1. Not finding help here? Sign up for a free 30min 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!

Two-photon collision process

  1. Apr 17, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=58027&stc=1&d=1366394833.png


    2. Relevant equations

    Zero-momentum frame is defined as the frame in which total momentum is zero.


    3. The attempt at a solution

    (a) In terms of the zero momentum variables, (e1 and e1 are to distinguish the two resultant particles)

    [itex] \vec{p'}_{HE} + \vec{p'}_{CBR} = \vec{p'}_{e1} + \vec{p'}_{e2} = 0 [/itex]

    This says that the photons have equal and opposite momenta in the zero-momentum frame, and so do the resultant particles. So the energy of both photons is equal, say to E'.

    [itex] E'_{HE} + E'_{CBR} = 2E' = E_{e1} + E_{e2} = 2m_0c^2 [/itex]

    I'm not sure if this right...I think this follows from the fact that we are dealing with the minimum photon energy situation, so the resultant particles would be stationary in the zero-momentum frame. Or is this the equation for the other frame in the question? (With the energies replaced)

    (b) Based on the information given,

    [itex] E_{HE} = \frac{1}{\Gamma} E'_{HE} [/itex]

    And similar for the CMB photons.

    I'm not sure how to write energy and momentum for the resultant particles in this frame. I'm guessing I have to substitute the above photon energy into the answer of part (a). But then I still don't know the energies/momenta of the resultant particles.

    Feeling very confused now! Any help would be appreciated.
     

    Attached Files:

    Last edited by a moderator: Apr 19, 2013
  2. jcsd
  3. Apr 17, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    (a) is right.
    For (b), consider that one energy will increase, one will decrease if you switch to the frame of the galaxy.
    You can assume that they have the same Γ as their Γ-factor.
     
  4. Apr 17, 2013 #3
    Do I need the explicit form of the doppler factor? The only thing I can think of is that the particles are moving in opposite directions maybe. Also, If (a) is correct, is the energy conservation equation sought in (b) given by:

    [itex]E_{HE}+E_{CMB} = 2\Gamma m_0 c^2[/itex]

    ?
     
  5. Apr 17, 2013 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    They move in opposite directions in the frame of the galaxy, right.
    Your last equation looks good, but you will need one more to calculate ##\Gamma##.
     
  6. Apr 17, 2013 #5
    So would the energies of the resultant particles be

    [itex]\Gamma m_o c^2[/itex]

    and

    [itex]\frac{1}{\Gamma} m_0 c^2[/itex]

    ?

    If I have the energies in the galaxy frame, I'm not sure where exactly to put them.
     
  7. Apr 18, 2013 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The energies of the electron and positron are ##\Gamma m c^2##, they have no relative velocity (at the threshold), and their rest frame is moving such that we have a factor of ##\Gamma##.
     
  8. Apr 18, 2013 #7
    Ok so an equation for the electrons in the galaxy frame would be

    [itex]\Gamma m_0 c^2 = \sqrt{(m_0 c^2)^2 + (p_ec)^2}[/itex]

    I still dont understand by what you mean by saying that "one energy will decrease while the other will increase". Does this refer to the photon energies?
     
  9. Apr 18, 2013 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.

    By the way, m0=m. The index 0 is not used any more in physics, masses are always "rest masses".
     
  10. Apr 18, 2013 #9
    So is

    [itex]E_{CBR} = \frac{1}{\Gamma}E'_{CBR}[/itex]

    and

    [itex]E_{HE} = \Gamma E'_{HE}[/itex]

    ?
     
  11. Apr 19, 2013 #10

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
     
  12. Apr 19, 2013 #11
    So the identity they want can proved as

    [itex]E_{HE}E_{CMB} = \frac{\Gamma}{\Gamma} E'_{HE}E'_{CMB} = (mc^2)^2[/itex]

    So I have got all the equation except the momentum conservation one in the galaxy frame. I guess the photon momenta can be obtained by

    [itex]p_{HE} = \frac{E_{HE}}{c}[/itex]

    But what about the electron momenta?
     
  13. Apr 19, 2013 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Eβ=pc
    I don't think you need the electron momenta in the galaxy frame.
     
  14. Apr 19, 2013 #13
    Indeed, its not needed to prove the expression, but the question asked for the two conservation expression in both frames.

    By β are you referring to v/c? I was thinking of using the "mass shell" formula.
     
  15. Apr 19, 2013 #14

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
    That is equivalent to my equation.
     
  16. Apr 19, 2013 #15
    I don't see how...assuming the expression is true

    [itex]E = \frac{pc}{\beta} = \sqrt{(mc^2)^2 + (pc)^2}[/itex]

    I haven't really seen this anywhere.

    Anyway, are the products moving in the same direction in the galaxy frame? (as neither have any momentum in the zero-momentum frame)
     
  17. Apr 20, 2013 #16

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    They are.

    Using ##p=\Gamma m v##, $$E^2
    = m^2c^4 + \frac{m^2v^2c^2}{1-\frac{v^2}{c^2}}
    = \frac{m^2c^4(1-\beta^2)}{1-\beta^2} + \frac{m^2\beta^2c^4}{1-\beta^2}
    = \frac{m^2c^4}{1-\beta^2}$$
    Therefore,
    $$E^2-E^2\beta^2=m^2c^4$$
    or
    $$E^2-m^2c^4=E^2\beta^2$$
    The left side is just ##p^2c^2##. Taking the square root on both sides, we get Eβ=pc. Now I used that relation twice... there is probably some easier way to get the same result.
     
  18. Apr 20, 2013 #17
    All clear! I understand better now, and I will keep that expression in mind.

    Thanks mfb!
     
  19. Apr 20, 2013 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    One thing to note is that the Doppler factor ##\Gamma## is not the same as the relativistic gamma factor ##\gamma = 1/\sqrt{1-\beta^2}##. So, the momentum of an electron would be written ##p = \gamma mv## rather than ##p = \Gamma mv##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Two-photon collision process
  1. Collision of two Balls (Replies: 1)

Loading...