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Homework Help: Relativity: Conversion of Mass and Energy

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A lambda particle decays into a proton and a pion, and it is observed that the proton is at left at rest. A. What is the energy of the pion? B. What was the energy of the original lambda? (The masses involved are m[tex]_{}\lambda[/tex] = 1116, m[tex]_{}p[/tex] = 938, and m[tex]_{}\pi[/tex] = 140, all in MeV/c^2

    2. Relevant equations
    E=[tex]\gamma[/tex](m)(c^2) P=[tex]\gamma[/tex](m)(u) E^2=(pc)^2+(mc^2)^2
    When u=0, E=mc^2

    3. The attempt at a solution
    I realize that this is largely a conservation of momentum and energy problem. The proton is at rest, so it has zero momentum. This also means that P[tex]\lambda[/tex]=P[tex]\pi[/tex]


    I have plugged in and tried to solve it down along with the momentum, but no matter what I do I can't do the algrebra. Any help is greatly appreciated.

    And the answers are 184 MeV and 1122 MeV, respectively
  2. jcsd
  3. Feb 1, 2010 #2


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    I always find it more straightforward to work with space-time vectors.

    (I'm using [tex] (ct;\mathbf{r})[/tex] coordinates.) Here are the four momenta of the three particles (multiplied by c to yield units of energy):
    [tex]P_\lambda = (E_\lambda,cp_\lambda) = \gamma_\lambda m_\lambda(c^2,c v_\lambda)[/tex]
    [tex]P_p = (E_p,0)=(m_pc^2,0)[/tex]
    [tex]P_\pi = (E_\pi,cp_\pi) = \gamma_\pi m_\pi(c^2,c v_\pi)[/tex]

    Now you can do conservation of momentum and energy in the same equation:
    [tex] P_\lambda = P_p + P_\pi[/tex]

    Note that the lambda particle and pion will not have the same velocities so they will not have the same gamma factors. It is better to work with the relativistic energy and momenta than with velocities directly.

    Also note that we have auxiliary equations if we need them:
    [tex] P_\lambda \cdot P_\lambda = m^2_\lambda c^4[/tex]
    [tex] P_\pi \cdot P_\pi = m^2_\pi c^4[/tex]
    (with the dot product being the SR metric [tex](a,b)\cdot(a,b)=a^2 - b^2[/tex]).
    (also with the proton but that's at rest so the mass equation is trivial.)

    Try working the problem from there and seeing if you can solve it.
  4. Feb 1, 2010 #3
    Thank you very much, I believe I was able to figure it out.
  5. Oct 3, 2010 #4
    I am working on the same problem and I can't seem to figure it out. I'm not familiar with Jambuagh's notation and I want to believe there is not enough information to solve the first part of the problem. You could solve for the E(pi) if you knew the momentum for the pion, but it is not given. If you could elaborate more on the math, I would appreciate greatly.
  6. Oct 3, 2010 #5
    I used the pythagorean relation, and it does indeed become quite troublesome. Ill try my best to keep this as clean as possible. With the assumptions that William has made (which to my novice knowledge is correct), the conservation of Energy states E(lamb) = E(pro) + E(pion). Utilizing the assumptions, we have the sqrt[((p(lamb)c)^2) + (m(lamb)c^2)^2]. Equate that to m(proton)times c^2 (proton is at rest) + sqrt(everything in the above sqrt but with the values corresponding to the pion). The trick here is to square both sides. This kills off the square root on the left side, leaving you to foil the right side. This should get you something like m(proton)c^2 whole term squared + 2m(proton)c^2 times the square root that was there on the right before foiling + (p(pion)c)^2 + (mpionc^2)^2. The two (p(pion)c)^2 cancel eachother out since their momentums are equivalent (Conservation of momentum). The rest should be clear from there and you end up being able to solve for the momentum of the pion, which can be thrown into its pythag relation since the term (mc^2)^2 is basically given.

    I did this problem a long time ago, but above is the post from my class forum which helped me solve it. We (assuming you are taking the same class as I did haha) don't learn what jambaugh posted, and everyone on this forum will post that unfortunately. You will just have to mess with the algrebra until it gets solved. Good luck!
  7. Oct 4, 2010 #6
    i just figured out exactly what you explained today. thanks for responding!
  8. Oct 4, 2010 #7


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    It's worth learning to use four-vectors. Using four-vectors, the answer pops out in literally two or three lines of algebra with nary a square root in sight. As jambaugh noted, conservation of energy and momentum gives you

    [tex]P_\Lambda = P_\pi+P_p[/tex]

    If you rearrange this slightly and square it, you get

    P_\Lambda-P_p &= P_\pi \\
    (P_\Lambda-P_p)^2 &= P_\pi^2 \\
    P_\Lambda^2 - 2P_\Lambda\cdot P_p + P_p^2 &= P_\pi^2

    where, for example, with c=1 and [itex]p_\pi[/itex] representing the 3-momentum of the pion,

    [tex]P_\pi^2 = (E_\pi, p_\pi)\cdot(E_\pi, p_\pi) = E_\pi^2-p_\pi^2 = m_\pi^2[/tex]

    When you work out the [itex]P_\Lambda\cdot P_p[/itex] term, you should see why I rearranged the original equation before squaring to make it easy to solve for the energy of the lambda particle.
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