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Confused over units

  1. Aug 10, 2011 #1

    BWV

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    if using hbar=c=1 and ev for mass/energy what are appropriate distance units? Wikipedia talks about inverse energy without really explaining them
     
  2. jcsd
  3. Aug 10, 2011 #2

    vanhees71

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    If you set [itex]c=1[/itex], lengths and time have the same unit. In HEP the most convenient one is [itex]1 \; \mathrm{fm}=10^{-15} \mathrm{m}[/itex] (femtometer, usually also called Fermi). Also obviously mass, energy, and momentum have the same unit. For this one uses [itex]\mathrm{MeV}[/itex] or [itex]\mathrm{GeV}[/itex]. Of course, [itex]c=1[/itex] is the "natural setting" for all relativistic physics, be it in the classical or the quantum description.

    In quantum physics one more fundamental constant enters the game, Planck's "quantum of action", [itex]h[/itex]. Since it turned out that it is much more convenient not to use frequency but angular velocity, nowadays we only use [itex]\hbar[/itex]. The natural choice in quantum physics is then of course [itex]\hbar=1[/itex]. An action has the dimension of [itex]\mathrm{length} \times \mathrm{momentum}[/itex] or [itex]\mathrm{time} \times \mathrm{energy}[/itex]. Now, if [itex]\hbar=1[/itex] these products are dimensionless, and thus, we need only one unit for energies, momenta, and masses. Times and distances are then measured in inverse energy units, or you argue the other way around and use a length and distance unit and measure energies, masses, and momenta in inverse length units. It depends on the context what's more convenient. Often one uses a mixed system, using fm for lengths and times and GeV for energy, momenta, and masses.

    To convert from these natural units to good old standard units, you only need the "conversion factor", [itex]\hbar c \approx 197 \; \mathrm{MeV} \, \mathrm{fm}=0.197 \; \mathrm{GeV} \, \mathrm{fm}[/itex].
     
  4. Aug 10, 2011 #3

    clem

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    The combination hbar*c=197.32 MeV-fm=1.
    You can use this to convert an answer in MeV to fm^-1.
    eg: 100 MeV=0.5 fm^-1.
     
  5. Aug 10, 2011 #4

    BWV

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    Thanks for the responses, one more followup

    is this an OK set of units? (it replicated the answer of an electron tunneling probability calculation in a book at least)

    h-bar = 197.32 MeV
    mass in MeV (used .5 MeV for electron mass)
    distance in FM

    there was no momentum in the calculation - would that be?
     
  6. Aug 11, 2011 #5

    vanhees71

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    You have [itex]\hbar=197 \; \mathrm{MeV}/c[/itex]. Don't mix natural units and SI units! What's you concrete problem. Perhaps the issue becomes more clear on your example.
     
  7. Aug 11, 2011 #6

    BWV

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    As I said, it a calculation of the probability of electron tunnelling

    d 200000
    m 0.5
    v-e 0.000001
    h 197
    e mass 0.5

    all in MeV except distance (d) which is in FM

    the probability is given as

    P = [itex]exp[-2d\sqrt{2m(V-E)/\hbar^2}][/itex]

    for the inputs above I get 0.13, which is the answer in the book (Cox - Intro to Quantum Theory and Atomic Structure chp 3)
     
  8. Aug 11, 2011 #7

    clem

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    In natural units, the hbar wouldn't be there. The square root would be in the unit MeV, which you divide by 197 MeV-fm to get the answer in fm^-1.
     
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