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Work of the electrostatic force equals charge times the potential drop and is path independent

Physical scope

The work done by the electric force of a prescribed static external field on a test charge of constant parameter $q$ carried along a prescribed piecewise $C^{1}$ path in a star-shaped pure electrostatic region: the work equals $q$ times the potential drop, depends only on the endpoints, vanishes on closed paths and is unchanged by the additive constant of the potential. No equation of motion, radiative loss, work on sources or empirical claim is included.

Statement

Let $Ω \subseteq R^{3}$ be open, star-shaped (Star-shaped open subsets of Euclidean space) and the spatial region of a pure electrostatic regime as in Scalar potential and the Poisson equation in a star-shaped electrostatic region, with fields of The electric and magnetic fields as fixed-time spatial vector fields, charge parameters of Charge parameter, frame charge and current densities, volume charge and flux, and the four-current and the units of Framework-local SI units, the vacuum constants and the exact/measured status of their values; thus $E : Ω \to R^{3}$ is $C^{1}$ , $B = 0$ and $J = 0$ (The static and pure electrostatic regimes and their fixed-time reading), and by Scalar potential and the Poisson equation in a star-shaped electrostatic region there is $φ \in C^{2} (Ω)$ with $E = - \nabla φ$ . Let $q \in R$ (coulombs) be the charge of a test particle and let $γ : [t_{0}, t_{1}] \to Ω$ be a piecewise $C^{1}$ path (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, $C^{k}$ Euclidean maps and diffeomorphisms). Define the work of the electric force along the prescribed path by the vector line integral $W (γ) : = \int_{γ} q E \cdot d r$ of Scalar line integrals with respect to arc length and vector-field line integrals; $q E$ is the laboratory three-force of The Lorentz coupling: external force on a test charge and the continuum force and power densities(i) when $B = 0$ , using the tensor components of The electromagnetic two-form, its observer decomposition and the four-current(a) and the four-force/three-force relation of Four-force identities and the work-energy relation in an inertial frame(c), as computed in [F5] below. Then:

(a) [Value] $W (γ) = q (φ (γ (t_{0})) - φ (γ (t_{1})))$ ;

(b) [Path independence] if $γ_{1}, γ_{2}$ are piecewise $C^{1}$ paths in $Ω$ with the same initial and the same terminal point, then $W (γ_{1}) = W (γ_{2})$ ; in particular $W (γ) = 0$ for every closed piecewise $C^{1}$ path;

(c) [Potential energy] with the potential energy $U : = q φ$ (joules) the identity reads $W (γ) = - (U (γ (t_{1})) - U (γ (t_{0})))$ ;

(d) [Gauge invariance of the work] if $φ^{'} \in C^{2} (Ω)$ is any other potential with $\nabla φ^{'} = - E$ , then $φ^{'} = φ + c$ with a constant $c \in R$ and the formula of (a) is unchanged (Two potentials of the same field differ by a constant on each piecewise-C1 path component).

The statement is a line-integral identity for a prescribed path in a prescribed static external field. It asserts no equation of motion, no radiation, no work done on the sources and no claim that the test particle actually follows $γ$ ; the interpretation of $W$ as work done by the adopted external coupling is the physical reading supplied by The Lorentz coupling: external force on a test charge and the continuum force and power densities, not a dynamical theorem.

Facts & Assumptions

Given: A star-shaped open $Ω \subseteq R^{3}$ carrying a pure electrostatic regime with $C^{1}$ field $E$ and $B = J = 0$ , a charge parameter $q \in R$ , and a piecewise $C^{1}$ path $γ : [t_{0}, t_{1}] \to Ω$ .

[F1]

The pure electrostatic regime has $B = 0$ , $J = 0$ , time-independent fields and the reduced equations; on a star-shaped region the potential theorem provides $φ \in C^{2} (Ω)$ with $\nabla φ = - E$ (The static and pure electrostatic regimes and their fixed-time reading, Scalar potential and the Poisson equation in a star-shaped electrostatic region, The electric and magnetic fields as fixed-time spatial vector fields, Charge parameter, frame charge and current densities, volume charge and flux, and the four-current, Framework-local SI units, the vacuum constants and the exact/measured status of their values, Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields, Star-shaped open subsets of Euclidean space).

[F2]

The vector line integral is $\int_{γ} F \cdot d r = \sum_{i} \int_{t_{i}}^{t_{i + 1}} ⟨ F (γ (t)), v_{i} (t) ⟩ d t$ over an admissible partition, the inner product is the Euclidean one and the integral is linear in the field (Scalar line integrals with respect to arc length and vector-field line integrals, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , $C^{k}$ Euclidean maps and diffeomorphisms, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, 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 $g (c) \neq 0$ ).

