# 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 ${\displaystyle {\vec {r}}}$ 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,

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

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

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

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,

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

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

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

so if the torque vanishes, the angular momentum is conserved. Specifically if the force is always directed along the position vector, ${\displaystyle {\vec {r}}}$, 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

${\displaystyle W_{12}=\int _{1}^{2}{\vec {F}}\cdot d{\vec {s}}.}$

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

${\displaystyle W_{12}=\int _{1}^{2}{\vec {F}}\cdot d{\vec {s}}=m\int _{1}^{2}{\frac {d{\vec {v}}}{dt}}\cdot {\vec {v}}dt={\frac {m}{2}}\int _{1}^{2}{\frac {d}{dt}}\left(v^{2}\right)dt={\frac {m}{2}}\left(v_{2}^{2}-v_{1}^{2}\right)=T_{2}-T_{1}}$

where we have defined the kinetic energy,

${\displaystyle T={\frac {1}{2}}mv^{2}}$.

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

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

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

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

call the potential energy ${\displaystyle V}$. In this case we find that

${\displaystyle W_{12}=V_{1}-V_{2}=T_{2}-T_{1}}$

so

${\displaystyle T_{1}+V_{1}=T_{2}+V_{2}}$

This is called the conservation of mechanical energy.