Physlib.Electromagnetism.PointParticle.ThreeDimension

Given the speed of light $c$, a charge $q \in \mathbb{R}$, and a position $\mathbf{r}_0$ in 3D space, the 4-current density $J$ of a point particle stationary at $\mathbf{r}_0$ is defined as the distribution $$J = cq \delta(\mathbf{r} - \mathbf{r}_0) \mathbf{e}_0$$ where $\mathbf{e}_0$ is the basis vector corresponding to the temporal component of the Lorentz vector space and $\delta(\mathbf{r} - \mathbf{r}_0)$ is the Dirac delta distribution centered at $\mathbf{r}_0$. In component form, this corresponds to a 4-vector field $J(\mathbf{r}) = (c q \delta(\mathbf{r} - \mathbf{r}_0), 0, 0, 0)$.

For a point particle with charge $q \in \mathbb{R}$ stationary at position $\mathbf{r}_0$ in 3D space, let $c$ be the speed of light. The 4-current density $J$ of this particle is equal to the spatial translation by $\mathbf{r}_0$ of a distribution centered at the origin. Specifically, $J$ is the translation of the distribution $cq \delta(\mathbf{r}) \mathbf{e}_0$, where $\delta(\mathbf{r})$ is the Dirac delta distribution in 3D space and $\mathbf{e}_0$ is the basis vector corresponding to the temporal component of the Lorentz 4-vector space.

For a point particle with charge $q \in \mathbb{R}$ stationary at position $\mathbf{r}_0$ in 3D space, the distributional charge density $\rho$ (derived from the 4-current density $J$ as $\rho = \frac{1}{c} J^0$) is a time-independent distribution equal to $q \delta(\mathbf{r} - \mathbf{r}_0)$, where $\delta$ is the Dirac delta distribution centered at $\mathbf{r}_0$.

Given the speed of light $c$, a charge $q \in \mathbb{R}$, and a position $\mathbf{r}_0$ in 3D space, the spatial 3-current density $\vec{j}$ of a point particle stationary at $\mathbf{r}_0$ is equal to zero.

Given the physical constants of free space $\mathcal{F}$ (including the permeability $\mu_0$ and the speed of light $c$), a charge $q \in \mathbb{R}$, and a position $\vec{r}_0 \in \mathbb{R}^3$, the electromagnetic 4-potential $A$ of a stationary point particle is a time-independent distribution. Its temporal component (the 0-th component in the Lorentz basis) is given by the function: $$A^0(\vec{r}) = \frac{q \mu_0 c}{4\pi |\vec{r} - \vec{r}_0|}$$ while its spatial components $\vec{A}(\vec{r})$ are zero. Here, $|\vec{r} - \vec{r}_0|$ denotes the Euclidean distance from a point $\vec{r}$ to the particle's position $\vec{r}_0$.

For a point particle with charge $q$ at position $\vec{r}_0 \in \mathbb{R}^3$ in a free space $\mathcal{F}$ (with permeability $\mu_0$ and speed of light $c$), its electromagnetic 4-potential $A$ is equal to the time-independent spacetime distribution obtained by translating the fundamental potential at the origin by $\vec{r}_0$. Specifically, it is the translation by $\vec{r}_0$ of the distribution defined by the function: $$\vec{x} \mapsto \frac{q \mu_0 c}{4 \pi \|\vec{x}\|} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the temporal basis vector (the 0-th component in the Lorentz basis) and $\|\vec{x}\|$ is the Euclidean norm in $\mathbb{R}^3$.

Given the physical constants of free space $\mathcal{F}$ (including the vacuum permittivity $\epsilon_0$), a charge $q \in \mathbb{R}$, and a position $\vec{r}_0 \in \mathbb{R}^3$, the scalar potential $V$ of the electromagnetic 4-potential for a stationary point particle is a time-independent distribution. Its value at a point $\vec{r}$ is given by: $$V(\vec{r}) = \frac{q}{4 \pi \epsilon_0 |\vec{r} - \vec{r}_0|}$$ where $|\vec{r} - \vec{r}_0|$ denotes the Euclidean distance between $\vec{r}$ and the particle's position $\vec{r}_0$.

Consider a stationary point particle in 3D space with electric charge $q$ located at position $\vec{r}_0$. Given the physical constants of free space $\mathcal{F}$ (including the speed of light $c$), the vector potential $\vec{A}$ associated with the electromagnetic 4-potential of this particle is zero.

Given the physical constants of free space $\mathcal{F}$ (including the vacuum permittivity $\epsilon_0$), a charge $q \in \mathbb{R}$, and a position $\vec{r}_0 \in \mathbb{R}^3$, the electric field $\vec{E}$ associated with the electromagnetic potential of a stationary point particle is a time-independent distribution. Its value at a point $\vec{r}$ is given by the expression: $$\vec{E}(\vec{r}) = \frac{q}{4 \pi \epsilon_0} \frac{\vec{r} - \vec{r}_0}{|\vec{r} - \vec{r}_0|^3}$$ where $|\vec{r} - \vec{r}_0|$ denotes the Euclidean distance between the observation point $\vec{r}$ and the particle's position $\vec{r}_0$.

Given the physical constants of free space $\mathcal{F}$ (including the speed of light $c$), a charge $q \in \mathbb{R}$, and a position $\vec{r}_0 \in \mathbb{R}^3$, the electric field distribution $\vec{E}$ produced by a stationary point particle is time-independent. Specifically, the partial derivative of the electric field with respect to time is zero: $$\frac{\partial \vec{E}}{\partial t} = 0$$

Consider a stationary point particle with electric charge $q \in \mathbb{R}$ located at position $\vec{r}_0 \in \mathbb{R}^3$. Given the physical constants of free space $\mathcal{F}$ (including the speed of light $c$), the magnetic field distribution, represented by the magnetic field matrix associated with the electromagnetic 4-potential of this particle, is zero.

Given the physical constants of free space $\mathcal{F}$ (including the vacuum permittivity $\epsilon_0$), a charge $q \in \mathbb{R}$, and a position $\vec{r}_0 \in \mathbb{R}^3$, the divergence of the electric field $\vec{E}$ produced by a stationary point particle, interpreted as a distribution, is given by: $$\nabla \cdot \vec{E} = \frac{1}{\epsilon_0} q \delta(\vec{r} - \vec{r}_0)$$ where $\delta(\vec{r} - \vec{r}_0)$ is the Dirac delta distribution centered at $\vec{r}_0$. Both the electric field and the charge density are treated as time-independent distributions.

Given a free space environment $\mathcal{F}$ (defined by constants such as the speed of light $c$ and the magnetic permeability $\mu_0$), consider a point particle with charge $q \in \mathbb{R}$ stationary at position $\vec{r}_0 \in \mathbb{R}^3$. Let $A$ be the distributional electromagnetic 4-potential of this stationary point particle and $J$ be its distributional 4-current density. Then $A$ is an extremum of the electromagnetic action with respect to $J$, which means it satisfies the source-dependent Maxwell equations. This is expressed by the vanishing of the variational gradient of the Lagrangian density $\mathcal{L}$ with respect to $A$: $$\frac{\delta \mathcal{L}}{\delta A} = 0$$ where $\frac{\delta \mathcal{L}}{\delta A}$ is the distribution whose temporal component corresponds to Gauss's Law ($\nabla \cdot \mathbf{E} = \rho/\epsilon_0$) and whose spatial components correspond to the Ampère-Maxwell Law ($\nabla \times \mathbf{B} - \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t} = \mu_0 \mathbf{j}$).