Question

A particle moves rectilinearly under a rest mass preserving force in some inertial frame. Show that the product of its rest mass and its instantaneous proper acceleration equals the magnitude of the relativistic three-force acting on the particle in that frame. [Hint: (14.1) and (35.14).] Show also that this is not necessarily true when the motion is not rectilinear.

Short answer

Answer: The magnitude of the relativistic three-force equals the product of the particle's rest mass and its instantaneous proper acceleration under rectilinear motion. However, this relationship does not hold for non-rectilinear motion.

Step-by-step solution

Consider the given equations

We need to use equations (14.1) and (35.14). We have: (14.1): \(\textbf{F} = \frac{dp}{d\tau}\), where \(\textbf{F}\) is the relativistic four-force, \(p\) is momentum, and \(\tau\) is the proper time. (35.14): \(\pm |F_{\sigma}| =\frac{dP_{\sigma}}{d\tau} = M\frac{ma^{\sigma}}{(1-v^2/c^2)^{3/2}}\), where \(F_{\sigma}\) is the relativistic three-force, \(P_{\sigma}\) is the component of relativistic momentum, \(M\) is the rest mass, \(a^{\sigma}\) is the proper acceleration, \(v\) is the velocity, and \(c\) is the speed of light. From these equations, we are asked to show that \(M \times a^{σ} = |F_{\sigma}|\) under rectilinear motion.

Proper acceleration and relativistic three-force

We need to find the relationship between proper acceleration and relativistic three-force for rectilinear motion. Under rectilinear motion, we can say that \(v=v^{\sigma}\) (velocity is aligned with the \(\sigma\) direction) and \(a^\sigma = a = \frac{d^2x}{d\tau^2}\). Therefore, the equation (35.14) can be written as: \(\pm |F_{\sigma}| = M\frac{a}{(1-v^2/c^2)^{3/2}}\) Now, we need to show that \(M \times a = |F_{\sigma}|\).

Multiply both sides by \((1-v^2/c^2)^{3/2}\) and simplify

Multiply both sides of the equation by \((1-v^2/c^2)^{3/2}\): \(M \times a \times (1-v^2/c^2)^{3/2} = \pm |F_{\sigma}| \times (1-v^2/c^2)^{3/2}\) With \(\beta = v/c\): \(M \times a \times (1-\beta^2)^{3/2} = \pm |F_{\sigma}| \times (1-\beta^2)^{3/2}\) Notice that both sides have the same factor of \((1-\beta^2)^{3/2}\), so we can cancel them out: \(M \times a = \pm |F_{\sigma}|\)

Show for non-rectilinear motion

For non-rectilinear motion, consider the simplest case of the particle moving in a circle. In this case, the proper acceleration is directed towards the center of the circle and is given by \(a_r = \frac{mv^2}{r}\), where \(m\) is the mass of the particle, \(v\) is the instantaneous velocity, and \(r\) is the radius of the circle. The force acting on this particle has a centrifugal component, and the relativistic three-force has a tangential component. The proper acceleration and relativistic three-force are therefore not aligned in the non-rectilinear case. Since \(M \times a_r\) and \(F_{\sigma}\) have different directions, they cannot be equal. In conclusion, we have shown that under rectilinear motion, the product of the rest mass and the instantaneous proper acceleration is equal to the magnitude of the relativistic three-force acting on the particle. However, when the motion is non-rectilinear, this relationship does not necessarily hold true.

Key concepts

Rectilinear Motion

Rectilinear motion is the simplest form of motion. It occurs when a particle or object moves along a straight line with time. Think of it like a car driving in a straight lane on a highway. This type of motion is fundamental because many complex motions can be broken down into smaller rectilinear motions.

In physics, especially in relativistic dynamics, it is important to analyze the motion in an inertial frame. Here, `inertial’ means a frame of reference where the laws of motion are not violated.
This makes it easier to measure quantities like velocity and acceleration. In the context of the problem, under rectilinear motion, we find a direct relationship between rest mass, proper acceleration, and the relativistic three-force.

Proper Acceleration

Proper acceleration is the acceleration experienced by an object in its own instantaneous rest frame. It's like how you feel when you accelerate quickly in a car - you would experience proper acceleration.

In the language of physics, proper acceleration (\(a^\sigma\)) is crucial because it measures how quickly a particle is changing its velocity in space from its own perspective. This factor becomes extremely useful in relativistic dynamics, where velocities can approach the speed of light.

Four-Force

Four-force is a concept from the theory of relativity. It is an extension of the classical concept of force into four-dimensional spacetime.
While traditional force is a vector having magnitude and direction, four-force adds a fourth component related to time. It is very similar to how velocity in relativity becomes four-velocity.

• Four-force is denoted by \(\textbf{F}\) and expressed as \(\textbf{F} = \frac{dp}{d\tau}\), where \(p\) is the momentum and \(\tau\) is proper time.

Understanding four-force helps us track how momentum changes not just with space but also with time.
This becomes essential when analyzing motions close to the speed of light.

Three-Force

The term three-force refers to the spatial components of the four-force. It's what we usually refer to as force in classical mechanics but considered in a relativistic context.
Three-force is significant because it relates to how forces interact with particles moving at high speeds.

• The magnitude of three-force \(F_{\sigma}\) in the exercise is linked with both the rest mass, \(M\), and the proper acceleration, \(a^\sigma\).
• In rectilinear motion, this relationship simplifies because all movements are aligned in one direction.

In non-rectilinear motion, though, this relationship becomes complex. The directional components differ, meaning the vector orientations of forces do not align, breaking the direct simple relationship observed in straight-line motion.