Physlib.Electromagnetism.Vacuum.IsPlaneWave

An electromagnetic potential $A$ in a $d$-dimensional free space $\mathcal{F}$ is defined to be a **plane wave** in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$) if there exist functions $\mathbf{E}_0: \mathbb{R} \to \mathbb{R}^d$ and $\mathbf{B}_0: \mathbb{R} \to \text{Mat}_{d \times d}(\mathbb{R})$ such that, for all time $t$ and position $\mathbf{x}$, the electric field $\mathbf{E}$ and the magnetic field matrix $\mathbf{B}$ satisfy: 1. $\mathbf{E}(t, \mathbf{x}) = \mathbf{E}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$ 2. $\mathbf{B}(t, \mathbf{x}) = \mathbf{B}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$ where $c$ is the speed of light in the free space $\mathcal{F}$, and $\mathbf{x} \cdot \hat{\mathbf{s}}$ denotes the standard Euclidean inner product.

Given a $d$-dimensional free space $\mathcal{F}$ and an electromagnetic potential $A$ that satisfies the condition of being a plane wave in direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), the function $\mathbf{E}_0: \mathbb{R} \to \mathbb{R}^d$ is the electric field function. It represents the profile of the electric field such that at any time $t$ and position $\mathbf{x}$, the electric field $\mathbf{E}$ is given by $\mathbf{E}(t, \mathbf{x}) = \mathbf{E}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$, where $c$ is the speed of light in the free space $\mathcal{F}$.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$. If $A$ is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), then the electric field $\mathbf{E}$ at any time $t$ and position $\mathbf{x}$ is given by the electric field function $\mathbf{E}_0$ of the plane wave as: $$\mathbf{E}(t, \mathbf{x}) = \mathbf{E}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$$ where $c$ is the speed of light in the free space $\mathcal{F}$ and $\mathbf{x} \cdot \hat{\mathbf{s}}$ denotes the standard Euclidean inner product.

For an electromagnetic potential $A$ in a $d$-dimensional free space $\mathcal{F}$ that is a plane wave in direction $\mathbf{s}$, the **magnetic function** is the function $\mathbf{B}_0: \mathbb{R} \to \text{Mat}_{d \times d}(\mathbb{R})$ that characterizes the profile of the magnetic field. It is defined such that for any time $t$ and position $\mathbf{x}$, the magnetic field matrix $\mathbf{B}$ satisfies $\mathbf{B}(t, \mathbf{x}) = \mathbf{B}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$, where $\hat{\mathbf{s}}$ is the unit vector in direction $\mathbf{s}$ and $c$ is the speed of light.

For an electromagnetic potential $A$ in a $d$-dimensional free space $\mathcal{F}$ that is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), the magnetic field matrix $\mathbf{B}$ at any time $t$ and position $\mathbf{x}$ is given by: $$\mathbf{B}(t, \mathbf{x}) = \mathbf{B}_0(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$$ where $\mathbf{B}_0$ is the magnetic function associated with the plane wave, $c$ is the speed of light in $\mathcal{F}$, and $\mathbf{x} \cdot \hat{\mathbf{s}}$ denotes the standard Euclidean inner product.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ that is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$). Then the electric field function $\mathbf{E}_0: \mathbb{R} \to \mathbb{R}^d$ characterizing the plane wave can be expressed in terms of the electric field $\mathbf{E}$ evaluated at the spatial origin $\mathbf{0}$ and time $t = -u/c$ as: $$\mathbf{E}_0(u) = \mathbf{E}\left(-\frac{u}{c}, \mathbf{0}\right)$$ for any $u \in \mathbb{R}$, where $c$ is the speed of light in the free space $\mathcal{F}$.

