Physlib.Electromagnetism.Dynamics.IsExtrema

Given the physical parameters of free space $\mathcal{F}$ (such as the vacuum permeability $\mu_0$ and permittivity $\varepsilon_0$), an electromagnetic 4-potential field $A$, and a Lorentz 4-current density field $J$ in $d$ spatial dimensions, the property $\text{IsExtrema}(\mathcal{F}, A, J)$ holds if the variational gradient of the electromagnetic Lagrangian with respect to $A$ is zero. This signifies that the potential $A$ satisfies the Euler–Lagrange equations for the electromagnetic action, corresponding to a stationary point of the action functional.

Let $\mathcal{F}$ be the physical parameters of free space, $A$ be an electromagnetic $4$-potential in $d$ spatial dimensions, and $J$ be a Lorentz $4$-current density. The potential $A$ is an extremum of the electromagnetic Lagrangian (denoted $\text{IsExtrema}(\mathcal{F}, A, J)$) if and only if the variational gradient of the Lagrangian with respect to $A$ is zero, i.e., $\frac{\delta \mathcal{L}}{\delta A} = 0$.

Let $d$ be the spatial dimension and $\mathcal{F}$ denote the physical parameters of free space with magnetic permeability $\mu_0$. For an infinitely differentiable ($C^\infty$) electromagnetic potential field $A$ and an infinitely differentiable Lorentz current density field $J$, the potential $A$ is an extremum of the electromagnetic Lagrangian (denoted $\text{IsExtrema}(\mathcal{F}, A, J)$) if and only if for every spacetime point $x$ and every index $\nu \in \{0, \dots, d\}$, the following equation holds: $$\sum_{\mu} \partial_\mu F^{\mu\nu}(x) = \mu_0 J^\nu(x)$$ where $F^{\mu\nu}$ are the components of the electromagnetic field strength matrix (Faraday tensor) and $J^\nu$ is the $\nu$-th component of the current density.

Let $\mathcal{F}$ be the physical parameters of free space with permeability $\mu_0$. Let $A$ be an infinitely differentiable electromagnetic potential field and $J$ be an infinitely differentiable Lorentz current density field in $d$ spatial dimensions. The potential $A$ is an extremum of the Lagrangian density if and only if for every point in spacetime $x$, the following tensor equation holds: $$\frac{1}{\mu_0} \partial_\kappa F^{\kappa \nu'}(x) - J^{\nu'}(x) = 0$$ where $F^{\kappa \nu'}$ is the electromagnetic field strength tensor and the indices are contracted and summed according to the Einstein notation.

Let $\mathcal{F}$ be the physical parameters of free space, let $A$ be an infinitely differentiable electromagnetic 4-potential field, and let $J$ be an infinitely differentiable Lorentz 4-current density field in $d$ spatial dimensions. For any Lorentz transformation $\Lambda$ in the Lorentz group, the transformed potential $A'(x) = \Lambda \cdot A(\Lambda^{-1} \cdot x)$ is an extremum of the Lagrangian density with respect to the transformed current density $J'(x) = \Lambda \cdot J(\Lambda^{-1} \cdot x)$ if and only if the original potential $A$ is an extremum of the Lagrangian density with respect to $J$. This demonstrates the Lorentz covariance of the Maxwell equations.

Let $d$ be the number of spatial dimensions and $\mathcal{F}$ be the physical parameters of free space, with electric permittivity $\varepsilon_0$ and magnetic permeability $\mu_0$. Let $A$ be an infinitely differentiable electromagnetic 4-potential and $J$ be an infinitely differentiable Lorentz 4-current density. The potential $A$ is an extremum of the electromagnetic Lagrangian if and only if for all times $t$ and spatial positions $\mathbf{x}$, the following two conditions (Gauss's law and Ampère's law) hold: 1. The divergence of the electric field $\mathbf{E}$ satisfies: $$\nabla \cdot \mathbf{E}(t, \mathbf{x}) = \frac{\rho(t, \mathbf{x})}{\varepsilon_0}$$ 2. For each spatial component $i \in \{1, \dots, d\}$, the time derivative of the electric field and the curl-like derivatives of the magnetic field matrix $M$ satisfy: $$\mu_0 \varepsilon_0 \frac{\partial E_i(t, \mathbf{x})}{\partial t} = \sum_{j=1}^d \frac{\partial M_{ji}(t, \mathbf{x})}{\partial x_j} - \mu_0 j_i(t, \mathbf{x})$$ where $\rho$ is the charge density and $j_i$ are the components of the current density.

