PhyslibAlpha.ClassicalMechanics.NortonDome.PhysicalSpace

For a Norton dome characterized by a parameter $g$ (representing gravitational acceleration), the horizontal distance $x(s)$ from the apex at a given arc length $s$ is defined by the function: $$x(s) = \frac{2g^2}{3} \left( 1 - \left( 1 - \frac{s}{g^2} \right)^{3/2} \right)$$ The profile of the dome is valid for arc lengths $0 \le s \le g^2$.

For a Norton dome with gravitational acceleration $g$, the horizontal distance $x(s)$ from the apex at arc length $s$ is given by the formula: $$x(s) = \frac{2g^2}{3} \left( 1 - \left( 1 - \frac{s}{g^2} \right)^{3/2} \right)$$

For a Norton dome, the horizontal distance $x(s)$ from the apex evaluated at arc length $s = 0$ is $0$.

Let $x(s)$ be the horizontal distance of the Norton dome profile from the apex as a function of the arc length $s$, and let $g$ denote the characteristic gravitational parameter of the dome. For any $s \in \mathbb{R}$, the derivative of the horizontal distance with respect to the arc length is given by: $$\frac{dx}{ds} = \sqrt{1 - \frac{s}{g^2}}$$

Let $x(s)$ be the horizontal distance of the Norton dome profile from the apex, considered as a function of the arc length $s \in \mathbb{R}$. The function $x$ is differentiable on $\mathbb{R}$.

For any real number $s$ representing the arc length such that $0 \le s \le g^2$, where $g$ is the characteristic parameter of the Norton dome, the following identity holds: $$\left( \sqrt{1 - \frac{s}{g^2}} \right)^2 + \left( \frac{\sqrt{s}}{g} \right)^2 = 1$$ In the context of the dome's geometry, this identity represents the unit-speed property $x'(s)^2 + h'(s)^2 = 1$, where $x'(s) = \sqrt{1 - s/g^2}$ is the derivative of the horizontal distance and $h'(s) = \sqrt{s}/g$ is the derivative of the vertical depth with respect to the arc length $s$.

The profile of the Norton dome is a curve in $\mathbb{R}^2$ (representing the vertical plane through the apex) parameterized by the arc length $s \in \mathbb{R}$. The position at arc length $s$ is given by the vector $(x(s), -h(s))$, where $x(s)$ is the horizontal distance from the apex and $h(s)$ is the depth of the dome surface below the apex.

For any real number $s$ representing the arc length, the first component (index 0) of the Norton dome's profile curve $P(s)$ is equal to the horizontal distance $x(s)$ from the apex. Specifically, if the profile is given by $P(s) = (x(s), -h(s))$, then $P(s)_0 = x(s)$.

For any arc length $s \in \mathbb{R}$, the vertical coordinate (the second component) of the Norton dome's profile curve at $s$ is equal to $-h(s)$, where $h(s)$ is the depth of the dome surface below the apex.

The profile curve $\text{profile}(s)$ of the Norton dome evaluated at arc length $s = 0$ is the origin $0$. This identifies the apex of the dome with the origin of the coordinate system.

The Norton dome is defined as the subset of the vertical plane $\mathbb{R}^2$ through its apex formed by the image of the profile curve $\text{profile}(s)$ for arc lengths $s$ in the closed interval $[0, g^2]$. That is, $\text{dome} = \{ \text{profile}(s) \mid 0 \le s \le g^2 \}$, where $g$ is the gravitational acceleration parameter.

For any real number $s$ representing the arc length such that $0 \le s \le g^2$, where $g$ is the gravitational acceleration, the point $\text{profile}(s)$ on the profile curve of the Norton dome is an element of the set $\text{dome}$.

The function maps a chart trajectory $r(t)$, which represents the arc length of a particle on the Norton dome as a function of time, to its corresponding trajectory in 2-dimensional physical space. At each time $t$, the position in the vertical plane is given by the profile curve evaluated at the current arc length $r(t)_0$, i.e., $t \mapsto \text{profile}(r(t)_0)$.

For a chart trajectory $r(t)$ representing the arc length $s = r(t)_0$ of a particle on a Norton dome at time $t$, the horizontal component (the 0-th coordinate) of the particle's physical position vector in 2D space is equal to the horizontal distance function $x(s)$ evaluated at the current arc length. That is, $$(\text{spaceTrajectory } r)(t)_0 = x(r(t)_0)$$ where $x(s) = \frac{2g^2}{3} \left( 1 - \left( 1 - \frac{s}{g^2} \right)^{3/2} \right)$ is the horizontal coordinate of the dome's profile.

For a trajectory $r(t)$ representing the arc length of a particle on the Norton dome at time $t$, the vertical coordinate (the second component) of its position in 2D physical space is the negative of the depth of the dome at that arc length: $$(\text{spaceTrajectory}(r, t))_1 = -h(r(t)_0)$$ where $h$ is the depth function of the Norton dome and $r(t)_0$ is the scalar value of the arc length at time $t$.

Let $r(t)$ be a trajectory representing the arc length of a particle as a function of time $t$. If at time $t$ the arc length $r(t)$ satisfies $0 \le r(t) \le g^2$ (where $g$ is the gravitational acceleration), then the particle's position in 2-dimensional physical space, given by the mapping $\text{spaceTrajectory}(r, t)$, is an element of the set defining the Norton dome.

Let $r: \text{Time} \to \mathbb{R}$ be an arc length trajectory on the Norton dome. If $r$ is differentiable with respect to time, then the corresponding physical trajectory in 2-dimensional space, given by $t \mapsto \text{spaceTrajectory}(r, t)$, is also differentiable with respect to time.

Let $r: \text{Time} \to \mathbb{R}$ be a differentiable arc length trajectory of a particle on a Norton dome, and let $g$ denote the gravitational acceleration. The velocity of the particle in the vertical 2D plane at time $t$ is the derivative of its physical space trajectory, given by: $$\frac{d}{dt} \text{spaceTrajectory}(r, t) = \left( \sqrt{1 - \frac{r(t)}{g^2}} \dot{r}(t), -\frac{\sqrt{r(t)}}{g} \dot{r}(t) \right)$$ where $\dot{r}(t)$ is the rate of change of the arc length with respect to time. This velocity vector corresponds to the unit tangent vector to the dome's profile curve multiplied by the scalar speed $\dot{r}$.

Let $r(t)$ be a differentiable trajectory representing the arc length of a particle on the Norton dome. If at time $t$ the arc length is within the valid range $0 \le r(t) \le g^2$, where $g$ is the acceleration due to gravity, then the square of the speed of the particle in 2D physical space (the norm of the time derivative of the physical trajectory) is equal to the square of the time derivative of the arc length: $$\|\frac{d}{dt} \mathbf{x}(t)\|^2 = (\dot{r}(t))^2$$ where $\mathbf{x}(t)$ is the position of the particle in the vertical plane (the `spaceTrajectory`).

Let $r: \text{Time} \to \mathbb{R}^1$ be a differentiable trajectory representing the arc length of a particle of mass $m$ on a Norton dome with gravitational acceleration $g$. For any time $t$ such that the arc length $s = r(t)_0$ is within the valid range $0 \le s \le g^2$, the chart kinetic energy is equal to the kinetic energy in 2D physical space: $$T(t) = \frac{1}{2} m \left\| \frac{d}{dt} \mathbf{x}(t) \right\|^2$$ where $\mathbf{x}(t)$ (the `spaceTrajectory`) is the position of the particle in the vertical plane.

For any trajectory $r(t)$ representing the arc length of a particle on the Norton dome, and for any time $t$, the chart potential energy $V(r(t))$ is equal to the gravitational potential energy in physical space, $m g y(t)$, where $m$ is the mass, $g$ is the gravitational acceleration, and $y(t)$ is the vertical coordinate (height) of the particle in 2D physical space at time $t$.

Consider a particle of mass $m$ on a Norton dome with gravitational acceleration $g$. Let $r(t)$ be a differentiable trajectory representing the arc length of the particle from the apex. For any time $t$ such that the arc length $s = r(t)_0$ is within the range $0 \le s \le g^2$, the chart Lagrangian $L$ evaluated at the current state $(t, s, \dot{s})$ is equal to the constrained Lagrangian of the particle in physical space: $$L(t, s, \dot{s}) = \frac{1}{2} m \|\mathbf{v}\|^2 - m g y$$ where $\mathbf{v}$ is the velocity vector $\frac{d}{dt} \mathbf{q}(t)$ of the particle in 2D physical space and $y$ is its vertical coordinate (height).

Consider a particle of mass $m$ on a Norton dome under gravitational acceleration $g$. Let $r(t)$ be a differentiable trajectory where $r(t)_0$ represents the arc length from the apex at time $t$. For any time $t$ such that the arc length $s = r(t)_0$ is within the valid range $0 \le s \le g^2$, the total chart energy $E(t)$ is equal to the total mechanical energy of the particle in 2D physical space: $$E(t) = \frac{1}{2} m \|\mathbf{v}(t)\|^2 + m g y(t)$$ where $\mathbf{v}(t)$ is the velocity vector of the particle in physical space (the derivative of the spatial trajectory) and $y(t)$ is its vertical coordinate (height).