Physlib.ClassicalMechanics.Lagrangian.TotalDerivativeEquivalence

A function $\delta L: \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$ (representing a change in a Lagrangian) is a total time derivative if there exists a differentiable function $F: \mathbb{R}^n \times \mathbb{R} \to \mathbb{R}$, depending only on position $r$ and time $t$, such that for all $r, v, t$: $$\delta L(r, v, t) = \text{D}F|_{(r, t)}(v, 1)$$ By the chain rule, this condition is equivalent to: $$\delta L(r, v, t) = \frac{\partial F}{\partial t}(r, t) + \nabla_r F(r, t) \cdot v$$ where $v$ is the velocity vector.

Let $\delta L : \mathbb{R}^n \to \mathbb{R}$ be a function of velocity $v$ such that $\delta L(0) = 0$. Suppose that $\delta L$ is a total time derivative, meaning the function $(r, v, t) \mapsto \delta L(v)$ satisfies the condition that there exists a differentiable function $F(r, t)$ depending only on position and time such that: $$\delta L(v) = \frac{d}{dt} F(r, t) = \frac{\partial F}{\partial t}(r, t) + \nabla_r F(r, t) \cdot v$$ Then, $\delta L$ must be linear in velocity. That is, there exists a constant vector $g \in \mathbb{R}^n$ such that for all $v \in \mathbb{R}^n$: $$\delta L(v) = \langle g, v \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the standard Euclidean inner product. This reflects the physical requirement that for $\delta L$ to be independent of position $r$ and time $t$, the gradient $\nabla_r F$ must be a constant vector $g$ and the partial derivative $\frac{\partial F}{\partial t}$ must vanish.

If a function $\delta L(t, q, v)$ is a total time derivative, meaning there exists a differentiable function $F(q, t)$ such that $\delta L(t, q, v) = \frac{d}{dt} F(q, t)$, then its negation $-\delta L$ is also a total time derivative.

Let $\delta L(t, q, v)$ be a function representing a change in a Lagrangian, where $t \in \text{Time}$ and $q, v \in X$. If $\delta L$ is a total time derivative—meaning there exists a smooth function $F(q, t)$ such that $\delta L(t, q, v) = \frac{d}{dt} F(q, t)$—then $\delta L$ is infinitely differentiable ($C^\infty$).

Let $X$ be a complete space. Suppose $\delta L: \text{Time} \to X \to X \to \mathbb{R}$ is a total time derivative (meaning there exists a function $F(q, t)$ such that $\delta L(t, q, v) = \frac{d}{dt} F(q, t)$). For any infinitely differentiable path $q: \text{Time} \to X$, the variational gradient of the functional mapping $q'$ to the density $\delta L(t, q'(t), \partial_t q'(t))$ at the path $q$ is the zero function. That is, the functional derivative of the action corresponding to $\delta L$ is zero: $$\frac{\delta}{\delta q} \int \delta L(t, q(t), \dot{q}(t)) \, dt = 0$$

Let $X$ be a complete space. Let $L$ and $\delta L$ be Lagrangians mapping $\text{Time} \to X \to X \to \mathbb{R}$, where $L(t, q, \dot{q})$ and $\delta L(t, q, \dot{q})$ represent the Lagrangian densities. Suppose $\delta L$ is a total time derivative (i.e., there exists a function $F(q, t)$ such that $\delta L = \frac{d}{dt}F$). For any $C^\infty$ trajectory $q: \text{Time} \to X$, if the functional $S[q] = \int L(t, q(t), \dot{q}(t)) \, dt$ has a variational gradient $\text{grad}$ at $q$, then the functional $S'[q] = \int (L + \delta L)(t, q(t), \dot{q}(t)) \, dt$ also has the same variational gradient $\text{grad}$ at $q$.

Let $X$ be a complete space. Let $L, L' : \text{Time} \times X \times X \to \mathbb{R}$ be two Lagrangians such that their difference $L' - L$ is a total time derivative. For any infinitely differentiable trajectory $q : \text{Time} \to X$, the variational gradient of the action functional $S'[q] = \int L'(t, q(t), \dot{q}(t)) \, dt$ is equal to the variational gradient of the action functional $S[q] = \int L(t, q(t), \dot{q}(t)) \, dt$ at $q$: $$\frac{\delta S'}{\delta q} = \frac{\delta S}{\delta q}$$ where $\dot{q}(t)$ denotes the time derivative of $q$.

Let $X$ be a complete space and let $L, L' : \text{Time} \to X \to X \to \mathbb{R}$ be two Lagrangians. If $L' - L$ is a total time derivative (that is, there exists a function $F(q, t)$ such that $(L' - L)(t, q, v) = \frac{\partial F}{\partial t} + \nabla_q F \cdot v$), and assuming that either $L$ or $L'$ is infinitely differentiable ($C^\infty$) and the trajectory $q: \text{Time} \to X$ is infinitely differentiable, then the Euler-Lagrange operators for $L$ and $L'$ are identical when evaluated along $q$: $$\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} = \frac{\partial L'}{\partial q} - \frac{d}{dt} \frac{\partial L'}{\partial \dot{q}}$$