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Local and fixed-volume momentum balance for smooth sources

Physical scope

The conditional momentum balance of the classical vacuum electromagnetic field: the local identity linking the time derivative of the momentum density and the divergence of the momentum flux tensor to the adopted continuum force density, and its fixed-volume traction form with the boundary stress retained, deduced from the smooth branch of the adopted vacuum Maxwell equations and the adopted continuum coupling. The statement is conditional on those adopted postulates, excludes the weak branch, singular source support and non-integrable products, and includes no self-force, radiation reaction or monopole term. No empirical claim is made.

Statement

Assume the smooth branch of The microscopic vacuum Maxwell equations: smooth and weak branches and clause (ii) of The Lorentz coupling: external force on a test charge and the continuum force and power densities on the laboratory region $U ≅ I \times Ω$ of Laboratory geometry: a restricted inertial chart, its spatial slices and fixed-time fields: the fields $E, B : Ω \times I \to R^{3}$ are jointly $C^{1}$ of The electric and magnetic fields as fixed-time spatial vector fields, the source fields $ρ, J$ are continuous of Charge parameter, frame charge and current densities, volume charge and flux, and the four-current, the constants are those of Framework-local SI units, the vacuum constants and the exact/measured status of their values, the force density is $f : = ρ E + J \times B$ of that clause (ii), in $N m^{- 3}$ , and $g, σ, Π = - σ$ are the momentum density, Maxwell stress and momentum flux of Momentum density, Maxwell stress, momentum flux and orbital angular density in vacuum. Then:

(a) [Local identity] On $Ω \times I$ , componentwise,

$\partial_{t} g - div σ = - f, equivalently \partial_{t} g + div Π = - f,$

that is, $\partial_{t} g_{i} - \sum_{j = 1}^{3} \partial_{j} σ_{i j} = - (ρ E_{i} + (J \times B)_{i})$ for every $i$ , with $\partial_{t}$ the partial derivative at fixed position and $div$ taken at fixed time.

(b) [Fixed-volume identity] Let $Ω^{'} \subseteq Ω$ be open with $\overline{Ω^{'}} \subseteq Ω$ carrying the finite piecewise $C^{1}$ presentation of Divergence for finite piecewise C1 presentations, let $t_{0} < t_{1}$ with $(t_{0}, t_{1}) \subseteq I$ , assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and let $n$ be the outward unit normal. Then for every $t \in (t_{0}, t_{1})$ and every $i$ ,

$\frac{d}{d t} \int_{Ω^{'}} g_{i} d x = \oint_{\partial Ω^{'}} (\sum_{j = 1}^{3} σ_{i j} n_{j}) d S - \int_{Ω^{'}} f_{i} d x = - \oint_{\partial Ω^{'}} (\sum_{j = 1}^{3} Π_{i j} n_{j}) d S - \int_{Ω^{'}} f_{i} d x,$

the closed-surface integrals being the finite face sums of that divergence theorem.

(c) [Scope] The identities are conditional on the two adopted postulates. The force density $f$ is defined only for the matter model of clause (ii), so no claim is made about a prescribed current without such a matter model. No conservation of field-plus-matter momentum follows without explicit matter balances and their boundary fluxes. The field-momentum rate in (b) is the boundary traction minus the integrated matter force; these terms can cancel, and neither may be discarded without justification. The weak branch, singular sources and non-integrable products are excluded as in Vacuum electromagnetic energy density and energy flux (the Poynting vector); no self-force, radiation reaction or magnetic-monopole coupling is included, in accordance with the exclusions of The Lorentz coupling: external force on a test charge and the continuum force and power densities. This is not an empirical claim.

Facts & Assumptions

Given: The laboratory region $U ≅ I \times Ω$ ; the smooth branch of The microscopic vacuum Maxwell equations: smooth and weak branches with the jointly $C^{1}$ fields of The electric and magnetic fields as fixed-time spatial vector fields and the continuous sources of Charge parameter, frame charge and current densities, volume charge and flux, and the four-current; clause (ii) of The Lorentz coupling: external force on a test charge and the continuum force and power densities; the constants of Framework-local SI units, the vacuum constants and the exact/measured status of their values; and $g, σ, Π$ of Momentum density, Maxwell stress, momentum flux and orbital angular density in vacuum. In clause (b) also ${A C}_{ω}$ , the admissible $Ω^{'}$ and $(t_{0}, t_{1}) \subseteq I$ .

