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Question about 3 momentum

  1. Apr 5, 2008 #1
    Hey guys, what exactly is a 3 momentum? I can't any references to it anywhere on the net, which actually tells me what is it, I know a energy 4 momentum is, px , py, pz e/c, but thats not much help!
  2. jcsd
  3. Apr 5, 2008 #2


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    Three momentum is simply a vector containing all three momenta,

    [tex]\boldmath P \unboldmath = \left(\begin{array}{c}p_x \\ p_y \\ p_z\end{array}\right)[/tex]
    Last edited: Apr 5, 2008
  4. Apr 5, 2008 #3


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    3-momentum is the "spatial-part" of the 4-momentum.

    Here is one place to consult:
    http://www2.maths.ox.ac.uk/~nwoodh/sr/ [Broken]
    Last edited by a moderator: May 3, 2017
  5. Apr 5, 2008 #4
    In a 3D reference frame a line of arbitary length and direction from the origin of the frame can be described in terms of x,y and z component. Using pythagorous theorem we can find the length of that arbitary line from [tex] L =\sqrt{x^2+y^2+z^2}[/tex] You could call that length the 3 length sometimes abreviated to ||L||. When we include an additional dimension of time to the 3 spatial dimensions then we have the 4 length [tex] \sqrt{x^2+y^2+z^2-(ct)^2}[/tex]

    or [tex]\sqrt{||L||^2-(ct)^2}[/tex]

    Similarly 3 velocity ||v|| = [tex] \sqrt{v_x^2+v_y^2+v_z^2}[/tex]

    and 3 momentum ||p|| = [tex] \sqrt{p_x^2+p_y^2+p_z^2}[/tex]

    (Edited to fix the typo pointed out by ehj)
    Last edited: Apr 5, 2008
  6. Apr 5, 2008 #5


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    Shouldn't it be?
  7. Apr 5, 2008 #6
    Yes, sorry about the typo. :(
  8. Apr 5, 2008 #7


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    I should clarify that, given a 4-momentum vector,
    the 3-momentum is essentially the "spatial-part" according to a given observer.
    That is, the 3-momentum is [obtained from] the vector-component of the 4-momentum that
    is [Minkowski-]perpendicular to an observer's 4-velocity.
    From a given 4-momentum vector, different observers will determine different 3-momentum vectors.

    Given a 4-momentum [tex]\tilde p[/tex] and an observer's 4-velocity [tex]\tilde u[/tex] (with [tex]\tilde u \cdot \tilde u=1[/tex] in the [tex]+---[/tex] convention),
    Write out this identity [a decomposition of [tex]\tilde p[/tex] into a part parallel to [tex]\tilde u[/tex], and the rest perpendicular to [tex]\tilde u[/tex]]:
    [tex]\tilde p = (\tilde p \cdot \tilde u)\tilde u + (\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex].

    The 4-vector [tex](\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex] is "purely spatial" according to the [tex] \tilde u [/tex] observer [check it by dotting with u], and can be thought of as a three-component vector in [tex] \tilde u [/tex]'s "space" by projection. That projected vector is the 3-momentum of the object according to [tex]\tilde u [/tex].

    (Note, however, that the 4-vector [tex](\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex] is generally NOT "purely spatial" according to another observer [tex] \tilde w [/tex]. To [tex] \tilde w [/tex], that 4-vector has both nonzero spatial- and temporal-parts.)
    Last edited by a moderator: May 3, 2017
  9. Apr 6, 2008 #8
    Hey guys, feel a bit silly, of course its just the partial part! Whilst we're on that topic though, why is it -(ct)^2 and not positive? Is it because CT=Distance, L^2+D^2 =p^2 ? Surely not though because, L ^2 = x^2 +y^2 +z^2
  10. Apr 6, 2008 #9


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    With the "-" sign in that position, the "magnitude" of a 4-vector is invariant between different inertial reference frames. With a "+" sign instead, the "magnitude" is not invariant.
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