For an electromagnetic potential $A$ in a $d$-dimensional free space $\mathcal{F}$ that is a plane wave in the direction $\mathbf{s}$, the magnetic function $\mathbf{B}_0: \mathbb{R} \to \text{Mat}_{d \times d}(\mathbb{R})$ is related to the magnetic field matrix $\mathbf{B}(t, \mathbf{x})$ by the identity: $$\mathbf{B}_0(u) = \mathbf{B}\left(-\frac{u}{c}, \mathbf{0}\right)$$ for all $u \in \mathbb{R}$, where $c$ is the speed of light in $\mathcal{F}$ and $\mathbf{0}$ is the origin of the $d$-dimensional space.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$) and let $\mathbf{E}_0$ be its associated electric field function. If there exists a function $\mathbf{E}_1: \mathbb{R} \to \mathbb{R}^d$ such that the electric field $\mathbf{E}$ satisfies $\mathbf{E}(t, \mathbf{x}) = \mathbf{E}_1(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$ for all time $t$ and position $\mathbf{x}$, then $\mathbf{E}_1 = \mathbf{E}_0$.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ that is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), and let $\mathbf{B}_0$ be its associated magnetic function. If $\mathbf{B}_1: \mathbb{R} \to \text{Mat}_{d \times d}(\mathbb{R})$ is any function such that for all time $t$ and position $\mathbf{x}$, the magnetic field matrix $\mathbf{B}$ satisfies $\mathbf{B}(t, \mathbf{x}) = \mathbf{B}_1(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$, then $\mathbf{B}_1 = \mathbf{B}_0$.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ that is twice continuously differentiable ($C^2$). If $A$ is a plane wave in the direction $\mathbf{s}$ with an associated electric field function $\mathbf{E}_0: \mathbb{R} \to \mathbb{R}^d$, then $\mathbf{E}_0$ is differentiable.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ that is twice continuously differentiable ($C^2$). If $A$ is a plane wave in the direction $\mathbf{s}$, then its associated magnetic function $\mathbf{B}_0: \mathbb{R} \to \text{Mat}_{d \times d}(\mathbb{R})$ is component-wise differentiable. That is, for any indices $i, j$, the function mapping $u \in \mathbb{R}$ to the $(i, j)$-th component of $\mathbf{B}_0(u)$ is differentiable.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$. If $A$ is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$) and has an associated electric field function $\mathbf{E}_0: \mathbb{R} \to \mathbb{R}^d$, then the partial derivative of the electric field $\mathbf{E}$ with respect to time $t$ at any position $\mathbf{x}$ is given by: $$\frac{\partial \mathbf{E}(t, \mathbf{x})}{\partial t} = -c \mathbf{E}_0'(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$$ where $c$ is the speed of light in the free space $\mathcal{F}$, and $\mathbf{E}_0'$ denotes the derivative of the function $\mathbf{E}_0$ with respect to its scalar argument.

For an electromagnetic potential $A$ in a $d$-dimensional free space $\mathcal{F}$ that is a twice continuously differentiable ($C^2$) plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), the time derivative of the $(i, j)$-th component of the magnetic field matrix $\mathbf{B}$ at time $t$ and position $\mathbf{x}$ is given by: $$\frac{\partial}{\partial t} B_{ij}(t, \mathbf{x}) = -c \frac{d}{du} (B_0)_{ij}(u) \bigg|_{u = \mathbf{x} \cdot \hat{\mathbf{s}} - ct}$$ where $c$ is the speed of light in $\mathcal{F}$, $\mathbf{B}_0$ is the magnetic function associated with the plane wave, and $\mathbf{x} \cdot \hat{\mathbf{s}}$ denotes the standard Euclidean inner product.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$. If $A$ is a plane wave in direction $\mathbf{s}$ with unit vector $\hat{\mathbf{s}}$, then for any time $t$ and position $\mathbf{x}$, the partial derivative of the electric field $\mathbf{E}$ with respect to the $i$-th spatial coordinate is given by: $$\frac{\partial \mathbf{E}(t, \mathbf{x})}{\partial x_i} = \hat{s}_i \mathbf{E}_0'(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$$ where $\mathbf{E}_0$ is the electric field function associated with the plane wave, $\mathbf{E}_0'$ is its derivative, and $c$ is the speed of light in the free space.

Let $A$ be an electromagnetic potential in $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is twice continuously differentiable and is a plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$). Then, for any time $t$, position $\mathbf{x}$, and indices $i, j, k \in \{1, \dots, d\}$, the partial derivative of the $(i, j)$-th component of the magnetic field matrix $\mathbf{B}$ with respect to the $k$-th spatial coordinate $x_k$ is given by: $$\frac{\partial}{\partial x_k} B_{ij}(t, \mathbf{x}) = \hat{s}_k B'_{0, ij}(\mathbf{x} \cdot \hat{\mathbf{s}} - ct)$$ where $B_{0, ij}$ is the $(i, j)$-th component of the magnetic function $\mathbf{B}_0$ associated with the plane wave, $B'_{0, ij}$ denotes its derivative with respect to its scalar argument, and $\mathbf{x} \cdot \hat{\mathbf{s}}$ is the standard Euclidean inner product.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. If $A$ is a plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$), then for any time $t$, position $\mathbf{x}$, and indices $i, k \in \{1, \dots, d\}$, the partial derivative of the $i$-th component of the electric field $E_i$ with respect to the $k$-th spatial coordinate $x_k$ is related to its time derivative by the following equation: $$\frac{\partial E_i(t, \mathbf{x})}{\partial x_k} = -\frac{\hat{s}_k}{c} \frac{\partial E_i(t, \mathbf{x})}{\partial t}$$ where $\hat{s}_k$ is the $k$-th component of the unit vector in the direction of propagation.

