The single four-dimensional tensor equation $\partial_\mu F^{\mu\nu} = K^\nu$ unifies Gauss's Law ($\nabla \cdot \mathbf{E} = \rho / \epsilon_0$) and the Ampère-Maxwell Law ($\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$) into one compact relativistic expression.
Here explains this concept by breaking down the four-dimensional tensor equation $\partial_\mu F^{\mu\nu} = K^\nu$ into its time and space components:
The equation $\partial_\mu F^{\mu 0} = K^0$ is expanded as:
$$ \partial_0 F^{00} + \sum_{i=1}^3 \partial_i F^{i 0} = K^0 $$
Since $F^{\mu\nu}$ is anti-symmetric, $F^{00} = 0$.
Using the definitions $F^{i 0} = E^i and K^0 = \rho / \epsilon_0$, the equation becomes:
$$ 0 + \sum_{i=1}^3 \frac{\partial E^i}{\partial x^i} = \frac{\rho}{\epsilon_0} $$
This is recognized as Gauss's Law: $\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$.
The equation $\partial_\mu F^{\mu j} = K^j$ is expanded as:
$$ \partial_0 F^{0 j} + \sum_{i=1}^3 \partial_i F^{i j} = K^j $$
Using the definitions $F^{0j} = -E^j, F^{ij} = -c \epsilon^{ijk} B^k, \partial_0 = \frac{1}{c} \frac{\partial}{\partial t}, and K^j = \frac{1}{c \epsilon_0} J^j$, the equation simplifies to:
Multiplying by c and rearranging gives the j-th component of:
$$ \nabla \times \mathbf{B} = \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t} + \frac{1}{c^2 \epsilon_0} \mathbf{J} $$
Using the identity $c^2 = 1/(\mu_0 \epsilon_0)$ (which means $1/c^2 = \mu_0 \epsilon_0$ and $1/(c^2 \epsilon_0) = \mu_0$), this is shown to be the Ampère-Maxwell Law: $\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$.
Thus, the single four-dimensional tensor equation $\partial_\mu F^{\mu\nu} = K^\nu$ concisely combines the two inhomogeneous Maxwell's equations.
What two Maxwell's equations are unified by the single four-dimensional tensor equation-L.mp4
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