PhyslibAlpha.ClassicalMechanics.NortonDome.Basic

In the Norton dome system, the mass $m$ of the particle is non-zero, $m \neq 0$.

For a Norton dome system, the gravitational acceleration $g$ satisfies $g \neq 0$.

The function $h(r)$ represents the depth of the Norton dome's surface below its apex at an arc length $r$ from the apex. It is defined by the formula: $$h(r) = \frac{2}{3g} \sqrt{r}^3$$ where $g$ is the gravitational acceleration. The height $h(r)$ vanishes at the apex ($r=0$) and for all negative arguments $r < 0$.

For any real number $r$, the depth $h(r)$ of the Norton dome's surface below its apex at arc length $r$ is given by the formula $$h(r) = \frac{2}{3g} \sqrt{r}^3$$ where $g$ is the gravitational acceleration.

For any real number $r$, the depth of the Norton dome $h(r)$ below its apex is non-negative, i.e., $h(r) \geq 0$. Here, $h(r)$ is defined as $\frac{2}{3g} \sqrt{r}^3$, where $g$ is the gravitational acceleration.

For a Norton dome, the depth $h(r)$ of the dome's surface below its apex at arc length $r$ is equal to zero if and only if $r \leq 0$.

For a Norton dome with gravitational acceleration $g$, let $h(r)$ be the function representing the depth of the surface below the apex at arc length $r$. For any $r \in \mathbb{R}$, the function $h$ is differentiable at $r$ with derivative $$\frac{d}{dr} h(r) = \frac{\sqrt{r}}{g}$$ where $\sqrt{r}$ is the square root of $r$ (defined as $0$ for $r < 0$).

The height function $h(r)$ of the Norton dome, defined by $h(r) = \frac{2}{3g} \sqrt{r}^3$ where $g$ is the gravitational acceleration and $r$ is the arc length from the apex, is differentiable for all $r \in \mathbb{R}$.

For the Norton dome, the derivative of the height function $h(r)$ with respect to the arc length $r$ is given by $$h'(r) = \frac{\sqrt{r}}{g}$$ where $g$ is the gravitational acceleration.

Let $h(r)$ be the function representing the depth of the Norton dome's surface below its apex at an arc length $r$, defined as $h(r) = \frac{2}{3g} (\sqrt{r})^3$ where $g$ is the gravitational acceleration. Then $h$ is a continuously differentiable function (of class $C^1$) on $\mathbb{R}$.

For a particle of mass $m$ on the Norton dome, given a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length, the kinetic energy is the function of time $t$ defined as: $$T(t) = \frac{1}{2} m \langle \dot{r}(t), \dot{r}(t) \rangle$$ where $\dot{r}(t)$ denotes the time derivative (velocity) of the trajectory at time $t$, and $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space.

The potential energy of a particle of mass $m$ on the Norton dome at a position $x \in \mathbb{R}^1$ is defined as $$V(x) = -m g h(x_0)$$ where $g$ is the gravitational acceleration, $x_0$ is the arc length from the apex, and $h(x_0)$ is the depth of the dome's surface below the apex. The potential is normalized to vanish at the apex ($x_0 = 0$).

For a particle on the Norton dome, given a trajectory $r: \text{Time} \to \mathbb{R}^1$ where $r(t)$ represents the position (arc length from the apex), the energy at time $t$ is the sum of its kinetic energy $T(t)$ and its potential energy $V(r(t))$: $$E(t) = T(t) + V(r(t))$$

For a particle of mass $m$ on a Norton dome, given a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length from the apex, the kinetic energy at time $t$ is given by: $$T(r)(t) = \frac{1}{2} m \langle \dot{r}(t), \dot{r}(t) \rangle$$ where $\dot{r}(t)$ denotes the time derivative of the trajectory at time $t$, and $\langle \cdot, \cdot \rangle$ denotes the standard inner product on the 1-dimensional Euclidean space.

For a particle of mass $m$ on a Norton dome, given a position $x \in \mathbb{R}^1$ where $x_0$ represents the arc length $r$ from the apex, the potential energy $V(x)$ is given by: $$V(x) = -\frac{2m}{3} r^{3/2}$$

For a particle on the Norton dome with a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length from the apex, the total energy $E$ at time $t$ is the sum of the kinetic energy $T$ and the potential energy $V$: $$E(t) = T(t) + V(r(t))$$

For any position $x$ in the 1-dimensional Euclidean space (representing the arc-length coordinate on the dome), the potential energy $V(x)$ of the particle is non-positive, i.e., $V(x) \leq 0$. This reflects the fact that the apex (where $V = 0$) is the highest point of the dome.

For a particle on the Norton dome at position $x$ in the 1-dimensional Euclidean space $\mathbb{R}^1$, the potential energy $V(x)$ is equal to zero if and only if the arc-length coordinate $x_0$ satisfies $x_0 \le 0$. This condition corresponds to the particle being at or "behind" the apex of the dome.

For every point $x$ in the 1-dimensional Euclidean space $\mathbb{R}^1$ (representing the arc-length coordinate), the function $y \mapsto (\sqrt{y_0})^3$ is differentiable at $x$, where $y_0$ is the coordinate of $y$.

The potential energy $V: \mathbb{R}^1 \to \mathbb{R}$ of a particle on the Norton dome is a differentiable function of the arc-length coordinate $x$.

For a particle of mass $m$ on the Norton dome, let $x \in \mathbb{R}^1$ denote its position, where $x_0 = r$ is the arc-length coordinate measured from the apex. The gradient of the potential energy $V(x)$ at position $x$ is given by $$\nabla V(x) = -m \sqrt{r} \mathbf{\hat{e}}_0,$$ where $\mathbf{\hat{e}}_0$ is the unit basis vector of the 1-dimensional arc-length coordinate.

Let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length of a particle on the Norton dome. If the velocity $\dot{r}(t)$ is differentiable with respect to time, then the kinetic energy $T(t) = \frac{1}{2} m \|\dot{r}(t)\|^2$ is a differentiable function of time.

Let $r: \text{Time} \to \mathbb{R}^1$ be a differentiable curve representing the position (arc length) of a particle on the Norton dome. Then the potential energy $V(r(t))$ is a differentiable function of time $t$.

Let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the position (arc length) of a particle on the Norton dome. If $r$ is differentiable and its velocity $\partial_t r$ is also differentiable (meaning the curve is twice differentiable), then the total energy $E(t)$ is a differentiable function of time.

Let $r : \text{Time} \to \mathbb{R}^1$ be a curve representing the arc length of a particle of mass $m$ on a Norton dome. If the velocity $\partial_t r$ is differentiable, then the rate of change of the kinetic energy $T$ is given by the inner product of the velocity and the mass times the acceleration: $$\partial_t T(t) = \langle \partial_t r(t), m \partial_t^2 r(t) \rangle_{\mathbb{R}}$$ where $\partial_t$ denotes the time derivative and $\langle \cdot, \cdot \rangle_{\mathbb{R}}$ is the standard inner product on the one-dimensional Euclidean space.

Let $r : \text{Time} \to \mathbb{R}^1$ be a differentiable curve representing the position of a particle on the Norton dome (parameterized by arc length). The rate of change of the potential energy $V$ along this curve is given by the inner product of the velocity $\partial_t r$ and the gradient of the potential $\nabla V$: $$\partial_t (V(r(t))) = \langle \partial_t r(t), \nabla V(r(t)) \rangle_{\mathbb{R}}$$ where $\partial_t$ denotes the time derivative and $\langle \cdot, \cdot \rangle_{\mathbb{R}}$ is the standard inner product on the one-dimensional Euclidean space.

Let $r : \text{Time} \to \mathbb{R}^1$ be a curve representing the trajectory (arc length from the apex) of a particle of mass $m$ on the Norton dome. If $r$ is twice differentiable (i.e., both the position $r$ and the velocity $\dot{r}$ are differentiable), then the time derivative of the total energy $E$ is the inner product of the velocity and the sum of the mass-acceleration term $m \ddot{r}$ and the gradient of the potential energy $\nabla V$: $$\frac{d}{dt} E(t) = \langle \dot{r}(t), m \ddot{r}(t) + \nabla V(r(t)) \rangle_{\mathbb{R}}$$ where $\dot{r}(t) = \partial_t r(t)$ is the velocity, $\ddot{r}(t) = \partial_t^2 r(t)$ is the acceleration, and $\langle \cdot, \cdot \rangle_{\mathbb{R}}$ denotes the standard inner product on the one-dimensional Euclidean space.

