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The orthogonal observer decomposition of a symmetric two-tensor and its uniqueness

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold, let $U \subseteq M$ be a nonempty open subset, let $T$ be a smooth symmetric covariant $2$ -tensor field on $U$ and let $n$ be a smooth observer congruence on $U$ , and let $μ$ , $J$ , $Φ$ , $S$ and $h$ be as in The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress. Then:

(a) Smoothness and elementary properties. $μ$ is a smooth function, $J$ and $Φ$ are smooth vector fields and $S$ is a smooth symmetric contravariant $2$ -tensor field on $U$ ; moreover

$n_{a} Φ^{a} = 0, S^{a b} n_{b} = 0, μ = - J^{a} n_{a} .$

(b) Orthogonal decomposition. The identity of smooth tensor fields

$T^{a b} = μ n^{a} n^{b} + n^{a} Φ^{b} + n^{b} Φ^{a} + S^{a b}$

holds on $U$ , together with the contraction identity $T^{a b} n_{a} = - μ n^{b} - Φ^{b}$ .

(c) Uniqueness. If $μ^{'}$ is a smooth function, $Ψ$ a smooth vector field with $Ψ^{a} n_{a} = 0$ and $S^{'}$ a smooth symmetric contravariant $2$ -tensor field with $S^{' a b} n_{b} = 0$ on $U$ , and if

$T^{a b} = μ^{'} n^{a} n^{b} + n^{a} Ψ^{b} + n^{b} Ψ^{a} + S^{' a b},$

then $μ^{'} = μ$ , $Ψ = Φ$ and $S^{'} = S$ .

(d) Values on the rest space and frame components. For every $p \in U$ and every $v \in n_{p}^{⊥}$ ,

$T_{p} (n_{p}, n_{p}) = μ (p), T_{p} (n_{p}, v) = - g_{p} (Φ_{p}, v),$

The covariant tensor obtained by lowering both slots of $S_{p}$ is $S_{p}^{♭} (v, w) : = S_{p} (g_{p} (v, \cdot), g_{p} (w, \cdot))$ for $v, w \in T_{p} M$ , with components $S_{a b}^{♭} = g_{a c} g_{b d} S^{c d}$ . The restrictions of $T_{p}$ and $S_{p}^{♭}$ to $n_{p}^{⊥} \times n_{p}^{⊥}$ agree. In every ordered $g_{p}$ -orthonormal basis $(e_{0}, e_{1}, e_{2}, e_{3})$ of $T_{p} M$ with $e_{0} = n_{p}$ ,

$μ (p) = T_{00} (p), Φ^{i} (p) = T^{0 i} (p) = - T_{0 i} (p) (i = 1, 2, 3),$

and $S^{i j} (p) = T^{i j} (p)$ for $i, j \in {1, 2, 3}$ while $S^{0 a} (p) = 0$ . Finally $g_{a b} S^{a b} = g_{a b} T^{a b} + μ$ .

Facts & Assumptions

Given: The time-oriented smooth Lorentzian four-manifold $(M, g, N)$ , the open subset $U$ , the smooth symmetric covariant $2$ -tensor field $T$ on $U$ , the smooth observer congruence $n$ on $U$ , and the fields $μ$ , $J$ , $Φ$ , $S$ , $h$ of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress.

[L1]

Every $v \in T_{p} M$ has the unique decomposition $v = - g_{p} (v, n_{p}) n_{p} + h_{p} (v)$ with $h_{p} (v) \in n_{p}^{⊥}$ ; $h_{p}$ is linear, $h_{p} \circ h_{p} = h_{p}$ , its kernel is $R n_{p}$ , its image is $n_{p}^{⊥}$ , and $g_{p} (h_{p} (v), h_{p} (w)) = g_{p} (v, w) + g_{p} (v, n_{p}) g_{p} (n_{p}, w)$ (The observer rest-space decomposition and the spatial projection, Observers, rest spaces and the spatial projection at an event).

[L2]

The fields $n$ , $g$ , $g^{- 1}$ and $T$ are smooth, and a tensor field defined from them by the standard index operations and pointwise contractions is smooth: its chart components are obtained from the smooth component functions of $n$ , $g$ , $g^{- 1}$ and $T$ by sums of products, and raising and lowering are the contractions $g^{a c} g^{b d} T_{c d}$ and $g_{a c} g_{b d} X^{c d}$ with the smooth inverse-metric components (Tensor fields of type (r,s) on a Lorentzian four-manifold and their components, Smooth covariant 2-tensor fields on a smooth manifold, The inverse metric, musical isomorphisms and index raising, $C^{r}$ and smooth maps between smooth manifolds).

[L3]

$g_{p} (n_{p}, n_{p}) = - 1$ at every $p \in U$ (Observers, rest spaces and the spatial projection at an event, Observer congruences, their observer curves and adapted charts).

[L4]

The inverse-metric components satisfy $g^{a c} g_{c b} = δ^{a}_{b}$ ; raising and lowering an index with $g^{- 1}$ and $g$ are inverse linear operations on each tensor slot (The inverse metric, musical isomorphisms and index raising).

[L5]

At every $p \in U$ there is a $g_{p}$ -orthonormal ordered basis $(e_{0}, e_{1}, e_{2}, e_{3})$ of $T_{p} M$ ; when $e_{0} = n_{p}$ it satisfies $g_{p} (e_{0}, e_{0}) = - 1$ and $g_{p} (e_{i}, e_{j}) = δ_{i j}$ for $i, j \in {1, 2, 3}$ , and the components of a tensor in that basis are its contractions with the dual basis (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1), Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Tensor fields of type (r,s) on a Lorentzian four-manifold and their components).

Proof

technique · direct; compute the projections of the tensor in index form, read off the four pieces, and prove uniqueness by contracting the putative decomposition against $n$ and its orthogonal complement

1.1L1L2L3L4algebra

(Smoothness and the elementary identities.) Writing components in a chart, $μ = T_{a b} n^{a} n^{b}$ is a sum of smooth products, and $J^{a} = - g^{a b} T_{b c} n^{c}$ is a sum of smooth products; hence both are smooth, and $Φ = J - μ n$ and $S^{a b} = h^{a}_{c} h^{b}_{d} T^{c d}$ are smooth by [L2]; symmetry of $T^{c d}$ gives symmetry of $S^{a b}$ by interchanging the two contracted indices. The field $h$ of The observer split of a symmetric two-tensor: energy density, energy-momentum current, flux and stress has $h^{a}_{b} n^{b} = n^{a} + n^{a} (n \cdot n) = 0$ and $n_{a} h^{a}_{b} = n_{b} + (n_{a} n^{a}) n_{b} = 0$ by [L3], so $n_{a} Φ^{a} = n_{a} J^{a} - μ (n \cdot n) = n_{a} J^{a} + μ$ and $S^{a b} n_{b} = h^{a}_{c} (n_{b} h^{b}_{d}) T^{c d} = 0$ ; finally $n_{a} J^{a} = - T^{a b} n_{a} n_{b} = - μ$ .

2.1step 1.1L3L4algebra

(The decomposition.) Expanding $h^{a}_{c} = δ^{a}_{c} + n^{a} n_{c}$ and $h^{b}_{d} = δ^{b}_{d} + n^{b} n_{d}$ in $S^{a b} = h^{a}_{c} h^{b}_{d} T^{c d}$ gives $S^{a b} = T^{a b} + n^{a} T^{c b} n_{c} + n^{b} T^{a d} n_{d} + n^{a} n^{b} T^{c d} n_{c} n_{d} = T^{a b} - n^{a} J^{b} - n^{b} J^{a} + μ n^{a} n^{b}$ , using $T^{c b} n_{c} = - J^{b}$ , $T^{a d} n_{d} = - J^{a}$ and $T^{c d} n_{c} n_{d} = μ$ . Substituting $J = μ n + Φ$ and solving for $T^{a b}$ gives $T^{a b} = μ n^{a} n^{b} + n^{a} Φ^{b} + n^{b} Φ^{a} + S^{a b},$ because $n^{a} (μ n^{b} + Φ^{b}) + n^{b} (μ n^{a} + Φ^{a}) - μ n^{a} n^{b} = μ n^{a} n^{b} + n^{a} Φ^{b} + n^{b} Φ^{a}$ . Contracting this identity with $n_{a}$ and using [L3] gives $T^{a b} n_{a} = - μ n^{b} - Φ^{b}$ , since $n_{a} Φ^{a} = 0$ and $n_{a} n^{a} = - 1$ .

