Problem 70 - University Physics with Modern Physics

Question

A spaceship moving at constant speed \(u\) relative to us broadcasts a radio signal at constant frequency \(f_{0}\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0}\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the time dilation formula to relate the periods in the stationary and moving frames) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758 c relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Step-by-Step Solution

Answer

(a) Frequency when passing is lower than \\(f_0\\): \\(f = \frac{f_0}{\gamma}\\). (b) Approaching \\(f \approx 907.528 \, \text{MHz}\\), away \\(f \approx 131.128 \, \text{MHz}\\). (c) Passing frequency \\(f \approx 227.228 \, \text{MHz}\\), with smaller shift than approaching.

Step 1: Understanding Doppler Effect for Waves

The Doppler effect explains how the frequency of a wave changes for an observer moving relative to the source of the wave. In this case, the frequency observed when the spaceship is approaching us differs from when it is receding.

Step 2: Relating Time Periods Using Time Dilation

According to time dilation from Einstein's theory of relativity, the period of the wave in the spaceship's frame \(T_0\) and in the observer's frame \(T\) are related by \(T = \gamma T_0\), where \(\gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}\) is the Lorentz factor, and \(u\) is the speed of the spaceship.

Step 3: Expressing Observed Frequency When Passing By

At the instant the spaceship passes by, the frequency observed \(f\) is given by \(f = \frac{1}{T} = \frac{1}{\gamma T_0} = \frac{f_0}{\gamma}\). This shows that the observed frequency is lower than \(f_0\).

Step 4: Calculating Observed Frequencies When Approaching and Moving Away

When the spaceship approaches, the frequency is given by \(f = f_0 \gamma (1 + \frac{u}{c})\) and when it moves away, \(f = f_0 \gamma (1 - \frac{u}{c})\). Substitute \(f_0 = 345 \, \text{MHz}\) and \(u = 0.758 c\) to calculate these frequencies.

Step 5: Calculating Lorentz Factor \( \gamma\)

Calculate the Lorentz factor: \(\gamma = \frac{1}{\sqrt{1 - (0.758)^2}} = \frac{1}{\sqrt{1 - 0.574764}} \approx 1.515\).

Step 6: Calculating Frequencies and Shifts When Approaching and Away

When approaching: \(f = 345 \, \text{MHz} \times 1.515 \times (1 + 0.758) = 907.528 \, \text{MHz}\). Frequency shift is: \(f - f_0 = 907.528 - 345 = 562.528 \, \text{MHz}\). When moving away: \(f = 345 \, \text{MHz} \times 1.515 \times (1 - 0.758) = 131.128 \, \text{MHz}\). Frequency shift is: \(131.128 - 345 = -213.872 \, \text{MHz}\).

Step 7: Calculating Frequency and Shift When Passing By

Using part (a), frequency observed as it passes by is \(f = \frac{345 \, \text{MHz}}{1.515} \approx 227.228 \, \text{MHz}\). The frequency shift is: \(f - f_0 = 227.228 - 345 = -117.772 \, \text{MHz}\).

Step 8: Analyzing the Frequency Shifts

The observed frequency when passing by is lower than the emitted frequency. The shift magnitude is less than when approaching but resembles the shift when moving away in sign (since it is negative, indicating lower frequency than emitted).

Key Concepts

Relativity and Time Dilation

To understand how frequency changes for an observer when the source moves, we need to delve into Einstein's theory of relativity. According to this theory, time for a moving object, like our spaceship, passes differently compared to a stationary observer. This difference is called time dilation.
In simple words, if two twins are separated, and one travels at high speed in a spaceship, upon returning, this twin would be younger than the one who stayed behind. This happens because of how time behaves at high speeds.
For waves emitted from a spaceship, this time dilation affects how the frequency is perceived. If the original period (the inverse of frequency) in the spaceship's frame is denoted as \(T_0\), the observed period \(T\) is given by \(T = \gamma T_0\). This factor \(\gamma\) alters the way we perceive time and, consequently, the frequency the observer receives.

Lorentz Factor

The Lorentz factor \(\gamma\) is a crucial component when analyzing relativistic effects such as time dilation. It quantifies how much time, length, and relativistic mass change for objects moving at speeds close to the speed of light. The formula for the Lorentz factor is:
\[\gamma = \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}\]
Where:

• \(u\) is the velocity of the object.
• \(c\) is the speed of light.

As speed \(u\) increases and approaches the speed of light, \(\gamma\) grows significantly. In our spaceship example, a speed \(u = 0.758c\) yields a \(\gamma\) of approximately 1.515, showing how time, lengths, and energy perceived deviate significantly from what we'd expect at everyday speeds. This deviation is crucial when explaining phenomena like the Doppler effect in a relativistic setting.

Frequency Shift

The frequency shift is a key result of the relativistic Doppler effect. Simply put, it describes how the frequency of a wave changes for an observer due to the motion of the source. Here's how this happens:
When the spaceship approaches, waves are compressed from the observer's perspective, leading to higher frequencies. Conversely, as it recedes, waves are stretched, resulting in lower frequencies.
Calculating this shift involves the observed and emitted frequencies. For an approaching spaceship:

• Observed frequency \( f = f_0 \gamma (1 + \frac{u}{c}) \)

And when moving away:

• \( f = f_0 \gamma (1 - \frac{u}{c}) \)

This formula ties the Lorentz factor \(\gamma\) and the ship's speed to the frequency change. For instance, calculating with \(f_0 = 345 \, \text{MHz}\) and \(u = 0.758c\), frequencies experienced by observers both approaching and moving away, are distinctively shifted.

Electromagnetic Waves

Electromagnetic waves are crucial for understanding how signals like radio waves behave when emitted from moving sources, such as a spaceship. They include visible light, radio waves, and other forms of radiation. These waves travel at the constant speed of light, \(c\).
These waves are affected by the relativistic Doppler effect, similar to sound waves, but there's more complexity due to relativistic influences at high velocities. This complexity becomes evident with fast-moving objects, where electromagnetic signals exhibit significant relativistic shifts.
As the spaceship moves relative to us at a high speed, the frequency and, consequently, the color or pitch of these waves is altered. This means the radio signals we receive will vary in frequency, becoming higher as the ship approaches and lower as it departs. Understanding these shifts is essential for interpreting signals correctly from rapidly moving, star-bound spaceships or distant astronomical objects.