The equation unifies the two laws by representing them as different components ($\nu=0$ and $\nu=j$) of a single four-dimensional tensor equation, demonstrating the elegant and compact nature of electromagnetism within the framework of special relativity.
The single four-dimensional tensor equation $\partial_\mu F^{\mu\nu} = K^\nu$ unifies Gauss's Law and the Ampère-Maxwell Law by combining the electric and magnetic fields and their sources into relativistic four-vectors and a rank-2 tensor.
This unification is revealed by expanding the single tensor equation for its two different components ($\nu=0$ and $\nu=j$):
1. Time Component ($\nu=0$): Gauss's Law
When the index $\nu$ is set to 0 (the time component), the equation $\partial_\mu F^{\mu 0} = K^0$ expands to:
$$
\partial_0 F^{00} + \sum_{i=1}^3 \partial_i F^{i 0} = K^0
$$
- $F^{00}$ Term: Since the Faraday tensor ($F^{\mu\nu}$) is anti-symmetric ($F^{\mu\nu} = -F^{\nu\mu}$), its diagonal elements are zero, so $F^{00} = 0$
- $F^{i0}$ Term: The components $F^{i0}$ are defined to be the components of the electric field $\mathbf{E}$ (i.e., $F^{i0} = E^i$).
- $K^0$ Term: The time component of the four-current ($K^\nu$) is defined in terms of the charge density $\rho$ (i.e., $K^0 = \rho / \epsilon_0$).
Substituting these definitions and noting that $\sum_{i=1}^3 \partial_i E^i$ is the divergence $\nabla \cdot \mathbf{E}$, the equation becomes:
$$
0 + \nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}
$$
This is precisely Gauss's Law ($\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$).
2. Spatial Component ($\nu=j$): Ampère-Maxwell Law
When the index $\nu$ is set to $j = 1, 2, 3$ (the spatial components), the equation $\partial_\mu F^{\mu j} = K^j$ expands to:
$$
\partial_0 F^{0 j} + \sum_{i=1}^3 \partial_i F^{i j} = K^j
$$
- $\partial_0 F^{0j}$ Term (Displacement Current):
- By anti-symmetry, $F^{0j} = -F^{j0}$. Since $F^{j0} = E^j$, we have $F^{0j} = -E^j$.
- The time derivative is $\partial_0 = \frac{\partial}{\partial x^0} = \frac{\partial}{\partial(ct)} = \frac{1}{c} \frac{\partial}{\partial t}$.
- This term becomes $\frac{1}{c} \frac{\partial}{\partial t}(-E^j) = -\frac{1}{c} \frac{\partial E^j}{\partial t}$.
- $\sum \partial_i F^{ij}$ Term (Curl of B):
- The spatial components $F^{ij}$ are defined in terms of the magnetic field $\mathbf{B}$ and the Levi-Civita tensor $\epsilon^{ijk}$ (i.e., $F^{ij} = -c \epsilon^{ijk} B^k$).
- Substituting this definition and performing the summation shows that this term is the $j$-th component of $c(\nabla \times \mathbf{B})$ (i.e., $\sum_{i=1}^3 \partial_i F^{ij} = c(\nabla \times \mathbf{B})^j$).
- $K^j$ Term (Conduction Current):
- The spatial components of the four-current ($K^j$) are defined in terms of the current density $\mathbf{J}$ (i.e., $K^j = \frac{1}{c\epsilon_0} J^j$).