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Hodge decomposition of an alternating two-form for an observer, invariants and the optional field identification

Statement

Let $(V, η)$ be the oriented Lorentz vector space of the affine Minkowski model with orientation and volume form $ε$ of The oriented volume form epsilon and the permutation symbols of an inertial frame, and let $F : V \times V \to R$ be an alternating bilinear form (Alternating $k$ -linear maps, Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms). Define the Hodge dual by $(* F)_{a b} : = \frac{1}{2} ε_{a b}^{c d} F_{c d}$ and set $(α \land β)_{a b} : = α_{a} β_{b} - α_{b} β_{a}$ . Use the standard form normalization $F = \frac{1}{2} \sum_{a, b} F_{a b} θ^{a} \land θ^{b}$ in the dual coframe $θ^{a} (e_{b}) = δ_{b}^{a}$ of an oriented orthonormal frame, and define $(F \land F) (v_{0}, v_{1}, v_{2}, v_{3}) : = 2 [F (v_{0}, v_{1}) F (v_{2}, v_{3}) - F (v_{0}, v_{2}) F (v_{1}, v_{3}) + F (v_{0}, v_{3}) F (v_{1}, v_{2})]$ , the alternating four-form exterior square. Let $n \in V$ be future unit, $η (n, n) = - 1$ , and put $e_{a} : = F_{a b} n^{b}$ , $b_{a} : = (* F)_{a b} n^{b}$ . (a) $e (n) = b (n) = 0$ , and $F = n^{♭} \land e - * (n^{♭} \land b)$ with $n^{♭} : = η (n, \cdot)$ ; also $* (* F) = - F$ , so the Hodge dual squares to $- 1$ on two-forms. (b) In an oriented orthonormal frame with $n = e_{0}$ : $e_{i} = F_{i 0}$ , $b_{i} = - \frac{1}{2} ϵ_{i j k} F_{j k}$ , $F_{0 i} = - e_{i}$ and $F_{i j} = - ϵ_{i j k} b_{k}$ ; hence the six components of $F$ are exactly the six components of the pair $(e, b)$ and the reconstruction in (a) is equivalent to these component formulae. (c) The quadratic invariants are $\frac{1}{2} F_{a b} F^{a b} = ∣ b ∣^{2} - ∣ e ∣^{2}$ and $F_{a b} (* F)^{a b} = - 4 (e \cdot b)$ , where $∣ \cdot ∣$ and $\cdot$ are those of the spatial structure of the frame (The Euclidean spatial structure and cross product of an inertial frame); also $F \land F = 2 (e \cdot b) ε$ . The two scalar expressions $I_{1} : = \frac{1}{2} F_{a b} F^{a b}$ and $I_{2} : = F_{a b} (* F)^{a b}$ form a basis of the homogeneous quadratic polynomials in the six components of $F$ invariant under the proper orthochronous Lorentz group; the square-wedge expression is dependent, $F \land F = - \frac{1}{2} I_{2} ε$ . $I_{2}$ is an oriented pseudoscalar and changes sign if the chosen orientation is reversed with the metric and $F$ held fixed. (d) (Optional SI identification.) Let $c > 0$ . If in an inertial frame one identifies $F^{i 0} = E_{i} / c$ and $F^{i j} = - ϵ_{i j k} B_{k}$ for numerical spatial vectors $E, B \in R^{3}$ , then $e_{i} = - E_{i} / c$ and $b_{i} = B_{i}$ , the invariants of (c) read $\frac{1}{2} F_{a b} F^{a b} = ∣ B ∣^{2} - ∣ E ∣^{2} / c^{2}$ and $F_{a b} (* F)^{a b} = 4 (E \cdot B) / c$ , and $F \land F = - 2 (E \cdot B) ε / c$ . If $κ : = (F \land F) (e_{0}, e_{1}, e_{2}, e_{3}) = (F \land F)_{0123}$ , then $E \cdot B = - c κ / 2$ . Under the passive boost of The passive Lorentz boost with parameter beta along $e_{1}$ with parameter $β \in (- 1, 1)$ one has $E_{1}^{'} = E_{1}$ , $B_{1}^{'} = B_{1}$ , $E_{⊥}^{'} = γ (E + v \times B)_{⊥}$ and $B_{⊥}^{'} = γ (B - v \times E / c^{2})_{⊥}$ where the cross product is the right-handed product (The cross product in $R^{3}$ ), with $γ = (1 - β^{2})^{- 1 / 2}$ and $v = β c$ along $e_{1}$ . In an SI reading, $c$ has units $m / s$ , $E$ has units $V / m$ and $B$ has units tesla, so $E / c$ and all components of $F$ have units tesla; these are conditional unit assignments, not measurements. (e) The theorem is algebra on the oriented Lorentz vector space with the stated index and duality conventions; clause (d) is an explicitly optional identification of the components of $F$ with fields the electromagnetic framework names $E, B$ , and the theorem asserts no field equation, no dynamics, no measurement and no transformation beyond the stated component law. No choice principle is used.

