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Adding a total time derivative and rescaling a Lagrangian: the Euler-Lagrange expression and the stationary curves

Statement

Let $(O, L)$ be a time-dependent Lagrangian chart over $[a, b] \times U$ (A time-dependent Lagrangian chart, its action functional and its fixed-endpoint first variations), let $A \subseteq R \times R^{n}$ be open and contain the time-configuration projection of $O$ , and let $F : A \to R$ be of class $C^{3}$ , and define

$\tilde{L} (t, q, v) : = L (t, q, v) + F_{t} (t, q) + F_{q} (t, q) \cdot v .$

Then:

$\tilde{L}$ is $C^{2}$ on $O$ , and for every $C^{2}$ curve $q : [a, b] \to U$ ,

$E_{\tilde{L}} (q) (t) = E_{L} (q) (t) (t \in [a, b]);$

for every such curve, $\tilde{S} (q) = S (q) + F (b, q (b)) - F (a, q (a))$ , where $S$ and $\tilde{S}$ are the actions of $L$ and $\tilde{L}$ ;
consequently $q$ is stationary for $\tilde{S}$ if and only if $q$ is stationary for $S$ ;
for every real $c \neq 0$ , $E_{c L} (q) = c E_{L} (q)$ and hence the stationary curves of $c L$ and of $L$ coincide; for $c = 0$ one has $E_{0} = 0$ , so every $C^{2}$ curve is stationary for the zero Lagrangian, and setting $c = 0$ does not preserve the stationary set unless every $C^{2}$ curve was already stationary for $L$ .

Facts & Assumptions

Given: The chart data $(O, L)$ , a $C^{3}$ function $F$ on $A$ covering the chart projection, the modified Lagrangian $\tilde{L}$ , a real $c$ and a $C^{2}$ curve $q : [a, b] \to U$ .

[F1]

The chain rule and the equality of mixed partial derivatives (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Clairaut--Schwarz theorem for continuous second partial derivatives, $C^{k}$ Euclidean maps and diffeomorphisms, $C^{k}$ maps and multi-index derivative notation in Euclidean space): the functions $F_{t}, F_{q}$ are $C^{2}$ ; their mixed second derivatives commute, so $\partial_{q_{i}} (F_{t}) = \partial_{t} (F_{q_{i}})$ and $\partial_{q_{i}} (F_{q_{j}}) = \partial_{q_{j}} (F_{q_{i}})$ on the domain.

[F2]

Along a $C^{2}$ curve $q$ , the function $t \mapsto F (t, q (t))$ is $C^{2}$ with derivative $F_{t} (t, q (t)) + F_{q} (t, q (t)) \cdot q^{'} (t)$ and second derivative $F_{t t} + 2 \sum_{i} F_{t q_{i}} q_{i}^{'} + \sum_{i, j} F_{q_{i} q_{j}} q_{i}^{'} q_{j}^{'} + F_{q} \cdot q^{''}$ ; this is the chain rule applied twice (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[F3]

If a $C^{1}$ function on an interval has derivative $G$ , then its increment is the integral of $G$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ); integrals are linear (Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ , The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

Proof technique: direct; compute the Euler–Lagrange expression of the added total derivative and integrate it exactly.

1.1F1F2algebra

Put $G (t, q, v) : = F_{t} (t, q) + F_{q} (t, q) \cdot v$ , so $\tilde{L} = L + G$ . Since $A$ covers the chart projection and $F$ is $C^{3}$ , $G$ and $\tilde{L}$ are defined and $C^{2}$ throughout $O$ . By [F1], $G_{q} (t, q, v) = F_{t q} + F_{q q} [v], G_{v} (t, q, v) = F_{q} (t, q),$ where the first display abbreviates the vector with entries $\partial_{q_{i}} F_{t} + \sum_{j} \partial_{q_{i}} F_{q_{j}} v_{j}$ . Hence for any $C^{2}$ curve $q$ , $\frac{d}{d t} G_{v} (t, q, q^{'}) = \frac{d}{d t} F_{q} (t, q) = F_{q t} + F_{q q} [q^{'}] = F_{t q} + F_{q q} [q^{'}] = G_{q} (t, q, q^{'}),$ using [F1] in the penultimate equality. Therefore $E_{G} (q) = G_{q} - \frac{d}{d t} G_{v} = 0$ along every $C^{2}$ curve.

1.2F2F3algebra

Clause 2: by [F2] and [F3], $\tilde{S} (q) - S (q) = \int_{a}^{b} \frac{d}{d t} F (t, q (t)) d t = F (b, q (b)) - F (a, q (a)),$ which is the displayed identity.

2.1step 1.1algebra

The Euler–Lagrange expression is additive and homogeneous in the Lagrangian: from the definition, $E_{L + c G} (q) = E_{L} (q) + c E_{G} (q)$ for every real $c$ , because $L + c G$ has gradients $L_{q} + c G_{q}$ and $L_{v} + c G_{v}$ and the derivative is linear. By step 1.1, $E_{\tilde{L}} (q) = E_{L} (q)$ ; this proves clause 1 and the identity in clause 4. For $c \neq 0$ , $E_{c L} = c E_{L}$ vanishes exactly when $E_{L}$ does, so the stationarity theorem gives the same stationary curves.

3.1step 2.1step 1.2given

Clause 3: stationary curves are characterised by the vanishing of the Euler–Lagrange expression (Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression). By clause 1 the two expressions agree, so the stationary sets of $S$ and $\tilde{S}$ coincide; equivalently, the boundary term of clause 2 is unchanged by the variations, whose endpoint values are zero by definition of admissible variation.

4.1step 2.1given∎

Clause 4, degenerate case: for $c = 0$ the expression $E_{0}$ vanishes identically, so by Stationarity of the time-dependent action is equivalent to the vanishing of the Euler-Lagrange expression every $C^{2}$ curve is stationary for the zero Lagrangian; if $L$ has a non-stationary curve, the stationary sets of $L$ and $0 \cdot L$ differ.

Remarks

Clause 2 is the exact statement of the common phrase that two Lagrangians differing by a total time derivative give the same equations of motion: the actions differ by endpoint values, which are held fixed in the variational problem, and the Euler–Lagrange expressions agree pointwise.
The rescaling clause is stated for the stationary set only. Multiplying the action by $c \neq 0$ also rescales the first variation and the Euler–Lagrange expression, but it does not change which curves are stationary; the case $c = 0$ is genuinely exceptional and is recorded because the zero Lagrangian is the degenerate witness used in the companion item on regular and hyperregular Lagrangians and in Scope of the Legendre map: regular versus hyperregular, the meaning of the covector datum, and the limits of the branch description.

Adding a total time derivative and rescaling a Lagrangian: the Euler-Lagrange expression and the stationary curves

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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