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Explicit local conduction-energy calculation

Example

Let $M = R^{4}$ with global length coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ and metric components $g_{a b} = η_{a b} = diag (- 1, 1, 1, 1)$ in the coordinate frame $\partial_{a} : = \partial / \partial x^{a}$ . Use this metric to raise and lower indices and its Levi–Civita connection $\nabla$ . Take orientation $d x^{0} \land d x^{1} \land d x^{2} \land d x^{3}$ , future time-orientation $N = \partial_{0}$ , and the positive parameters $c, μ_{0}$ of Smooth formal matter, EM tensors and observer quantities, with $x^{0} = c t$ . Choose constants with SI units $[E_{0}] = V m^{- 1}$ , $[χ] = S m^{- 1}$ , and $[ϵ_{i}] = [p_{0}] = J m^{- 3}$ , with $E_{0} \neq 0$ , $χ > 0$ , $ϵ_{i} > 0$ , and $p_{0} \in R$ . Let $I = (- ϵ_{i} / (2 χ E_{0}^{2}), ϵ_{i} / (2 χ E_{0}^{2}))$ , which has units of seconds; put $t = x^{0} / c$ and restrict all fields to the open spacetime domain $D : = {x \in R^{4} : t (x) \in I} ≅ R^{3} \times I$ . The maps $E, B : D \to R^{3}$ below give the spatial components relative to the positively oriented orthonormal frame $(\partial_{0}, \partial_{1}, \partial_{2}, \partial_{3})$ , with $e_{x} = (1, 0, 0)$ . Define

$U = c \partial_{0}, E = E_{0} e_{x}, B = 0, ϵ (t) = ϵ_{i} + χ E_{0}^{2} t, p = p_{0}, σ = 0, q^{a} = (0, χ E_{0}, 0, 0) .$

Let $F$ be the antisymmetric covariant two-tensor with $F_{01} = - E_{0} / c$ and all other independent components zero, and set $j = q$ . All the displayed scalar, vector and tensor fields are smooth on $D$ . The interval gives $ϵ_{i} / 2 < ϵ (t) < 3 ϵ_{i} / 2$ throughout $D$ . The imposed field is a fixed background for the matter calculation; the nonzero current is not claimed to solve its Maxwell equation.

Facts & Assumptions

Given: The stated fields on $R^{3} \times I$ , with $χ, ϵ_{i} > 0$ , $E_{0} \neq 0$ , and $U = (c, 0, 0, 0)$ .

[F1]

Smooth formal matter, EM tensors and observer quantities: $T_{m}^{a b} = (ϵ + p) U^{a} U^{b} / c^{2} + p g^{a b}$ , $E_{U}^{a} = F^{a}_{b} U^{b}$ , and the formal exchange force is $f^{b} = F^{b}_{c} j^{c}$ .

[F2]

Perfect-fluid exchange projections and conduction heating: if $q \cdot U = 0$ and $q = χ E_{U}$ , then the comoving exchange term is $- U_{b} f^{b} = χ g (E_{U}, E_{U})$ .

[A1]

The current is imposed as $j = q$ with $q^{a} U_{a} = 0$ ; the fixed $F$ is not asserted to satisfy the sourced Maxwell equation.

Verification

1.1F1A1algebra

The constant diagonal metric has inverse $g^{a b} = η^{a b} = diag (- 1, 1, 1, 1)$ and zero Levi–Civita Christoffel symbols. Thus $U_{a} = (- c, 0, 0, 0)$ , $g (U, U) = - c^{2}$ , and $g (U, N) = - c < 0$ , so $U$ is future timelike with the required normalization. The chosen $q$ is orthogonal to $U$ , and $E_{U}^{1} = F^{1}_{0} U^{0} = (E_{0} / c) c = E_{0}$ , with its other components zero. Thus $q = χ E_{U}$ . The stress tensor has $T_{m}^{00} = ϵ (t)$ , $T_{m}^{i i} = p_{0}$ , and zero off-diagonal components, so $\nabla_{a} T_{m}^{a 0} = \partial_{0} ϵ = (1 / c) d ϵ / d t = χ E_{0}^{2} / c$ while $\nabla_{a} T_{m}^{a i} = 0$ .

2.1F1F2step 1.1algebra

Raising the first index with $η$ gives $F^{0}_{1} = E_{0} / c$ and $F^{i}_{1} = 0$ for $i = 1, 2, 3$ . Hence $f^{0} = F^{0}_{1} q^{1} = χ E_{0}^{2} / c$ and $f^{i} = 0$ . The component equations verify $\nabla_{a} T_{m}^{a b} = f^{b}$ , and [F2] gives $- U_{b} f^{b} = χ E_{0}^{2} > 0$ .

3.1F1A1algebra∎

Since $F$ is constant, $\nabla_{a} F^{a b} = 0$ , whereas $- μ_{0} j^{1} = - μ_{0} χ E_{0} \neq 0$ . Thus the matter exchange and positive heating calculation is conditional on the imposed background and is not a sourced Maxwell solution.

Remarks

The calculation is local and includes no equation of state, entropy balance, electromagnetic backreaction, or global fluid solution. The units $[χ] = S m^{- 1}$ and $[E_{0}] = V m^{- 1}$ make $χ E_{0}^{2}$ a power density, but the example reports no measurement.

Explicit local conduction-energy calculation

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources