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Particles in a tube

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data
    One way to produce a (muon-) neutrino beam is to produce a beam of pions and direct them down an evacuated pipe letting the pions decay to a neutrino (i.e., Muon + neutrino) Assume the neutrino is massless, what is the expression for the energy of the neutrino as a function of the pion’s mass m, energy E, and of the angle of emission of the neutrino in the lab frame (i.e., the angle between the neutrino and pion directions in the lab)?
    Assume that the particles are moving at relativistic speeds

    2. The attempt at a solution
    The attached diagram is not part of hte problem's given. The diagram is my own work. Is it wrong to assume that the pion is going to split in a way analogous to Compton scattering?
    Until the diagram is approved here is the description. assume the right and upward to be positive. The pion moves to the right and splits into the muon which goes off at angle phi and the neutrino which goes off at angle theta.

    the momentum and the energy of the pion is
    [tex] p_{\pi} = \gamma_{\pi} m_{\pi} v_{\pi} [/tex] (1)
    [tex] E_{\pi} = \gamma m_{\pi} c^2 [/tex] (2)

    the momentum in the X and Y directions of the muon
    [tex] p_{x} = \gamma_{\mu} m_{\mu} v_{\mu} \cos\phi [/tex] (3)
    [tex] p_{y} = \gamma_{\mu} m_{\mu} v_{\mu} \sin\phi [/tex] (4)
    Energy is
    [tex] E_{\mu} = \gamma_{\mu} m_{\mu}c^2 [/tex] (5)

    The energy of the neutrino is [itex] E_{\nu} [/itex]
    the momentum in the X and Y directions of the neutrino
    [tex] p_{x} = \frac{E_{\nu}}{c} \cos\theta [/tex] (6)
    [tex] p_{y} = -\frac{E_{\nu}}{c} \sin\theta [/tex] (7)

    so are my equations correct??

    For conservation of momentum in the X direction
    [tex] \gamma_{\pi} m_{\pi} v_{\pi} = \gamma_{\mu} m_{\mu} v_{\mu} \cos\phi + \frac{E_{\nu}}{c} \cos\theta [/tex]

    For conservation of momentum in the Y direction
    [tex] \gamma_{\mu} m_{\mu} v_{\mu} \sin\phi = \frac{E_{\nu}}{c} \sin\theta [/tex]
    [tex] E_{\nu} = \frac{\gamma_{\mu} m_{\mu} v_{\mu}c}{sin\theta} \sin\phi [/tex]

    For conservation of energy
    [tex] \gamma_{\pi} m_{\pi} c^2 = \gamma_{\mu} m_{\mu}c^2 + E_{\nu} [/tex]
    [tex]E_{\nu} = E_{\pi} - \gamma_{\mu} m_{\mu}c^2 [/tex]

    But energy of the muon is [tex] E_{\mu}^2 = p_{\mu}^2 c^2 - m_{\mu}^2 c^4 [/tex]

    [tex] E_{\nu} = E_{\pi}- (p_{\mu}^2 c^2+ m_{\mu}^2 c^4[/tex]

    using conservation of momentum we can get [tex]p_{\mu} = \frac{E_{\nu}\sin\theta}{c\sin\phi}[/tex]

    So i have the Energy of the neturino is terms of the eneryg of hte pion, the emission angle of the neutrino theta and the mass of the muon... How do i get rid of the mass of the muon and the emission angle of the muon... phi? I ve been trying for quite some time please help!

    Thanks for any and all of your help!
     

    Attached Files:

    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 30, 2008 #2
    I posted this separate because the equations above look really complicating when really they arent

    For conservation of momentum in the X direction
    [tex] \gamma_{\pi} m_{\pi} v_{\pi} = \gamma_{\mu} m_{\mu} v_{\mu} \cos\phi + \frac{E_{\nu}}{c} \cos\theta [/tex]

    For conservation of momentum in the Y direction
    [tex] \gamma_{\mu} m_{\mu} v_{\mu} \sin\phi = \frac{E_{\nu}}{c} \sin\theta [/tex]
    [tex] E_{\nu} = \frac{\gamma_{\mu} m_{\mu} v_{\mu}c}{sin\theta} \sin\phi [/tex]

    For conservation of energy
    [tex] \gamma_{\pi} m_{\pi} c^2 = \gamma_{\mu} m_{\mu}c^2 + E_{\nu} [/tex]
    [tex]E_{\nu} = E_{\pi} - \gamma_{\mu} m_{\mu}c^2 [/tex]

    But energy of the muon is [tex] E_{\mu}^2 = p_{\mu}^2 c^2 - m_{\mu}^2 c^4 [/tex]

    [tex] E_{\nu} = E_{\pi}- (p_{\mu}^2 c^2+ m_{\mu}^2 c^4[/tex]

    using conservation of momentum we can get [tex]p_{\mu} = \frac{E_{\nu}\sin\theta}{c\sin\phi}[/tex]

    So i have the Energy of the neturino is terms of the eneryg of hte pion, the emission angle of the neutrino theta and the mass of the muon... How do i get rid of the mass of the muon and the emission angle of the muon... phi? I ve been trying for quite some time please help!
     
  4. Jan 31, 2008 #3
    OK i worked on it a bit more
    here's what i have

    [tex] E_{\nu} = E_{\pi} - E_{\mu} [/tex]
    [tex] E_{\nu} = E_{\pi} - p_{\mu}^2 c^2 + m_{\mu}^2 c^4 [/tex]

    Conservation of energy in the X direction

    [tex] p_{\pi} = p_{\nu}\cos\theta + p_{\mu}\cos\phi[/tex]
    [tex] \frac{E_{\pi}^2}{c^2} - m_{\pi}^2 c^2 = \frac{E_{nu}}{c}\cos\theta +\left(\frac{E_{\mu}^2}{c^2} - m_{\mu}^2 c^2\right) \cos\phi[/tex]

    Conservation of momentum in the Y direction
    [tex] \frac{E_{\nu}}{c}\sin\theta = \left(\frac{E_{\mu}^2}{c^2}-m_{\mu}^2 c^2\right)\sin\phi [/tex]


    dividing the two above equations we get [itex] \tan\phi[/itex] and we can solve for phi
    One we know Phi we can solve for the momentum of the muon. Once we have that we can sub this into the energy conservation eqatuions above and we are done

    i get this after some solving (also i make c=1, 'natural units')
    [tex] E_{\nu} = E_{\pi} - E_{\nu}^2 \sin^2 \theta \left(\sin \left[\arctan\left(\frac{E_{\nu}\sin\theta}{E_{\pi}^2 - m_{\pi}^2 - E_{\nu}\cos\theta}}\right)\right]\right)^2 + m_{\mu}^2 [/tex]

    But now i have the neutrino energy everywhere! Is there a simpler way of doing this?
     
  5. Jan 31, 2008 #4

    kdv

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    Looks complicated.

    I haven't checked your steps but have you thought about using four-momentum? soundsl ike it woudl be easy.

    writing [tex] P_{muon} = P_{pion} - P_{\nu} [/tex] and squaring both sides seems like it would directly give the result
     
  6. Feb 1, 2008 #5
    unsure how to do this...
    Is the 4 momentum like (E,px,py,pz)

    The mometum of the pion is
    [tex]P_{\pi}= (E_{\pi},p_{\pi},0,0)
    [tex]P_{\mu}= (E_{\mu},p_{\mu}\cosphi,p_{\mu}\sin\phi,0)
    [tex]P_{\mu}= (E_{\nuu},-p_{\nu}\cos\theta,-p_{\nu}\sin\phi,0)
    [tex] P_{muon} = P_{pion} - P_{\nu} [/tex]

    But wouldnt squaring the above give us
    [tex] P_{\mu}^2 = (P_{\pi} - P_{\nu})^2 [/tex]

    the other temrs are fine except for the [tex] 2 P_{\pi}\cdot P_{\nu} [/tex]

    plus how will i get rid of the momentum (not 4 momentum) of the muon?
     
  7. Feb 1, 2008 #6

    kdv

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    The square of the four-momentum of any particle gives [tex] P^2 = m^2 [/tex] (that's not true for so-called virtual particles but that' irrelevant here). So [tex] P_{\mu}^2 = m_{\mu}^2 [/tex]

    Also, you should write the components of the four-momentum of the neutrino explicitly. this is where the direction of the neutrino relative to the pion will enter. Note that you don't have to introduce a phi and a theta angle. You only need to introduce one angle by working in the plane containing the three-momenta of all three particles.
     
  8. Feb 1, 2008 #7
    If this is true then doesnt [tex]P_{\pi}^2 = m_{pi}^2 [/tex]? So then how do we get the energy of the pion into our solution

    [tex]P_{\pi}= (E_{\pi},p_{\pi},0,0) [/tex]
    [tex]P_{\mu}= (E_{\mu},p_{\mu}\cosphi,p_{\mu}\sin\phi,0)[/tex]

    neutrino's four momentum
    [tex]P_{\nu}= (E_{\nu},-p_{\nu}\cos\theta,-p_{\nu}\sin\phi,0)[/tex]

    consevaton of 4 momentum
    [tex] P_{\nu} = P_{pion} - P_{\mu} [/tex]

    do u mean like that?
     
  9. Feb 1, 2008 #8

    kdv

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    I am not sure what your theta and your phi are. But you may choose the axis such that the components of the neutrino four-momentum are simply [tex]P_{\nu}= (E_{\nu},p_{\nu}\sin\theta,p_{\nu}\cos\theta,0)[/tex] with theta being the angle between the three momentum of the decaying pion and the three-momentum of the neutrino.

    You will get the energy of the pion when you calculate the product [tex] P_{\pi} \cdot P_{\nu} [/tex]
     
  10. Feb 1, 2008 #9

    kdv

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    Like I said in one of my previous posts, start instead from
    [tex] P_{\mu} = P_{\pi} - P_{\nu} [/tex] and square both sides. This way you won't have to deal with the energy of the muon which you don't want in your equation (right? You want the energy of the neutrino interms of the energy of the pion, correct?)
     
