How can I calculate the speed of the kaon using four-momentum conservation?

[Shiz]

Homework Statement

So a kaon moving at some speed in the +x direction spontaneously decays into one pion and one anti-pion. The anti-pion moves away with velocity of 0.8c, and the pion moves away with velocity of 0.9c.

Mass of kaon = 498 MeV/c^2
Mass of pion/anti-pion = 140 MeV/c^2

Homework Equations

I understand that momentum of the kaon qualms the momentums of the two pions. p(kaon) = p(pion) + p(anti-pion). I can then square both sides and use the principle of invariance.

The Attempt at a Solution

What I'm having issues with is how to calculate the term of +2 p(pion)•p(anti-pion).

I don't understand how to multiply the vector parts. p(pion) = ( E(pion)/c , vector p(pion) )
p(anti-pion) = ( E(anti-pion)/c , vector p(anti-pion) )

I'll get a cos(theta) term out of this dot product on the vector side but how do I use the velocities I have to get a dot product of those two momentum vectors? They're both moving so I am guessing I have to substitute in (gamma)(mass of the particle)(velocity vector) but I just don't understand how to do the math after that.

Help please!

 

[vanhees71]

What's the question? I guess you should get the original speed of the kaon. Let's also set $c=1$, which makes everything a lot easier. Now we measure masses, energies, and momenta in the same unit, MeV.

You have energy-momentum conservation
$$p=p_1'+p_2',$$
where $p$ is the four-momentum of the kaon in the initial state, and $p_1'$, $p_2'$ are the four momenta of the pions in the final state.

Further you have the on-shell conditions
$$p^2=m_{\text{K}}^2, \quad p_1'^2=p_2'^2=m_{\pi}^2.$$
The relation between three-velocity and energy and momentum is
$$\vec{v}=\frac{\vec{p}}{E},$$
for all the particles.

Now you can calculate the momenta of the pions and then use the above properties of the four-vectors. From this you should be able to get the three-momentum of the kaon and then its speed.