The question
An observer, S, has measured the cooordinates of two events, a, and b, in space-time:
$a^\mu = \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \end{pmatrix} cm$
$b^\mu = \begin{pmatrix} 4 \\ 0 \\ 4 \\ 0 \end{pmatrix} cm$.
(1)
A particle, q, propagates from a to b. What is the proper velocity of this particle? Work
out numerical values for all components of the four-vector, with at least three
significant digits.
Solution:
?y = 4 ? 2 = 2 cm in a time ?(ct) = 4 ? 1 = 3 cm. (I wonder how
many will get the unit right, here, and remember that the zero component
is ct.).
So its ordinary 3-velocity is ? = (0, 2/3, 0) (dimensionless). Equivalent for `v is also OK
with correct dimension either via multiplication by c or explicitly in cm/s
the ? factor is $1/\sqrt{1 - 4/9} = 3/\sqrt{5} \sim 1.34$.
thus, the result is u = (1.342, 0.00, 0.894, 0.00)c.
