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.