Physlib.Electromagnetism.Kinematics.EMPotential

For a given spatial dimension $d$, an electromagnetic potential $A$ can be treated as a function that maps a spacetime point $x \in \text{SpaceTime}_d$ to a contravariant Lorentz vector $A(x) \in \text{Vector}_d$. This allows the potential to be evaluated directly at any point in spacetime.

For a given spatial dimension $d$, the addition of two electromagnetic potentials $A$ and $B$ is defined pointwise. The sum $A + B$ is the electromagnetic potential that maps each point $x$ in spacetime to the sum of the Lorentz vectors $A(x)$ and $B(x)$, where the addition is performed in the space of Lorentz vectors $\text{Vector}_d$.

For any two electromagnetic potentials $A$ and $B$ in $d$ spatial dimensions, the underlying function of the sum $A + B$ is equal to the sum of the underlying functions of $A$ and $B$. Here, the electromagnetic potential is considered as a mapping from spacetime to contravariant Lorentz vectors, and the addition of potentials is defined pointwise.

For any two electromagnetic potentials $A$ and $B$ in $d$ spatial dimensions, and for any point $x$ in spacetime, the evaluation of their sum at $x$ is equal to the sum of the individual potentials evaluated at $x$, written as $(A + B)(x) = A(x) + B(x)$.

For a given spatial dimension $d$, the electromagnetic potential $A$, which is a function from spacetime to Lorentz vectors, can be multiplied by a real scalar $r \in \mathbb{R}$. This operation is defined pointwise: for any point $x$ in spacetime, $(r \cdot A)(x) = r \cdot A(x)$, where the right-hand side denotes the scalar multiplication of the Lorentz vector $A(x)$ by the real number $r$.

For any spatial dimension $d$, real scalar $r \in \mathbb{R}$, and electromagnetic potential $A$, the underlying function (or value) of the scaled potential $r \cdot A$ is equal to the scalar $r$ multiplied by the underlying function of $A$, expressed as $(r \cdot A). \text{val} = r \cdot A. \text{val}$.

For any real scalar $r \in \mathbb{R}$, electromagnetic potential field $A$, and point $x$ in spacetime, the value of the scalar-multiplied potential at $x$ is equal to the scalar $r$ multiplied by the value of the potential at $x$, i.e., $(r \cdot A)(x) = r \cdot A(x)$.

In a spacetime with $d$ spatial dimensions, let $A$ be a differentiable electromagnetic potential field and $\Lambda$ be a Lorentz transformation. For any point $x$ in spacetime and indices $\mu, \nu \in \{0, 1, \dots, d\}$, the $\nu$-th component of the partial derivative with respect to the $\mu$-th coordinate of the transformed potential $A'(x) = \Lambda \cdot A(\Lambda^{-1} x)$ is given by: $$\partial_\mu \left( \Lambda \cdot A(\Lambda^{-1} x) \right)^\nu = \sum_{\kappa} \sum_{\rho} \Lambda_{\nu \kappa} (\Lambda^{-1})_{\rho \mu} \partial_\rho A^\kappa (\Lambda^{-1} x)$$ where $\Lambda_{\nu \kappa}$ denotes the components of the Lorentz transformation matrix, $(\Lambda^{-1})_{\rho \mu}$ denotes the components of its inverse matrix, and $\partial_\rho A^\kappa$ is the $\kappa$-th component of the $\rho$-th partial derivative of the original potential $A$ evaluated at the point $\Lambda^{-1} x$.

Let $d$ be the number of spatial dimensions and $A$ be an electromagnetic potential field. If $A$ is differentiable as a function from spacetime to the space of Lorentz vectors, then for every index $\mu$ in the spacetime index set $\{0, 1, \dots, d\}$, the component function $x \mapsto A^\mu(x)$ is also differentiable.

