# Homework Help: Issue calculating the velocity of pions from K-meson decay

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1. Mar 29, 2017

### mountevans

1. The problem statement, all variables and given/known data
We were supplied with several images of hydrogen bubble chamber tracks, from which we measured the length of the (invisible) K0 track, angles of departure from the K0 meson's track and curvature. As a way to limit the amount of data we're talking about, I'll only discuss the first particle. The length of the K0's track was 0.016 m, the radius of curvature of the π+ was 0.65 m, the π- 0.50 m, the angles of departure (π+-) were 44 and 24 degrees respectively, and the magnetic field was 1.4 Teslas.

The momentum of the positive k meson tracks is given as 535 MeV/c. Rest mass of positive and negative pions is 139.57 MeV/c2. c = 3*108 m/s, of course.

2. Relevant equations
p = q*r*B
T = 1/2 mv2
γ = 1/√(1-v2/c2)
p [in MeV/c] = 3 ∗ 102 ∗ r [in m] ∗ B [in T]
F = mv2/r

3. The attempt at a solution
Using the formula for p above, 3 ∗ 102 * (0.65 m) * (1.4 T), I get 382.6 MeV/c for the π+ and 294.3 MeV/c for the π- total momenta.

However, I'm now at the step where I think I need to determine the total energy of the pions so I can apply a gamma correction to their speed. When I take the above equations,

F = mv2/r and p = mv = q*r*B

I find that v must equal q*r*B/m, where m is the mass(-energy?) of the pions. So, here appears to lie my mistake: I divided 382.6 MeV/c by 139.57 MeV/c2, and got 2.74, which I would assume is in units of c (which can't be right for v, since nothing exceeds v least of all a pion). So it seems like I'm missing a factor here - should I have converted the momenta in MeV/c to eV/c?

For the later components of this homework, we simply use 2D kinematics to find the momentum vectors of the pions - the sum of the components in-axis with the path of the kaon should equal the momentum of the neutral kaon, and the total momentum in the normal axis should sum to zero.

2. Mar 29, 2017

### vela

Staff Emeritus

Those equations are non-relativistic. You can use the relationship $E^2 = (mc^2)^2 + (pc)^2$ where $E$ is the total energy of the particle and $p$ is the relativistic momentum. Also, your expression for the kinetic energy is should be $T=(\gamma-1)mc^2 = E - mc^2$.

3. Mar 30, 2017

### mountevans

Is $p=γmv$ correct in this instance? (where m = rest mass). Would the rest energy of the pion be 139.57 * (3*108)2 MeV, rather than the mass (in MeV/c2)? When going from mass to energy, if I'm not using c=1, I need to actually multiply by c2, right?

Last edited: Mar 30, 2017
4. Apr 1, 2017

### mountevans

It was simply a problem with converting between MeV and J, if you can believe it. This page was very helpful.