Edit to “Solution”

JAMF edited Jun 14, 2026, 3:04 PM

View solution History

revision #18768 parent #18732 ← older

en/14.3.16.md +2 −2

			### Statement

			$14.3.16$ Solve problem $14.3.15$ for plane wave propagation in a dielectric medium with refractive index $n$.

			$14.3.15$ How many times will the amplitude of a plane electromagnetic wave change
			when it passes into a coordinate system moving at a speed $\beta c$ in the direction
			of wave propagation?

			### Solution

			To solve this problem we need to use the Lorentz transformation for fields, taking into account that the fields of the electromagnetic wave are
			perpendicular to the velocity of the moving frame. Also, in each frame, the relation $\vec{E} = [\vec{v} \times \vec{B}]$ holds, where $\vec{v}$ is the
		@@ -13,13 +13,13 @@Solution
			velocity vector of the electromagnetic wave; in this case inside the dielectric media $\vec{v} = \frac{c}{n}$.

			\begin{equation}
		−	E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1-\beta n}{\sqrt{1-\beta^2}}
		+	E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1+\beta n}{\sqrt{1-\beta^2}}
			\end{equation}

			And using the relation between fields:

			\begin{equation}
		−	B' = B \frac{1-\beta n}{\sqrt{1-\beta^2}}
		+	B' = B \frac{1+\beta n}{\sqrt{1-\beta^2}}
			\end{equation}

			So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.

			#### Answer

			So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.
		unchanged lines 4

	### Statement		### Statement

	$14.3.16$ Solve problem $14.3.15$ for plane wave propagation in a dielectric medium with refractive index $n$.		$14.3.16$ Solve problem $14.3.15$ for plane wave propagation in a dielectric medium with refractive index $n$.

	$14.3.15$ How many times will the amplitude of a plane electromagnetic wave change		$14.3.15$ How many times will the amplitude of a plane electromagnetic wave change
	when it passes into a coordinate system moving at a speed $\beta c$ in the direction		when it passes into a coordinate system moving at a speed $\beta c$ in the direction
	of wave propagation?		of wave propagation?

	### Solution		### Solution

	To solve this problem we need to use the Lorentz transformation for fields, taking into account that the fields of the electromagnetic wave are		To solve this problem we need to use the Lorentz transformation for fields, taking into account that the fields of the electromagnetic wave are
	perpendicular to the velocity of the moving frame. Also, in each frame, the relation $\vec{E} = [\vec{v} \times \vec{B}]$ holds, where $\vec{v}$ is the		perpendicular to the velocity of the moving frame. Also, in each frame, the relation $\vec{E} = [\vec{v} \times \vec{B}]$ holds, where $\vec{v}$ is the
	@@ -13,13 +13,13 @@Solution
	velocity vector of the electromagnetic wave; in this case inside the dielectric media $\vec{v} = \frac{c}{n}$.		velocity vector of the electromagnetic wave; in this case inside the dielectric media $\vec{v} = \frac{c}{n}$.

	\begin{equation}		\begin{equation}
	E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1-\beta n}{\sqrt{1-\beta^2}}		E' = \frac{E - \beta c B}{\sqrt{1-\beta^2}} = E \frac{1+\beta n}{\sqrt{1-\beta^2}}
	\end{equation}		\end{equation}

	And using the relation between fields:		And using the relation between fields:

	\begin{equation}		\begin{equation}
	B' = B \frac{1-\beta n}{\sqrt{1-\beta^2}}		B' = B \frac{1+\beta n}{\sqrt{1-\beta^2}}
	\end{equation}		\end{equation}

	So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.		So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.

	#### Answer		#### Answer

	So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.		So, the amplitude becomes $\frac{1-\beta n}{\sqrt{1-\beta^2}}$ times smaller.
	unchanged lines 4