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 make use of conservation of energy and momentum.
Let's start by finding the energy of each gamma ray. Since the pion has rest energy of 135 MeV, its total energy can be calculated using the relativistic energy equation:
E_pion = γ * m_pion * c^2
where γ is the Lorentz factor given by γ = 1 / sqrt(1 - (v^2 / c^2)), m_pion is the rest mass of the pion, and c is the speed of light.
Plugging in the values, we have:
E_pion = (1 / sqrt(1 - (0.98c)^2 / c^2)) * 135 MeV
Next, we know that the pion decays into two gamma rays of equal energy. Let's call the energy of each gamma ray E_gamma.
E_pion = 2 * E_gamma
Solving for E_gamma:
E_gamma = E_pion / 2
We can substitute the expression for E_pion into this equation to obtain the energy of each gamma ray.
E_gamma = [(1 / sqrt(1 - (0.98c)^2 / c^2)) * 135 MeV] / 2
Next, let's find the angle θ between the direction of motion and the gamma rays. We can use the conservation of momentum to find the relationship between the momentum of the pion and the momentum of the gamma rays.
Since the pion and gamma rays move at the speed of light, we can write:
γ * m_pion * v_pion = γ_gamma * m_gamma * v_gamma
where γ_gamma is the Lorentz factor for the gamma rays, m_gamma is the mass of a gamma ray (which is zero), and v_gamma is the velocity of the gamma rays.
Since the pion and gamma rays are moving along the same direction, we can write:
m_pion * v_pion = 2 * γ_gamma * m_gamma * v_gamma
Now, we know that the momentum of a photon (gamma ray) is given by p = E / c, where E is the energy of the photon.
Let's substitute E_gamma into the equation:
m_pion * v_pion = 2 * γ_gamma * (E_gamma / c)
Simplifying, we get:
v_pion = 2 * γ_gamma * (E_gamma / (m_pion * c))
Finally, let's solve for γ_gamma:
γ_gamma = (v_pion * m_pion * c) / (2 * E_gamma)
Plugging in the values, we can calculate γ_gamma. The angle θ can also be calculated using the expression:
cos(θ) = v_gamma / v_pion
Simplifying, we get:
cos(θ) = 1 / (2 * γ_gamma)
Using these formulas, we can find the angle θ and the energies of the two gamma rays.
To find the angle θ and the energies of the two gamma rays, we need to use the conservation of momentum and energy.
Let's start by considering the conservation of energy. The pion initially has a rest energy of 135 MeV, and it decays into two gamma rays. Each gamma ray has an energy of Eγ.
From the conservation of energy, we can write:
Rest energy of pion = Energy of gamma ray 1 + Energy of gamma ray 2
135 MeV = Eγ + Eγ
Simplifying this equation, we get:
135 MeV = 2Eγ
Dividing both sides by 2, we find:
Eγ = 67.5 MeV
So each gamma ray has an energy of 67.5 MeV.
Now, let's consider the conservation of momentum. The pion is initially moving through the lab frame at v=0.98c. After the decay, the two gamma rays are emitted in opposite directions and make equal angles θ with the direction of motion.
Using the conservation of momentum, we can write:
Initial momentum of pion = Momentum of gamma ray 1 + Momentum of gamma ray 2
The initial momentum of the pion is given by:
Initial momentum of pion = (mass of pion) * (velocity of pion)
Since we know the rest energy of the pion is 135 MeV, we can use the equation E = mc^2 to find the mass of the pion. The rest energy is equal to the total energy (including kinetic energy) when the pion is at rest.
135 MeV = (mass of pion) * (c^2)
Mass of pion = 135 MeV / c^2
To find the initial momentum of the pion, we can use the relativistic momentum formula:
p = m * v / sqrt(1 - (v^2 / c^2))
Plugging in the values, we get:
Initial momentum of pion = (mass of pion) * (velocity of pion) / sqrt(1 - (velocity of pion)^2 / c^2)
Now that we know the initial momentum of the pion, we can set it equal to the momentum of the two gamma rays. Since they are emitted in opposite directions, their momenta cancel each other out.
Let's denote θ as the angle between one of the gamma rays and the original direction of motion.
The momentum of a gamma ray can be calculated using the relativistic momentum formula:
Momentum of gamma ray = (Energy of gamma ray) / c
We have two gamma rays, so we have:
2 * (Energy of gamma ray) * cos(θ) / c = Initial momentum of pion
Substituting the values:
2 * (67.5 MeV) * cos(θ) / c = [(mass of pion) * (velocity of pion) / sqrt(1 - (velocity of pion)^2 / c^2)].
To solve for θ, rearrange the equation:
cos(θ) = [(mass of pion) * (velocity of pion) * c / (2 * (67.5 MeV) * sqrt(1 - (velocity of pion)^2 / c^2))]
Once you have the value of cos(θ), you can find θ itself using inverse cosine (also known as arccos) function.
Finally, substituting the value of θ in the equation for the momentum of the gamma ray, we can find the momentum and energy of each gamma ray.
Momentum of gamma ray = (67.5 MeV) * cos(θ) / c
Energy of gamma ray = (momentum of gamma ray) * c
Calculating these values will give you the angle θ and the energies of the two gamma rays.