Новое решение
en/14.1.7.md
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| + | ### Statement | ||
| + | |||
| + | $14.1.7.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | \documentclass[12pt,a4paper]{article} | ||
| + | \usepackage[english]{babel} | ||
| + | \usepackage{float} | ||
| + | \usepackage{wrapfig} | ||
| + | \usepackage{lmodern} | ||
| + | \usepackage[T1]{fontenc} | ||
| + | \usepackage[utf8]{inputenc} | ||
| + | \usepackage{microtype} | ||
| + | \usepackage{graphicx} | ||
| + | \usepackage{booktabs} | ||
| + | \usepackage{amsmath,amssymb} | ||
| + | \usepackage{hyperref} | ||
| + | \usepackage{csquotes} | ||
| + | \usepackage{geometry} | ||
| + | \usepackage{subcaption} | ||
| + | \usepackage{tikz} | ||
| + | \usepackage{array} | ||
| + | \usepackage{pgfplots} | ||
| + | \usepackage{wrapfig} | ||
| + | \usepackage{subcaption} | ||
| + | |||
| + | \begin{document} | ||
| + | |||
| + | \begin{center} | ||
| + | \Large \textbf{Statement} | ||
| + | \end{center} | ||
| + | |||
| + | $14.1.7$: Along the line connecting two stationary stations relative to each other, a | ||
| + | spacecraft was moving at a speed of $v$ relative to the stations. ”The stations | ||
| + | were at the same distance from our ship when our light signal was reflected | ||
| + | on them at the same time, since the light signals were sent simultaneously | ||
| + | to the stations and they returned after being reflected from the stations at | ||
| + | the same time,” says the observer from the ship. Employees of the station | ||
| + | observed that the signals were reflected from the stations at different times. | ||
| + | How can these differences be explained? What difference in reflection times | ||
| + | was observed by the station staff if the distance between the stations (in their | ||
| + | system) is equal to l? At what distances did they fix the ship at the moments | ||
| + | of reflections of signals from stations? | ||
| + | |||
| + | \begin{center} | ||
| + | \Large \textbf{Solution} | ||
| + | \end{center} | ||
| + | |||
| + | The key idea in this problem is that light takes a finite, non-zero time to travel from one place to another. To solve this problem we can use | ||
| + | Lorentz transformations or some direct calculations analyzing the process. Let's see both; let's begin with the first way: | ||
| + | |||
| + | \begin{center} | ||
| + | \large \textbf{First Solution} | ||
| + | \end{center} | ||
| + | |||
| + | First, the signals were sent by the spaceship when the spaceship was at distances $x_1$ and $l-x_1$ from the stations. Henceforth, we will work in the frame where the stations are at rest for this solution. Also, the signal traveling to the left is signal $1$ and is reflected by the first station, and the signal traveling to the right is signal $2$ and is reflected by station $2$. | ||
| + | |||
| + | To calculate $x_1$ we need to use the idea that the signals, after being reflected, reach the ship at the midpoint between the stations; otherwise the signals would not reach the ship at the same time. | ||
| + | |||
| + | The time taken by ray $1$ to reach the spaceship at the midpoint is: | ||
| + | |||
| + | \begin{equation} | ||
| + | t_1 = \frac{x_1 + l/2}{c} | ||
| + | \end{equation} | ||
| + | |||
| + | On the other hand, the spaceship will move a distance $l/2 - x_1 = v t_1$ in that time. Substituting the time into this equation yields: | ||
| + | |||
| + | \begin{equation} | ||
| + | \frac{x_1 + l/2}{c} = \frac{l/2 - x_1}{v} \;\rightarrow\; x_1 = \frac{l}{2} \frac{1-v/c}{1+v/c} | ||
| + | \end{equation} | ||
| + | |||
| + | Knowing this, we can find the time difference between signals at one of the stations. The difference $\Delta t$ arises because ray $1$ takes a time | ||
| + | $t_{11} = \frac{x_1}{c} = \frac{l}{2c} \frac{1-v/c}{1+v/c}$ to travel from the emission point to the first station. On | ||
| + | the other hand, ray $2$ takes a time $t_{22} = \frac{l-x_1}{c} = \frac{l}{2c} \frac{1+3v/c}{1+v/c}$ to reach the second station, and a time $t_{21} = \frac{l}{c}$ | ||
| + | to travel from the first station to the second station (or vice versa). Thus, the difference between the reflection times as seen from station $1$ is: | ||
| + | |||
| + | \begin{equation} | ||
| + | \Delta t = t_{22} + t_{21} - t_{11} = \frac{l}{c} + \frac{2l (v/c)}{1+v/c} = \frac{l}{c} \frac{1 + 3v/c}{1+v/c} | ||
| + | \end{equation} | ||
| + | |||
| + | Finally, when the spaceship sends the signals, it is at a distance $x_1$ from station $1$. The stations will observe the spaceship's position when the signals reach each station. Note that the signals take time to travel from the emission point to the stations. When signal $1$ reaches station $1$, the spaceship will be at a distance: | ||
| + | |||
| + | \begin{equation} | ||
| + | x'_1 = x_1 + v t_{11} = \frac{l}{2} \left(1-\frac{v}{c}\right) | ||
| + | \end{equation} | ||
| + | |||
| + | And when ray $2$ reaches station $2$, the spaceship will be at: | ||
| + | |||
| + | \begin{equation} | ||
