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Tensors And Observers — Examples

1 · Prerequisites

2 · Summary

These examples and counterexamples work the objects of gr-tensors-and-observers on explicit components. On the tensor side, a mixed type $(1, 1)$ field on flat four-space is transformed from the identity chart to an affine chart in which one Jacobian entry is constant and the offending component becomes the smooth function $y^{1} - y^{0}$ ; the trace computed in the two charts is verified to agree, and the example records that the individual diagonal components are not invariant (A mixed tensor field: components in two charts and the invariance of its trace).

On the observer side, the constant boosted fields $E_{0} = γ (\partial_{0} + β \partial_{1})$ , $E_{1} = γ (β \partial_{0} + \partial_{1})$ , $E_{2} = \partial_{2}$ , $E_{3} = \partial_{3}$ on flat four-space are verified to be a smooth future-directed tetrad field, and the chart $u = (x^{0}, x^{1} - β x^{0}, x^{2}, x^{3})$ is exhibited as an adapted chart for the parallel congruence $n = E_{0}$ in which $n = γ \partial_{u^{0}}$ , so an adapted chart's coordinate field need not equal the congruence (A constant boosted tetrad field on flat spacetime and an adapted chart). A second, shearing congruence $n = γ (\partial_{0} + Ω \partial_{2})$ with $Ω = λ x^{1}$ and its tetrad are verified on the strip $∣ λ x^{1} ∣ < 1 / 2$ , an adapted chart is produced with explicit inverse, and the bracket $[E_{0}, E_{1}] = - γ^{'} \partial_{0} - (γ Ω)^{'} \partial_{2}$ shows that this future-directed tetrad field is not the coordinate frame of any chart (A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame).

The observer-relative constructions are checked on the boost family. For $u = γ (n + β E_{1})$ the example computes $g (u, u) = - 1$ , $γ_{o} = - g (n, u) = γ$ , $h_{n} (u) = γ β E_{1}$ and $g (w, w) = β^{2} c^{2} = c^{2} (1 - γ^{- 2})$ , verifying the relative-velocity decomposition theorem in an explicit family (The relative-velocity decomposition of a boost, checked in an explicit tetrad); for $P = m c u$ it then computes rest mass $m (P) = m$ , energy $E_{n} (P) = m c^{2} γ$ , spatial momentum $q_{n} (P) = m c γ β E_{1}$ and verifies the mass shell $E_{n} (P)^{2} = m^{2} c^{4} + c^{2} g (q_{n} (P), q_{n} (P))$ (Observer energy, spatial momentum and mass shell of a boosted test body).

Two counterexamples delimit the definitions. The shearing tetrad's nonvanishing bracket contradicts the necessary commuting condition, refuting the universal claim that every smooth future-directed tetrad field on an open subset of flat Lorentzian four-space is the coordinate frame of a chart (A tetrad field that is not the coordinate frame of any chart). For the same congruence the flat form $α = γ (- d x^{0} + Ω d x^{2})$ satisfies $α \land d α = - γ^{2} λ d x^{0} \land d x^{1} \land d x^{2} \neq 0$ , so the rest spaces are not tangent to any local hypersurfaces: existence of an adapted chart and integrability of the rest spaces are different properties (A congruence whose rest spaces are not tangent to any local hypersurfaces).

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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A mixed tensor field: components in two charts and the invariance of its trace

Example

Let $M = R^{4}$ carry the flat Lorentzian metric $g = diag (- 1, 1, 1, 1)$ and the time orientation $N = \partial_{0}$ , so that $(M, g, N)$ is a time-oriented smooth Lorentzian four-manifold (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model); here $g$ and $N$ enter only to place the charts and the tensor field in the setting of the cited definition and take no part in the computation. Let $x = (x^{0}, x^{1}, x^{2}, x^{3})$ be the identity chart and let the chart $y = (y^{0}, y^{1}, y^{2}, y^{3})$ be defined by

$y^{0} = x^{0}, y^{1} = x^{1} + x^{0}, y^{2} = x^{2}, y^{3} = x^{3} .$

Define a type $(1, 1)$ tensor field $T$ on $M$ by its components in the $x$ chart,

$T^{0}_{1} : = 1, T^{2}_{3} : = x^{1}, T^{a}_{b} : = 0 otherwise .$

Then $T$ is a smooth type $(1, 1)$ tensor field. Its components in the chart $y$ are

$T^{' 0}_{0} = - 1, T^{' 0}_{1} = 1, T^{' 1}_{0} = - 1, T^{' 1}_{1} = 1, T^{' 2}_{3} = y^{1} - y^{0}, others 0,$

and its trace computed in the two charts agrees: it is $0$ in both, as the chart-independence of contraction requires.

Facts & Assumptions

Given: The flat Lorentzian four-manifold $(R^{4}, g, N)$ with time orientation $N = \partial_{0}$ , the charts $x$ and $y$ on $R^{4}$ , and the component data $T^{0}_{1} = 1$ , $T^{2}_{3} = x^{1}$ , all other $T^{a}_{b} = 0$ .

[L1]

A smooth type $(1, 1)$ tensor field has component functions $T^{a}_{b} (q) = T_{q} (d x^{a} ∣_{q}, \partial_{b} ∣_{q})$ in each chart, and it is smooth exactly when these components are $C^{\infty}$ in every smooth chart (Tensor fields of type (r,s) on a Lorentzian four-manifold and their components).

