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A one-label model with a time-dependent potential: the time-translation identity fails and the mechanical energy changes at the rate of the explicit time dependence

Physical scope

A worked model check inside the framework: a one-label global potential model (d=0) in the three-dimensional force branch whose potential depends explicitly on time through a prescribed vector drive f(t), with force -k x + f(t); the time-translation identity of the pair holds exactly when the drive is constant, the mechanical energy changes at the rate -f'(t) . x along motions, and for the sinusoidally driven specialisation an explicit global motion is exhibited whose energy derivative at a stated time is nonzero, together with a numerical instance. All values are model values of the declared idealization; the drive is adopted external data, and the example reports no measurement.

Example

Work in the one-label force model of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N = 1$ , mass $m > 0$ , time window $I_{0} = R$ , configuration domain $Q = R^{3}$ , and let $f : R \to R^{3}$ be of class $C^{2}$ , of dimension $μ ℓ τ^{- 2}$ (newtons), be a prescribed external drive. Adopt the time-dependent potential

$U (t, x) : = \frac{k}{2} ∣ x ∣^{2} - f (t) \cdot x, k > 0 with [k] = μ τ^{- 2},$

of class $C^{2}$ and dimension $μ ℓ^{2} τ^{- 2}$ , and check that the presentation $(U, 0)$ of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets has force map $F (t, x, v) = - k x + f (t)$ of the standing regularity (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance clause (iii)), so that the model is a global potential model ( $d \equiv 0$ ) with mechanical energy

$E (t, x, v) = \frac{m}{2} ∣ v ∣^{2} + \frac{k}{2} ∣ x ∣^{2} - f (t) \cdot x .$

Verify:

For time translations take the family parameter to be measured in seconds; then the generator value $τ \equiv 1$ is dimensionless and the charge $- E$ has units of joules.

presentation and identity. $\nabla_{x} U (t, x) = k x - f (t)$ , so $- \nabla_{x} U = - k x + f (t) = F$ ; and $\partial_{t} U (t, x) = - f^{'} (t) \cdot x$ , so the time-translation datum $τ \equiv 1, ξ \equiv 0, B \equiv 0$ satisfies the model identity of Continuous symmetries of the mechanical Lagrangian of a potential presentation: symmetry data, the Noether identity, the charge and its dimensions if and only if $f^{'} \equiv 0$ on $R$ .
energy rate. Along every motion $x : J \to Q$ of the model,

$E^{'} (t) = \partial_{t} U (t, x (t)) = - f^{'} (t) \cdot x (t);$

in particular the mechanical energy is not conserved along any motion on which $t \mapsto f^{'} (t) \cdot x (t)$ does not vanish identically, and if $f^{'} (t_{0}) \cdot x_{0} \neq 0$ then the maximal motion through $(t_{0}, x_{0}, 0)$ has nonzero energy derivative at $t_{0}$ . 3. explicit sinusoidally driven motion. Specialise to $f (t) : = f_{0} \cos (Ω t) \hat{e}$ with $f_{0} > 0$ of force dimension, $Ω > 0$ of dimension $τ^{- 1}$ and a dimensionless unit vector $\hat{e} \in R^{3}$ , and assume $k \neq m Ω^{2}$ . Then

$x_{p} (t) : = \frac{f_{0} \cos (Ω t)}{k - m Ω^{2}} \hat{e} (t \in R)$

is a motion of the model on $R$ , and along it

$E (t) = \frac{f_{0}^{2} [m Ω^{2} - (k - m Ω^{2}) \cos^{2} (Ω t)]}{2 (k - m Ω^{2})^{2}}, E^{'} (t) = \frac{f_{0}^{2} Ω \sin (Ω t) \cos (Ω t)}{k - m Ω^{2}};$

at $t_{0} = π / (4 Ω)$ one has $E^{'} (t_{0}) = f_{0}^{2} Ω / (2 (k - m Ω^{2})) \neq 0$ . 4. numerical instance. For $m = 1 k g$ , $k = 1 k g s^{- 2}$ , $Ω = 1 / 2 s^{- 1}$ , $f_{0} = 1 N$ and $\hat{e} = e_{1}$ one has $k - m Ω^{2} = 3 / 4$ and $x_{p} (t) = \frac{4}{3} \cos (t / 2) e_{1}$ ; at $t = π / 2 s$ the values are $x_{p} = \frac{2 \sqrt{2}}{3} m$ , $x_{p}^{'} = - \frac{\sqrt{2}}{3} m s^{- 1}$ , $E = - \frac{1}{9} J$ and $E^{'} = \frac{1}{3} J s^{- 1}$ .

All values are model values of the declared idealization; the drive is adopted external data and no measurement is reported.

Facts & Assumptions

Given: The model, the time-dependent potential $U$ , the drive $f$ , the sinusoidal specialisation, the curve $x_{p}$ and the numerical instance.

[F1]

Model and presentation (Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance, Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets, The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions, Configuration domain of $N$ labeled particles: open admissible sets, regular holonomic constraints and the mass operator, Dimensions and SI unit conventions of the non-relativistic classical mechanics framework): $U$ is $C^{2}$ with $\nabla_{x} U = k x - f (t)$ and $\partial_{t} U = - f^{'} (t) \cdot x$ ; $F = - \nabla_{x} U$ has the standing regularity; the model is a global potential model with mechanical energy $E = \frac{m}{2} ∣ v ∣^{2} + U (t, x)$ ; motions satisfy $m x^{''} = F (t, x, x^{'}) = - k x + f (t)$ (The associated first-order system, motions, maximal motions and the Newtonian flow of a force model).

[F2]

Energy balance and the time-translation statement (Mechanical energy balance for a potential presentation: $E^{'} = \partial_{t} U$ plus the residual power, conservation for a static potential with no residual forces, and the integrated work form clause 1, Time translation and the mechanical energy of the model: conservation in the autonomous global potential branch and the sharp failure for a time-dependent potential): along a motion of the presentation the mechanical energy is differentiable with $E^{'} = \partial_{t} U + \sum_{i} ⟨ d_{i}, x_{i}^{'} ⟩$ , which with $d \equiv 0$ is $E^{'} = \partial_{t} U$ ; the time-translation datum satisfies the model identity exactly when $\partial_{t} U \equiv 0$ , and then the energy is conserved, while a state with $\partial_{t} U (t_{0}, x_{0}) \neq 0$ produces a maximal motion with nonzero energy derivative there.

[F3]

Calculus (The derivatives of sine and cosine are cosine and minus sine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , $C^{k}$ Euclidean maps and diffeomorphisms, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): the derivatives of sine and cosine with the chain rule give $f^{'} (t) = - f_{0} Ω \sin (Ω t) \hat{e}$ and the derivatives of $x_{p}$ ; the Euclidean inner product is bilinear and $∣ \hat{e} ∣ = 1$ .

Verification

technique · direct; verify the presentation and the identity condition, differentiate the energy along the explicit motion and evaluate the numerical instance

1.1F1F2F3algebra

Clause 1: differentiating $U (t, x) = \frac{k}{2} ∣ x ∣^{2} - f (t) \cdot x$ in the configuration slot gives $\nabla_{x} U (t, x) = k x - f (t)$ , hence $F = - \nabla_{x} U = - k x + f (t)$ , which is continuous and locally Lipschitz in $(x, v)$ locally uniformly in $t$ because $f$ is continuous and the map is affine in the state; differentiating in $t$ gives $\partial_{t} U (t, x) = - f^{'} (t) \cdot x$ , which vanishes identically exactly when $f^{'} \equiv 0$ ; by [F2] the time-translation datum satisfies the model identity exactly then.

1.2F2algebra

Clause 2: along a motion, $d \equiv 0$ and [F2] gives $E^{'} (t) = \partial_{t} U (t, x (t)) = - f^{'} (t) \cdot x (t)$ ; if the right-hand side does not vanish identically, then $E$ is not constant on $J$ ; and if $f^{'} (t_{0}) \cdot x_{0} \neq 0$ , the maximal motion through $(t_{0}, x_{0}, 0)$ exists with $E^{'} (t_{0}) = - f^{'} (t_{0}) \cdot x_{0} \neq 0$ .

1.3F1F3algebra

Clause 3, the curve is a motion: with $f (t) = f_{0} \cos (Ω t) \hat{e}$ one has $x_{p}^{''} (t) = - Ω^{2} x_{p} (t)$ by [F3], so $m x_{p}^{''} = - m Ω^{2} x_{p}$ , while $- k x_{p} (t) + f (t) = f_{0} \cos (Ω t) (1 - \frac{k}{k - m Ω^{2}}) \hat{e} = \frac{f_{0} \cos (Ω t)}{k - m Ω^{2}} (k - m Ω^{2} - k) \hat{e} = - m Ω^{2} x_{p} (t),$ so the equation $m x_{p}^{''} = - k x_{p} + f$ holds on $R$ .

2.1F2F3step 1.3algebra

Clause 3, the energy values: by [F2] and step 1.3, $E^{'} = - f^{'} (t) \cdot x_{p} (t) = f_{0} Ω \sin (Ω t) (x_{p} (t) \cdot \hat{e}) = f_{0}^{2} Ω \sin (Ω t) \cos (Ω t) / (k - m Ω^{2})$ , since $x_{p} (t) \cdot \hat{e} = f_{0} \cos (Ω t) / (k - m Ω^{2})$ ; substituting the displayed curve into $E = \frac{m}{2} ∣ x_{p}^{'} ∣^{2} + \frac{k}{2} ∣ x_{p} ∣^{2} - f \cdot x_{p}$ gives $E (t) = \frac{m f_{0}^{2} Ω^{2} \sin^{2} (Ω t) + k f_{0}^{2} \cos^{2} (Ω t)}{2 (k - m Ω^{2})^{2}} - \frac{f_{0}^{2} \cos^{2} (Ω t)}{k - m Ω^{2}} = \frac{f_{0}^{2} [m Ω^{2} - (k - m Ω^{2}) \cos^{2} (Ω t)]}{2 (k - m Ω^{2})^{2}},$ where the second form uses $\sin^{2} + \cos^{2} = 1$ ; differentiating this expression with [F3] returns the displayed derivative, and at $t_{0} = π / (4 Ω)$ it equals $f_{0}^{2} Ω / (2 (k - m Ω^{2})) \neq 0$ .

3.1step 2.1F3algebra

Clause 4, numerical instance: with $m = 1$ , $k = 1$ , $Ω = 1 / 2$ , $f_{0} = 1$ and $\hat{e} = e_{1}$ one has $k - m Ω^{2} = 3 / 4$ and $x_{p} (t) = \frac{4}{3} \cos (t / 2) e_{1}$ , so at $t = π / 2$ : $x_{p} = \frac{4}{3} \cos (π / 4) e_{1} = \frac{2 \sqrt{2}}{3} e_{1}$ , $x_{p}^{'} = - \frac{2}{3} \sin (π / 4) e_{1} = - \frac{\sqrt{2}}{3} e_{1}$ , $f (t) \cdot x_{p} = \cos (π / 4) \cdot \frac{2 \sqrt{2}}{3} = \frac{2}{3}$ , $∣ x_{p}^{'} ∣^{2} = \frac{2}{9}$ , $∣ x_{p} ∣^{2} = \frac{8}{9}$ ; hence $E = \frac{1}{2} \cdot \frac{2}{9} + \frac{1}{2} \cdot \frac{8}{9} - \frac{2}{3} = \frac{1}{9} + \frac{4}{9} - \frac{2}{3} = - \frac{1}{9} J$ , and $E^{'} = f_{0}^{2} Ω \sin (Ω t) \cos (Ω t) / (k - m Ω^{2}) = \frac{1}{2} \cdot \frac{1}{2} / (\frac{3}{4}) = \frac{1}{3} J s^{- 1} \neq 0$ .

4.1step 1.1step 1.2step 1.3step 2.1step 3.1∎

Clauses 1 and 2 are steps 1.1 and 1.2, and clause 3 with its numerical instance is steps 1.3, 2.1 and 3.1; the example exhibits a model with an explicit time dependence in its potential along which the mechanical energy is not conserved, in agreement with the sharpness clause of the time-translation statement, and it reports no measurement.

Remarks

The driven model is adopted external data. The drive $f (t)$ is prescribed and does not depend on the state; the object producing it is not a label of the model. The example therefore shows that the energy balance of the pair and the failure of the time-translation identity are properties of the declared presentation, not of the framework's kinematics; the same mechanical equation with a constant force would have an autonomous representative and a conserved energy.
Representative dependence. Adding a function $c (t)$ to $U$ shifts the energy by $c (t)$ and its rate by $c^{'} (t)$ , as recorded in clause (c) of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets; the representative declared here is the one in which the drive is visible as $\partial_{t} U = - f^{'} (t) \cdot x$ , and the numerical values of $E$ depend on that choice of representative.

A one-label model with a time-dependent potential: the time-translation identity fails and the mechanical energy changes at the rate of the explicit time dependence

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources