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Neutrino mass

  1. Mar 19, 2007 #1
    I tried and I got a ridiculous large number for the neutrino mass. I basically used [tex]E=\gamma m c[/tex] and then, for 10 MeV neutrinos, time taken is t + 10 seconds to reach Earth, for 50 MeV neutrinos, time taken is t seconds.
    [tex]Speed = \frac{100 000}{(t+10)}[/tex] for 10 MeV and [tex]Speed = \frac{100 000}{t}[/tex] for 50 MeV. Here's the question:

    Neutrinos were detected from a supernova (distance ~ 100 000 light years) over a time period of ~ 10 seconds. The neutrinos have an energy range from 10 MeV to 50 MeV.
    What's the mass of the neutrino, assuming all neutrinos were produced at the same time and their mass is small compared to their energy.
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2

    Meir Achuz

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    Your speed is in LY/sec. Did you convert units?
    Speed in LY/Y is so close to one that you have to use a Taylor expansion.
  4. Mar 19, 2007 #3
    Yes, I did unit conversions. A light year = [tex]9.46 \times 10^{15} meters [/tex]

    Here's what I did:

    for 10 MeV neutrinos, [tex] 10\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t+10})^2/c^2}}[/tex]

    and hence, [tex]t+10 = \frac{10\;M\;e\;d}{10\;M\;e\;c}(1-\frac{m^2\;c^2}{10^2\;M^2\;e^2})^{-1/2}[/tex]

    Binomial expansion, keep only first two terms yield:
    [tex]t+10 = \frac{d}{c}(1+\frac{m^2\;c^2}{2\times10^2\;\M^2\;e^2})[/tex]

    do the same for 50 Mev neutrinos,

    for 50 MeV neutrinos, [tex] 50\;Me = \frac{m\;c}{\sqrt{1 - (\frac{d}{t})^2/c^2} }[/tex]

    simultaneous equations, cancelling t, yield:

    [tex]10 = \frac{d\;m^2\;c}{2\;M^2\;e^2}(\frac{1}{10^2}-\frac{1}{50^2})[/tex]

    Solving for m (neutrino mass), I got 7.7 GeV !?
    Last edited: Mar 19, 2007
  5. Mar 19, 2007 #4

    Meir Achuz

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    I get: 10 sec=(10^5 yr X m^2/2)(1/100-1/2500), with m in MeV.
    I don't see where your e^2 came from, why your c is i the numerator.
    What is your M?
    Use 1LY/c=1 year, keep all masses in MeV.
    Use t=(L/c)\sqrt{m^+p^2}/p~L/c-(m^2/2E)(L/c).
  6. Mar 19, 2007 #5
    You also must take into account how long the neutrino sees itself moving as compared to what the earth sees as how long the neurtino is moving from a to b
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