Found a way around relativity of simultaneity

[Ich]

left rope event occurs at (0.75, -1.25) in train frame and perceives at (12.5, -7.5). So, to travel (- 1.25 - 7.5) = 8.75 spatial distance with 0.53 speed requires 16.51 sec.

-7.5 - (-1.25) = -6.25

 

[Janus]

Ich has pointed out what I was going to. ( I had a longer reply ready but then lost it).

You transposed the distances traveled by the wave.

The left wave travels a distance of 6.25 ls at a speed of ~0.53c, which takes 11.75 sec (after correcting for the rounding error introduced by the inaccurate value for the speed.)

And in turn, the right wave travels 8.75 ls at a speed of ~0.66c, which takes 13.25 sec (after making the same rounding correction.)

Using the correct values you get the same answer as above. The right wave leaves at -0.75 sec, and the left leaves at 0.75 sec, and both reach the platform at 12.5 sec according to the rest frame of the train.

 

[mananvpanchal]

Ok, so that is the error. I am very sorry for that.

The calculation perfectly shows 0.75 for left event and -0.75 for right event.

I have checked the maths by putting watchmen in both frame. Both observer sees that platform ropes meet at platform observer simultaneously and both observer sees that train ropes meet at train observer unsimultaneously if the events is simultaneous in platform frame and unsimultaneous in train frame.

Thanks to all.