[F1]

The four smooth-branch equations hold pointwise on $Ω \times I$ : $div E = ρ / ε_{0}$ , $div B = 0$ , $curl E = - \partial_{t} B$ , $curl B = μ_{0} J + c^{- 2} \partial_{t} E$ , with $ε_{0} = (μ_{0} c^{2})^{- 1}$ , $c > 0$ and $μ_{0} > 0$ (The microscopic vacuum Maxwell equations: smooth and weak branches, Framework-local SI units, the vacuum constants and the exact/measured status of their values).

[F2]

The fields are jointly $C^{1}$ on the open $Ω \times I$ : for fixed position the component maps $s \mapsto E_{i} (x, s)$ , $s \mapsto B_{i} (x, s)$ are differentiable with derivatives $\partial_{t} E_{i}$ , $\partial_{t} B_{i}$ ; sums, scalar multiples and products of differentiable functions of the time variable obey the usual rules, and components of a cross product are finite sums of component products (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , 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 $g (c) \neq 0$ , The cross product in $R^{3}$ ).

[F3]

The objects $g, σ, Π$ of Momentum density, Maxwell stress, momentum flux and orbital angular density in vacuum are of class $C^{1}$ on $Ω \times I$ , with $g = ε_{0} E \times B$ , $σ_{i j} = ε_{0} (E_{i} E_{j} - \frac{1}{2} δ_{i j} ∣ E ∣^{2}) + μ_{0}^{- 1} (B_{i} B_{j} - \frac{1}{2} δ_{i j} ∣ B ∣^{2})$ symmetric, $Π = - σ$ , and $(div σ)_{i} = \sum_{j = 1}^{3} \partial_{j} σ_{i j}$ at fixed time; the dot and cross products are those of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and The cross product in $R^{3}$ .

[F4]

Quadratic stress identity: for a $C^{1}$ field $F : U \to R^{3}$ on an open $U \subseteq R^{3}$ and $Q_{i j} = F_{i} F_{j} - \frac{1}{2} δ_{i j} ∣ F ∣^{2}$ , $div Q = F div F - F \times curl F$ pointwise (The quadratic stress identity and the local angular-balance identity).

[F5]

Fixed-volume tools: for compact measurable $K$ with finite Lebesgue measure and $f, \partial_{t} f$ continuous on a neighbourhood of $K \times [a, b]$ , $t \mapsto \int_{K} f d λ_{n}$ is differentiable on $(a, b)$ with derivative $\int_{K} \partial_{t} f d λ_{n}$ (Differentiation under the integral sign over a compact measurable set and over a compact $C^{1}$ surface); if $Ω$ carries the finite piecewise $C^{1}$ presentation and $F \in C^{1} (\overline{Ω}; R^{n})$ , then $\int_{Ω} div F = \sum_{j} \int_{S_{j}} F \cdot ν_{j} d S$ with faces counted once off the edge set, under ${A C}_{ω}$ (Divergence for finite piecewise C1 presentations); and the boundary of such an $Ω$ is Lebesgue-null by the graph-coordinate Fubini and rigid-change-of-variables argument in Proof step 4.1 of that divergence theorem, followed by finite subadditivity over its faces; a nonnegative measurable integral over a null set vanishes (A nonnegative integral over a null set vanishes), so signed integrals over $Ω$ and over its closure agree.

[F6]

The force density is $f = ρ E + J \times B$ , a continuous map $Ω \times I \to R^{3}$ with unit $N m^{- 3}$ (The Lorentz coupling: external force on a test charge and the continuum force and power densities, Charge parameter, frame charge and current densities, volume charge and flux, and the four-current).

Proof

technique · direct; apply the quadratic stress identity with coefficients $\varepsilon_0$ and $\mu_0^{-1}$, substitute the four smooth Maxwell equations, identify the time derivative of the momentum density by the product rule, and integrate the resulting pointwise identity over an admissible fixed volume with the divergence theorem

1.1givenF3F4algebra

Divergence of the stress tensor: define $Q_{i j}^{(E)} = E_{i} E_{j} - \frac{1}{2} δ_{i j} ∣ E ∣^{2}$ and $Q_{i j}^{(B)} = B_{i} B_{j} - \frac{1}{2} δ_{i j} ∣ B ∣^{2}$ at fixed time. Both are $C^{1}$ on the open $Ω$ , and [F4] applied to $F = E$ and to $F = B$ gives $div Q^{(E)} = E div E - E \times curl E$ and $div Q^{(B)} = B div B - B \times curl B$ ; since $σ = ε_{0} Q^{(E)} + μ_{0}^{- 1} Q^{(B)}$ and $div$ is linear in its argument, $div σ = ε_{0} (E div E - E \times curl E) + μ_{0}^{- 1} (B div B - B \times curl B)$ pointwise on $Ω \times I$ .

1.2givenF2F3algebra

Time derivative of the momentum density: because $g = ε_{0} E \times B$ and the components of $E \times B$ are finite sums of products of the component functions, the product rule and linearity [F2] give $\partial_{t} (E \times B) = \partial_{t} E \times B + E \times \partial_{t} B$ componentwise, hence $\partial_{t} g = ε_{0} (\partial_{t} E \times B + E \times \partial_{t} B)$ pointwise.

1.3givenF3F5algebra

Fixed-volume setup: let $Ω^{'}$ , $t_{0} < t_{1}$ be as in clause (b), fix $t \in (t_{0}, t_{1})$ and choose $a, b$ with $t_{0} < a < t < b < t_{1}$ . The closure $\overline{Ω^{'}}$ is compact and contained in the open $Ω$ , and each $g_{i}$ together with $\partial_{t} g_{i}$ is continuous on a neighbourhood of $\overline{Ω^{'}} \times [a, b]$ by [F3]; the boundary is Lebesgue-null by the finite piecewise $C^{1}$ presentation and [F5], so integrals over $Ω^{'}$ and over its closure agree, and [F5] makes $t \mapsto \int_{Ω^{'}} g_{i} (x, t) d x$ differentiable at $t$ with derivative $\int_{Ω^{'}} \partial_{t} g_{i} (x, t) d x$ .

2.1step 1.1step 1.2F1F6algebra

Local identity: substitute [F1] into step 1.1: $div σ = ρ E - ε_{0} E \times (- \partial_{t} B) - μ_{0}^{- 1} B \times (μ_{0} J + c^{- 2} \partial_{t} E) = ρ E + ε_{0} E \times \partial_{t} B - B \times J - (μ_{0} c^{2})^{- 1} B \times \partial_{t} E$ . By bilinearity and alternation of the cross product, $- B \times J = J \times B$ and $- (μ_{0} c^{2})^{- 1} B \times \partial_{t} E = ε_{0} \partial_{t} E \times B$ ; hence $div σ = ρ E + J \times B + ε_{0} (E \times \partial_{t} B + \partial_{t} E \times B) = f + \partial_{t} g$ by [F6] and step 1.2. Therefore $\partial_{t} g - div σ = - f$ , equivalently $\partial_{t} g + div Π = - f$ because $Π = - σ$ ; this is clause (a).

3.1step 1.3step 2.1F5algebra

Fixed-volume identity: by steps 1.3 and 2.1, for each $i$ we have $\frac{d}{d t} \int_{Ω^{'}} g_{i} d x = \int_{Ω^{'}} \partial_{t} g_{i} d x = \int_{Ω^{'}} (div σ)_{i} d x - \int_{Ω^{'}} f_{i} d x$ ; the fixed-time vector field $x \mapsto (σ_{i 1}, σ_{i 2}, σ_{i 3}) (x, t)$ is $C^{1}$ on a neighbourhood of $\overline{Ω^{'}}$ by [F3], so the divergence theorem of [F5] gives $\int_{Ω^{'}} (div σ)_{i} d x = \oint_{\partial Ω^{'}} \sum_{j} σ_{i j} n_{j} d S$ as a finite face sum. Substituting $Π = - σ$ turns this into the two displayed equalities of clause (b) under the stated ${A C}_{ω}$ .

4.1step 2.1step 3.1F1F3F6∎

Clause (c) records the postulational status of the two inputs, the model dependence of $f$ and of the boundary term, and the exclusions inherited from the weak branch and from the coupling; the argument used only the adopted postulates, the definitions of $g, σ, Π$ , the quadratic stress identity and the integration interface, and made no empirical claim.

Remarks

The two coefficient applications. The identity is applied once with coefficient $ε_{0}$ to the electric quadratic form and once with coefficient $μ_{0}^{- 1}$ to the magnetic one; equivalently one applies it to the pair $(\sqrt{ε_{0}} E, μ_{0}^{- 1 / 2} B)$ , but the displayed form keeps the SI constants in their physical places.
Why the magnetic curl term survives. Unlike the energy balance, the momentum balance does not cancel the magnetic terms: $div Q^{(B)} = - B \times curl B$ contributes the Ampere-law current and the displacement term, whose combination with $\partial_{t} g$ produces the force density $J \times B$ .
Boundary term and traction. On a fixed volume the identity reads $d P_{Ω^{'}, i} / d t = \oint_{\partial Ω^{'}} σ_{i j} n_{j} d S - \int_{Ω^{'}} f_{i} d x$ , the first term being the traction of the field on the boundary; the minus sign in $Π = - σ$ fixes the sign convention of every later material balance.

Local and fixed-volume momentum balance for smooth sources

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources