Androcles, you seem to have the attributions totally screwed
up. Since Randy's post do not appear here (is this a private
mail discussion, or what?), it is almost impossible to find
out who said what.


Androcles wrote:
"Randy Poe" wrote in message
oups.com...

Androcles wrote:

Below is Randy Poe's attempt to handle a coordinate system
which the imbecile Dirk Van de moortel says I do not understand.

Randy Poe:
I'm going to take another tack, one which may dispense with
all this endless origin silliness. The kind of argument that
could result from trying to introduce and then solve Einstein's
differential equation could keep us here till the heat death
of the universe.

Let's look at the transit time of the light signals.


Now, from the point of view of S, the emitter and mirror
are some distance d apart.


d1 = (c+v)t.
d2 = (c-v)t.


d = d1 = d2.

What are d1 and d2, and why should they be equal to each
other and d?



[snip]


Left in for the shame of it:

The light signal leaves the
emitter at x1, at time t1. It is traveling toward the
mirror at speed c, but at the same time the mirror is
receding at speed v. So from the point of view of the S
observer, the transmit time from emitter to mirror is
d/(c-v). Which means t2 = t1 + d/(c-v).


On the return trip, the light signal is coming back at
c and the apparatus at the back of the truck is rushing
to meet it at speed v. They converge at a rate of c+v,
and the transit time is d/(c+v). Which means t3 =


t1 + d/(c-v) + d/(c+v)
t3 = t1 + d/(c-v) + d/(c+v)= t1 + d*[1/(c-v) + 1/(c+v)]= t1 +
d*[(c+v)/(c^2-v^2) + (c-v)/(c^2-v^2)]= t1 + d*[(c + v + c - v)/(c^2 -


v^2)]= t1 + d*[2c/(c^2 - v^2)]Or


t3-t1 = d*[2c/(c^2-v^2)] - Randy


LOL!

Read: "I did not understand a word, so I'll simply
ridicule it."



[snip]


Bye,
Bjoern