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An explanation of special relativity I'm struggling with, goes like this:

A rod traveling by a "stationary" observer has its length measured by use of two stationary synchronized clocks (synchronized from the perspective of the observer). The resulting length will appear "shorter" than if the rod were stationary and measured more conventionally.

I don't even understand the premise. How would two stationary clocks be used to measure the length of a moving rod? Instead I can imagine using a single clock with a "split" metric. That is, capture time when the head of the rod passes the clock, and capture the time when the tail of the rod passes. If the velocity is known, the length can be determined.

But my formulation only requires one clock, and isn't pertinent to the point of the original presentation. Can someone explain to me how two stationary "synchronized" clocks can be used to measure the length of a rod in motion?

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If you are struggling with relativity, a little bit of space-time geometry will make it completely intuitive, to the point where it is no more mysterious than Euclid's geometry (although half the psychological effect is to make Euclid's geometry more mysterious): physics.stackexchange.com/questions/12435/… –  Ron Maimon Dec 18 '11 at 10:27

1 Answer 1

You need two clocks if you don't know the velocity of the object a priori. Using a single clock will tell you only the time it takes to get past a particular point (that is, the rod's length divided by its velocity).

If you have two clocks with a known distance between them, you can determine the velocity (given by the time difference between arrival at the first and second clock), which you can then use to find the length.

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Ok, I see. Now I know the velocity. But I don't see how I can find the length, unless I take two time measurements at one of the clocks (as the head and tail of the rod passes by). Is that the idea, or no? –  Brent Arias Oct 19 '11 at 6:45
    
The point is that you know the velocity. Your goal is to measure the length –  valdo Oct 19 '11 at 7:29
    
Yes, you would register the time of the head passing and the time of the tail passing for both clocks. –  Adam Strandberg Oct 19 '11 at 7:50
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True, but... if you can be assured constant (though unknown) velocity, you only need make three measurements. The fourth value can be derived from the other three if you like, but it is unnecessary. –  AdamRedwine Oct 19 '11 at 12:09

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