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Mechanical energy balance for a potential presentation: E′=∂tU plus the residual power, conservation for a static potential with no residual forces, and the integrated work form

Physical scope

The mechanical energy balance of a model presented through a potential with declared residual forces: the derivative of the mechanical energy equals the partial time derivative of the potential plus the power of the residual forces; a time-independent potential with no residual forces conserves the mechanical energy along every motion; and the integrated form expresses the work of the total force through the change of the potential and the residual work. The result is a conditional deduction from the adopted potential presentation, with the normalization of the potential carried along; it asserts no measured value and no conservation outside the stated hypotheses.

Statement

Adopt the force model, one inertial frame, and the potential presentation of Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (a): a C2 potential U:I0×Q⟶R of dimension μℓ2τ−2 and residual forces di of the standing regularity, with Fi(t,x,v)=−∇xiU(t,x)+di(t,x,v)(i=1,…,N) on the state domain. Let x:J→Q be a motion of the model on an interval J with nonempty interior, write di(t):=di(t,x(t),w(t)) for the evaluated residual forces and E(t):=K(t)+U(t,x(t)) for the mechanical energy of the definition (clause (b)). Then:

  1. E is differentiable on J and E′(t)=∂tU(t,x(t))+∑i⟨di(t), wi(t)⟩(t∈J): the mechanical energy changes at the rate of the explicit time dependence of the potential plus the power of the residual forces;
  2. if ∂tU≡0 on I0×Q and each evaluated residual force di vanishes identically along this motion, then E is constant on J. If, more strongly, every residual-force map di vanishes on its entire state domain, the model is a global potential model with the declared representative U;
  3. for all t0<t1 in J, K(t1)−K(t0)=U(t0,x(t0))−U(t1,x(t1))+∫t0t1[∂tU(t,x(t))+∑i⟨di(t),wi(t)⟩]dt, equivalently W([t0,t1])=U(t0,x(t0))−U(t1,x(t1))+∫t0t1[∂tU+∑i⟨di,wi⟩]dt: the work of the total force is the decrease of the potential energy plus the work of the explicit time dependence and of the residual forces.

The representative U is the one declared by the model; replacing U by U+c(t) changes E by c(t) and E′ by c′(t), as recorded in Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clause (c). The statement is read in the fixed frame; no empirical content is claimed.

Facts & Assumptions

Given: The potential presentation of the model, one inertial frame, a motion x:J→Q with velocities wi=xi′, evaluated residual forces di, and the mechanical energy E=K+U(t,x(t)).

[L1]

Notation (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets clauses (a)-(b), Total mass, total momentum, centre of mass, angular momentum, torque, kinetic energy, power and work of a particle model in an inertial frame): K, W and P are the kinetic energy, work and power of the definition; Q is nonempty and open; U is C2 hence C1; t↦U(t,x(t)) is defined on J; the equation of motion is mix¨i=Fi(t,x,w) with Fi=−∇xiU+di, so that mix¨i(t)=−∇xiU(t,x(t))+di(t) with di(t)=di(t,x(t),w(t)).

[L2]

Mathematical supplier (Work-energy balance ddt(K+U(t,x(t)))=∂tU(t,x(t))+∑i⟨di(t),xi′(t)⟩ for forces mixi′′=−∇xiU+di): under the displayed hypotheses with U of class C1 on the open I0×Q and maps di along the curve, if mixi′′(t)=−∇xiU(t,x(t))+di(t) on I, then K′=∑i⟨mixi′′,xi′⟩; ddtU(t,x(t))=∂tU(t,x(t))+∑i⟨∇xiU(t,x(t)),xi′(t)⟩; ddt(K+U(t,x(t)))=∂tU(t,x(t))+∑i⟨di(t),xi′(t)⟩; and if di≡0 for all i and ∂tU≡0 then t↦K+U(t,x(t)) is constant on I.

[L3]

Work-energy theorem (Power of the total force and the work-energy theorem: K′=P and K(t1)−K(t0)=W([t0,t1]) along a motion): K(t1)−K(t0)=W([t0,t1])=∫t0t1P for t0<t1 in J, and P=∑i⟨Fi(t,x,w),wi⟩. The Euclidean inner product is bilinear (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn) and finite sums are linear (Laws of finite sums and finite products).

Proof

technique · direct; instantiate the mathematical work-energy balance for the potential presentation and integrate the identity to obtain the work form
1.1givenL1L2algebra

The hypotheses of [L2] hold: I:=J is an interval with nonempty interior, Q is nonempty and open, U is C1 on I0×Q (indeed C2), x is twice differentiable, and the equation of motion together with the presentation gives mix¨i(t)=−∇xiU(t,x(t))+di(t) with the evaluated residual forces of the statement. Applying [L2] gives that E=K+U(t,x(t)) is differentiable on J with E′(t)=∂tU(t,x(t))+∑i⟨di(t),wi(t)⟩, which is clause 1.

2.1step 1.1L2

Clause 2: if ∂tU≡0 and di≡0, then the right side of step 1.1 vanishes at every t∈J; the final clause of [L2] gives directly that E is constant on J.

2.2step 1.1L1L2L3algebra

Clause 3: fix t0<t1 in J. By [L3], K(t1)−K(t0)=∫t0t1P, and on J P=∑i⟨−∇xiU(t,x(t))+di(t),wi(t)⟩=−∑i⟨∇xiU(t,x(t)),wi(t)⟩+∑i⟨di(t),wi(t)⟩. By the chain-rule clause of [L2], ∑i⟨∇xiU(t,x(t)),wi(t)⟩=ddtU(t,x(t))−∂tU(t,x(t)), so P=∂tU(t,x(t))+∑i⟨di(t),wi(t)⟩−ddtU(t,x(t)).

3.1step 2.2L1L3L4algebra∎

The function t↦U(t,x(t)) is C1 on J (composition of the C1 map U with the C1 curve t↦(t,x(t))), so [L4] gives ∫t0t1ddtU(t,x(t)) dt=U(t1,x(t1))−U(t0,x(t0)). Integrating the identity of step 2.2 over [t0,t1] and substituting this value, K(t1)−K(t0)=∫t0t1P=U(t0,x(t0))−U(t1,x(t1))+∫t0t1[∂tU(t,x(t))+∑i⟨di(t),wi(t)⟩]dt, which is the first form of clause 3; the second form is the definition of W([t0,t1]) by [L1] and [L3].

Remarks

Depends on

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