Let $d$ be the number of spatial dimensions. Let $A : \text{SpaceTime}_d \to \text{Vector}_d$ be an infinitely differentiable ($C^\infty$) electromagnetic potential field. For any spacetime index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, the variational adjoint derivative of the component evaluation map $A \mapsto A^\mu$ (which extracts the $\mu$-th component of the potential) is the operator that maps a scalar field $\psi$ to the vector field $x \mapsto \psi(x) \mathbf{e}_\mu$, where $\mathbf{e}_\mu$ is the $\mu$-th standard basis vector of $\text{Vector}_d$.

Let $d$ be the number of spatial dimensions. Let $A : \text{SpaceTime}_d \to \text{Vector}_d$ be an infinitely differentiable ($C^\infty$) electromagnetic potential field. For any spacetime indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$, the variational adjoint derivative of the map $A \mapsto \partial_\mu A^\nu$ (the partial derivative of the $\nu$-th component of the potential with respect to the $\mu$-th coordinate) is the operator that maps a scalar field $\psi$ to the vector field $x \mapsto -(\partial_\mu \psi(x)) \mathbf{e}_\nu$, where $\partial_\mu \psi$ is the partial derivative of $\psi$ in the direction of the $\mu$-th coordinate and $\mathbf{e}_\nu$ is the $\nu$-th standard basis vector of $\text{Vector}_d$.

For an electromagnetic potential $A$ defined on a $(1+d)$-dimensional spacetime, the derivative $\text{deriv} A$ is a map from spacetime to the tensor product space $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$. At any point $x$ in spacetime, the tensor is defined by the sum $$\sum_{\mu, \nu} (\partial_\mu A^\nu(x)) (e^\mu \otimes e_\nu)$$ where $\partial_\mu A^\nu(x)$ denotes the partial derivative of the $\nu$-th component of the potential $A$ with respect to the $\mu$-th coordinate, and $e^\mu$ and $e_\nu$ are the standard basis elements for the space of Lorentz covectors and vectors, respectively.

In a $(1+d)$-dimensional spacetime, let $A$ be a differentiable electromagnetic potential field and $\Lambda$ be a Lorentz transformation. Let the transformed potential field be defined by $A'(x) = \Lambda \cdot A(\Lambda^{-1} x)$. The derivative tensor of the transformed potential at point $x$ is equal to the Lorentz transformation of the derivative tensor of the original potential evaluated at $\Lambda^{-1} x$: $$\text{deriv } A'(x) = \Lambda \cdot \text{deriv } A(\Lambda^{-1} x)$$ where $\text{deriv } A$ is the tensor field $x \mapsto \partial_\mu A^\nu(x)$ and the action of $\Lambda$ on the right-hand side is the standard Lorentz transformation for a $(1,1)$-tensor in $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$.

For an electromagnetic potential $A$ on a $(1+d)$-dimensional spacetime and a point $x$ in spacetime, the $(\mu, \nu)$-th component of the derivative tensor $\text{deriv } A(x)$ in the standard basis (formed by the tensor product of the Lorentz covector basis $\{e^\mu\}$ and the Lorentz vector basis $\{e_\nu\}$) is given by the partial derivative of the $\nu$-th component of the potential $A$ with respect to the $\mu$-th coordinate, denoted $\partial_\mu A^\nu(x)$.

In a $(1+d)$-dimensional spacetime, let $A$ be an electromagnetic potential and $x$ be a point in spacetime. Let $\text{deriv } A(x)$ be the derivative tensor at $x$, which is an element of the tensor product space $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$. Let $\Phi$ be the linear equivalence (`toTensor`) that maps this object to the formal tensor space of the species `realLorentzTensor d` with index colors $[\text{down}, \text{up}]$. For any multi-index $b = (b_0, b_1)$ in the set of component indices, the representation of the tensor $\Phi(\text{deriv } A(x))$ in the canonical tensor basis is given by: $$[\Phi(\text{deriv } A(x))]_b = \partial_{\mu} A^{\nu}(x)$$ where $\mu = \psi(b_0)$ and $\nu = \psi(b_1)$ are the spacetime indices in $\{0, \dots, d\}$ corresponding to the component indices $b_0$ and $b_1$ via the canonical equivalence $\psi$.

