# Homework Help: Need Help Choosing Proper Time

1. Aug 30, 2011

1. The problem statement, all variables and given/known data
How fast must a meter stick be moving if its length is observed to shrink to 0.5 m?

3. 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'

L' = Lp*y (WRONG)

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

L' = Lp/y (RIGHT)

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

2. Aug 30, 2011

### edgepflow

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

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

3. Aug 30, 2011

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

4. Aug 30, 2011

### SteamKing

Staff Emeritus
That's not what you said in the OP.

5. Aug 30, 2011

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.

6. Aug 31, 2011

### vela

Staff Emeritus
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.

7. Aug 31, 2011