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## Homework Statement

An antiproton of energy 35 ##GeV## from a source outside the solar system interacts with a proton in the upper atmosphere traveling on a trajectory which is radial with respect to the centre of the Earth. The antiproton annihilates the proton with the final outcome that two muons each of mass 106 ##MeV/c^2## are produced and nothing else. The interaction takes negligible time and occurs at a height of 20 ##km## above the Earth’s surface. For the case that both muons have the same energy, as measured from the Earth, what is the probability that they will both reach the surface? The lifetime of a muon is 2.2 ##μs##.

## Homework Equations

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$$γ=\frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}$$

$$E^2-p^2c^2=m_0^2c^4$$

$$p=γmv$$ and $$E=γmc^2$$

hence: $$p/E=v/c^2$$

$$KE=mc^2(γ-1)$$

And for later on in the question...

$$τ'=γτ$$

$$N/N_0=e^{-t/τ}$$

The associated probability that a particle survives/decays over a time t are: $$P(Survive)=e^{-t/τ}$$ $$P(Decay)=1 - e^{-t/τ}$$

## The Attempt at a Solution

##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2##

##E_{\bar p}^2 + m_p^2 + 2E_{\bar p}m_p - P_{\bar p}^2 = 4E^2##

As ##E_{\bar p}^2-P_{\bar p}^2=m_{{\bar p}/p}##

##2m_p^2+2E_{\bar p}m_p=4E^2##

##E=\sqrt {{\frac 1 2}(m_p^2+E_{\bar p}m_p)}## = 4106.15MeV

So each muon has energy 2053MeV

##KE=E - m_0=2053-106=1947MeV##

##KE=(γ-1)mc^2##

so ##γ=19.368##

In the muon's frame the 20km distance is Lorentz contracted,

##20km/19.368=1.0326km##

##t=d/s=1032.63/0.99866c=3.44914μs##

##P(Survive)=e^{-3.44914/2.2}=0.2085##

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Issues I have:

Firstly, I'm pretty sure that I've ignored some of the content of the question that should be included to get an appropriate answer, such as the proton being on a radial trajectory.

Secondly, I'm pretty sure that the two muons produced should be traveling at angles to the normal, and so would have flight paths of distances greater than 20km, however I am unsure how to account for this in my calculations.

Finally, I don't have a lot of faith in the methods I have used, so apologies in advance if they're just straight up wrong.

Thanks for any help, I really appreciate it,

Cait