Let $d$ be the number of spatial dimensions and $\mathcal{F}$ represent the physical parameters of free space, including the vacuum permittivity $\varepsilon_0$ and magnetic permeability $\mu_0$. Let $A$ be an infinitely differentiable electromagnetic 4-potential and $J$ be an infinitely differentiable Lorentz 4-current density. If $A$ is an extremum of the electromagnetic Lagrangian associated with $J$, then for any time $t$, spatial position $\mathbf{x}$, and spatial component $i \in \{1, \dots, d\}$, the time derivative of the $i$-th component of the electric field $\mathbf{E}$ satisfies: $$\frac{\partial E_i(t, \mathbf{x})}{\partial t} = \frac{1}{\mu_0 \varepsilon_0} \sum_{j=1}^d \frac{\partial M_{ji}(t, \mathbf{x})}{\partial x_j} - \frac{1}{\varepsilon_0} j_i(t, \mathbf{x})$$ where $M$ is the magnetic field matrix and $j_i$ is the $i$-th component of the current density.

Let $d$ be the number of spatial dimensions and $\mathcal{F}$ represent the physical parameters of free space, including the speed of light $c$ and vacuum permittivity $\varepsilon_0$. Let $A$ be an infinitely differentiable electromagnetic 4-potential and $J$ be an infinitely differentiable Lorentz 4-current density. If $A$ is an extremum of the electromagnetic Lagrangian associated with $J$, then for any time $t$, spatial position $\mathbf{x}$, and spatial indices $i, j \in \{1, \dots, d\}$, the components of the magnetic field matrix $B_{ij}$ satisfy the following wave-like equation: $$\frac{\partial^2 B_{ij}(t, \mathbf{x})}{\partial t^2} = c^2 \sum_{k=1}^d \frac{\partial^2 B_{ij}(t, \mathbf{x})}{\partial x_k^2} + \frac{1}{\varepsilon_0} \left( \frac{\partial j_i(t, \mathbf{x})}{\partial x_j} - \frac{\partial j_j(t, \mathbf{x})}{\partial x_i} \right)$$ where $j_i$ and $j_j$ are the spatial components of the current density.

Let $d$ be the number of spatial dimensions and $\mathcal{F}$ represent the physical parameters of free space, including the speed of light $c$, the vacuum permittivity $\varepsilon_0$, and the magnetic permeability $\mu_0$. Let $A$ be an infinitely differentiable electromagnetic 4-potential and $J$ be an infinitely differentiable Lorentz 4-current density. If $A$ is an extremum of the electromagnetic Lagrangian associated with $J$, then for any time $t$, spatial position $\mathbf{x}$, and spatial component $i \in \{1, \dots, d\}$, the second time derivative of the $i$-th component of the electric field $\mathbf{E}$ satisfies the following wave-like equation: $$\frac{\partial^2 E_i(t, \mathbf{x})}{\partial t^2} = c^2 \sum_{j=1}^d \frac{\partial^2 E_i(t, \mathbf{x})}{\partial x_j^2} - \frac{c^2}{\varepsilon_0} \frac{\partial \rho(t, \mathbf{x})}{\partial x_i} - c^2 \mu_0 \frac{\partial j_i(t, \mathbf{x})}{\partial t}$$ where $\rho$ is the charge density and $j_i$ is the $i$-th component of the current density.

For a spatial dimension $d$, a free space environment $\mathcal{F}$ (containing physical constants such as $\mu_0$ and $c$), a distributional electromagnetic potential $A$, and a distributional Lorentz current density $J$, the property `IsExtrema` is the proposition that $A$ is an extremum of the electromagnetic Lagrangian density $\mathcal{L}$. This condition is defined by the vanishing of the variational gradient of the Lagrangian density with respect to the potential $A$: $$\frac{\delta \mathcal{L}}{\delta A} = 0$$ where $\frac{\delta \mathcal{L}}{\delta A}$ represents the distributional gradient (the functional derivative) of the Lagrangian density associated with $A$ and $J$ in the environment $\mathcal{F}$.

For a spatial dimension $d$, a free space environment $\mathcal{F}$ (containing physical constants), a distributional electromagnetic potential $A$, and a distributional Lorentz current density $J$, the potential $A$ is an extremum of the electromagnetic Lagrangian density $\mathcal{L}$ if and only if the variational gradient of the Lagrangian density with respect to $A$ vanishes: $$\frac{\delta \mathcal{L}}{\delta A} = 0$$ where $\frac{\delta \mathcal{L}}{\delta A}$ represents the functional derivative of the Lagrangian density with respect to the potential.

