Physlib.Electromagnetism.Kinematics.FieldStrength

Given an electromagnetic potential $A$ defined on a $(1+d)$-dimensional spacetime, the function `toFieldStrength` assigns to $A$ its corresponding field strength tensor field $F$ (also known as the Faraday tensor). At each point $x$ in spacetime, the value $F(x)$ is an element of the tensor product of two Lorentz vector spaces $V \otimes V$. The tensor is defined as the antisymmetrized derivative of the potential: \[ F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu \] This is constructed by taking the difference between the derivative of the potential contracted with the Minkowski metric $\eta$ and its transpose (the same term with indices swapped).

For an electromagnetic potential $A$ in $(1+d)$-dimensional spacetime, the value of the field strength tensor $F$ at a point $x$ is given by the difference: $$F^{\mu\nu}(x) = \eta^{\mu\rho} \partial_\rho A^\nu(x) - \eta^{\nu\rho} \partial_\rho A^\mu(x)$$ where $\eta$ is the Minkowski metric and $\partial_\rho A^\nu$ denotes the derivative of the potential. In the formal expression, this is represented by taking the tensor $T^{\mu\nu} = \eta^{\mu\rho} \partial_\rho A^\nu$ and subtracting its index-permuted version $T^{\nu\mu}$.

For an electromagnetic potential $A$ on a $(1+d)$-dimensional spacetime and a point $x$ in spacetime, the representation of the field strength tensor $F(x)$ (the Faraday tensor) in the formal tensor index notation is equal to the antisymmetrized raised-index derivative of the potential: $$\text{toTensor}(F(x)) = T^{\mu\nu} - T^{\nu\mu}$$ where $T^{\mu\nu} = \eta^{\mu\rho} \partial_\rho A^\nu(x)$ is the tensor constructed by contracting the Minkowski metric $\eta$ with the derivative of the potential $\partial A$ at $x$. Here, $\text{toTensor}$ denotes the linear equivalence between the physical tensor product space and the formal index-based tensor species representation.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $F$ be the associated field strength tensor (Faraday tensor). For any point $x$ in spacetime and any multi-index $b = (b_0, b_1)$ corresponding to the contravariant components of a rank-2 tensor, the component of the field strength tensor $F$ at index $b$ in the canonical basis is given by the formula: $$[F(x)]^{\mu\nu} = \sum_{\kappa} \left( \eta^{\mu\kappa} \partial_\kappa A^\nu(x) - \eta^{\nu\kappa} \partial_\kappa A^\mu(x) \right)$$ where $\mu$ and $\nu$ are the spacetime indices in $\text{Fin } 1 \oplus \text{Fin } d$ corresponding to the multi-index components $b_0$ and $b_1$, $\eta$ denotes the Minkowski metric, and $\partial_\kappa A^\nu(x)$ is the partial derivative of the $\nu$-th component of the potential with respect to the $\kappa$-th coordinate at point $x$.

Let $d$ be the number of spatial dimensions. For an electromagnetic potential $A$ and a point $x$ in spacetime, let $F(x)$ be the field strength tensor (Faraday tensor) at $x$, which is an element of the tensor product space $V \otimes V$ of Lorentz vectors. Let $\mathcal{B}_{\text{tensor}}$ be the canonical basis for the formal $(2,0)$-tensor space of the species `realLorentzTensor d`, and let $\mathcal{B}_V \otimes \mathcal{B}_V$ be the basis for $V \otimes V$ induced by the standard Lorentz vector basis $\mathcal{B}_V$. For any multi-index $b = (b_0, b_1)$ in the component index set of the formal tensor, the coordinate of $F(x)$ in the formal tensor basis at $b$ is equal to the coordinate of $F(x)$ in the tensor product basis at the index pair $(\mu, \nu)$, where $\mu$ and $\nu$ are the Lorentz indices in $\text{Fin } 1 \oplus \text{Fin } d$ corresponding to $b_0$ and $b_1$ via the standard index equivalence.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $F(x)$ be the field strength tensor (Faraday tensor) at a point $x$. For any pair of spacetime indices $(\mu, \nu)$, the component of the field strength tensor in the standard tensor product basis is given by: $$[F(x)]^{\mu\nu} = \sum_{\kappa} \left( \eta^{\mu\kappa} \partial_\kappa A^\nu(x) - \eta^{\nu\kappa} \partial_\kappa A^\mu(x) \right)$$ where $\eta$ is the Minkowski metric and $\partial_\kappa A^\nu(x)$ denotes the partial derivative of the $\nu$-th component of the potential with respect to the $\kappa$-th coordinate at point $x$.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $x$ be a point in spacetime. For any pair of spacetime indices $(\mu, \nu)$, the $(\mu, \nu)$-th component of the field strength tensor $F$ at $x$ is given by: $$F^{\mu\nu}(x) = \eta^{\mu\mu} \partial_\mu A^\nu(x) - \eta^{\nu\nu} \partial_\nu A^\mu(x)$$ where $\eta$ is the Minkowski metric, and $\partial_\mu A^\nu(x)$ denotes the partial derivative of the $\nu$-th component of the potential with respect to the $\mu$-th coordinate evaluated at point $x$.

