Edits to “Statement”, “Solution”, “Answer”

Alexphysics edited Jun 23, 2026, 3:19 PM

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revision #19107 parent #19106 ← older

en/14.4.1.md +19 −8

		@@ -1,7 +1,18 @@
			### Statement

		−	$14.4.1.$ ~~[Insert the problem statement]~~
		+	$14.4.1.$

		+	An electron entering an extended stationary and uniform electric field at the
		+	velocity βc departs from it after a time τ The electron velocity is directed
		+	along the field. How long will an electron stay in the field if, on the contrary,
		+	the field hits a stationary electron with the same speed? Solve this problem
		+	in two ways, using: a) relativistic deceleration effect of time, b) the Lorentz
		+	formula, according to which the mass of a particle moving at a speed of βc,m =
		+	mi 1 − β2, where miis the rest mass of the particle.
		+	∗) If the problem does not require a numerical answer, denote the rest mass
		+	of the electron me, the charge e.
		+
		+
			### Solution

			First situation (system S): A uniform electric field $\mathbf{E}$ is at rest. An electron enters with speed$ v = \beta c$ parallel to $\mathbf{E}$, remains for a time$ \tau$ and exits.

			Second situation (system S'): The electron is initially at rest and it is the field that moves toward it with speed $v = \beta c$. The electron interacts with the field for a time T (to be determined).

			Method a) Time dilation

			In the second situation, we place ourselves in the reference frame that moves with the field (S). There the electron enters with speed $ \beta c $ and takes $\tau $ to exit

			In system S' (electron initially at rest), the moving field passes over it between two events (arrival of the leading edge and departure of the trailing edge of the field). The time interval in S' between these two events is

			$T = \gamma \tau, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$

			$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}}}$

			Method b) Lorentz force and relativistic Newton's second law

			First situation: Electron moving at $\beta c $ inside the field at rest.
			Force on the electron $ F = eE$
			The equation of motion in relativity

			$\frac{dp}{dt} = F \quad\Rightarrow\quad \Delta p = eE\,\tau$

			where $ p = \gamma m_e v = \dfrac{m_e v}{\sqrt{1-\beta^2}}$

			The electron reverses its speed (enters with $+v$ and exits with$ -v$). The total change in momentum is

			$\Delta p = \gamma m_e v - (-\gamma m_e v) = 2\gamma m_e v = 2\gamma m_e c\beta$

			Therefore:

			$\tau = \frac{2\gamma m_e c\beta}{eE}$

			Second situation: The field "hits" the electron at rest with speed $\beta c.
			$In the electron's reference frame, it starts from rest and is accelerated by the moving field. At the end, the electron acquires a speed $\beta_1 c $ given by the relativistic velocity addition between the initial speed of the field ($\beta c$) and the final speed in the field's frame ($-\beta c$)

			$\beta_1 c = \frac{\beta c + \beta c}{1 + \beta^2} = \frac{2\beta c}{1+\beta^2}$

			The change in momentum of the electron (from rest to $\beta_1 c$) is

			$\Delta p = \gamma_1 m_e \beta_1 c, \quad \gamma_1 = \frac{1}{\sqrt{1-\beta_1^2}}$

			The force on the electron is the same $eE $
			The time T that the interaction lasts satisfies
		@@ -50,27 +61,27 @@Solution

			$T = \frac{\Delta p}{eE} = \frac{\gamma_1 m_e \beta_1 c}{eE}$

		−	Using the relations between$ \gamma_1, \beta_1 and \gamma, \beta$
		+	Using the relations between$ \gamma_1, \beta_1$ and $ \gamma, \beta$

			it can be shown that

		−	$\gamma_1 = \gamma(1+\beta^2), \quad \beta_1 = \frac{2\beta}{1+\beta^2},
		+	$\gamma_1 = \gamma(1+\beta^2), \quad \beta_1 = \frac{2\beta}{1+\beta^2}$

		−	\gamma_1 \beta_1 = 2\gamma \beta$
		+	$\gamma_1 \beta_1 = 2\gamma \beta$

			Therefore

			$T = \frac{2\gamma m_e c\beta}{eE} = \gamma \tau$

			$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$

		−	4. Final result
		−
			Both methods yield the same result:

		−	\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}.
		+	$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$.


			#### Answer

		−	~~[Insert a concise answer or boxed~~ result]
		+	Both methods yield the same result:
		+
		+	$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$.

	@@ -1,7 +1,18 @@
	### Statement		### Statement

	$14.4.1.$ ~~[Insert the problem statement]~~		$14.4.1.$

			An electron entering an extended stationary and uniform electric field at the
			velocity βc departs from it after a time τ The electron velocity is directed
			along the field. How long will an electron stay in the field if, on the contrary,
			the field hits a stationary electron with the same speed? Solve this problem
			in two ways, using: a) relativistic deceleration effect of time, b) the Lorentz
			formula, according to which the mass of a particle moving at a speed of βc,m =
			mi 1 − β2, where miis the rest mass of the particle.
			∗) If the problem does not require a numerical answer, denote the rest mass
			of the electron me, the charge e.


	### Solution		### Solution

	First situation (system S): A uniform electric field $\mathbf{E}$ is at rest. An electron enters with speed$ v = \beta c$ parallel to $\mathbf{E}$, remains for a time$ \tau$ and exits.		First situation (system S): A uniform electric field $\mathbf{E}$ is at rest. An electron enters with speed$ v = \beta c$ parallel to $\mathbf{E}$, remains for a time$ \tau$ and exits.

	Second situation (system S'): The electron is initially at rest and it is the field that moves toward it with speed $v = \beta c$. The electron interacts with the field for a time T (to be determined).		Second situation (system S'): The electron is initially at rest and it is the field that moves toward it with speed $v = \beta c$. The electron interacts with the field for a time T (to be determined).

	Method a) Time dilation		Method a) Time dilation

	In the second situation, we place ourselves in the reference frame that moves with the field (S). There the electron enters with speed $ \beta c $ and takes $\tau $ to exit		In the second situation, we place ourselves in the reference frame that moves with the field (S). There the electron enters with speed $ \beta c $ and takes $\tau $ to exit

	In system S' (electron initially at rest), the moving field passes over it between two events (arrival of the leading edge and departure of the trailing edge of the field). The time interval in S' between these two events is		In system S' (electron initially at rest), the moving field passes over it between two events (arrival of the leading edge and departure of the trailing edge of the field). The time interval in S' between these two events is

	$T = \gamma \tau, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$		$T = \gamma \tau, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$

	$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}}}$		$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}}}$

	Method b) Lorentz force and relativistic Newton's second law		Method b) Lorentz force and relativistic Newton's second law

	First situation: Electron moving at $\beta c $ inside the field at rest.		First situation: Electron moving at $\beta c $ inside the field at rest.
	Force on the electron $ F = eE$		Force on the electron $ F = eE$
	The equation of motion in relativity		The equation of motion in relativity

	$\frac{dp}{dt} = F \quad\Rightarrow\quad \Delta p = eE\,\tau$		$\frac{dp}{dt} = F \quad\Rightarrow\quad \Delta p = eE\,\tau$

	where $ p = \gamma m_e v = \dfrac{m_e v}{\sqrt{1-\beta^2}}$		where $ p = \gamma m_e v = \dfrac{m_e v}{\sqrt{1-\beta^2}}$

	The electron reverses its speed (enters with $+v$ and exits with$ -v$). The total change in momentum is		The electron reverses its speed (enters with $+v$ and exits with$ -v$). The total change in momentum is

	$\Delta p = \gamma m_e v - (-\gamma m_e v) = 2\gamma m_e v = 2\gamma m_e c\beta$		$\Delta p = \gamma m_e v - (-\gamma m_e v) = 2\gamma m_e v = 2\gamma m_e c\beta$

	Therefore:		Therefore:

	$\tau = \frac{2\gamma m_e c\beta}{eE}$		$\tau = \frac{2\gamma m_e c\beta}{eE}$

	Second situation: The field "hits" the electron at rest with speed $\beta c.		Second situation: The field "hits" the electron at rest with speed $\beta c.
	$In the electron's reference frame, it starts from rest and is accelerated by the moving field. At the end, the electron acquires a speed $\beta_1 c $ given by the relativistic velocity addition between the initial speed of the field ($\beta c$) and the final speed in the field's frame ($-\beta c$)		$In the electron's reference frame, it starts from rest and is accelerated by the moving field. At the end, the electron acquires a speed $\beta_1 c $ given by the relativistic velocity addition between the initial speed of the field ($\beta c$) and the final speed in the field's frame ($-\beta c$)

	$\beta_1 c = \frac{\beta c + \beta c}{1 + \beta^2} = \frac{2\beta c}{1+\beta^2}$		$\beta_1 c = \frac{\beta c + \beta c}{1 + \beta^2} = \frac{2\beta c}{1+\beta^2}$

	The change in momentum of the electron (from rest to $\beta_1 c$) is		The change in momentum of the electron (from rest to $\beta_1 c$) is

	$\Delta p = \gamma_1 m_e \beta_1 c, \quad \gamma_1 = \frac{1}{\sqrt{1-\beta_1^2}}$		$\Delta p = \gamma_1 m_e \beta_1 c, \quad \gamma_1 = \frac{1}{\sqrt{1-\beta_1^2}}$

	The force on the electron is the same $eE $		The force on the electron is the same $eE $
	The time T that the interaction lasts satisfies		The time T that the interaction lasts satisfies
	@@ -50,27 +61,27 @@Solution

	$T = \frac{\Delta p}{eE} = \frac{\gamma_1 m_e \beta_1 c}{eE}$		$T = \frac{\Delta p}{eE} = \frac{\gamma_1 m_e \beta_1 c}{eE}$

	Using the relations between$ \gamma_1, \beta_1 and \gamma, \beta$		Using the relations between$ \gamma_1, \beta_1$ and $ \gamma, \beta$

	it can be shown that		it can be shown that

	$\gamma_1 = \gamma(1+\beta^2), \quad \beta_1 = \frac{2\beta}{1+\beta^2},		$\gamma_1 = \gamma(1+\beta^2), \quad \beta_1 = \frac{2\beta}{1+\beta^2}$

	\gamma_1 \beta_1 = 2\gamma \beta$		$\gamma_1 \beta_1 = 2\gamma \beta$

	Therefore		Therefore

	$T = \frac{2\gamma m_e c\beta}{eE} = \gamma \tau$		$T = \frac{2\gamma m_e c\beta}{eE} = \gamma \tau$

	$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$		$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$

	4. Final result

	Both methods yield the same result:		Both methods yield the same result:

	\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}.		$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$.


	#### Answer		#### Answer

	~~[Insert a concise answer or boxed~~ result]		Both methods yield the same result:

			$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$.