meopemuk said:
I don't see any contradiction here. You, basically, need to solve a system of two equations. Denoting d=24mm the traveled distance, t_0= 8.01x10^-11 sec the lifetime at rest, and
t = t_0 (1 - v^2/c^2)^{-1/2}
the lifetime of the moving particle, we can find the velocity from equation
d = vt
This velocity should be lower than the speed of light, of course.
Eugene.
Isn't t_0 the rest lifetime of the particle? It's the observed time in the lab-frame that should be longer (\gamma > 1). The difficulty is that, in the expression t = t_0 (1 - v^2/c^2)^{-1/2}, it is t_0 that equals 82 picoseconds, while the lab-frame time is about 80 picoseconds.
The muon illustration is representative. Its rest lifetime is 2.2 microseconds; if it were to travel at the speed of light, it would decay in a distance of 0.66 km. The fact that muons from high-altitude cosmic ray showers are detected on the ground is taken to indicate the effect of time dilation, in which \gamma is in the range of at least 50 to 100, giving it an Earth-frame lifetime sufficient to travel some tens of kilometers, in accordance with d ~ ct. In the rest frame of the muon, it is the distance from its point of creation to the ground that appears contracted by a factor of at least 50 to 100, so it only needs to cover a fraction of a kilometer to the apparently approaching ground, again in accordance with d ~ ct.
If the omega were observed
in the lab frame to have a lifetime of 82 picoseconds and covers 24 mm in that time, then its lab-frame velocity would be 0.9763 c, giving
\gamma = 4.617 . But isn't it the moving omega which has the "dilated" lifetime? We would then find a rest lifetime of 82/4.617 = 17.8 picoseconds. If it's the rest lifetime that is "dilated", we should see a lifetime of (4.617)·82 = 379 picoseconds for the lab-frame lifetime and the track would be
(379 picoseconds)(0.9763)(2.998x10^10 cm/sec) = 11.1 cm.
I think there is something in the problem statement that is not right. (The lifetime issue would be fine if this were the charmed-omega, but the rest energy is wrong.)
(After some further thought...) I finally decided to break down and "reverse engineer" the given answer. When you find gamma from the ratio total energy/rest energy, solve for v, time-dilate the rest lifetime, and now calculate the track length using
d = v · ( \gammat_0 ) , you get 23.87 mm.
SO, the issue appears to be that the precision of the track length is not given at a sufficiently high level to resolve the needed values accurately. The observed track length is too close to d = ct_0 to get a clear result at low precision.
To sum up, t_0 in the dilation equation is the
rest lifetime and the method meopemuk describes is otherwise correct in principle. However, the measurement given for the track length needs to be given to at least three significant figures (four would be better) in order to find v with enough precision to solve the problem.