Given an electromagnetic potential $A$ defined on a $(1+d)$-dimensional spacetime and a point $x$ in that spacetime, the function `fieldStrengthMatrix` returns the components of the field strength tensor $F$ (the Faraday tensor) at $x$ as a matrix. This matrix is obtained by taking the coordinate representation of the tensor $F(x) = \partial^\mu A^\nu - \partial^\nu A^\mu$ with respect to the standard basis of the spacetime, where indices $\mu, \nu$ range over the temporal and spatial components $\{0, 1, \dots, d\}$.

For an electromagnetic potential $A$ in a spacetime with $d$ spatial dimensions, the field strength matrix at a point $x$ is equal to the coordinate representation of the field strength tensor $F(x)$ with respect to the basis formed by the tensor product of the standard Lorentz covector basis and the standard Lorentz vector basis.

For a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $x$ be a point in spacetime. Let $F(x)$ be the field strength tensor (Faraday tensor) at $x$, and let $\mathbf{F}(x)$ be its matrix representation (the field strength matrix). For any indices $\mu, \nu \in \{0, 1, \dots, d\}$, the $(\mu, \nu)$-th entry of the field strength matrix is equal to the coordinate of the formal tensor representation of $F(x)$ with respect to the canonical tensor basis at the corresponding multi-index $(\mu, \nu)$.

For an electromagnetic potential $A$ defined on a $(1+d)$-dimensional spacetime, the field strength tensor $F$ (the Faraday tensor) at a point $x$ can be expressed as a linear combination of the standard basis tensors $e_\mu \otimes e_\nu$. Specifically, the tensor is given by: \[ F(x) = \sum_{\mu} \sum_{\nu} F_{\mu\nu}(x) (e_\mu \otimes e_\nu) \] where $F_{\mu\nu}(x)$ are the components of the field strength matrix at $x$, and $\{e_\mu\}$ is the standard basis for the space of Lorentz vectors $\text{Vector}_d$.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential. If $A$ is twice continuously differentiable ($C^2$), then for any pair of spacetime indices $\mu$ and $\nu$, the $(\mu, \nu)$-th component of the field strength matrix $F_{\mu\nu}$ (representing the Faraday tensor) is a differentiable function with respect to the spacetime coordinates.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential. If $A$ is twice continuously differentiable ($C^2$), then the associated field strength tensor field $F$ (the Faraday tensor) is differentiable with respect to the spacetime coordinates.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential. If $A$ is twice continuously differentiable (of class $C^2$), then for any fixed time $t$, speed of light $c$, and indices $\mu, \nu \in \{0, 1, \dots, d\}$, the $(\mu, \nu)$-th component of the field strength matrix $F_{\mu\nu}$ is differentiable with respect to the spatial coordinates $x$. Specifically, the function $x \mapsto F_{\mu\nu}(t, x)$, where $(t, x)$ is the spacetime point reconstructed from time $t$ and spatial position $x$, is differentiable.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential that is twice continuously differentiable (of class $C^2$). For any fixed spatial point $x$, speed of light $c$, and any pair of spacetime indices $\mu$ and $\nu$, the $(\mu, \nu)$-th component of the field strength matrix $F_{\mu\nu}$ evaluated at the spacetime point corresponding to time $t$ and position $x$ is a differentiable function with respect to time $t$.

For a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential. If $A$ is $n+1$ times continuously differentiable (of class $C^{n+1}$), then for any spacetime indices $\mu$ and $\nu$, the $(\mu, \nu)$-th component of the field strength matrix of $A$, mapping a spacetime point $x$ to $F^{\mu\nu}(x)$, is $n$ times continuously differentiable (of class $C^n$).

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential. If $A$ is infinitely differentiable (of class $C^\infty$), then for any spacetime indices $\mu$ and $\nu$, the $(\mu, \nu)$-th component of the field strength matrix, mapping a spacetime point $x$ to the value $F_{\mu\nu}(x)$, is also infinitely differentiable (of class $C^\infty$).

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $F$ be the associated field strength tensor (Faraday tensor) defined by $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$. For any point $x$ in spacetime and any indices $\mu$ and $\nu$ representing spacetime coordinates, the field strength tensor is antisymmetric: $$F^{\mu\nu}(x) = -F^{\nu\mu}(x)$$

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $F(x)$ be the field strength matrix (Faraday matrix) at a point $x$. For any spacetime indices $\mu$ and $\nu$, the components of the field strength matrix satisfy: $$F(x)_{\mu\nu} = -F(x)_{\nu\mu}$$

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $F(x)$ be its corresponding field strength matrix (Faraday matrix) at a point $x$. For any spacetime index $\mu \in \{0, 1, \dots, d\}$, the diagonal entry of the field strength matrix is zero: $$F_{\mu\mu}(x) = 0$$

Let $d$ be a natural number representing spatial dimensions. Let $A$ be a differentiable electromagnetic potential field on a $(1+d)$-dimensional spacetime. For any Lorentz transformation $\Lambda \in \mathcal{L}$ and any spacetime point $x$, let $A'$ be the Lorentz-transformed potential defined by $A'(x) = \Lambda A(\Lambda^{-1} x)$. Then the field strength tensor $F$ (the Faraday tensor) satisfies the following equivariance property: \[ F[A'](x) = \Lambda \cdot F[A](\Lambda^{-1} x) \] where $F[A]$ denotes the field strength tensor associated with potential $A$, and the operation $\cdot$ represents the action of the Lorentz group on a rank-2 tensor.

Let $d$ be a natural number representing spatial dimensions. Let $A$ be a differentiable electromagnetic potential field on a $(1+d)$-dimensional spacetime. For any Lorentz transformation $\Lambda$ in the Lorentz group $\mathcal{L}$ and any spacetime point $x$, let $A'$ be the Lorentz-transformed potential defined by $A'(x) = \Lambda A(\Lambda^{-1} x)$. Then, for any spacetime indices $\mu, \nu \in \{0, 1, \dots, d\}$, the $(\mu, \nu)$-th component of the field strength matrix $F$ associated with $A'$ at point $x$ satisfies: \[ F^{\mu\nu}[A'](x) = \sum_{\kappa} \sum_{\rho} \Lambda_{\mu\kappa} \Lambda_{\nu\rho} F^{\kappa\rho}[A](\Lambda^{-1} x) \] where $F^{\kappa\rho}[A](\Lambda^{-1} x)$ is the $(\kappa, \rho)$-th component of the field strength matrix associated with $A$ evaluated at the point $\Lambda^{-1} x$, and $\Lambda_{ij}$ denotes the $(i, j)$-th entry of the matrix representation of the Lorentz transformation.

In a $(1+d)$-dimensional spacetime, let $A_1$ and $A_2$ be two differentiable electromagnetic potentials. For any point $x$ in spacetime, the field strength tensor $F$ (also known as the Faraday tensor) satisfies the additivity property: \[ F[A_1 + A_2](x) = F[A_1](x) + F[A_2](x) \] where $F[A]$ denotes the field strength tensor derived from the potential $A$.

