Re: The velocity of light going pass a moving train.

On Jun 21, 12:50 pm, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
Dono <s...@xxxxxxxxxxx> wrote in news:1182450172.908373.143140
@a26g2000pre.googlegroups.com:

right, for the train's frame.

cos(pi) = -1.

The beam misses the mirror, in which direction?
cos(0) = 1.

The beam misses the mirror, in which direction?

In the track frame the beam hits the mirror a little behind the
center, in the direction opposite to the train movement.

No. I am asking you to interpret the cos(pi) = -1 and cos(0) = 1 figures
for me. Assume that there is a laser that can be configured to aim in the
zero direction and the pi direction. Just in the train frame, which
directions are the 0 and the pi and what toes the -1 and 1 mean, exactly,
to you?

You may draw a diagram, if you like.

I am trying to find out how you have your protractor oriented and how you
are defining your angle theta.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+...@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap





I did this earlier for Jeckyl


y
k ^
^ |
\ |
\ |
\ --------------- x (cos(theta)=0

K is the wave vector in the track frame, theta_track is the angle
between k and the semipositive x axis. Check the Einstein paper.

.

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