# Homework Help: Problem, where is the mistake ?

1. Nov 9, 2007

### 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 , thats 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 isnt according to the explained in ( 2 ). Where is the mistake ?

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• ###### Dibujo.bmp
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Last edited: Nov 9, 2007
2. Nov 9, 2007

### Staff: Mentor

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

t' = γ(t-vx/c²)

Your formula neglects the relativity of simultaneity.

3. Nov 10, 2007

### alvaros

Thank you for the answer. I made a google search and I found your formula, but I dont 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 ?