Question

Two identical particles, each of mass \(1.30 \mathrm{mg}\), moving with equal but opposite velocities of \(0.580 \mathrm{c}\) in the laboratory reference frame, collide and stick together. Find the mass of the resulting particle.

Short answer

The mass of the resulting particle after the collision when calculated using the total energy, is found to be \(M = \frac{E}{c^2}\), where \(E\) is the total energy of the system.

Step-by-step solution

Understand the collision

In the problem, two particles of identical mass \(m_0 = 1.30~\mathrm{mg}\), and equal but opposite velocities collide and stick together. Therefore, this is a problem of two-body collision where two bodies become one after impacting.

Use the relativistic equations of momentum and energy

In special relativity, four-momentum is preserved. So, calculate the components of four-momentum separately for each particle and add them together. In the laboratory frame, the momentum of one of the particles is given by \(p_1 = \gamma m_0 v\), where \(v = 0.580c\) and \(\gamma = \frac{1}{\sqrt{1-v^2/c^2}}\). The momentum of the other particle is \(-p_1\) since it moves in the opposite direction with the same speed.

Calculate the total energy

The mass-energy \(E_1\) of one of the particles is given by \(E_1^2 = (pc)^2 + (m_0 c^2)^2\) where \(E = \gamma m_0 c^2\). The total energy \(E\) of the system is the sum of \(E_1\) and \(E_2\). Since they are identical particles, \(E = 2E_1\).

Calculate the mass of the resulting particle

The total four-momentum of the system after the collision is \(P = \sqrt{(pc)^2 + (E/c)^2}\). Since the particles stick together after collision, the momentum \(p\) is zero, \(P = E/c = M c\), where \(M\) is the mass of the resulting particle. Dividing both sides by \(c\), the resulting mass \(M = \frac{E}{c^2}\).

Key concepts

Special Relativity

Special relativity, a theory proposed by Albert Einstein, fundamentally revises our understanding of physics at high velocities near the speed of light, denoted by the symbol 'c'. This paramount theory changed the long-held notions of space, time, mass, and energy. It posits that the laws of physics are the same in all inertial frames of reference and that the speed of light in vacuum is the same for all observers, regardless of the motion of the light source.

A key point to remember is that as an object's speed approaches the speed of light, relativistic effects become significant, causing time dilation and length contraction. In our exercise, particles moving at velocities of 0.580c, which is 58% the speed of light, require a relativistic approach for accurate calculations.

Momentum Conservation

Conservation laws are cornerstones in physics, providing symmetry and fundamental predictability. In the context of collisions, the conservation of momentum principle states that the total momentum of a closed system remains constant if external forces are absent.

In relativistic collisions, it's essential to consider the relativistic momentum, where momentum 'p' is given by the relation \( p = \gamma m_0 v \), with \( \gamma \) (gamma factor) being \( \frac{1}{\sqrt{1-(v/c)^2}} \) and accounting for the effects of traveling at speeds close to 'c'. In the provided exercise, despite the high velocities, the initial and final momentum must balance, ensuring momentum conservation in the relativistic regime.

Rest Mass

Rest mass, often denoted as \( m_0 \), is the mass that an object has when it is not in motion relative to an observer. This concept is particularly important in special relativity, where an object's mass can appear to increase as its speed approaches the speed of light. However, rest mass remains invariant regardless of the reference frame or the object's velocity.

Understanding rest mass allows us to differentiate it from the relativistic mass, which includes the energy contribution due to an object's motion. In our exercise, the rest mass of each particle is given, enabling us to tackle the problem with a clear understanding of how mass behaves at high speeds.

Energy-Momentum Relation

The energy-momentum relation is a critical concept in relativistic physics, revealing the intimate connection between an object's energy, momentum, and rest mass. The famous equation \( E^2 = (pc)^2 + (m_0c^2)^2 \) encapsulates this relationship. The term \( m_0c^2 \) is the rest energy, and \( pc \) is the relativistic momentum component of the total energy 'E'.

In our exercise, we use this relation to find the combined energy of the two particles when they collide and become a single entity with rest mass 'M'. The fact that after the collision the momentum is zero simplifies the task, allowing the use of the equation to find the rest mass of the resulting particle by calculating the total energy and then using the relation \( M = \frac{E}{c^2} \).