Relativity: disagreement between two results

[P3X-018]

I have this problem, which I think I can solve in 2 ways, but only 1 of them seems to be correct, but I don't understand why the other isn't correct. The problem is as follows:

"A bar with the rest length L_0 moves in its length direction with the velocity v through a lab. A particle moves along the same line but in opposite direction with the same speed v [velocity -v]. How long will it take the particle to pass the bar, [time] mearsured from the lab."

The easy way to do this, is by saying that if L is the length of the bar seen from the lab, then the time the particle will take to pass the bar will be

$$T = \frac{L}{2v} = \frac{L_0}{2v\gamma}$$
(this is also the result given in the solutions manual)

But if we look at the situation from the bars reference frame, then the particle will be moving with a velocity

$$v' = \frac{2v}{1+v^2/c^2}$$

So the time the particle will take to pass the bar seen from the bars intertial frame will be

$$T' = \frac{L_0}{v'} = \frac{L_0(1+v^2/c^2)}{2v}$$

This time T' seen from the lab most be

$$T = \gamma T' = \gamma(1+v^2/c^2)\frac{L_0}{2v}$$

But this last result ain't equal to the first one. Why is that so? Where did I make a mistake in my last calculations?

 

[Andrew Mason]

I have this problem, which I think I can solve in 2 ways, but only 1 of them seems to be correct, but I don't understand why the other isn't correct. The problem is as follows:

"A bar with the rest length L_0 moves in its length direction with the velocity v through a lab. A particle moves along the same line but in opposite direction with the same speed v [velocity -v]. How long will it take the particle to pass the bar, [time] mearsured from the lab."

The easy way to do this, is by saying that if L is the length of the bar seen from the lab, then the time the particle will take to pass the bar will be

$$T = \frac{L}{2v} = \frac{L_0}{2v\gamma}$$
(this is also the result given in the solutions manual)

But if we look at the situation from the bars reference frame, then the particle will be moving with a velocity

$$v' = \frac{2v}{1+v^2/c^2}$$

So the time the particle will take to pass the bar seen from the bars intertial frame will be

$$T' = \frac{L_0}{v'} = \frac{L_0(1+v^2/c^2)}{2v}$$

This time T' seen from the lab most be

$$T = \gamma T' = \gamma(1+v^2/c^2)\frac{L_0}{2v}$$

But this last result ain't equal to the first one. Why is that so? Where did I make a mistake in my last calculations?

No mistake. The observers in different inertial systems disagree on when the particle passes the bar. This is an illustration of the relative nature of simultanaeity.

AM

 

[Galileo]

I must say this kept me stumped for a few minutes, but the solution is simple.

You used the time dilation formula to find T ($$T=\gamma T'$$), but this you may not do here.
Remember how time dilation can be derived from the Lorentz transform:
$$\Delta t=\gamma(\Delta t'+\frac{v}{c^2}\Delta x')$$
For a clock stationary in S' we have $$\Delta x'=0$$, i.e. two subsequent ticks of the clock occur at the same location in S'. However, the positions of the events also play a role in determining the time interval between the events in S.

Let's say event 1 is that the particle crosses the front of the bar and event 2 is when it crosses the back end of the bar. Then $$\Delta x'=-L_0$$ and $$\Delta t'=\frac{L_0}{2v}(1+v^2/c^2)$$

Plug that into the Lorentz transform above and you get the same answer.