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Em Minkowski Mathematical Prerequisites — Examples

1 · Prerequisites

2 · Summary

The companion page records the hypothesis boundaries of the main page in one worked example. Explicit mathematical hypothesis/convention examples checks that the vector $e_{0}$ is future timelike and unit; that the boost sends it to $(γ, - γ β, 0, 0)$ with quadratic value $η (Λ_{β} e_{0}, Λ_{β} e_{0}) = - 1$ and positive time component while the factor $γ$ is undefined at $β = 1$ ; that the null line $(s, s, 0, 0)$ makes the proper-time construction identically zero and is therefore not a parameter; that the symmetric form $F = η$ has $Φ = i d$ and violates $η (u, Φ (u)) = 0$ , exhibiting the antisymmetry hypothesis as genuine; and that the skew witness $F_{01} = 1$ sends $e_{0}$ to $- e_{1}$ with the invariant pairing and the raised-index formula agreeing in sign. No physical clock premise and no experimental claim is used.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Explicit mathematical hypothesis/convention examples

Example

With $η = diag (- 1, 1, 1, 1)$ : (i) the vector $v$ with components $(1, 0, 0, 0)$ is future timelike and unit, $η (v, v) = - 1$ . (ii) For $0 < β < 1$ the boost $Λ_{β}$ of The explicit Lorentz boost preserves $η$ , has determinant $1$ and preserves the future cone sends $v$ to the vector with components $(γ, - γ β, 0, 0)$ , whose quadratic value $η (Λ_{β} v, Λ_{β} v)$ is $- γ^{2} (1 - β^{2}) = - 1$ and whose time component $γ$ is positive; at $β = 1$ the factor $γ$ is undefined, so the hypothesis $0 \leq β < 1$ cannot be dropped. (iii) The null line $z (s) = (s, s, 0, 0)$ , $s \in R$ , has $η (z^{'}, z^{'}) = 0$ , so the proper-time construction of Proper time along a $C^{1}$ future timelike curve: strictly increasing reparametrization with $η (U, U) = - c^{2}$ gives $τ \equiv 0$ , which is not an invertible parameter: the timelike hypothesis cannot be weakened to causal. (iv) For the symmetric nonzero form $F : = η$ one has $Φ = i d$ and $η (v, Φ (v)) = η (v, v) = - 1 \neq 0$ , so the antisymmetry hypothesis of Antisymmetric contractions vanish and $m U^{'} = q Φ (U)$ preserves $η (U, U)$ is genuine. (v) For the skew form with $F_{01} = 1$ , $F_{10} = - 1$ and other coefficients zero, the operator of Antisymmetric contractions vanish and $m U^{'} = q Φ (U)$ preserves $η (U, U)$ sends $e_{0}$ to $- e_{1}$ : $η (- e_{1}, w) = - w^{1} = F (w, e_{0})$ . Thus its invariant pairing and standard raised-index formula have the same sign, and $η (e_{0}, Φ (e_{0})) = 0$ . No physical clock premise or experimental claim is used.

Facts & Assumptions

Given: The Minkowski vector space $(V, η)$ of The Minkowski affine four-space: the form $η$ , causal cones, frames and Lorentz maps with its basis $(e_{0}, e_{1}, e_{2}, e_{3})$ , the vector $v = e_{0}$ , whose components are $(1, 0, 0, 0)$ , a real $β$ with $0 < β < 1$ , the boost $Λ_{β}$ and the number $γ = (1 - β^{2})^{- 1 / 2}$ of The explicit Lorentz boost preserves $η$ , has determinant $1$ and preserves the future cone, the null line $z (s) = (s, s, 0, 0)$ on $J : = R$ , and the forms $F : = η$ and the skew form with $F_{01} = 1$ , $F_{10} = - 1$ and other coefficients zero.

[F1]

On $V = R^{4}$ the form $η (v, w) = - v^{0} w^{0} + v^{1} w^{1} + v^{2} w^{2} + v^{3} w^{3}$ is symmetric and bilinear with matrix $diag (- 1, 1, 1, 1)$ , it is nondegenerate, the vectors $e_{0}, e_{1}, e_{2}, e_{3}$ are the standard ordered basis, and $C^{+}$ consists of the nonzero causal vectors with positive time coordinate (The Minkowski affine four-space: the form $η$ , causal cones, frames and Lorentz maps).

[L1]

For $0 \leq β < 1$ the map $Λ_{β}$ with $(Λ_{β} v)^{0} = γ (v^{0} - β v^{1})$ , $(Λ_{β} v)^{1} = γ (v^{1} - β v^{0})$ , $(Λ_{β} v)^{2} = v^{2}$ , $(Λ_{β} v)^{3} = v^{3}$ is a proper orthochronous Lorentz map, its inverse is $Λ_{- β}$ , and at $β = 1$ the factor $γ$ is undefined (The explicit Lorentz boost preserves $η$ , has determinant $1$ and preserves the future cone).

