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The Killing energy of a static dust in flat space and its conservation across slices

Physical scope

An exact flat-space instance of the framework's Killing energy: a constant static dust with the length-normalized translation Killing field of the identity chart, the current three-form of its Killing current, the value of the Killing energy on three-chains carried by constant-time slices, and its conservation up to the side flux. The example is an exact model calculation under the framework's unit reading; it asserts no measured value and no empirical mass, and the identification of its numbers with any physical energy requires separate empirical premises.

Example

Let M=R4 carry the identity chart, the flat metric g=diag⁡(−1,1,1,1), the time orientation N=∂0 and the chart orientation, with metric-selected connection ∇ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Metric compatibility, torsion freedom and the metric-selected connection class). Let ξ:=∂0, let ρ>0 be constant with the framework reading ρ=ε in J m−3 (rest-mass density ρm=ρ/c2), let u:=∂0 and let Tab:=ρ uaub, the tensor of the dust model (ρ,u) (The dust matter model: rest density, four-velocity, stress-energy and dust system). Then:

  1. ξ is a Killing field on M that is everywhere timelike and future-directed, with X=−g(ξ,ξ)=1 and congruence n=ξ; and the dust system holds, so ∇aTab=0 on M.
  2. The Killing current of The timelike Killing current and the charge of a three-chain is Jξa=Tabξb=−ρ ∂0a, and its current 3-form is ιJξvolg=−ρ dx1∧dx2∧dx3.
  3. For every finite smooth singular 3-chain Σ of M whose simplices have chart velocity vectors (∂1,∂2,∂3) at every parameter point — in particular for simplices parametrized by the spatial chart coordinates on a constant-x0 slice — the Killing energy of Σ is

Qξ(Σ)=ρ∫Σdx1∧dx2∧dx3,

the integral of the model's rest-energy-density form over the chain in the framework's unit reading. 4. For every finite smooth singular 4-chain C of M with ∂C=Σ2−Σ1+B,

Qξ(Σ2)−Qξ(Σ1)=−Qξ(B);

in particular Qξ(Σ2)=Qξ(Σ1) whenever the velocity vectors of every simplex of B span a hyperplane containing ∂0, since then the side flux vanishes. 5. Consequently this prescribed flat-source calculation realizes a conserved Killing charge: it is ρ times the coordinate-volume reading of the slice region, read as the model's rest energy. The prescribed nonzero dust source is not asserted to solve the gravitational field law on this flat metric. No measured value is asserted.

Facts & Assumptions

Given: The flat Lorentzian four-manifold (R4,g) with identity chart, g=diag⁡(−1,1,1,1), chart orientation, metric-selected connection ∇, volume form volg and current flux Φ; the vector fields ξ=u=∂0; the constant ρ>0; the tensor Tab=ρ uaub.

[L1]

On the identity chart the Christoffel symbols vanish, ∇ is the coordinate connection, volg=dx0∧dx1∧dx2∧dx3, and for a smooth vector field J, ιJvolg=∑a(−1)aJa dx0∧⋯dxa^⋯∧dx3 (The metric volume form, current flux through 3-faces and finite-chain charges, The derivative of the metric density and the chart formula for the divergence of a vector field, Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields, Interior product of a form by a vector field, The wedge product of differential forms).

[L2]

ξ=∂0 has constant components in the identity chart, so it is Killing, g(ξ,ξ)=−1, g(ξ,N)=−1 and X=1; hence n=ξ is a smooth observer congruence (Killing fields of a Lorentzian metric, Killing fields and conserved currents, and on-shell conservation of the Hilbert stress-energy tensor clause (a), Observer congruences, their observer curves and adapted charts, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L3]

(ρ,∂0) is a dust model with tensor Tab=ρ uaub; the dust system holds for constant ρ and constant u, and it is equivalent to ∇aTab=0 (The dust matter model: rest density, four-velocity, stress-energy and dust system, Divergence-free dust is exactly the dust system, and divergence-free perfect fluid is exactly the fluid equations clause (a)).

[L4]

The Killing current and Killing energy are Jξa=Tabξb and Qξ(Σ)=−ΦJξ(Σ); if ξ is timelike with X=−g(ξ,ξ) and n=ξ/X, then Jξ=−XJ with J the measured current (The timelike Killing current and the charge of a three-chain clauses (a)-(c), The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress).

[L5]

If T is divergence-free and ξ is timelike Killing on the oriented domain, Jξ=−XJ is divergence-free by Kinematics of a timelike Killing congruence and conservation of the redshift-weighted energy current clause (d). Finite-chain Stokes gives the Killing energy balance Qξ(Σ2)−Qξ(Σ1)=−Qξ(B) for every finite smooth singular 4-chain with ∂C=Σ2−Σ1+B (Divergence, Stokes flux balance and the non-null face formula for a Lorentzian current clause (b)).

Verification

technique · direct; check the Killing and dust hypotheses, compute the Killing current and its three-form in the identity chart, evaluate the flux on slice chains, and apply the divergence-free Killing-current lemma and finite-chain Stokes
1.1L2L3given

(Hypotheses.) By [L2], ξ is a Killing field that is everywhere timelike and future-directed, with X=1 and n=ξ; by [L3] the constant dust (ρ,∂0) satisfies the dust system and ∇aTab=0. Hence the Killing-current construction applies and the conservation and symmetry hypotheses of [L5] are met; no gravitational field equation is used.

2.1step 1.1L1L4algebra

(The Killing current and its three-form.) By [L1], ξb=ηb0=−δ0b, so Jξa=Tabξb=ρ uaubξb=ρ ∂0a(u0ξ0)=ρ ∂0a(1⋅(−1))=−ρ ∂0a. Since ι∂0volg=dx1∧dx2∧dx3 by [L1], linearity of the interior product gives ιJξvolg=−ρ dx1∧dx2∧dx3. This proves clause 2.

3.1step 2.1L1L4algebra

(Killing energy on slice chains.) Let σ be a smooth singular 3-simplex of M whose chart velocity vectors at every parameter point are σ∗∂i=∂i for i=1,2,3 (so that its image lies in a constant-x0 slice and it is parametrized by the slice coordinates). Then σ∗(dx1∧dx2∧dx3)=dt1∧dt2∧dt3, so by the definition of the flux through a simplex as the integral of the pulled-back current form, ΦJξ(σ)=−ρ∫Δ3dt1∧dt2∧dt3. By additivity of the flux over finite chains the same formula holds for every finite smooth singular 3-chain Σ with the stated velocity vectors, and Qξ(Σ)=−ΦJξ(Σ)=ρ∫Σdx1∧dx2∧dx3 by [L4]. This proves clause 3.

4.1step 2.1L1L4L5algebra∎

(Conservation and side flux.) By [L5] applied to the present Killing field, Qξ(Σ2)−Qξ(Σ1)=−Qξ(B) for every finite smooth singular 4-chain C with ∂C=Σ2−Σ1+B. If the velocity vectors of every simplex of B span a hyperplane containing ∂0, then the four arguments Jξ,v1,v2,v3 of volg are linearly dependent — Jξ is a constant multiple of ∂0 — so ιJξvolg vanishes on them and ΦJξ(B)=0, whence Qξ(B)=0 and Qξ(Σ2)=Qξ(Σ1). This proves clause 4 and completes the verification.

Remarks

  • Reading of the value. In the framework's stress-energy reading, T has components in J m−3. Here ξ=∂0 is length-normalized because x0=ct is a length coordinate, and X=1; hence this Qξ(Σ) is read in joules as the rest energy assigned to the chain. The parameter ρ is the dust model's rest-energy-density parameter in the SI reading; ρm=ρ/c2 is the corresponding mass density. The identification of ρ∫Σdx1∧dx2∧dx3 with a physical mass or energy requires separate empirical premises and is not asserted.

  • Why conservation needs the side flux. The current Jξ has no spatial components, so a side face swept out along ∂0 carries no flux; a side face tilted so that its velocity vectors span a hyperplane containing ∂0 also carries none, by the linear-dependence argument of step 4.1. For a general side chain the flux need not vanish and the equality of clause 4 is a balance identity of bookkeeping, exactly as in Divergence, Stokes flux balance and the non-null face formula for a Lorentzian current.

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