Constant of Motion

A conserved quantity for a given Trajectory. This is true of the Total Energy of a system, however other quantities can also be constants of motion

Properties

For a velocity independent force, an energy function is conserved iff the potential energy gradient is the symmetric of the force

$E(x,\dot{x})=\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x)$ is a Constant of Motion $\iff -\nabla V=F$, where

• $E(x,\dot{x})=\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x)$ is the Energy function aka the Total Energy
• $V(x):{\mathbb{R}}^n\to {\mathbb{R}}^n$ is the Potential energy
• $F(x(t))=m\ddot{x}$ is the velocity-independent Force

Proof

We differentiate the potential energy to get

$$\begin{array}{rll}\frac{d}{dt}\left(\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x)\right)& =m{\sum }_{j=1}^n{\dot{x}}_j{\ddot{x}}_j+{\sum }_{j=1}^n\frac{\partial V}{\partial x_j}{\dot{x}}_j& \\ & =\dot{x}\cdot \Vert m\ddot{x}+\nabla V\Vert & \\ & =\dot{x}\cdot \Vert F(x)+\nabla V\Vert & \end{array}$$

Since $\dot{x}(t)$ changes over time, the derivative is zero iff $F(x)+\nabla V=0$

For a velocity dependent force, an energy function is conserved iff the force can be decomposed into a component resulting from potential energy, and a component where the force is orthogonal to the velocity

$E(x,\dot{x})=\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x)$ is a Constant of Motion $\iff v\cdot F_2(x,\dot{x})=0$, where

• $E(x,\dot{x})=\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x)$ , the Energy function aka the Total Energy
• $V(x):{\mathbb{R}}^n\to {\mathbb{R}}^n$ , a smooth function representing the Potential energy
• $F(x,v):=-\nabla V(x)+F_2(x,v)$ , the velocity-dependent Force
• $F_2:{\mathbb{R}}^n\times {\mathbb{R}}^n\to {\mathbb{R}}^n$, the Force component not explained by the Potential energy
• $v=\dot{x}$ is the velocity

Proof

$$\begin{array}{rl}\frac{d}{dt}\left(\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x(t))\right)& =m{\sum }_{j=1}^n{\dot{x}}_j(t){\ddot{x}}_j(t)+{\sum }_{j=1}^n\frac{\partial V}{\partial x_j}{\dot{x}}_j(t)\\ & =\dot{x}(t)\cdot \left[m\ddot{x}(t)+\nabla V\right]\end{array}$$

Since the force $m\ddot{x}=F(x,v)$ is decomposed as $F(x,v)=-\nabla V+F_2(x,v)$ , we get

$$\begin{array}{rl}\frac{d}{dt}\left(\frac{1}{2}m\Vert \dot{x}{\Vert }^2+V(x(t))\right)& =\dot{x}(t)\cdot \left[F(x)+\nabla V\right]\\ & =\dot{x}(t)\cdot [-\nabla V+F_2(x,v)+\nabla V]\\ & =\dot{x}(t)\cdot F_2(x,v)\end{array}$$

Hence the Energy is constant iff $F_2$ is orthogonal to the velocity $v=\dot{x}$ .

• I need to rename the results above so that they fit into nice notes.