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Relativistic collision

  1. Nov 2, 2006 #1
    At LEP, electrons and positrons, each of energy E= 45.6 GeV (I guess this is total since it's a particle physics course), are collided head-on, and have exactly the right energy to produce Zo particles at rest.
    [tex]e^+ e^- = Z_0 [/tex]

    This is simple but I'm having a bit of trouble with collisions.

    The particles initially have total energy [tex]2E= 2(\gamma)m_ec^2 [/tex], since Z will be be created at rest all of this energy will go into the creation of mass of Z.

    To what energy would a positron have to be accelerated if in collision with a stationary atomic electron it were to produce a Z0?

    The kinetic energy [tex] T = (\gamma - 1) m_ec^2 [/tex]
    [tex]2(\gamma)m_ec^2 = m_Zc^2 [/tex]

    [tex]2E = (\gamma - 1) m_ec^2 + 2m_ec^2 [/tex]
    [tex]2E = T + 2m_ec^2 [/tex]
    [tex]T = 2E - 2m_ec^2 [/tex]
    = 2(45.6 eV) + 2(0.511) = 92.2 ev

    I'm not really sure if my thinking is right here, I've basically ignored Z.
     
    Last edited: Nov 2, 2006
  2. jcsd
  3. Nov 2, 2006 #2
    Ok, just realised I totally ignored conservation of momentum I'll have another go.
     
  4. Nov 2, 2006 #3
    [tex](2E)^2 = E_t^2 - p^2c^2 , E_t = E_1 + m_ec^2 [/tex]
    [tex] p = \sqrt((E_1^2 / c^2) - m_e^2c^2 ) [/tex]
    [tex] 4E^2 = 2E_1m_e c^2 [/tex]
    [tex]=> E_1 = (4E^2 ) / (2m_e c^2 ) [/tex]
    E_1 = energy of incident positron.
    which would give me a huge energy. Any ideas?
     
  5. Nov 3, 2006 #4

    OlderDan

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    I haven't worked through the calculation, but surely you expect a lot more energy in the second case because the Zo has to fly off with all the momentum of the incident positron. Are you unsure of your calculation?
     
    Last edited: Nov 3, 2006
  6. Nov 3, 2006 #5
    I'd expect more energy but this increases at roughly the square of the initial energy I guess that's why they smash particles into each other. I wasn't reall sure of the step I used was correct it seems I may have oversimplified somewhere.
     
  7. Nov 3, 2006 #6

    OlderDan

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    I'll take a look at it soon and compare results. Of course if somebody else wants to do it.....
     
  8. Nov 3, 2006 #7

    OlderDan

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    I think you left out the rest energy of the target electron, but that is of little consequence. I got

    [tex] E_1 = (4E^2 ) / (2m_e c^2 ) - m_e c^2 [/tex]

    Your conclusion about why they smash the particles together keeping the CM at rest is well founded. You might want to go a step further and find the velocity of the Zo that is created in this scenario.
     
    Last edited: Nov 3, 2006
  9. Nov 3, 2006 #8
    Could you write down the steps you did, if it's not too long. I can't find the [tex]m_ec^2[/tex] and although it clearly makes zero difference my Prof would mark it wrong.
     
  10. Nov 3, 2006 #9

    OlderDan

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    Just add the rest energy of the electron to the energy of the positron in the energy equality.

    [tex] E = \sqrt {p^2 c^2 + m^2 c^4 } = 45.6{\rm{ GeV}} [/tex]

    [tex] E_{Zo} = Mc^2 = 2E = 2\sqrt {p^2 c^2 + m^2 c^4 } = 91.2{\rm{ GeV}} [/tex]

    [tex] E_Z = \sqrt {P^2 c^2 + M^2 c^4 } = \sqrt {p_1 ^2 c^2 + M^2 c^4 } = E_1 + mc^2 [/tex] energy and momentum conservation [tex]\left( {P = p_1 } \right) [/tex]

    [tex] E_Z ^2 = p_1 ^2 c^2 + M^2 c^4 = \left( {E_1 + mc^2 } \right)^2 = E_1 ^2 + 2E_1 mc^2 + m^2 c^4 [/tex]

    [tex] p_1 ^2 c^2 + \left( {2E} \right)^2 = p_1 ^2 c^2 + m^2 c^4 + 2E_1 mc^2 + m^2 c^4 [/tex]

    [tex] 4E^2 = 2m^2 c^4 + 2E_1 mc^2 [/tex]

    [tex] E_1 = \frac{{2E^2 }}{{mc^2 }} - mc^2 [/tex]

    I had used different subscripts and changed them all to match yours. I think I got them all, but check up on me.
     
