# Simple special relativity, length from duration g

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1. Jul 19, 2014

### CuriousParrot

I'm thinking about a very basic scenario in special relativity, but I've got something backwards, and I could use help understanding why it's wrong.

Suppose a rod R is moving past an observer O at speed v.

The observer wants to know how long the rod is, so he starts his stopwatch as soon as the front of the rod reaches him, and stops his watch as soon as the end of the rod passes him by.

The observer measures To seconds. Knowing the rod is moving at v, he calculates its length as he sees it, Lo, as: $L_o = v T_o$

From the rod's perspective, the rod is sitting still and the observer is moving. If the rod measured the same way, starting a stopwatch as soon as the front edge reached the observer and stopping when the rear edge passed the observer, the rod would say it took Tr seconds to pass, making the rod's length $L_r = v T_r$

The observer should find the duration of the crossing to be inflated by gamma, $T_o = \gamma T_r$

Substituting yields: $L_o = \gamma L_r$

... that's not right, though. The time should increase by gamma, but the length should decrease by gamma. But length and time are proportional to each other.

What's wrong here?

Last edited: Jul 19, 2014
2. Jul 19, 2014

### TSny

What is the reason behind this statement?

Are you familiar with the term "proper time" between two events? Which of $T_o$ and $T_r$ is the proper time?

3. Jul 19, 2014

### CuriousParrot

Suppose there was a light clock on the rod (bouncing a beam of light back and forth, perpendicular to the direction of motion between the rod and the observer.) In the rod's frame of reference, each tick of the clock takes $T_r$ seconds. But our observer, O, would see a longer path-length for those bounces, due to $v$, and would measure a time-between-bounce of $T_o > T_r$, with the dilation factor gamma.

I think that also describes the situation with the rod flying past the observer. Perhaps the clock started a tick just as the front of the rod met the observer, and finished a tick just as the rear of the rod passed the observer. R and O should agree that exactly one tick happened, but they disagree about how long a tick takes.

I thought $T_r$ was the proper time, being the time measured in the frame of the moving object.

Even if I had that backwards, the reverse conclusion would be that $\gamma T_o = T_r$ and hence $\gamma L_o = L_r$, hence both time and length would decrease in frame O relative to frame R, which still isn't right.

4. Jul 19, 2014

### TSny

That's correct. But this is not too relevant to your problem. A light clock attached to a point of the rod measures time between when a light signal leaves the point of the rod and when the light signal returns to the same point of the rod. But in your problem, you are considering the time between an event that occurs at one of the rod and another event at the other end of the rod.

You have an observer O who watches the rod pass a fixed point in the O-frame. There is the event when the front of the rod pass the fixed point and there is the event when the back of the rod passes the fixed point. These two events occur at the same place in the O-frame. So, you could have a light clock in the O-frame oriented perpendicular to the motion of the rod such that the light leaves the lower end of the clock when the front of the rod passes the fixed O-point and the light returns to the lower end of the clock when the back of the rod passes the fixed O-point. From the rod's point of view, the light clock in the O-frame is in motion.

For which frame is the time between ticks of this clock the greatest?

The rod and observer have relative motion, but you can't say which is "actually" moving. So, you can't define proper time as the time measured in the frame of the "moving object".

The inertial reference frame that measures the proper time between two events is the frame for which the two events occur at the same place.

Reconsider whether this is right or not after thinking more about which time is the "proper" time.

5. Jul 20, 2014

### CuriousParrot

Thank you, I appreciate your response.

I think I understand where I was getting confused:

The proper length is a property of objects, not events, and belongs to the object's rest frame: $L_r = \gamma L_o$.

The proper time between the two events belongs to the frame where they are not separated in space: $T_o = T_r / \gamma$.

I was incorrectly thinking that the proper length and proper time should necessarily come from the same frame.

Last edited: Jul 20, 2014
6. Jul 20, 2014

### TSny

OK. Sounds like you've got it. Good work.