[L2]

In overlapping charts the components of a smooth type $(r, s)$ tensor field satisfy $T^{' a_{1} \dots a_{r}}_{b_{1} \dots b_{s}} (y (q)) = \sum_{c, d} (\prod_{i} \frac{\partial y^{a_{i}}}{\partial x^{c_{i}}}) (\prod_{j} \frac{\partial x^{d_{j}}}{\partial y^{b_{j}}}) T^{c_{1} \dots c_{r}}_{d_{1} \dots d_{s}} (x (q))$ (The component transformation law of a smooth tensor field).

[L3]

The trace $tr T$ of a smooth type $(1, 1)$ field is a smooth function whose value in every smooth chart is $\sum_{a} T^{a}_{a}$ (Contraction of a smooth tensor field is a smooth tensor field).

Verification

technique · direct

1.1givenalgebra

The two maps $y$ and $x = (y^{0}, y^{1} - y^{0}, y^{2}, y^{3})$ are mutually inverse smooth maps between open subsets of $R^{4}$ , so $x$ and $y$ are smooth charts on $R^{4}$ . Their Jacobians are constant: $\frac{\partial y^{a}}{\partial x^{c}}$ is $1$ for $(a, c) \in {(0, 0), (1, 0), (1, 1), (2, 2), (3, 3)}$ and $0$ otherwise, while $\frac{\partial x^{d}}{\partial y^{b}}$ is $1$ for $(d, b) \in {(0, 0), (1, 1), (2, 2), (3, 3)}$ and $- 1$ for $(d, b) = (1, 0)$ , and $0$ otherwise.

1.2givenL1

The components $T^{a}_{b}$ are $1$ , $x^{1}$ and $0$ , hence $C^{\infty}$ on $R^{4}$ ; by [L1] the field $T$ is a smooth type $(1, 1)$ tensor field.

2.1step 1.1L2algebra

By [L2] and step 1.1, $T^{' a}_{b} = \sum_{c, d} \frac{\partial y^{a}}{\partial x^{c}} \frac{\partial x^{d}}{\partial y^{b}} T^{c}_{d} = \frac{\partial y^{a}}{\partial x^{0}} \frac{\partial x^{1}}{\partial y^{b}} + (y^{1} - y^{0}) \frac{\partial y^{a}}{\partial x^{2}} \frac{\partial x^{3}}{\partial y^{b}}$ , where $x^{1} = y^{1} - y^{0}$ . Evaluating the two terms: at $(a, b) = (0, 0)$ we get $1 \cdot (- 1) + 0 = - 1$ ; at $(a, b) = (0, 1)$ we get $1 \cdot 1 + 0 = 1$ ; at $(a, b) = (1, 0)$ we get $1 \cdot (- 1) + 0 = - 1$ ; at $(a, b) = (1, 1)$ we get $1 \cdot 1 + 0 = 1$ ; at $(a, b) = (2, 3)$ we get $0 + (y^{1} - y^{0}) \cdot 1 \cdot 1 = y^{1} - y^{0}$ ; all other entries vanish because the corresponding Jacobian entries are $0$ .

3.1step 2.1L3algebra∎

In the $x$ chart the diagonal components are $T^{0}_{0} = T^{1}_{1} = T^{2}_{2} = T^{3}_{3} = 0$ , so $\sum_{a} T^{a}_{a} = 0$ ; in the $y$ chart, step 2.1 gives diagonal entries $T^{' 0}_{0} = - 1$ and $T^{' 1}_{1} = 1$ with $T^{' 2}_{2} = T^{' 3}_{3} = 0$ , so $\sum_{a} T^{' a}_{a} = - 1 + 1 + 0 + 0 = 0$ . The two values agree, as required by [L3].

Remarks

The example also shows that the diagonal components individually are not invariant: the trace is the chart-independent quantity. The mixed field $T$ has both contravariant and covariant slots, and step 2.1 uses one Jacobian factor of each kind, in contrast to the purely covariant case.

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A constant boosted tetrad field on flat spacetime and an adapted chart

Example

Let $M = R^{4}$ carry the identity chart $x = (x^{0}, x^{1}, x^{2}, x^{3})$ and the flat Lorentzian metric $g = diag (- 1, 1, 1, 1)$ , with time orientation $N = \partial_{0}$ . Let $β \in (- 1, 1)$ and let $γ : = (1 - β^{2})^{- 1 / 2} > 0$ . Define the constant-coefficient fields

$E_{0} : = γ (\partial_{0} + β \partial_{1}), E_{1} : = γ (β \partial_{0} + \partial_{1}), E_{2} : = \partial_{2}, E_{3} : = \partial_{3} .$

Then $(E_{0}, E_{1}, E_{2}, E_{3})$ is a smooth future-directed tetrad field on $M$ , and $n : = E_{0}$ is an observer congruence whose integral curves have $x^{1} - β x^{0}$ , $x^{2}$ and $x^{3}$ constant; hence $u : = (x^{0}, x^{1} - β x^{0}, x^{2}, x^{3})$ is an $n$ -adapted chart. In that chart $n = γ \partial_{u^{0}}$ , so the coordinate field of an adapted chart need not equal the congruence.

Facts & Assumptions

Given: The flat Lorentzian four-manifold $(R^{4}, g)$ with identity chart $x$ , time orientation $N = \partial_{0}$ , parameter $β \in (- 1, 1)$ and $γ = (1 - β^{2})^{- 1 / 2}$ .

