ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Zero force leaves an exponentially growing acceleration branch

Physical scope

A constructed mathematical example in the adopted Abraham–Lorentz ODE: zero prescribed force admits an exponentially growing acceleration branch, and position plus velocity at one time do not select it. This is a model property, not an observed particle trajectory or an empirical claim.

Example

In the adopted equation, set $f = 0$ and fix any $t_{0} \in R$ . For every constant acceleration vector $C \in R^{3}$ with units $m s^{- 2}$ , the acceleration $a_{C} (t) = C \exp ((t - t_{0}) / τ)$ is a solution. For any fixed position $x_{0}$ and velocity $v_{0}$ at $t_{0}$ , a trajectory with that acceleration is

$x_{C} (t) = x_{0} + v_{0} (t - t_{0}) + C τ^{2} (\exp ((t - t_{0}) / τ) - 1 - \frac{t - t_{0}}{τ}) .$

Thus the same position and velocity at $t_{0}$ admit different acceleration branches; $C \neq 0$ grows exponentially as $t \to \infty$ . This does not establish that any physical charge follows such a trajectory.

Example references

Runaway and bounded-future branches of the Abraham–Lorentz equation

Facts & Assumptions

Given: The adopted equation with $f = 0$ , $m, τ > 0$ , and prescribed data $x (t_{0}) = x_{0}$ , $\dot{x} (t_{0}) = v_{0}$ .

[A1]

For zero force, every acceleration solution is $a_{*} (t) + C \exp ((t - t_{0}) / τ)$ , and the bounded-future branch is $a_{*} = 0$ . (Runaway and bounded-future branches of the Abraham–Lorentz equation)

[F2]

The real exponential is defined by its convergent power series and has derivative $\exp^{'} = \exp$ ; the usual product and chain rules hold. (The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , 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 chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ )

Verification

1.1A1F2algebra

Since $\frac{d}{d t} \exp ((t - t_{0}) / τ) = τ^{- 1} \exp ((t - t_{0}) / τ)$ , the acceleration $a_{C}$ satisfies $a_{C} - τ {\dot{a}}_{C} = 0$ , hence $m (a_{C} - τ {\dot{a}}_{C}) = f$ .

2.1A1F2step 1.1algebra∎

Differentiating $x_{C}$ twice gives ${\ddot{x}}_{C} = a_{C}$ ; at $t = t_{0}$ the exponential factor is $1$ , so $x_{C} (t_{0}) = x_{0}$ and ${\dot{x}}_{C} (t_{0}) = v_{0}$ . Choosing $C = 0$ and any $C \neq 0$ therefore gives different solutions with the same position and velocity data, and the nonzero branch is unbounded in the future by exponential growth.

Zero force leaves an exponentially growing acceleration branch

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources