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Worked orthogonal complement of a timelike displacement is positive definite

Example

Worked computation. In an inertial frame of an oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ , with $η = diag (- 1, 1, 1, 1)$ in the frame's coordinates, take the displacement

$u = (2, 1, 0, 0),$

a future timelike displacement with $η (u, u) = - 3$ . A displacement $v = (v^{0}, v^{1}, v^{2}, v^{3})$ is $η$ -orthogonal to $u$ exactly when $- 2 v^{0} + v^{1} = 0$ , that is, exactly when $v^{1} = 2 v^{0}$ . On that hyperplane

$η (v, v) = - (v^{0})^{2} + (2 v^{0})^{2} + (v^{2})^{2} + (v^{3})^{2} = 3 (v^{0})^{2} + (v^{2})^{2} + (v^{3})^{2},$

which is nonnegative and vanishes exactly at $v = 0$ . Hence the restriction of $η$ to $u^{⊥}$ is positive definite, and every nonzero displacement orthogonal to a timelike displacement is spacelike.

Facts & Assumptions

Given: An inertial frame with $η = diag (- 1, 1, 1, 1)$ in its coordinates, and the displacement $u = (2, 1, 0, 0)$ .

[L1]

$η (v, w) = - v^{0} w^{0} + v^{1} w^{1} + v^{2} w^{2} + v^{3} w^{3}$ in an $η$ -orthonormal basis, so $η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ (Inertial frames and inertial charts of the affine Minkowski model).

[L2]

$u ⊥ v$ means $η (u, v) = 0$ , and $u^{⊥} = {v : η (u, v) = 0}$ is a linear subspace (eta-orthogonality and orthogonal complements in a Lorentz vector space).

[L3]

A vector is timelike, null or spacelike according as $η (v, v)$ is negative, zero with $v \neq 0$ , or positive (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

[L4]

The future closed cone is ${v : v^{0} \geq ∣ v_{v e c} ∣}$ , so a timelike vector with $v^{0} > ∣ v_{v e c} ∣$ is future timelike (Time orientation, future and past timelike/null/causal vectors, and the closed cones, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L5]

For $x, y \in R^{3}$ one has $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ , where $⟨ \cdot, \cdot ⟩$ and $∥ \cdot ∥_{2}$ are the Euclidean inner product and norm of $R^{3}$ (Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Verification

technique · direct computation with the diagonal form

1.1L1L2algebra

Evaluating the form on $u$ against $v$ , $η (u, v) = - 2 v^{0} + v^{1}$ , so the orthogonality condition $η (u, v) = 0$ is $v^{1} = 2 v^{0}$ and $u^{⊥}$ is that hyperplane by [L2].

2.1step 1.1L1algebra

On $u^{⊥}$ , substituting $v^{1} = 2 v^{0}$ into the quadratic form gives $η (v, v) = - (v^{0})^{2} + 4 (v^{0})^{2} + (v^{2})^{2} + (v^{3})^{2} = 3 (v^{0})^{2} + (v^{2})^{2} + (v^{3})^{2}$ , a sum of nonnegative terms; it vanishes exactly when $v^{0} = v^{2} = v^{3} = 0$ , and then also $v^{1} = 2 v^{0} = 0$ , so $v = 0$ . Hence the restriction of $η$ to $u^{⊥}$ is positive definite.

3.1step 1.1step 2.1L1L2L3L4algebra

Every nonzero $v \in u^{⊥}$ is spacelike by step 2.1 and [L3]; moreover $u$ itself is future timelike by [L4], since $η (u, u) = - 3 < 0$ and $u^{0} = 2 > 1 = ∣ u_{v e c} ∣$ . Thus the orthogonal complement of this timelike displacement is genuinely positive definite, in contrast with the null case.

4.1step 3.1L1L2L3L5algebra∎

For the general clause of the Example, let $\tilde{u} = ({\tilde{u}}^{0}, {\tilde{u}}_{v e c})$ be any timelike displacement and let $w = (w^{0}, w_{v e c}) \neq 0$ satisfy $η (\tilde{u}, w) = 0$ . Timelikeness gives $∣ {\tilde{u}}^{0} ∣ > ∣ {\tilde{u}}_{v e c} ∣$ by [L1] and [L3], and orthogonality gives ${\tilde{u}}_{v e c} \cdot w_{v e c} = {\tilde{u}}^{0} w^{0}$ . Suppose, for contradiction, that $η (w, w) \leq 0$ ; then $∣ w_{v e c} ∣ \leq ∣ w^{0} ∣$ , and $w_{v e c} \neq 0$ , because $w_{v e c} = 0$ would give ${\tilde{u}}^{0} w^{0} = 0$ and hence $w = 0$ , contradicting $w \neq 0$ . Cauchy–Schwarz [L5] now yields the contradictory chain $∣ {\tilde{u}}^{0} ∣ ∣ w^{0} ∣ = ∣ {\tilde{u}}_{v e c} \cdot w_{v e c} ∣ \leq ∣ {\tilde{u}}_{v e c} ∣ ∣ w_{v e c} ∣ < ∣ {\tilde{u}}^{0} ∣ ∣ w_{v e c} ∣ \leq ∣ {\tilde{u}}^{0} ∣ ∣ w^{0} ∣$ , where the strict inequality uses $∣ {\tilde{u}}_{v e c} ∣ < ∣ {\tilde{u}}^{0} ∣$ and $w_{v e c} \neq 0$ and the last inequality uses $∣ w_{v e c} ∣ \leq ∣ w^{0} ∣$ . Hence $η (w, w) > 0$ , that is, every nonzero displacement orthogonal to a timelike displacement is spacelike by [L3].

Depends on

Dependency tree · 2 levels

Sources

K. Conrad, Bilinear Forms (orthogonality, radicals, inertia)