[L1]

On $R^{4}$ with the flat metric, the coordinate fields $\partial_{a}$ are $g$ -orthonormal with $g (\partial_{a}, \partial_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and a nonzero causal vector $v$ is future-directed exactly when $g (v, N) < 0$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Local frame fields, coordinate frames and tetrad fields).

[L2]

A tetrad field on an open set is a tuple of four pointwise independent smooth vector fields with $g (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ pointwise, and it is future-directed when $E_{0}$ is future-directed at every point (Local frame fields, coordinate frames and tetrad fields).

[L3]

A vector field on a chart domain is smooth exactly when its coordinate components are smooth; constant components are smooth (Smoothness of a vector field is equivalent to smooth coordinate components).

[L4]

An observer congruence on an open set is a smooth field $n$ with $g_{q} (n_{q}, n_{q}) = - 1$ whose values are future-directed; a chart is $n$ -adapted when its spatial coordinates are constant along every integral curve of $n$ (Observer congruences, their observer curves and adapted charts).

Verification

technique · direct

1.1givenL3

The fields $E_{0}, E_{1}, E_{2}, E_{3}$ have constant coordinate components by their definition, so they are smooth vector fields on $R^{4}$ by [L3].

2.1step 1.1L2algebra

Evaluating the flat metric on the constant fields gives $g (E_{0}, E_{0}) = γ^{2} (- 1 + β^{2}) = - 1$ , $g (E_{1}, E_{1}) = γ^{2} (- β^{2} + 1) = 1$ , $g (E_{0}, E_{1}) = γ^{2} (- β + β) = 0$ , $g (E_{2}, E_{2}) = g (E_{3}, E_{3}) = 1$ and all remaining pairings $0$ because the fields involve only the $\partial_{0}, \partial_{1}$ directions or are the orthonormal fields $\partial_{2}, \partial_{3}$ . Hence $g (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ pointwise, and the tuple is an ordered basis at every point; it is a tetrad field by [L2].

3.1step 2.1L1L2algebra

The first member satisfies $g (E_{0}, N) = g (E_{0}, \partial_{0}) = - γ < 0$ , so $E_{0}$ is future-directed by [L1] and $(E_{0}, E_{1}, E_{2}, E_{3})$ is a future-directed tetrad field.

4.1step 3.1L4algebra

The field $n : = E_{0} = γ (\partial_{0} + β \partial_{1})$ is a smooth future unit timelike field, hence an observer congruence by [L4]. Its integral curves satisfy ${\dot{x}}^{0} = γ$ , ${\dot{x}}^{1} = γ β$ , ${\dot{x}}^{2} = {\dot{x}}^{3} = 0$ , so $x^{2}$ and $x^{3}$ are constant and $\frac{d}{d s} (x^{1} - β x^{0}) = γ β - β γ = 0$ ; therefore $x^{1} - β x^{0}$ , $x^{2}$ and $x^{3}$ are constant along every integral curve.

5.1step 4.1L4algebra∎

The map $u = (u^{0}, u^{1}, u^{2}, u^{3}) : = (x^{0}, x^{1} - β x^{0}, x^{2}, x^{3})$ has the smooth inverse $x^{0} = u^{0}$ , $x^{1} = u^{1} + β u^{0}$ , $x^{2} = u^{2}$ , $x^{3} = u^{3}$ , so it is a chart, and by step 4.1 its spatial coordinates $u^{1}, u^{2}, u^{3}$ are constant along the integral curves of $n$ ; the chart is $n$ -adapted by [L4]. In these coordinates $\partial_{u^{0}} = \partial_{0} + β \partial_{1} = n / γ$ , so $n = γ \partial_{u^{0}}$ . It equals $\partial_{u^{0}}$ exactly when $β = 0$ (then $γ = 1$ ), and differs when $β \neq 0$ ; thus an adapted chart need not have the congruence as its coordinate field. No uniqueness of adapted charts is claimed.

Remarks

The example shows that an adapted chart exists for this congruence and that its coordinate time need not be the flow parameter of $n$ ; the existence corollary on the tensors page produces a different adapted chart in which $n = \partial_{u^{0}}$ exactly. The congruence is parallel in the identity chart, so its observers are straight lines; no curvature or acceleration claim is made.

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A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame

Example

Let $M = R^{4}$ carry the identity chart $x$ and the flat Lorentzian metric $g = diag (- 1, 1, 1, 1)$ with time orientation $N = \partial_{0}$ . Fix $λ > 0$ and put

$W : = {(x^{0}, x^{1}, x^{2}, x^{3}) : ∣ λ x^{1} ∣ < 1 / 2}, Ω : = λ x^{1}, γ : = (1 - Ω^{2})^{- 1 / 2} > 0.$

Define the smooth fields on $W$

$E_{0} : = γ (\partial_{0} + Ω \partial_{2}), E_{1} : = \partial_{1}, E_{2} : = γ (Ω \partial_{0} + \partial_{2}), E_{3} : = \partial_{3} .$

Then $(E_{0}, E_{1}, E_{2}, E_{3})$ is a future-directed tetrad field on $W$ , its first member $n : = E_{0}$ is an observer congruence, the chart $u : = (x^{0}, x^{1}, x^{2} - Ω (x^{1}) x^{0}, x^{3})$ is $n$ -adapted with $\partial_{u^{0}} = n / γ$ , and the tetrad is not the coordinate frame of any chart, because $[E_{0}, E_{1}] = - γ^{'} \partial_{0} - (γ Ω)^{'} \partial_{2}$ is nonzero at every point of $W$ with $x^{1} \neq 0$ , where $γ^{'} = γ^{3} λ^{2} x^{1}$ .

Facts & Assumptions

Given: The flat Lorentzian four-manifold $(R^{4}, g)$ with identity chart $x$ , time orientation $N = \partial_{0}$ , parameters $λ > 0$ , $Ω = λ x^{1}$ , $γ = (1 - Ω^{2})^{- 1 / 2}$ on the open set $W = {∣ Ω ∣ < 1 / 2}$ , and the four fields $E_{0}, E_{1}, E_{2}, E_{3}$ .

[L1]

On $R^{4}$ with the flat metric the coordinate fields satisfy $g (\partial_{a}, \partial_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , and a nonzero causal vector is future-directed exactly when $g (v, N) < 0$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

[L2]

A tetrad field on an open set is a tuple of four pointwise independent smooth vector fields with $g (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ pointwise; the coordinate frame of a chart has pairwise commuting members, so a frame with some nonvanishing bracket is not the coordinate frame of any chart (Local frame fields, coordinate frames and tetrad fields).

[L3]

An observer congruence on an open set is a smooth field $n$ with $g (n, n) = - 1$ and future-directed values; a chart is $n$ -adapted when its spatial coordinates are constant along every integral curve of $n$ (Observer congruences, their observer curves and adapted charts).

[L4]

A vector field is smooth exactly when its coordinate components are smooth (Smoothness of a vector field is equivalent to smooth coordinate components).

[L5]

Finite sums, products and composites of $C^{\infty}$ functions are $C^{\infty}$ on their domains ( $C^{k}$ Euclidean maps are closed under componentwise algebra and composition).

[L6]

The squaring map $u \mapsto u^{2}$ is injective on $(0, \infty)$ with derivative $2 u \neq 0$ there, so by the injective-regular-map theorem its inverse $u \mapsto u^{1 / 2}$ is $C^{\infty}$ on $(0, \infty)$ ; the reciprocal $u \mapsto u^{- 1}$ is $C^{\infty}$ on $(0, \infty)$ by the iterated product and quotient rules (An injective regular $C^{k}$ map is a $C^{k}$ diffeomorphism onto its image, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L7]

In a chart, $[X, Y] = \sum_{j} (\sum_{i} X^{i} \partial_{i} Y^{j} - Y^{i} \partial_{i} X^{j}) \partial_{j}$ (Coordinate formula for the Lie bracket).

Verification

technique · direct

1.1L5L6given

The set $W$ is open and $1 - Ω^{2} > 3 / 4 > 0$ on $W$ . By [L6] the function $u \mapsto u^{- 1 / 2}$ is $C^{\infty}$ on $(0, \infty)$ , so $γ$ is $C^{\infty}$ on $W$ as a composite with the polynomial $1 - Ω^{2}$ , and $Ω, γ, γ Ω$ are $C^{\infty}$ on $W$ by [L5].

2.1step 1.1L1L3L4algebra

The fields $E_{0}, \dots, E_{3}$ have $C^{\infty}$ components in the chart $x$ by step 1.1, hence are smooth vector fields on $W$ by [L4]. Moreover $g (n, n) = γ^{2} (- 1 + Ω^{2}) = - 1$ and $g (n, N) = g (γ (\partial_{0} + Ω \partial_{2}), \partial_{0}) = - γ < 0$ , so $n = E_{0}$ is future unit timelike; by [L3] it is an observer congruence on $W$ .

3.1step 2.1L2algebra

Evaluating $g$ on the fields gives: $g (E_{0}, E_{0}) = γ^{2} (- 1 + Ω^{2}) = - 1$ ; $g (E_{2}, E_{2}) = γ^{2} (Ω^{2} g (\partial_{0}, \partial_{0}) + g (\partial_{2}, \partial_{2})) = γ^{2} (1 - Ω^{2}) = 1$ ; $g (E_{0}, E_{2}) = γ^{2} (Ω g (\partial_{0}, \partial_{0}) + Ω g (\partial_{2}, \partial_{2})) = γ^{2} (- Ω + Ω) = 0$ ; $g (E_{1}, E_{1}) = g (E_{3}, E_{3}) = 1$ ; and every pairing involving $E_{1} = \partial_{1}$ or $E_{3} = \partial_{3}$ vanishes because $\partial_{1}, \partial_{3}$ are orthogonal to $\partial_{0}, \partial_{2}$ . Hence $g (E_{A}, E_{B}) = diag (- 1, 1, 1, 1)_{A B}$ pointwise, and the tuple is a tetrad field by [L2]; it is future-directed because $E_{0} = n$ is.

3.2step 2.1algebra

An integral curve of $n$ satisfies ${\dot{x}}^{0} = γ$ , ${\dot{x}}^{1} = 0$ , ${\dot{x}}^{2} = γ Ω$ , ${\dot{x}}^{3} = 0$ ; along such a curve $x^{1}$ is constant, so the values $γ$ , $Ω$ are constant on it, and therefore $x^{3}$ is constant and $\frac{d}{d s} (x^{2} - Ω x^{0}) = γ Ω - Ω γ = 0$ . Thus $x^{1}$ , $x^{3}$ and $x^{2} - Ω (x^{1}) x^{0}$ are constant along every integral curve.

4.1step 3.2L3algebra

The map $u = (u^{0}, u^{1}, u^{2}, u^{3}) : = (x^{0}, x^{1}, x^{2} - Ω (x^{1}) x^{0}, x^{3})$ has the explicit smooth inverse $x^{0} = u^{0}$ , $x^{1} = u^{1}$ , $x^{2} = u^{2} + Ω (u^{1}) u^{0}$ , $x^{3} = u^{3}$ on its image, so it is a chart; by step 3.2 its spatial coordinates are constant along the integral curves of $n$ , so the chart is $n$ -adapted by [L3]. In these coordinates $\partial_{u^{0}} = \partial_{0} + Ω \partial_{2} = n / γ$ , so the coordinate field of the adapted chart is $n / γ$ , not $n$ .

5.1step 4.1L2L7algebra∎

By [L7], the bracket of $E_{0} = γ \partial_{0} + γ Ω \partial_{2}$ and $E_{1} = \partial_{1}$ is $[E_{0}, E_{1}]^{j} = \sum_{i} (E_{0}^{i} \partial_{i} E_{1}^{j} - E_{1}^{i} \partial_{i} E_{0}^{j}) = - \partial_{1} E_{0}^{j}$ , so $[E_{0}, E_{1}] = - \partial_{1} (γ) \partial_{0} - \partial_{1} (γ Ω) \partial_{2} = - γ^{'} \partial_{0} - (γ Ω)^{'} \partial_{2}$ . Since $γ = (1 - Ω^{2})^{- 1 / 2}$ and $Ω = λ x^{1}$ , the chain and power rules give $γ^{'} = γ^{3} Ω Ω^{'} = γ^{3} λ^{2} x^{1}$ , which is nonzero at every point of $W$ with $x^{1} \neq 0$ ; hence $[E_{0}, E_{1}] \neq 0$ at such points. By the commuting necessary condition of [L2], the tetrad field is not the coordinate frame of any chart.

Remarks

The congruence is a shearing motion: its observers move in the $\partial_{2}$ direction with a speed $Ω$ that depends on $x^{1}$ . The example shows in one place that congruence, adapted chart, tetrad field and coordinate frame are four different objects: an adapted chart exists while the tetrad is not a coordinate frame, and the adapted chart does not have $n$ as a coordinate field. Integrability of the rest spaces of this congruence is decided by the separate counterexample on this page.

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The relative-velocity decomposition of a boost, checked in an explicit tetrad

Example

Let $(R^{4}, g)$ and the constant boosted tetrad field $(E_{0}, E_{1}, E_{2}, E_{3})$ be as in A constant boosted tetrad field on flat spacetime and an adapted chart, with $E_{0} = n$ future unit timelike, $E_{1}$ unit spacelike and $g (E_{0}, E_{1}) = 0$ . Let $β \in (- 1, 1)$ , $γ = (1 - β^{2})^{- 1 / 2}$ and $c > 0$ , and set

$w : = β c E_{1}, u : = γ (n + \frac{w}{c}) = γ (E_{0} + β E_{1}) .$

Then $u$ is future unit timelike, $- g (n, u) = γ \geq 1$ , $h_{n} (u) = γ β E_{1}$ , $w = c h_{n} (u) / γ$ , and $g (w, w) = β^{2} c^{2} = c^{2} (1 - γ^{- 2}) < c^{2}$ whenever $β \neq 0$ . This verifies the relative-velocity decomposition theorem of the tensors page in an explicit family.

Facts & Assumptions

Given: The constant boosted tetrad $(E_{0}, E_{1}, E_{2}, E_{3})$ on $R^{4}$ with flat metric $g$ , its first member $n = E_{0}$ , parameters $β \in (- 1, 1)$ , $γ = (1 - β^{2})^{- 1 / 2}$ , $c > 0$ , and the vectors $w = β c E_{1}$ , $u = γ (n + w / c)$ .

[L1]

The fields $E_{0} = γ (\partial_{0} + β \partial_{1})$ , $E_{1} = γ (β \partial_{0} + \partial_{1})$ satisfy $g (E_{0}, E_{0}) = - 1$ , $g (E_{1}, E_{1}) = 1$ , $g (E_{0}, E_{1}) = 0$ ; $n = E_{0}$ is future-directed; and $E_{2} = \partial_{2}$ , $E_{3} = \partial_{3}$ (A constant boosted tetrad field on flat spacetime and an adapted chart).

[L2]

If $n$ is an observer, $c > 0$ , $u$ is future unit timelike and $γ_{o} = - g (n, u)$ , $w_{o} = c h_{n} (u) / γ_{o}$ , then $γ_{o} \geq 1$ , $u = γ_{o} (n + w_{o} / c)$ and $g (w_{o}, w_{o}) = c^{2} (1 - γ_{o}^{- 2})$ (Relative velocity decomposition of a future unit timelike vector).

[L3]

An observer is a future-directed unit timelike vector; the spatial projection is $h_{n} (v) = v + g_{p} (n, v) n$ (Observers, rest spaces and the spatial projection at an event), and a nonzero causal vector $v$ is future-directed exactly when $g (v, N) < 0$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model).

