New solution

Valter edited
View solution History
revision #22175
en/14.3.8.md +737 −0
@@ -0,0 +1,737 @@
+### Statement
+
+$14.3.8.$ a. The formula for the transformation of the $\overrightarrow{E}$ and $\overrightarrow{B}$ fields when they move at the speed $\overrightarrow{\beta} c$ has the following form:
+
+$\overrightarrow{E}' = \overrightarrow{E}_{\parallel} + \gamma(\overrightarrow{E}_{\perp} - [\overrightarrow{\beta} \times \overrightarrow{B} ]), \overrightarrow{B}' = \overrightarrow{B}_{\parallel} + \gamma(\overrightarrow{B}_{\perp} + (\overrightarrow{\beta} \times \overrightarrow{E} ]), \gamma = \frac{1}{\sqrt{1 - \beta^2}}$,
+
+where $\overrightarrow{E}'$ and $\overrightarrow{B}'$ are the electric and magnetic fields in the drift; $\overrightarrow{E}_{\parallel}$, $\overrightarrow{E}_{\perp}$ and $\overrightarrow{B}_{\parallel}$, $\overrightarrow{B}_{\perp}$ — components electric and magnetic fields fields, parallel services and perpendicular lines $- c \overrightarrow{\beta}$ in the initial system. The movement of the $\overrightarrow{E}'$ and $\overrightarrow{B}'$ fields at a speed of $- c \beta$ returns the previous state. Check it out.
+
+b. Using the field transformation formulas given in point $a$, solve the following problems: 14.3.1–14.3.3, 14.3.5.
+
+c. Using the field transformation formulas given in point $a$, solve problems 14.3.6 a, b, and 14.3.7.
+
+d. Prove that for $\beta \to 1$, the fields $\overrightarrow{E}'$ and $B'$ are perpendicular.
+
+### Solution
+
+Introduce
+
+$$
+\boldsymbol{\beta}=\frac{\mathbf v}{c},\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}.
+$$
+
+The components of the electric and magnetic fields parallel and perpendicular to $\boldsymbol{\beta}$ will be denoted by the subscripts $\parallel$ and $\perp$, respectively.
+
+The field transformation formulas are
+
+$$
+\mathbf E'=\mathbf E_{\parallel}+\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),
+$$
+
+$$
+\mathbf B'=\mathbf B_{\parallel}+\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).
+$$
+
+---
+
+#### a) Verification of the inverse transformation
+
+After the first transformation with velocity $c\boldsymbol{\beta}$, the parallel components remain unchanged:
+
+$$
+\mathbf E'_{\parallel}=\mathbf E_{\parallel},\qquad \mathbf B'_{\parallel}=\mathbf B_{\parallel}.
+$$
+
+The transverse components are
+
+$$
+\mathbf E'_{\perp}=\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),
+$$
+
+$$
+\mathbf B'_{\perp}=\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).
+$$
+
+Now perform the inverse transformation by replacing $\boldsymbol{\beta}$ with $-\boldsymbol{\beta}$.
+
+For the transverse electric field,
+
+$$
+\mathbf E''_{\perp}=\gamma\left(\mathbf E'_{\perp}+[\boldsymbol{\beta}\times\mathbf B']\right).
+$$
+
+Since the parallel component of $\mathbf B'$ gives no contribution to the cross product,
+
+$$
+[\boldsymbol{\beta}\times\mathbf B']=\gamma\left([\boldsymbol{\beta}\times\mathbf B_{\perp}]+[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]\right).
+$$
+
+Therefore,
+
+$$
+\mathbf E''_{\perp}=\gamma^2\left(\mathbf E_{\perp}+[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]\right).
+$$
+
+Using
+
+$$
+[\mathbf a\times[\mathbf a\times\mathbf b]]=\mathbf a(\mathbf a\cdot\mathbf b)-a^2\mathbf b,
+$$
+
+and noting that only the transverse part of $\mathbf E$ contributes, we obtain
+
+$$
+[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]=-\beta^2\mathbf E_{\perp}.
+$$
+
+Hence,
+
+$$
+\mathbf E''_{\perp}=\gamma^2(1-\beta^2)\mathbf E_{\perp}.
+$$
+
+Since
+
+$$
+\gamma^2(1-\beta^2)=1,
+$$
+
+we find
+
+$$
+\mathbf E''_{\perp}=\mathbf E_{\perp}.
+$$
+
+The parallel component is unchanged as well:
+
+$$
+\mathbf E''_{\parallel}=\mathbf E_{\parallel}.
+$$
+
+Thus,
+
+$$
+\boxed{\mathbf E''=\mathbf E}.
+$$
+
+The same calculation for the magnetic field gives
+
+$$
+\boxed{\mathbf B''=\mathbf B}.
+$$
+
+Therefore, two successive transformations with velocities $c\boldsymbol{\beta}$ and $-c\boldsymbol{\beta}$ return the fields to their original values.
+
+---
+
+### b) Applications of the transformation formulas
+
+#### Problem 14.3.1
+
+In the rest frame of the capacitor,
+
+$$
+\mathbf B=0.
+$$
+
+The capacitor moves parallel to its plates, whereas the electric field is perpendicular to them. Hence,
+
+$$
+\mathbf E\perp\boldsymbol{\beta}.
+$$
+
+Therefore,
+
+$$
+\mathbf E'=\gamma\mathbf E,
+$$
+
+and
+
+$$
+\boxed{E'=\gamma E}.
+$$
+
+Because of Lorentz contraction, the dimension of each plate along the direction of motion decreases by a factor $\gamma$. Since the charge of the plate is invariant, its surface charge density becomes
+
+$$
+\boxed{\sigma'=\gamma\sigma}.
+$$
+
+The magnetic field is
+
+$$
+\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].
+$$
+
+Since $\mathbf E'=\gamma\mathbf E$,
+
+$$
+\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.
+$$
+
+In magnitude,
+
+$$
+\boxed{B'=\beta E'=\gamma\beta E}.
+$$
+
+Thus,
+
+$$
+\boxed{\sigma'=\gamma\sigma,\qquad E'=\gamma E,\qquad B'=\beta E'}.
+$$
+
+---
+
+#### Problem 14.3.2
+
+The capacitor now moves at an angle $\alpha$ to the planes of its plates.
+
+Since the electric field is perpendicular to the plates,
+
+$$
+E_{\parallel}=E\sin\alpha,
+$$
+
+$$
+E_{\perp}=E\cos\alpha.
+$$
+
+The parallel component is unchanged:
+
+$$
+\boxed{E'_{\parallel}=E\sin\alpha}.
+$$
+
+The transverse component is multiplied by $\gamma$:
+
+$$
+\boxed{E'_{\perp}=\gamma E\cos\alpha}.
+$$
+
+Therefore,
+
+$$
+\boxed{E'=E\sqrt{\sin^2\alpha+\gamma^2\cos^2\alpha}}.
+$$
+
+The magnetic field is
+
+$$
+\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E],
+$$
+
+so that
+
+$$
+\boxed{B'=\gamma\beta E\cos\alpha}.
+$$
+
+Since
+
+$$
+E'_{\perp}=\gamma E\cos\alpha,
+$$
+
+we have
+
+$$
+\boxed{B'=\beta E'_{\perp}}.
+$$
+
+Equivalently, in vector form,
+
+$$
+\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.
+$$
+
+If the transformed surface charge density is also required, the area of a plate transforms as
+
+$$
+S'=S\sqrt{1-\beta^2\cos^2\alpha},
+$$
+
+and therefore
+
+$$
+\boxed{\sigma'=\frac{\sigma}{\sqrt{1-\beta^2\cos^2\alpha}}}.
+$$
+
+---
+
+#### Problem 14.3.3
+
+Let $\rho$ be the charge per unit length of the wire in its rest frame.
+
+The electric field of an infinite charged wire in Gaussian units is
+
+$$
+E_r=\frac{2\rho}{r},
+$$
+
+where $r$ is the distance from the wire.
+
+There is no magnetic field in the rest frame:
+
+$$
+\mathbf B=0.
+$$
+
+Since the wire moves along its own direction, the electric field is perpendicular to $\boldsymbol{\beta}$. Hence,
+
+$$
+E'_r=\gamma E_r.
+$$
+
+Therefore,
+
+$$
+\boxed{E'_r=\frac{2\gamma\rho}{r}}.
+$$
+
+Lorentz contraction also gives the transformed linear charge density
+
+$$
+\boxed{\rho'=\gamma\rho}.
+$$
+
+The magnetic field is
+
+$$
+\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].
+$$
+
+Thus,
+
+$$
+\boxed{B'=\beta E'_r=\frac{2\gamma\beta\rho}{r}}.
+$$
+
+Hence,
+
+$$
+\boxed{\rho'=\gamma\rho,\qquad E'_r=\frac{2\gamma\rho}{r},\qquad B'=\frac{2\gamma\beta\rho}{r}}.
+$$
+
+---
+
+#### Problem 14.3.5
+
+In the original frame, the straight conductor is electrically neutral.
+
+Let the volume charge densities of the ions and electrons be
+
+$$
+\rho_i=\rho,\qquad \rho_e=-\rho.
+$$
+
+The ions are at rest, while the conduction electrons move with speed $\beta c$ relative to the conductor.
+
+Consider a frame in which the conductor moves with speed
+
+$$
+\beta_1c,
+$$
+
+where
+
+$$
+\beta_1=k\beta.