Facts & Assumptions

Given: The oriented Lorentz vector space $(V, η)$ of the affine Minkowski model with its fixed orientation and oriented volume form $ε$ , normalized by $ε (e_{0}, e_{1}, e_{2}, e_{3}) = + 1$ in every positively oriented orthonormal frame (The oriented volume form epsilon and the permutation symbols of an inertial frame, Oriented time-oriented affine Minkowski space (the spacetime primitives), Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1)); the raised volume form is $ε^{a b c d} = η^{a a^{'}} η^{b b^{'}} η^{c c^{'}} η^{d d^{'}} ε_{a^{'} b^{'} c^{'} d^{'}}$ , so $ε^{0123} = η^{00} η^{11} η^{22} η^{33} ε_{0123} = - 1$ . An alternating bilinear form $F$ with components $F_{a b} = - F_{b a}$ (Alternating $k$ -linear maps, Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms); a future unit $n \in V$ with $η (n, n) = - 1$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Time orientation, future and past timelike/null/causal vectors, and the closed cones); and the conventions $(* F)_{a b} : = \frac{1}{2} ε_{a b}^{c d} F_{c d}$ , $(α \land β)_{a b} : = α_{a} β_{b} - α_{b} β_{a}$ , the dual coframe $θ^{a} (e_{b}) = δ_{b}^{a}$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ), and $(F \land F) (v_{0}, \dots, v_{3}) : = 2 [F (v_{0}, v_{1}) F (v_{2}, v_{3}) - F (v_{0}, v_{2}) F (v_{1}, v_{3}) + F (v_{0}, v_{3}) F (v_{1}, v_{2})]$ . The spatial inner product of the frame is the restriction of $η$ to $n^{⊥}$ (The Euclidean spatial structure and cross product of an inertial frame, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), and $e_{a} : = F_{a b} n^{b}$ , $b_{a} : = (* F)_{a b} n^{b}$ .

[F1]

Observer rest space: for future unit $n$ the subspace $n^{⊥} = {v : η (n, v) = 0}$ is three-dimensional and $η$ restricts to a positive definite form on it; the map $h (v) = v + η (n, v) n$ has image $n^{⊥}$ and satisfies $h^{a}_{b} n^{b} = 0 = n_{a} h^{a}_{b}$ . (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(c))

[F2]

Gram–Schmidt: every finite-dimensional inner product space has an orthonormal basis. (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans)

[F3]

Tensor transformation and invariance of the volume form: for a proper orthochronous Lorentz map $L$ the components of a $(0, 2)$ form satisfy $F_{a b}^{'} = (Λ^{- 1})^{c}_{a} (Λ^{- 1})^{d}_{b} F_{c d}$ for the chart transition matrix $Λ$ , the alternating covariant laws hold on each slot, and $ε_{a b c d}^{'} = ε_{a b c d}$ (the oriented volume form is invariant); the boosts $B (β)$ are proper orthochronous with $B (β)^{- 1} = B (- β)$ and every spatial rotation extension $diag (1, R)$ is a proper orthochronous Lorentz map. (Transformation of tensor, covector, derivative and current components under affine Lorentz maps,(iii),(v), The passive boost is a proper orthochronous Lorentz map with inverse B(-beta), Unique boost-rotation factorization of proper orthochronous Lorentz maps,(iii))

[F4]

