陶哲轩在假期为一群高中学生举办了讲座，内容涉及多普勒效应和爱因斯坦的质能方程.并且把他的讲义上传到了他的博客.
我学习了他的讲义,在此记录.
Exercise 2.1 (Airport puzzle). Alice is trying to get from one end of a large airport to another. The airport is 500 meters long, and she walks at one meter per second. In the middle of the airport,there is a moving walkway that is 100 meters long, and moves in the direction Alice wants to go in at the speed of one meter per second;thus, if she is on the walkway while still walking, her net speed will in fact be two meters per second. On the other hand, Alice’s shoelaces are undone, and at some point before she arrives at the end of the airport, she has to spend ten seconds to stop walking and tie her shoelaces. She wants to arrive at the other end of the airport as soon as possible, and has to decide whether it is better for Alice to tie one’s shoelaces before getting on the walkway,during, or afterwards.
(i) Draw spacetime diagrams to illustrate the three differentchoices Alice has (tying before, tying during, and tying after the walkway). (An example of the first choice is given in Figure 3.)
Simple.
(ii) What should Alice do?
Alice can do whatever she likes.
(iii) Does the answer to (ii) change if one changes some of the numbers in the exercise (e.g. length of walkway, Alice’s walking speed, time taken to tie shoelaces)?
No.
Exercise 3.1. An ambulance $A$, driven by Alice, has a siren that operates at the frequency of 500 Hertz;this means that the sound waves that the siren generates oscillate 500 times a second. We assume that the frequency that the siren emits at remains the same at 500 Hertz, regardless of whether the ambulance is moving or is stationary. It is a calm day with no wind, and sound travels (in any direction) through the air at a speed of 300 meters per second
(about 670 miles per hour). An observer Bob (B) is standing still by a road. At noon 12 : 00 : 00, Alice is 300 meters away from Bob,but approaching Bob at the speed of 30 meters per second (about 67 miles per hour). Thus, for instance,one second later at 12 : 00 :01, Alice is now only 270 meters away from Bob, Alice passes Bob at 12 : 00 : 10, and by 12 : 00 : 20, Alice is again 300 meters from Bob but is now receding from Bob at 30 meters per second. (See
Figure 4.)
(a) When does the sound emitted at 12:00:00 reach Bob? When does the sound emitted at 12:00:01 reach Bob? How many oscillations of the siren will Bob hear in that time interval? What is the frequency that Bob hears of the siren?
Answer: The speed of the sound is 300m/s,so the sound emitted at 12:00:00 reach Bob at 12:00:01.The sound emitted at 12:00:01 reach Bob at 12:00:01:54.
Now I compute the number of oscillations Bob will hear in that time interval.
\begin{align*}
\frac{500}{\frac{270}{300}}=555.555....
\end{align*}
\begin{align*}
1.9\times 555.555...\approx 1055
\end{align*}
The frequency of the siren that Bob hears is 555.
(b) When does the sound emitted at 12 : 00 : 20 reach Bob? When does the sound emitted at 12 : 00 : 21 reach Bob? How many oscillations of the siren will Bob hear in that time interval? What is the frequency that Bob hear of the siren?
Answer:The sound emitted at 12:00:20 reach Bob at 12:00:21.The sound emitted at 12:00:21 reach Bob at 12:00:21:54.The frequency of the siren that Bob hears is 450,the number of the ocillations Bob will here in that interval is
\begin{align*}
1.9\times 450=855
\end{align*}
Exercise 3.2 (One-way Doppler shift from Alice to Bob). Suppose that Alice (A) is emitting radiation at some frequency $f$ (i.e. $f$ cycles per second), regardless of whether she is moving or stationary. This radiation travels at speed $c$ in all directions. Another observer Bob (B) is at rest.
(a) If Alice is moving at speed $v$ directly towards Bob (for some $0<v<c$), at what frequency does Bob perceive Alice’s radiation to be? Is this frequency higher or lower than $f$ ?
Answer:Bob receive Alice's rediation at the frequency of
\begin{align*}
\frac{f}{1-\frac{v}{c}}
\end{align*}
This frequency is higher than $f$.
(b) If Alice is moving at speed $v$ directly away from Bob (for some $0<v<c$), at what frequency does Bob perceive Alice’s radiation to be? Is this frequency higher or lower than $f$ ?
Answer:
\begin{align*}
\frac{f}{1+\frac{v}{c}}
\end{align*}
Lower than $f$.
(c) What happens in (a) or (b) if Alice travels at a speed $v$ that is faster than $c$? Draw a spacetime diagram to illustrate this situation.
Answer: It will form shock wave.
