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I'm having trouble reconciling the quantitative and conceptual aspects of length contraction. This example is taken out of a book:

Say a particle is moving toward us at 0.99c, relative to us. If at rest, this particle typically decays within 0.2 microseconds, then we will measure 1.4 microseconds (I understand this part). Then, relative to us, the particle travels 1.4 * 0.99c ≈ 420 meters. One way to find the distance relative to the particle is to simply do: 0.2 * 0.99c ≈ 60 meters. This doesn't make sense to me. Shouldn't what "we" measure be the smaller length? For this situation to mathematically work out we have to do L = Lo/γ, where L = 60 and Lo = 420. Shouldn't it be the other way around? Isn't the distance measured in the rest frame 60 (the smaller value), not 420?

This is not a homework question. I'm just reviewing these concepts and was stumped by this one.

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I am not sure why you are having the confusion. Your calculations are correct. The distance you measure in your reference frame are the actual distances, so you would measure 420 meters. The "clock" for the particle in your reference frame would be running slower so the particle would decay in 1.4 microsecond in your reference frame.

In the particle's reference frame the clock would show 0.2 microseconds for the decay since there is no time dialation for a clock traveling with the particle in the particle's reference frame. However, the particle would see the distance it is traveling as 60 meters. The distances in your reference frame would appear contracted in the particle's reference frame. Therefore, the particle would travel 60 meters in 0.2 microseconds and decay in the particle's reference frame.

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  • $\begingroup$ I thought that length contraction would cause "my" measurement of the length to be smaller than the particle's measurement of itself. Is this incorrect? If so, then is the Lo in L = Lo/γ the length relative to the non-moving reference frame? $\endgroup$ – uncreative May 21 '16 at 2:09
  • $\begingroup$ L= Lo/y is the length the other reference frame is measuring. Lo is what you are measuring. $\endgroup$ – Peter R May 21 '16 at 3:54
  • $\begingroup$ In this example (phy.olemiss.edu/HEP/QuarkNet/length.html) it seems the other way around though. How to explain this? Also, shouldn't the observer find the distance that the particle moves to be shorter than the length the particle itself measures? $\endgroup$ – uncreative May 21 '16 at 4:03
  • $\begingroup$ I have a feeling the issue is that the endpoints of the path of the particle are not associated with events happening at the same time. I wonder what would happen if you Lorentz transformed using two events that are simultaneous in the particle's frame, like the instant of decay and another 60 meters "back" but at that same 0.2 microsecond time coordinate. (I haven't taken the time to seriously think this through yet, so I may be a bit off track...) $\endgroup$ – user55515 May 21 '16 at 4:56
  • $\begingroup$ Hm, good point w/ the difference between instantaneous measurements over time...I didn't realize the difference. Not sure what you mean by looking at an even in the particle's frame that is 60 meters "back"? $\endgroup$ – uncreative May 21 '16 at 7:27

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