1. Dec 3, 2014

### StonedPhysicist

• Member warned about lack of template
Here is the question:
If 1000 muons are incident from a height of 10 km above the earths surface, how much slower than the velocity of light must they be travelling in the rest frame of the Earths surface for 990 of them to be expected to arrive at the ground undecayed? The mean life of a muon is 2.2 microseconds. You can use equation: (gamma)(beta)=sqrt((gamma)^2+1) for (gamma) much greater than 1.

I cant seem to work out how to do this. So far I used the decay equation to find the time taken for the 1000 muons to decay to 990 muons as 2.21x10^-8 seconds but i am not sure where to go after this, help will be greatly appreciated.

2. Dec 3, 2014

### BvU

The 22.1 ns you found is in the rest frame of the muons. You need a Lorentz transform to the earth frame, where that time is 10 km / v

Must admit I don't understand the fun of using $\gamma\beta\approx \sqrt{\gamma^2+1}$ when I know that $\gamma\beta=\sqrt{\gamma^2-1}$

3. Dec 3, 2014

### StonedPhysicist

can you explain how to do the transformation? , i have the formula ct'= (gamma)(ct-(beta)x) but dont see how to do it

4. Dec 3, 2014

### BvU

Muons are at rest in their own frame, so x = 0. Makes it easy, doesn't it ?
More nice info here

5. Dec 3, 2014

### StonedPhysicist

using ct=(gamma)(ct'+(beta)x') subbing in x'=0 , t=10km/v i end up with: v=10/(gamma)t' BUT I do not know gamma? so cannot complete the calculation? or am i missing something?

6. Dec 3, 2014

### BvU

$\gamma = \sqrt{1\over {1-\beta^2}}\$ and $\ \beta = {v\over c}$

And the distance covered by the muons in the earth frame is a lot more than 10 m !

7. Dec 3, 2014

### StonedPhysicist

yes thankyou! finally i got the answer that the velocity of the muon is 66 m/s less than c (in the rest frame)!!! woo!!

8. Dec 3, 2014

### BvU

In the earth frame...

and, as you see, $\beta\gamma = \gamma$ for all practical purposes

Last edited: Dec 3, 2014
9. Nov 28, 2015

### Isaac Pepper

I follow the reasoning for this.
The radioactive decay equation gives us $N=N_0e^{\frac{-t'}{t_0}} \rightarrow t'=2.21*10^{-8}s$
This is the time in the rest frame of the muon.
Then using the Lorentz transformations : $t = \gamma t'$
$v=\frac{d}{t} \rightarrow v=\frac{d}{\gamma t'}$

How has he managed to find v from $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$?
I've tried many different rearrangements and haven't managed to pull v out of it, I must be doing something wrong.

$\gamma v = \frac{d}{t'} = \frac{10*10^3}{2.21*10^{-8}}$
...

EDIT:: $\frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{d}{t'}$

$v(1-\frac{v^2}{c^2})^{\frac{-1}{2}} = \frac{d}{t'}$
$v^{-2}(1-\frac{v^2}{c^2})=(\frac{d}{t'})^{-2}=\frac{t'^2}{d^2}$
$v^{-2}-\frac{1}{c^2}=\frac{t'^2}{d^2}$
$v^{-2}=\frac{t'^2}{d^2}+\frac{1}{c^2}$
Skipping a few steps of algebra...
Therefore $v=\sqrt{\frac{d'^2*c^2}{c^2t'^2+d'^2}} = 299999934.1$

And $c-v = 66 m.s^{-1}$
I have finally worked it out too!
Of course this is all assuming that that speed of light is $3*10^8 m.s^{-1}$ which it isn't... so I'll have to work something different out

Last edited: Nov 28, 2015