Question

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to ours. Assume that the two reference frames have parallel axes, and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by distance \(l\) in the \(x^{\prime}-\) direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in our reference frame.

Short answer

Answer: The expressions for the readings of the two synchronized clocks in their own reference frame as functions of the time coordinate in the observer's reference frame are: $$ t'_1 = \frac{t}{\gamma} $$ and $$ t'_2 = \frac{t}{\gamma} - \frac{vl}{c^2\gamma} $$

Step-by-step solution

Set up the clock coordinates in their frame

According to the problem, the two clocks are separated by a distance \(l\) in the \(x'\) direction. The coordinates for the first clock are \((0,0,0)\), and the coordinates for the second clock are \((l,0,0)\). Both clocks have the same \(y'\) and \(z'\), so we can focus only on the \(x'\) coordinate.

Invert the Lorentz transformation

To find how the clock readings \(t'\) depend on the unprimed frame's time coordinate \(t\), we need to invert the Lorentz transformation. The Lorentz transformation is given by: $$ x' = \gamma (x - vt), \:\: y' = y, \:\: z' = z, \:\: t' = \gamma (t - \frac{vx}{c^2}) $$ where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\). Inverting the Lorentz transformation, we obtain: $$ x = \gamma (x' + vt'), \:\: y = y', \:\: z = z', \:\: t = \gamma (t' + \frac{vx'}{c^2}) $$

Find \(t'\)

For the first clock, we need to express \(t'\) in terms of \(t\). From the first clock's coordinates, we have \(x'=0\), \(y'=0\), and \(z'=0\). Using the inverse Lorentz transformation, we have \(x = \gamma vt'\) and \(t = \gamma t'\). Solving for \(t'\) in terms of \(t\), we get: $$ t'_1 = \frac{t}{\gamma} $$ For the second clock, we have \(x'=l\), \(y'=0\), and \(z'=0\). Again using the inverse Lorentz transformation, we get \(t = \gamma(t' + \frac{vl}{c^2})\). Solving for \(t'\) in terms of \(t\), we obtain: $$ t'_2 = \frac{t}{\gamma} - \frac{vl}{c^2\gamma} $$

Write the final expressions

We found the expressions for \(t'_1\) and \(t'_2\) for the two clocks as functions of the time coordinate \(t\) in our reference frame. These expressions are: $$ t'_1 = \frac{t}{\gamma} $$ and $$ t'_2 = \frac{t}{\gamma} - \frac{vl}{c^2\gamma} $$

Key concepts

Special Relativity

Special relativity, formulated by Albert Einstein in 1905, transformed our understanding of space, time, and motion. At the heart of special relativity is the idea that the laws of physics are the same for all observers, regardless of their relative motion, provided they are not influenced by gravity, which is the realm of general relativity. Special relativity introduces several groundbreaking concepts, including the constancy of the speed of light in a vacuum for all observers, regardless of their relative motion, and the relativity of simultaneity—events that are simultaneous in one frame of reference may not be so in another.

Special relativity uses the Lorentz transformation equations to relate the physical quantities measured in different inertial frames of reference—that is, frames of reference that are moving at constant velocities relative to each other. These equations account for how measurements of time, length, and mass appear altered to an observer in motion relative to the observed object or event. One of the most counterintuitive and fascinating outcomes is that time and space are not absolute but are interwoven in a four-dimensional space-time continuum.

Time Dilation

Time dilation is a direct consequence of the postulates of special relativity. It describes how, according to an observer in one inertial frame, a clock that is moving relative to them seems to tick more slowly. The faster the relative speed, the more significant the effect. This phenomenon is not due to any mechanical failures or peculiarities in the moving clock but because of the very nature of time itself as per special relativity.

To put it mathematically, the time interval \( \Delta t' \) between two ticks as measured in the clock's own rest frame is different from the time interval \( \Delta t \) as measured by an observer for whom the clock is moving at speed \( v \). The relationship between these two times is given by the Lorentz factor \( \gamma \), where \( \Delta t = \gamma \Delta t' \) and \( \gamma = \frac{1}{{\sqrt{1 - \frac{v^2}{c^2}}}} \), with \( c \) representing the speed of light. For speeds much less than the speed of light, time dilation is negligible, but it becomes significant as \( v \) approaches \( c \).

Reference Frames

Reference frames are essential in physics for describing the position, velocity, and acceleration of objects. They are like coordinate systems that can move relative to one another. When we discuss movement and speeds, it's critical to specify the reference frame from which these are being measured. Inertial reference frames, in particular, are those that either remain stationary or move at a constant velocity—they are not accelerating.

In the context of the exercise solution, we discussed two different reference frames: our frame and the frame where the clocks are moving at a speed \( v \). The Lorentz transformation allows us to translate physical quantities between these two inertial frames. Careful use of the correct reference frame is crucial when applying the Lorentz transformation to ensure we properly understand how events in one frame are viewed from another.

By comparing the clock readings \( t' \) in the moving frame with the time coordinate \( t \) in our frame, we were able to derive their relationship, which showcases how time dilation occurs due to their relative motion. Understanding reference frames enables us to grasp why time is not absolute but is instead dependent on the observer's frame of reference, a notion that was unthinkable before the advent of special relativity.