The elementary particle known as the positive kaon `(K^(+))` is unstable in that it can decay (transform) into other particles. Although the decay occurs randomly, we find that, on average, a positive kaon has a lifetime of 0.1237 `mus` when stationary-that is, when the lifetime is measured in the rest frame of the kaon. If a positive kaon has a speed of 0.990c relative to a laboratory reference frame when the kaon is produced, how far can it travel in that frame during its lifetime according to classical physics (which is a reasonable approximation for speeds much less than c) and according to special relativity (which is correct for all physically possible speeds)?

Explanation

KEY IDEAS
1. We have two (inertial) reference frames, one attached to the kaon and the other attached to the laboratory.
2. This problem also involves two events: the start of the kaon.s travel (when the kaon is produced) and the end of that travel (at the end of the kaon.s lifetime).
3. The distance traveled by the kaon between those two events is related to its speed v and the time interval for the travel by
`v=("distance")/("time interval")" "`(36-10)
With these ideas in mind, let us solve for the distance first with classical physics and then with special relativity.
Classical physics: In classical physics we would find the same distance and time interval whether we measured them from the kaon frame or frame or from the laboratory frame. Thus, we need not be careful about the frame in which the measurements are made. To find the kaon.s travel distance `d_(cp)` according to classical physics, we first rewrite Eq. 36-10 as
`d_(cp)=v Deltat, " "`(36-11)
where `Deltat` is the time interval between the two events in either frame. Then, substituting 0.990c for v and 0.1237 `mus` for `Deltat` in Eq. 36-11, we find
`d_(cp)=(0.990c)Deltat`
`=(0.990)(299 458 m//s) (0.1237 xx 10^(-6)s)`
`=36.7m`.
This is how far the kaon would travel if classical physics were correct at speeds close to c.
Special relativity: In special relativity we must be very careful that both the distance and the time interval in Eq. 36-10 are measured in the same reference frame - especially when the speed is close to c, as here. Thus, to find the actual travel distance `d_(sr)` of the kaon as measured from the laboratory frame and according to special relativity, we rewrite Eq. 36-10 as
`d_(sr)=v Deltat, " "`(36-12)
where `Deltat` is the time interval between the two events as measured from the laboratory frame.
Before we can evaluate `d_(sr)` in Eq. 36-12, we must find `Deltat`. The 0.1237 `mus` time interval is a proper time because the two events occur at the same location in the kaon frame - namely, at the kaon itself. Therefore, let `Deltat_(0)` represent this proper time interval. Then we can use Eq. 36-9 `(Deltat=gamma Deltat_(0))` for time dilation to find the time interval `Deltat` as measured from the laboratory frame. Using Eq. 36-8 to substitute for `gamma` in Eq. 36-9 leads to
`Deltat=(Deltat_(0))/(sqrt(1-(v//c)^(2)))=8.769xx10^(-7)s`.
This is about seven times longer than the kaon.s proper lifetime. That is, the kaon.s lifetime is about seven times longer in the laboratory than in its own frame - the kaon.s lifetime is dilated. We can now evaluate Eq. 36-12 for the travel distance `d_(sr)` in the laboratory frame as
`d_(sr)=v Deltat=(0.990c)Deltat`
`=(0.990)(299 792 458 m//s) (8.769xx 10^(-7)s)`
`=260m`.
This is about seven times `d_(cp)`. Experiments like the one outlined here, which verify special relativity, became routine in physics laboratories decades ago. The engineering design and the construction of any scientific or medical facility that employs highspeed particles must take relativity into account.