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Neutrino Oscillation

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    I doubt that my problem is supposed to be in the Advanced Physics forum, but the subject is advanced. Okay, my problem is pretty stupid. The assignment is deriving a probability using some approximations and such, for 2-neutrino oscillation. I can do the entire calculation (which isn't difficult), except for one thing and that is the unit conversion :cry:

    The problem statement is rewriting the left hand side to the right hand:
    [tex]\frac{\Delta m^2 c^3 L}{4 \hbar E} \cong 1.27\frac{ \Delta m^2 L}{E}[/tex]
    , with m in [tex]\mathrm{eV} / c^2[/tex], E in [tex]\mathrm{GeV}[/tex] and L in [tex]\mathrm{km}[/tex]

    2. Relevant equations

    [tex]\frac{\Delta m^2 c^3 L}{4 \hbar E} \cong 1.27\frac{ \Delta m^2 L}{E}[/tex]

    3. The attempt at a solution

    So okay, I assume the terms on the left hand side are in SI units. So mass can be expressed as J / c^2. One eV is equal to the e*J with e the charge of the electron. J is therefor equal to the reciprocal of e multiplied with eV. Insert that into the term and I thought I had converted J successfully to eV with factor of 1/e. But here is the problem, this becomes approximately 10E19 squared, and dividing it to Planck's constant, gives an enormous factor, inconsistent with the answer.

    Now I know this problem is stupid, but my brain is not cooperating. I know what to do to get the factor 1.27, but I don't understand why! You can get the 1.27 factor, simply by replacing the units with the new units and calculating that:

    [tex]\frac{(1.6*10^{-19})^2 10^3}{4\hbar 10^9 1.6*10^{-19} c} \cong 1.2669[/tex]

    I find the use of units and conversions in HEP in general rather confusing.
  2. jcsd
  3. Jan 10, 2010 #2
    a) You should work in natural units.

    b) You think in natural units.

    a) This means that you regard c, hbar as mere conversion factors.

    b) This means that you know that mass = energy = inverse length, etc.

    So, you should simply put hbar = c = 1 first as these constants are irrelevant scaling constants that have nothing to do with physics. Then, all you need to do is to write the expression for the probability as:

    P = 1/4 (delta m)^2/E L

    In this expression both m, E and L have to be expressed in mutually consistent units. The moment you choose an arbitrary unit for the mass m, the unit for the energy is fixed (this is the same unit as the unit you have chosen for the mass), also the unit for L is fixed. What you then want to do is to invent your own scaling constants to suit your own interests. So, you want to modify the above equation because you want to insert values for m, E and L that are measured in your own favorite units (which is not the SI unit system). You want to measure m in eV, E in GeV and L in km.

    The above equation assumes that if m is given in eV that E is also given in eV and that L is given in inverse eV. So, the only non-trivial thing to do here is to convert kilometers to inverse eV. This is most easily done by hijacking some formula relating energy or momentum to distances, like the formula for the Compton-wavelength that are valid in Si units and rewriting it in a SI-dimensionally consistent way until you get the desired result. E.g.:

    lambda = hbar/(mc) = hbar c/(m c^2) = hbar c/energy

    So, if we used primed symbols to denote the values expressed in your favorite units, we have:

    delta m = delta m' (because we can freely choose the units of one of the three quantities)

    E = E' 10^9 (E has to be in eV because we choose to express m in eV, so we must multiply E' by 10^9)

    L = L'*10^3/(hbar c) e = 10^3/[hbar c ] 1.6*10^(-19) L' (we first convert L' to meters, then we see from the above equation that we need to divide meters by hbar c to convert that to inverse Joules, then we need to convert that to eV. Since eV = 1.6*10^(-19) Joules,we have Joules^(-1) = 1.6*10^(-19) eV^(-1)).

    So, the numerical factor is:

    1/4 * 10^(-9) *10^3/[hbar c] 1.6*10^(-19)

    Note that hbar and c are regarded as dimensionless scaling constants in this derivation. In the above expression we give then the numerical value they have in SI units when we take meter = second = kilogram = 1
  4. Jan 11, 2010 #3
    Note that since you already got an equation with the conversion factors c and hbar in it, you could simply have converted units in the standard way as any high school student should be able to do to answer this particular question. But then, that's not a good exercise to do unit conversions as we do in high energy physics.

    In such cases I prefer tio mix natural units with Si units (that's allowed because hbar and c are equal to 1 in natural units anyway). You can substitute:

    delta m = (delta m/eV) eV

    E = (E/GeV) GeV

    L = (L/Km) Km

    Then you express the quantities outside of the brackets in SI units, the expressions inside the brackets are dimensionless (we use natural units here). The eV, GeV and Km outside the brackets are combined with the c and hbar and the 1/4 to produce the desired factor.
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