Adventures in Parallel Dimension #1729 I

Relevant wiki: Special Relativity

This problem deals with the Lorentz Contraction and Time Dilation concepts from Special Relativity :

The Biggs Hoson's lifespan at rest, $t_{0}$ , is the same thing as the lifespan of the particle traveling at $v_{0}$ from the particle's frame of reference. Similarly, the measured distance $d$ that the Biggs Hoson travels is from the frame of the scientists taking the data.

Let's define $t_{0}'$ to be the lifespan of the Biggs Hoson from the frame of the scientists and $d'$ to be the distance traveled by the Biggs Hoson during its lifespan in the frame of the particle.

Then $t_{0}'=t_{0} \gamma$ and $d'=\dfrac{d}{\gamma}$ for $\gamma=\dfrac{c}{\sqrt{c^{2}-v^{2}}}.$

If we calculate average velocity in the frame of the scientists then:

$v_{0}= \frac{d}{t_{0}'}=\frac{d}{t_{0} \gamma}=\frac{d \sqrt{c^{2}-v_{0}^{2}}}{t_{0}c} \Rightarrow v_{0}^{2}t_{0}^{2}c^{2}=d^{2}c^{2}-d^{2}v_{0}^{2} \Rightarrow v_{0}=\frac{dc}{\sqrt{c^{2}t_{0}^{2}+d^{2}}}.$

It follows that $v_{0}c^{-1} |_{t_{0}=10^{-9}s, \: d=1m} \approx \boxed{.958}.$

Note that it would not matter whose frame we calculate average velocity from, which is to be expected. By doing it from the Biggs Hoson's perspective distance is $d'$ and time is $t_{0}$ , which would give rise to the exact same equation.