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]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!
Yes, sorry about the typo. :(Shouldn't it be?
[tex]\sqrt{||L||^2-(ct)^2}[/tex]
I should clarify that, given a 4-momentum vector,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]
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.why is it -(ct)^2 and not positive?