PhyslibAlpha.ClassicalFieldTheory.Local.TotalDivergenceEquivalence

For two local Lagrangians $\mathcal{L}_{\text{source}}$ and $\mathcal{L}_{\text{target}}$ of order $k+1$ and a packaged total divergence $T$ of current order $k$, the proposition states that for every field $f: \mathbb{R}^d \to \mathbb{R}^m$, the action density of $\mathcal{L}_{\text{target}}$ is equal to the sum of the action density of $\mathcal{L}_{\text{source}}$ and the action density of the Lagrangian $\mathcal{L}_T$ associated with the total divergence $T$. In terms of the $k+1$-jet $j^{k+1} f(x)$ of the field at point $x$: $$\mathcal{L}_{\text{target}}(j^{k+1} f(x)) = \mathcal{L}_{\text{source}}(j^{k+1} f(x)) + \mathcal{L}_T(j^{k+1} f(x))$$

Two local Lagrangians $L_{\text{source}}$ and $L_{\text{target}}$ of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$ are **Euler-Lagrange equivalent** if they determine the same local Euler-Lagrange operator $\mathcal{E}(L)$ for every field $f$: $$\mathcal{E}(L_{\text{target}})[f] = \mathcal{E}(L_{\text{source}})[f]$$ where $\mathcal{E}(L)[f]$ is the function mapping $x \in \mathbb{R}^d$ to the vector of variational derivatives of the Lagrangian density.

For any local Lagrangian $L$ of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$, $L$ is Euler-Lagrange equivalent to itself. That is, for every field $f$, the local Euler-Lagrange operator satisfies $\mathcal{E}(L)[f] = \mathcal{E}(L)[f]$.

Let $L_1$ and $L_2$ be two local Lagrangians of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$. If $L_1$ and $L_2$ are Euler-Lagrange equivalent, then $L_2$ and $L_1$ are also Euler-Lagrange equivalent.

Let $L_1, L_2,$ and $L_3$ be local Lagrangians of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$. If $L_1$ is Euler-Lagrange equivalent to $L_2$, and $L_2$ is Euler-Lagrange equivalent to $L_3$, then $L_1$ is Euler-Lagrange equivalent to $L_3$. Two local Lagrangians are considered Euler-Lagrange equivalent if they determine the same local Euler-Lagrange operator $\mathcal{E}(L)$, such that $\mathcal{E}(L_i)[f] = \mathcal{E}(L_j)[f]$ for every field $f$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be two local Lagrangians of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$. If $L_{\text{source}}$ and $L_{\text{target}}$ are Euler-Lagrange equivalent, then for any field $f$, the resulting local Euler-Lagrange operators are equal: $$\mathcal{E}(L_{\text{target}})[f] = \mathcal{E}(L_{\text{source}})[f]$$ where $\mathcal{E}(L)[f]$ denotes the local Euler-Lagrange operator applied to the field $f$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be local Lagrangians of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$. If $L_{\text{source}}$ and $L_{\text{target}}$ are Euler-Lagrange equivalent, then for any field $f$, the Euler-Lagrange equations for the target Lagrangian vanish if and only if the Euler-Lagrange equations for the source Lagrangian vanish: $$\mathcal{E}(L_{\text{target}})[f] = 0 \iff \mathcal{E}(L_{\text{source}})[f] = 0$$ where $\mathcal{E}(L)[f]$ denotes the local Euler-Lagrange operator applied to the field $f$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be two local $k$-th order Lagrangians that are Euler-Lagrange equivalent, meaning they determine the same Euler-Lagrange operator $\mathcal{E}(L_{\text{source}}) = \mathcal{E}(L_{\text{target}})$. Suppose both Lagrangians are smooth in their jet coordinates and a field $f: \mathbb{R}^d \to \mathbb{R}^m$ is admissible for the action functional for both $L_{\text{source}}$ and $L_{\text{target}}$ (i.e., $f$ is $C^\infty$ and has finite action). Then $f$ is a critical field for the action of $L_{\text{target}}$ if and only if it is a critical field for the action of $L_{\text{source}}$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be two $k$-th order local Lagrangians that are Euler-Lagrange equivalent, meaning they determine the same Euler-Lagrange operator $\mathcal{E}(L_{\text{source}}) = \mathcal{E}(L_{\text{target}})$. Suppose both Lagrangians are smooth in their jet coordinates. For any field $f: \mathbb{R}^d \to \mathbb{R}^m$ that is admissible for the action functionals of both Lagrangians, if $f$ is a critical point for the action of $L_{\text{source}}$, then $f$ is also a critical point for the action of $L_{\text{target}}$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be two local Lagrangians of order $k$ for fields $f \colon \mathbb{R}^d \to \mathbb{R}^m$. Suppose $L_{\text{source}}$ and $L_{\text{target}}$ are Euler-Lagrange equivalent, such that their Euler-Lagrange operators satisfy $\mathcal{E}(L_{\text{source}})[f] = \mathcal{E}(L_{\text{target}})[f]$ for every field $f$. Furthermore, assume that $L_{\text{source}}$ and $L_{\text{target}}$ are smooth in their jet coordinates, and the field $f$ is admissible for the action functionals of both Lagrangians. If $f$ is a critical field for the action functional of $L_{\text{target}}$, then $f$ is also a critical field for the action functional of $L_{\text{source}}$.

