Правка разделов «Statement», «Solution», «Answer»
en/14.3.17.md
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| ### Statement | |||
| − | $14.3.17 | ||
| + | $14.3.17$ A plane electromagnetic wave is incident perpendicularly on a metal wall moving at the speed $\beta c$. How many times will the wave amplitude change during reflection? | ||
| ### Solution | |||
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| − | \begin{document} | ||
| − | |||
| − | $14.3.17$ A plane electromagnetic wave is incident perpendicularly on a metal wall moving at the speed $\beta c$. How many times will the wave amplitude change during reflection? | ||
| − | |||
| To solve this problem we need to use the idea that due to the relativistic Doppler effect, the frequency of the electromagnetic wave in the moving frame is $\nu' = \nu \sqrt{\frac{1+\beta}{1-\beta}}$. We can obtain this result using the Lorentz | |||
| transformation for momentum and energy and $E = h \nu$, $p = \frac{h}{\lambda} = \frac{h \nu}{c}$: | |||
| \begin{equation} | |||
| E' = \frac{E -(-\beta c) p}{\sqrt{1-\beta^2}} \rightarrow h \nu' = \frac{h \nu + \beta c \frac{h \nu}{c} }{\sqrt{1-\beta^2}} = \nu \sqrt{\frac{1+\beta}{1-\beta}} | |||
| \end{equation} | |||
| The negative sign is because the velocity of the frame points in the negative direction. This same frequency will be that of the reflected wave. Returning to the Earth frame using the same transformation: | |||
| \begin{equation} | |||
| @@ -42,12 +17,8 @@Solution | |||
| h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c}}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | |||
| \end{equation} | |||
| + | #### Answer | ||
| + | |||
| \begin{equation} | |||
| \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | |||
| \end{equation} | |||
| − | |||
| − | \end{document} | ||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,34 +1,9 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.17 |
$14.3.17$ A plane electromagnetic wave is incident perpendicularly on a metal wall moving at the speed $\beta c$. How many times will the wave amplitude change during reflection? | ||
| ### Solution | ### 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} | |||
| $14.3.17$ A plane electromagnetic wave is incident perpendicularly on a metal wall moving at the speed $\beta c$. How many times will the wave amplitude change during reflection? | |||
| To solve this problem we need to use the idea that due to the relativistic Doppler effect, the frequency of the electromagnetic wave in the moving frame is $\nu' = \nu \sqrt{\frac{1+\beta}{1-\beta}}$. We can obtain this result using the Lorentz | To solve this problem we need to use the idea that due to the relativistic Doppler effect, the frequency of the electromagnetic wave in the moving frame is $\nu' = \nu \sqrt{\frac{1+\beta}{1-\beta}}$. We can obtain this result using the Lorentz | ||
| transformation for momentum and energy and $E = h \nu$, $p = \frac{h}{\lambda} = \frac{h \nu}{c}$: | transformation for momentum and energy and $E = h \nu$, $p = \frac{h}{\lambda} = \frac{h \nu}{c}$: | ||
| \begin{equation} | \begin{equation} | ||
| E' = \frac{E -(-\beta c) p}{\sqrt{1-\beta^2}} \rightarrow h \nu' = \frac{h \nu + \beta c \frac{h \nu}{c} }{\sqrt{1-\beta^2}} = \nu \sqrt{\frac{1+\beta}{1-\beta}} | E' = \frac{E -(-\beta c) p}{\sqrt{1-\beta^2}} \rightarrow h \nu' = \frac{h \nu + \beta c \frac{h \nu}{c} }{\sqrt{1-\beta^2}} = \nu \sqrt{\frac{1+\beta}{1-\beta}} | ||
| \end{equation} | \end{equation} | ||
| The negative sign is because the velocity of the frame points in the negative direction. This same frequency will be that of the reflected wave. Returning to the Earth frame using the same transformation: | The negative sign is because the velocity of the frame points in the negative direction. This same frequency will be that of the reflected wave. Returning to the Earth frame using the same transformation: | ||
| \begin{equation} | \begin{equation} | ||
| @@ -42,12 +17,8 @@Solution | |||
| h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c}}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c}}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | ||
| \end{equation} | \end{equation} | ||
| #### Answer | |||
| \begin{equation} | \begin{equation} | ||
| \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | ||
| \end{equation} | \end{equation} | ||
| \end{document} | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||