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4-vector - momentum^2

  1. Oct 25, 2007 #1
    I'm having a problem understanding this:


    If we take c=1.

    Here is what bothers me:

    [tex]P(E, \vec{p})=E^2-(\vec{p})^2[/tex]

    Now, I assume that E=mc^2, and for c=1, E^2=m^2? Is that correct?

    And I don't know what p^2 is, I look at it as:


    What am I doing wrong?
  2. jcsd
  3. Oct 25, 2007 #2


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    I am not exactly sure what bothers you but one point: [tex] E = \gamma m c^2 [/tex], not mc^2 (which is valid only in the rest frame of the particle). Also the momentum is the relativistic three-momentum so it's [itex] \gamma m \vec{v} [/itex]
  4. Oct 25, 2007 #3

    Meir Achuz

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    Your first equation shows m^2=E^2-p^2.
    Your assumption E=m is wrong.
  5. Oct 25, 2007 #4


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    The symbol 'E' in this equation is the total energy of the particle. The popularly-known equation E = mc^2 refers to the "rest mass-energy of the particle" and is really an incorrect use of the symbol.

    'p' here is just the regular ol' 3-momentum, mass times the 3-dimensional velocity vector for the particle. The so-called "4-momentum" [tex]P^{\mu}[/tex] is a 4-dimensional vector whose components are

    ( iE, p_x, p_y, p_z ) [or flip signs depending on whose notational convention you use].

    Your original relation then,


    is then just taking the "dot product" of P with itself to get the square of the magnitude,

    [tex]P^2 = E^2 - p^2 = (mc^2)^2[/tex] , again with appropriate adjustments for local notational practice.
    Last edited: Oct 25, 2007
  6. Oct 26, 2007 #5


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    As already said, the popular formula E = mc^2 -- mostly misquoted -- is a) not generally applicable and b) not even a main result of special relativity, it's more like a small remark buried somewhere deep inside the text.

    To stick with Ene Dene's approach: if you plug in the correct formula
    [tex]E = \sqrt{ (\gamma m c^2)^2 + (m p^2)^2 }[/tex]
    you will get a consistent result.
    In units where c = 1, it'd be
    [tex]E^2 = \gamma m^2 + p^4[/tex]
    [tex]p = \gamma m v[/tex].
    Also note that you can deduce where E = mc^2 is applicable; the general formula reduces to it in the rest frame (p = 0) at non-relativistic speeds ([itex]\gamma \approx 1[/itex]).
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