In a $(1+d)$-dimensional spacetime, let $A_1$ and $A_2$ be two differentiable electromagnetic potentials. For any point $x$ in spacetime, the field strength matrix (the coordinate representation of the Faraday tensor $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$) of the sum $A_1 + A_2$ is equal to the sum of the field strength matrices of $A_1$ and $A_2$. That is: \[ [F(A_1 + A_2)](x) = [F(A_1)](x) + [F(A_2)](x) \] where $[F(A)](x)$ denotes the matrix representation of the field strength tensor derived from the potential $A$ at point $x$.

For any $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $c \in \mathbb{R}$ be a real scalar. If $A$ is differentiable, then at any point $x$ in spacetime, the field strength tensor $F$ (also known as the Faraday tensor) of the scaled potential $c \cdot A$ is equal to $c$ times the field strength tensor of $A$. That is: \[ F[c \cdot A](x) = c \cdot F[A](x) \] where $F[A]$ denotes the field strength tensor derived from the potential $A$ via the relation $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$.

For a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $c \in \mathbb{R}$ be a real scalar. If $A$ is differentiable, then at any point $x$ in spacetime, the field strength matrix (the coordinate representation of the Faraday tensor $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$) of the scaled potential $c \cdot A$ is equal to $c$ times the field strength matrix of $A$. That is: \[ [F(c \cdot A)](x) = c \cdot [F(A)](x) \] where $[F(A)](x)$ denotes the matrix representation of the field strength tensor derived from the potential $A$ at point $x$.

Given a distributional electromagnetic potential $A$ in $(1+d)$-dimensional spacetime and a Schwartz test function $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$, this function defines an auxiliary rank-2 contravariant tensor (an element of $\text{Vector}_d \otimes \mathbb{R} \text{Vector}_d$). The tensor is constructed by antisymmetrizing the distributional derivative of the potential. Specifically, let $\eta^{\mu\nu}$ be the Minkowski metric and $\partial_\nu A_{\mu'}$ represent the derivative of the distribution $A_{\mu'}$ acting on the test function $\epsilon$. The auxiliary field strength tensor $F^{\mu\nu}$ is given by the expression: $$F^{\mu\nu} = \eta^{\mu\mu'} \partial_\nu A_{\mu'} - \eta^{\nu\nu'} \partial_\mu A_{\nu'}$$ This represents $\partial_\nu A^\mu - \partial_\mu A^\nu$ in distributional form. As an auxiliary definition, it provides the raw tensor value without additional linearity or continuity structures.

Let $d$ be the number of spatial dimensions. For a distributional electromagnetic potential $A$ and a Schwartz test function $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$, the auxiliary field strength tensor $F^{\mu\nu}$ (denoted by `fieldStrengthAux A ε`) is equal to the difference between the tensor $T^{\mu\nu} = \eta^{\mu\rho} \partial_\nu A_\rho$ and its index-permuted version $T^{\nu\mu}$. Specifically, $$F^{\mu\nu} = T^{\mu\nu} - T^{\nu\mu}$$ where $T^{\mu\nu}$ is the tensor resulting from the contraction of the Minkowski metric $\eta$ with the distributional derivative of the potential $A$ evaluated at $\epsilon$.

Let $d$ be the number of spatial dimensions. For a distributional electromagnetic potential $A$ and a Schwartz test function $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$, the formal tensor representation (given by the map `Tensorial.toTensor`) of the auxiliary field strength $F^{\mu\nu}$ is equal to the difference between the tensor $T^{\mu\nu} = \eta^{\mu\rho} \partial_\nu A_\rho$ and its index-permuted version $T^{\nu\mu}$. Specifically, $$\text{toTensor}(F^{\mu\nu}) = T^{\mu\nu} - T^{\nu\mu}$$ where $T^{\mu\nu}$ is the tensor resulting from the contraction of the Minkowski metric $\eta$ with the distributional derivative of the potential $A$ evaluated at $\epsilon$.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. Let $F^{\mu\nu}$ be the auxiliary rank-2 contravariant field strength tensor (defined as `fieldStrengthAux A ε`). For a multi-index $b = (b_0, b_1)$ corresponding to spacetime indices $\mu$ and $\nu$ via the canonical isomorphism $\phi: \text{ComponentIdx} \cong \text{Fin}(1+d)$, the component of the tensor in the canonical basis is given by: $$[F^{\mu\nu}]_b = \sum_{\kappa=0}^{d} \left( \eta^{\mu\kappa} \partial_\kappa A_\nu[\epsilon] - \eta^{\nu\kappa} \partial_\kappa A_\mu[\epsilon] \right)$$ where $\eta^{\mu\kappa}$ denotes the components of the Minkowski metric, and $\partial_\kappa A_\nu[\epsilon]$ is the $\kappa$-th distributional derivative of the $\nu$-th component of the potential $A$ evaluated at the test function $\epsilon$.

Let $A$ be a distributional electromagnetic potential in a spacetime with $d$ spatial dimensions and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. Let $F = \text{fieldStrengthAux}(A, \epsilon)$ be the auxiliary rank-2 contravariant field strength tensor. For any tensor multi-index $b$ consisting of two contravariant indices, the component of $F$ (viewed as a formal tensor) in the canonical tensor basis at index $b$ is equal to the component of $F$ (viewed as an element of the tensor product of Lorentz vector spaces) in the product basis at indices $\mu = \phi(b_0)$ and $\nu = \phi(b_1)$, where $\phi$ is the canonical equivalence between tensor component indices and the spacetime index set $\text{Fin } 1 \oplus \text{Fin } d$.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. Let $F^{\mu\nu}$ be the auxiliary rank-2 contravariant field strength tensor (defined as `fieldStrengthAux A ε`). For a pair of spacetime indices $(\mu, \nu)$, the component of this tensor in the product basis of Lorentz vectors is given by: $$[F^{\mu\nu}]_{\mu, \nu} = \sum_{\kappa=0}^{d} \left( \eta^{\mu\kappa} \partial_\kappa A_\nu[\epsilon] - \eta^{\nu\kappa} \partial_\kappa A_\mu[\epsilon] \right)$$ where $\eta$ denotes the Minkowski metric and $\partial_\kappa A_\nu[\epsilon]$ represents the $\kappa$-th distributional derivative of the $\nu$-th component of the potential $A$ evaluated at the test function $\epsilon$.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. The component of the auxiliary rank-2 contravariant field strength tensor $F^{\mu\nu}$ (defined by `fieldStrengthAux A ε`) at the indices $(\mu, \nu)$ is given by: $$[F^{\mu\nu}]_{\mu,\nu} = \eta^{\mu\mu} \partial_\mu A_\nu[\epsilon] - \eta^{\nu\nu} \partial_\nu A_\mu[\epsilon]$$ where $\eta$ denotes the Minkowski metric and $\partial_\kappa A_\lambda[\epsilon]$ represents the $\kappa$-th distributional derivative of the $\lambda$-th component of the potential $A$ evaluated at the test function $\epsilon$. This formula simplifies the general sum over indices by leveraging the fact that the Minkowski metric is diagonal.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. The auxiliary rank-2 contravariant field strength tensor $F_{\text{aux}}(A, \epsilon)$ can be expressed as a sum over the tensor products of the standard basis vectors $\{e_\mu\}$ and $\{e_\nu\}$ as follows: $$F_{\text{aux}}(A, \epsilon) = \sum_{\mu, \nu} \left( \eta^{\mu\mu} \partial_\mu A_\nu[\epsilon] - \eta^{\nu\nu} \partial_\nu A_\mu[\epsilon] \right) e_\mu \otimes e_\nu$$ where $\eta^{\mu\mu}$ are the diagonal components of the Minkowski metric and $\partial_\kappa A_\lambda[\epsilon]$ denotes the $\kappa$-th distributional derivative of the $\lambda$-th component of the potential $A$ evaluated at the test function $\epsilon$.

