Relativistic kinematics - particle physics Q

Question

I need some help with this relativity question. Just some hints would be good. I have given the question and my attempt.

$D^0$ mesons can decay to $K^+$ and $π^-$ mesons. By assuming $c=1$, show that the energy $K^+$ in the rest frame of the $D^0$ meson is given by $$E_K = \frac{1}{2} \left(M_D + \frac{m^2_K-m^2_{\pi}}{M_D} \right) $$

I have tried the following:

Conserving energy overall: $M_D = m^2_{\pi} + m^2_K$ (because the momenta have to be equal and opposite).

Then I considered the energy for the pion: $E^2_{\pi} = m^2_{\pi} + p^2$

and the k meson $E^2_K = m^2_K + p^2$, but I don't know what to do from here. When I combine them, I get square roots and horrible quantities. If someone could show me how this problem is done and give some tips for how to approach such problems, that'd be really helpful!

Answer 1

Your energy conservation equation $M_D = m_\pi^2 + m_K^2$ is wrong because the mass of the $D_0$ meson needs to provide the kinetic energy as well: $M_D = E_\pi + E_K = \sqrt{m_\pi^2 + p_\pi^2} + \sqrt{m_K^2 + p_K^2}$. Also, the units don't match.

An easier way to start is with momentum conservation, $p_\pi^2 = p_K^2$, and make substitutions from $E^2 = m^2 + p^2$ for each particle.

Answer 2

From a linear-algebra point of view, write the 4-momentum conservation as $$\tilde D=\tilde K + \tilde \pi.$$

(You may want to also draw this as triangle in an energy-momentum diagram [akin to a spacetime diagram] and do the problem using components, possibly with rapidities. A diagram will at least guide you in a geometric/trigonometric strategy... rather than merely guessing at how to solve a system of equations.)

Note that $E_K=\hat D \cdot \tilde K$ (the time-component of $\tilde K$ in the $D$-frame, using the dot-product of 4-vectors---the time-axis in the $D$-frame is given by $\hat D$, the unit 4-vector along $\tilde D$).

So, we wish to form $\tilde D \cdot \tilde K$, then $\hat D \cdot \tilde K$,

Here we go: $$\tilde\pi=\tilde D-\tilde K$$ So, $$\begin{align*} \tilde\pi\cdot\tilde\pi &=(\tilde D-\tilde K)\cdot(\tilde D-\tilde K)\\ &=\tilde D\cdot\tilde D+ \tilde K\cdot\tilde K - 2(\tilde D\cdot\tilde K)\\ \end{align*} $$ Thus, $$\tilde D\cdot \tilde K=\frac{1}{2}\left( \tilde D\cdot\tilde D+ \tilde K\cdot\tilde K - \tilde\pi \cdot\tilde\pi \right),$$ which is called a "polarization identity", expressing a dot-product in terms of a quadratic-form. Since $M_D^2=\tilde D\cdot\tilde D$, and similarly for the other particles, we have: $$\begin{align*} \frac{\tilde D}{M_D}\cdot \tilde K&=\frac{1}{2M_D}\left( \tilde D\cdot\tilde D+ \tilde K\cdot\tilde K - \tilde\pi \cdot\tilde\pi \right)\\ E_K&=\frac{1}{2M_D}\left( M_D^2+ M_K^2 - M_{\pi}^2 \right)\\ \end{align*} $$