Lagrangian - Aetherwisp

The Lagrangian $L$ is the difference between the kinetic $T$ and Potential $V$ energy of a conservative system:

L(q,\dot{q},t)=T(q,\dot{q},t)-V(q,t)

$q$ are the generalized coordinates.

Cyclical coordinates#

Consider the case of a Lagrangian in $n$ dimensions that does not explicitly depend on some coordinates $q_{m+1},\ldots,q_{n}$ , such that $L\equiv L(q_{1},\ldots,q_{m},\dot{q}_{1},\ldots,\dot{q}_{m},t)$ . These are known as cyclical coordinates (the remaining ones are free coordinates). These have special properties that make them useful to study. For one, $L$ is invariant over translations of cyclical coordinates. If we send $q_{l}$ to $q_{l}+c$ for some constant $c$ and $l=m+1,\ldots,n$ , the Lagrangian does not change[^1]. Also, $\partial L/ \partial q_{l} =0$ and therefore

\frac{d}{dt}\frac{ \partial L }{ \partial \dot{q}_{l} }=0

This means that all conjugate momenta $P_{l}$ of cyclical coordinates are constants of motion. This is very useful, as it means we can knock out $n-m$ variables and just solve the remaining $m$ ; the rest will follow suit.

For the sake of brevity, for the rest of this article we'll use $q_\text{free}\equiv(q_{1},\ldots,q_{m})$ to refer to the free coordinates alone. Similarly, $\dot{q}_\text{free}\equiv(\dot{q}_{1},\ldots,\dot{q}_{m})$ is the free velocities alone.

Effective Lagrangians#

The Lagrangian does not depend on cyclical coordinates. This means that, in theory, we can rewrite the Lagrangian in a form that explicitly does not include them. Such a Lagrangian is known as the effective Lagrangian of the system: it describes the same system, but with only variables that actually matter (i.e. have an effect).

To do so, consider the level set given by the conjugate momenta

P_{l}(q_\text{free},\dot{q}_{\text{free}},t,\underbrace{ \dot{q}_{m+1},\ldots,\dot{q}_{n} }_{ n-m\text{ cyclical unknowns} })=\tilde{P}_{l}=\text{constant}\tag{*}

This inverts to another level set

\dot{q}_{l}=u_{l}(q_\text{free},\dot{q}_\text{free},t,\tilde{P}_{m+1},\ldots,\tilde{P}_{n})\tag{**}

where $u_{l}$ is some function that inverts the $\tilde{P}_{l}$ relation above.

We can determine the effective Lagrangian over this level set as

\begin{align} &L_\text{eff}(q_\text{free},\dot{q}_\text{free},t,\tilde{P}_{m+1},\ldots,\tilde{P}_{n}) \\ &\equiv L(q_\text{free},\dot{q}_\text{free},t,u_{m+1}(q_\text{free},\dot{q}_\text{free},t,\tilde{P}),\ldots,u_{n}(q_\text{free},\dot{q}_\text{free},t,\tilde{P}))-\sum_{l=m+1}^{n} \tilde{P}_{l}u_{l}(q_\text{free},\dot{q}_\text{free},t,\tilde{P}) \end{align}

Without all of the dependencies it reads

\boxed{L_\text{eff}=\left.{\left( L-\sum_{l=m+1}^{n} \tilde{P}_{l}\dot{q}_{l} \right)}\right|_{\dot{q}_{l}=u_{l}(q_\text{free},\dot{q}_\text{free},t,\tilde{P})}}

This is great and all, but nowhere does it say that this is a valid solution to the system. We now look for the derivatives of $L_\text{eff}$ in order to prove this:

\frac{ \partial L_\text{eff} }{ \partial q_{h} } =\frac{ \partial L }{ \partial q_{h} } +\sum_{k=m+1}^{n} \underbrace{ \frac{ \partial L }{ \partial \dot{q}_{k} } }_{ P_{k} } \frac{ \partial u_{k} }{ \partial q_{h} } -\sum_{l=m+1}^{n} \tilde{P}_{l}\frac{ \partial u_{l} }{ \partial q_{h} }

and now the time derivatives:

