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Muon production and decay from relativistic annihilation

  • Thread starter caitphys
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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 travelling 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


[/B]
$$γ=\frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}$$
$$E^2-p^2c^2=m_0^2c^4$$
$$p=γmv$$ 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##

-----

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 travelling 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
 

Answers and Replies

  • #2
TSny
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Welcome to PF!

Homework Statement



An antiproton of energy 35 ##GeV## from a source ....
When I see statements like this, I'm not sure whether or not the energy given is the kinetic energy of the particle or the total energy (including rest mass energy). In this problem, it probably won't matter much since the rest mass energy of an antiproton is pretty small compared to 35 Gev.

3. The Attempt at a Solution
##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2##
This doesn't look right to me. Can you explain in words what this equation represents?

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.
This is related to your second question below.

Secondly, I'm pretty sure that the two muons produced should be travelling 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.
What can you get from conservation of linear momentum?
 
  • #3
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Thanks for the response!

This doesn't look right to me. Can you explain in words what this equation represents?
Okay, so my thinking for ##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2## was that using ##A^2=(\sum\nolimits E)^2-(\sum\nolimits P)^2##,

Choosing my frame as that of the proton, since this seemed to minimise the impact of my confusion associated with its trajectory, for ##A^2## the first term is the energy of the antiproton and the rest mass of the proton, and the second just the momentum of the antiproton, given that in its rest frame the momentum of the proton shall be zero. Then, since ##A^2## is invariant, working in the ZMF frame after collision, as the muons have equal energies, the value of ##A^2## is merely the total energy squared, as total momentum will be zero, so ##(2E)^2##.

What can you get from conservation of linear momentum?
As ##P=\sqrt {E^2-m^2}=34987.4MeV##, conservation of momentum would suggest the two muons have momentums of half this, ##17493,71##, although I am ignoring any momentum the proton may have. From this I could calculate the velocity of the muons, as by conservation of energy, they have half of the antiproton's energy each, ##17.5MeV##, and so using ##v=\frac {pc^2} {E}##, ##v=0.99964c##

Finding γ and then using this to Lorentz contracting the distance (to 536.6m), I get a new survival probability of 0.44311

I've tried to use conservation of momentum here, but I'm not sure if it is fully realised, certainly the accompanying script for the problem stated that, "the main problem was that the students, almost uniformly, did not notice that the two muons would have to be going at an angle to the normal and that the flight path would therefore be longer".
 
  • #4
vela
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Okay, so my thinking for ##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2## was that using ##A^2=(\sum\nolimits E)^2-(\sum\nolimits P)^2##,

Choosing my frame as that of the proton, since this seemed to minimise the impact of my confusion associated with its trajectory, for ##A^2## the first term is the energy of the antiproton and the rest mass of the proton, and the second just the momentum of the antiproton, given that in its rest frame the momentum of the proton shall be zero. Then, since ##A^2## is invariant, working in the ZMF frame after collision, as the muons have equal energies, the value of ##A^2## is merely the total energy squared, as total momentum will be zero, so ##(2E)^2##.
This is fine.

As ##P=\sqrt {E^2-m^2}=34987.4MeV##, conservation of momentum would suggest the two muons have momentums of half this, ##17493,71##, although I am ignoring any momentum the proton may have. From this I could calculate the velocity of the muons, as by conservation of energy, they have half of the antiproton's energy each, ##17.5MeV##, and so using ##v=\frac {pc^2} {E}##, ##v=0.99964c##
Given the energy of the antiproton, it's reasonable to approximation to assume the proton is essentially at rest.

You've already switched back to working in the Earth's frame, and you're thinking of motion along just one dimension. I suggest you finish analyzing the collision in the center-of-mass frame, where it's relatively easy to do the calculations, and then transform those results back to the Earth's frame.

You reasoned that the two muons will have the same energy in the center-of-mass frame, and the problem statement says the muons also have the same energy as measured in the Earth's rest frame. Think about what this means about the direction the muons are moving in the center-of-mass frame.
 
  • #5
TSny
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Okay, so my thinking for ##(E_{\bar p}+m_ p)^2 - P_{\bar p}^2 = 4E^2## was that using ##A^2=(\sum\nolimits E)^2-(\sum\nolimits P)^2##,

Choosing my frame as that of the proton, since this seemed to minimise the impact of my confusion associated with its trajectory, for ##A^2## the first term is the energy of the antiproton and the rest mass of the proton, and the second just the momentum of the antiproton, given that in its rest frame the momentum of the proton shall be zero. Then, since ##A^2## is invariant, working in the ZMF frame after collision, as the muons have equal energies, the value of ##A^2## is merely the total energy squared, as total momentum will be zero, so ##(2E)^2##.
OK, this is correct as long as ##E## on the right side of the equation is interpreted as the energy of each muon in the ZMF. Personally, for this problem I find it easier to stick with the earth frame. Using conservation of energy, it is easy to determine the energy of each muon in the earth frame. The magnitude of the 3-momentum of each muon in the earth frame is then easily obtained.

As ##P=\sqrt {E^2-m^2}=34987.4MeV##, conservation of momentum would suggest the two muons have momentums of half this, ##17493,71##
The momentum of each muon in the earth frame is not equal to half the momentum of the antiproton. If you let the x-direction be the direction of motion of the antiproton, then you can easily use conservation of x-component of momentum to show that the x-component of momentum of each muon in the earth frame is half the momentum of the antiproton.

Think about how to use this to determine the direction of motion of each muon in the earth frame.
 

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