Problem 118
Question
Use Einstein's special-relativity equation $$R_{a}=R_{f} \sqrt{1-\left(\frac{v}{c}\right)^{2}}$$ described in the Blitzer Bonus on page \(47,\) to solve this exercise. You are moving at \(90 \%\) of the speed of light. Substitute \(0.9 c\) for \(v,\) your velocity, in the equation. What is your aging rate, correct to two decimal places, relative to a friend on Earth? If you are gone for 44 weeks, approximately how many weeks have passed for your friend?
Step-by-Step Solution
Verified Answer
The rate at which you age relative to your friend on Earth will be calculated from the special-relativity equation eventually. Then, based on this aging rate, the time passed for your friend can be determined while you were gone for 44 weeks.
1Step 1: Substitution of values
First, substitute \(v = 0.9c\) into the equation. \(R_{f}\) is 1 because time for a stationary observer passes at a normal rate of 1 second per seconds. Thus we get: \[R_{a}= 1 \sqrt{1-\left(\frac{0.9c}{c}\right)^{2}}\]
2Step 2: Simplification of equation
Simplify the equation. Cancel out the \(c\) from the numerator and the denominator, and calculate the value inside the square root: \[R_{a}= 1 \sqrt{1-0.9^{2}}\]
3Step 3: Calculating the aging rate
Calculate the square root expression and multiply the result by 1 to get the aging rate \(R_{a}\). Round the answer to two decimal places.
4Step 4: Calculate the time passing for a friend
Next, use the aging rate \(R_{a}\) and the time you were gone (44 weeks) to calculate the time passing for a friend. The time for the friend can be calculated as \(R_{a}\) times the time you were gone.
Key Concepts
Relativistic Aging RateTime DilationVelocity Near the Speed of Light
Relativistic Aging Rate
The concept of the relativistic aging rate is a fascinating consequence of Einstein's theory of Special Relativity. Simply put, it refers to the difference in observed aging when comparing two people moving relative to each other at high speeds.
In the realm of special relativity, time behaves unusually. When an object moves at speeds that are a significant fraction of the speed of light, relative to an observer, time for the moving object will pass more slowly when observed from the stationary observer's point of view. In the exercise, the equation \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\) is used to calculate how much an individual ages (or 'runs slow' time-wise) when traveling at a velocity \(v\) close to the speed of light \(c\).
Here's a visual: imagine you have a twin; you board a spacecraft while your twin stays on Earth. If your spacecraft speeds off close to the speed of light and then returns, you would find that you have aged less than your twin. This rate at which you age, as compared to your twin, is what we call the relativistic aging rate. It's not just a thought experiment; this phenomenon has been confirmed by particle decay rates in particle accelerators and experiments involving precise clocks flown in aircraft.
In the realm of special relativity, time behaves unusually. When an object moves at speeds that are a significant fraction of the speed of light, relative to an observer, time for the moving object will pass more slowly when observed from the stationary observer's point of view. In the exercise, the equation \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\) is used to calculate how much an individual ages (or 'runs slow' time-wise) when traveling at a velocity \(v\) close to the speed of light \(c\).
Here's a visual: imagine you have a twin; you board a spacecraft while your twin stays on Earth. If your spacecraft speeds off close to the speed of light and then returns, you would find that you have aged less than your twin. This rate at which you age, as compared to your twin, is what we call the relativistic aging rate. It's not just a thought experiment; this phenomenon has been confirmed by particle decay rates in particle accelerators and experiments involving precise clocks flown in aircraft.
Time Dilation
Time dilation is one of those mind-bending concepts that seems more like science fiction than reality. Yet, it is a very real aspect of our universe, predicted by Einstein's theory of special relativity and confirmed by numerous experiments.
Time dilation refers to the stretching of time observed in a system in motion relative to an observer. As the velocity of the system approaches the speed of light \(c\), time within the moving system seems to slow down when I've been observed from an outside stationary frame of reference. This slow down is quantified by the equation used in our exercise \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\), which determines how much time has 'dilated' or stretched.
For instance, if you're looking at a clock on a spaceship moving at 90% of the speed of light compared to your stationary position, you would notice that the clock ticks more slowly than a similar clock by your side. It's this time stretching effect that is at the heart of time dilation. Yet keep in mind, for the person on the spaceship, time feels normal—time dilation is only apparent from an outside perspective.
Time dilation refers to the stretching of time observed in a system in motion relative to an observer. As the velocity of the system approaches the speed of light \(c\), time within the moving system seems to slow down when I've been observed from an outside stationary frame of reference. This slow down is quantified by the equation used in our exercise \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\), which determines how much time has 'dilated' or stretched.
For instance, if you're looking at a clock on a spaceship moving at 90% of the speed of light compared to your stationary position, you would notice that the clock ticks more slowly than a similar clock by your side. It's this time stretching effect that is at the heart of time dilation. Yet keep in mind, for the person on the spaceship, time feels normal—time dilation is only apparent from an outside perspective.
Velocity Near the Speed of Light
Approaching the speed of light brings to life the counterintuitive predictions of special relativity. As we deal with velocities nearing light-speed, we enter a realm where time and space look and behave differently from our everyday experiences.
Mathematically, when we express velocity as a fraction of the speed of light like \(v = 0.9c\), we clearly see the relative nature of motion. The letter \(c\) represents the speed of light, approximately \(299,792,458\) meters per second, a constant that is the same in all frames of reference. Factoring this velocity into our relativistic equations, as in the example \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\), reveals the magnitude by which time and aging are affected.
Indeed, as objects speed up to significant fractions of \(c\), they exhibit decreasing aging rates and expanded time intervals (time dilation) compared to slower-moving objects. This fundamental limit of \(c\) is not just a speed limit but is woven into the very fabric of space and time, making velocities near light-speed a cornerstone of modern physics discussions.
Mathematically, when we express velocity as a fraction of the speed of light like \(v = 0.9c\), we clearly see the relative nature of motion. The letter \(c\) represents the speed of light, approximately \(299,792,458\) meters per second, a constant that is the same in all frames of reference. Factoring this velocity into our relativistic equations, as in the example \(R_a = R_f \sqrt{1-(\frac{v}{c})^2}\), reveals the magnitude by which time and aging are affected.
Indeed, as objects speed up to significant fractions of \(c\), they exhibit decreasing aging rates and expanded time intervals (time dilation) compared to slower-moving objects. This fundamental limit of \(c\) is not just a speed limit but is woven into the very fabric of space and time, making velocities near light-speed a cornerstone of modern physics discussions.
Other exercises in this chapter
Problem 117
The early Greeks believed that the most pleasing of all rectangles were golden rectangles, whose ratio of width to height is $$\frac{w}{h}=\frac{2}{\sqrt{5}-1}$
View solution Problem 117
Use the order of operations to simplify each expression. \(\frac{2(-2)-4(-3)}{5-8}\)
View solution Problem 118
Use the order of operations to simplify each expression. \(\frac{6(-4)-5(-3)}{9-10}\)
View solution Problem 119
The mass of one oxygen molecule is \(5.3 \times 10^{-23}\) gram. Find the mass of \(20,000\) molecules of oxygen. Express the answer in scientific notation.
View solution