"Greg Egan - Foundations 1 - Special Relativity" - читать интересную книгу автора (Egan Greg)

Suppose we want to determine the x-coordinate of the event P in Figure 7. We
want to pin down its location in space, by marking it on the x-axis. In space alone, we'd
do this simply by drawing a line through P at right angles to the x-axis тАФ but is there
any justification for doing this on a spacetime diagram?
 Egan: "Foundations 1"/p.14


DP and EP are the world lines of two pulses of light, aimed at each other, which
collide precisely at event P. Assuming that the two pulses leave тАЬat the same timeтАЭ, they
must travel equal distances in order to arrive together. So the event Q, mid-way between
D and E, marks P's location on the x-axis. Because the light rays DP and EP both make
45┬░ angles with the x-axis, QP must be perpendicular to the axis.
But now suppose we use the same method to project OP onto some arbitrary
spacelike vector OG, instead of the x-axis. The same two light rays intersect OG at B
and C, and the event S is mid-way between them. So according to an observer who
considers that B and C (rather than D and E) happen тАЬat the same timeтАЭ, S and P (rather
than Q and P) happen тАЬat the same placeтАЭ. The two concepts are bound together, just like
тАЬleft-rightтАЭ and тАЬforwards-backwardsтАЭ; exactly what you mean by each one depends on
what you mean by the other.
OS is the projection of OP onto OG, in exactly the same sense as OQ is the
projection of OP onto the x-axis. Does the spacetime metric, Eqn (13), give us the length
of OS? First, since SP isn't perpendicular to OG, it will help to know what its direction
is. By drawing in another 45┬░ line, SH, parallel to BP, we can make a triangle SHC, the
same shape as BPC but half the size (since BS and SC are equal). That means H must
bisect PC, and the angles either side of it will be equal.
SP, then, must make exactly the same angle with the t-axis as OG does with the
x-axis. Since OG is the vector (u,w), and SP has coordinates (TQ, UR):


TQ/UR = tan A
= w/u
TQ u = UR w
TQ u тАУ UR w = 0
g[(TQ, UR),(u,w)] = 0


Applying the Euclidean metric to perpendicular vectors in Euclidean space gives
zero. By the same criterion, this shows that SP and OG really are perpendicular, in
spacetime, even though the lines we draw for them on paper are not. In a perspective
drawing of a room where a certain wall is viewed face-on, right angles on that wall will
be right angles in the drawing тАФ but right angles on the floor, the ceiling, and other walls
will not. Don't take this analogy too seriously тАФ the details in each case are quite
different тАФ but when you're drawing a spacetime diagram it would be a great surprise if
you could show everything without distortion.
Now, to find OS, we make use of Eqn (14a), the new version of Pythagoras's
theorem: