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The perfect-fluid energy equation recovered from the local energy balance

Example

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold with metric-selected connection $\nabla$ , let $U \subseteq M$ be a nonempty open subset and let $(ε, p, n)$ be a perfect-fluid model on $U$ with stress-energy tensor $T^{a b} = (ε + p) n^{a} n^{b} + p g^{a b}$ (The perfect-fluid matter model: rest energy density, pressure, four-velocity and stress-energy); suppose $\nabla^{a} T_{a b} = 0$ on $U$ . Take the observer congruence of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress to be the fluid field $n$ itself, and write $D = n^{a} \nabla_{a}$ and $θ = \nabla_{a} n^{a}$ . Then:

For this congruence $μ = ε$ , $J = ε n$ , $Φ = 0$ and $S^{a b} = p h^{a b}$ with $h^{a b} = g^{a b} + n^{a} n^{b}$ .
$T^{a b} \nabla_{a} n_{b} = p θ$ , so the local balance of Divergence of the measured energy-momentum current and the finite-chain balance identity clause (b) reads

$\nabla_{a} (ε n^{a}) = - p θ, equivalently D ε + (ε + p) θ = 0.$

This is exactly the energy equation of Divergence-free dust is exactly the dust system, and divergence-free perfect fluid is exactly the fluid equations clause (b) under the hypothesis $\nabla^{a} T_{a b} = 0$ ; the momentum (Euler) equation is the complementary projection of $\nabla^{a} T_{a b} = 0$ and is not implied by the energy balance alone.

Facts & Assumptions

Given: The time-oriented smooth Lorentzian four-manifold $(M, g, N)$ with metric-selected connection $\nabla$ , the open subset $U$ , the perfect-fluid model $(ε, p, n)$ on $U$ with tensor $T^{a b} = (ε + p) n^{a} n^{b} + p g^{a b}$ , and the hypothesis $\nabla^{a} T_{a b} = 0$ on $U$ .

[L1]

A perfect-fluid model consists of smooth functions $ε, p : U \to R$ and a smooth future-directed unit timelike field $n$ with $g (n, n) = - 1$ ; its stress-energy tensor is $T^{a b} = (ε + p) n^{a} n^{b} + p g^{a b}$ , and no positivity of $ε$ or $p$ is required (The perfect-fluid matter model: rest energy density, pressure, four-velocity and stress-energy, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L2]

The measured fields of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress satisfy $μ = T_{a b} n^{a} n^{b}$ , $J^{a} = - T^{a b} n_{b}$ , $Φ = J - μ n$ , $S^{a b} = h^{a}_{c} h^{b}_{d} T^{c d}$ and the decomposition $T^{a b} = μ n^{a} n^{b} + n^{a} Φ^{b} + n^{b} Φ^{a} + S^{a b}$ (The orthogonal observer decomposition of a symmetric two-tensor and its uniqueness clause (b)).

[L3]

With $\nabla^{a} T_{a b} = 0$ , the identities $\nabla_{a} J^{a} = - T^{a b} \nabla_{a} n_{b}$ and $D μ + μ θ + \nabla_{a} Φ^{a} = \nabla_{a} J^{a}$ hold (Divergence of the measured energy-momentum current and the finite-chain balance identity clauses (a) and (b)).

[L4]

$θ = \nabla_{a} n^{a} = h^{a b} \nabla_{a} n_{b}$ (Acceleration, expansion and the along-congruence derivative of an observer congruence clause (b), Spatiality and curve identity of the acceleration, the divergence form of the expansion, and the along-congruence derivative clause (c)); and $n^{b} \nabla_{a} n_{b} = 0$ , because $\nabla$ is $g$ -compatible and $n \cdot n = - 1$ is constant, so $0 = \nabla_{a} (n \cdot n) = 2 n^{b} \nabla_{a} n_{b}$ (Metric compatibility, torsion freedom and the metric-selected connection class clause (a), Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields).

[L5]

$\nabla^{a} T_{a b} = 0$ holds if and only if $D ε + (ε + p) θ = 0$ and $(ε + p) a^{b} + h^{b a} \partial_{a} p = 0$ , where $a^{b} = n^{a} \nabla_{a} n^{b}$ (Divergence-free dust is exactly the dust system, and divergence-free perfect fluid is exactly the fluid equations clause (b)).

[L6]

The extension of $\nabla$ to tensor fields satisfies the product rule and $\nabla g^{- 1} = 0$ , so indices may be raised and lowered across $\nabla$ (Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields clause (c), The inverse metric, musical isomorphisms and index raising).

Verification

technique · direct; substitute the perfect-fluid tensor into the definitions of the measured fields, contract the divergence with the fluid field, and compare with the known conservation equivalence

1.1L1L4given

(The congruence data.) By [L1], $n$ is a smooth future-directed unit timelike field, hence an observer congruence; its acceleration is $a = \nabla_{n} n$ , its expansion is $θ = \nabla_{a} n^{a}$ , and $n^{b} \nabla_{a} n_{b} = 0$ by [L4].

2.1step 1.1L2L6algebra

(Measured fields.) By [L6] and [L2], $μ = T_{a b} n^{a} n^{b} = (ε + p) (n \cdot n)^{2} + p (n \cdot n) = ε$ , and $J^{a} = - T^{a b} n_{b} = - (ε + p) n^{a} (n \cdot n) - p n^{a} = ε n^{a}$ , so $J = ε n$ and $Φ = J - μ n = 0$ . Similarly $S^{a b} = h^{a}_{c} h^{b}_{d} [(ε + p) n^{c} n^{d} + p g^{c d}] = p h^{a b}$ , since $h^{a}_{c} n^{c} = 0$ . This proves clause 1.

2.2step 1.1L4L6algebra

(The source term.) By [L6], $T^{a b} \nabla_{a} n_{b} = (ε + p) n^{a} n^{b} \nabla_{a} n_{b} + p g^{a b} \nabla_{a} n_{b} = 0 + p θ$ by [L4], where the first term vanishes because $n^{b} \nabla_{a} n_{b} = 0$ .

3.1step 2.1step 2.2L3L5algebra∎

(The balance.) By [L3], $\nabla_{a} J^{a} = - T^{a b} \nabla_{a} n_{b} = - p θ$ , while $\nabla_{a} J^{a} = \nabla_{a} (ε n^{a}) = n^{a} \nabla_{a} ε + ε \nabla_{a} n^{a} = D ε + ε θ$ by step 2.1; hence $D ε + (ε + p) θ = 0$ . This is the energy equation of [L5] under the present hypothesis $\nabla^{a} T_{a b} = 0$ ; the Euler equation of [L5] is the projection orthogonal to $n$ and is not derived from the balance used here. This proves clauses 2 and 3.

Remarks

Consistency across pages. The example connects the new balance identity of Divergence of the measured energy-momentum current and the finite-chain balance identity with the conservation equivalence already proved in Divergence-free dust is exactly the dust system, and divergence-free perfect fluid is exactly the fluid equations; it does not reprove the latter and uses no equation of state.
No dynamics. Only the algebraic identities and the hypothesis $\nabla^{a} T_{a b} = 0$ are used; the example asserts no existence of a fluid solution and no measured value. A model that adds an equation of state restricts the admissible tuples $(ε, p, n)$ without changing the identities.

The perfect-fluid energy equation recovered from the local energy balance

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources