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

"Dono" <sa_ge@xxxxxxxxxxx> wrote in message
news:1182577727.883469.143760@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Jun 22, 5:16 pm, "Jeckyl" <n...@xxxxxxxxxxx> wrote:
"Dono" <s...@xxxxxxxxxxx> wrote in message
news:1182552473.841010.125410@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I have already done all that in the posts you ignored. But, just to keep
you
amused...

1. The distance the center of the mirror moved to the right by the
time the light hit the ceiling

vt, where t = gamma.(h/c) = 1/sqrt(1-v^2/c^2).h/c = h/sqrt(c^2-v^2)
vt = vh/sqrt(c^2-v^2) = h/sqrt(c^2/v^2-1)
Correct.
so the aberration angle is a = atan(h/vt) = atan(sqrt(c^2/v^2-1))
No, you need to get the aberration angle independent from the movement
of the mirror, you are simply rigging the results in order to show
that the two distances are the same.

That is what the aberration is .. you are again showing your lack of
understanding.

2. The distance the light is aberrated to the right.
aberration is an angle, not a distance.

I asked you for the distance, i.e. where does the light intercept the
ceiling.

We know that .. it is vt

So, you need to calculate a distance

I did

and to show that it is
equal to the one at point 1. You are simply cheating at this point,
and you are getting caught.

The angle is acos(v/c).

How do you know that ?

I used the math. I've shown you the working previously.

The problem is that you don't understand the math because you do not
understand what aberration actually is.

When you learn what aberration is, come back and ask again.



.

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