Energy distribution of photon from neutral pion ($\pi^0$) decay

Question

The laboratory-frame energy $E_\gamma$ of either photon is uniformly distributed between $E_{min} = \frac{1}{2}E_{\pi}(1-\beta)$ and $E_{max} = \frac{1}{2}E_{\pi}(1+\beta)$, where $E_\pi$ is the energy of the pion and $\beta=\frac{v}{c}$ is the pion's velocity divided by the velocity of light.

I read this paragraph on Statistical data analysis by Glen Cowan in pg.31 and tried to study why. And below is what I've thought.

In CM frame, pi0 decays to two photons isotropically and each photon has energy of $\frac{1}{2}E_{\pi}$. But in the lab frame, pi0 is boosted in a certain direction with kinetic energy $(\gamma-1)E_{\pi}$ We can approximate this as... \begin{align} \ \gamma-1=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\simeq1+\frac{1}{2}\frac{v^2}{c^2}-1=\frac{1}{2}\frac{v^2}{c^2} \end{align} with an assumption that v is smaller than c. Then the photon can have its maximal energy of $\frac{1}{2}E_\pi(1+\frac{v^2}{c^2}$) (if photon's direction is aligned with pi0's direction) and its minimal energy of $\frac{1}{2}E_\pi(1-\frac{v^2}{c^2})$(if photon's direction is anti-aligned with pi0's direction). Since two photon decay isotropically in CM frame, energy distribution of photon from neutral pion decay shows uniform distribution between [$\frac{1}{2}E_\pi(1-\frac{v^2}{c^2})$,$\frac{1}{2}E_\pi(1+\frac{v^2}{c^2}$)].

But here, I got $β^2$ not just β. Is there anything wrong with my idea?

Answer 1

The issue is with your Lorentz transformation. In the CM frame, each photon has energy: $$E_\gamma^{CM}=\frac{m_\pi}{2}$$ with $m_\pi$ the rest mass of the pion. It is related to its energy in the lab frame by: $$E_\pi=\gamma m_\pi$$ Back to the lab frame, each photon goin in direction $\theta,\phi$ (spherical coordinates with respect to the lab’s velocity) in the CM frame has energy: $$ E_\gamma^L=E_\gamma^{CM}(\gamma-\beta\cos\theta) $$ which you can rewrite: $$ E_\gamma^L=\frac{1}{2}(E_\pi-m_\pi\beta\cos\theta) $$ You recover the extremal values at: $$ E_\pm=\frac{1}{2}(E_\pi\pm m_\pi\beta) $$ and the distribution is uniform, since if the photons scatter isotropically in the CM frame, $\cos\theta$ is distributed uniformly.

In the non relativistic limit, $\beta\ll 1$ so $E_\pi=m_\pi$ and you recover Cowan’s claim.

Hope this helps.