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

Eqn (8a) is identical to Eqn (4), which is what we set out to prove.
Setting B equal to 90┬░, Eqn (8b) shows that the metric function for two
perpendicular vectors is zero, since the cosine of 90┬░ is zero. That's why the metric is
able to тАЬpick outтАЭ the part of one vector that's parallel to another; the part that's
perpendicular simply yields zero.




Calculating the way a point's coordinates change when the reference frame is
 Egan: "Foundations 1"/p.9


rotated in space is easy, now. In Figure 4, imagine you're standing at the origin, O,
looking straight ahead in the direction of the axis marked y1, with x1 pointing directly to
your right. P marks some fixed object in front of you, say a tree. OQ measures how far
to the right of you the tree is, and OR measures how far ahead it is. (Negative numbers
would be used if it was to the left, or behind you.)
Now suppose you turn your entire body through the angle A, so you're looking
in the direction y2, and x2 points directly to your right. The new coordinates you'd give
P are OS and OT. We already have OS for exactly this situation, Eqn (6), and all that's
needed to work out OT are the coordinates of a unit vector that points along the y2 axis.
From Figure 4 it's clear that (тАУsin A, cos A) does the job, so:


OT = g[(OQ, OR),(тАУsin A, cos A)] (9)


Writing (x1,y1) for the coordinates of any point in the original reference frame,
and (x2,y2) for the coordinates of the same point in the rotated reference frame, Eqns (6)
and (9) become:


x2 = g[(x1,y1),(cos A, sin A)]
= x1cos A + y1sin A (10a)
y2 = g[(x1,y1),(тАУsin A, cos A)]
= y1cos A тАУ x1sin A (10b)


This is the standard way of expressing the change of coordinates for a rotation in
space. There's another way, though, which is worth writing down because of its
similarity with the most common form of the equivalent spacetime equations. Put


s = tan A
= sin A / cos A