The function `deriv` is a linear map that takes an electromagnetic potential $A$, treated as a distribution on a $(1+d)$-dimensional spacetime, and returns its distributional derivative. The result is a distribution taking values in the tensor product space $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$, which corresponds to the $(1,1)$-tensor field $\partial_\mu A^\nu$.

Let $A$ be an electromagnetic potential distribution on a $(1+d)$-dimensional spacetime and $\varepsilon$ be a test function in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$. The distributional derivative of $A$ evaluated at $\varepsilon$, denoted $(\text{deriv } A)(\varepsilon)$, is equal to the sum: $$(\text{deriv } A)(\varepsilon) = \sum_{\mu, \nu} (\partial_\mu A^\nu)(\varepsilon) (e_\mu \otimes e_\nu)$$ where $e_\mu$ and $e_\nu$ are the standard basis elements for Lorentz covectors and vectors respectively, and $(\partial_\mu A^\nu)(\varepsilon)$ represents the distributional derivative of the $\nu$-th component of the potential $A$ with respect to the $\mu$-th coordinate acting on the test function $\varepsilon$.

Let $A$ be an electromagnetic potential distribution on a $(1+d)$-dimensional spacetime. For any test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$ and any pair of spacetime indices $(\mu, \nu)$, the $(\mu, \nu)$-component of the distributional derivative $(\text{deriv } A)(\varepsilon)$ with respect to the standard tensor product basis of $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$ is equal to the distributional derivative of the $\nu$-th component of $A$ with respect to the $\mu$-th coordinate evaluated at $\varepsilon$. Mathematically, this is expressed as: $$[(\text{deriv } A)(\varepsilon)]_{\mu\nu} = (\partial_\mu A^\nu)(\varepsilon)$$

Let $A$ be an electromagnetic potential distribution on a $(1+d)$-dimensional spacetime and $\varepsilon$ be a test function in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$. Let $\text{deriv } A$ denote the distributional derivative of $A$, which is a distribution taking values in the tensor product space $\text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d)$, corresponding to the field $\partial_\mu A^\nu$. Let $\Phi: \text{CoVector}(d) \otimes_{\mathbb{R}} \text{Vector}(d) \cong S.\text{Tensor}([\text{down}, \text{up}])$ be the canonical linear equivalence that maps the physical tensor product to the formal tensor space of the species $S = \text{realLorentzTensor}(d)$. For any multi-index $b = (b_0, b_1)$ identifying a component of a $(1,1)$-tensor, the component of the formal tensor $\Phi((\text{deriv } A)(\varepsilon))$ with respect to the canonical tensor basis is given by: $$[\Phi((\text{deriv } A)(\varepsilon))]_b = (\partial_\mu A^\nu)(\varepsilon)$$ where $\mu$ and $\nu$ are the spacetime indices in $\text{Fin } 1 \oplus \text{Fin } d$ corresponding to the component indices $b_0$ and $b_1$ respectively.

For an electromagnetic potential $A$ treated as a distribution on a $(1+d)$-dimensional spacetime and a Lorentz transformation $\Lambda$ from the Lorentz group $\mathcal{L}$, the distributional derivative operator $\partial$ (which maps the potential to the $(1,1)$-tensor $\partial_\mu A^\nu$) is equivariant under the action of the Lorentz group. That is, $$\partial(\Lambda \cdot A) = \Lambda \cdot (\partial A)$$ where $\Lambda \cdot A$ denotes the action of the Lorentz transformation on the potential distribution and $\Lambda \cdot (\partial A)$ denotes the action on the resulting tensor distribution.

Given a spatial dimension $d$, the speed of light $c$, and a time-dependent vector potential $\mathbf{A} : \text{Time} \to \text{Space}(d) \to \mathbb{R}^d$, this definition constructs an electromagnetic 4-potential $A^\mu$. In this construction, the scalar potential $\phi$ is set to zero, resulting in a 4-potential of the form $A^\mu = (0, \mathbf{A})$.

Given the spatial dimension $d$, the speed of light $c$, and a static (time-independent) vector potential $\mathbf{A} : \text{Space}(d) \to \mathbb{R}^d$, this definition constructs an electromagnetic 4-potential $A^\mu$. In this construction, the scalar potential $\phi$ is set to zero and $\mathbf{A}$ is treated as constant in time, resulting in a 4-potential of the form $A^\mu = (0, \mathbf{A})$.

Given the speed of light $c$, a scalar potential $\phi: \text{Time} \times \text{Space}^d \to \mathbb{R}$, and a vector potential $\mathbf{A}: \text{Time} \times \text{Space}^d \to \mathbb{R}^d$, this definition constructs the electromagnetic 4-potential $A^\mu$ in $d$ spatial dimensions.

For a given spatial dimension $d$ and speed of light $c$, let $\phi: \text{Time} \times \text{Space}^d \to \mathbb{R}$ be a scalar potential and $\mathbf{A}: \text{Time} \times \text{Space}^d \to \mathbb{R}^d$ be a vector potential. The electromagnetic $4$-potential $A^\mu$ constructed from both $\phi$ and $\mathbf{A}$ is equal to the sum of the $4$-potential constructed from the scalar potential $\phi$ (with a vanishing vector potential) and the $4$-potential constructed from the vector potential $\mathbf{A}$ (with a vanishing scalar potential): $$A^\mu(\phi, \mathbf{A}) = A^\mu(\phi, \mathbf{0}) + A^\mu(0, \mathbf{A})$$

Given a spatial dimension $d$, the speed of light $c$, a time-independent (static) scalar potential $\phi: \text{Space}^d \to \mathbb{R}$, and a static vector potential $\mathbf{A}: \text{Space}^d \to \mathbb{R}^d$, this definition constructs the electromagnetic 4-potential $A^\mu$. The resulting 4-potential is defined by treating $\phi$ and $\mathbf{A}$ as constant in time, such that $A^\mu(t, \mathbf{x}) = (\phi(\mathbf{x})/c, \mathbf{A}(\mathbf{x}))$.

For a given spatial dimension $d$ and speed of light $c$, let $\phi: \text{Space}^d \to \mathbb{R}$ be a static scalar potential and $\mathbf{A}: \text{Space}^d \to \mathbb{R}^d$ be a static vector potential. The electromagnetic $4$-potential $A^\mu$ constructed from these static potentials using `ofStaticPotentials` is equal to the $4$-potential constructed using the general `ofPotentials` definition where the potentials are taken to be constant functions of time: $$\text{ofStaticPotentials}(c, \phi, \mathbf{A}) = \text{ofPotentials}(c, t \mapsto \phi, t \mapsto \mathbf{A})$$

Given the speed of light $c$, a time-dependent electric field $\mathbf{E}: \text{Time} \times \text{Space}^3 \to \mathbb{R}^3$, and a time-dependent magnetic field $\mathbf{B}: \text{Time} \times \text{Space}^3 \to \mathbb{R}^3$, this definition constructs the corresponding electromagnetic 4-potential $A^\mu$ in 3-dimensional space. The potential is uniquely determined by these fields using the Poincaré gauge (also known as the multipole gauge).

This definition establishes the action of the Lorentz group $\mathcal{L}$ on the space of electromagnetic potentials. For a given Lorentz transformation $\Lambda \in \mathcal{L}$ and an electromagnetic potential $A$, which is a function from spacetime to contravariant Lorentz vectors, the transformed potential $\Lambda \cdot A$ is defined such that at any spacetime point $x$, $(\Lambda \cdot A)(x) = \Lambda (A(\Lambda^{-1}x))$. This corresponds to the standard transformation law for a vector field under the Lorentz group.

For a given number of spatial dimensions $d$, let $\Lambda$ be a Lorentz transformation in the Lorentz group $\mathcal{L}$ and $A$ be an electromagnetic potential. The transformed potential $\Lambda \cdot A$ is defined as the function mapping each spacetime point $x$ to the vector $\Lambda A(\Lambda^{-1}x)$.

