Hey guys,
So I know this is a stupidly easy problem, but when I did the practice test I got hung up, I'm wondering if someone can resolve this:
The question goes: A meter stick with a speed of 0.8c moves past an observer. In the observer’s reference frame, how long does it take the stick to pass the observer ?
So I calculate gamma = 5/3. Here's where I got messed:
I figured there are two ways to approach the time, one way is to do (where T and L = proper time/length, t, l are not):
v = l/T = l(gamma)/t, since t = T(gamma). If you plug and chug with this though, you get the wrong answer.
If you use v = L/t = l/[(gamma)t], you get the right answer, since gamma is in the denominator here but in the numerator with the other equation. What makes you decide to use this one over the other one?? Is there something I'm messing up and they should yield the same answer??
Thanks,
Ari
0877, Problem 24

 Posts: 80
 Joined: Tue Oct 12, 2010 8:00 am
Re: 0877, Problem 24
It's hard to follow your equations. Please, rewrite them using latex
Nevertheless, the answer is $$\frac{\text{the length of the stick in the observer's frame}}{\text{the speed of the stick in the oberver's frame}} = \frac{l_0 /\gamma}{v} \equiv \frac{l_0}{\gamma v}$$. All you have to do is to convert the proper length of the stick to its length in the observer's frame.
Details: http://physicsworks.wordpress.com/2011/10/10/24/
Nevertheless, the answer is $$\frac{\text{the length of the stick in the observer's frame}}{\text{the speed of the stick in the oberver's frame}} = \frac{l_0 /\gamma}{v} \equiv \frac{l_0}{\gamma v}$$. All you have to do is to convert the proper length of the stick to its length in the observer's frame.
Details: http://physicsworks.wordpress.com/2011/10/10/24/