+$$
+
+Define
+
+$$
+\gamma_1=\frac{1}{\sqrt{1-\beta_1^2}}.
+$$
+
+##### Ion charge density
+
+The ions, initially at rest, move with speed $\beta_1c$ in the new frame. Their longitudinal separations are Lorentz-contracted, so
+
+$$
+\boxed{\rho'_i=\gamma_1\rho}.
+$$
+
+##### Electron charge density
+
+First pass to the rest frame of the electrons.
+
+Since their charge density in the original frame is $-\rho$, their proper charge density is
+
+$$
+\rho_{e0}=-\frac{\rho}{\gamma},
+$$
+
+where
+
+$$
+\gamma=\frac{1}{\sqrt{1-\beta^2}}.
+$$
+
+The relative velocity between the electron rest frame and the required frame is
+
+$$
+\beta_2=\frac{\beta_1-\beta}{1-\beta_1\beta}.
+$$
+
+The corresponding Lorentz factor satisfies
+
+$$
+\gamma_2=\gamma\gamma_1(1-\beta\beta_1).
+$$
+
+Hence,
+
+$$
+\rho'_e=\gamma_2\rho_{e0}.
+$$
+
+Therefore,
+
+$$
+\boxed{\rho'_e=-\gamma_1(1-\beta\beta_1)\rho}.
+$$
+
+The total charge density in the moving conductor is
+
+$$
+\rho'_{\Sigma}=\rho'_i+\rho'_e.
+$$
+
+Thus,
+
+$$
+\boxed{\rho'_{\Sigma}=\gamma_1\beta\beta_1\rho}.
+$$
+
+The conductor is therefore no longer electrically neutral in this frame.
+
+##### Magnetic field
+
+In the original frame,
+
+$$
+\mathbf E=0.
+$$
+
+The magnetic field of a straight conductor is perpendicular to the conductor and therefore perpendicular to $\boldsymbol{\beta}_1$.
+
+The transformation law gives
+
+$$
+\mathbf B_1=\gamma_1\mathbf B.
+$$
+
+Hence,
+
+$$
+\boxed{B_1=\gamma_1B}.
+$$
+
+##### Electric field
+
+Since $\mathbf E=0$,
+
+$$
+\mathbf E_1=-\gamma_1[\boldsymbol{\beta}_1\times\mathbf B].
+$$
+
+Using $\mathbf B_1=\gamma_1\mathbf B$, we obtain
+
+$$
+\boxed{\mathbf E_1=-[\boldsymbol{\beta}_1\times\mathbf B_1]}.
+$$
+
+Therefore, in magnitude,
+
+$$
+\boxed{E_1=\beta_1B_1}.
+$$
+
+The final results are
+
+$$
+\boxed{\rho'_i=\gamma_1\rho},
+$$
+
+$$
+\boxed{\rho'_e=-\gamma_1(1-\beta\beta_1)\rho},
+$$
+
+$$
+\boxed{\rho'_{\Sigma}=\gamma_1\beta\beta_1\rho},
+$$
+
+$$
+\boxed{B_1=\gamma_1B},
+$$
+
+$$
+\boxed{E_1=\beta_1B_1}.
+$$
+
+---
+
+### c) Problems 14.3.6 and 14.3.7
+
+#### Problem 14.3.6
+
+![|841x241, 70%](../../img/14.3.8/Снимок экрана 2026-10-01 191732.png)
+
+Suppose that in the original frame there is only a magnetic field:
+
+$$
+\mathbf E=0.
+$$
+
+The transformed electric field is
+
+$$
+\mathbf E'=-\gamma[\boldsymbol{\beta}\times\mathbf B].
+$$
+
+The transformed magnetic field is
+
+$$
+\mathbf B'=\mathbf B_{\parallel}+\gamma\mathbf B_{\perp}.
+$$
+
+Since
+
+$$
+[\boldsymbol{\beta}\times\mathbf B_{\parallel}]=0,
+$$
+
+we have
+
+$$
+[\boldsymbol{\beta}\times\mathbf B']=\gamma[\boldsymbol{\beta}\times\mathbf B].
+$$
+
+Therefore,
+
+$$
+\boxed{\mathbf E'=-[\boldsymbol{\beta}\times\mathbf B']}.
+$$
+
+At low velocities,
+
+$$
+\beta\ll1,
+$$
+
+so that
+
+$$
+\gamma\approx1,\qquad \mathbf B'\approx\mathbf B.
+$$
+
+Hence,
+
+$$
+\boxed{\mathbf E'\approx-[\boldsymbol{\beta}\times\mathbf B]}.
+$$
+
+Thus, a field that is purely magnetic in one inertial frame generally contains an electric component in another frame.
+
+---
+
+#### Problem 14.3.7
+
+Now suppose that in the original frame there is only an electric field:
+
+$$
+\mathbf B=0.
+$$
+
+Then
+
+$$
+\mathbf E'=\mathbf E_{\parallel}+\gamma\mathbf E_{\perp},
+$$
+
+and
+
+$$
+\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].
+$$
+
+Since the parallel component of $\mathbf E'$ does not contribute to the cross product,
+
+$$
+[\boldsymbol{\beta}\times\mathbf E']=\gamma[\boldsymbol{\beta}\times\mathbf E].
+$$
+
+Therefore,
+
+$$
+\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.
+$$
+
+For
+
+$$
+\beta\ll1,
+$$
+
+we have
+
+$$
+\gamma\approx1,\qquad \mathbf E'\approx\mathbf E.
+$$
+
+Hence,
+
+$$
+\boxed{\mathbf B'\approx[\boldsymbol{\beta}\times\mathbf E]}.
+$$
+
+Thus, a field that is purely electric in one frame generally contains a magnetic component in another frame.
+
+---
+
+### d) The limit $\beta\to1$
+
+Consider the transverse components
+
+$$
+\mathbf E'_{\perp}=\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),
+$$
+
+$$
+\mathbf B'_{\perp}=\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).
+$$
+
+Let
+
+$$
+\mathbf n=\frac{\boldsymbol{\beta}}{\beta}
+$$
+
+be a unit vector along the direction of motion.
+
+As $\beta\to1$, define
+
+$$
+\mathbf A=\mathbf E_{\perp}-[\mathbf n\times\mathbf B_{\perp}].
+$$
+
+Then the leading transverse electric field is
+
+$$
+\mathbf E'_{\perp}\sim\gamma\mathbf A.
+$$
+
+Now,
+
+$$
+[\mathbf n\times\mathbf A]=[\mathbf n\times\mathbf E_{\perp}]-[\mathbf n\times[\mathbf n\times\mathbf B_{\perp}]].
+$$
+
+Since $\mathbf B_{\perp}\perp\mathbf n$,
+
+$$
+[\mathbf n\times[\mathbf n\times\mathbf B_{\perp}]]=-\mathbf B_{\perp}.
+$$
+
+Therefore,
+
+$$
+[\mathbf n\times\mathbf A]=\mathbf B_{\perp}+[\mathbf n\times\mathbf E_{\perp}].
+$$
+
+Hence the leading transverse magnetic field is
+
+$$
+\mathbf B'_{\perp}\sim\gamma[\mathbf n\times\mathbf A].
+$$
+
+Thus,
+
+$$
+\mathbf B'_{\perp}\sim[\mathbf n\times\mathbf E'_{\perp}].
+$$
+
+A cross product is perpendicular to the vector from which it is constructed. Therefore, in the ultrarelativistic limit,
+
+$$
+\boxed{\mathbf E'\perp\mathbf B'}.
+$$
+
+The same conclusion follows from the Lorentz invariant
+
+$$
+\boxed{\mathbf E'\cdot\mathbf B'=\mathbf E\cdot\mathbf B}.
+$$
+
+The scalar product remains finite, whereas in the generic ultrarelativistic case
+
+$$
+E'\sim\gamma,\qquad B'\sim\gamma.
+$$
+
+Thus,
+
+$$
+E'B'\sim\gamma^2\to\infty.
+$$
+
+If $\theta'$ is the angle between the transformed fields,
+
+$$
+\cos\theta'=\frac{\mathbf E'\cdot\mathbf B'}{E'B'}.
+$$
+
+Therefore,
+
+$$
+\cos\theta'\to0,
+$$
+
+and hence
+
+$$
+\boxed{\theta'\to\frac{\pi}{2}}.
+$$
+
+This statement refers to the non-degenerate case in which the leading transverse terms do not cancel.
+
+---
+
+#### Final results
+
+The Lorentz transformations of the fields are mutually inverse:
+
+$$
+\boxed{\mathbf E''=\mathbf E,\qquad \mathbf B''=\mathbf B}.
+$$
+
+For a purely magnetic field,
+
+$$
+\boxed{\mathbf E'=-[\boldsymbol{\beta}\times\mathbf B']}.
+$$
+
+For a purely electric field,
+
+$$
+\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.
+$$
+
+At low velocities,
+
+$$
+\boxed{\mathbf E'\approx-[\boldsymbol{\beta}\times\mathbf B]},
+$$
+
+$$
+\boxed{\mathbf B'\approx[\boldsymbol{\beta}\times\mathbf E]}.
+$$
+
+In the generic ultrarelativistic limit,
+
+$$
+\boxed{\beta\to1\quad\Longrightarrow\quad\mathbf E'\perp\mathbf B'}.
+$$