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A smooth pointwise subluminal curve with no retarded root at an event

Physical scope

Mathematical example delimiting the kinematic hypothesis of the retarded point geometry: a smooth curve with $∣ z^{'} ∣ < c$ pointwise but no uniform bound, for which the retarded root equation fails at one event. Excluded: any claim about physical charges, any contradiction with theorems (the curve lies outside their hypotheses) and every empirical statement.

Example

Let $c > 0$ and $a > 0$ , and define $z : R \to R^{3}$ by $z (s) : = (\sqrt{c^{2} s^{2} + a^{2}}, 0, 0)$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Then: (a) $z$ is of class $C^{\infty}$ on $R$ and $∣ z^{'} (s) ∣ < c$ for every $s \in R$ ; nevertheless there is no constant $v_{*} < c$ with $∣ z^{'} (s) ∣ \leq v_{*}$ for all $s$ , because $∣ z^{'} (s) ∣ \to c$ as $s \to \pm \infty$ ; (b) at the event $(x, t) = (0, 0)$ the retarded equation $t - s = ∣ x - z (s) ∣ / c$ of Uniformly subluminal prescribed worldline, retarded root and retarded geometric data off the worldline has no solution in $R$ : the equation reads $- s = \sqrt{c^{2} s^{2} + a^{2}} / c$ , whose right-hand side is at least $a / c > 0$ , and squaring both sides would give $c^{2} s^{2} + a^{2} = c^{2} s^{2}$ , i.e. $a^{2} = 0$ , contrary to $a > 0$ . Hence pointwise subluminality of a smooth curve does not by itself produce the retarded root, and the uniform bound $v_{*} < c$ of Uniformly subluminal prescribed worldline, retarded root and retarded geometric data off the worldline, or an explicit cone-intersection hypothesis, cannot be dropped; the supplier Global existence, uniqueness and differentiability of the retarded root indeed assumes such a uniform bound. (c) [Scope] This is a mathematical example about the prescribed curve and the root equation. The curve is future timelike pointwise ( $∣ z^{'} (s) ∣ < c$ for every $s$ ) but not uniformly subluminal; it is not asserted to be the worldline of any physical charge, and no measurement is reported. The example contradicts no theorem: it exhibits a curve lying outside the hypotheses of the root lemma.

Facts & Assumptions

Given: constants $c > 0$ , $a > 0$ and the curve $z (s) = (\sqrt{c^{2} s^{2} + a^{2}}, 0, 0)$ ; the Euclidean norm of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and the absolute value of Absolute value in an ordered field.

[L1]

$c^{2} s^{2} + a^{2} \geq a^{2} > 0$ for every $s$ , and $\sqrt{\cdot}$ is continuous and differentiable on $(0, \infty)$ with $\frac{d}{d t} \sqrt{t} = \frac{1}{2 \sqrt{t}} = \frac{1}{2} t^{- 1 / 2}$ ; general power derivatives and continuity are as in Continuity and derivatives of positive-base real powers. (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Continuity and derivatives of positive-base real powers)

[L2]

Chain rule and algebra of derivatives for composites of differentiable maps, and the notion of the derivative of a vector-valued curve with its componentwise computation. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , 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 $g (c) \neq 0$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, $C^{k}$ Euclidean maps and diffeomorphisms)

[L3]

Monotonicity of squares and square roots on nonnegative reals: for $p, q \geq 0$ , $p < q$ if and only if $p^{2} < q^{2}$ . (Squaring is monotone on the nonnegatives, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ )

[L4]

The retarded equation and the uniform-bound hypothesis are those of Uniformly subluminal prescribed worldline, retarded root and retarded geometric data off the worldline; the cited root lemma states its existence and uniqueness result for curves with $∣ z^{'} (s) ∣ \leq v_{*} < c$ for all $s$ . (Uniformly subluminal prescribed worldline, retarded root and retarded geometric data off the worldline, Global existence, uniqueness and differentiability of the retarded root)

Verification

technique · direct computation and contradiction

1.1givenL1L2L3algebra

Smoothness and speed (a). The map $s \mapsto c^{2} s^{2} + a^{2}$ is a polynomial, hence $C^{\infty}$ , with values in $[a^{2}, \infty) \subseteq (0, \infty)$ ; by [L1] the square root is $C^{\infty}$ on $(0, \infty)$ , so the composite $s \mapsto \sqrt{c^{2} s^{2} + a^{2}}$ is $C^{\infty}$ and $z$ is $C^{\infty}$ into $R^{3}$ [L2]. Its derivative is $z^{'} (s) = (c^{2} s / \sqrt{c^{2} s^{2} + a^{2}}, 0, 0)$ by the chain rule [L1, L2], so $∣ z^{'} (s) ∣ = c^{2} ∣ s ∣ / \sqrt{c^{2} s^{2} + a^{2}}$ [L2]. If $s = 0$ this is $0 < c$ ; if $s \neq 0$ , dividing numerator and denominator by $c ∣ s ∣ > 0$ gives $∣ z^{'} (s) ∣ = c / \sqrt{1 + a^{2} / (c^{2} s^{2})} < c$ because $a^{2} / (c^{2} s^{2}) > 0$ [L3]. Hence $∣ z^{'} (s) ∣ < c$ for every $s$ .

2.1step 1.1L1L3algebra

No uniform bound (a, continued). For $s > 0$ , $∣ z^{'} (s) ∣ = c / \sqrt{1 + a^{2} / (c^{2} s^{2})} \to c$ as $s \to \infty$ , since $a^{2} / (c^{2} s^{2}) \to 0$ and the square root and reciprocal are continuous on their domains [L1]; the same limit holds as $s \to - \infty$ by symmetry. Consequently, for every $v_{*} < c$ there is $s$ with $∣ z^{'} (s) ∣ > v_{*}$ : choose $s$ so large that $∣ z^{'} (s) ∣ > v_{*}$ , possible because the values approach $c$ from below and $v_{*} < c$ ; thus no $v_{*} < c$ bounds $∣ z^{'} ∣$ on all of $R$ .

3.1step 1.1step 2.1L3L4algebra∎

No retarded root at $(0, 0)$ (b) and scope. At $(x, t) = (0, 0)$ the equation $t - s = ∣ x - z (s) ∣ / c$ of [L4] reads $- s = ∣ z (s) ∣ / c = \sqrt{c^{2} s^{2} + a^{2}} / c$ , and the right-hand side is $\geq a / c > 0$ ; hence any solution would satisfy $s < 0$ , so both sides are nonnegative and squaring is equivalent. Squaring gives $s^{2} = (c^{2} s^{2} + a^{2}) / c^{2}$ , that is $c^{2} s^{2} = c^{2} s^{2} + a^{2}$ and then $a^{2} = 0$ , contradicting $a > 0$ ; therefore no solution $s \in R$ exists at this event. Since $z$ is pointwise subluminal by step 1.1 but carries no uniform bound by step 2.1, this exhibits a smooth curve outside the hypotheses of the root lemma [L4] for which the retarded equation fails: the uniform bound cannot be dropped, and no theorem is contradicted because the curve is not in its hypothesis class. Clause (c) records that the curve is a mathematical object, not the worldline of a physical charge, and that no measurement is reported.

A smooth pointwise subluminal curve with no retarded root at an event

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources