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.
1. Conservation of energy: The rest energy of the pion is converted into the energy of the two gamma rays. Since the pion has a rest energy of 135 MeV, each gamma ray will have an energy of 67.5 MeV.
2. Conservation of momentum: In the lab frame, the total momentum of the system is zero before and after the decay. The momentum of the pion is given by its relativistic momentum:
p_pion = γ*m_pion*v
where γ is the Lorentz factor and m_pion is the rest mass of the pion. Since the pion is moving at v = 0.98c, we have:
v = 0.98*c = 0.98*3.0e8 m/s = 2.94e8 m/s
m_pion = 135 MeV/c^2 = 135*10^6 eV/(3.0e8 m/s)^2 = 1.5e-23 eV s^2/m^2
Using these values, we can calculate the Lorentz factor:
γ = 1/sqrt(1-(v/c)^2) = 1/sqrt(1-(0.98)^2) = 5
Now we can calculate the momentum of the pion:
p_pion = γ*m_pion*v = 5*1.5e-23 eV s^2/m^2*2.94e8 m/s ≈ 8.77e-15 eV s
Since the total momentum of the system is zero, the gamma rays should travel in opposite directions and have equal magnitudes of momentum. Therefore, the magnitude of the momentum of each gamma ray is:
p_gamma = p_pion/2 = 8.77e-15 eV s/2 ≈ 4.38e-15 eV s
Finally, we can calculate the velocities of the gamma rays by dividing their momenta by their energies:
v_gamma = p_gamma/E_gamma = (4.38e-15 eV s) / (67.5 MeV) ≈ 6.48e-23 s / 1 = 6.48e-23 m/s
The angle θ between the direction of motion and the direction of one of the gamma rays can be found using the formula:
cos(θ) = v_gamma / (c * γ)
cos(θ) = (6.48e-23 m/s) / (3.0e8 m/s * 5) ≈ 4.32e-32
θ = acos(4.32e-32) ≈ 1.57079 radians ≈ 90 degrees
So, the angle θ is approximately 90 degrees, and the energy of each gamma ray is approximately 67.5 MeV.
To find the angle θ and the energies of the two gamma rays, we can use the principle of conservation of momentum and energy.
1. Conservation of momentum:
Since the pion and the gamma rays travel in the same direction, the total momentum before and after the decay should be conserved.
The momentum of the pion before the decay can be calculated using its rest energy:
p_pion = E_pion / c
where E_pion is the rest energy of the pion and c is the speed of light.
The total momentum after the decay should be zero, as gamma rays have no mass. Therefore, the momentum of each gamma ray is equal in magnitude but opposite in direction:
p_gamma = -p_gamma = -E_gamma / c
where E_gamma is the energy of each gamma ray.
Considering the angles they make with the direction of motion, we can write the momentum conservation equation as:
p_pion = 2 * p_gamma * cos(θ)
2. Conservation of energy:
The total energy before and after the decay should also be conserved.
The energy of the pion before the decay is its rest energy:
E_pion = rest energy = 135 MeV
The total energy after the decay is the sum of the energies of the two gamma rays:
2 * E_gamma = E_pion
Now, let's solve the equations to find the unknowns.
First, rearrange the momentum conservation equation:
p_pion = 2 * p_gamma * cos(θ)
E_pion / c = 2 * (-E_gamma / c) * cos(θ)
Substitute the values:
135 MeV / c = 2 * (-E_gamma / c) * cos(θ)
Next, rearrange the energy conservation equation:
2 * E_gamma = E_pion
2 * E_gamma = 135 MeV
Solve this equation for E_gamma:
E_gamma = 135 MeV / 2
Finally, substitute the value of E_gamma into the momentum conservation equation and solve for θ:
135 MeV / c = 2 * (-E_gamma / c) * cos(θ)
135 MeV / c = 2 * (-135 MeV / 2c) * cos(θ)
Simplify the equation:
1 = -cos(θ)
To solve for θ, take the inverse cosine (arccos) of both sides:
θ = arccos(1)
Since arccos(1) is zero, the angle θ is zero.
Therefore, the angle θ between the direction of motion and the two gamma rays is zero, and the energies of the two gamma rays are both 135 MeV / 2.