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Equations for the initial and final momentum

  1. Jun 24, 2007 #1
    1. The problem statement, all variables and given/known data

    How does one use the equations for the conservation of momentum and energy (see my last post in this forum)

    [tex] cp' = cp +\hbar(\omega+\omega') [/tex]
    [tex] E' = E +\hbar(\omega-\omega') [/tex]

    to derive the following

    [tex] cp = -(\hbar\omega + \hbar\omega')/2 + sqrt(1+(m^2c^4/(h-bar\omega\omega')) (\hbar\omega - \hbar\omega')/2) [/tex]
    and
    [tex] cp = (\hbar\omega + \hbar\omega')/2 + sqrt(1+(m^2c^4/(h-bar\omega\omega')) (\hbar\omega - \hbar\omega')/2) [/tex]



    2. Relevant equations

    with Einstein's equation [tex] E^2 = (cp)^2 + (mc^2)^2 [/tex]


    3. The attempt at a solution

    I got to a quadratic equation with
    a = 4*omega*omega'
    b = 4*\hbar*(omega*omega')(omega + omega')
    c = 4*\hbar^2*omega^2*omega'^2 - m^2*c^4*(omega - omega')^2 but that does not work. If I work backwards, I get something that is very close to this except without the first term in c. But I double-checked everything and could not find any mistakes.

    By the way, how can I make those h-bar's with TeX syntax?
     
    Last edited: Jun 24, 2007
  2. jcsd
  3. Jun 24, 2007 #2

    malawi_glenn

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    \hbar

    [tex] \hbar [/tex]
     
  4. Jun 29, 2007 #3
    Does my question make sense to people?
     
  5. Jun 29, 2007 #4

    Dick

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    What's not clear to me is what is the difference between E and omega. For a photon for example, E=hbar*omega. So what's with an equation with both E and omega in it?
     
  6. Jun 29, 2007 #5
    That relationship between E and omega is used here to describe the conservation of energy. The conservation of energy equation here can be interpreted as the energy of the particle after the collision is equal to its energy before the collision plus h-bar times the change in frequency of the photon. That equation, with the conservation of momentum and Einstein's relation should allow you to express cp and cp' in terms of the frequencies. The math just doesn't seem to work out for me though :(

    Sorry the second one should be cp'.
     
    Last edited: Jun 29, 2007
  7. Jun 29, 2007 #6

    Dick

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    Can you describe exactly what the collision problem is? If it's a photon then m=0. Since you've got a m in there I assume there is a third particle? See how confused I am?!
     
  8. Jun 29, 2007 #7
    You're right. I changed my last post so it makes sense. The experiment involves a particle (of mass m) with unknown energy (E) and unkown momentum (p). A leftward-moving photon with a known frequency (w) collides with the particle and returns rightward and the photon's new frequency (w') is measured.
     
  9. Jun 30, 2007 #8

    Dick

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    In this case you have four unknown states to work with, the initial state of the particle and the photon and the final state of particle and photon. But I only see one 'p' in your solution. Are you sure the particle isn't initially at rest?
     
  10. Jun 30, 2007 #9
    The initial (w) and final (w') frequencies of the photon are known. You can calculate the initial (p_photon) and final (p_photon') momenta of the photon if you wanted to via the equation cp_photon = hw.
     
  11. Jun 30, 2007 #10

    Dick

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    Your final state cp involves only the energy momenta of the photons. There is no room for an initial state on the massive particle. The question is badly stated.
     
  12. Jul 1, 2007 #11
    I do not understand your last post. Firstly there is only one photon. Secondly, what do you mean "There is no room for an initial state on the massive particle"? The goal is to express the initial and final momentum of the massive particle in terms of the initial and final frequency of the photon.
     
  13. Jul 1, 2007 #12

    Dick

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    The two photons refer to incoming and outgoing. My point is that in your final equation there is only the energy of these and 'p', which I assume to be the final momentum of the massive particle. Where is the initial momentum of the massive particle? You can't determine outgoing without it. Is it zero!?
     
  14. Jul 2, 2007 #13

    Dick

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    Uh, I take that last comment back. Sorry. After sitting down and actually working it out, you can eliminate the initial particle state and final particle energy in favor of the photon energies. And the result is very like what you posted except that I get the +/- on the sqrt term.
     
  15. Jul 3, 2007 #14
    Yes--if it is a quadratic it should be +/- (but I'm not sure how multiple solutions makes sense in the context of this problem). But which a, b, and c did you plug into the quadratic formula? Mine are not correct are they?
     
    Last edited: Jul 3, 2007
  16. Jul 3, 2007 #15

    Dick

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    Ok. Here's what I got:

    a=1
    b=-(gi+gf)
    c=m^2/2+gi*gf-(gi^2+gf^2)/(4*gi*gf)

    where I've put c=1 and gi and gf are the photon energies. Which looks sort of like your results divided by gi*gf. I'm not suprised you are having problems, that's about my sixth try. It's really easy to make a mistake. But I don't think you are doing anything fundamentally wrong. Just algebra.
     
  17. Jul 3, 2007 #16

    Dick

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    BTW the two solutions to the quadratic correspond to reversing the initial and final states.
     
  18. Jul 4, 2007 #17
    Sorry, when I plug in your abc to the quadratic formula I get this for the square root term:


    [tex]\frac{\sqrt{\hbar^2(\omega+\omega')^2-2m^2c^4+4\hbar^2\omega\omega'-\frac{\omega+\omega'}{\omega\omega'}}}{2}[/tex]

    I do not see how this reduces to the square root term in the originial equation:

    [tex]\sqrt{1+\frac{m^2c^4}{\hbar^2\omega\omega'}} \frac{\hbar\omega - \hbar\omega'}{2}[/tex]
     
  19. Jul 4, 2007 #18

    Dick

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    I'm sorry too. Left out a factor of m^2 on the last term of c.

    c=m^2/2+gi*gf-(gi^2+gf^2)*m^2/(4*gi*gf).
     
    Last edited: Jul 4, 2007
  20. Jul 6, 2007 #19
    I see. Thanks.
     
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