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A particle, traveling at $0.5c$ relative to a stationary observer, travels $3.95 \rm ~cm$ in its frame of reference.

What is the distance the particle travel in the observer's frame of reference?

Since $3.95 cm$ is the proper length (the distance it travels in its rest frame (which has to be its own frame since in its own frame it is at rest) is proper length by definition) thus the distance measured in the laboratory is $$ L_O = \frac{L_P}{\gamma}$$ where $\gamma = \frac{2}{\sqrt3}$

Thus $L_O = 3.42\rm ~cm$

But isn't it contradictory since the measured length in the laboratory should be higher?

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mathnoob123
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2 Answers2

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There is a contradiction in the question because in the particle's rest frame it is at rest by definition. So in its rest frame the particle doesn't travel any distance and can't travel $3.95$ cm.

I suspect what the question means is that in the particle's rest frame the rest of the world travels $3.95$ cm towards it, or to put it another way the particle lifetime is:

$$ t = \frac{0.0395}{0.5c} \approx 0.263 \,\text{ns} $$

The task is then to work out how far the particle travels in the lab frame. The risky way to do this is to start throwing factors of gamma around i.e. to argue that the particle lifetime as measured in the lab is $t = \gamma t$ and the distance travelled in the lab is then:

$$ d_\text{lab} = 0.5 c t' = 0.5 c \gamma t = \gamma \times 0.0395 \approx 0.0456 \,\text{m} $$

The safe way to do the calculation is to define the two spacetime events marking the creation then decay of the particle, us e the Lorentz transformations to transform them into the lab frame then find the distance travelled. Thuis is done in detail in my answer to The real meaning of time dilation.

John Rennie
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Special relativity predicts contraction so there is no contradiction.

"Length contraction is the phenomenon that a moving object's length is measured to be shorter than its proper length ..." https://en.wikipedia.org/wiki/Length_contraction

my2cts
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