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

On Jun 19, 11:05 pm, "Jeckyl" <n...@xxxxxxxxxxx> wrote:
"Dono" <s...@xxxxxxxxxxx> wrote in message

news:1182303768.824683.175520@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Yes, I kept trying to get dr.Jeckyll to understand it ......I already
showed him the formula, to no avail (at least, so far).

You were simply misunderstanding the problem and so coming up with the wrong
solution.

No, idiot. You simply don't understand aberration, that is all.
If you stopped talking and you started using math you would
understand. But since you avoid using math like the plague, you keep
repeating the same errors.



It helps when you are actually talking about the same problem as
everyone else (G, Harry, myself). As I said .. you were using the right
formulas but misapplying it (as far as the problem the rest of us were
talking about).

Again, no, idiot. The description of the problem in math terms is not
as ambigous as you keep making it to be.
Here it is, one more time, mr. Jackasss:

-In the traincar frame theta_car=pi/2
-In the track frame

cos(theta_track)=(cos(theta_car)-v/c)/(1-v/c*cos(theta_car))

So, can you calculate cos(theta_track)? I asked you 5 times, why are
you so shy about using a little math?

Since you don't get the math and you didn't get the "separation speed"
explanation, I will give you a third explanation: since in relativity
all frames are equivalent, instead of having the train moving Left to
Right with respect to the tracks, imagine that the tracks move Right
to Left while the light bounces vertically in the car frame. How is
the light inclined in the track frame? If you still don't get it, look
at these pictures:

http://www.fourmilab.ch/cship/aberration.html




.

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