Question

Need help understanding how an answer was achieved: (my worked out answer is inside)
Particle physicists use particle track detectors to determine the lifetime of short-lived particles. A muon has a mean lifetime of 2.2 micro/seconds and makes a track 9.5 cm long before decaying into an electron and two neutrinos. What was the speed of the muon?

Answer 1

\[T_0 = 2.2 µs\]
\[L = 9.5cm\]

Using,
\[\beta = \frac{L*\sqrt{1-\beta^2}}{c*T_0}\]
Solving for
\[\beta\]
\[\beta=\frac{L}{\sqrt{L^2+c^2*T^2}} \]

You can neglect c for the mean time?

I get
\[\beta = 0.974218\]

However, the solution is supposedly
\[\beta = 1.4*10^{-4}\]

Answer 2

Question reminds me of the "Mt Washington expt/demo" done by MIT and printed in a book by AP FRENCH. I've had a look at this and am confused. You want the SPEED of the muon, but you quote an "answer" of 0.0001 no units for something (beta according to the post). I doubt that that's the SPEED of the muon, 'cos it's going fairly fast. An approximation based on LOW SPEED mechanics gives me 40, 000 m/s for the speed. This is well below c, and MAY make this a "classical" problem. Maybe the muon speed is much higher than that approx, but its a lot bigger for a SPEED numbe than is the "beta value" in the post.
http://perendis.webs.com

Answer 3

Yeah,

\[\beta = \frac{v}{c}\]

Solving for v will get you

\[v = 1.4*10^{-4}*c\]

Answer 4

At the END of the muon's journey it may very well be going at the lower speed. Problem seems to be to work backwards to what "caused" that particular speed. Being a high speed muon, special rel is implied ?

Answer 5

I'll show you the solution from the text book:

In the muon’s frame
\[T_0 =  2.2 μs \]
In the lab frame the time is longer; see Equation (2.19)*:
\[T' = T_0.\]
In the lab the distance traveled is
\[9.5cm = v*T' = v * \gamma * T_0 = \beta * c * \gamma * T_0 \]
since
\[v = \beta c.\]
Therefore,
\[\beta = \frac{9.5cm(\sqrt{1- \beta ^2})}{cT_0}, so\]
\[\beta = \frac{v}{c} =\frac{9.5cm(\sqrt{1- \beta ^2})}{c (2.2\mu s)}.\]
Now all quantities are known except
\[\beta .\]
Solving for \[\beta \] we find
\[\beta = 1.4*10^{-4}\]
or
\[ v = 1.4*10^{-4}c\]
*For more context this is how Equation 2.19 is defined in the text book:

\[T' = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma*T_0\]

Answer 6

Correction to line in the above comment:

"In the lab frame the time is longer; see Equation (2.19)*: \[T' = T_0\]"

The equation is incorrect I forgot to type the gamma.
It should be:

In the lab frame the time is longer; see Equation (2.19)*:
\[T' = \gamma * T_0\]