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Massive square-root Legendre map and its future shell

Statement

Let $(M, g, T)$ be a smooth time-oriented Lorentzian four-manifold. At $x \in M$ , for $p \in T_{x}^{*} M$ , define $p^{♯}$ as the unique vector satisfying $g_{x} (p^{♯}, W) = p (W)$ for every $W \in T_{x} M$ , and set $g_{x}^{- 1} (p, r) : = g_{x} (p^{♯}, r^{♯})$ . In a chart, $p^{♯} = g^{a b} p_{b} \partial_{a}$ , where $(g^{a b})$ is the inverse matrix of $g_{a b} = g (\partial_{a}, \partial_{b})$ . A covector on this background is future when its metric dual is future-directed. Let $m, c > 0$ and $L (x, V) = - m c \sqrt{- g (V, V)}$ on the future timelike tangent cone. Its fibre derivative is

$p_{a} = m c \frac{g_{a b} V^{b}}{s}, s = \sqrt{- g (V, V)} > 0.$

The velocity Hessian is $m c (g_{a b} / s + V_{a} V_{b} / s^{3})$ , has kernel exactly $R V$ , and rank three. The image consists exactly of the future shell $Σ_{m}^{+} = {(x, p) \in T^{*} M : g_{x}^{- 1} (p, p) = - m^{2} c^{2}, p^{♯} future}$ . Each fibre is a connected positive ray, and the Legendre map is a surjective submersion onto this embedded image. Its energy $p (V) - L$ is zero; the map has no local inverse in the four-velocity direction.

Facts & Assumptions

Given: A smooth nondegenerate Lorentz metric of signature $(- + + +)$ , its time orientation, $m > 0$ , $c > 0$ , and a future timelike $V \neq 0$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Assume countable choice ACω, meaning every sequence of nonempty sets has a choice function; no full Axiom of Choice is assumed.

[F1]

Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model supplies a smooth time-oriented Lorentzian four-manifold with signature $(-, +, +, +)$ and its future cone. The constants $m, c > 0$ are parameters of the explicitly specified mathematical function $L$ .

[F2]

Particle spacetime, observers and worldlines: a massive worldline is future timelike, and the background metric and time orientation are fixed data.

[F3]

A regular level set is an embedded submanifold: a smooth level with nonzero differential is an embedded hypersurface.

[F4]

Positive homogeneity forces Legendre degeneracy: positive degree-one homogeneity forces a null velocity-Hessian direction and zero energy.

[F5]

Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1): every four-dimensional Lorentz vector space admits an orthonormal basis with metric $diag (- 1, 1, 1, 1)$ .

[F6]

Norms of four-velocity and four-acceleration, and the observer rest-space decomposition: in a Lorentz vector space with signature $(- + + +)$ , the orthogonal complement of a future unit timelike vector is positive definite.

Proof

1.1givenA1F1algebra

Use the supplied smooth tangent/cotangent bundle structure under [A1] and the Lorentzian data [F1]. Differentiate $L = - m c \sqrt{- g_{i j} V^{i} V^{j}}$ with respect to $V^{a}$ . Because $s > 0$ , ordinary chain rule gives $p_{a} = m c V_{a} / s$ . Differentiating again gives $L_{V^{a} V^{b}} = m c (g_{a b} / s + V_{a} V_{b} / s^{3})$ .

2.1givenF1F2F5F6step 1.1algebra

Contract the Hessian with $V^{b}$ : $m c (V_{a} / s + V_{a} g (V, V) / s^{3}) = 0$ , since $g (V, V) = - s^{2}$ . For $W ⊥_{g} V$ , contraction with $W$ has quadratic value $(m c / s) g (W, W)$ . The normalized vector $u = V / s$ is future unit timelike by [F2]; [F5] identifies this tangent Lorentz space with the standard Lorentz vector space, and [F6] makes its orthogonal complement positive definite. Thus the Hessian is positive on every nonzero $W ⊥_{g} V$ , so its kernel is precisely $R V$ and its rank is three.

2.2givenF1F2step 1.1algebra

The momentum in step 1.1 satisfies $g^{- 1} (p, p) = (m c / s)^{2} g (V, V) = - m^{2} c^{2}$ and is future because it is a positive multiple of $V^{♭}$ . Conversely, for any future covector on this shell, set $V = p^{♯}$ . Then $s = m c$ , and the same formula returns $p$ ; every positive multiple of $V$ returns the same covector. This proves exact image and positive-ray fibres.

3.1givenF3step 2.1step 2.2algebra

The defining function $C = \frac{1}{2} (g^{- 1} (p, p) + m^{2} c^{2})$ has fibre differential $d_{p} C = p^{♯} \neq 0$ on the shell. By [F3] the shell is embedded. The Legendre map fixes the base point and has fibre differential of rank three by step 2.1, so its differential has rank $4 + 3 = 7$ , the shell dimension; it is a submersion onto that image.

4.1F4step 2.1∎

Euler's identity from [F4] gives $p (V) - L = 0$ , while the nonzero kernel direction in step 2.1 rules out a local diffeomorphism and any four-velocity inverse.

Massive square-root Legendre map and its future shell

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources