PhyslibAlpha.ClassicalMechanics.NortonDome.Solution

For a given departure time $T \in \mathbb{R}$ and an instant $\tau \in \mathbb{R}$, the delayed quartic function is defined as $$\frac{(\max(\tau - T, 0))^4}{144}$$ This function is zero for all $\tau \le T$ and follows the power law $\frac{(\tau - T)^4}{144}$ for $\tau > T$. It represents the position of a particle on the Norton dome that remains at rest at the apex until time $T$.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$, the delayed quartic function is given by the expression $$\text{delayedQuartic}(T, \tau) = \frac{(\max(\tau - T, 0))^4}{144}.$$

For any real numbers $T$ and $\tau$, the delayed quartic function $\frac{(\max(\tau - T, 0))^4}{144}$ is non-negative, i.e., $$0 \leq \frac{(\max(\tau - T, 0))^4}{144}$$ where $T$ represents the instant the particle leaves the apex and $\tau$ is the current time.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$ such that $\tau \leq T$, the delayed quartic function $\text{delayedQuartic}(T, \tau)$, defined as $\frac{(\max(\tau - T, 0))^4}{144}$, is equal to $0$.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$, if $\tau > T$, then the delayed quartic function $\frac{\max(\tau - T, 0)^4}{144}$ is strictly positive.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$, the square root of the delayed quartic function $r(\tau) = \frac{(\max(\tau - T, 0))^4}{144}$ is given by $$\sqrt{r(\tau)} = \frac{(\max(\tau - T, 0))^2}{12}.$$ This identity shows that the square root of the position function matches its second derivative $\ddot{r}(\tau)$, thereby satisfying the equation of motion $\ddot{r} = \sqrt{r}$ for a particle on the Norton dome.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$, the delayed quartic function $r_T(\tau) = \frac{(\max(\tau - T, 0))^4}{144}$ is differentiable at $\tau$ with derivative $$\frac{d}{d\tau} r_T(\tau) = \frac{(\max(\tau - T, 0))^3}{36}$$

For a given departure time $T \in \mathbb{R}$, let the delayed quartic function be defined as $f_T(\tau) = \frac{\max(\tau - T, 0)^4}{144}$. The derivative of this function with respect to time $\tau$ is given by: $$\frac{d}{d\tau} f_T(\tau) = \frac{(\max(\tau - T, 0))^3}{36}$$ This represents the velocity of a particle on the Norton dome that begins to slide off the apex at time $T$.

For any departure time $T \in \mathbb{R}$ and any instant $\tau \in \mathbb{R}$, the second derivative of the delayed quartic function $f_T(\tau) = \frac{\max(\tau - T, 0)^4}{144}$ exists and is given by: $$\frac{d^2}{d\tau^2} \left( \frac{\max(\tau - T, 0)^4}{144} \right) = \frac{\max(\tau - T, 0)^2}{12}$$

For any $T \in \mathbb{R}$, the delayed quartic function $f_T(\tau) = \frac{\max(\tau - T, 0)^4}{144}$ is of class $C^2$ (twice continuously differentiable) on $\mathbb{R}$.

For a given departure time $T \in \mathbb{R}$, the function maps an instant $t$ to the position of a particle on the Norton dome in a one-dimensional Euclidean space $\mathbb{R}^1$. The position is given by the vector: $$r_T(t) = \frac{(\max(t - T, 0))^4}{144} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the unit basis vector of $\mathbb{R}^1$. This trajectory describes a particle that remains at rest at the apex (arc length $0$) until time $T$, and subsequently slides away such that its arc length follows the quartic path $\frac{(t - T)^4}{144}$.

For any departure time $T \in \mathbb{R}$ and any instant $t$, the arc length of the motion leaving the apex of the Norton dome at time $T$ (represented as the $0$-th component of the position vector $r_T(t)$) is equal to the delayed quartic function evaluated at $t$: $$(r_T(t))_0 = \frac{(\max(t - T, 0))^4}{144}$$

Let $r_T(t)$ be the trajectory of a particle on the Norton dome that leaves the apex at time $T \in \mathbb{R}$. For any time $t$ such that $t \le T$, the particle remains at the apex, that is, $r_T(t) = 0$.

Let $r_T(t)$ denote the trajectory of a particle on the Norton dome that is scheduled to leave the apex at time $T \in \mathbb{R}$. For any departure time $T \ge 0$, the position of the particle at the initial time $t = 0$ is at the apex, i.e., $r_T(0) = 0$.

For any departure time $T \in \mathbb{R}$ and any time $t > T$, the position of the particle on the Norton dome $r_T(t)$ has a strictly positive arc-length coordinate, i.e., $r_T(t)_0 > 0$.

For any departure time $T \in \mathbb{R}$, the motion $r_T: \text{Time} \to \mathbb{R}^1$ leaving the apex of the Norton dome at time $T$ is twice continuously differentiable (i.e., it is a $C^2$ function).

For any departure time $T \in \mathbb{R}$, the velocity of the motion $r_T(t)$ leaving the apex of the Norton dome at time $T$ is given by the time derivative: $$\frac{d}{dt} r_T(t) = \frac{(\max(t - T, 0))^3}{36} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the standard unit basis vector in the one-dimensional Euclidean space $\mathbb{R}^1$.

For any departure time $T \in \mathbb{R}$, the second time derivative (acceleration) of the motion $r_T(t)$ on the Norton dome is given by $$\frac{d^2}{dt^2} r_T(t) = \frac{(\max(t - T, 0))^2}{12} \mathbf{e}_0$$ where $t$ is the time and $\mathbf{e}_0$ is the unit basis vector in the one-dimensional Euclidean space $\mathbb{R}^1$.

For any departure time $T \ge 0$, the motion $r_T(t)$ on the Norton dome, which describes a particle leaving the apex at instant $T$, has a velocity of zero at the initial time $t = 0$. That is, $\dot{r}_T(0) = 0$.

For any departure time $T \in \mathbb{R}$, the motion $r_T(t)$ defined as the trajectory leaving the apex of the Norton dome at instant $T$ satisfies the equation of motion $\ddot{r} = \sqrt{r}$. Specifically, its acceleration $\frac{d^2}{dt^2} r_T(t) = \frac{(\max(t - T, 0))^2}{12}$ is equal to the square root of its arc length $\sqrt{r_T(t)}$.

For any real number $T$, the trajectory $r_T(t)$ defined by $$r_T(t) = \frac{(\max(t - T, 0))^4}{144} \mathbf{e}_0$$ (where $\mathbf{e}_0$ is the unit basis vector in $\mathbb{R}^1$) is a solution to the equation of motion for the Norton dome. This represents a particle that remains at rest at the apex until time $T$ and then begins to slide down the dome.

Let $r_T$ denote the trajectory of a particle on the Norton dome that leaves the apex at time $T$, defined as $$r_T(t) = \frac{(\max(t - T, 0))^4}{144} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the unit basis vector. For any two departure times $T, T' \in \mathbb{R}$ such that $T < T'$, the corresponding trajectories $r_T$ and $r_{T'}$ are distinct.

Let $r_T$ denote the trajectory of a particle on the Norton dome that remains at the apex until time $T \in \mathbb{R}$ and then slides off, defined by the function: $$r_T(t) = \frac{(\max(t - T, 0))^4}{144} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the unit basis vector in $\mathbb{R}^1$. The mapping from the departure time $T$ to the trajectory $r_T$ is injective; that is, distinct departure times $T_1 \neq T_2$ result in distinct physical motions $r_{T_1} \neq r_{T_2}$.

For any departure time $T \in \mathbb{R}$, the motion $r_T$ representing a particle that leaves the apex of the Norton dome at time $T$ is not identical to the zero motion (the constant trajectory $0$ where the particle remains at rest at the apex).

The constant trajectory $r(t) = 0$, representing a particle remaining at rest at the apex of the Norton dome, satisfies the equation of motion for the system $S$.

For the Norton dome system, the motion that consists of the particle remaining at rest at the apex (represented by the constant zero function) has total energy equal to $0$ for all times $t$.

There exist two trajectories $x, y : \text{Time} \to \mathbb{R}^1$ that are solutions to the equations of motion for the Norton dome system $S$ such that they have the same initial position $x(0) = y(0)$ and the same initial velocity $\dot{x}(0) = \dot{y}(0)$, but are distinct trajectories ($x \neq y$).

