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Relativistic decay problem

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi everyone, I'm having a bit of trouble with solving this problem:
    A ∏0 meson with rest mass m has a kinetic energy K. It decays in flight into two photons whose paths are along the direction of motion of the meson. Find the energies of the two photons.

    2. Relevant equations
    E2=p2c2+m2c4,
    which for a photon reduces to
    E=pc

    3. The attempt at a solution
    I am using 4-vectors to solve this problem.
    Before the decay, the pion has the four momenta P=(E/c,p,0,0). After we have two photons with four momenta P1=(E1/c,p1,0,0) and P2=(E2/c,p2,0,0).

    By conservation of momentum and energy, E1+E2=E and p1+p2=p.

    I would like to take the following approach - equate the four momenta and square both sides, using the fact that the quantity E2/c2-p2 is invariant.

    P=P1+P2. Squaring and using the relevant equation above gives m2c2=E1E2/c2-p1p2.

    But Using the fact that E1=p1c and E2=p2c reduces the right hand side to zero, which is very wrong... I believe my problem may be with the signs of the momenta - from the centre of mass frame it is clear the photons have to travel in opposite directions. But I don't expect that I would have to account for this in my working - as for example when solving collision problems in classical mechanics using conservation of momentum, I would put all my unknown velocities in as symbols, and the maths would simply give me their direction (by the sign). Also note I can solve this is the centre of mass frame and then transform the results into the desired frame, but the fact that I can't do it directly in this frame must be a sign of misunderstanding. So can anybody help please, thankyou :)
     
    Last edited: Dec 29, 2013
  2. jcsd
  3. Dec 29, 2013 #2

    PeroK

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    Are you sure you haven't forgotten the pion's kinetic energy?
     
  4. Dec 29, 2013 #3
    P.P=(E/c,p,0,0).(E/c,p,0,0)=E2/c2-p2=m2c2 is how I got to that...
     
  5. Dec 29, 2013 #4

    PeroK

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    You've got the energy-momentum equation in your post. Rearranged it should be:

    [itex]p^2 = \frac{E^2}{c^2} - m^2c^2[/itex]

    If you were doing this in the pion frame, p would be 0 here. But not in the lab frame. The E here is not just mc^2 but includes the kinetic energy as well.
     
  6. Dec 29, 2013 #5
    I understand this, but I am not sure how this means my working is incorrect. I have just taken the 4-momentum inner product of each side with itself. I believe that if I do this for a single particle it should just return the mass squared x c squared, regardless of whether the particle has momentum or not, or kinetic energy or not.
     
  7. Dec 29, 2013 #6

    TSny

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    I think that's correct. [Edit: On second thought, this equation needs a 2 placed in it somewhere.]

    Yes, you have to be careful with the signs here. Note that your p's are actually x-components of momentum px, which could have a positive or negative value. But E is always positive.

    If you write E1=p1xc, then you are forcing p1x to be positive since the left side is positive.
     
    Last edited: Dec 29, 2013
  8. Dec 29, 2013 #7

    PeroK

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    It looks like you've got an answer.
     
  9. Dec 29, 2013 #8
    I was thinking that - the relation E^2=p^2c^2-m^2c^4 uses the magnitude of the momentum vector (is that right?).

    The only way around this I can see then is to call what I had as p2, -p2, as I know one of the momenta has to be negative anyway, and everything should work out. However it doesn't seem like a very nice method - I'm having to play around with the symbols to make the maths work. Is there a better way to do it?
     
  10. Dec 29, 2013 #9

    TSny

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    Off hand, I don't see a better way. Putting the negative sign in "by hand" doesn't seem bad to me. But someone else might chime in.

    By the way, I believe that your equation m2c2=E1E2/c2-p1p2 is not quite correct after all. See if you might have dropped one or more factors of 2.
     
  11. Dec 29, 2013 #10

    Ah yeah, forgot the factor of two from the square on the RHS when typing it up. Thanks for your help.

    If anybody has a more elegant method please let me know - otherwise it seems the CM frame is the way to go.
     
  12. Dec 30, 2013 #11

    PeroK

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    I assume you're trying to get the energy of the photons in terms of the mass and KE of the pion. If γ1 is the photon in the direction of the original motion and γ2 the other then (using unsigned momenta):

    [itex]p_1 - p_2 = p \ ∴ \ E_1 - E_2 = pc[/itex]

    [itex]E_1 + E_2 = E[/itex]

    So:

    [itex]E_1 = \frac12 (E + pc) \ and \ E_2 = \frac12 (E - pc)[/itex]

    And:

    [itex]p^2c^2 = E^2 - m^2c^4 = m^2c^4 + 2Kmc^2 + K^2 -m^2c^4 = 2Kmc^2 + K^2 [/itex]

    So:

    [itex]E_{1/2} = \frac12 (mc^2 + K \pm \sqrt{2Kmc^2 + K^2} )[/itex]
     
  13. Dec 30, 2013 #12
    I think this is the method I had already mentioned, i.e calling the second momentum -p2 as opposed to p2 so that the calculations worked out, unless I'm misunderstanding something...

    What do mean by 'using unsigned momenta'?
     
  14. Dec 30, 2013 #13

    PeroK

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    I meant take the positive value, so that p_1 and p_2 > 0. Rather than p_1 being positive and p_2 being negative.
     
  15. Dec 30, 2013 #14
    Does anybody have any ideas on dealing with things like this:

    I have a velocity in one frame which is very close to c, v, which is the CM frame, and I want to transform it into a velocity in a frame in which one of the two particles is at rest. So this would be given by u=2v/(1+v^2/c^2). But that just comes out as c on my calculator (when I need the exact number obviously), which is very annoying...
     
    Last edited: Dec 30, 2013
  16. Dec 30, 2013 #15

    vela

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    Right. As you noted, it comes from squaring the four-vector ##p^\mu = (E, \vec{p})##. When you do that, you get
    $$m^2 = E^2 - \vec{p}\cdot\vec{p} = E^2 - \lvert \vec{p} \rvert^2.$$
    As you noted, in the rest frame of the pion, the photons go off in opposite directions. There's no Lorentz boost that can change the direction of the photon from going left to going right or vice versa. You'd have to be able to outrun one of the photons to do that — that is, go to a frame that's moving faster than ##c##. So in the lab frame, you know that one photon went left and one went right.

    From a math standpoint, you actually don't have ##E = pc##; you have ##E^2 = (\vec{p}c)^2## from which you can get the two solutions ##E = \lvert \pm \vec{p} c \rvert##. So you can look at it as if you're simply throwing out the solution that doesn't work rather than arbitrarily putting in a negative sign to make it work.

    Usually, it's easiest to simply use ##p## to denote the magnitude of the momentum and then put in the signs as appropriate based on physical reasoning. For this problem, you could have
    \begin{align*}
    p_\pi &= (E, p) \\
    p_1 &= (E_1, p_1) \\
    p_2 &= (E_2, -p_2).
    \end{align*} Conservation of energy and momentum gives you
    $$p_\pi = p_1 + p_2,$$ and when you square it, you get
    $$m_\pi^2 = 2[E_1 E_2 - p_1(- p_2)] = 2(E_1 E_2 + p_1 p_2).$$ You have the plus sign on the RHS like you want.
     
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