For a particle of mass $m$ on the Norton dome, the Lagrangian $L$ is a function of time $t$, position $x \in \mathbb{R}^1$ (represented as the arc length from the apex), and velocity $v \in \mathbb{R}^1$, defined as the difference between the kinetic energy and the potential energy: $$L(t, x, v) = \frac{1}{2} m \|v\|^2 - V(x)$$ where $\|v\|^2 = \langle v, v \rangle_{\mathbb{R}}$ is the square of the norm of the velocity and $V(x)$ is the potential energy at position $x$. The Lagrangian is independent of the time $t$.

For any trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length from the apex of the Norton dome and for any time $t$, the Lagrangian $L$ evaluated at position $r(t)$ and velocity $\dot{r}(t)$ is equal to the kinetic energy $T$ at time $t$ minus the potential energy $V$ at position $r(t)$: $$L(t, r(t), \dot{r}(t)) = T(t) - V(r(t))$$ where $\dot{r}(t)$ denotes the time derivative of $r$ at $t$.

For a particle of mass $m$ on a Norton dome, let $L(t, x, v)$ be the Lagrangian of the system, where $t$ is the time, $x \in \mathbb{R}^1$ is the position vector (with $x_0$ representing the arc length from the apex), and $v \in \mathbb{R}^1$ is the velocity vector. The gradient of the Lagrangian with respect to the position $x$ is given by $$\nabla_x L(t, x, v) = m \sqrt{x_0} \mathbf{e}_1$$ where $\mathbf{e}_1$ is the unit basis vector in $\mathbb{R}^1$. This gradient is equal to the generalized force $F = -\nabla V$, where $V$ is the potential energy.

For a particle of mass $m$ on the Norton dome, let $L(t, x, v)$ be the Lagrangian at time $t$, position $x \in \mathbb{R}^1$, and velocity $v \in \mathbb{R}^1$. The gradient of the Lagrangian with respect to the velocity $v$ is equal to the momentum $m v$, which is expressed as: $$\nabla_v L(t, x, v) = m v$$

For a position $x$ in the one-dimensional configuration space $\mathbb{R}^1$ (representing the arc length from the apex), the generalized force $\mathbf{F}(x)$ acting on the particle on the Norton dome is defined as the negative gradient of the potential energy $V(x)$, namely $\mathbf{F}(x) = -\nabla V(x)$.

For a Norton dome with mass $m$, the generalized force $\mathbf{F}(x)$ acting on a particle at a position $x$ in the one-dimensional configuration space $\mathbb{R}^1$ (where the coordinate $x_0$ represents the arc length from the apex) is given by $$\mathbf{F}(x) = (m \sqrt{x_0}) \mathbf{e}_1$$ where $\mathbf{e}_1$ is the unit basis vector in $\mathbb{R}^1$. The force points away from the apex and vanishes at $x=0$.

For a Norton dome with mass $m$, the $0$-th component of the generalized force $F(x)$ at a position $x \in \mathbb{R}^1$ is given by $$F(x)_0 = m \sqrt{x_0},$$ where $x_0$ is the coordinate of $x$ representing the arc length from the apex.

For a Norton dome $S$, the generalized force $\mathbf{F}$ acting on the particle at the apex ($r=0$) is zero, i.e., $\mathbf{F}(0) = \mathbf{0}$.

The generalized force $\mathbf{F}$ acting on a particle on the Norton dome is continuous with respect to the arc length $r$.

In a one-dimensional Euclidean space, let $\mathbf{e}_0$ be the unit vector (represented by `EuclideanSpace.single 0 1`). For any two real numbers $a$ and $b$, the distance between the vectors $a \mathbf{e}_0$ and $b \mathbf{e}_0$ is equal to the absolute difference between the coefficients, $|a - b|$.

For any non-negative Lipschitz constant $K$ and any radius $\epsilon > 0$, the generalized force $\mathbf{F}$ of the Norton dome is not Lipschitz continuous with constant $K$ on the closed ball $\bar{B}_\epsilon(0)$ centered at the apex.

A trajectory $r: \text{Time} \to \mathbb{R}^1$ satisfies the equation of motion for a particle of mass $m$ on the Norton dome if, at every instant $t$, the product of the mass and the acceleration $\ddot{r}(t)$ equals the generalized force $\mathbf{F}(r(t))$. That is, $$m \ddot{r}(t) = \mathbf{F}(r(t))$$ where $r$ represents the arc length coordinate from the apex and $\ddot{r}$ denotes the second derivative with respect to time.