[L2]

For a nonempty open interval $J$ and a $C^{1}$ curve $z$ with $η (z^{'}, z^{'}) < 0$ and $z^{0}^{'} > 0$ the function $τ (s) = c^{- 1} \int_{s_{0}}^{s} \sqrt{g}$ with $g = - η (z^{'}, z^{'})$ has $τ^{'} > 0$ and gives the tangent $U (τ (s)) = c z^{'} (s) / \sqrt{g (s)}$ with $η (U, U) = - c^{2}$ ; and if $η (z^{'}, z^{'}) = 0$ on $J$ , then $g = 0$ , $τ \equiv 0$ and $τ$ is not injective (Proper time along a $C^{1}$ future timelike curve: strictly increasing reparametrization with $η (U, U) = - c^{2}$ ).

[L3]

For a skew-symmetric bilinear form $F$ and each $u$ there is exactly one $Φ (u)$ with $η (Φ (u), w) = F (w, u)$ for all $w$ , the map $Φ$ is linear, $η (u, Φ (u)) = 0$ , and in a frame $Φ (u)^{μ} = η^{μ α} F_{α β} u^{β}$ with $F_{α β} = F (e_{α}, e_{β})$ and $(η^{μ α})$ the inverse of $(η_{α β})$ (Antisymmetric contractions vanish and $m U^{'} = q Φ (U)$ preserves $η (U, U)$ ).

[L4]

$R$ is a nonempty open interval, and a nonempty open interval is order-convex (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

For $a > 0$ and rationals $r, s$ one has $a^{r} > 0$ , $a^{- r} = (a^{r})^{- 1}$ and $a^{r + s} = a^{r} a^{s}$ (Rational powers $a^{r}$ of a positive base, Laws of rational exponents).

Verification

technique · direct

1.1givenF1algebra

(i) $η (v, v) = η (e_{0}, e_{0}) = - 1 < 0$ and $v^{0} = 1 > 0$ , so $v$ is future timelike and unit in the sense $η (v, v) = - 1$ .

1.2givenL1algebra

(ii, coordinates) For $0 < β < 1$ , [L1] gives $Λ_{β} e_{0} = (γ (1 - β \cdot 0), γ (0 - β \cdot 1), 0, 0) = (γ, - γ β, 0, 0)$ .

1.3givenF1L2algebra

(iii, null tangent) The line $z (s) = (s, s, 0, 0)$ has the constant derivative $z^{'} = (1, 1, 0, 0)$ , so $η (z^{'}, z^{'}) = - 1 + 1 = 0$ ; its positive time component also makes the tangent future causal.

1.4givenF1algebra

(iv, symmetric form) For $F : = η$ , symmetry gives $η (u, w) = F (w, u)$ for every $w$ , so $Φ (u) : = u$ satisfies the defining pairing. If another vector $y$ does so, $η (y - u, w) = 0$ for every $w$ , hence $y = u$ by nondegeneracy [F1]. Thus $Φ = id$ without using the skew hypothesis of [L3], and $η (v, Φ (v)) = - 1 \neq 0$ .

1.5givenF1L3algebra

(v, skew sign witness) For the skew form with $F_{01} = 1$ , $F_{10} = - 1$ and other coefficients zero, $F (w, e_{0}) = - w^{1} = η (- e_{1}, w)$ for every $w$ , so $Φ (e_{0}) = - e_{1}$ by [L3]. The raised-index formula gives $Φ (e_{0})^{μ} = \sum_{α} η^{μ α} F_{α 0} = - η^{μ 1}$ , again the components of $- e_{1}$ . Thus both signs agree and $η (e_{0}, Φ (e_{0})) = 0$ .

1.6givenL1

(ii, excluded endpoint) At $β = 1$ , $1 - β^{2} = 0$ , so $γ = (1 - β^{2})^{- 1 / 2}$ is undefined by [L1]; the endpoint cannot be included in this formula.

2.1step 1.2L1L5algebra

(ii, norm) By step 1.2, $η (Λ_{β} v, Λ_{β} v) = - γ^{2} + γ^{2} β^{2} = - γ^{2} (1 - β^{2}) = - 1$ . Indeed [L5] with $a : = 1 - β^{2} > 0$ gives $γ^{2} a = a^{- 1 / 2} a^{- 1 / 2} a = a^{- 1} a = 1$ , and $γ > 0$ by [L1] and [L5].

2.2step 1.3L2L4

(iii, conclusion) $J : = R$ is a nonempty open interval by [L4]. The line of step 1.3 has constant, hence continuous, derivative, so is $C^{1}$ . The null clause of [L2] gives $g = 0$ and $τ \equiv 0$ , which is not injective; its future causal tangent therefore does not suffice for proper-time reparametrization.

3.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 2.1step 2.2∎

All five checks are established — (i) in step 1.1, (ii) in steps 1.2, 2.1 and 1.6, (iii) in steps 1.3 and 2.2, (iv) in step 1.4, and (v) in step 1.5 — with no physical clock premise and no experimental claim.