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Question about units, if the units are given in the equation or as

  1. Feb 3, 2014 #1
    ... part of the value?

    first example:

    v (MHz) = 110 B(nT) E^2 (GeV)

    could someone confirm that when an equation is like this you simply change E into GeV before squaring it and making the calculation? slightly confused.

    second example:

    rest mass of a Pion is stated by Wiki as 139.57018(35) MeV/c^2, why is this? are there units of c^2? what if you want to use the mass in a calculation, do you divide 139MeV by the speed of light squared? not sure what to do with this.

    Thanks for any help.
     
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 3, 2014 #2

    HallsofIvy

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    If E is given in "GeV" then squaring will give units of "GeV squared". Of course, they should then combine with whatever units you have for "B(nT)" to give the correct units for v.

    MeV is a measure of a type of energy which has basic MKS units "kilogram meters per second squared" so that dividing by a speed, squared, like c^2, leaves mass units: kilograms.
     
  4. Feb 3, 2014 #3
    thanks for that!!

    so why is it given over c^2 and not in kilograms? and also if there is a calculation where the pion mass is multiplied by c^2, do the c^2's simply cancel?

    thanks again.
     
  5. Feb 3, 2014 #4

    D H

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    Correction: "kilogram meters per second squared" -- that's the Newton (i.e., units of force). Energy has dimensionality mass*velocity2, or units of kilogram*meters squared / seconds squared.

    Metric units can be rather inconvenient in particle physics. Expressing mass in terms of energy/c2 tells a much better story. For example, how much energy is released in an annihilation event? Expressing rest mass in energy/c2 is much more informative than expressing it in kilograms. In fact, in mass is expressed in electron-volts rather than electron-volts/c2; that missing division by c2 is implied.
     
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