Question

A famous result in Newtonian dynamics is that if a particle in motion collides elastically with an identical particle at rest, the two particles emerge from the collision on perpendicular trajectories. Does the same hold in the special theory of relativity? Suppose a particle of rest mass \(m\) and total energy \(E\) collides with an identical particle at rest, the same two particles emerging from the collision with new velocities. Are those velocities necessarily perpendicular? Explain.

Short answer

Answer: No, the velocities after the collision of particles are not necessarily perpendicular, as special relativity shows it isn't guaranteed by the conservation of energy and momentum.

Step-by-step solution

Understand the problem and its givens

In this problem, a particle of rest mass 'm' and total energy 'E' collides with an identical particle at rest. We need to find out if their velocities after the collision are necessarily perpendicular.

Conservation of momentum and energy

During an elastic collision, both momentum and energy are conserved. Since one of the particles is initially at rest, its momentum is zero. Let's denote the momentum of the moving particle as 'p' and its velocity as 'v'. Then, the total momentum before the collision is 'p'. From the conservation of momentum, the total momentum after the collision also needs to be 'p'. Similarly, the total energy before the collision is the sum of the energies of both particles 'E + mc^2' (mc^2 being the rest mass energy of the second particle). From the conservation of energy, the total energy after the collision should also be 'E + mc^2'.

Analyze the problem using the special theory of relativity

For the motion analysis, we will use the 4-vector components: The time-like component (energy divided by the speed of light) and the space-like component (momentum). The 4-momentum of the initial moving particle can be represented as \(p_1^\mu = (\frac{E}{c}, p)\) and the 4-momentum of the second particle (at rest) as \(p_2^\mu = (\frac{mc^2}{c}, 0)\). The total 4-momentum before the collision is \(P^\mu = (\frac{E}{c}+\frac{mc^2}{c}, p)\). Lorentz invariant quantity for the total 4-momentum can be calculated using \(P^\mu P_\mu = m^2c^2\). Now, let the velocities after the collision be \((v_1', v_2')\). Energy and momentum conservation implies that the 4-momenta after the collision must add up to the 4-momentum before the collision. We can represent \(p_1^{'\mu} = (\frac{E_1'}{c}, p_1')\) and \(p_2^{'\mu} = (\frac{E_2'}{c}, p_2')\). According to the conservation of energy and momentum, we have: $\begin{pmatrix} \frac{E_1'}{c}\\ p_1' \end{pmatrix} + \begin{pmatrix} \frac{E_2'}{c}\\ p_2' \end{pmatrix} \P= \begin{pmatrix} \frac{E}{c} + \frac{mc^2}{c}\\ p \end{pmatrix}$

Calculating new energies and perpendicular velocities

From the equations obtained in step 3, we can calculate the new energies and velocities after the collision. With these values, we can compute the dot product of the velocity vectors \(v_1'\) and \(v_2'\) as \(v_1' \cdot v_2' = p_1' \cdot p_2' + \frac{E_1'E_2'}{c^2}\). Then, if the velocities are perpendicular, the dot product should equal zero. However, it's not possible to guarantee that the dot product will always be zero for arbitrary initial conditions and under the conservation laws provided. This conclusion states that velocities after the collision are not necessarily perpendicular. So, the answer to the question is "No", the velocities after the collision of particles in motion are not necessarily perpendicular, as special relativity shows it isn't guaranteed by the conservation of energy and momentum.

Key concepts

Elastic Collision

In the realm of physics, an elastic collision is a type of encounter between two bodies in which no kinetic energy is lost. It's essential in problems involving collisions to determine the final state of the participating objects. In an elastic collision, both the total kinetic energy and total momentum are conserved, which provides us with key tools to solve such problems.

For example, when two billiard balls collide on a pool table, they bounce off each other. If the collision is perfectly elastic and friction is ignored, the total energy the balls carry before and after the collision remains the same. However, the directions and speeds of the balls will change, depending on the angle and velocities at which they struck each other.

In the context of the special theory of relativity, the concept of an elastic collision is more complex because the conservation laws apply to energy and momentum in all frames of reference, not just the kinetic energy observed in classical physics. This leads us to a deeper analysis involving 4-momenta rather than classical mechanics.

Conservation of Momentum

The conservation of momentum principle states that, within an isolated system, the total momentum remains constant assuming no external forces act upon it. This principle is pivotal not only in classical mechanics but also in relativistic mechanics.

In our given problem, we initially have a moving particle and one at rest. Prior to collision, the total momentum of the system is simply the momentum of the moving particle since the resting particle contributes zero. After collision, the individual momenta of the particles may change, but their sum must equal the initial total momentum. This fact remains true regardless of whether we are using classical velocities or relativistic 4-momenta to describe the motion.

When we approach this exercise, understanding the conservation of momentum provides a robust starting point for further calculations and helps us chart the course of particles post-collision. By applying conservation laws, we can predict the outcome of the collision even before considering other relativistic effects.

4-momentum

4-momentum revolutionizes the way we comprehend motion in relativistic physics. It extends the classical three-dimensional momentum vector into a four-dimensional vector, which incorporates both the energy and momentum of an object. This four-vector includes a time-like component, representing energy, and space-like components, representing the three-dimensional momentum.

The notation for 4-momentum is usually \(p^\mu = (\frac{E}{c}, \vec{p})\), where \(E\) is the total energy of the particle (including its rest mass energy \(mc^2\)), \(c\) is the speed of light, and \(\vec{p}\) is the three-dimensional momentum. In the special theory of relativity, the 4-momentum obeys the conservation laws, much like its classical counterpart, but with the critical condition that it applies across all frames of reference.

The significance of 4-momentum manifests when we try to solve the exercise problem, as it allows for the conservation laws to be applied in the context of relativistic velocities and energy, leading to the conclusion that the velocities won't necessarily be perpendicular after the collision.

Rest Mass Energy

A compelling contribution of Einstein's special theory of relativity to our understanding of matter and energy is expressed in the equation \(E = mc^2\). This equation shows us that mass (\(m\)) has an inherent energy associated with it called rest mass energy, even when it's not in motion. The parameter \(c\) represents the speed of light and indicates that a small amount of mass can correspond to a vast amount of energy due to the constant's \(c^2\) tremendous value.

This core concept is instrumental while analyzing collisions in a relativistic framework, as seen in the homework problem. The particle at rest has energy by virtue of its mass, and this energy plays a crucial role in the conservation equations post-collision. Importantly, the rest mass energy forms a part of the total energy that's conserved, along with the kinetic energy of moving bodies, a concept that’s distinctly different from classical physics’ energy conservation.