1. Not finding help here? Sign up for a free 30min 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!

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

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    Here is one place to consult:
    http://www2.maths.ox.ac.uk/~nwoodh/sr/
     
    Last edited: Apr 5, 2008
  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

    ehj

    User Avatar

    Shouldn't it be?
    [tex]\sqrt{||L||^2-(ct)^2}[/tex]
     
  7. Apr 5, 2008 #6
    Yes, sorry about the typo. :(
     
  8. Apr 5, 2008 #7

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.)
     
  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

    jtbell

    User Avatar

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?