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Math  /  Calculus

QuestionProblem 1: When a cosmic ray (e.g. a proton) hits a molecule in the atmosphere, muons are created during the particle shower. Muons decay with a mean lifetime ⟨τ⟩=\langle\tau\rangle= 2.20μ s2.20 \mu \mathrm{~s}. (a) What is the decay rate, Γ\Gamma, for the muon?
Let us now model our cosmic ray as a particle of mass M. Assume that the particle of mass MM decays into two particles of mass m≠M:M→mmm \neq M: M \rightarrow m m. (a) Write down the conservation of the quadrimomentum in the reference frame of the particle of mass MM (hint: the particle MM is at rest in its reference frame) (b) Using the conservation equation above, compute the energies and momentum of the two particles of mass mm after the decay. (c) Is there a maximum value for mm over which the decay is no longer admissible? If the answer is yes, write such value.

Studdy Solution

STEP 1

1. The mean lifetime of a muon is given as ⟨τ⟩=2.20 μs \langle \tau \rangle = 2.20 \, \mu \mathrm{s} .
2. The decay rate Γ\Gamma is the inverse of the mean lifetime.
3. Particle MM is initially at rest in its reference frame.
4. The decay process is M→m+m M \rightarrow m + m .

STEP 2

1. Calculate the decay rate Γ\Gamma for the muon.
2. Write the conservation of quadrimomentum for the decay M→m+m M \rightarrow m + m .
3. Compute the energies and momentum of the two particles of mass mm.
4. Determine the maximum value for mm for which the decay is admissible.

STEP 3

Calculate the decay rate Γ\Gamma for the muon:
The decay rate Γ\Gamma is the inverse of the mean lifetime:
Γ=1⟨τ⟩ \Gamma = \frac{1}{\langle \tau \rangle}
Substitute the given mean lifetime:
Γ=12.20×10−6 s \Gamma = \frac{1}{2.20 \times 10^{-6} \, \mathrm{s}}
Calculate Γ\Gamma:
Γ≈4.545×105 s−1 \Gamma \approx 4.545 \times 10^5 \, \mathrm{s}^{-1}

STEP 4

Write the conservation of quadrimomentum for the decay M→m+m M \rightarrow m + m :
In the rest frame of particle MM, the initial quadrimomentum is:
PM=(Mc,0) P_M = (Mc, \mathbf{0})
The final quadrimomentum for the two particles of mass mm is:
P1=(E1/c,p1),P2=(E2/c,p2) P_1 = (E_1/c, \mathbf{p}_1), \quad P_2 = (E_2/c, \mathbf{p}_2)
Conservation of quadrimomentum gives:
PM=P1+P2 P_M = P_1 + P_2

STEP 5

Compute the energies and momentum of the two particles of mass mm:
Since MM is at rest initially, and assuming the decay is symmetric, we have:
E1=E2=E E_1 = E_2 = E p1=−p2=p \mathbf{p}_1 = -\mathbf{p}_2 = \mathbf{p}
Using energy conservation:
Mc2=2E Mc^2 = 2E
Thus:
E=Mc22 E = \frac{Mc^2}{2}
Using the relation E2=(pc)2+(mc2)2E^2 = (pc)^2 + (mc^2)^2, we find:
(Mc22)2=(pc)2+(mc2)2 \left(\frac{Mc^2}{2}\right)^2 = (pc)^2 + (mc^2)^2
Solving for pp:
(pc)2=(M2c44)−(m2c4) (pc)^2 = \left(\frac{M^2c^4}{4}\right) - (m^2c^4)
p=c2M2−4m2 p = \frac{c}{2} \sqrt{M^2 - 4m^2}

STEP 6

Determine the maximum value for mm for which the decay is admissible:
For the decay to be admissible, the expression under the square root must be non-negative:
M2−4m2≥0 M^2 - 4m^2 \geq 0
Solving for mm:
M2≥4m2 M^2 \geq 4m^2
m≤M2 m \leq \frac{M}{2}
Thus, the maximum value for mm is:
M2 \boxed{\frac{M}{2}}

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