Passive boost: for $β \in (- 1, 1)$ and $γ = (1 - β^{2})^{- 1 / 2}$ the boost along the spatial axis $e_{1}$ has the matrix $Λ^{0}_{0} = Λ^{1}_{1} = γ$ , $Λ^{0}_{1} = Λ^{1}_{0} = - γ β$ , and identity on the other two spatial directions; it is proper orthochronous with $B (β)^{- 1} = B (- β)$ . (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta))

[F5]

The right-handed cross product of $R^{3}$ : $(u \times v)_{i} = ϵ_{i j k} u_{j} v_{k}$ with $ϵ_{123} = + 1$ . (The cross product in $R^{3}$ )

Proof

technique · direct

1.1givenF1F2

Adapted oriented frame. By [F1] the rest space $n^{⊥}$ is a three-dimensional inner product space; [F2] gives an orthonormal basis $(u_{1}, u_{2}, u_{3})$ of $n^{⊥}$ . The ordered frame $(n, u_{1}, u_{2}, u_{3})$ is orthonormal; if its orientation class is negative, replace $u_{3}$ by $- u_{3}$ , which keeps orthonormality and makes the frame positively oriented. Writing $e_{0} : = n$ , $e_{1}, e_{2}, e_{3}$ for this frame, the components satisfy $η_{a b} = diag (- 1, 1, 1, 1)$ and $ε_{0123} = + 1$ . Only finitely many choices are made.

1.2given

Hodge dual on the six basis forms. Let $θ^{a}$ be the dual coframe and set $a_{i} : = θ^{0} \land θ^{i}$ and $s_{1} : = θ^{2} \land θ^{3}$ , $s_{2} : = θ^{3} \land θ^{1}$ , $s_{3} : = θ^{1} \land θ^{2}$ . Evaluating the definition $(* F)_{a b} = \frac{1}{2} ε_{a b}^{c d} F_{c d}$ on the basis forms, using $ε^{0123} = - 1$ and the antisymmetry of $ε$ , gives $(* a_{i}) = - s_{i}, (* s_{i}) = a_{i} (i = 1, 2, 3) .$ Indeed the only nonzero contributions come from the complementary index pair, and $ε_{0 i}^{j k} = + ϵ_{i j k}$ and $ε_{i j}^{0 k} = - ϵ_{i j k}$ for spatial indices. Consequently $* (* F) = - F$ on every two-form, since $*^{2} = - 1$ on each of the six basis forms. For $F = \frac{1}{2} F_{a b} θ^{a} \land θ^{b}$ the dual is $(* F) = e_{i} s_{i} - b_{i} a_{i}$ whenever the components $e_{i} = F_{i 0}$ and $b_{i} = - \frac{1}{2} ϵ_{i j k} F_{j k}$ are used, by direct substitution into the definition.

2.1step 1.2given

Components of $e$ and $b$ and reconstruction (a),(b). With $n = e_{0}$ one has $n^{a} = δ_{0}^{a}$ , $n_{a} = - δ_{a}^{0}$ , so $e_{0} = F_{00} = 0$ , $e_{i} = F_{i 0}$ and $b_{0} = (* F)_{00} = 0$ , $b_{i} = (* F)_{i 0} = - \frac{1}{2} ϵ_{i j k} F_{j k}$ by step 1.2. Hence $e (n) = e_{0} = 0$ and $b (n) = b_{0} = 0$ . Antisymmetry gives $F_{0 i} = - F_{i 0} = - e_{i}$ , and inverting the relation $b_{i} = - \frac{1}{2} ϵ_{i j k} F_{j k}$ with the identity $ϵ_{i j k} ϵ_{i j l} = 2 δ_{k l}$ gives $F_{i j} = - ϵ_{i j k} b_{k}$ . Comparing components with $n^{♭} \land e$ (whose only nonzero components are $(n^{♭} \land e)_{0 i} = - e_{i}$ ) and with $* (n^{♭} \land b)$ (whose only nonzero components are $* (n^{♭} \land b)_{i j} = ϵ_{i j k} b_{k}$ , computed as in step 1.2) gives $F = n^{♭} \land e - * (n^{♭} \land b)$ , and $* (* F) = - F$ by step 1.2. Identifying the horizontal covectors $e, b$ with their metric dual vectors $e^{♯}, b^{♯} \in n^{⊥}$ , the map $F \mapsto (e^{♯}, b^{♯})$ is thus a linear isomorphism from the six-dimensional space of alternating bilinear forms onto $n^{⊥} \oplus n^{⊥}$ , with inverse given by these component formulae.

