Physlib.SpaceAndTime.SpaceTime.Derivatives

Given a function $f : \text{SpaceTime}_d \to M$, where $M$ is a topological vector space over $\mathbb{R}$, and a coordinate index $\mu \in \{0\} \oplus \{1, \dots, d\}$, the function $\partial_\mu f$ is the partial derivative of $f$ along the $\mu$-th coordinate. For each point $y$ in spacetime, the value $(\partial_\mu f)(y)$ is defined as the Fréchet derivative of $f$ at $y$ evaluated in the direction of the standard basis vector $e_\mu$.

The notation $\partial_\mu$ denotes the partial derivative operator on spacetime. Given a coordinate index $\mu$ (representing either the temporal dimension or one of the $d$ spatial dimensions) and a function $f: \text{SpaceTime } d \to M$, $\partial_\mu f$ is the derivative of $f$ along the $\mu$-th coordinate.

For any natural number $d$ representing the number of spatial dimensions, let $f : \text{SpaceTime}_d \to M$ be a function into a topological vector space $M$ over $\mathbb{R}$. For any coordinate index $\mu \in \{0\} \oplus \{1, \dots, d\}$ and any point $y \in \text{SpaceTime}_d$, the partial derivative $\partial_\mu f(y)$ is equal to the Fréchet derivative of $f$ at $y$ evaluated in the direction of the standard basis vector $e_\mu$.

For a given spatial dimension $d \in \mathbb{N}$, a function $f: \text{SpaceTime}_d \to \text{Vector}_d$ is differentiable if and only if for every coordinate index $\nu \in \text{Fin } 1 \oplus \text{Fin } d$, the component function $x \mapsto f(x)_\nu$ is differentiable.

For a spatial dimension $d \in \mathbb{N}$, a function $f: \text{SpaceTime}_d \to \text{Vector}_d$ is $n$-times continuously differentiable ($C^n$) over $\mathbb{R}$ if and only if for every coordinate index $\nu \in \{0, \dots, d\}$, the scalar-valued component function $f_\nu: \text{SpaceTime}_d \to \mathbb{R}$ (defined by $f_\nu(x) = f(x)_\nu$) is $n$-times continuously differentiable.

For a given spatial dimension $d$, let $f: \text{SpaceTime}_d \to \text{Vector}_d$ be a differentiable function over $\mathbb{R}$. For any coordinate indices $\mu, \nu \in \{0\} \oplus \{1, \dots, d\}$ and any point $y \in \text{SpaceTime}_d$, the $\nu$-th component of the partial derivative $\partial_\mu f$ at $y$ is equal to the Fréchet derivative of the $\nu$-th component function $f_\nu$ (where $f_\nu(x) = f(x)_\nu$) at point $y$ evaluated in the direction of the $\mu$-th standard basis vector $e_\mu$: $$(\partial_\mu f(y))_\nu = D(f_\nu)(y)(e_\mu)$$

Let $d$ be a natural number and let $f: \text{SpaceTime}_d \to \text{Vector}_d$ be a differentiable function over the real numbers $\mathbb{R}$. For any point $y \in \text{SpaceTime}_d$, any tangent vector $\Delta t \in \text{SpaceTime}_d$, and any coordinate index $\nu \in \text{Fin } 1 \oplus \text{Fin } d$, the $\nu$-th component of the Fréchet derivative of $f$ at $y$ evaluated at $\Delta t$ is equal to the Fréchet derivative of the component function $f_\nu(x) = (f(x))_\nu$ at $y$ evaluated at $\Delta t$: $$(Df(y)(\Delta t))_\nu = D(f_\nu)(y)(\Delta t)$$

For any natural number $d$ and coordinate indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$ in spacetime, the partial derivative of the $\nu$-th coordinate function $x \mapsto x_\nu$ with respect to the $\mu$-th coordinate is given by: $$\partial_\mu x_\nu = \begin{cases} 1 & \text{if } \mu = \nu \\ 0 & \text{if } \mu \neq \nu \end{cases}$$

For any natural number $d$ representing the spatial dimensions and any coordinate index $\mu \in \{0\} \oplus \{1, \dots, d\}$ in spacetime, the partial derivative $\partial_\mu$ of the constant zero function $f: \text{SpaceTime}_d \to \mathbb{R}$ (defined by $f(x) = 0$ for all $x$) is itself the constant zero function.

