Problem 84 - Fundamentals of Physics

Question

A radar transmitter \(T\) is fixed to a reference frame \(S^{\prime}\) that is moving to the right with speed \(v\) relative to reference frame \(S\) (Fig. 37-33). A mechanical timer (essentially a clock) in frame \(S^{\prime}\), having a period \(\tau_{0}\) (measured in \(S^{\prime}\) ), causes transmitter \(T\) to emit timed radar pulses, which travel at the speed of light and are receive fixed in frame \(S\). (a) What is the period \(\tau\) of the timer as detected by observer \(A\), who is fixed in frame \(S ?\) (b) Show that at receiver \(R\) the time interval between pulses arriving from \(T\) is not \(\tau\) or \(\tau_{0}\), but $$ \tau_{R}=\tau_{0} \sqrt{\frac{c+v}{c-v}} $$ (c) Explain why receiver \(R\) and observer \(A\), who are in the same reference frame, measure a different period for the transmitter. (Hint: A clock and a radar pulse are not the same thing.)

Step-by-Step Solution

Answer

(a) \( \tau = \tau_0 / \sqrt{1 - v^2/c^2} \). (b) \( \tau_R = \tau_0 \sqrt{\frac{c+v}{c-v}} \). (c) Different effects apply: time dilation vs. Doppler shift.

Step 1: Understand Time Dilation

In special relativity, time dilation refers to the fact that time passes at different rates in different reference frames. If a clock is moving with respect to an observer at a speed \( v \), the time interval \( \tau \) measured by the observer is related to the proper time \( \tau_0 \) (the time interval measured by the clock itself in its own rest frame) by the equation: \( \tau = \tau_0 / \sqrt{1 - v^2/c^2} \).

Step 2: Apply Time Dilation for Observer A

Observer \( A \) in frame \( S \) detects the period of the timer as \( \tau = \tau_0 / \sqrt{1 - v^2/c^2} \) since the transmitter \( T \) is moving relative to observer \( A \). This is the effect of time dilation as seen from observer \( A \), who measures everything with respect to frame \( S \).

Step 3: Determine the Time Interval at Receiver R

Receiver \( R \) measures the time interval between arriving radar pulses from \( T \) as \( \tau_R \). Using the relativistic Doppler effect, this interval is given by \( \tau_{R} = \tau_0 \sqrt{\frac{c+v}{c-v}} \). This accounts for both the time dilation and the Doppler effect as it measures the frequency of radar pulses.

Step 4: Explain the Difference in Measured Periods

Observer \( A \) and receiver \( R \) measure different periods because they experience different relativistic effects. Observer \( A \) measures time dilation while receiver \( R \) measures the Doppler shift of radar waves. The radar pulse frequency is affected by both the movement of \( T \) and the finite speed of light, leading to these disparities.

Key Concepts

Time Dilation

Time dilation is a phenomenon in special relativity where time appears to move differently depending on the relative motion between observers. Imagine you have a clock that ticks at a regular interval when it's at rest, known as the proper time, denoted by \( \tau_0 \).
Now, let's say this clock is moving at a velocity \( v \) relative to another observer. According to the observer who is not moving with the clock, the time interval \( \tau \) between ticks will seem longer than \( \tau_0 \).
A simple example is when astronauts in space, moving at high speeds, experience less passage of time compared to people on Earth.

The formula that represents this time dilation is:

• \( \tau = \frac{\tau_0}{\sqrt{1 - \frac{v^2}{c^2}}} \)

Here, \( c \) is the speed of light, and the term \( \sqrt{1 - \frac{v^2}{c^2}} \) is called the Lorentz factor. As the speed \( v \) approaches the speed of light, the Lorentz factor becomes significant, increasing the time measured by the stationary observer. This means the moving clock ticks slower from their perspective.

Doppler Effect

The Doppler effect refers to the change in frequency or wavelength of waves in relation to an observer who is moving relative to the wave source. In the context of special relativity, this effect becomes more pronounced because of the high speeds involved, such as those close to the speed of light.
When the source of the wave, say a radar transmitter, moves towards the observer, the waves are compressed, leading to a higher frequency or pitch. Conversely, if the source moves away, the waves are stretched out, resulting in a lower frequency.

For moving radar pulses in special relativity, the observed time interval for an observer at receiver \( R \) who looks at the moving transmitter is not the same as either \( \tau \) or \( \tau_0 \). Instead, it is given by:

• \( \tau_R = \tau_0 \sqrt{\frac{c+v}{c-v}} \)

This formula accounts for both time dilation and the Doppler effect's impact on the frequency of radar pulses you receive. It reflects the combined result of motion and the finite speed of light on the observation of these signals.

Reference Frames

In physics, a reference frame is a perspective from which an observer measures and records physical events. It's like the vantage point or "stage" from where you view motion and events.

There are two reference frames in play here:

• \( S \) - where observer \( A \) is stationary, a fixed viewpoint.
• \( S^{\prime} \) - where the transmitter \( T \) is moving at velocity \( v \) relative to \( S \).

In special relativity, different observers see the time and space differently based on their relative motion. A great analogy is considering how people in different moving cars can perceive each other's speeds and timing.
The differing observations made by \( A \) and \( R \) arise because of these separate reference frames.
While \( A \) measures \( \tau \) purely as a consequence of time dilation, \( R \) incorporates the Doppler effect because it involves the movement and relative speed of the radar pulse signals, meaning they are not simply observing the passing of time but also the frequency and timing of waves from a dynamic source.