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Many neutrinos are traveling through the earth

  1. Apr 15, 2004 #1
    I'm a newb so forgive me if this question is ridiculous. In theory(cuz we have no idea of really knowing) how many neutrinos are traveling through the earth at any given time. Or better yet, how many are traveling through lets say a 1 m x 1m square in any arbitrary position on earth. I'm not really looking for a number, but maybe a general order of magnitude. Like are neutrinos coming from all over the universe and passing through us or are they relatively rare?
  2. jcsd
  3. Apr 15, 2004 #2


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    The number density of each neutrino species is theorized to be about 56 per cubic centimeter, for a total number density of around 150 neutrinos per cubic centimeter.

    Given that the upper limit on their mass is ~3 eV and their characteristic temperature now is about 2 K, you can calculate their average velocity.

    The average kinetic energy per particle is [itex]E = \frac{3}{2}kT[/itex], where T is the temperature and k is Boltzmann's constant. For a temperature of 2 K, the neutrinos have an average kinetic energy of 4.1419509 x 10-23 J.

    The velocity at which a particle of mass [itex]m_0[/itex] will have energy E is given by:

    [tex]E = (\gamma - 1) \, m_0 c^2[/tex]


    [tex]\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

    Plugging and chugging gives us an average velocity of about 0.01 c.

    The number of particles crossing a 1 m2 plane is just the number density (in #/m^3) times the velocity (in m/s). There are 150,000,000 neutrinos per cubic meter, and they're moving at about 3,935,435.55 meters per second, for a grand total of 5.90315332 * 1014 crossing a one-meter square area per second. Quite a lot!

    - Warren
  4. Apr 15, 2004 #3
    Wow, thank you for that. I could never have expected such a perfect response.
  5. Apr 15, 2004 #4


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    Welcome to physicsforums, Zaphod. Don't forget your towel!

    - Warren
  6. Jun 13, 2004 #5
    Given that the upper limit on their mass is ~3 eV

    Given that particles are fields and that each field consists of a force and an anti-force. Then in any field obeying the Inverse Square Law there must be a difference between the total force and anti-force acting on any given radii. This difference is the neutrino mass and can in theory be calculated. It arises because it is what remains when the force field is removed. Just as infinity has a minimum energy level so also do particles have a minimum energy level it stands out from the background level only because of its greater density.
    In Cassels Laws os Nature the author gives a list of theoretical neutrino masses.
  7. Jun 29, 2004 #6


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    There are several ways to give the neutrino a mass, all of them go beyond the standard model.

    You can add adhoc Dirac terms to your electroweak lagrangian, but you're stuck in the sense that you have to couple it to a higgs field, and without knowledge about the VeV (from say a scalar higgs), you don't even know what type of suppression factor you need to write in by hand, much less explain why it has to be so finely tuned and unnatural.

    Typically, there is a theoretical prejudice to adding a Majorana Mass, since its neater. The usual natural theories that output such a mass is say SO(10), the simplest is when you live in a 16 dimensional representation, and you have a remaining singlet field (called sterile). You can identify the right handed neutrino as the single set of dimension 5 operators. Moreover, it will undergo the seesaw mechanism, which naturally gives it a small mass relative to electroweak symmetry breaking scale.

    In SUSY models, you can do even better, though of course the amount of free parameters in the theory starts growing.
  8. Jun 29, 2004 #7


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    chroot: as far as I understand you did calculate the flux of cosmic background neutrinos.

    But I think this flux does not have a specific direction though any surface on earth because the cosmic neutrinos are in thermal equilibrium. Thus same amount of neutrinos crossing in one direction will cross in the other direction.

    Anyway, I assume (I do not really know) this is negligible against the flux of solar neutrinos crossing any surface on earth. Isn't it?

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