14.4 : Zeno's paradoxes---Flying arrow is at rest

Paradox 14.9 [Zeno's paradox] [Flying arrow is at rest]

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Consider a flying arrow. In any one instant of time, the arrow is at rest. Therefore, If the arrow is motionless at every instant, and time is entirely composed of instants, then motion is impossible.

[Achilles and a tortoise]

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I consider competition of Achilles and a tortoise. Let the start point of a tortoise (a late runner) be the front from the starting point of Achilles (a quick runner). Suppose that both started simultaneously. If Achilles tries to pass a tortoise, Achilles has to go to the place in which a tortoise is present now. However, then, the tortoise should have gone ahead more. Achilles has to go to the place in which a tortoise is present now further. Even Achilles continues this infinite, he can never catch up with a tortoise.

Figure 14.10 [=Figure 1.1 in $\S$1.1: The location of quantum language in the history of world-description] Fig.1.1: the location of "quantum language" in the world-views

Problem 14.11 [The meaning of Zeno's paradox]

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Describe "flying arrow" and "Achilles an a tortoise" in (classical) quantum language!

Formulation 14.12 [The formulation of dynamical system theory in the narrow sense]
Dynamical system theory is formulated as follows. \begin{align} \underset{\mbox{}}{\fbox{Dynamical system theory}} = \underset{\mbox{}}{\fbox{①:State equation}} + \underset{\mbox{}}{\fbox{②:Measurement equation}} \tag{14.9} \end{align}

①: $ \underset{\mbox{}}{\fbox{State equation}} $ is as follows. Let $T={\mathbb R}$ be the time axis. For each $t ( \in T)$, consider the state space $\Omega_t = {\mathbb R }^n$ ($n$-dimensional real space). The state equation (Chap. 13 (13.2)) is defined by the following simultaneous ordinary differential equation of the first order

\begin{align} & \underset{\mbox{}}{\fbox{State equation}} = \left\{\begin{array}{ll} \frac{d\omega_1}{dt}{} (t)=v_1(\omega_1(t),\omega_2(t),\ldots,\omega_n(t),\epsilon_1(t), t) \\ \frac{d\omega_2}{dt}{} (t)=v_2(\omega_1(t),\omega_2(t),\ldots,\omega_n(t),\epsilon_2(t), t) \\ \cdots \cdots \\ \frac{d\omega_n}{dt}{} (t)=v_n (\omega_1(t),\omega_2(t),\ldots,\omega_n(t), \epsilon_n(t),t) \end{array}\right. \tag{14.10} \end{align} where $\epsilon_k(t)$ is a noise ($k=1,2, \cdots, n $).

②: $ \underset{\mbox{}}{\fbox{Measurement equation}} $ is as follows. Consider the measured value space $X = {\mathbb R }^m$ ($m$-dimensional real space). The measurement equation (Chap.13 (13.2)) is defined by

\begin{align} & \underset{\mbox{}}{\fbox{Measurement equation}} = \left\{\begin{array}{ll} x_1(t)=g_1(\omega_1(t),\omega_2(t),\ldots,\omega_n(t),\eta_1(t), t) \\ x_2(t)=g_2(\omega_1(t),\omega_2(t),\ldots,\eta_n(t),\eta_2(t), t) \\ \cdots \cdots \\ x_m(t) =g_m (\omega_1(t),\omega_2(t),\ldots,\eta_n(t), \eta_n(t),t) \end{array}\right. \tag{14.11} \end{align}

where $g(=(g_1, g_2, \cdots, g_n)): \Omega \times {\mathbb R}^2 \to X$ is the system quantity and $\eta_k(t)$ is a noise ($k=1,2, \cdots, m $). Here, $x(t)(=(x_1(t), x_2(t), \cdots, x_n(t)))$ is called a motion function.

Answer 14.13[The dynamical system theoretical answer to "flying arrow (in Paradox 14.9)"]

Let $q(t)$ be the position of the flying arrow at time $t$. That is, consider the motion function $q(t)$.

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Note that the following logic (i.e., Zeno's logic ) is wrong: $\bullet$ for each time $t$, the position $q(t)$ of the flying arrow is determined. $\Longrightarrow$ the motion function $q$ is a constant function

Thus, Zeno's logic is wrong.



[The dynamical system theoretical answer to "Achilles and a tortoise (in Paradox 14.9)"]

For example, assume that the velocity $v_q$ [resp. $v_s$] of the quickest [resp. slowest] runner is equal to $v(>0)$ [resp. ${\gamma}v \; ( 0<{\gamma}<1)$]. And further, assume that the position of the quickest [resp. slowest] runner at time $t=0$ is equal to $0$ [resp. $a \; (>0)$]. Thus, we can assume that the position ${\xi (t)}$ of the quickest runner and the position $\eta (t)$ of the slowest runner at time $t$ $( \ge 0)$ is respectively represented by

\begin{eqnarray} \left\{\begin{array}{ll} \xi (t) =vt \\ \eta(t) = {\gamma}vt + a \end{array}\right. \tag{14.12} \end{eqnarray}

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Calculations

The formula (14.8) can be calculated as follows (i.e., (i) or (ii)):

[(i): Algebraic calculation of (14.8)]:

Solving $\xi(s_0)=\eta(s_0)$, that is, \begin{align} vs_0 = \gamma v s_0 + a \end{align} we get $s_0= \frac{a}{{(1-\gamma)} v}$. That is, at time $s_0= \frac{a}{{(1-\gamma)} v}$, the fast runner catches up with the slow runner.

[(ii): Iterative calculation of (14.8)]:

Define $t_k$ $(k=0,1,...)$ such that, $t_0=0$ and

\begin{align} t_{k+1}= \gamma v t_k + a \;\;\;\; (k=0,1,2,...) \end{align}

Thus, we see that $t_k=\frac{(1-{\gamma}^k)a}{(1-{\gamma})v}$ $(k=0,1,...)$. Then, we have that

\begin{align} \big( \xi (t_k), \eta (t_k) \big) & = \big( \frac{(1-{\gamma}^k)a}{1-{\gamma}}, \frac{(1-{\gamma}^{k+1})a}{1-{\gamma}} \big) \nonumber \\ & \to \big( \frac{a}{1-{\gamma}},\frac{a}{1-{\gamma}} \big) \tag{14.13} \end{align}

as $k \to \infty $. Therefore, the quickest runner catches up with the slowest at time $s_0 =\frac{a}{(1-{\gamma})v}$.

[(iii): Conclusion]:

After all, by the above (i) or (ii), we can conclude that

$(\sharp):$

the quickest runner can overtake the slowest at time $s_0 =\frac{a}{(1-{\gamma})v}$.

Answer 14.14 [The answer to Problem14.11 ] or [Answer to Problem 14.9: Zeno's paradox(flying arrow)]

In Corollary 14.7, putting \begin{align} q(t)=y_t( = g_t(\phi_{{t_0},t } (\omega_{t_0} ))) \end{align} we get the time-position function $q(t)$.