Exercise 3.3 (One-way Doppler shift from Bob to Alice). Suppose that Bob (B) is at rest, but is emitting radiation at some frequency $f$ while remaining stationary. This radiation travels at speed $c$ in all directions. Another observer Alice (A) is moving (but agrees with Bob on how fast time is passing). (See Figure 5)
(a) If Alice is moving at speed $v$ directly towards Bob (for some $0<v<c$), at what frequency does Alice perceive Bob’s radiation to be? Is this frequency higher or lower than $f$ ? Draw a spacetime diagram to illustrate this situation.
Answer:
\begin{align*}
f(1+\frac{v}{c})
\end{align*}
Higher than $f$.
(b) If Alice is moving at speed $v$ directly away from Bob (for some $0<v<c$), at what frequency does Alice perceive Bob’s radiation to be? Is this frequency higher or lower than $f$ ? Draw a spacetime diagram to illustrate this situation.
Answer:
\begin{align*}
f(1-\frac{v}{c})
\end{align*}
lower.
Exercise 3.4 (Two-way Doppler shifts (optional)). Suppose that Bob (B) is emitting radiation at some frequency $f$ while remaining stationary. This radiation travels at speed $c$ in all directions. Another observer Alice (A) is moving. If the radiation hits Alice, it immediately bounces back in the opposite direction(but still traveling at $c$), until it returns to Bob.
(a) If Alice is moving at speed $v$ directly towards Bob (for some $0<v<c$), at what frequency does Bob perceive the reflection of his radiation off of Alice to be? Draw a spacetime diagram to illustrate this situation.
Answer:
\begin{align*}
\frac{f(1+\frac{v}{c})}{1-\frac{v}{c}}
\end{align*}
(b) If Alice is moving at speed $v$ directly away from Bob (for some $0<v<c$), at what frequency does Bob perceive the reflection of his radiation off of Alice to be? Draw a spacetime diagram to illustrate this situation.
Answer:
\begin{align*}
\frac{f(1-\frac{v}{c})}{1+\frac{v}{c}}
\end{align*}
(c) What if instead of having the radiation emitting from Bob,bouncing from Alice, and returning to Bob, we consider the reverse situation of radiation emitting from Alice, bouncing from Bob, and returning to Alice?
Answer:
When Alice is moving towards Bob,
\begin{align*}
\frac{f(1+\frac{v}{c})}{1-\frac{v}{c}}
\end{align*}
When Alice is moving away from Bob,
\begin{align*}
\frac{f(1-\frac{v}{c})}{1+\frac{v}{c}}
\end{align*}
Exercise 4.1 Let Alice move at a speed $v$ while Bob is at rest,where $0<v<c$.
(d)From
\begin{align*}
\lambda \frac{c+v}{c}=\frac{1}{\lambda}\frac{c}{c-v}
\end{align*}
we know that
\begin{align*}
\lambda=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}
\end{align*}
(e)
\begin{align*}
\frac{\frac{1}{1-\frac{v}{c}}}{\frac{c}{\sqrt{c^2-v^2}}}=\frac{\sqrt{c+v}}{\sqrt{c-v}}
\end{align*}
\begin{align*}
\frac{\frac{c}{c+v}}{\frac{c}{\sqrt{c^2-v^2}}}=\frac{\sqrt{c-v}}{\sqrt{c+v}}
\end{align*}
(g)
\begin{align*}
\left(\frac{\sqrt{c+v}}{\sqrt{c-v}}\right)^2=\frac{c+v}{c-v}
\end{align*}
\begin{align*}
\left(\frac{\sqrt{c-v}}{\sqrt{c+v}}\right)^2=\frac{c-v}{c+v}
\end{align*}
Exercise 4.2. (a) What fraction of the speed of light does Alice has to travel at before her time dilation factor λ becomes 2 (so that Alice does everything twice as slowly as Bob)?
Answer:
\begin{align*}
\frac{1}{\sqrt{1-(\frac{v}{c})^2}}=2
\end{align*}
So
\begin{align*}
\frac{v}{c}=\frac{\sqrt{3}}{2}
\end{align*}
(b) What fraction of the speed of light does Alice has to travel at before her time dilation factor λ becomes 10?
\begin{align*}
\frac{1}{\sqrt{1-(\frac{v}{c})^2}}=10
\end{align*}
So
\begin{align*}
\frac{v}{c}=\frac{3 \sqrt{11}}{10}
\end{align*}
(c) What fraction of the speed of light does Alice has to travel at before her time dilation factor λ becomes 100?
\begin{align*}
\frac{3 \sqrt{1111}}{100}\approx 99.99\%
\end{align*}
Exercise 5.1.(b)Let $\frac{v}{c}=l$,It is easy to verify that
\begin{align*}
\frac{K}{E}=\frac{1}{\sqrt{1-l^2}}-1
\end{align*}
Done.
转载于:https://www.cnblogs.com/yeluqing/archive/2013/01/23/3827852.html