Advanced Classical Mechanics/Energy and Angular Momentum

Let's start with some basic definitions. First m is the inertial mass of a particle -- we will consider only particles with constant masses (although the physics of rockets provides an important counter-example). A particle for us is something that can be described completely by its position $$\vec{r}$$ and mass (it doesn't have any important internal degrees of freedom).

Momentum and Force
We can define the velocity and momentum of the particle,

$$ \vec{v} = \frac{d \vec {r}}{dt} $$ and $$ \vec{p} = m \vec{v} $$.

Because the acceleration is the rate of change of velocity we have

$$ \vec{F} = m \vec{a} = m \frac{d \vec{v}}{dt} = \frac{d\vec{p}}{dt}, $$

so if the force vanishes, then the momentum is constant in time or conserved.

Angular Momentum and Torque
Let's also define the angular momentum and the torque,

$$ \vec{L} = \vec{r} \times \vec{p} $$ and $$ \vec{N} = \vec{r} \times \vec{F} = \vec{r} \times \frac{d \vec{p}}{d t}. $$

Let's calculate the rate of change of the angular momentum

$$ \frac{d}{dt} \vec{L} = \frac{d \vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d \vec{p}}{dt } = \vec{N} $$

so if the torque vanishes, the angular momentum is conserved. Specifically if the force is always directed along the position vector, $$\vec r$$, the torque vanishes and the angular momentum is conserved. This is called a central force.

Work and Energy
We can define the work done on a particle while moving it from position #1 to position #2 to be

$$ W_{12} = \int_1^2 \vec{F} \cdot d \vec{s}. $$

Again let's assume that the mass of the particle is constant so

$$ W_{12} = \int_1^2 \vec{F} \cdot d \vec{s} = m \int_1^2 \frac{d \vec{v}}{dt} \cdot \vec{v} d t = \frac{m}{2} \int_1^2 \frac{d}{dt} \left ( v^2 \right ) dt = \frac{m}{2} \left ( v_2^2 - v_1^2 \right ) = T_2 - T_1 $$

where we have defined the kinetic energy,

$$T=\frac{1}{2} m v^2$$.

If the integral $$W_{12}$$ depends only on the locations of the points 1 and 2 and not the path, then the force or system is conservative. In particular,

$$ \oint \vec{F} \cdot d \vec{s} = 0 $$ for such a system.

This means that the force is the gradient of some other function

$$ \vec{F} = -\vec{\nabla} V(r) $$

call the potential energy $$V$$. In this case we find that

$$ W_{12} = V_1 - V_2 = T_2 - T_1 $$

so

$$ T_1 + V_1 = T_2 + V_2 $$

This is called the conservation of mechanical energy.