# Muon decay and Special Relativity

JackFlash

## Homework Statement

Cosmic muons are produced when protons in cosmic rays hit the atmosphere about 10km above us. How fast do muons have to travel in order to reach the Earth before decaying if they live 2.2μs before decaying? Consider the analysis for the rest frame of a) the Earth and b) the muon.

## Homework Equations

γ = $\frac{1}{\sqrt{1-β^{2}}}$ = $\frac{c}{\sqrt{c^{2}-v^{2}}}$
β = $\frac{v}{c}$
t' = γ(t - $\frac{βx}{c}$)

## The Attempt at a Solution

Really, I'd just like for someone to double check my work to see if I'm on the right track.
So, for a)
t' = 2.2μs
β = $\frac{x}{ct}$
γ = $\frac{1}{\sqrt{1-(x/ct)^{2}}}$ = $\frac{ct}{\sqrt{(ct)^{2}-x^{2}}}$

SPOILER shows me solving for t.
Scroll down to "EDIT:" for a neater method.

t' = $\frac{ct}{\sqrt{(ct)^{2}-x^{2}}}$(t - $\frac{x^{2}}{c^{2}t}$) = $\frac{c}{\sqrt{(ct)^{2}-x^{2}}}$($t^{2} - \frac{x^{2}}{c^{2}}$)

Mostly plugging in and subbing. I factored a 1/t out of the parentheses in that last step to cancel out the t in the numerator.

${\sqrt{(ct)^{2}-x^{2}}}$ * t' = ct$^{2}$ - $\frac{x^{2}}{c}$

Square both sides and use 2.2 for t'

4.84($c^{2}t^{2} - x^{2}) = c^{2}t^{4} - 2x^{2}t^{2} + \frac{x^{4}}{c^{2}}$

Then I solve for t. Using v = x/t, I can find the speed needed.

t' = γ(t - $\frac{βx}{c}$)
to
t' = γ(t + $\frac{βx}{c}$)

EDIT:
Alternatively, solving directly for v:
γ = $\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
t = x/v
t' = γ($\frac{x}{v} - \frac{vx}{c^{2}}$)
Divide both sides by γ, square both sides...
$t'^{2}(1-\frac{v^{2}}{c^{2}}) = (\frac{x}{v})^{2} - 2(\frac{x}{c})^{2} - (\frac{vx}{c^{2}})^{2}$

Some very creative factoring and canceling later...

$v^{4}(\frac{x}{c})^{2} - t'^{2}) + v^{2}(2x^{2} + (ct')^{2}) - (cx)^{2} = 0$

Getting Mathematica to solve for me, where:
$t' = 2.2*10^{-6}$
$x = 1*10^{4}$
$c = 3*10^{8}$

Gives three imaginary values and:
$v -> 1.92991*10^{8}$ for part a)

I assume part b) would be just a sign change for velocity when solving?

Last edited:

bossman27
I think you made this a bit harder than need be. Keep in mind that there are essentially two "observers" moving towards each other, thus the velocity from the earth rest frame and muon rest frame should agree (even though the muon "thinks" the earth is moving). Two objects moving towards or away from each other will each attribute the same velocity to the other object. I will use "e" and "u" subscripts to denote the particular frame rather than primes, since it's just easier to keep track of that way:

Known variables:
$d_{e} = 10km$
$t_{u} = 2.2\mu s$
due to time dilation: $t_{e} = \gamma t_{u}$
due to length contraction: $d_{u} = \frac{d_{e}}{\gamma}$

From earth's rest frame:
$\frac{d_{e}}{t_{e}} = \frac{10km}{\gamma 2.2\mu s} = v_{e}$

for the sake of not having to keep typing it out, I'm going to define $R = \frac{10km}{2.2\mu s} = 4.\overline{54}\times 10^{9} \frac{m}{s}$

Then we simplify:
$R^{2} = \frac{v_{e}^{2}}{1 - \beta ^2} \Rightarrow (1 - B^{2})R^{2} = v_{e}^{2}$
Latex is being difficult so I can't type the final equation, but the second to last step is: $v_{e}^{2} = \frac{R^{2}}{1 - (\frac{R}{C})^{2}}$

We end up with $v_{e} = 2.99349 \times 10^{8} \frac{m}{s}$ which sounds about right because atmospheric muons normally travel at around $\beta \approx .994$

In the Muon's Frame:
$\frac{d_{u}}{t_{u}} = \frac{\frac{10km}{\gamma}}{2.2\mu s} = v_{u}$

you can see that this will reduce to the same equation as before, thus:
$v_{e} = v_{u}$ (even though strictly speaking the velocity in the muon's rest frame is defined qualitatively as the velocity of earth's surface)