Verification

technique · direct

1.1L1L3algebra

Since $g (n, n) = - 1$ , $g (E_{1}, E_{1}) = 1$ and $g (n, E_{1}) = 0$ by [L1], bilinearity gives $g (u, u) = γ^{2} (g (n, n) + 2 β g (n, E_{1}) + β^{2} g (E_{1}, E_{1})) = γ^{2} (- 1 + β^{2}) = - 1$ , so $u$ is unit timelike. Moreover $g (n, N) = g (E_{0}, \partial_{0}) = - γ$ and $g (E_{1}, \partial_{0}) = - γ β$ by [L1], so $g (u, N) = γ (g (n, N) + β g (E_{1}, N)) = - γ^{2} (1 + β^{2}) < 0$ ; by [L3] the vector $u$ is future-directed.

2.1step 1.1L1L3algebra

By [L3] and [L1], $h_{n} (u) = u + g (u, n) n = u - γ n = γ (E_{0} + β E_{1}) - γ E_{0} = γ β E_{1}$ . Moreover $g (u, n) = γ (g (n, n) + β g (E_{1}, n)) = - γ$ , since $g (n, E_{1}) = 0$ ; hence $γ_{o} : = - g (n, u) = γ$ and $w_{o} : = c h_{n} (u) / γ_{o} = c β E_{1} = w$ .

3.1step 1.1step 2.1L2algebra∎

The identities $u = γ (n + w / c)$ and $g (w, w) = β^{2} c^{2}$ hold by definition of $w$ and [L1]; since $γ^{- 2} = 1 - β^{2}$ , we have $g (w, w) = c^{2} (1 - γ^{- 2}) < c^{2}$ for $β \neq 0$ , while for $β = 0$ the vector $w$ vanishes and $u = n$ . Thus the explicit pair $(γ, w)$ is exactly the decomposition produced by [L2] for $u$ , and $γ \geq 1$ .

Remarks

The example is the flat-spacetime boost written in the abstract split of a single tetrad: no coordinate chart enters, and the only data are the observer $n$ , the spatial direction $E_{1}$ , the speed parameter $β$ and the conversion constant $c$ .

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Observer energy, spatial momentum and mass shell of a boosted test body

Example

Keep the setting of The relative-velocity decomposition of a boost, checked in an explicit tetrad: the flat metric $g$ on $R^{4}$ , the tetrad $(E_{0}, E_{1}, E_{2}, E_{3})$ , the observer $n = E_{0}$ , the parameters $β \in (- 1, 1)$ , $γ = (1 - β^{2})^{- 1 / 2}$ , $c > 0$ , and the future unit timelike vector $u = γ (E_{0} + β E_{1})$ with $g (u, u) = - 1$ and $- g (n, u) = γ$ . Let $m > 0$ and set

$P : = m c u .$

Then $P$ is future timelike, its rest mass is $m (P) = m$ , and relative to the observer $n$ its energy and spatial momentum are

$E_{n} (P) = m c^{2} γ, q_{n} (P) = m c γ β E_{1},$

so that $g (q_{n} (P), q_{n} (P)) = m^{2} c^{2} γ^{2} β^{2}$ and the mass shell $E_{n} (P)^{2} = m^{2} c^{4} + c^{2} g (q_{n} (P), q_{n} (P))$ holds. This is the standard boosted test-body kinematics in the abstract observer split.

Facts & Assumptions

Given: The flat metric $g$ and tetrad $(E_{A})$ of the boosted-frame example, the observer $n = E_{0}$ , parameters $β, γ, c$ , the vector $u = γ (E_{0} + β E_{1})$ , a mass $m > 0$ , and the vector $P = m c u$ .

[L1]

$u$ is future unit timelike, $- g (n, u) = γ$ , $h_{n} (u) = γ β E_{1}$ , and $g (E_{1}, E_{1}) = 1$ with $g (n, E_{1}) = 0$ (The relative-velocity decomposition of a boost, checked in an explicit tetrad).

[L2]

For $P \in T_{p} M$ the observer energy and spatial momentum are $E_{n} (P) = - c g (P, n)$ and $q_{n} (P) = h_{n} (P)$ ; for $g (P, P) \leq 0$ the rest mass is $m (P) = \sqrt{- g (P, P)} / c \geq 0$ , and if $m (P) > 0$ then $U (P) = P / m (P)$ satisfies $g (U (P), U (P)) = - c^{2}$ (Observer-relative energy, spatial momentum and rest mass of a tangent vector).

[L3]

If $P$ is future timelike with rest mass $m$ , then with $u = P / (m c)$ and $γ = - g (n, u)$ one has $E_{n} (P) = m c^{2} γ$ , $q_{n} (P) = m c h_{n} (u)$ , $g (q_{n} (P), q_{n} (P)) = m^{2} c^{2} (γ^{2} - 1)$ and $E_{n} (P)^{2} = m^{2} c^{4} + c^{2} g (q_{n} (P), q_{n} (P))$ (Positivity, relative-velocity identities and the mass shell of observer-relative energy and momentum).

[L4]

For $a > 0$ the nonnegative square root satisfies $(\sqrt{a})^{2} = a$ and $\sqrt{a} > 0$ (Rational powers $a^{r}$ of a positive base).

Verification

technique · direct

1.1L1L2L4algebra

Since $g (u, u) = - 1$ by [L1], we get $g (P, P) = m^{2} c^{2} g (u, u) = - m^{2} c^{2}$ , so $- g (P, P) = m^{2} c^{2} > 0$ ; by [L4] and [L2] the rest mass is $m (P) = \sqrt{m^{2} c^{2}} / c = m$ . In particular $P = m c u$ is future timelike and $P / (m (P) c) = u$ .