[F3]

The gradient theorem evaluates the line integral of a gradient of a $C^{1}$ function along a piecewise $C^{1}$ path as the endpoint increment (The gradient theorem: the line integral of a gradient is the endpoint increment).

[F4]

Two $C^{1}$ potentials with the same gradient differ by a constant on each piecewise- $C^{1}$ path component, and a star-shaped set has a single such component through its centre (Two potentials of the same field differ by a constant on each piecewise-C1 path component, Star-shaped open subsets of Euclidean space, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[F5]

For a massive test worldline in the prescribed external field, The Lorentz coupling: external force on a test charge and the continuum force and power densities(i) gives the four-force $K^{μ} = q F^{μ ν} U_{ν}$ . Write $w$ for its laboratory coordinate velocity, $∣ w ∣ < c$ , and $γ : = (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2} > 0$ . In the laboratory frame $U^{0} = γ c$ and $U_{0} = - γ c$ , while $K^{i} = γ f_{i}$ with $f = d p / d t$ the laboratory three-force (Four-force identities and the work-energy relation in an inertial frame, Statement and (c)). When $B = 0$ , The electromagnetic two-form, its observer decomposition and the four-current(a) gives $F^{i 0} = - E_{i} / c$ and $F^{i j} = 0$ , hence $γ f_{i} = q (- E_{i} / c) (- γ c) = q γ E_{i}$ . Since $γ > 0$ , $f = q E$ . This identifies the force used by the work definition whenever the external particle model applies; the prescribed path itself is not assumed to be a particle worldline.

Proof

technique · direct

1.1givenF1F2

By [F1] the potential $φ \in C^{2} (Ω)$ satisfies $\nabla φ = - E$ ; the field $q E$ is $C^{1}$ on $Ω$ for the constant $q$ , and the work $W (γ) = \int_{γ} q E \cdot d r$ is defined by the vector line integral of [F2], which is finite because the integrand is continuous on each smooth piece of the compact parameter interval.

2.1step 1.1F1F2F3

Clause (a): by the linearity of the line integral in the field and the constancy of $q$ , $W (γ) = q \int_{γ} E \cdot d r$ ([F2]); substituting $E = - \nabla φ$ and applying the gradient theorem [F3] gives $W (γ) = - q \int_{γ} \nabla φ \cdot d r = - q (φ (γ (t_{1})) - φ (γ (t_{0}))) = q (φ (γ (t_{0})) - φ (γ (t_{1}))),$ which is the displayed formula, including the degenerate case $t_{0} = t_{1}$ where the line integral is $0$ and the right side vanishes too.

3.1step 2.1

Clause (b): if $γ_{1}, γ_{2}$ are piecewise $C^{1}$ paths in $Ω$ with the same initial point $x$ and the same terminal point $y$ , then by step 2.1 applied to each path $W (γ_{1}) = q (φ (x) - φ (y)) = W (γ_{2})$ ; for a closed path, where $x = y$ , both formulas give $W (γ) = 0$ .

4.1step 3.1

Clause (c): with $U : = q φ$ , the identity of step 2.1 reads $W (γ) = U (γ (t_{0})) - U (γ (t_{1})) = - (U (γ (t_{1})) - U (γ (t_{0})))$ , the potential energy being the constant multiple $q φ$ of the potential.

5.1step 4.1F4

Clause (d): let $φ^{'} \in C^{2} (Ω)$ satisfy $\nabla φ^{'} = - E = \nabla φ$ . Then $\nabla (φ^{'} - φ) = 0$ on $Ω$ , and by [F4] the difference is constant on every piecewise- $C^{1}$ path component of $Ω$ ; since $Ω$ is star-shaped, every point is joined to the centre by a $C^{1}$ segment, so all of $Ω$ is one path component ([F4]) and $φ^{'} = φ + c$ for a constant $c \in R$ . Substituting into the formula of step 2.1 replaces $φ$ by $φ + c$ and leaves the right side unchanged, so the work is independent of the additive normalisation of the potential.

6.1step 3.1step 4.1step 5.1F5∎

Clauses (a)-(d) are steps 2.1, 3.1, 4.1 and 5.1. The force whose work was computed is the electric term $q E$ of the adopted external coupling [F5] along the prescribed path; the computation used only the potential representation, the gradient theorem and the constancy of the potential up to an additive constant, and asserted no equation of motion, no radiation, no work done on the sources and no claim that the particle follows $γ$ , which is exactly the scope stated.

Work of the electrostatic force equals charge times the potential drop and is path independent

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources