Why is the 3 momentum represented with a negative sign in the 4-vector?

[Davio]

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 that's not much help!

 

[Hootenanny]

Three momentum is simply a vector containing all three momenta,

\boldmath P \unboldmath = \left(\begin{array}{c}p_x \\ p_y \\ p_z\end{array}\right)

 

[robphy]

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

Here is one place to consult:
http://www2.maths.ox.ac.uk/~nwoodh/sr/

 

[yuiop]

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 that's not much help!

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 L =\sqrt{x^2+y^2+z^2} 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 \sqrt{x^2+y^2+z^2-(ct)^2}

or \sqrt{||L||^2-(ct)^2}

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

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

(Edited to fix the typo pointed out by ehj)

 

[ehj]

Shouldn't it be?
\sqrt{||L||^2-(ct)^2}

 

[yuiop]

Shouldn't it be?
\sqrt{||L||^2-(ct)^2}

Yes, sorry about the typo. :(

 

[robphy]

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

Here is one place to consult:
http://www2.maths.ox.ac.uk/~nwoodh/sr/

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

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

(Note, however, that the 4-vector (\tilde p - (\tilde p \cdot \tilde u)\tilde u) is generally NOT "purely spatial" according to another observer \tilde w. To \tilde w, that 4-vector has both nonzero spatial- and temporal-parts.)

 

[Davio]

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

 

[jtbell]

why is it -(ct)^2 and not positive?

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.