Let $A$ be an electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is a twice continuously differentiable ($C^2$) plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$). Then, for any time $t$, position $\mathbf{x}$, and indices $i, j, k \in \{1, \dots, d\}$, the partial derivative of the $(i, j)$-th component of the magnetic field matrix $\mathbf{B}$ with respect to the $k$-th spatial coordinate $x_k$ is related to its time derivative by: $$\frac{\partial}{\partial x_k} B_{ij}(t, \mathbf{x}) = -\frac{\hat{s}_k}{c} \frac{\partial}{\partial t} B_{ij}(t, \mathbf{x})$$ where $\hat{s}_k$ is the $k$-th component of the unit vector $\hat{\mathbf{s}}$.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. If $A$ is a plane wave in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$), then for any time $t$, position $\mathbf{x}$, and spatial indices $i, j \in \{1, \dots, d\}$, the time derivative of the $(i, j)$-th component of the magnetic field matrix $B_{ij}$ is equal to the time derivative of a linear combination of the electric field components $E_i$ and $E_j$: $$\frac{\partial B_{ij}(t, \mathbf{x})}{\partial t} = \frac{\partial}{\partial t} \left( \frac{\hat{s}_j E_i(t, \mathbf{x}) - \hat{s}_i E_j(t, \mathbf{x})}{c} \right)$$ where $E_k(t, \mathbf{x})$ is the $k$-th component of the electric field at time $t$ and position $\mathbf{x}$, and $\hat{s}_k$ is the $k$-th component of the unit vector in the direction of propagation.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. If $A$ is a plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$), then for any time $t$, position $\mathbf{x}$, and spatial indices $i, j, k \in \{1, \dots, d\}$, the partial derivative of the $(i, j)$-th component of the magnetic field matrix $B_{ij}$ with respect to the $k$-th spatial coordinate $x_k$ is equal to the partial derivative of a specific linear combination of electric field components $E_i$ and $E_j$: $$\frac{\partial B_{ij}(t, \mathbf{x})}{\partial x_k} = \frac{\partial}{\partial x_k} \left( \frac{\hat{s}_j E_i(t, \mathbf{x}) - \hat{s}_i E_j(t, \mathbf{x})}{c} \right)$$ where $E_m(t, \mathbf{x})$ is the $m$-th component of the electric field at time $t$ and position $\mathbf{x}$, and $\hat{s}_m$ is the $m$-th component of the unit vector in the direction of propagation.

Let $A$ be a twice continuously differentiable ($C^2$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. If $A$ is a plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$), then for any spatial indices $i, j \in \{1, \dots, d\}$, there exists a constant $C$ such that for all time $t$ and position $\mathbf{x}$, the $(i, j)$-th component of the magnetic field matrix $B_{ij}$ is related to the components of the electric field $E$ by: $$B_{ij}(t, \mathbf{x}) = \frac{1}{c} \left( \hat{s}_j E_i(t, \mathbf{x}) - \hat{s}_i E_j(t, \mathbf{x}) \right) + C$$ where $E_m(t, \mathbf{x})$ denotes the $m$-th component of the electric field at time $t$ and position $\mathbf{x}$.

Let $A$ be an infinitely differentiable electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is a plane wave in the direction $\mathbf{s}$ (represented by the unit vector $\hat{\mathbf{s}}$) and is an extremum of the electromagnetic Lagrangian with zero current density ($J=0$). Then, for any time $t$, position $\mathbf{x}$, and spatial component $i \in \{1, \dots, d\}$, the time derivative of the $i$-th component of the electric field $\mathbf{E}$ is equal to the time derivative of a weighted sum of the magnetic field matrix components: $$\frac{\partial E_i(t, \mathbf{x})}{\partial t} = \frac{\partial}{\partial t} \left( c \sum_{j=1}^d B_{ij}(t, \mathbf{x}) \hat{s}_j \right)$$ where $B_{ij}$ are the components of the magnetic field matrix.

Let $A$ be an infinitely differentiable ($C^\infty$) electromagnetic potential in $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is a plane wave propagating in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$) and satisfies the vacuum field equations (i.e., it is an extremum of the electromagnetic action with zero current density). Then, for any time $t$, position $\mathbf{x}$, and spatial indices $i, k \in \{1, \dots, d\}$, the partial derivative of the $i$-th component of the electric field $E_i$ with respect to the $k$-th spatial coordinate $x_k$ is equal to the partial derivative of a weighted sum of the magnetic field matrix components: $$\frac{\partial E_i(t, \mathbf{x})}{\partial x_k} = \frac{\partial}{\partial x_k} \left( c \sum_{j=1}^d B_{ij}(t, \mathbf{x}) \hat{s}_j \right)$$ where $B_{ij}$ represents the components of the magnetic field matrix.

Let $A$ be an infinitely differentiable ($C^\infty$) electromagnetic potential in a $d$-dimensional free space $\mathcal{F}$ with speed of light $c$. Suppose $A$ is a plane wave propagating in the direction $\mathbf{s}$ (with unit vector $\hat{\mathbf{s}}$) and satisfies the vacuum field equations (it is an extremum of the electromagnetic action with zero current density $J=0$). Then, for each spatial index $i \in \{1, \dots, d\}$, there exists a constant $C$ such that for all time $t$ and position $\mathbf{x}$, the $i$-th component of the electric field $E_i$ is given by: $$E_i(t, \mathbf{x}) = c \sum_{j=1}^d B_{ij}(t, \mathbf{x}) \hat{s}_j + C$$ where $B_{ij}$ are the components of the magnetic field matrix.