- #1

JackFlash

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## 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 = [itex]\frac{1}{\sqrt{1-β^{2}}}[/itex] = [itex]\frac{c}{\sqrt{c^{2}-v^{2}}}[/itex]

β = [itex]\frac{v}{c}[/itex]

t' = γ(t - [itex]\frac{βx}{c}[/itex])

## 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 = [itex]\frac{x}{ct}[/itex]

γ = [itex]\frac{1}{\sqrt{1-(x/ct)^{2}}}[/itex] = [itex]\frac{ct}{\sqrt{(ct)^{2}-x^{2}}}[/itex]

SPOILER shows me solving for t.

**Scroll down to "EDIT:" for a neater method.**

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.

[itex]{\sqrt{(ct)^{2}-x^{2}}}[/itex] * t' = ct[itex]^{2}[/itex] - [itex]\frac{x^{2}}{c}[/itex]

Square both sides and use 2.2 for t'

4.84([itex]c^{2}t^{2} - x^{2}) = c^{2}t^{4} - 2x^{2}t^{2} + \frac{x^{4}}{c^{2}} [/itex]

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

For part b), I'd just change the sign in:

t' = γ(t - [itex]\frac{βx}{c}[/itex])

to

t' = γ(t + [itex]\frac{βx}{c}[/itex])

**EDIT:**

Alternatively, solving directly for v:

γ = [itex]\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

t = x/v

t' = γ([itex]\frac{x}{v} - \frac{vx}{c^{2}}[/itex])

Divide both sides by γ, square both sides...

[itex]t'^{2}(1-\frac{v^{2}}{c^{2}}) = (\frac{x}{v})^{2} - 2(\frac{x}{c})^{2} - (\frac{vx}{c^{2}})^{2} [/itex]

Some very creative factoring and canceling later...

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

Getting Mathematica to solve for me, where:

[itex]t' = 2.2*10^{-6}[/itex]

[itex]x = 1*10^{4}[/itex]

[itex]c = 3*10^{8}[/itex]

Gives three imaginary values and:

[itex]v -> 1.92991*10^{8}[/itex] for part a)

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

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