3.1step 1.1step 2.1L3L4algebra

(Uniqueness.) Let $μ^{'}, Ψ, S^{'}$ be as in (c). Contracting the assumed decomposition with $n_{a} n_{b}$ kills every term containing $Ψ$ (because $Ψ^{a} n_{a} = 0$ ) or $S^{'}$ (because $S^{' a b} n_{b} = 0$ ) and gives $μ^{'} = T^{a b} n_{a} n_{b} = μ$ by [L3]. Contracting with $n_{a}$ alone and using $n_{a} n^{a} = - 1$ gives $T^{a b} n_{a} = - μ^{'} n^{b} - Ψ^{b}$ , so $Ψ^{b} = - T^{a b} n_{a} - μ^{'} n^{b} = J^{b} - μ n^{b} = Φ^{b}$ , where $- T^{a b} n_{a} = J^{b}$ is the definition of $J$ and $Φ = J - μ n$ . Solving the decomposition for $S^{'}$ now gives $S^{' a b} = T^{a b} - μ n^{a} n^{b} - n^{a} Φ^{b} - n^{b} Φ^{a} = S^{a b}$ by step 2.1.

4.1step 2.1L3L4L5algebra∎

(Rest-space values and frame components.) By definition $μ = T (n, n)$ . For $v \in n_{p}^{⊥}$ , step 2.1 gives $T_{p} (n_{p}, v) = T^{a b} n_{a} v_{b} = (- μ n^{b} - Φ^{b}) v_{b} = - g_{p} (Φ_{p}, v)$ , since $n^{b} v_{b} = 0$ . For $v, w \in n_{p}^{⊥}$ , contract the decomposition of step 2.1 with $v_{a} w_{b}$ , where $v_{a} = g_{a c} v^{c}$ and $w_{b} = g_{b d} w^{d}$ . Every term containing $n^{a}$ or $n^{b}$ vanishes, giving $T_{p} (v, w) = S^{a b} v_{a} w_{b} = S_{p}^{♭} (v, w)$ . In an orthonormal basis with $e_{0} = n_{p}$ , [L5] gives $n^{a} = δ_{0}^{a}$ and $n_{a} = - δ_{a}^{0}$ , so $h^{0}_{b} = δ_{b}^{0} + n^{0} n_{b} = δ_{b}^{0} - δ_{b}^{0} = 0$ and $h^{i}_{b} = δ_{b}^{i}$ ; consequently $S^{0 a} = 0$ , $S^{i j} = T^{i j}$ and $Φ^{i} = J^{i} - μ n^{i} = J^{i} = - T^{i b} n_{b} = - T^{i 0} n_{0} = T^{i 0} = - T_{0 i}$ , while $μ = T_{00}$ . Contracting the decomposition with $g_{a b}$ gives $g_{a b} S^{a b} = g_{a b} T^{a b} - μ (n \cdot n) - 2 g (n, Φ) = g_{a b} T^{a b} + μ$ ; that is, $g_{a b} S^{a b} = g_{a b} T^{a b} + μ$ .

Remarks

Minimal hypotheses. Only symmetry of $T$ and the congruence property $g (n, n) = - 1$ are used; no equation of motion enters the pointwise algebraic decomposition; smoothness of the fields uses the smooth inputs and inverse metric in [L2]. The sign convention is $J = - ♯ (T (n, \cdot))$ and $T (n, v) = - g (Φ, v)$ on the rest space.
Origin of the names. The names measured energy density, energy flux and stress are attached to $μ$ , $Φ$ and $S$ only through the framework's physical reading on this page; this lemma proves the algebraic identities alone.

The orthogonal observer decomposition of a symmetric two-tensor and its uniqueness

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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