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Homework Help: Neutrino Communication Problem

  1. Jun 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I need some help understanding the following problem:

    A distant, advanced civilization wishes to make contact with us using neutrinos rather than photons as the transmitting medium to avoid problems of obscuration along the line of sight. Suppose they use an [itex]\bar{\nu_e}[/itex] antineutrino beam to generate ##W^-## on Earth via the process

    [itex]\bar{\nu_e} + e^- \to W^-[/itex]​

    Calculate the energy of the antineutrinos that would be required, in eV.

    2. Relevant equations

    3. The attempt at a solution

    I'm not sure what concepts/equations are required here.

    We have a modulated beams of antineutrino as information carriers that converts into electrons, resulting in the emission of a ##W^-## (or absorption of ##W^+##). Does this mean an antineutrino must have at least an energy equal to the rest energy of an electron (##m_ec^2##)? :confused:

    Any help is greatly appreciated.
  2. jcsd
  3. Jun 23, 2013 #2


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    I think the generated W's should be real.
    If you have a neutrino with 511 keV of energy, and an electron, can they react to produce a (real) W?

    Think of energy/momentum conservation.
  4. Jun 25, 2013 #3
    Thank you for your input.

    So the e- and the neutrino must have the same magnitude of energy? (I thought neutrinos do not interact with matter so that they would annihilate :confused:)

    The energy/momentum conservation is ##E_{\bar{\nu}} + E_{e^-} = E_{W^-}##. If the electron is at rest (##P_{e^-}=0##) the invariant becomes ##E_{e^-} = m_{e^-} c^2 = 510.7 \ keV##.

    Therefore for the 511 keV case we can produce a W boson. Is that right?

    Also, I think the W's will be real not virtual, because in the following diagram I drew, the W- has an external line coming out:

    Last edited by a moderator: May 6, 2017
  5. Jun 25, 2013 #4
    So you should just get [itex]E_{\bar{\nu}} = 80.4 GeV - 511keV[/itex], that is, if Wolfram is giving me the correct value for the W boson
  6. Jun 25, 2013 #5


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    They can be created, and the time-reversed process is always possible, too. It is just rare.

    That is a good approach, but 4-momenta would be even better here.
    No. How did you get that conclusion?

    With an electron at rest (511 keV) and a neutrino of 511 keV, your total energy is just ~1MeV. How should that give a W-boson of ~80 GeV, even without its kinetic energy?

    @spaderdabomb: That is wrong, the produced W-boson will not be at rest. In addition, please do not (try to) post full solutions to homework problems.
  7. Jun 25, 2013 #6
    I don't see how I put was incorrect. I mean, energy must be conserved. The kinetic energy plus the rest mass energy of the incoming neutrinos, plus the rest mass energy of the electron it collides with must be equal to or greater than the rest mass energy of the W boson.
  8. Jun 25, 2013 #7


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    Sure, but momentum has to be conserved, too.

    The electron has no momentum, the neutrino has a momentum. To conserve momentum, the W boson has to move, so some of the energy has to go into its motion.
  9. Jun 25, 2013 #8
    Hmmm you're right. Crap shoulda used four vectors =/ i always get lazy
  10. Jun 27, 2013 #9
    I'm not sure if this is correct, but I assumed that the neutrino comes in along the x-axis, and the emitted W- also moves in that direction (to conserve momentum). And since ##m_{W^\pm} = 80.403 \ GeV/c^2##, I got these four momentum vectors:

    ##\begin{pmatrix} 80.403 \ GeV/c \\ P_x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 511 \ KeV/c \\ 0 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} E_{\nu}/c \\ P_x \\ 0 \\ 0 \end{pmatrix}##​

    or ##p_{\nu} = p_{W^-} \ - \ p_{e^-}##. So how does the energy that goes into W's motion factor in this equation? :confused:

    I'm not also sure how to calculate the momentum of the antineutrinos in the beam, but here is what I did so far using the energy-momentum relationship for a relativistic particle:

    ##E_{\nu}^2 = p_{\nu}^2 c^2 + (m_{\nu} c^2)^2##

    ##\implies p_{\nu}= 80 \ GeV - 511 \ KeV=\sqrt{E^2_{\nu}-m_{\nu}^2}## (in units where c = 1)

    Is this right?
  11. Jun 27, 2013 #10


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    The first entry for the W is wrong. The mass of the W boson is related to its 4-vector, but not in that way...

    You can neglect the neutrino mass in this problem.
  12. Jun 29, 2013 #11
    So what goes in the first entry for the W? I was just following my textbook (it says the zeroth component of the 4 vector is simply E/c). :confused:
  13. Jun 29, 2013 #12


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    Simply E/c, exactly. You do not know E yet, but you know a relation between E and p of the W boson.
  14. Jun 30, 2013 #13
    Yes, so the relationship between E and p of W boson is


    And for the ν it is

    ##E_{\nu}=|p_{\nu}|c## (treating it as a massless particle)

    Putting these into the equation for conservation of energy, we have

    ##|p_{\nu}|c + 511 \ KeV = c\sqrt{m_W^2c^2+p_W^2}##

    But this is an equation for two unknowns, how do I solve this for the momentum of the W or the neutrino?
  15. Jun 30, 2013 #14


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    You already used ##p_\nu=p_W## in post 9.
  16. Jul 1, 2013 #15
    Thank you very much for reminding me, here is what I've got:

    ##|p_{\nu}|c + 511 \ KeV = c\sqrt{m_W^2c^2+p_\nu^2}## or

    ##(p_\nu c + 511 \ KeV)^2 = c^2 (m^2c^2+p_\nu^2) ##

    Solving for |pν|,

    ##\implies p_\nu = \frac{m_W^2 c^4-511 \ MeV}{2(511 \ KeV)c}##

    Meanwhile, the energy of the neutrino (from E2-p2c2=m2c4) is

    ##E_\nu = p_\nu \times c = \frac{(80.403\times10^9)^2 (3\times10^8)^2- (511 \times 10^6)}{2(511 \times 10^3)} = 5.69 \times 10^{32} \ eV##

    So that's a huge amount of energy to deliver to a particle... did I make a mistake or is that a realistic answer (for this situation)?
  17. Jul 1, 2013 #16


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    The factors of c are wrong. You need more "c" at the momenta, or less at the masses. Just check the units of the result, they do not match.
    In addition, I think you forgot a square at 511 in the numerator, but that won't change the result.
    Yes, it is a huge number (just not as huge as you calculated), and the conclusion is that you don't want to use that method of communication ;).
  18. Jul 3, 2013 #17
    I agree. It makes perfect sense now. Thank you for taking the time to help with the problem, I really appreciate it. :)
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