Problem 15 - University Physics with Modern Physics

Question

An observer in frame \(S^{\prime}\) is moving to the right \((+x-\text { direction })\) at speed \(u=0.600 \mathrm{c}\) away from a stationary observer in frame \(S\) . The observer in \(S^{\prime}\) measures the speed \(v^{\prime}\) of a particle moving to the right away from her. What speed \(v\) does the observer in S measure for the particle if \((a) v^{\prime}=0.400 c ;(b) v^{\prime}=0.900 c\) (c) \(v^{\prime}=0.990 c ?\)

Step-by-Step Solution

Answer

(a) \(v \approx 0.806c\), (b) \(v \approx 0.974c\), (c) \(v \approx 0.997c\).

Step 1: Identify the Relativistic Velocity Transformation Formula

In special relativity, to find the velocity of a particle in another frame, we use the relativistic velocity transformation formula: \[ v = \frac{v' + u}{1 + \frac{v'u}{c^2}} \] where \(v\) is the velocity as measured by the stationary observer in frame \(S\), \(v'\) is the velocity in the moving frame \(S'\), and \(u\) is the velocity of the moving frame \(S'\) relative to \(S\). \(c\) is the speed of light.

Step 2: Solve for (a) \(v' = 0.400c\)

Substitute \(v' = 0.400c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.400c + 0.600c}{1 + \frac{0.400c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.400c + 0.600c = 1.000c\). Simplify the denominator: \(1 + \frac{0.240c^2}{c^2} = 1.240\). Thus, \[ v = \frac{1.000c}{1.240} \approx 0.806c \].

Step 3: Solve for (b) \(v' = 0.900c\)

Substitute \(v' = 0.900c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.900c + 0.600c}{1 + \frac{0.900c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.900c + 0.600c = 1.500c\). Simplify the denominator: \(1 + \frac{0.540c^2}{c^2} = 1.540\). Thus, \[ v = \frac{1.500c}{1.540} \approx 0.974c \].

Step 4: Solve for (c) \(v' = 0.990c\)

Substitute \(v' = 0.990c\) and \(u = 0.600c\) into the formula: \[ v = \frac{0.990c + 0.600c}{1 + \frac{0.990c \times 0.600c}{c^2}} \] Simplify the numerator: \(0.990c + 0.600c = 1.590c\). Simplify the denominator: \(1 + \frac{0.594c^2}{c^2} = 1.594\). Thus, \[ v = \frac{1.590c}{1.594} \approx 0.997c \].

Key Concepts

Special Relativity

Albert Einstein introduced the theory of special relativity in 1905. It fundamentally changes how we understand time, space, and motion. At its core, special relativity states that the laws of physics are the same for all non-accelerating observers, no matter their speed relative to each other. A surprising result of this theory is that the speed of light is always constant — about 299,792,458 meters per second — no matter the observer's frame of reference.

• The theory combines space and time into a single four-dimensional fabric called spacetime.
• It challenges our classical notions, showing that time can slow down and lengths can contract for objects moving close to the speed of light.
• Core principles include time dilation and length contraction, which become noticeable at high velocities.

This theory underpins modern physics, influencing fields from cosmology to quantum mechanics and even GPS technology. It's crucial to process these abstract concepts first to understand phenomena such as relativistic velocity transformation.

Frame of Reference

A frame of reference is essentially a viewpoint from which an observer measures and perceives physical quantities like position, velocity, and acceleration. In the context of special relativity, frames of reference become even more significant.

• Inertial frames are those where an object either remains at rest or moves at a constant velocity unless acted upon by an external force.
• Different observers in varying frames can measure the same event differently, particularly when dealing with high speeds close to the speed of light.
• The concept is crucial for understanding relativistic effects because the measurements can change based on the observer's movement relative to what they are observing.

In the original exercise, the stationary observer holds the frame \(S\), and the moving observer holds the frame \(S'\). To find how a moving observer's perceived velocity compares to a stationary observer's, we use the relativistic velocity transformation formula, showing how these frames interact at high speeds.

Speed of Light

The speed of light, denoted by \(c\), is one of the most critical constants in physics. It represents the ultimate speed limit in the universe and is fundamental in the theory of special relativity. In the relativistic velocity transformation exercise, the speed of light plays a vital role.

• The constancy of light's speed leads to many peculiar effects described by special relativity, such as time dilation and the relativity of simultaneity.
• No information or physical object can travel faster than \(c\), providing a universal speed limit.
• The speed of light ensures that the velocity equations, like the relativistic velocity transformation, do not result in speeds surpassing \(c\).

Understanding why \(c\) is constant requires abandoning the notion of absolute space and time, embracing instead that measurements of time and distance can change based on the observer's motion. This is pivotal for solving exercises involving high velocities, where seemingly strange outcomes result from these relativistic effects.