# Simplifying KE/Momentum Relativistic Problem

1. Nov 20, 2011

### Vorde

1. The problem statement, all variables and given/known data

A D0 meson (with a rest mass of 1.86 GeV/c2), initially at rest, decays into a K0 meson (with a rest mass of .51 GeV/c2) and a $\pi$0 meson (with a rest mass of .12 GeV/c2). What is the Kinetic Energy of the $\pi$0 meson?

2. Relevant equations

E=$\gamma$m0c2

p=$\gamma$m0v

3. The attempt at a solution

Subtracting final rest energy from intial energy its easy to see that the KE(K0+$\pi$0)=1.23 GeV

Also: $\gamma$K0mK0c2+$\gamma$$\pi$0m$\pi$0c2 = 1.86 GeV/c^2 (The total energy of the two Mesons must equal the first meson)
Without going through all the horrible steps me and my friend did, eventually we got a equation that solved: VK0=f(V$\pi$0) (I don't have the final equation we got on me)

When then plugged that back into: $\gamma$K0mK0c2+$\gamma$$\pi$0m$\pi$0c2 = 1.86 GeV/c^2

Now at this point we had a solvable equation, the only variable was V$\pi$0, but it was a hellish equation. After about 20 minutes working on trying to solve it we ran out of time and had to give up- the algebra was just to hard.
Had we had enough time, I'm completely confident we could have solved it, as it was simply a matter of foiling again and again (I think we needed to do it three levels down).
My question is: Is there a way of solving this problem without having to go in and solve for one of the variables in terms of the other, or doing that in a simpler equation?

Thank you, especially if you made it through this.

2. Nov 20, 2011

### Staff: Mentor

If you don't need the velocities (i.e. you're not asked for them explicitly), don't take that route to find the energies of the outgoing particles. Instead, use

$$E^2 = (pc)^2 + (m_0 c^2)^2$$

which is true for each particle, together with conservation of energy and conservation of momentum. You never have to see a single "v".

3. Nov 20, 2011

### vela

Staff Emeritus
Two suggestions:

1. Stick with energy (total, not kinetic) and momentum.
2. Take advantage of $E^2 - (pc)^2 = (mc^2)^2$.

If you learn how to use four-vectors, you can get the answer in about three or four lines of simple algebra. Read Griffith's chapter on special relativity in his Introduction to Particle Physics.

4. Nov 20, 2011

### Vorde

Thank you to both of you, while I haven't actually gotten the answer yet, I have gotten numerical values for both the momentum and velocity of one of the particles, I'll be able to get the answer with about 5 more minutes of work.

In addition, I know the momentum energy four vector, but I've never been given a problem requiring it so using it didn't occur to me, I am going to look at its usefulness more carefully.