For a spacetime with $d$ spatial dimensions, let $\Lambda \in \mathcal{L}$ be a Lorentz transformation and $A$ be an electromagnetic potential. The value of the transformed potential $\Lambda \cdot A$ at a spacetime point $x$ is given by the relation $(\Lambda \cdot A)(x) = \Lambda (A(\Lambda^{-1}x))$.

For a given number of spatial dimensions $d$, the Lorentz group $\mathcal{L}$ acts on the space of electromagnetic potentials $A^\mu$. An electromagnetic potential is a function from spacetime to contravariant Lorentz vectors. For a Lorentz transformation $\Lambda \in \mathcal{L}$ and a potential $A$, the action $\Lambda \cdot A$ is defined by $(\Lambda \cdot A)(x) = \Lambda (A(\Lambda^{-1}x))$ for every spacetime point $x$. This statement establishes that this operation satisfies the axioms of a multiplicative group action, specifically the identity law $I \cdot A = A$ and the associativity law $(\Lambda_1 \Lambda_2) \cdot A = \Lambda_1 \cdot (\Lambda_2 \cdot A)$.

For a spacetime with $d$ spatial dimensions, the Lorentz group $\mathcal{L}$ acts distributively on the space of electromagnetic potentials $A^\mu$. An electromagnetic potential is defined as a function from spacetime to contravariant Lorentz vectors. For a Lorentz transformation $\Lambda \in \mathcal{L}$ and potentials $A, A_1, A_2$, the action is defined by $(\Lambda \cdot A)(x) = \Lambda (A(\Lambda^{-1}x))$. This statement asserts that this action preserves the additive structure of the potentials, satisfying $\Lambda \cdot (A_1 + A_2) = \Lambda \cdot A_1 + \Lambda \cdot A_2$, as well as the zero element $\Lambda \cdot 0 = 0$, in addition to the standard group action axioms.

Let $A$ be an electromagnetic potential in a spacetime with $d$ spatial dimensions. If $A$ is differentiable over $\mathbb{R}$, then for any Lorentz transformation $\Lambda \in \mathcal{L}$, the Lorentz-transformed potential $x \mapsto \Lambda \cdot A(\Lambda^{-1} \cdot x)$ is also differentiable.

Let $A$ be an electromagnetic potential on a spacetime with $d$ spatial dimensions. If $A$ is $n$-times continuously differentiable (of class $C^n$) over $\mathbb{R}$, then for any Lorentz transformation $\Lambda$ in the Lorentz group $\mathcal{L}$, the transformed potential $x \mapsto \Lambda \cdot A(\Lambda^{-1} \cdot x)$ is also $n$-times continuously differentiable.

Let $A$ be an electromagnetic potential in a spacetime with $d$ spatial dimensions. If $A$ is twice continuously differentiable (of class $C^2$) over $\mathbb{R}$, then for any spacetime indices $\mu$ and $\nu$, the component of the partial derivative $\partial_\mu A^\nu$ is a differentiable function of the spacetime coordinates.

Let $A$ be an electromagnetic potential on a spacetime of spatial dimension $d$. If $A$ is a smooth function ($C^\infty$), then for any spacetime indices $\mu$ and $\nu$, the function $x \mapsto \partial_\mu A^\nu(x)$, representing the $\nu$-th component of the partial derivative of $A$ with respect to the $\mu$-th coordinate, is differentiable.

Let $A$ be an electromagnetic potential in $d$ spatial dimensions, defined as a function from spacetime to Lorentz vectors. If $A$ is $C^{n+1}$ smooth, then for any spacetime indices $\mu, \nu \in \{0, 1, \dots, d\}$, the function mapping a spacetime point $x$ to the $\mu$-th partial derivative of the $\nu$-th component of the potential, $x \mapsto \partial_\mu A^\nu(x)$, is $C^n$ smooth.

Let $d$ be the spatial dimension and $c$ denote the speed of light. Let $\phi: \text{Time} \times \text{Space } d \to \mathbb{R}$ be a scalar potential. If $\phi$ is differentiable as a function of spacetime, then the electromagnetic potential $A^\mu$ constructed from this scalar potential is also differentiable.

