Generalized Momentum and when it's conserved

Ex: a particle moving through space with speed x'

T=$\frac{1}{2}mx’^2$

Momentum $mx’$ can be expressed as $\frac{\partial T}{\partial x’}$

Under the condition that V is not a function of velocity x’, then momentum can also be expressed as:

$$ p=\frac{\partial L}{\partial x’} $$

Generalized/“conjugate”/“canonical” momentum #

In general, generalized momentum $p_k$ corresponding to generalized coordinate $q_k$:

$$ p_k=\frac{\partial L}{\partial q_k’} $$

Utilizing this in the lagrangian equation for a conservative system:

$$ p_k’=\frac{\partial L}{\partial q_k} $$

Cyclical/ignorable coordinate #

Let’s say there’s a generalized coordinate $q_K$ for which ‘L’ is not dependent on.

$$ \begin{align} \frac{\partial L}{\partial q_K}=0\Rightarrow p_K’=0\\ \frac{d}{dt}(p_K)=0\\ p_K\\ or\\ \frac{\partial L}{\partial q_k’}\\ is\\ constant \end{align} $$

Example #

Lagriangian of a particle moving in a central force field, in polar coordinates:

$L=T-V=\frac{1}{2}m(r’^2+r^2\theta ‘^2)-V(r)$

Since L doesn’t contain θ,

$$ p_\theta=\frac{\partial L}{\partial \theta ‘}=mr^2\theta’=constant $$

Generalized momentum for this coordinate, p_θ is a constant of motion in time and is a “first integral” since it is gotten by integrating p_θ’=0

Takeaway #

If a coordinate corresponding to a translation along a direction is not appearing in the expression of the lagrangian of a system, the coordinate is cyclic and the corresponding conjugate momentum is constant.