Particle Collision, conservation of momentum

[bobblo]

Homework Statement

Consider the reaction pion(+) + neutron --> Kaon(+) + lambda particle(0). The rest masses of the particles are m(pion) = 140MeV,m(n) = 940MeV, m(K) = 494MeV, and m(lambda) = 1115MeV. What is the threshold kinetic energy of the pion particle to create a Kaon at an angle of 90degrees (with respect to its initial direction of motion) in the lab (in which the n is at rest)?

Homework Equations

Conservation of momentum, P(initial) = P(final) and P² is invariant and = m0²c². Conservation of energy across a certain frame of reference.

The Attempt at a Solution

Well knowing conservation of momentum and P² is invariant, i get an equation P(pion)² + P(kaon)² + P(neutron)² - P(lambda)² = 2[P(n) dot P(K)] - 2[P(pi) dot P(n)] + 2[P(pi) dot P(K)]. The LHS is constant across all frames. I deduce the RHS to be in the lab frame:
2/c²[E(K)E(n) - E(pi)E(n) + E(pi)E(K)]. I figure that because the neutron is at rest, while pion and kaon should be 90degrees apart and thus only their 4th component matter when using the dot product. Here i am stuck because i do not know how to find the energy of the kaon in the lab frame, E(K), in order to find the threshold energy of the pion. Thanks for any help.

 

[vela]

Conservation of momentum gives you

$$p_\pi^\mu + p_n^\mu = p_K^\mu + p_\Lambda^\mu$$

It might help to write out explicitly what the components of the four-momenta are. For example, assume the pion is moving in the +x direction with momentum p, you have

$$\begin{align*} p_\pi^\mu & = (E_\pi, p, 0, 0) \\ p_n^\mu & = (m_n,0,0,0) \end{align*}$$

I think if you write out the four-momenta for the kaon and lambda, it'll become apparent what condition leads to the threshold energy.