3.1step 2.1given

Invariants (c). Raising indices with the diagonal metric, $F^{0 i} = e_{i}$ , $F^{i 0} = - e_{i}$ and $F^{i j} = - ϵ_{i j k} b_{k}$ , while $(* F)^{0 i} = b_{i}$ , $(* F)^{i 0} = - b_{i}$ and $(* F)^{i j} = ϵ_{i j k} e_{k}$ ; counting both orders in each antisymmetric pair, $\frac{1}{2} F_{a b} F^{a b} = \frac{1}{2} (- 2 ∣ e ∣^{2} + 2 ∣ b ∣^{2}) = ∣ b ∣^{2} - ∣ e ∣^{2}, F_{a b} (* F)^{a b} = - 2 (e \cdot b) - 2 (b \cdot e) = - 4 (e \cdot b) .$ Evaluating the normalized alternating four-form on $(e_{0}, \dots, e_{3})$ , only the three terms $2 [F_{01} F_{23} - F_{02} F_{13} + F_{03} F_{12}] = 2 [e_{1} b_{1} + e_{2} b_{2} + e_{3} b_{3}]$ survive, so $F \land F = 2 (e \cdot b) ε$ .

4.1F3F4step 2.1step 3.1

Invariance and observer components. For a proper orthochronous inertial transition with matrix $Λ$ , [F3] gives $F'_{a b} = (Λ^{- 1})^{c}_{a} (Λ^{- 1})^{d}_{b} F_{c d}$ . The contractions $I_{1}, I_{2}$ and the coefficient of $F \land F$ are unchanged, because inverse index factors cancel against the metric factors and the oriented volume form is preserved. For the component lists $(e, b)$ in this paragraph, each frame uses its own time-axis observer, whose components in that frame are $(1, 0, 0, 0)$ ; these lists compare different observers. Under $Λ = diag (1, R)$ , $R \in S O (3)$ , the covariant law and step 2.1 give $(e', b') = (R e, R b)$ . For the passive boost along $e_{1}$ , substitution of the inverse matrix from [F4] gives $e'_{1} = e_{1}$ , $b'_{1} = b_{1}$ and $e'_{2} = γ (e_{2} + β b_{3}), e'_{3} = γ (e_{3} - β b_{2}), b'_{2} = γ (b_{2} - β e_{3}), b'_{3} = γ (b_{3} + β e_{2}) .$ For a fixed physical observer, its primed vector components instead satisfy $n' = Λ n$ , and the associated horizontal covectors transform by the covariant law; that is a distinct comparison.

4.2F3F4F5step 2.1step 3.1

Optional identification (d). Let $c > 0$ and suppose that in the frame one has $F^{i 0} = E_{i} / c$ , $F^{i j} = - ϵ_{i j k} B_{k}$ for spatial data $E, B$ . Then $F_{i 0} = η_{i i} η_{00} F^{i 0} = - E_{i} / c$ , so $e_{i} = - E_{i} / c$ ; and $b_{i} = - \frac{1}{2} ϵ_{i j k} (- ϵ_{j k l} B_{l}) = B_{i}$ by step 2.1. Substituting into step 3.1 gives $\frac{1}{2} F_{a b} F^{a b} = ∣ B ∣^{2} - ∣ E ∣^{2} / c^{2}$ , $F_{a b} (* F)^{a b} = 4 (E \cdot B) / c$ and $F \land F = - 2 (E \cdot B) ε / c$ , so $κ = (F \land F)_{0123} = - 2 E \cdot B / c$ and $E \cdot B = - c κ / 2$ . Under the passive boost along $e_{1}$ the component law $F^{' a b} = Λ^{a}_{c} Λ^{b}_{d} F^{c d}$ of [F3] and [F4], evaluated directly on the six components of $F$ , gives $F^{' 10} = E_{1} / c$ , $F^{' 20} = γ (E_{2} / c - β B_{3})$ , $F^{' 30} = γ (E_{3} / c + β B_{2})$ , $F^{' 23} = - B_{1}$ , $F^{' 13} = γ (B_{2} + β E_{3} / c)$ and $F^{' 12} = γ (- B_{3} + β E_{2} / c)$ ; translating back by the identification and writing $v = β c$ this is exactly $E_{1}^{'} = E_{1}$ , $B_{1}^{'} = B_{1}$ , $E_{⊥}^{'} = γ (E + v \times B)_{⊥}$ , $B_{⊥}^{'} = γ (B - v \times E / c^{2})_{⊥}$ , which are the stated vector formulae under the right-handed cross product [F5]. At $β = 0$ , $γ = 1$ and the identities are immediate.

