[Jaaanosik (OP)]

OK, no frame dragging, no v^2.
c^2 = (c/2)^2
c!=c
How do you solve this?
Jano

[Kryptid]

OK, no frame dragging, no v^2.
c^2 = (c/2)^2
c!=c
How do you solve this?
Jano

Since you admit that you are no longer talking about frame-dragging, please put that in the appropriate thread ("Is relativity reciprocal?" seems like a good one).

[Jaaanosik (OP)]

...
As Kryptid has also said, No, this is not frame dragging.  Frame dragging is a GR effect that involves a rotating mass.
It has has nothing to do with your example.  In fact, your example isn't even a good one for aberration.

I am aware of the GR frame dragging.
I am presenting an argument that the frame dragging exists in SR as well.
The light moves up on the train car but it is 'dragged' to the right from platform point of view.
The aberration stands. It is a good example.
If the platform observer wants to hit the same spot where the red arrow hits the train middle point on the other side then the platform observer cannot shoot at 90 degrees.
cos(theta) = v/c = 0.866 ===> theta=30deg
The simplified formula,
Jano

[Jaaanosik (OP)]

Since you admit that you are no longer talking about frame-dragging, please put that in the appropriate thread ("Is relativity reciprocal?" seems like a good one).

How do we solve this?

c^2 = (c/2)^2
c!=c

[Kryptid]

The light moves up on the train car but it is 'dragged' to the right from platform point of view.

For the last time, what you are talking about is not frame-dragging. Did you even watch the video I posted?

Since you missed it the first time...

How do we solve this?

c^2 = (c/2)^2
c!=c

Did you not just read what I posted about putting this in the appropriate thread?