Reduction to Standard Maxwell Equations and Lorentz Force Law
1. Standard Inhomogeneous Maxwell Equations ($\partial_{\mu}F^{\mu\nu} = \mu_{0}J^{\nu}$):
Using the Minkowski metric tensor with signature $(+,-,-,-)$, the coordinate four-vector and partial derivative are defined as:
$$x^{\mu} = (ct, x, y, z), \quad \partial_{\mu} = \left(\frac{1}{c}\frac{\partial}{\partial t}, \vec{\nabla}\right)$$
The four-current vector is $J^{\nu} = (c\rho, \vec{J})$, and the antisymmetric field tensor $F^{\mu\nu}$ is:
$$F^{\mu\nu} = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}$$
- For $\nu = 0$ (Time component):
$$\partial_{\mu}F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0} = 0 + \vec{\nabla} \cdot \left(\frac{\vec{E}}{c}\right) = \mu_0 J^0 = \mu_0 (c\rho)$$
Multiplying by $c$ and using $\mu_0 \epsilon_0 c^2 = 1$, this simplifies to Gauss's Law:
$$\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$
- For $\nu = i$ (Spatial components $i = 1, 2, 3$):
$$\partial_0 F^{0i} + \partial_j F^{ji} = \frac{1}{c}\frac{\partial}{\partial t}\left(-\frac{E^i}{c}\right) + (\vec{\nabla} \times \vec{B})^i = \mu_0 J^i$$
Rearranging the terms gives the Ampère-Maxwell Law:
$$\vec{\nabla} \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$
2. Standard Homogeneous Maxwell Equations ($\partial_{\mu}\tilde{F}^{\mu\nu} = 0$):
The dual field tensor $\tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$ is obtained by replacing $\vec{E}/c \to \vec{B}$ and $\vec{B} \to -\vec{E}/c$:
$$\tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix}$$
- For $\nu = 0$:
$$\partial_i \tilde{F}^{i0} = \vec{\nabla} \cdot \vec{B} = 0 \quad \text{(Gauss's Law for Magnetism)}$$
- For $\nu = i$:
$$\partial_0 \tilde{F}^{0i} + \partial_j \tilde{F}^{ji} = \frac{1}{c}\frac{\partial (-B^i)}{\partial t} + \left(-\frac{1}{c}\vec{\nabla} \times \vec{E}\right)^i = 0$$
Multiplying by $-c$ yields Faraday's Law of Induction:
$$\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$
3. Standard Lorentz Force Law ($f^{\mu} = q F^{\mu\nu} V_{\nu} = \frac{dP^{\mu}}{d\tau}$):
The four-velocity is $V_{\nu} = \eta_{\nu\alpha} V^{\alpha} = \gamma (c, -\vec{v})$ and four-momentum is $P^{\mu} = (\gamma m c, \gamma m \vec{v})$. Evaluating the spatial component $\mu = i$:
$$f^i = q \left( F^{i0}V_0 + F^{ij}V_j \right) = q \gamma \left[ \left(\frac{E^i}{c}\right)(c) + (\vec{v} \times \vec{B})^i \right] = \gamma q \left( \vec{E} + \vec{v} \times \vec{B} \right)^i$$
Since $f^i = \frac{dP^i}{d\tau} = \gamma \frac{d\vec{p}}{dt}$, canceling $\gamma$ from both sides gives the standard 3D Lorentz Force Law:
$$\frac{d\vec{p}}{dt} = q(\vec{E} + \vec{v} \times \vec{B})$$
(b) Transformation of Electric and Magnetic Fields
For a boost with relative speed $u$ along the $x$-direction with $\gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}$ and $\beta = u/c$, the transformation matrix is:
$$\Lambda^{\mu}{}_{\nu} = \begin{pmatrix} \gamma_u & -\gamma_u \beta & 0 & 0 \\ -\gamma_u \beta & \gamma_u & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
Applying tensor transformation $F'^{\mu\nu} = \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} F^{\alpha\beta}$, the transformed field components in frame $S'$ are:
- Electric Field Components:
$$E'_x = E_x$$
$$E'_y = \gamma_u (E_y - u B_z)$$
$$E'_z = \gamma_u (E_z + u B_y)$$
- Magnetic Field Components:
$$B'_x = B_x$$
$$B'_y = \gamma_u \left(B_y + \frac{u}{c^2} E_z\right)$$
$$B'_z = \gamma_u \left(B_z - \frac{u}{c^2} E_y\right)$$
(c) Evaluation of Invariants
1. Evaluation of $F^{\mu\nu}F_{\mu\nu}$:
Expanding the double sum over four dimensions:
$$F^{\mu\nu}F_{\mu\nu} = F^{0i}F_{0i} + F^{i0}F_{i0} + F^{ij}F_{ij} = 2 F^{0i}F_{0i} + F^{ij}F_{ij}$$
Using $F^{0i} = -E^i/c$, $F_{0i} = E_i/c$, and $F^{ij}F_{ij} = 2\vec{B}^2$:
$$F^{\mu\nu}F_{\mu\nu} = 2\left(-\frac{\vec{E}^2}{c^2}\right) + 2\vec{B}^2 = 2\left( \vec{B}^2 - \frac{\vec{E}^2}{c^2} \right)$$
2. Evaluation of $\tilde{F}^{\mu\nu}F_{\mu\nu}$:
Expanding the contraction:
$$\tilde{F}^{\mu\nu}F_{\mu\nu} = \tilde{F}^{0i}F_{0i} + \tilde{F}^{i0}F_{i0} + \tilde{F}^{ij}F_{ij}$$
Substituting $\tilde{F}^{0i} = -B^i$, $F_{0i} = E^i/c$, and $\tilde{F}^{ij} = -\frac{1}{c}\epsilon^{ijk}E_k$:
$$\tilde{F}^{\mu\nu}F_{\mu\nu} = 2\left(-B^i \frac{E^i}{c}\right) + 2\left(-\frac{B^i E^i}{c}\right) = -\frac{4}{c}(\vec{E} \cdot \vec{B})$$
Proof of Orthochronous Lorentz Transformations ($\Lambda_0^0 \ge 1$) Form a Subgroup
Proof: Orthochronous Lorentz Transformations ($\Lambda_0^0 \ge 1$) Form a Subgroup
Any general Lorentz transformation matrix $\Lambda$ satisfies the orthogonality condition for Minkowski metric $\eta = \text{diag}(1, -1, -1, -1)$:
$$\Lambda^T \eta \Lambda = \eta$$
Evaluating the $(0,0)$ component of this metric condition gives:
$$(\Lambda_0^0)^2 - \sum_{i=1}^3 (\Lambda_i^0)^2 = 1 \implies (\Lambda_0^0)^2 = 1 + \sum_{i=1}^3 (\Lambda_i^0)^2 \ge 1$$
This implies either $\Lambda_0^0 \ge 1$ (orthochronous) or $\Lambda_0^0 \le -1$ (antichronous). To show that the set of orthochronous transformations $\mathcal{L}^+ = \{ \Lambda \in \mathcal{L} \mid \Lambda_0^0 \ge 1 \}$ forms a subgroup, we verify the four group axioms:
- Identity Element: The identity transformation matrix $I$ has $I_0^0 = 1 \ge 1$. Thus, $I \in \mathcal{L}^+$.
- Closure under Multiplication: Let $\Lambda_1, \Lambda_2 \in \mathcal{L}^+$, meaning $(\Lambda_1)_0^0 \ge 1$ and $(\Lambda_2)_0^0 \ge 1$. Let $\Lambda_3 = \Lambda_1 \Lambda_2$. The time component is:
$$(\Lambda_3)_0^0 = (\Lambda_1)_0^0 (\Lambda_2)_0^0 + \sum_{i=1}^3 (\Lambda_1)_i^0 (\Lambda_2)_0^i$$
By Cauchy-Schwarz inequality for spatial vectors:
$$\left| \sum_{i=1}^3 (\Lambda_1)_i^0 (\Lambda_2)_0^i \right| \le \sqrt{\sum_{i=1}^3 ((\Lambda_1)_i^0)^2} \sqrt{\sum_{i=1}^3 ((\Lambda_2)_0^i)^2}$$
Using $((\Lambda_1)_i^0)^2 = ((\Lambda_1)_0^0)^2 - 1$ and $((\Lambda_2)_0^i)^2 = ((\Lambda_2)_0^0)^2 - 1$:
$$(\Lambda_3)_0^0 \ge (\Lambda_1)_0^0 (\Lambda_2)_0^0 - \sqrt{((\Lambda_1)_0^0)^2 - 1}\sqrt{((\Lambda_2)_0^0)^2 - 1}$$
For any $x, y \ge 1$, $xy - \sqrt{x^2 - 1}\sqrt{y^2 - 1} \ge 1$. Hence $(\Lambda_3)_0^0 \ge 1$, which proves closure.
- Inverse Element: For any $\Lambda \in \mathcal{L}^+$, the inverse matrix is $\Lambda^{-1} = \eta \Lambda^T \eta$. Its $(0,0)$ component is:
$$(\Lambda^{-1})_0^0 = \eta_{00} \Lambda_0^0 \eta^{00} = (1) \Lambda_0^0 (1) = \Lambda_0^0 \ge 1$$
Therefore, $\Lambda^{-1} \in \mathcal{L}^+$.
- Associativity: Matrix multiplication is inherently associative.
Since all subgroup criteria are met, the orthochronous Lorentz transformations form a subgroup of the Lorentz group.
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