For a given spatial dimension $d$, the field strength $F$ is a linear map that associates a distributional electromagnetic potential $A$ with a rank-2 contravariant tensor distribution. Specifically, for a distributional potential $A$, the field strength $F(A)$ is a distribution that maps a Schwartz test function $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ to a rank-2 tensor in $\text{Vector}_d \otimes \text{Vector}_d$. The components of this tensor are given by: $$(F(A)(\epsilon))^{\mu\nu} = \eta^{\mu\mu} (\partial_\nu A_\mu)(\epsilon) - \eta^{\nu\nu} (\partial_\mu A_\nu)(\epsilon)$$ where $\eta$ denotes the Minkowski metric and $(\partial_\nu A_\mu)(\epsilon)$ represents the distributional derivative of the $\mu$-th component of the potential acting on the test function $\epsilon$. The map $A \mapsto F(A)$ is linear, and for each $A$, the map $\epsilon \mapsto F(A)(\epsilon)$ is linear and continuous.

For a distributional electromagnetic potential $A$ in $(1+d)$-dimensional spacetime and a Schwartz test function $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$, the evaluation of the field strength tensor distribution $F(A)$ at $\epsilon$ is equal to the auxiliary field strength tensor $F_{\text{aux}}(A, \epsilon)$ defined by the antisymmetrized distributional derivatives of the potential.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. The field strength tensor distribution $F(A)$ evaluated at $\epsilon$ is given by the basis expansion: $$F(A)(\epsilon) = \sum_{\mu, \nu} \left( \eta^{\mu\mu} (\partial_\mu A_\nu)(\epsilon) - \eta^{\nu\nu} (\partial_\nu A_\mu)(\epsilon) \right) e_\mu \otimes e_\nu$$ where $\{e_\mu\}$ is the standard basis for the space of Lorentz vectors, $\eta^{\mu\mu}$ are the diagonal components of the Minkowski metric, and $(\partial_\kappa A_\lambda)(\epsilon)$ denotes the $\kappa$-th distributional derivative of the $\lambda$-th component of the potential $A$ acting on the test function $\epsilon$.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. The $(\mu, \nu)$ component of the rank-2 field strength tensor distribution $F(A)$ evaluated at $\epsilon$ is given by: $$(F(A)(\epsilon))^{\mu\nu} = \eta^{\mu\mu} (\partial_\mu A_\nu)(\epsilon) - \eta^{\nu\nu} (\partial_\nu A_\mu)(\epsilon)$$ where $\eta$ denotes the diagonal Minkowski metric and $(\partial_\kappa A_\lambda)(\epsilon)$ represents the $\kappa$-th distributional derivative of the $\lambda$-th component of the potential $A$ acting on the test function $\epsilon$.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. The diagonal components of the rank-2 field strength tensor distribution $F(A)$ evaluated at $\epsilon$ are zero. That is, for any spacetime index $\mu$: $$(F(A)(\epsilon))^{\mu\mu} = 0$$ where $(F(A)(\epsilon))^{\mu\mu}$ denotes the $(\mu, \mu)$ component of the tensor in the standard coordinate basis.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. For any spacetime indices $\mu$ and $\nu$ in the standard coordinate basis, the diagonal components of the $\nu$-th distributional derivative of the field strength tensor $F(A)$ evaluated at $\epsilon$ are zero. That is: $$(\partial_\nu F(A)(\epsilon))^{\mu\mu} = 0$$ where $(\partial_\nu F(A)(\epsilon))^{\mu\mu}$ denotes the $(\mu, \mu)$ component of the tensor-valued distribution derivative in the standard basis.

In $(1+d)$-dimensional spacetime, let $A$ be a distributional electromagnetic potential and $\epsilon \in \mathcal{S}(\mathbb{R}^{1+d}, \mathbb{R})$ be a Schwartz test function. Let $F(A)(\epsilon)$ be the rank-2 field strength tensor evaluated at $\epsilon$. For any spacetime indices $\mu$ and $\nu$ in the standard basis, the components of the tensor satisfy the antisymmetry property: $$(F(A)(\epsilon))^{\mu\nu} = -(F(A)(\epsilon))^{\nu\mu}$$

For a given number of spatial dimensions $d$, let $A$ be a distributional electromagnetic potential and $\Lambda$ be an element of the Lorentz group $\mathcal{L}$. The operation of taking the field strength tensor $F$ is equivariant under the Lorentz group action, meaning: $$F(\Lambda \cdot A) = \Lambda \cdot F(A)$$ In other words, the field strength of a Lorentz-transformed potential is equal to the Lorentz transformation of the field strength of the original potential.