5.1step 3.1step 4.1

The invariant quadratic polynomials. Let $P$ be a homogeneous quadratic polynomial in the six components of $F$ that is invariant under all proper orthochronous maps; by the isomorphism of step 2.1, $P$ is a quadratic form in $(e, b)$ , say $P = e^{T} A e + b^{T} B b + e^{T} C b$ with $A, B$ symmetric. Invariance under the rotations $R \oplus R$ of step 4.1 means $R^{T} A R = A$ , $R^{T} B R = B$ and $R^{T} C R = C$ for every $R \in S O (3)$ . The half-turns $R = diag (1, - 1, - 1)$ and $R = diag (- 1, - 1, 1)$ are proper rotations, and since $R$ is diagonal the conditions read $R_{i i} M_{i j} R_{j j} = M_{i j}$ entrywise; together they force every off-diagonal entry of $A$ and of $B$ to vanish, while the cyclic permutation matrix (a proper rotation) forces the surviving diagonal entries to be equal, so $A = a I_{3}$ and $B = b I_{3}$ . The same two half-turns force every off-diagonal entry of $C$ to vanish, and the cyclic permutation forces its diagonal entries to agree, so $C = c I_{3}$ . Hence $P = a ∣ e ∣^{2} + b ∣ b ∣^{2} + c (e \cdot b)$ for real $a, b, c$ . Invariance under the boost of step 4.1 applied to $e = (0, 1, 0)$ , $b = 0$ , which is sent to $e^{'} = (0, γ, 0)$ , $b^{'} = (0, 0, γ β)$ , gives $a = a γ^{2} + b γ^{2} β^{2}$ , i.e. $(a + b) γ^{2} β^{2} = 0$ , hence $a + b = 0$ for $β \neq 0$ (and by continuity of the polynomial identity in $β$ also for $β = 0$ ). Thus $P = a (∣ e ∣^{2} - ∣ b ∣^{2}) + c (e \cdot b) = - a I_{1} - \frac{c}{4} I_{2}$ , so $I_{1}$ and $I_{2}$ span the invariant quadratic polynomials. They are linearly independent: at $b = 0$ , $e \neq 0$ one has $I_{1} = - ∣ e ∣^{2} \neq 0$ , $I_{2} = 0$ , and at $e = b \neq 0$ one has $I_{1} = 0$ , $I_{2} = - 4 ∣ e ∣^{2} \neq 0$ . Finally $F \land F = 2 (e \cdot b) ε = - \frac{1}{2} I_{2} ε$ , and $I_{2} = F_{a b} (* F)^{a b}$ changes sign if the orientation is reversed with $η$ and $F$ held fixed, because then $ε$ is replaced by $- ε$ and hence $* F$ by $- * F$ .

6.1step 2.1step 3.1step 5.1step 4.2∎

Clauses (a) and (b) are steps 1.2 and 2.1, clause (c) is steps 3.1 and 5.1, and clause (d) is step 4.2. The proof is finite-dimensional linear algebra on the oriented Lorentz vector space with the stated index conventions; the only choices are the finite Gram–Schmidt and orientation adjustments of step 1.1; clause (d) is an optional identification of components with fields of the electromagnetic framework and the theorem asserts no field equation, no dynamics, no measurement, no energy condition and no transformation beyond the stated component law. No choice principle is used.

Hodge decomposition of an alternating two-form for an observer, invariants and the optional field identification

Statement

Facts & Assumptions

Proof

Depends on

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