For any spatial dimension $d$ and any natural number $n$, let $c$ be the speed of light and $\phi : \text{Time} \times \text{Space } d \to \mathbb{R}$ be a scalar potential. If the uncurried version of the scalar potential $\phi$ is $n$-times continuously differentiable ($C^n$) over $\mathbb{R}$, then the electromagnetic potential $A$ constructed from $\phi$, denoted as `ofScalarPotential c φ`, is also $n$-times continuously differentiable.

Let $d$ be the spatial dimension and $c$ be the speed of light. Let $\mathbf{A} : \text{Time} \times \text{Space } d \to \mathbb{R}^d$ be a time-dependent vector potential. If the mapping $(t, \mathbf{x}) \mapsto \mathbf{A}(t, \mathbf{x})$ is differentiable, then the electromagnetic 4-potential $A^\mu$ defined by setting the scalar potential $\phi = 0$ and the vector part to $\mathbf{A}$ (i.e., $A^\mu = (0, \mathbf{A})$) is also differentiable.

Let $d$ be the spatial dimension and $c$ be the speed of light. Let $\mathbf{A} : \text{Time} \times \text{Space } d \to \mathbb{R}^d$ be a time-dependent vector potential. If $\mathbf{A}$ is $n$-times continuously differentiable (of class $C^n$) with respect to time and space, then the electromagnetic 4-potential $A^\mu = (0, \mathbf{A})$ (constructed by setting the scalar potential $\phi = 0$) is also $n$-times continuously differentiable.

Let $d$ be the spatial dimension and $c$ be the speed of light. Consider a scalar potential $\phi: \text{Time} \times \text{Space}^d \to \mathbb{R}$ and a vector potential $\mathbf{A}: \text{Time} \times \text{Space}^d \to \mathbb{R}^d$. If $\phi$ and $\mathbf{A}$ are differentiable as functions of spacetime (i.e., their uncurried forms mapping from $\text{Time} \times \text{Space}^d$ are differentiable), then the electromagnetic 4-potential $A^\mu$ constructed from $\phi$ and $\mathbf{A}$ is also differentiable.

For any spatial dimension $d$ and speed of light $c$, let $\phi: \text{Time} \to \text{Space}^d \to \mathbb{R}$ be a scalar potential and $\mathbf{A}: \text{Time} \to \text{Space}^d \to \mathbb{R}^d$ be a vector potential. If both $\phi$ and $\mathbf{A}$ are $C^n$ differentiable (as functions from the product space $\text{Time} \times \text{Space}^d$), then the electromagnetic 4-potential $A^\mu$ constructed from $\phi$ and $\mathbf{A}$ is also $C^n$ differentiable.

For any spatial dimension 3 and natural number $n$, let $c$ be the speed of light, $\mathbf{E}: \text{Time} \times \text{Space}^3 \to \mathbb{R}^3$ be the electric field, and $\mathbf{B}: \text{Time} \times \text{Space}^3 \to \mathbb{R}^3$ be the magnetic field. If both $\mathbf{E}$ and $\mathbf{B}$ are $C^n$ differentiable as functions of spacetime, then the electromagnetic 4-potential $A^\mu$ constructed from $\mathbf{E}$ and $\mathbf{B}$ (using the Poincaré gauge) is also $C^n$ differentiable.

For a differentiable electromagnetic potential $A$ on a $(1+d)$-dimensional spacetime, the derivative tensor evaluated at a point $x$, denoted as $\text{deriv} A(x)$, is equal to the tensor derivative of the underlying function of $A$ at that point, $\text{tensorDeriv}(A)(x)$.

For any differentiable electromagnetic potential $A$ on a $(1+d)$-dimensional spacetime and any point $x$ in spacetime, the $(\mu, \nu)$-component of the tensor derivative of $A$ at $x$ (representing the tensor in $\text{CoVector} \otimes \text{Vector}$ formed by the derivatives) is equal to the partial derivative of the $\nu$-th component of the potential with respect to the $\mu$-th coordinate, denoted by $\partial_\mu A^\nu(x)$.