2.1step 1.1L1L2algebra

By [L2] and [L1], $E_{n} (P) = - c g (m c u, n) = - m c^{2} g (u, n) = m c^{2} γ$ because $g (u, n) = - g (n, u) = - γ$ ; and $q_{n} (P) = h_{n} (P) = m c h_{n} (u) = m c γ β E_{1}$ .

3.1step 2.1L1L3algebra∎

Using $g (q_{n} (P), q_{n} (P)) = m^{2} c^{2} γ^{2} β^{2}$ and $γ^{2} (1 - β^{2}) = 1$ , we compute $E_{n} (P)^{2} - c^{2} g (q_{n} (P), q_{n} (P)) = m^{2} c^{4} γ^{2} (1 - β^{2}) = m^{2} c^{4}$ , which is the mass shell relation of [L3]; equivalently $E_{n} (P)^{2} = m^{2} c^{4} + c^{2} g (q_{n} (P), q_{n} (P))$ . The energy is positive because $m, c, γ > 0$ .

Remarks

The example computes the same numbers as the elementary boost formula in an inertial frame, but only through the observer split of a single tetrad: the energy $m c^{2} γ$ and momentum $m c γ β$ are vector and scalar data in $T_{p} M$ and its dual, not coordinate components. No measurement or empirical claim is made; the physical reading of $E_{n}$ and $q_{n}$ belongs to the framework's units and postulates.

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A tetrad field that is not the coordinate frame of any chart

Statement refuted

Statement refuted: "every smooth future-directed tetrad field on a nonempty open subset of flat Lorentzian four-space is the coordinate frame of some smooth chart" (Local frame fields, coordinate frames and tetrad fields).

Facts & Assumptions

Given: The shearing congruence example on the open set $W = {∣ λ x^{1} ∣ < 1 / 2} \subseteq R^{4}$ with its future-directed tetrad field $E = (E_{0}, E_{1}, E_{2}, E_{3})$ .

[L1]

On $W$ the tuple $(E_{0}, E_{1}, E_{2}, E_{3})$ with $E_{0} = γ (\partial_{0} + Ω \partial_{2})$ , $E_{1} = \partial_{1}$ , $E_{2} = γ (Ω \partial_{0} + \partial_{2})$ , $E_{3} = \partial_{3}$ , $Ω = λ x^{1}$ , $γ = (1 - Ω^{2})^{- 1 / 2}$ , is a smooth future-directed tetrad field, and $[E_{0}, E_{1}] = - γ^{'} \partial_{0} - (γ Ω)^{'} \partial_{2}$ with $γ^{'} = γ^{3} λ^{2} x^{1}$ , nonzero at every point of $W$ with $x^{1} \neq 0$ (A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame).

[L2]

The coordinate frame $(\partial_{v^{0}}, \partial_{v^{1}}, \partial_{v^{2}}, \partial_{v^{3}})$ of a smooth chart has pairwise commuting members: $[\partial_{v^{a}}, \partial_{v^{b}}] = 0$ for all $a, b$ ; consequently a frame field with a nonvanishing bracket is not the coordinate frame of any chart (Local frame fields, coordinate frames and tetrad fields).

Counterexample

1.1L1algebra

By [L1], $E$ is a smooth future-directed tetrad field on the nonempty open subset $W$ of flat Lorentzian four-space, and $[E_{0}, E_{1}] = - γ^{'} \partial_{0} - (γ Ω)^{'} \partial_{2}$ with $γ^{'} = γ^{3} λ^{2} x^{1}$ . At every point of $W$ with $x^{1} \neq 0$ we have $γ^{'} \neq 0$ , so the coefficient of $\partial_{0}$ in $[E_{0}, E_{1}]$ is nonzero and $[E_{0}, E_{1}] \neq 0$ there; such points exist in $W$ .

2.1step 1.1L2∎

Suppose, for contradiction, that $E$ were the coordinate frame of a smooth chart $(V, v)$ on some open subset of $W$ : that is, $E_{a} = \partial_{v^{a}}$ on the chart domain for $a = 0, 1, 2, 3$ . Then by [L2] we would have $[E_{a}, E_{b}] = [\partial_{v^{a}}, \partial_{v^{b}}] = 0$ for all $a, b$ , contradicting step 1.1: every nonempty open chart domain $V \subseteq W$ contains a point with $x^{1} \neq 0$ , since the hyperplane ${x^{1} = 0}$ has empty interior in $R^{4}$ , and the bracket is nonzero at that point. Hence no such chart exists, and the displayed general statement is refuted.

Remarks

This counterexample uses the necessary commutation condition to refute the universal claim; it makes no assertion about the sufficiency of that condition. The companion example A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame exhibits an adapted chart for the same congruence, so "adapted chart" and "coordinate frame equal to the tetrad" are different requirements.

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A congruence whose rest spaces are not tangent to any local hypersurfaces

Statement refuted

Statement refuted: "for every observer congruence on a nonempty open subset of flat Lorentzian four-space, the family of rest spaces is tangent to local hypersurfaces" (Smoothness of the rest-space family and its Frobenius integrability criterion).

