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.