Let $E$ be a total divergence equivalence between a source Lagrangian $L_{\text{source}}$ and a target Lagrangian $L_{\text{target}}$, where $L_{\text{div}}$ denotes the Lagrangian representing the total divergence. For any field $f: \mathbb{R}^d \to \mathbb{R}^m$, the action density of the target Lagrangian is the sum of the action density of the source Lagrangian and the action density of the divergence term. That is, for every point $x \in \mathbb{R}^d$: $$\mathcal{L}_{\text{target}}[f](x) = \mathcal{L}_{\text{source}}[f](x) + \mathcal{L}_{\text{div}}[f](x)$$ where $\mathcal{L}[f](x)$ is the action density defined by $L(j^k f(x))$, and $j^k f(x)$ is the $k$-jet of the field $f$ at $x$.

Let $E$ be a total divergence equivalence between a source Lagrangian $L_{\text{source}}$ and a target Lagrangian $L_{\text{target}}$ of dimension $d$, field dimension $m$, and jet order $k$. For any field $f: \mathbb{R}^d \to \mathbb{R}^m$, the local Euler-Lagrange operators of the target and source Lagrangians are equal: $$\mathcal{E}(L_{\text{target}})[f] = \mathcal{E}(L_{\text{source}})[f]$$ where $\mathcal{E}(L)[f]$ denotes the local Euler-Lagrange operator associated with Lagrangian $L$ evaluated on the field $f$.

Let $E$ be a total divergence equivalence between a source Lagrangian $L_{\text{source}}$ and a target Lagrangian $L_{\text{target}}$ of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$. For any field $f$, the Euler-Lagrange equations for the target Lagrangian are satisfied if and only if the Euler-Lagrange equations for the source Lagrangian are satisfied. That is, $$\mathcal{E}(L_{\text{target}})[f] = 0 \iff \mathcal{E}(L_{\text{source}})[f] = 0$$ where $\mathcal{E}(L)$ is the local Euler-Lagrange operator defined such that its $a$-th component at point $x$ is: $$(\mathcal{E}(L)[f](x))_a = \sum_{|I| \le k} (-1)^{|I|} D_I \left( \frac{\partial L}{\partial u^a_I} \right) [f](x)$$

Let $L_{\text{source}}$ and $L_{\text{target}}$ be local Lagrangians of order $k$ for fields $f: \mathbb{R}^d \to \mathbb{R}^m$ that differ by a total divergence. Suppose that the field $f$ is admissible for the action functional of both Lagrangians (meaning $f$ is $C^\infty$ and the action densities $x \mapsto L(j^k f(x))$ are integrable) and that both Lagrangians are smooth in their jet coordinates. Then $f$ is a critical point for the action of $L_{\text{target}}$ if and only if it is a critical point for the action of $L_{\text{source}}$: $$\text{IsCritical}(L_{\text{target}}, f) \iff \text{IsCritical}(L_{\text{source}}, f)$$ where $\text{IsCritical}(L, f)$ denotes that the first variation of the action $S(L, f) = \int L(j^k f(x)) dx$ vanishes at $f$.

Let $L_{\text{source}}$ and $L_{\text{target}}$ be two local Lagrangians of order $k$ that differ by a total divergence. Let $f: \mathbb{R}^d \to \mathbb{R}^m$ be a field such that $f$ is admissible for the action functionals of both $L_{\text{source}}$ and $L_{\text{target}}$ (i.e., $f$ is smooth and the action densities are integrable). Assuming that both $L_{\text{source}}$ and $L_{\text{target}}$ are smooth in their jet coordinates, if $f$ is a critical point for the action functional of $L_{\text{source}}$, then $f$ is also a critical point for the action functional of $L_{\text{target}}$.

Let $E$ be a total divergence equivalence between two local Lagrangians $L_{\text{source}}$ and $L_{\text{target}}$ of order $k$. Let $f : \mathbb{R}^d \to \mathbb{R}^m$ be a field. Suppose that $(L_{\text{source}}, f)$ and $(L_{\text{target}}, f)$ are both admissible for the action functional (meaning $f$ is smooth and the resulting action densities are integrable) and that both $L_{\text{source}}$ and $L_{\text{target}}$ are smooth in their respective jet coordinates. If $f$ is a critical point for the action associated with $L_{\text{target}}$, then $f$ is also a critical point for the action associated with $L_{\text{source}}$.