For a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length of a particle of mass $m$ on a Norton dome, the equation of motion is satisfied if and only if at every instant $t$, the sum of the mass multiplied by the acceleration $\ddot{r}(t)$ and the gradient of the potential energy $\nabla V(r(t))$ vanishes: $$m \ddot{r}(t) + \nabla V(r(t)) = 0.$$ Here, $\ddot{r}$ denotes the second derivative with respect to time, and $V$ is the potential energy of the system.

For a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length from the apex of a Norton dome, the equation of motion is satisfied if and only if for every instant $t$, the second time derivative of the position satisfies the scalar relation: $$\ddot{r}(t) = \sqrt{r(t)}$$ In this form, the mass $m$ and the gravitational acceleration $g$ have cancelled out.

A trajectory $r: \text{Time} \to \mathbb{R}^1$, representing the arc length coordinate from the apex, is a solution for the Norton dome $S$ if it satisfies the following conditions: 1. The position function $r$ is differentiable with respect to time. 2. The velocity function $\dot{r} = \partial_t r$ is differentiable with respect to time (making $r$ twice differentiable). 3. The trajectory satisfies the equation of motion for the Norton dome, $m \ddot{r}(t) = \mathbf{F}(r(t))$, where $m$ is the mass and $\mathbf{F}$ is the generalized force.

For a Norton dome $S$, let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length coordinate from the apex. If $r$ is a solution for the system $S$, then the position function $r$ is differentiable with respect to time.

Let $S$ be a Norton dome and $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length from the apex. If $r$ is a solution to the equations of motion for the dome, then its velocity $\dot{r} = \partial_t r$ is differentiable with respect to time.

Let $S$ be a Norton dome with mass $m$ and generalized force $\mathbf{F}$. If a trajectory $r : \text{Time} \to \mathbb{R}^1$, representing the arc length coordinate from the apex, is a solution for the Norton dome, then it satisfies the equation of motion $$m \ddot{r}(t) = \mathbf{F}(r(t))$$ at every instant $t$, where $\ddot{r}$ is the second derivative of $r$ with respect to time.

Consider a particle of mass $m$ on a Norton dome. Let $r: \text{Time} \to \mathbb{R}$ be a curve representing the arc length from the apex, and suppose that the velocity $\dot{r}$ is differentiable. Then the Euler-Lagrange operator applied to the Lagrangian $L$ along the trajectory $r$ is equal to the generalized force $F(r)$ minus the product of the mass and the acceleration $\ddot{r}$: $$\left( \text{EL}(L, r) \right)(t) = F(r(t)) - m \ddot{r}(t)$$ where the Euler-Lagrange operator is defined as $\frac{\partial L}{\partial r} - \frac{d}{dt} \frac{\partial L}{\partial \dot{r}}$.

Let $S$ be a Norton dome with mass $m$. For a trajectory $r: \text{Time} \to \mathbb{R}^1$ representing the arc length from the apex, suppose that the velocity $\dot{r}$ is differentiable. Then the trajectory satisfies the equation of motion $m \ddot{r}(t) = \mathbf{F}(r(t))$ if and only if the Euler-Lagrange operator of the Lagrangian $L$ vanishes along the curve, that is: $$\frac{\partial L}{\partial r} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = 0$$ where $\dot{r}$ and $\ddot{r}$ are the first and second time derivatives of $r$, $\mathbf{F}$ is the generalized force, and the Lagrangian is defined as $L = T - V$.

Let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length from the apex of a Norton dome $S$. Suppose that $r$ is twice differentiable (specifically, that $r$ and its velocity $\dot{r}$ are differentiable). If $r$ satisfies the equation of motion $m \ddot{r}(t) = \mathbf{F}(r(t))$, then the time derivative of the total energy $E(t)$ of the system is zero for all $t$: $$\frac{d}{dt} E(t) = 0$$ where $E(t)$ is the sum of the kinetic and potential energies at time $t$.

Consider a particle on a Norton dome $S$. Let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length from the apex. Suppose that $r$ is twice differentiable and satisfies the equation of motion $m \ddot{r}(t) = \mathbf{F}(r(t))$. Then for any time $t$, the total energy $E(t)$ of the particle is equal to its initial value $E(0)$.

For a Norton dome $S$, let $r: \text{Time} \to \mathbb{R}^1$ be a trajectory representing the arc length from the apex. If $r$ is a solution to the equation of motion (as defined by `IsSolution`), then for any time $t$, the total energy $E(t)$ of the particle is equal to its initial energy at time $t = 0$: $$E(t) = E(0)$$