Particle Decay Momentum Analysis

This problem involves the decay of a heavy particle into two lighter particles. We are asked to find the maximum possible momentum component of one of the lighter particles in a direction perpendicular to the initial direction of motion of the heavy particle.

Applying Conservation Laws

The decay process must obey the fundamental principles of conservation of energy and conservation of momentum. Let the heavy particle be denoted by \(M\) and the two identical light particles by \(m_1\) and \(m_2\) (both with rest mass \(m\)). The heavy particle is initially moving along the positive z-direction.

• Initial State: Heavy particle \(M\) moving in +z direction. Let its energy be \(E_M\) and momentum be \(\vec{P}_M\). Since it's moving along the z-axis, \(\vec{P}_M = (0, 0, P_M)\).
• Final State: Two light particles \(m_1\) and \(m_2\). Let their energies be \(E_1, E_2\) and momenta be \(\vec{p}_1, \vec{p}_2\). \(\vec{p}_1 = (p_{1x}, p_{1y}, p_{1z})\), \(\vec{p}_2 = (p_{2x}, p_{2y}, p_{2z})\).

The relativistic energy-momentum relation for a particle with rest mass \(m_0\) and momentum \(\vec{p}\) is \(E^2 = (pc)^2 + (m_0c^2)^2\), where \(p = |\vec{p}|\).

Conservation of momentum states:

\(\vec{P}_M = \vec{p}_1 + \vec{p}_2\)

In components:

• \(0 = p_{1x} + p_{2x} \implies p_{2x} = -p_{1x}\)
• \(0 = p_{1y} + p_{2y} \implies p_{2y} = -p_{1y}\)
• \(P_M = p_{1z} + p_{2z}\)

Conservation of energy states:

\(E_M = E_1 + E_2\)

Analyzing in the Rest Frame (Centre of Mass Frame)

The problem is simplified by considering the decay in the rest frame of the heavy particle \(M\). In this frame, the total momentum before decay is zero.

• Initial State (Rest Frame): Heavy particle \(M\) is at rest. Energy \(E_M^* = Mc^2\), momentum \(\vec{P}_M^* = \vec{0}\).
• Final State (Rest Frame): Two light particles \(m_1\) and \(m_2\) with energies \(E_1^*, E_2^*\) and momenta \(\vec{p}_1^*, \vec{p}_2^*\).

Conservation of momentum in the rest frame:

\(\vec{0} = \vec{p}_1^* + \vec{p}_2^* \implies \vec{p}_2^* = -\vec{p}_1^*\)

This means the two light particles are emitted in opposite directions with equal momentum magnitude. Let \(p^* = |\vec{p}_1^*| = |\vec{p}_2^*|\).

Conservation of energy in the rest frame:

\(E_M^* = E_1^* + E_2^*\)

\(Mc^2 = \sqrt{(p_1^*c)^2 + (mc^2)^2} + \sqrt{(p_2^*c)^2 + (mc^2)^2}\)

Since \(p_1^* = p_2^* = p^*\) and their rest masses are the same, their energies are equal: \(E_1^* = E_2^* = E^* = \sqrt{(p^*c)^2 + (mc^2)^2}\).

So, \(Mc^2 = 2E^* = 2\sqrt{(p^*c)^2 + (mc^2)^2}\).

Squaring both sides:

\((Mc^2)^2 = 4((p^*c)^2 + (mc^2)^2)\)

\(M^2c^4 = 4p^{*2}c^2 + 4m^2c^4\)

Divide by \(c^2\) (assuming \(c \ne 0\)):

\(M^2c^2 = 4p^{*2} + 4m^2c^2\)

\(4p^{*2} = M^2c^2 - 4m^2c^2\)

\(p^{*2} = \frac{(M^2 - 4m^2)c^2}{4}\)

Taking the square root, the magnitude of the momentum of each light particle in the rest frame is:

\(p^* = \frac{c}{2}\sqrt{M^2 - 4m^2}\)

The condition \(M > 2m\) ensures that \(M^2 > 4m^2\), so the value under the square root is positive and the decay is possible.

Maximum Perpendicular Momentum in Lab Frame

We need to find the maximum momentum of a light particle in a direction perpendicular to the z-direction in the lab frame. The lab frame is where the heavy particle was initially moving along the +z axis.

The momentum vector of a light particle in the rest frame is \(\vec{p}^* = (p_x^*, p_y^*, p_z^*)\), where \(p_x^{*2} + p_y^{*2} + p_z^{*2} = p^{*2}\).

The perpendicular momentum in the lab frame is \(p_\perp = \sqrt{p_x^2 + p_y^2}\). According to Lorentz transformations (with the velocity between frames along the z-axis), the components of momentum perpendicular to the relative velocity are invariant:

\(p_x = p_x^*\)

\(p_y = p_y^*\)

Thus, the perpendicular momentum in the lab frame is \(p_\perp = \sqrt{p_x^{*2} + p_y^{*2}}\).

We want to maximize \(p_\perp\). This occurs when the entire momentum \(p^*\) in the rest frame is directed perpendicular to the z-axis (which is the direction of motion of the heavy particle in the lab frame, and hence the relative velocity between frames). In this scenario, \(p_z^* = 0\), and \(p_x^{*2} + p_y^{*2} = p^{*2}\).

The maximum value of \(p_\perp\) is therefore equal to \(p^*\).

Maximum \(p_\perp = p^* = \frac{c}{2}\sqrt{M^2 - 4m^2}\).

Conclusion

The maximum value of the momentum that any one of the lighter particles can have in a direction perpendicular to the z-direction (the initial direction of the heavy particle) is the magnitude of the momentum of the light particle in the rest frame of the heavy particle.

Maximum perpendicular momentum \( = \frac{c}{2}\sqrt{M^2 - 4m^2}\).

This corresponds to Option 1.