# Relativistic kinematics - particle physics Q

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!

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Jun 6 '17 at 12:13
• @ACuriousMind Yes, I know that. This question asks for a general way to approach relativistic questions like these with this as an example, as The person gave. – PhysicsMathsLove Jun 6 '17 at 13:36

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.

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*}

• please note that the policy of the site is not to solve homework problems completely, but help – anna v Jun 8 '17 at 3:45