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: Modern Physics Problem

  1. Feb 3, 2005 #1
    The neutral pion is an elementary particle of the meson family that has a rest mass energy of 135 MeV. This particle is unstable and decays into two photons ("light particles" of no rest mass and energy E=pc). Consider now the following situation: a neutral pion has a kinetic energy of 270 MeV as measured in a given frame. Find the momentum, the direction of propagation and the energy of each of the two emitted photons, given that one of the photons is emitted in a direction perpendicular to the initial velocity of the neutral pion.

    My solution: by conservation of energy Kpion+Mpion*c^2=(p1+p2)c^2

    by conservation of momentum:

    x-direction: Ppion=p1cos(theta1) + p2cos(theta2)=p2cos(theta2)
    y-direction: 0=p1-p2sin(theta2)

    For the pion we have Ppion*c^2=(E^2-mc^2)^0.5=4.498E-15*c^2

    My problem now is solving for theta2. theta2=arctan(P2y/P2x)
    I already know P2x=Ppion, but I can't get P2y.

  2. jcsd
  3. Feb 3, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    You have a system of 3 equations with 3 unknowns
    [tex] p_{2}\cos \vartheta=P [/tex] (1)

    [tex]p_{2}\sin\theta=p_{1} [/tex] (2)

    [tex] p_{1}+p_{2}=\frac{K}{c}+Mc [/tex] (3)

    ,where M is the rest mass of the neutral pion,K is the KE of the pion & P is the pion's (relativistic) momentum...

    Solve the system & find the 3 unknowns...

  4. Feb 4, 2005 #3
    Hi Daniel I am getting the following,

    from (1): p2=P/cos(theta2)
    from (2): p1=p2sin(theta2)

    substitution in (3) yields p2sin(theta2)+P/cos(theta2)=K/c+Mc
    and after a few manipulations I get tan(theta2)+
    sec(theta2)=(K/c+Mc)/P I am not sure how to solve
    for theta2.
  5. Feb 4, 2005 #4


    User Avatar
    Homework Helper

    Hint : take equation (1) squared plus equation (2) squared.
    Last edited: Feb 4, 2005
  6. Feb 4, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    That doesn't work...The "theta's" are different... :tongue2: Cf. [itex] \theta [/itex] to [itex] \vartheta [/itex].

    It was a joke...They're the same.

    Good advice... :smile:

  7. Feb 4, 2005 #6
  8. Feb 4, 2005 #7


    User Avatar
    Homework Helper

    Were the "different" thetas a typo ? :wink:

    Anyway, let me just work this out as I would do it.

    Using Daniel's notation,

    Square (1) and (2) and add :

    [tex]p^2_2 = p^2_1 + P^2[/tex]

    [tex]p^2_2 - p^2_1 = P^2[/tex]

    [tex](p_2 - p_1)(p_2 + p_1) = P^2[/tex] ---(4)

    Put equation (3) into (4) and rearrange,

    [tex]p_2 - p_1 = \frac{P^2}{\frac{K}{c} + Mc}[/tex] ---eqn(5)

    Take (3) + (5) :

    [tex]2p_2 = \frac{P^2}{\frac{K}{c} + Mc} + \frac{K}{c} + Mc[/tex]

    and you can find [itex]p_2[/itex] and then [itex]p_1[/itex]

    How to get [itex]P^2[/itex] in terms of what's given in the question ? For that, I would use [tex]E^2 = P^2c^2 + m^2c^4[/tex]

    where [tex]E = K + mc^2[/tex]. You're given [itex]K[/itex] and the rest energy.

    I'll leave the orig. poster to do the final simplifications.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook