Special Relativity & Electromagnetism

Step 1: Define the 4-potential

$A^\mu = (\Phi/c, A_x, A_y, A_z)$ with the Minkowski metric $\eta_{\mu\nu} = \text{diag}(+1,-1,-1,-1)$. The covariant derivative is $\partial_\mu = (\partial_0, \partial_i) = (c^{-1}\partial_t, \nabla)$.

Step 2: Compute $F^{0i}$ components

$F^{0i} = \partial^0 A^i - \partial^i A^0 = -\frac{1}{c}\frac{\partial A^i}{\partial t} - \frac{\partial(\Phi/c)}{\partial x^i}(-1) = -E_i/c$. Note the sign conventions with the metric.

Step 3: Compute $F^{ij}$ components

$F^{ij} = \partial^i A^j - \partial^j A^i = -(\partial_i A_j - \partial_j A_i) = -\epsilon_{ijk}B_k$. For example, $F^{12} = -B_z$.

Step 4: Antisymmetry

$F^{\mu\nu} = -F^{\nu\mu}$ by construction. This means $F^{\mu\nu}$ has 6 independent components: 3 from $\mathbf{E}$ and 3 from $\mathbf{B}$.

Step 5: The dual tensor

The dual tensor $\tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$ is obtained by $\mathbf{E} \to c\mathbf{B}$, $\mathbf{B} \to -\mathbf{E}/c$. It encodes the same physics with E and B roles swapped, crucial for expressing the homogeneous Maxwell equations.

Step 1: Lorentz boost matrix

For a boost along $x$: $\Lambda^0_{\ 0} = \gamma$, $\Lambda^0_{\ 1} = -\gamma\beta$, $\Lambda^1_{\ 0} = -\gamma\beta$, $\Lambda^1_{\ 1} = \gamma$, with $\Lambda^2_{\ 2} = \Lambda^3_{\ 3} = 1$.

Step 2: Compute $F'^{01} = E_x'/c$

$F'^{01} = \Lambda^0_{\ \alpha}\Lambda^1_{\ \beta}F^{\alpha\beta} = \gamma^2(F^{01} - \beta F^{11} - \beta F^{00} + \beta^2 F^{10})$. Since $F^{00} = F^{11} = 0$ and $F^{10} = -F^{01}$: $F'^{01} = F^{01}(1-\beta^2)\gamma^2 = F^{01}$.

Step 3: Compute $F'^{02} = E_y'/c$

$F'^{02} = \Lambda^0_{\ \alpha}\Lambda^2_{\ \beta}F^{\alpha\beta} = \gamma(F^{02} - \beta F^{12})$. This gives $E_y' = \gamma(E_y - vB_z)$.

Step 4: Compute $F'^{12} = B_z'$

$F'^{12} = \Lambda^1_{\ \alpha}\Lambda^2_{\ \beta}F^{\alpha\beta} = \gamma(F^{12} - \beta F^{02})$. This gives $B_z' = \gamma(B_z - vE_y/c^2)$.

Step 5: General compact form

$$\boxed{\mathbf{E}' = \gamma(\mathbf{E}+\mathbf{v}\times\mathbf{B}) - (\gamma-1)(\mathbf{E}\cdot\hat{v})\hat{v}}$$

This shows that what appears as a pure magnetic field in one frame has an electric component in another. Magnetism is a relativistic effect of moving charges.

Step 1: The $\nu = 0$ component

$\partial_\mu F^{\mu 0} = \partial_i F^{i0} = \partial_i(E_i/c) = (\nabla\cdot\mathbf{E})/c$. The right side is $\mu_0 J^0 = \mu_0 c\rho = \rho/(\epsilon_0 c)$. This gives Gauss's law: $\nabla\cdot\mathbf{E} = \rho/\epsilon_0$.

Step 2: The $\nu = i$ components

$\partial_\mu F^{\mu i} = \partial_0 F^{0i} + \partial_j F^{ji}$. The first term gives $-\frac{1}{c}\frac{\partial E_i}{\partial t}\frac{1}{c}$. The second gives $(\nabla\times\mathbf{B})_i$.

Step 3: Combine

$(\nabla\times\mathbf{B})_i - \frac{1}{c^2}\frac{\partial E_i}{\partial t} = \mu_0 J_i$, which is the Ampere-Maxwell law.

Step 4: Charge conservation as an identity

Taking $\partial_\nu$ of both sides: $\partial_\nu\partial_\mu F^{\mu\nu} = 0$ (antisymmetry), giving $\partial_\nu J^\nu = 0$, i.e., $\partial\rho/\partial t + \nabla\cdot\mathbf{J} = 0$. Charge conservation is a mathematical consequence of the field equations.

Step 5: Gauge invariance

The definition $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is invariant under $A^\mu \to A^\mu + \partial^\mu\chi$. In the Lorenz gauge $\partial_\mu A^\mu = 0$, the equation of motion becomes: $$\boxed{\Box A^\mu = -\mu_0 J^\mu}$$

where $\Box = \partial_\mu\partial^\mu$ is the d'Alembertian operator. Each component satisfies an independent wave equation.

Step 1: Compute $F_{\mu\nu}F^{\mu\nu}$

Expanding the contraction: $F_{\mu\nu}F^{\mu\nu} = 2(F_{01}F^{01} + F_{02}F^{02} + F_{03}F^{03} + F_{12}F^{12} + F_{13}F^{13} + F_{23}F^{23})$. Using $F_{0i}F^{0i} = -E_i^2/c^2$ and $F_{ij}F^{ij} = 2B^2$:

$$\mathcal{F} = 2(B^2 - E^2/c^2)$$

Step 2: Compute $F_{\mu\nu}\tilde{F}^{\mu\nu}$

This pseudo-scalar invariant equals $-(4/c)\mathbf{E}\cdot\mathbf{B}$.

Step 3: Electric-type fields ($\mathcal{F} < 0$)

When $E^2 > c^2 B^2$, there exists a frame where $\mathbf{B} = 0$ (pure electric field). Example: a static charge.

Step 4: Magnetic-type fields ($\mathcal{F} > 0$)

When $c^2 B^2 > E^2$, there exists a frame where $\mathbf{E} = 0$ (pure magnetic field). Example: a steady current.

Step 5: Null fields ($\mathcal{F} = \mathcal{G} = 0$)

When $E = cB$ and $\mathbf{E}\perp\mathbf{B}$, no Lorentz transformation can eliminate either field. This is the case for plane electromagnetic waves. Null fields are invariant under a special subgroup of Lorentz transformations.

Step 1: $T^{00}$ is the energy density

$T^{00} = \frac{1}{2}(\epsilon_0 E^2 + B^2/\mu_0) = u$, the familiar electromagnetic energy density.

Step 2: $T^{0i}$ is the momentum density (Poynting vector)

$cT^{0i} = S_i/c$ where $\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0$ is the Poynting vector. The momentum density is $\mathbf{g} = \mathbf{S}/c^2$.

Step 3: $T^{ij}$ is the Maxwell stress tensor

$T^{ij} = \epsilon_0(E_i E_j - \frac{1}{2}\delta_{ij}E^2) + \frac{1}{\mu_0}(B_i B_j - \frac{1}{2}\delta_{ij}B^2)$. This gives the electromagnetic momentum flux (radiation pressure and shear stress).

Step 4: Conservation equation

Using the Maxwell equations, one shows: $\partial_\mu T^{\mu\nu} = -F^{\nu\alpha}J_\alpha$. The right side is the Lorentz force density. In the absence of charges: $\partial_\mu T^{\mu\nu} = 0$.

Step 5: Physical content

The $\nu = 0$ component gives Poynting's theorem: $\partial u/\partial t + \nabla\cdot\mathbf{S} = -\mathbf{J}\cdot\mathbf{E}$. The $\nu = i$ components give the momentum conservation: the rate of change of field momentum plus the divergence of the stress tensor equals the Lorentz force density.