Правка разделов «Statement», «Solution», «Answer»
en/12.1.29.md
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| @@ -1,13 +1,11 @@ | |||
| ### Statement | |||
| − | $12.1.29.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| $12.1.29$ The frequency of a sinusoidal wave incident on a moving metal wall perpendicular | |||
| to its surface changes by $\Delta$ during reflection. The initial frequency of | |||
| the wave is $\nu_0$. Determine the wall speed. | |||
| + | ### Solution | ||
| + | |||
| The change in frequency of the electromagnetic wave is due to the relativistic Doppler effect. The frequency of the reflected wave is: | |||
| \begin{equation} | |||
| \nu = \nu_0 \frac{1+v/c}{1-v/c} | |||
| \end{equation} | |||
| We can obtain this result using the Lorentz transformation for energy and momentum: | |||
| \begin{equation} | |||
| h \nu = \frac{h \nu_0 - (-v) h \nu_0 / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu = \nu_0 \sqrt{\frac{1+v/c}{1-v/c}} | |||
| \end{equation} | |||
| After reflection in the moving frame, the wave will have the same frequency; coming back to the Earth frame: | |||
| \begin{equation} | |||
| h \nu_1 = \frac{h \nu + v h \nu / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu_1 = \nu_0 \frac{1+v/c}{1-v/c} | |||
| \end{equation} | |||
| So, the difference is: | |||
| \begin{equation} | |||
| \Delta = \nu_1 - \nu_0 = \nu_0 \left(\frac{1+v/c}{1-v/c} - 1\right) = \nu_0 \frac{2 v/c}{1-v/c} | |||
| \end{equation} | |||
| And finally: | |||
| \begin{equation} | |||
| v = \frac{c \Delta }{2 \nu_0 + \Delta} | |||
| \end{equation} | |||
| @@ -40,4 +38,6 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | \begin{equation} | ||
| + | v = \frac{c \Delta }{2 \nu_0 + \Delta} | ||
| + | \end{equation} | ||
| @@ -1,13 +1,11 @@ | |||
| ### Statement | ### Statement | ||
| $12.1.29.$ [Insert the problem statement] | |||
| ### Solution | |||
| $12.1.29$ The frequency of a sinusoidal wave incident on a moving metal wall perpendicular | $12.1.29$ The frequency of a sinusoidal wave incident on a moving metal wall perpendicular | ||
| to its surface changes by $\Delta$ during reflection. The initial frequency of | to its surface changes by $\Delta$ during reflection. The initial frequency of | ||
| the wave is $\nu_0$. Determine the wall speed. | the wave is $\nu_0$. Determine the wall speed. | ||
| ### Solution | |||
| The change in frequency of the electromagnetic wave is due to the relativistic Doppler effect. The frequency of the reflected wave is: | The change in frequency of the electromagnetic wave is due to the relativistic Doppler effect. The frequency of the reflected wave is: | ||
| \begin{equation} | \begin{equation} | ||
| \nu = \nu_0 \frac{1+v/c}{1-v/c} | \nu = \nu_0 \frac{1+v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| We can obtain this result using the Lorentz transformation for energy and momentum: | We can obtain this result using the Lorentz transformation for energy and momentum: | ||
| \begin{equation} | \begin{equation} | ||
| h \nu = \frac{h \nu_0 - (-v) h \nu_0 / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu = \nu_0 \sqrt{\frac{1+v/c}{1-v/c}} | h \nu = \frac{h \nu_0 - (-v) h \nu_0 / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu = \nu_0 \sqrt{\frac{1+v/c}{1-v/c}} | ||
| \end{equation} | \end{equation} | ||
| After reflection in the moving frame, the wave will have the same frequency; coming back to the Earth frame: | After reflection in the moving frame, the wave will have the same frequency; coming back to the Earth frame: | ||
| \begin{equation} | \begin{equation} | ||
| h \nu_1 = \frac{h \nu + v h \nu / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu_1 = \nu_0 \frac{1+v/c}{1-v/c} | h \nu_1 = \frac{h \nu + v h \nu / c}{\sqrt{1-v^2/c^2}} \rightarrow \nu_1 = \nu_0 \frac{1+v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| So, the difference is: | So, the difference is: | ||
| \begin{equation} | \begin{equation} | ||
| \Delta = \nu_1 - \nu_0 = \nu_0 \left(\frac{1+v/c}{1-v/c} - 1\right) = \nu_0 \frac{2 v/c}{1-v/c} | \Delta = \nu_1 - \nu_0 = \nu_0 \left(\frac{1+v/c}{1-v/c} - 1\right) = \nu_0 \frac{2 v/c}{1-v/c} | ||
| \end{equation} | \end{equation} | ||
| And finally: | And finally: | ||
| \begin{equation} | \begin{equation} | ||
| v = \frac{c \Delta }{2 \nu_0 + \Delta} | v = \frac{c \Delta }{2 \nu_0 + \Delta} | ||
| \end{equation} | \end{equation} | ||
| @@ -40,4 +38,6 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | \begin{equation} | ||
| v = \frac{c \Delta }{2 \nu_0 + \Delta} | |||
| \end{equation} | |||