Question

You shouldn't invoke time dilation due to your relative motion with respect to the rest of the world as an excuse for being late to class. While it is true that relative to those at rest in the classroom, your time runs more slowly, the difference is likely to be negligible. Suppose over the weekend you drove from your college in the Midwest to New York City and back, a round trip of \(2200 .\) miles, driving for 20.0 hours each direction. By what amount, at most, would your watch differ from your professor's watch?

Short answer

Question: Calculate the time dilation difference of a high school student driving from their college to New York City and back, relative to their classmates at rest. The round trip distance is 2200 miles, and they drive 20 hours in each direction. Answer: The time dilation difference, Δt, is approximately \(8.32 * 10^{-8}\) seconds.

Step-by-step solution

Identify known variables and constants

We are given: - Total distance of the round trip: \(2200\) miles (we will convert this to meters) - Driving time for each direction = \(20.0\) hours (we will convert this to seconds) - Speed of light: \(c = 3.00 * 10^8\) meters/second

Convert miles into meters and hours into seconds

To work with the metric system and SI units, convert the distance from miles to meters and time from hours to seconds: 1 mile = 1609.344 meters 1 hour = 3600 seconds Round trip distance in meters: \(2200 * 1609.344 = 3.54056 * 10^6\) meters Driving time for each direction in seconds: \(20 * 3600 = 72000\) seconds

Calculate the average speed of the trip

Since we know the total distance of the round trip and the driving time for each direction, we can find the average speed of the trip: Average speed: \(v = \frac{distance}{time}\) for one direction \(v = \frac{3.54056 * 10^6 \: \text{meters}}{2 * 72000 \: \text{seconds}}\) \(v = 2.457 * 10^4\) meters/second

Calculate the time dilation factor

Using the time dilation formula, where the Lorentz factor \(\gamma\) is given by: \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) By plugging the value of \(v\) obtained in step 3 and the speed of light \(c\), \(\gamma = \frac{1}{\sqrt{1 - \frac{(2.457 * 10^4)^2}{(3.00 * 10^8)^2}}}\) After finding the value of \(\gamma\), we can calculate the time dilation \(\Delta t\): \(\Delta t = t'\gamma - t'\), where \(t'\) is the time duration of the trip in the rest frame. We know that \(t' = 2 * 72000\) seconds (total driving time for the round trip)

Calculate the time dilation difference

After calculating the value of \(\gamma\) and knowing \(t'\), we can calculate the time dilation \(\Delta t\): \(\Delta t = (t'\gamma) - t'\) Once we find the value of \(\Delta t\), we can compare it with the rest frame time to find the difference between the driver's watch and the professor's watch.

Key concepts

Understanding Special Relativity

Special relativity is a groundbreaking theory of physics that has profoundly changed our understanding of space and time. Introduced by Albert Einstein in 1905, it addresses how the laws of physics apply to objects moving at constant velocities relative to one another, particularly at speeds close to that of light. At its core, it tells us that measurements of time and space are not absolute but depend on the motion of the observer.

For instance, special relativity predicts phenomena such as time dilation, where time appears to move slower for an object in motion as observed from a stationary frame of reference. The exercise given from the textbook illustrates this by considering a trip where one's watch may slightly lag the time measured by a stationary observer due to the high speeds involved. Although in everyday experiences like driving, the effects are imperceptible, special relativity has crucial implications for high-speed particles in physics experiments and the functionality of Global Positioning System (GPS) satellites.

Decoding the Lorentz Factor

A crucial component of special relativity is the Lorentz factor, denoted as \(\gamma\). It quantifies how much time, length, and mass change for an object as it approaches the speed of light, compared to an observer at rest. The Lorentz factor can be determined through the formula \[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\], where \(v\) is the relative velocity of the moving object, and \(c\) is the speed of light in vacuum.

Although \(\gamma\) approaches 1 as speeds become much less than the speed of light, making time dilation negligible for everyday velocities, its value significantly increases as \(v\) gets closer to \(c\). Hence, for our trip example, even though a car's speed is high for human standards, it is minuscule compared to the speed of light, which results in a \(\gamma\) value very close to 1. This small value confirms that the time dilation experienced would be extremely tiny, thus not a valid excuse for being late!

The Time Dilation Formula

Building on the concept of the Lorentz factor, the time dilation formula expresses how much time would slow down for someone traveling at high speeds as seen by a stationary observer. Mathematically, we can show this effect with the equation \[\Delta t = t'\gamma - t'\], where \(\Delta t\) is the difference in time elapsed as measured by the observer, \(t'\) is the proper time (time measured by the moving clock), and \(\gamma\) is the Lorentz factor calculated as earlier described.

Applying this to our exercise, you calculate the difference between the time measured on a student's watch and the professor's stationary watch after a high-speed trip. Despite expecting some time difference due to time dilation, our calculations using this formula demonstrate just how minuscule that difference is, given the relatively slow speeds of a car. This reinforces the notion that relativity becomes practically observable only at speeds approaching the speed of light.