  11. Feb 1, 2008 #10

    kdv

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    Sorry, I had the cos and sin theta switched in the neutrino four momentum It should have been

    [tex]P_{\nu}= (E_{\nu},p_{\nu}\cos\theta,p_{\nu}\sin\theta,0)[/tex]
     
  12. Feb 1, 2008 #11
    i see your point now

    Using the diagram i have posted in the first post i had it set up like this
    [tex]P_{\nu}= (E_{\nu},p_{\nu}\cos\theta,-p_{\nu}\sin\theta,0)[/tex]
    [tex]P_{\pi} = (E_{\pi},p_{\pi},0,0}[/tex]

    does that make sense using the diagram up there?

    OK so using
    [tex] P_{\mu} = P_{\pi} - P_{\nu} [/tex]

    and squaring both sides

    [tex] m_{\mu}^2 = m_{\pi}^2 - 2 P_{\pi}\cdot P_{\nu} + E_{\nu}^2 [/tex]

    and using the 4 momentum of the muon and the neutrino
    [tex] m_{\mu}^2 = m_{\pi}^2 - 2 \left(E_{\pi}E_{\nu}+p_{\pi}p_{\nu}\cos\theta\right) + P_{\nu}^2 [/tex]

    now what is the square of the 4 momentum of the neutrino? If, for a particle iwht mass, the square of the 4 momentum is the mass of the particle, then for a massless particle it should be equal to the E/c?

    and after this should i rewrite all the momenta of the muon and neutrino in terms of energy?

    And

    Thanks for your help so far !
     
  13. Feb 1, 2008 #12
    Oh and by the way when you said

    [tex] P^2 = m^2 [/tex]

    this is using 'natural' units right? where h=c=1?

    otherwise it would be
    [tex]P^2 = m^2 c^4 [/tex]?
     
  14. Feb 1, 2008 #13

    kdv

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    Yes...Being a particle physicist, I always unconsciously set c=1 :wink:
     
  15. Feb 1, 2008 #14

    kdv

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    Yes. Sorry, I had not even noticed the diagram!
    No, the square of the four-momentum of a massless particle is simply 0!
    Right, you have to reexpress the magnitude of the three-momenta (which is what you have in your final expression) in terms of the energies. For the neutrino it's simply [tex] p_{\nu} = E_{\nu}/c [/tex] since it's massless. For the pion, it will have the square root of E^2 minus m^2 but this is where you could use the fact that the pion is relativistic to neglect its mass compared to its total energy so that, in the relativistic limit, [tex] p_{\pi} \approx E_{\pi}/c [/tex] also.


    You are welcome.
     
  16. Feb 1, 2008 #15
    alright then

    [tex] m_{\mu}^2 = m_{\pi}^2 - 2 \left(E_{\pi}E_{\nu}+p_{\pi}p_{\nu}\cos\theta\right) + P_{\nu}^2 [/tex]

    rewriting [tex] p_{\nu} = E_{\nu} [/tex] where c=1 for 'natural' units

    and [tex] p_{\pi} = \sqrt{E_{\pi}^2 - m_{\pi}^2} [/tex]

    [tex] m_{\mu}^2 = m_{\pi}^2 - 2 \left(E_{\pi}E_{\nu}+E_{\nu}\sqrt{E_{\pi}^2 - m_{\pi}^2}\cos\theta\right) [/tex]
    [tex]m_{\mu}^2 = m_{\pi}^2 - 2E_{\nu} \left(E_{\pi}+\sqrt{E_{\pi}^2 - m_{\pi}^2}\cos\theta\right) [/tex]
    and then rearranging we get

    [tex] E_{\nu} = \frac{m_{\pi}^2 - m_{\mu}^2}{2\left(E_{\pi}+\sqrt{E_{\pi}^2 - m_{\pi}^2}\cos\theta\right)} [/tex]

    good?
     
  17. Feb 1, 2008 #16

    kdv

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    Looks good!!
     
  18. Feb 1, 2008 #17
    Thats awesome

    thank you so much
     
  19. Feb 1, 2008 #18

    kdv

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    You are welcome.

    Isn't that a nice formula? very compact and sweet. (Again, if you use the fact that teh pion is ultrarelativistic, it simplifies even more since you may neglect the mass compared to the total energy of the pion)

    and you see how using four-momentum conservation is a powerful approach. You can recover the result using separately energy and three-momentum conservation but it's a very long and painful calculation.

    I am glad I could help.

    Take care.
     
  20. Feb 1, 2008 #19

    kdv

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    Oops. sorry, I had read too quickly. There is a sign mistake. It should read


    [tex] m_{\mu}^2 = m_{\pi}^2 - 2 \left(E_{\pi}E_{\nu}-p_{\pi}p_{\nu}\cos\theta\right) + P_{\nu}^2 [/tex]

    recall that the dot product between two four vectors involves a relative minus sign.

    sorry if I missed it earlier, I was hurrying up because I was in the middel of helping students.
     
  21. Feb 1, 2008 #20
    alright thanks for that
    but why is there negative sign?
     
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