In the Norton dome system $S$, the uniqueness of solutions given initial conditions does not hold. Specifically, it is not the case that for any two solutions $x, y : \text{Time} \to \mathbb{R}^1$, the conditions $x(0) = y(0)$ and $\partial_t x(0) = \partial_t y(0)$ imply that $x = y$.

For any real number $T \geq 0$, there exists a trajectory $r : \text{Time} \to \mathbb{R}$ that is a solution to the Norton dome's equation of motion such that the particle starts from rest at the apex at time $t = 0$ (i.e., $r(0) = 0$ and its time derivative $\dot{r}(0) = 0$). This trajectory has the property that the particle remains at the apex for all times $t \leq T$ (satisfying $r(t) = 0$) and moves away from the apex for all times $t > T$ (satisfying $r(t) > 0$).

For any departure time $T \in \mathbb{R}$ and any instant $t$, the total energy of the motion $r_T$ (the trajectory where a particle leaves the apex of the Norton dome at time $T$) is equal to $0$.

Let $M$ be a normed real vector space and $f : \text{Time} \to M$ be a function representing a curve. If $f$ is differentiable and its time derivative $\partial_t f$ is also differentiable (i.e., $f$ is twice differentiable), then for any $t \in \text{Time}$, the second time derivative of the time-reversed curve $s \mapsto f(-s)$ evaluated at $t$ is equal to the second time derivative of $f$ evaluated at $-t$. In mathematical notation: $$\frac{d}{dt} \left( \frac{d}{dt} f(-t) \right) = \left. \frac{d^2 f}{d\tau^2} \right|_{\tau = -t}$$ where the left-hand side represents the second derivative of the composition of $f$ with the negation operation.

Let $S$ be a Norton dome system and $r: \text{Time} \to \mathbb{R}^1$ be a trajectory. If $r$ is twice differentiable (specifically, $r$ and its time derivative $\partial_t r$ are differentiable) and satisfies the equation of motion for $S$, then the time-reversed trajectory $t \mapsto r(-t)$ also satisfies the equation of motion. This symmetry arises because the equation of motion for the Norton dome has no velocity-dependent terms.

Let $S$ be a Norton dome system. If a motion $r: \text{Time} \to \mathbb{R}^1$ is a solution to the equations of motion for $S$, then the time-reversed motion defined by $t \mapsto r(-t)$ is also a solution for $S$.

For a given parameter $T \in \mathbb{R}$, this function describes the trajectory of a particle on the Norton dome that arrives at the apex at time $-T$. It is defined as the time reversal of the motion leaving the apex at time $T$. The position at time $t$ in the one-dimensional Euclidean space $\mathbb{R}^1$ is given by: $$\text{arrival}(T)(t) = \frac{(\max(-t - T, 0))^4}{144} \mathbf{e}_0$$ where $\mathbf{e}_0$ is the unit basis vector. This motion represents a particle sliding up the dome and coming to a permanent rest at the apex at the instant $t = -T$.

For any real number $T$, the trajectory $r: \text{Time} \to \mathbb{R}^1$ that arrives at the apex of the Norton dome at time $-T$, defined as $$r(t) = \frac{(\max(-t - T, 0))^4}{144} \mathbf{e}_0,$$ is a solution to the equations of motion for the Norton dome system $S$.

For any $T \in \mathbb{R}$, let $\text{arrival}(T)$ denote the trajectory of a particle on the Norton dome that arrives at the apex at time $-T$. For any time $t < -T$, the particle is off the apex, meaning its radial position coordinate satisfies $(\text{arrival}(T)(t))_0 > 0$.

For any real parameter $T$ and any time $t \geq -T$, the position of the particle in the motion $\text{arrival}(T)$ is $0$. This means that a particle arriving at the apex of the Norton dome at the instant $-T$ remains at rest at the apex for all subsequent times.

For any real parameter $T$, let $\text{arrival}(T)$ denote the trajectory of a particle on the Norton dome that arrives at the apex at time $-T$. For all times $t \geq -T$, the velocity of the particle (the time derivative of its position) is zero: $$\frac{d}{dt}(\text{arrival}(T))(t) = 0$$ This implies that a particle arriving at the apex at time $-T$ does so with zero velocity and remains at rest for all subsequent times.