\frac{d}{dt} \frac{ \partial L_\text{eff} }{ \partial \dot{q}_{h} } =\frac{d}{dt} \left[ \frac{ \partial L }{ \partial \dot{q}_{h} } +\sum_{k=m+1}^{n} \underbrace{ \frac{ \partial L }{ \partial \dot{q}_{k} } }_{ P_{k} } \frac{ \partial u_{k} }{ \partial \dot{q}_{h} } -\sum_{l=m+1}^{n} \tilde{P}_{l}\frac{ \partial u_{l} }{ \partial \dot{q}_{h} } \right]

If we subtract one from we get

E=\frac{d}{dt} \frac{ \partial L_\text{eff} }{ \partial \dot{q}_{h} } -\frac{ \partial L_\text{eff} }{ \partial q_{h} } \quad\text{(on the level set *)}

(???)

This takes care of the $m$ non-cyclical coordinates $q_{1},\ldots,q_{m}$ . We now need to figure out the remaining $q_{l}$ cyclical coordinates from these ones. To do so, we put the $q_{h}$ found from $E$ in the level set $(**)$ . This yields $\dot{q}_{l}(t)=F_{l}(t)$ , from which we can find $q_{l}(t)$ by integration.

Examples#

> which is a [[harmonic oscillator]] under a constant force $mg$, interpreted as [[Gravity|gravity]]. $L$ does not depend explicitly on neither $x$ nor $y$, which makes both cyclical coordinates. We should therefore be able to solve the system with just $z$. We find the conjugate momenta of $x$ and $y$ as > $$\tilde{P}_{x}=\frac{ \partial L }{ \partial \dot{x} } =m \dot{x}\quad\to \quad \dot{x}=\frac{\tilde{P}_{x}}{m}

> The conjugate momenta are constants of motion. We rewrite the Lagrangian as the effective one by substituting $\dot{x}$ and $\dot{y}$ with $\tilde{P}_{x}$ and $\tilde{P}_{y}$: > $$\begin{align} > L_\text{eff}(z,\dot{z},\tilde{P}_{x},\tilde{P}_{y})&=\frac{m}{2}\left( \left( \frac{\tilde{P}_{x}}{m} \right)^{2}+\left( \frac{\tilde{P}_{y}}{m} \right)^{2}+\dot{z}^{2} \right)-mgz- \frac{\tilde{P}_{x}^{2}}{m}- \frac{\tilde{P}_{z}^{2}}{m} \\ > &=\frac{m}{2}\dot{z}^{2}-mgz+\text{const.} > \end{align}

Take a time derivative and we get

> which is trivial to solve: > $$z(t)=- \frac{1}{2}gt^{2}+at+z_{0}

where $z_{0}$ is a constant. We have fully solved the system: $x$ and $y$ are determined by constants alone. In fact,

> $$\dot{y}=\frac{\tilde{P}_{y}}{m}\quad\to \quad y(t)=\frac{\tilde{P}_{y}t}{m}+y_{0}

> Both $\varphi$ and $z$ are cyclical coordinates. The conjugate momentum of $\varphi$ is > $$\frac{ \partial L }{ \partial \dot{\varphi} } =mr^{2}\dot{\varphi}+ \frac{eB}{2}r^{2}=l\quad\to \quad \dot{\varphi}=\frac{l- \frac{eB}{2}r^{2}}{mr^{2}}=\frac{l}{mr^{2}}- \frac{eB}{2m}

The effective Lagrangian is

> The last two terms becomes > $$\frac{eB}{2}r^{2}\left( \frac{l}{mr^{2}}- \frac{eB}{2m} \right)-l\left( \frac{l}{mr^{2}}- \frac{eB}{2m} \right)=-mr^{2}\left( \frac{l}{mr^{2}}- \frac{eB}{2m} \right)^{2}

The whole equation Lagrangian then is

> (TODO: Finish this; end of lesson 27/03/2025) ### Angular momentum There is a strong connection between [[rotation|rotations]] and [[Constant of motion|constants of motion]]. If $L$ is [[Transformation invariance|invariant under a rotation]] about an axis, then the [[angular momentum]] component on that axis is a constant of motion. This is true because of [[Nöther's theorem]]. Also see [[Rotation#Conservation of angular momentum]]. [^1]: This is rather obvious: if $L$ is not dependent on $q_{l}$, of course it wouldn't change when $q_{l}$ changes. It is however a useful property and worth stressing.