How Fast Must a Meter Stick Move to Appear Half Its Length?

[alksjadf]

Homework Statement

How fast must a meter stick be moving if its length is observed to shrink to 0.5 m?

The Attempt at a Solution

A meter stick moving at speed V to a stationary observer is the same thing as a stationary meter stick and an observer moving at speed V.

Lproper = Lp = 1 meter
Lobserved = L' = .5 meter
y = 1/sqrt(1- v^2/c^2)

Time it takes for observer to cover the distance of 1 meter, /\t:
/\t = Lp/V

Length moving observer sees, L':
L' = V*/\t'
where /\t' is the time it takes for the meter stick to fly by in observer's reference frame

Choosing Proper Time:
If I choose proper time to be /\t then:
/\t*y = /\t'

Through substitution this leads to:
L' = Lp*y (WRONG)

If i choose proper time to be /\t' then:
/\t = /\t'*y

Through substitution this leads to:
L' = Lp/y (RIGHT)

Why can't I choose the wrong one? What makes the other one right?

 

[edgepflow]

I would just start with:

L = L$_{}o$$\sqrt{}1-(v / c)^2$

and solve for v noting L / Lo = 1/2

 

[alksjadf]

Well sure I can just memorize formulas but I want to understand where they come from.

 

[SteamKing]

That's not what you said in the OP.

 

[alksjadf]

Sure it is. By doing it the right way, the formula is derived and by doing the wrong way, it isn't. I want to know why the right way is right and wrong way is wrong.

 

[vela]

This is an important detail that many students often overlook about time dilation. The time dilation formula $\Delta t = \gamma \Delta t'$ applies only when the clock is at rest in the primed frame, that is, when $\Delta t'$ is the proper time between two events.

To measure $\Delta t'$, the observer starts the timer when the front end of the meter stick passes by and stops the timer when the back end passes. Because the observer is at rest in his frame, $\Delta t'$ is the proper time between the two events, and the time dilation formula can be used to find the time elapsed between the same two events in the meter stick's rest frame. In the meter stick's rest frame, however, the two events are separated in both space and time, so $\Delta t$ isn't the proper time. Instead, you would calculate the proper time $\Delta\tau$ using $(c\Delta \tau)^2 = (c\Delta t)^2 - \Delta x^2$.

Are you familiar with the Lorentz transformations and space-time diagrams? I find it a lot clearer what's going on when I analyze the problem using those.

 

[alksjadf]

Yeah, I think I'm beginning to understand it better. And no, I haven't learned the Lorentz transformations yet, I will be learning it next lecture. Hopefully that will clear things up even more. Thanks.