Facts & Assumptions

Given: The shearing congruence $n = γ (\partial_{0} + Ω \partial_{2})$ on $W = {∣ λ x^{1} ∣ < 1 / 2} \subseteq R^{4}$ with its flat Lorentzian metric $g = diag (- 1, 1, 1, 1)$ , time orientation $N = \partial_{0}$ , and the one-form $α = n^{♭}$ .

[L1]

On $W$ the field $n$ is an observer congruence with $Ω = λ x^{1}$ , $γ = (1 - Ω^{2})^{- 1 / 2} > 0$ , $λ > 0$ , and its rest space at $q$ is $n_{q}^{⊥} = \ker α_{q}$ (A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame, Smoothness of the rest-space family and its Frobenius integrability criterion).

[L2]

The rest spaces of a congruence are tangent to local hypersurfaces if and only if $α \land d α = 0$ on the domain, where $α = n^{♭}$ (Smoothness of the rest-space family and its Frobenius integrability criterion).

[L3]

The flat map is $α_{q} (v) = g_{q} (n_{q}, v)$ , so in the chart $x$ the components of $α$ are $α_{a} = \sum_{b} g_{a b} n^{b}$ (Smoothness of the rest-space family and its Frobenius integrability criterion, A smooth differential $k$ -form).

[L4]

The exterior derivative is $R$ -linear of degree one and satisfies $d (β \land γ) = d β \land γ + (- 1)^{\deg β} β \land d γ$ ; the exterior derivative of a smooth function is its differential (The exterior derivative is a graded derivation, The exterior derivative of a function is its differential, The exterior derivative by the invariant vector-field formula).

[L5]

The wedge product of differential forms is defined pointwise by the wedge product of alternating covectors (The wedge product of differential forms).

Counterexample

1.1L1L3algebra

In the chart $x$ the components of $n$ are $(γ, 0, γ Ω, 0)$ and $g_{a b} = diag (- 1, 1, 1, 1)_{a b}$ , so by [L3] the components of $α$ are $α_{0} = - γ$ , $α_{2} = γ Ω$ and $α_{1} = α_{3} = 0$ ; that is, $α = γ (- d x^{0} + Ω d x^{2})$ , a smooth $1$ -form on $W$ .

2.1step 1.1L4algebra

Since $γ$ and $Ω$ depend only on $x^{1}$ , their differentials are $d γ = γ^{'} d x^{1}$ and $d Ω = Ω^{'} d x^{1}$ by [L4], with $Ω^{'} = λ$ and $γ^{'} = γ^{3} Ω Ω^{'}$ . Applying the Leibniz rule of [L4] twice gives $d α = d γ \land (- d x^{0} + Ω d x^{2}) + γ d (- d x^{0} + Ω d x^{2}) = γ^{'} d x^{1} \land (- d x^{0}) + γ^{'} Ω d x^{1} \land d x^{2} + γ Ω^{'} d x^{1} \land d x^{2} = γ^{'} d x^{0} \land d x^{1} + (γ^{'} Ω + γ Ω^{'}) d x^{1} \land d x^{2}$ .

3.1step 2.1L5algebra

By [L5] and step 2.1, $α \land d α = γ (- d x^{0} + Ω d x^{2}) \land [γ^{'} d x^{0} \land d x^{1} + (γ^{'} Ω + γ Ω^{'}) d x^{1} \land d x^{2}]$ . The terms containing a repeated coordinate differential vanish, and using $d x^{2} \land d x^{0} \land d x^{1} = d x^{0} \land d x^{1} \land d x^{2}$ we get $α \land d α = γ [- (γ^{'} Ω + γ Ω^{'}) + Ω γ^{'}] d x^{0} \land d x^{1} \land d x^{2} = - γ^{2} Ω^{'} d x^{0} \land d x^{1} \land d x^{2} = - γ^{2} λ d x^{0} \land d x^{1} \land d x^{2}$ . Since $λ > 0$ and $γ > 0$ on $W$ , the coefficient $- γ^{2} λ$ is nonzero at every point of $W$ , so $α \land d α$ is nowhere zero on $W$ .

4.1step 3.1L2∎

By [L2], the rest spaces of the congruence are tangent to local hypersurfaces if and only if $α \land d α = 0$ ; since $α \land d α \neq 0$ at every point of $W$ by step 3.1, they are not tangent to any local hypersurface. This refutes the displayed general statement.

Remarks

By the companion example A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame this same congruence does admit an adapted chart, so "existence of an adapted chart" and "integrability of the rest spaces" are genuinely different properties: the first always holds locally, the second fails here. The counterexample makes no claim about congruences with vanishing vorticity, and no global foliation claim.

Tensors And Observers — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

A mixed tensor field: components in two charts and the invariance of its trace

Example

Facts & Assumptions

Verification

Remarks

A constant boosted tetrad field on flat spacetime and an adapted chart

Example

Facts & Assumptions

Verification

Remarks

A shearing observer congruence on flat spacetime: tetrad, adapted chart and non-coordinate frame

Example

Facts & Assumptions

Verification

Remarks

The relative-velocity decomposition of a boost, checked in an explicit tetrad

Example

Facts & Assumptions

Verification

Remarks

Observer energy, spatial momentum and mass shell of a boosted test body

Example

Facts & Assumptions

Verification

Remarks

A tetrad field that is not the coordinate frame of any chart

Statement refuted

Facts & Assumptions

Counterexample

Remarks

A congruence whose rest spaces are not tangent to any local hypersurfaces

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources