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Electrostatic test-energy first integral

Physical scope

A finite-segment charged test probe in a prescribed, stationary electrostatic potential on Minkowski spacetime. The conserved energy is conditional on the adopted Lorentz-force law and excludes self-field, self-gravity, radiation reaction, and any coupled source claim.

Example

On Minkowski spacetime with signature $(-, +, +, +)$ , time orientation $\partial_{0}$ , and length coordinates $x^{0} = c t$ , let $ϕ (x)$ be a $C^{2}$ time-independent scalar potential on an open spatial region and set $A = - ϕ (x) d x^{0} / c$ . Use SI units $[ϕ] = V$ , $[q] = C$ , $[m] = k g$ , and $[c] = m s^{- 1}$ . For a $C^{2}$ future-timelike proper-time segment satisfying the charged-probe law of Separately adopted fixed-background Maxwell and charged probes, define $w = d x / d t$ and $γ = (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2}$ . Then

$E_{t e s t} = γ m c^{2} + q ϕ (x)$

is constant along the segment. The field and metric are prescribed data; this is neither a measurement nor a coupled point-source solution.

Facts & Assumptions

Given: Minkowski metric, the potential $A = - ϕ d x^{0} / c$ with $\partial_{0} ϕ = 0$ , constants $m > 0, q \in R$ , and a finite proper-time curve segment governed by the adopted test-body model.

[F1]

Separately adopted fixed-background Maxwell and charged probes: on a finite local-potential segment, the probe is a massive charged test body with fixed $m, q$ ; the metric and $A$ are independent prescribed inputs, and the force-law formulation is the exact conditional equivalent of the adopted action under its inherited ${A C}_{ω}$ hypothesis. No self-field or self-gravity is included.

[F2]

Killing exchange current and gauge-compensated charged first integral: for a supplied smooth local potential and an existing $C^{2}$ proper-time curve satisfying $m \nabla_{U} U = q F^{♯} (U)$ , a Killing field $ξ$ and compensator with $L_{ξ} A = d χ$ give the constant $C = m g (U, ξ) + q (A (ξ) - χ)$ . Its displayed expression is a template here; the direct calculation below proves constancy for this Example's $C^{2}$ potential without applying the smooth-potential theorem beyond its hypotheses.

[A1]

The potential is stationary, $ξ = \partial_{0}$ is the Minkowski time-translation Killing field, and $χ = 0$ . The curve remains inside the spatial potential domain.

Verification

1.1givenF1F2A1algebra

Take the candidate expression of [F2], $C = m g (U, ξ) + q A (ξ) = m U_{0} + q A_{0}$ , with $ξ = \partial_{0}$ and $χ = 0$ . The actual $C^{2}$ potential suffices for the following direct calculation. In Cartesian Minkowski coordinates the connection coefficients vanish; by [F1], the supplied $C^{2}$ curve obeys the Lorentz law, whose lowered time component is $m d U_{0} / d τ = q F_{0 i} U^{i}$ . Since $A_{0} = - ϕ / c$ , $A_{i} = 0$ and $\partial_{0} ϕ = 0$ , exterior differentiation gives $F_{0 i} = - \partial_{i} A_{0} = \partial_{i} ϕ / c$ . The chain rule therefore yields $d C / d τ = q F_{0 i} U^{i} + q \partial_{i} A_{0} U^{i} = 0$ at every interior parameter value. A real differentiable function with zero derivative is constant on each compact interior subinterval by the mean-value theorem; continuity extends the same value to the included endpoints of the finite segment. This proves constancy at the stated $C^{2}$ regularity without a smooth-potential application of [F2].

2.1givenA1step 1.1algebra

Since $\partial_{0}$ is future-directed, $U^{0} > 0$ . The relation $U^{i} / U^{0} = w^{i} / c$ and $g (U, U) = - c^{2}$ imply $U^{0} = c (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2} = γ c$ , hence $U_{0} = - γ c$ . Also $A (ξ) = A_{0} = - ϕ / c$ , so $C = - γ m c - q ϕ / c = - E_{t e s t} / c$ . Since $c > 0$ is constant, the displayed test energy is constant.

3.1F1A1step 2.1algebra∎

For an explicit work check, $F_{0 i} = - E_{i} / c$ and $A_{0} = - ϕ / c$ give $E = - \nabla ϕ$ and $B = 0$ , so the flat Lorentz law is $d (γ m w) / d t = q E$ . Writing $a = d w / d t$ , differentiation gives $d γ / d t = γ^{3} w \cdot a / c^{2}$ and $d (γ m w) / d t = m γ a + m γ^{3} (w \cdot a) w / c^{2}$ ; hence $w \cdot d (γ m w) / d t = m γ^{3} w \cdot a = d (γ m c^{2}) / d t$ . Thus $d (γ m c^{2}) / d t = q E \cdot w$ , whereas $d (q ϕ) / d t = q \nabla ϕ \cdot w = - q E \cdot w$ . The two work terms cancel, agreeing with step 2.1.

Electrostatic test-energy first integral

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources