what will be the events relative to the ground observer , a time-like or space like?

nitsuj

lol, yea, not quite "only a matter of".

To your point of the "common ancestor", I think you're missing the significance of this with regard to FTL information "transmission". If I give you a box, And you travel away from me to the other side of the world, and you open the box and find half a banana and deduce I have the other half that is not FTL transfer of information...

Adel Makram

Really I don`t. I am making up my mind but a bit slowly

Adel Makram

Firstly, I do not need it from the point of view of the train observer. But mathematically, I need a correlative tool with the ground frame so as to compare what will be the expected difference in time between A&B with an expected difference if light would have been emitted from A to B as a function of v! Exactly like the derivation of LT. so if every observer observes sequences related to light transmission from his frame point of view no LT would had ever come!

Secondly, this math could be considered as follow: The ground observer calculate what the train observer would see. So he must includes v

If the flash occurs at A, the light takes the same time to reach B that the source takes to reach the mid-point. In this case the v/c=0.5

There are 2 train observers at A and B
But there is just one ground observer to observe all events from the point where midpoint of the train coincides with the source

Adel Makram

Yes I guess that if all calculations would be done by the ground observer expecting what the train observer would see, this can also solve the problem raised up by Michel in my calculation that the events of receiving lights by A and the source reaching the midpoint of train are not simultaneous relative to the train observer but they are relative to the ground one!

Michael C

It doesn't! Your results for t`<sub>a</sub> and t`_b are incorrect.

There's a simple relation between t`a and t`b which you can use to check your result. I'll label with "C" the place on the train from where the light flash starts. It's clear that the distances add up like this:

[itex]CA`+ CB` = AB`[/itex]

[itex]CA`[/itex] is [itex]t`_{a}c[/itex] and [itex]CB`[/itex] is [itex]t`_{b}c[/itex], so:

[itex]t`_{a}c + t`_{b}c = AB`[/itex]

and therefore:

[itex]t`_{a} + t`_{b} = \frac{AB`}{c}[/itex]

Since [itex]t`_{a}[/itex] and [itex]t`_{b}[/itex] are both positive (light isn't going backwards in time!), it's clear that their difference can never exceed [itex]\frac{AB`}{c}[/itex].