$e^+e^-\rightarrow f\bar{f}$ process in the center of mass frame

Question

In Mark Thomson's Modern Particle Physics, page 152-153, writes the four-momenta of the particles involved in the $e^+e^-\rightarrow f\bar{f}$: $$p1=(E,0,0,+E)$$ $$p2=(E,0,0,-E)$$ $$p3=(E,+\beta E\sin\theta,0,+\beta E\cos\theta)$$ $$p4=(E,-+\beta E\sin\theta,0,-\beta E\cos\theta)$$

where $\beta=v/c$. For simplicity, assume natural units, so $c=1$. He further writes the following: $$\beta^2=\left(1-\frac{4m_f^2}{s}\right)$$ and I'm having trouble deducing this formula. So, in the CM frame, the available energy is $\sqrt{s}$ so each particle gets $\frac{\sqrt{s}}{2}$. Therefore, for a single fermion we should have $$\frac{\sqrt{s}}{2}=m_f+\frac{1}{2}m_f\beta^2$$, i.e, the kinetic energy plus the rest energy of the fermion. But after manipulation I get: $$\beta^2=\frac{\sqrt{s}}{2}\left(1-\frac{2m}{\sqrt{s}}\right)$$ Where am I wrong?

Answer 1

You went wrong in the right-hand-side of your next-to-the-last equation, by using a nonrelativistic expression for the energy. In Special Relativity, the kinetic energy is not $\frac{1}{2}mv^2$. The total energy (rest energy plus kinetic energy) of a particle of mass $m$ moving with speed $v$ is

$$E=\frac{mc^2}{\sqrt{1-v^2/c^2}}$$

which means that the kinetic energy is

$$K=mc^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)=\frac{1}{2}mv^2+O(v^4).$$

The Newtonian expression $\frac{1}{2}mv^2$ is actually just the leading term when you expand the relativistic kinetic energy in powers of the speed. The other terms become important as $v\rightarrow c$.

So the correct equation is

$$\frac{\sqrt{s}}{2}=\frac{m_f}{\sqrt{1-\beta^2}}.$$

Answer 2

The relativistic equation connecting energy, mass and momentum is $$ E^2 = |\vec p|^2 + m^2$$ Here, for $f$ we have $$ E^2 = E^2\beta^2 + m_f^2$$ $$ \beta^2 = 1-\frac{m_f^2}{E^2} = 1-\frac{4m_f^2}{s}$$