1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Relativistic Particle Decay

  1. Oct 31, 2013 #1
    1. The problem statement, all variables and given/known data
    A K_0 particle has a mass of 497.7 MeV/c^2. It decays into a -∏ and +∏, each having a mass of 139.6 MeV/c^2. Following the decay of the K_0, one of the pions is at rest in the laboratory. Determine the kinetic energy of the other pion after the decay and of the K_0 prior to the decay

    2. Relevant equations
    E = γmc^2 = mc^2 + K

    3. The attempt at a solution
    This is the exact wording of the problem from my text book. I feel like in a chapter meant to be an introduction to special relativity they would specify whether we are talking about rest mass's or relativistic masses. The semi-general consensus amongst my peers is to do the problem as if all mass's given are rest masses.

    Kinetic energy of K_0 particle before decay:
    (γ-1)mc^2 = Kinetic Energy
    and the total energy is γmc^2

    Since the total energy before the decay is γmc^2, and energy has to be conserved in all reference frames, in the reference frame of the lab where one of the ∏ particles is at rest
    I think I can put
    Rest mass energy + kinetic energy of K_0 particle = rest mass energy of both ∏ particles + the kinetic energy of one of them.

    So now I have two equations and three unknowns, being γ and the two kinetic energies...

    This may not be the most clearest thing I've ever typed but special relativity isn't exactly clear in my head either X_x.
  2. jcsd
  3. Oct 31, 2013 #2


    User Avatar
    2017 Award

    Staff: Mentor

    Mass is always "rest mass". Forget "relativistic mass". Scientists got rid of that concept some decades ago.

    Momentum conservation gives you the missing equation. I would not use three unknowns, it is sufficient to consider γ (or v, or p, does not matter) for the moving pion and the kaon.
  4. Oct 31, 2013 #3
    i'm still a bit confused, is the point your trying to make that I only need two equations to solve for the two kinetic energies and gamma because of how gamma and kinetic energy are naturally connected or something?
  5. Nov 1, 2013 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Yep, the only thing you need (on this level of discussion) are energy-momentum conservation and the on-shell conditions for the particles, i.e.,
    [tex]E_1=E_1'+E_2', \quad \vec{p}_1=\vec{p}_1'+\vec{p}_2', E_1^2=m_{\mathrm{K}}^2 c^4 + c^2 \vec{p}_1^2, \quad E_j'{}^2=m_{\pi}^2 c^4 + c^2 \vec{p}_j'^2.[/tex]
    Here [itex](E_1/c,\vec{p}_1[/itex] denotes the four-momentum of the Kaon and [itex]E_j'/c,\vec{p}_j'[/itex] for [itex]j \in \{1,2 \}[/itex] are the four momenta of the pions, and indeed relativistic mass is only a confusing idea from the very early times of relativity. This was abandoned in 1908 by Minkowski in analyzing the mathematical structure of special relativity. Unfortunately the old concepts still survived some (even otherwise very good) textbooks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted