Edits to “Statement”, “Solution”, “Answer”
en/14.3.18.md
+6 −10
| @@ -1,15 +1,13 @@ | |||
| ### Statement | |||
| − | $14.3.18.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| $14.3.18$ Solve problem 14.3.17 when an electromagnetic wave hits a moving wall at an | |||
| angle $\alpha$. | |||
| $14.3.17$ 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 | ||
| + | |||
| To solve this problem we can use the Lorentz transformation for momentum and energy in the horizontal direction. | |||
| The horizontal component of the momentum is: | |||
| \begin{equation} | |||
| p_x = \frac{h \nu }{c} \cos\alpha \qquad E = h \nu | |||
| \end{equation} | |||
| Everything else is the same as in problem $14.3.17$. | |||
| \begin{equation} | |||
| h \nu' = \frac{h \nu - (-\beta c) p_x}{\sqrt{1-\beta^2}} = \frac{h \nu + h \nu \beta \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu' = \nu \frac{1+\beta \cos \alpha}{\sqrt{1-\beta^2}} | |||
| \end{equation} | |||
| The negative sign appears because the velocity of the wall points in the negative direction. | |||
| @@ -27,13 +25,11 @@Solution | |||
| The reflected wave will have this frequency; then transforming back to the Earth frame: | |||
| \begin{equation} | |||
| − | h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta | ||
| + | h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2} | ||
| \end{equation} | |||
| − | \begin{equation} | ||
| − | \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | ||
| − | \end{equation} | ||
| − | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2} | ||
| + | \end{equation} | ||
| @@ -1,15 +1,13 @@ | |||
| ### Statement | ### Statement | ||
| $14.3.18.$ [Insert the problem statement] | |||
| ### Solution | |||
| $14.3.18$ Solve problem 14.3.17 when an electromagnetic wave hits a moving wall at an | $14.3.18$ Solve problem 14.3.17 when an electromagnetic wave hits a moving wall at an | ||
| angle $\alpha$. | angle $\alpha$. | ||
| $14.3.17$ 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 | $14.3.17$ 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? | reflection? | ||
| ### Solution | |||
| To solve this problem we can use the Lorentz transformation for momentum and energy in the horizontal direction. | To solve this problem we can use the Lorentz transformation for momentum and energy in the horizontal direction. | ||
| The horizontal component of the momentum is: | The horizontal component of the momentum is: | ||
| \begin{equation} | \begin{equation} | ||
| p_x = \frac{h \nu }{c} \cos\alpha \qquad E = h \nu | p_x = \frac{h \nu }{c} \cos\alpha \qquad E = h \nu | ||
| \end{equation} | \end{equation} | ||
| Everything else is the same as in problem $14.3.17$. | Everything else is the same as in problem $14.3.17$. | ||
| \begin{equation} | \begin{equation} | ||
| h \nu' = \frac{h \nu - (-\beta c) p_x}{\sqrt{1-\beta^2}} = \frac{h \nu + h \nu \beta \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu' = \nu \frac{1+\beta \cos \alpha}{\sqrt{1-\beta^2}} | h \nu' = \frac{h \nu - (-\beta c) p_x}{\sqrt{1-\beta^2}} = \frac{h \nu + h \nu \beta \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu' = \nu \frac{1+\beta \cos \alpha}{\sqrt{1-\beta^2}} | ||
| \end{equation} | \end{equation} | ||
| The negative sign appears because the velocity of the wall points in the negative direction. | The negative sign appears because the velocity of the wall points in the negative direction. | ||
| @@ -27,13 +25,11 @@Solution | |||
| The reflected wave will have this frequency; then transforming back to the Earth frame: | The reflected wave will have this frequency; then transforming back to the Earth frame: | ||
| \begin{equation} | \begin{equation} | ||
| h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta |
h \nu_{\text{reflected}} = \frac{h \nu' + \beta c \frac{h \nu'}{c} \cos\alpha}{\sqrt{1-\beta^2}} \rightarrow \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2} | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | |||
| \nu_{\text{reflected}} = \nu \frac{1+\beta}{1-\beta} | |||
| \end{equation} | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| \nu_{\text{reflected}} = \nu \frac{(1+\beta\cos\alpha)^2}{1-\beta^2} | |||
| \end{equation} | |||