Let $d$ be the number of spatial dimensions and $\text{SpaceTime}_d$ be the $(1+d)$-dimensional spacetime. Let $M$ be a real normed vector space. For a differentiable function $f: \text{SpaceTime}_d \to M$ and a Lorentz transformation $\Lambda$ from the Lorentz group $\mathcal{L}$, the partial derivative of the function $x \mapsto f(\Lambda x)$ with respect to the $\mu$-th coordinate at a point $x$ is given by: $$\partial_\mu (f(\Lambda x)) = \sum_{\nu} \Lambda_{\nu\mu} (\partial_\nu f)(\Lambda x)$$ where $\Lambda_{\nu\mu}$ denotes the component of the Lorentz transformation matrix at row $\nu$ and column $\mu$, and $\partial_\nu f$ is the partial derivative of $f$ along the $\nu$-th coordinate.

Let $d$ be the number of spatial dimensions and $\text{SpaceTime}_d$ be the $(1+d)$-dimensional spacetime. Let $M$ be a real normed vector space. For a differentiable function $f: \text{SpaceTime}_d \to M$, a Lorentz transformation $\Lambda$ from the Lorentz group, and a coordinate index $\mu$, the partial derivative of the Lorentz-transformed function $x \mapsto \Lambda \cdot f(\Lambda^{-1}x)$ at a point $x$ is given by: $$\partial_\mu (\Lambda \cdot f(\Lambda^{-1}x)) = \sum_{\nu} (\Lambda^{-1})_{\nu\mu} \Lambda \cdot (\partial_\nu f)(\Lambda^{-1}x)$$ where $(\Lambda^{-1})_{\nu\mu}$ denotes the component of the inverse Lorentz transformation matrix at row $\nu$ and column $\mu$, $\Lambda \cdot (\dots)$ denotes the action of the Lorentz group on $M$, and $\partial_\nu f$ is the partial derivative of $f$ along the $\nu$-th coordinate.

Let $d$ be a natural number and $M$ be a real normed space. For a given speed of light $c$, let $\Phi_c: \text{SpaceTime}_d \to \text{Time} \times \text{Space}_d$ be the isomorphism that decomposes a spacetime point into its temporal and spatial components. For a differentiable function $f: \text{SpaceTime}_d \to M$, a point $x \in \text{SpaceTime}_d$, and a spatial basis index $i \in \{0, \dots, d-1\}$, the partial derivative of $f$ along the $i$-th spatial coordinate of spacetime is equal to the spatial derivative of $f$ (holding time constant) at the spatial position of $x$: $$\partial_{\text{inr } i} f(x) = \text{deriv}_i \left( \mathbf{y} \mapsto f(\Phi_c^{-1}(t, \mathbf{y})) \right)(\mathbf{s})$$ where $(t, \mathbf{s}) = \Phi_c(x)$ are the time and space components of $x$, and $\text{deriv}_i$ denotes the spatial derivative in the direction of the $i$-th basis vector of $\text{Space}_d$.

Let $d$ be a natural number and $M$ be a real normed space. For a given speed of light $c$, let $\Phi_c: \text{SpaceTime}_d \to \text{Time} \times \text{Space}_d$ be the isomorphism that decomposes a spacetime point into its temporal and spatial components. For a differentiable function $f: \text{SpaceTime}_d \to M$ and a point $x \in \text{SpaceTime}_d$, let $(t, \mathbf{s}) = \Phi_c(x)$ be the time and space components of $x$ respectively. The partial derivative of $f$ along the temporal coordinate of spacetime at $x$ is equal to the reciprocal of the speed of light times the time derivative of $f$ (holding the spatial position constant) evaluated at $t$: $$\partial_{\text{inl } 0} f(x) = \frac{1}{c} \cdot \text{Time.deriv} \left( t' \mapsto f(\Phi_c^{-1}(t', \mathbf{s})) \right)(t)$$ where $\text{Time.deriv}$ denotes the derivative with respect to the temporal parameter.

For a given coordinate index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, the distributional derivative operator maps an $M$-valued distribution $f$ on $\text{SpaceTime } d$ to its partial derivative in the direction of the $\mu$-th basis vector $e_\mu$. This operator is a $\mathbb{R}$-linear map from the space of distributions $(\text{SpaceTime } d \to d[\mathbb{R}] M)$ to itself, defined by evaluating the distributional Fréchet derivative $Df$ at the basis vector $e_\mu$. For a test function $\eta$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the resulting distribution satisfies $(\partial_\mu f)(\eta) = (Df)(\eta)(e_\mu)$.

Let $M$ be a real normed vector space and $d$ be the spatial dimension. For any distribution $f \colon \text{SpaceTime } d \to d[\mathbb{R}] M$, coordinate index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, and test function $\varepsilon \in \mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the distributional partial derivative $\partial_\mu f$ applied to $\varepsilon$ is equal to the distributional Fréchet derivative $Df$ applied to $\varepsilon$ and evaluated at the $\mu$-th basis vector $e_\mu$: $$(\partial_\mu f)(\varepsilon) = (Df)(\varepsilon)(e_\mu)$$

Let $M$ be a real normed vector space and $d$ be the spatial dimension. For any distribution $f$ on $\text{SpaceTime } d$ valued in $M$, coordinate index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, and test function $\varepsilon \in \mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the evaluation of the distributional partial derivative $\partial_\mu f$ at $\varepsilon$ satisfies: $$(\partial_\mu f)(\varepsilon) = -f(\partial_\mu \varepsilon)$$ where $\partial_\mu \varepsilon$ denotes the partial derivative of the test function $\varepsilon$ in the direction of the $\mu$-th basis vector $e_\mu$.

Let $M$ be a real normed vector space and $d$ be the spatial dimension. For any $M$-valued distribution $f$ on $\text{SpaceTime } d$, any coordinate index $\mu \in \text{Fin } 1 \oplus \text{Fin } d$, and any test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, applying the distribution $f$ to the partial derivative of the test function $\partial_\mu \varepsilon$ is equal to the negative of the distributional partial derivative $\partial_\mu f$ applied to the test function $\varepsilon$: \[ f(\partial_\mu \varepsilon) = -(\partial_\mu f)(\varepsilon) \] where $\partial_\mu \varepsilon$ denotes the derivative of $\varepsilon$ in the direction of the $\mu$-th basis vector $e_\mu$.

Let $d$ be the spatial dimension and $M$ be a real normed vector space. For any $M$-valued distribution $f$ on $\text{SpaceTime } d$ and any coordinate indices $\mu, \nu \in \text{Fin } 1 \oplus \text{Fin } d$, the distributional partial derivatives commute: \[ \partial_\mu (\partial_\nu f) = \partial_\nu (\partial_\mu f) \] where $\partial_\mu$ denotes the distributional partial derivative operator `distDeriv μ` along the $\mu$-th coordinate direction.

Let $d$ be the spatial dimension and $M$ be a real normed vector space. For any $M$-valued distribution $f$ on $\text{SpaceTime}_d$ and any Lorentz transformation $\Lambda$ in the Lorentz group $\mathcal{L}$, the distributional partial derivative of the Lorentz-transformed distribution $\Lambda \cdot f$ with respect to the $\mu$-th coordinate is given by: \[ \partial_\mu (\Lambda \cdot f) = \sum_\nu (\Lambda^{-1})_{\nu\mu} (\Lambda \cdot \partial_\nu f) \] where $\partial_\mu$ and $\partial_\nu$ are distributional partial derivatives along the respective coordinate indices, and $(\Lambda^{-1})_{\nu\mu}$ is the component of the inverse Lorentz transformation matrix at row $\nu$ and column $\mu$.

Given a function $f : \text{SpaceTime}_d \to M$ from a $(1+d)$-dimensional spacetime to a real topological vector space $M$, the tensor derivative is a function mapping each spacetime point $x$ to an element of the tensor product $\text{CoVector}_d \otimes_{\mathbb{R}} M$. It is defined by the sum: \[ \text{tensorDeriv}(f)(x) = \sum_{\mu} e_\mu \otimes \partial_\mu f(x) \] where $\{e_\mu\}$ is the standard basis of the space of Lorentz covectors $\text{CoVector}_d$, and $\partial_\mu f(x)$ is the partial derivative of $f$ at the point $x$ along the $\mu$-th coordinate.

Let $d$ be the number of spatial dimensions and $\text{SpaceTime}_d$ be the $(1+d)$-dimensional spacetime. Let $M$ be a real normed vector space equipped with an action of the Lorentz group $\mathcal{L}$. For a differentiable function $f : \text{SpaceTime}_d \to M$ and a Lorentz transformation $\Lambda \in \mathcal{L}$, the tensor derivative of the Lorentz-transformed function $x \mapsto \Lambda \cdot f(\Lambda^{-1}x)$ at a point $x \in \text{SpaceTime}_d$ is given by: \[ \text{tensorDeriv} (y \mapsto \Lambda \cdot f(\Lambda^{-1} y))(x) = \Lambda \cdot \text{tensorDeriv}(f)(\Lambda^{-1} x) \] where $\text{tensorDeriv}(f)$ is the tensor-valued function mapping $x$ to $\sum_{\mu} e_\mu \otimes \partial_\mu f(x) \in \text{CoVector}_d \otimes_{\mathbb{R}} M$, and the action of $\Lambda$ on the right-hand side is the induced action on the tensor product space.

Let $d$ be the number of spatial dimensions and $M$ be a real vector space equipped with a tensorial structure for an index configuration $c$. Let $f : \text{SpaceTime}_d \to M$ be a differentiable function and $x \in \text{SpaceTime}_d$ be a point. For any multi-index $b$ belonging to the combined index set of $\text{CoVector}_d \otimes M$, the component of the tensor derivative at $x$ corresponding to $b$ is equal to the partial derivative of the $b_2$-th component of $f$ along the $\mu$-th coordinate: \[ [ \text{toTensor}(\text{tensorDeriv } f(x)) ]_b = \partial_\mu \left( y \mapsto [ \text{toTensor}(f(y)) ]_{b_2} \right)(x) \] where the multi-index $b$ is decomposed into a covector index $b_1$ and a tensor multi-index $b_2$ via the canonical isomorphism, and $\mu \in \{0, \dots, d\}$ is the spacetime coordinate index corresponding to $b_1$.

Let $d$ be the number of spatial dimensions and $M$ be a finite-dimensional real inner product space (typically representing a tensor space). The **distributional tensor derivative** is an $\mathbb{R}$-linear map that sends an $M$-valued distribution $f$ on $\text{SpaceTime } d$ to a distribution valued in the tensor product space $\text{CoVector}(d) \otimes_{\mathbb{R}} M$. For an $M$-valued distribution $f$ and a test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the value of the tensor derivative is given by the sum over spacetime indices: $$\text{distTensorDeriv}(f)(\varepsilon) = \sum_{\mu} e_\mu \otimes (\partial_\mu f)(\varepsilon)$$ where $\{e_\mu\}$ is the standard basis for the space of Lorentz covectors $\text{CoVector}(d)$, and $\partial_\mu f$ is the distributional partial derivative of $f$ in the direction of the $\mu$-th coordinate.

Let $M$ be a finite-dimensional real inner product space and $d$ be the number of spatial dimensions. For an $M$-valued distribution $f$ on $\text{SpaceTime } d$ and a test function $\varepsilon$ in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$, the value of the distributional tensor derivative $\text{distTensorDeriv}(f)$ applied to $\varepsilon$ is given by the sum: $$\text{distTensorDeriv}(f)(\varepsilon) = \sum_{\mu} e_\mu \otimes (\partial_\mu f)(\varepsilon)$$ where $\{e_\mu\}$ is the standard basis for the space of Lorentz covectors $\text{CoVector}(d)$, and $\partial_\mu f$ denotes the distributional partial derivative of $f$ in the direction of the $\mu$-th coordinate index.

Let $d$ be the number of spatial dimensions and $M$ be a finite-dimensional real inner product space that carries a tensorial representation of the Lorentz group $\mathcal{L}$ with index configuration $c$. For any $M$-valued distribution $f$ on $\text{SpaceTime}_d$ and any Lorentz transformation $\Lambda \in \mathcal{L}$, the distributional tensor derivative $\nabla f$ (which is a distribution valued in the tensor product space $\text{CoVector}(d) \otimes M$) satisfies the property: \[ \nabla (\Lambda \cdot f) = \Lambda \cdot (\nabla f) \] where $\Lambda \cdot f$ denotes the action of the Lorentz transformation on the distribution, and the action on the right-hand side is the induced Lorentz action on the tensor-valued distribution.

Let $d$ be the number of spatial dimensions and $M$ be a finite-dimensional real inner product space equipped with a "tensorial" structure corresponding to a sequence of index colors $c$. Let $f$ be an $M$-valued distribution on $\text{SpaceTime } d$ and $\varepsilon$ be a test function in the Schwartz space $\mathcal{S}(\text{SpaceTime } d, \mathbb{R})$. The distributional tensor derivative $\nabla f$ is a distribution valued in the space of tensors with index configuration $\text{down} \mathbin{+\!+} c$ (where $\text{down}$ represents the Lorentz covector index). For any multi-index $b$ in the component index set of this tensor space, let $b \cong (\mu, b_2)$ be the canonical decomposition where $\mu \in \{0, \dots, d\}$ is the spacetime index associated with the covector part and $b_2$ is the multi-index associated with $M$. The theorem states that the $b$-th component of the tensor derivative evaluated at $\varepsilon$ is equal to the $b_2$-th component of the partial distributional derivative $\partial_\mu f$ evaluated at $\varepsilon$: $$[\nabla f(\varepsilon)]_{(\mu, b_2)} = [(\partial_\mu f)(\varepsilon)]_{b_2}$$ where the components are taken with respect to the canonical bases of the respective tensor spaces.