $+z$ direction, electron in $-z$ direction)
(1) in center-of-mass frame:(as the figure in page 164)
$$
\begin{aligned}
k&=(E,0,0,E) \\
k^{\prime}&=(E,Esin\theta,0,Ecos\theta) \\
k\cdot k^{\prime}&=E^2 (1-cos\theta)
\end{aligned}
$$
(2) in the lab frame:(as the figure in page 162)
$$
\begin{aligned}
k&=(\bar{\omega},0,0,\bar{\omega}) \\
k^{\prime}&=(yE,0,0,-yE) \\
k\cdot k^{\prime}&=2yE\bar{\omega}\simeq 2yE^2
\end{aligned}
$$
then, if we let the $k\cdot k^{\prime}$ in center-of-mass frame equal to that in lab frame, we can get the expression of $y$, but why? here is my puzzle?
(1) why $k\cdot k^{\prime}$ is invariant in different reference system?
(2) why in lab frame we don't need to introduce the angle dependence?
However, that's just my personal thoughts.
If you have any comments, I am really appreciate it.
Best Answer
From the figure on page 166, in the CM, you have two relativistic invariants, $$ s= (\omega +\sqrt{\omega^2 +m^2})^2\approx 4\omega^2; ~~~~k\cdot k'=\omega^2 (1-\cos\theta) \approx \frac{s}{4} (1-\cos\theta ). $$
Instead, the low energy ($\varpi$) photon collides with a basically light-like higher energy (E) electron and therefore bounces right back (your question 2) $$ s=(\varpi +E)^2-\left(\varpi -\sqrt{E^2-m^2}\right )^2 =4\varpi E + m^2 +\varpi m^2 /E +...\approx 4\varpi E ; \\ k\cdot k'= 2\varpi yE= y s/2 ; $$ so that, comparing the relativistic invariants in each frame, $$ y= (1-\cos\theta)/2= (1+\cos\chi)/2 \approx 1-\chi^2/4+... $$
The linchpin of the process is the head-to-head collision of two light-like objects, the soft photon and the hard electron, a small pellet versus a fast truck. In that frame, as opposed to the CM one, there is no room for deflection: the soft object bounces right back. (Unlike your terrible misimpression, we actually have $E\gg \varpi$.)