For a spatial dimension $d$, a free space environment $\mathcal{F}$, a distributional electromagnetic potential $A$, and a distributional Lorentz current density $J$, the potential $A$ is an extremum of the electromagnetic Lagrangian density $\mathcal{L}$ if and only if for every test function $\varepsilon$ on space-time, both the time component and every spatial component of the variational gradient $\frac{\delta \mathcal{L}}{\delta A}$ evaluated at $\varepsilon$ are zero: $$\left( \frac{\delta \mathcal{L}}{\delta A}(\varepsilon) \right)_0 = 0 \quad \text{and} \quad \forall i \in \{1, \dots, d\}, \left( \frac{\delta \mathcal{L}}{\delta A}(\varepsilon) \right)_i = 0$$ where $\frac{\delta \mathcal{L}}{\delta A}$ denotes the functional derivative of the Lagrangian density with respect to the potential.

For a spatial dimension $d$, let $\mathcal{F}$ be a free space environment with electric permittivity $\epsilon_0$ and magnetic permeability $\mu_0$. Let $A$ be a distributional electromagnetic potential and $J$ be a distributional Lorentz current density. The potential $A$ is an extremum of the electromagnetic Lagrangian density if and only if for every test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the following two conditions hold: 1. **Gauss's Law:** The divergence of the distributional electric field $\mathbf{E}$ satisfies $$\langle \nabla \cdot \mathbf{E}, \varepsilon \rangle = \frac{1}{\epsilon_0} \langle \rho, \varepsilon \rangle$$ where $\rho$ is the distributional charge density associated with $J$. 2. **Ampère's Law:** For every spatial index $i \in \{1, \dots, d\}$, the $i$-th component of the fields satisfies $$\mu_0 \epsilon_0 \langle \frac{\partial E_i}{\partial t}, \varepsilon \rangle - \sum_{j=1}^d \langle \frac{\partial B_{ji}}{\partial x_j}, \varepsilon \rangle + \mu_0 \langle j_i, \varepsilon \rangle = 0$$ where $B_{ji}$ is the $(j, i)$-th component of the distributional magnetic field matrix and $j_i$ is the $i$-th component of the distributional current density associated with $J$.

For a spatial dimension $d$, let $\mathcal{F}$ be a free space environment with electric permittivity $\epsilon_0$ and magnetic permeability $\mu_0$. Let $A$ be a distributional electromagnetic potential and $J$ be a distributional Lorentz current density. The potential $A$ is an extremum of the electromagnetic Lagrangian density if and only if for every test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{Time} \times \text{Space}_d, \mathbb{R})$, the following two conditions hold: 1. **Gauss's Law:** The divergence of the distributional electric field $\mathbf{E}$ satisfies $$\langle \nabla \cdot \mathbf{E}, \varepsilon \rangle = \frac{1}{\epsilon_0} \langle \rho, \varepsilon \rangle$$ where $\rho$ is the distributional charge density associated with $J$. 2. **Ampère's Law:** For every spatial index $i \in \{1, \dots, d\}$, the components of the fields satisfy $$\mu_0 \epsilon_0 \langle \frac{\partial E_i}{\partial t}, \varepsilon \rangle - \sum_{j=1}^d \langle -\left(\frac{\partial^2 A_i}{\partial x_j^2} - \frac{\partial^2 A_j}{\partial x_j \partial x_i}\right), \varepsilon \rangle + \mu_0 \langle j_i, \varepsilon \rangle = 0$$ where $A_i$ is the $i$-th component of the distributional vector potential $\vec{A}$ and $j_i$ is the $i$-th component of the distributional current density associated with $J$.

Let $d$ be the number of spatial dimensions. For a free space environment $\mathcal{F}$ with magnetic permeability $\mu_0$, let $A$ be a distributional electromagnetic potential and $J$ be a distributional Lorentz current density. Then $A$ is an extremum of the electromagnetic Lagrangian density if and only if for every test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the following tensor equation holds: $$\frac{1}{\mu_0} (\partial_\kappa F^{\kappa\nu})(\varepsilon) - J^\nu(\varepsilon) = 0$$ where $F^{\kappa\nu}$ is the electromagnetic field strength tensor, $J^\nu$ is the $\nu$-th component of the current density distribution, and $\partial_\kappa$ denotes the distributional derivative with respect to the $\kappa$-th coordinate (with implicit summation over $\kappa$).

Let $d$ be the number of spatial dimensions and $\mathcal{F}$ be a free space environment. For any distributional electromagnetic potential $A$, any distributional Lorentz current density $J$, and any Lorentz transformation $\Lambda$ in the Lorentz group $\mathcal{L}$, the potential $A$ is an extremum of the electromagnetic Lagrangian density (which corresponds to satisfying Maxwell's equations) for the current density $J$ if and only if the Lorentz-transformed potential $\Lambda \cdot A$ is an extremum for the Lorentz-transformed current density $\Lambda \cdot J$.