Новое решение

JAMF правка от 13 июн. 2026 г., 21:24

правка #18736 позже →

en/14.3.18.md +39 −0

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		+	### 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?
		+
		+	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.
		+	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 \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]

	@@ -0,0 +1,39 @@
			### 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?

			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.
			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 \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]