1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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]).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook