Total-derivative Lagrangians, the potential representative and scaling: invariant stationary curves with endpoint terms

Adopt the force model with potential presentation $(U,d)$ and mechanical Lagrangian $L=K-U$ of The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, with masses $m_1,\dots ,m_N>0$, open configuration domain $Q\subseteq {\mathbb{R}}^{3N}$, time window $I_0$, mass operator $\mathsf{M}$ and $C^2$ potential $U:I_0\times Q\to \mathbb{R}$ (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets). Let $a<b$ with $[a,b]\subseteq I_0$, let $F:I_0\times Q\to \mathbb{R}$ be of class $C^3$ with dimension $[F]=[S]=\mu {\mathrm{\ell }}^2{\tau }^{-1}$, and define

$L^{\mathrm{\prime }}(t,x,v):=L(t,x,v)+F_t(t,x)+F_x(t,x)\cdot v\mathrm{.}$

Then:

1. $L^{\mathrm{\prime }}$ is $C^2$ on $I_0\times Q\times {\mathbb{R}}^{3N}$, has the same dimension as $L$, and for every $C^2$ admissible curve $x:[a,b]\to Q$ one has $E_{L^{\mathrm{\prime }}}(x)(t)=E_L(x)(t)$ for all $t\in [a,b]$ and

$S^{\mathrm{\prime }}(x)=S(x)+F(b,x(b))-F(a,x(a)),$

where $S^{\mathrm{\prime }}$ is the action of $L^{\mathrm{\prime }}$; consequently $x$ is stationary for $S^{\mathrm{\prime }}$ if and only if $x$ is stationary for $S$, and, by The stationary-action criterion and the potential-part Euler-Lagrange equation, both are equivalent to the model's potential-part equation $\mathsf{M}{x}^{\mathrm{\prime }\mathrm{\prime }}=-{\mathrm{\nabla }}_{x}U$ along $x$.

2. Potential representative. Let $c:I_0\to \mathbb{R}$ be of class $C^2$ with dimension $[c]=[U]=\mu {\mathrm{\ell }}^2{\tau }^{-2}$ and antiderivative $C$ (i.e. $C^{\mathrm{\prime }}=c$). Then $\tilde{U}:=U+c$ remains of class $C^2$ and is the representative of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (c), with Lagrangian $\tilde{L}=K-\tilde{U}$. Then $\tilde{L}=L+M$ where $M(t,x,v)=F_t+F_x\cdot v$ for $F(t,x):=-C(t)$, so clause 1 applies: the actions of $L$ and $\tilde{L}$ differ by $C(a)-C(b)$, independently of the curve, and the stationary curves and the model's equation are unchanged.

3. Scaling. For every dimensionless real $\lambda \mathrm{\ne }0$ one has $E_{\lambda L}(x)=\lambda E_L(x)$ pointwise, so the stationary curves of the Lagrangian $\lambda L$ coincide with those of $L$; the value $\lambda =0$ is excluded because the zero Lagrangian has every curve stationary.

The statement is conditional on the adopted force branch and presentation; it asserts no minimisation, no conservation law and no empirical content, and it does not remove the need to declare which representative and normalisation are used by an energy statement.

• The endpoint term is not a physical energy. The difference $F(b,x(b))-F(a,x(a))$ in clause 1 is an endpoint value of the added function; it depends on the endpoints and on the representative chosen, and no energy balance is asserted from it.

• Normalisation is a declared convention. Because the potential is determined only up to an independently chosen $c_C(t)$ on each connected component $C$ of $Q$ and the Lagrangian only up to total derivatives, every energy or action statement of the framework records the representative used; the model's equation of motion is insensitive to both choices.