    Last edited: Nov 3, 2006
  11. Nov 3, 2006 #10
    Yes I see what I missed thanks. Trying to get used to 4-momentas, I think this is right would be nice if somoene could check it for me.

    Q. Calculate the minimum antineutrino energy required for the following reaction to occur when an antineutrino hits a stationary proton, and creates a neutron and electron.

    The stationary proton has 4-momentum [tex]p_p = (m_p, 0, 0 ,0)[/tex]
    The antineutrino has 4-momentum [tex]p_2 = (E_{an}, E_{an}, 0, 0)[/tex]. At threshold the final frame will have 4-momentum [tex]p' = (m_e + m_n. 0)[/tex]

    [tex]p' = p_1 + p_2[/tex]
    [tex](m_e + m_n)^2 = (m_p + E_{an})^2 [/tex]
    [tex] m_e^2 + 2m_em_n + m_n^2 = E_{an}^2 + m_p^2 + 2E_{an}m_p [/tex]
    [tex] => E_{an}^2 + 2E_{an}m_p + m_p^2 - (m_e^2 + 2m_em_n + m_n^2 ) [/tex]
    E_an = 1.8MeV
    Trying to get used to 4-momentas, the answer just doesn't look correct though. edit: looked on google and it says typical neutrino energies are 1--10GeV
     
    Last edited: Nov 3, 2006
  12. Nov 4, 2006 #11

    OlderDan

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    I think I see a couple of problems here. I don't think you can lump the masses of the electron and neutron together as if they were a single particle, and I see no way the final momentum can only have an energy component. Also, is that x component of the antineutrino momentum the energy? Is that a typo?
     
  13. Nov 4, 2006 #12
    Was trying to leave it without the c's, am I right to assume the antineutrino momentum would be E_an / c as if it were a photon?
     
  14. Nov 4, 2006 #13
    I'm guess I'm confused as to what expression too use instead of [tex] E = \sqrt {p^2 c^2 + m^2 c^4 } [/tex] for a neutrino.
     
  15. Nov 4, 2006 #14

    OlderDan

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    Doh!!! I forgot that neutrinos are massless (well almost). I'm not a fan of leaving out c (i.e., c = 1), but if you can keep it straight without them, more power to you. I still think there has to be something to include the final 3-momenta terms. I'll look at this a bit more.
     
  16. Nov 5, 2006 #15

    OlderDan

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    That is the right expression for everything. The calculation you did earlier is good enough in this case, but you really made it more complicated than it is.

    See the annotations in blue in the quote above.

    I went round and round the barn trying to incorporate the final momentum into the problem. I finally decided there were only two choices. One is the equations I got with the proton at rest are transcendental equations that need to be solved graphically. The second is you could do the problem in the center of momentum frame so there is no final 3-momentum, and then transform back to the lab frame. Using the first approach I found

    E_an = 1.806054MeV

    compared to the result you get when ignoring the final 3-momentum

    E_an = 1.804319Mev

    So I was right about needing the final momentum.:rolleyes: It makes a difference of about 0.1%. Seems I have little intuition for the outcome of these events.:yuck:

    I have not done the second appraoach, but I probably should at some point.
     
    Last edited: Nov 5, 2006
  17. Nov 5, 2006 #16
    I think this works.

    [tex] E_t^2 - |p_t|^2c^2 = E_0^2; E_t = E_an + m_pc^2[/tex]
    [tex] (E_{an}^2 + 2E_an m_ec^2 +m_p^2c^4) - |E_{an} / c|^2c^2 = (m_e + m_n)^2c^4 [/tex]
    [tex] 2E_{an} m_pc^2 + m_p^2c^4 = (m_e^2 + 2m_em_n + m_n^2)c^4 [/tex]
    [tex] => E_{an} = (\frac{m_e^2 + 2m_em_n + m_n^2 - m_p^2}{2m_p})c^2 [/tex]

    E_an = 1.8 MeV; Particle masses: electron 0.511 MeV/c2, proton 938.3 MeV/c2, neutron 939.6 MeV/c2.
     
    Last edited: Nov 5, 2006
  18. Nov 5, 2006 #17

    OlderDan

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    Yeah, that's it exactly. I was hanging on to the solution for the breakdown of the individual final momenta and energies of the neutron and electron, but you don't need to do that to find E_an. The total final momentum and mass works just fine if you are willing to give up on the breakdown. You do still need to go out several decimal places to see the difference between your latest calculation and the one you did earlier. Your way gives an analytical result that exactly matches the result I got with considerably more effort.
     
    Last edited: Nov 5, 2006
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