A pion has rest energy 135MeV. It decays into two gamma rays that travel at the speed of light. A pion
moving through the lab frame at v=0.98c decays into two gamma rays of equal energies, making equal
angles θ with the direction of motion. Find the angle θ and the energies of the two gamma rays. Hint:
gamma rays are electromagnetic radiation with E=pc.
To solve this problem, we can use the conservation of energy and momentum.
First, let's consider the pion at rest. The rest energy of the pion is given as 135 MeV.
When the pion decays, it produces two gamma rays. Since the gamma rays are massless particles, their energy is given by E = pc, where p is their momentum.
Next, let's consider the pion in motion. It is moving through the lab frame at v = 0.98c.
By applying the Lorentz transformation, we can relate the energy and momentum of the gamma rays in the lab frame to their values in the pion rest frame.
Using conservation of momentum, we can derive the following equation:
p_g1 + p_g2 = p_pion
Since the gamma rays are emitted at equal angles θ with the direction of motion, their momenta can be expressed as:
p_g1 = E_g1 * cos(θ)
p_g2 = E_g2 * cos(θ)
Therefore, the conservation of momentum equation becomes:
E_g1 * cos(θ) + E_g2 * cos(θ) = p_pion
We also know that the total energy of the gamma rays must equal the rest energy of the pion:
E_g1 + E_g2 = 135 MeV
Now we have two equations that we can solve simultaneously to find the angle θ and the energies of the gamma rays.
From the second equation, we can express E_g1 in terms of E_g2:
E_g1 = 135 MeV - E_g2
Substituting this expression into the first equation, we get:
(135 MeV - E_g2) * cos(θ) + E_g2 * cos(θ) = p_pion
Simplifying this equation, we have:
135 MeV * cos(θ) = p_pion
Now we can solve for θ:
θ = cos^(-1)(p_pion / 135 MeV)
Knowing the value of p_pion is crucial to finding the angle θ. Unfortunately, the problem statement doesn't provide this information directly. Without it, we wouldn't be able to determine the exact value of θ.