| + | x'_2 = l - x_1 - v t_{22} = (l - x_1)\left(1-\frac{v}{c}\right) = \frac{l}{2} \frac{1+3v/c}{1+v/c} \left(1-\frac{v}{c}\right) | ||
| + | \end{equation} | ||
| + | |||
| + | \end{document} | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,98 @@ | |||
| ### Statement | |||
| $14.1.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| \documentclass[12pt,a4paper]{article} | |||
| \usepackage[english]{babel} | |||
| \usepackage{float} | |||
| \usepackage{wrapfig} | |||
| \usepackage{lmodern} | |||
| \usepackage[T1]{fontenc} | |||
| \usepackage[utf8]{inputenc} | |||
| \usepackage{microtype} | |||
| \usepackage{graphicx} | |||
| \usepackage{booktabs} | |||
| \usepackage{amsmath,amssymb} | |||
| \usepackage{hyperref} | |||
| \usepackage{csquotes} | |||
| \usepackage{geometry} | |||
| \usepackage{subcaption} | |||
| \usepackage{tikz} | |||
| \usepackage{array} | |||
| \usepackage{pgfplots} | |||
| \usepackage{wrapfig} | |||
| \usepackage{subcaption} | |||
| \begin{document} | |||
| \begin{center} | |||
| \Large \textbf{Statement} | |||
| \end{center} | |||
| $14.1.7$: Along the line connecting two stationary stations relative to each other, a | |||
| spacecraft was moving at a speed of $v$ relative to the stations. ”The stations | |||
| were at the same distance from our ship when our light signal was reflected | |||
| on them at the same time, since the light signals were sent simultaneously | |||
| to the stations and they returned after being reflected from the stations at | |||
| the same time,” says the observer from the ship. Employees of the station | |||
| observed that the signals were reflected from the stations at different times. | |||
| How can these differences be explained? What difference in reflection times | |||
| was observed by the station staff if the distance between the stations (in their | |||
| system) is equal to l? At what distances did they fix the ship at the moments | |||
| of reflections of signals from stations? | |||
| \begin{center} | |||
| \Large \textbf{Solution} | |||
| \end{center} | |||
| The key idea in this problem is that light takes a finite, non-zero time to travel from one place to another. To solve this problem we can use | |||
| Lorentz transformations or some direct calculations analyzing the process. Let's see both; let's begin with the first way: | |||
| \begin{center} | |||
| \large \textbf{First Solution} | |||
| \end{center} | |||
| First, the signals were sent by the spaceship when the spaceship was at distances $x_1$ and $l-x_1$ from the stations. Henceforth, we will work in the frame where the stations are at rest for this solution. Also, the signal traveling to the left is signal $1$ and is reflected by the first station, and the signal traveling to the right is signal $2$ and is reflected by station $2$. | |||
| To calculate $x_1$ we need to use the idea that the signals, after being reflected, reach the ship at the midpoint between the stations; otherwise the signals would not reach the ship at the same time. | |||
| The time taken by ray $1$ to reach the spaceship at the midpoint is: | |||
| \begin{equation} | |||
| t_1 = \frac{x_1 + l/2}{c} | |||
| \end{equation} | |||
| On the other hand, the spaceship will move a distance $l/2 - x_1 = v t_1$ in that time. Substituting the time into this equation yields: | |||
| \begin{equation} | |||
| \frac{x_1 + l/2}{c} = \frac{l/2 - x_1}{v} \;\rightarrow\; x_1 = \frac{l}{2} \frac{1-v/c}{1+v/c} | |||
| \end{equation} | |||
| Knowing this, we can find the time difference between signals at one of the stations. The difference $\Delta t$ arises because ray $1$ takes a time | |||
| $t_{11} = \frac{x_1}{c} = \frac{l}{2c} \frac{1-v/c}{1+v/c}$ to travel from the emission point to the first station. On | |||
| the other hand, ray $2$ takes a time $t_{22} = \frac{l-x_1}{c} = \frac{l}{2c} \frac{1+3v/c}{1+v/c}$ to reach the second station, and a time $t_{21} = \frac{l}{c}$ | |||
| to travel from the first station to the second station (or vice versa). Thus, the difference between the reflection times as seen from station $1$ is: | |||
| \begin{equation} | |||
| \Delta t = t_{22} + t_{21} - t_{11} = \frac{l}{c} + \frac{2l (v/c)}{1+v/c} = \frac{l}{c} \frac{1 + 3v/c}{1+v/c} | |||
| \end{equation} | |||
| Finally, when the spaceship sends the signals, it is at a distance $x_1$ from station $1$. The stations will observe the spaceship's position when the signals reach each station. Note that the signals take time to travel from the emission point to the stations. When signal $1$ reaches station $1$, the spaceship will be at a distance: | |||
| \begin{equation} | |||
| x'_1 = x_1 + v t_{11} = \frac{l}{2} \left(1-\frac{v}{c}\right) | |||
| \end{equation} | |||
| And when ray $2$ reaches station $2$, the spaceship will be at: | |||
| \begin{equation} | |||
| x'_2 = l - x_1 - v t_{22} = (l - x_1)\left(1-\frac{v}{c}\right) = \frac{l}{2} \frac{1+3v/c}{1+v/c} \left(1-\frac{v}{c}\right) | |||
| \end{equation} | |||
| \end{document} | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||