Where Is the Mistake in the Relativity Problem with Moving Rods?

[alvaros]

Referring to the attached drawing:
Two rods 3m long. Rod A is moving at 0,866 c from right to left.
The pictures are taken from the point of view of rod B.

( 1 )In 1 a ray of ligth reaches the end of both rods. It happens that the reading of the clocks at rod A and rod B is 0.
Because v = 0.866 *c -> l´= l * 0.5 , that's why the left end of rod A is at the middle point of rod B.

( 2 )In 2 after 10 ns ( at clocks on rod B ) the light reaches the left end of rod B, assuming c = 30 cm/ns.
Rod A has moved 0.866 * c * 30 ns/cm = 260 cm.

But, because t'= t *0.5, the reading on clocks at rod A is 5 ns. So the light has moved, in the frame of reference of A, 150 cm -> the light must be at the middle of rod A, and it isn't according to the explained in ( 2 ). Where is the mistake ?

 

[Dale]

Where is the mistake ?

You have the wrong formula for t'. The correct formula is:

t' = γ(t-vx/c²)

Your formula neglects the relativity of simultaneity.

 

[alvaros]

Thank you for the answer. I made a google search and I found your formula, but I don't know how to apply it ( what does x mean ? )

Could you resolve the problem with numbers ?
Is the drawing 1 correct ? ( the left end of rod A is at the middle of rod B ? )
Where is the right end of rod A when the ligth reaches the left end or rod B ?