✎ Lagrangian Mechanics [The Physics Travel Guide]

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====== Lagrangian Mechanics ======
//see also [[formalisms:lagrangian_formalism]] //
<tabbox Intuitive>

In Lagrangian mechanics we derive how a particle will evolve using the idea that ‘total amount that happened’ from
one moment to another as a particle traces out a path is minimal.

So formulated differently, the basic idea is that nature is lazy. All we have to do is find the laziest way to realize something and this is exactly how nature will behave.
<tabbox Concrete> 
In Lagrangian mechanics we define a quantity
\begin{equation}
 L\equiv K(t)-V(q(t))
\end{equation}
called the __Lagrangian__. In addition, for any trajectory 
$q$ with $q(t_0)=a$, $q(t_1)=b$, we 
we define the corresponding __action__
\begin{equation}
 S(q)\equiv \int_{t_0}^{t_1}L(t)\,dt.
\end{equation}

[{{ :theories:trajectories.png?nolink&400|Source: Lectures on Classical Mechanics by [[http://math.ucr.edu/home/baez/classical/texfiles/2005/book/classical.pdf|John C. Baez]]}}]

The basic idea is now that nature causes 
particles to follow the trajectories with the //least// amount of action.

In the image on the right-hand side this could be the solid line denotes by $q$. Another path is shown as a dashed line and denoted by $q(s)$. The Lagrangian approach assigns to each path a quantity called action as defined above and then tells us that the correct path that an object really follows is the path with minimal action. In our example the path $q$ could have an action of $3$ and the path $q_s$ an action of $5$. Hence, path $q$ is correct and not path $q_s$.

----

The mathematical toolset that we use to find the paths with minima action is known as [[basic_tools:variational_calculus|variational calculus]].

The final result using this machinery is that the path with minimal action is the one that satisfies the [[equations:euler_lagrange_equations|Euler-Lagrange equations]].

----

Using the Lagrangian approach, we can derive [[theories:newtonian_mechanics|Newtonian mechanics]]. Alternatively, we can start with Newtonian mechanics and derive Lagrangian mechanics.

-->Derivation of Newton's second law#

A point with minimal action is $S$ defined through the condition
\begin{equation}
 \left. \frac{d}{ds}S(q_s)\right|_{s=0} = 0
\end{equation}
where
\[
 q_s = q + s\delta q
\]
for all $\delta q \colon [t_0,t_1]\rightarrow\mathbb{R}^n$ with
\[
 \delta q(t_0)=\delta q(t_1)=0.
\]
We want to derive
\begin{equation}
 F=ma \quad \Leftrightarrow \quad 
\left.\frac{d}{ds} S(q_s)\right|_{s=0}=0 \;
 \textrm{for all} \; \delta q\colon [t_0,t_1]\rightarrow\mathbb{R}^n 
\; \textrm{with}\;  \delta q(t_0)=\delta q(t_1)=0.
\end{equation}
To do this, we now use the definition of the action and the chain rule:
\[
 \begin{split}
  \left.\frac{d}{ds}S(q_s)\right|_{s=0} &= \frac{d}{ds}\left.\int_{t_0}^{t_1}
  \frac{1}{2}m\dot{q}_s(t)\cdot\dot{q}_s(t)-V(q_s(t))\,dt\right|_{s=0} \\ \\
  &= \left.\int_{t_0}^{t_1}\frac{d}{ds}\left[
  \frac{1}{2}m\dot{q}_s(t) \cdot \dot{q}_s(t) -
  V(q_s(t))\right]dt\right|_{s=0} \\ \\
  &= \left.\int_{t_0}^{t_1}\left[
  m\dot{q}_s \cdot \frac{d}{ds}\dot{q}_s(t)- \nabla V(q_s(t)) \cdot \frac{d}{ds}q_s(t)\right]dt\right|_{s=0} \\
\end{split}
\]
Next we note that
\[    \frac{d}{ds} q_s (t) = \delta q(t)  \]
so
\[    \frac{d}{ds} \dot{q}_s (t) = \frac{d}{ds} \frac{d}{dt} {q}_s (t)
                                 = \frac{d}{dt} \frac{d}{ds} {q}_s (t) 
                                 = \frac{d}{dt} \delta{q}(t) .\]
Therefore we have
\[
\begin{split}
  \left.\frac{d}{ds}S(q_s)\right|_{s=0} &= 
  \int_{t_0}^{t_1}\left[ m\dot{q} \cdot \frac{d}{dt} \delta q(t) - 
 \nabla V(q(t)) \cdot \delta q(t)\right]dt .
 \end{split}
\]
Finally, we can integrate by parts and note that the boundary terms vanish 
because $\delta q=0$ at $t_1$ and $t_0$:
\[
 \begin{split}
  \frac{d}{ds}S(q_s)\rvert_{s=0} 
  &= \left.\int_{t_0}^{t_1}\left[ -m\ddot{q}(t)-\nabla V(q(t))\right]
\cdot \delta q(t) dt\right. . \\
 \end{split}
\]
Looking at this, we can see that the variation of the action vanishes for //all// possible
variations $\delta q$ if and only if the term in brackets is exactly
zero:
\[
 -m\ddot{q}(t)-\nabla V(q(t)) = 0.
\]
Therefore, the correct path $q$ with a minimal amount of action $S$ is the one that follows from the equation
\[
 F=ma.
\]
<--

----

**Examples**

--> Harmonic Oscillator#

The Lagrangian in classical mechanics is given by the kinetic energy $ \cal T$ minus the potential energy $ \cal U$:

$$  \cal L=T-U . $$

Therefore, a harmonic oscillator in the Lagrangian framework is characterized by the action

$$ S = \int dt \left( \frac{m}{2} \left(\frac{dx}{dt}\right) - \frac{k}{2}x^2 \right) ,$$
where $L= \frac{m}{2} \left(\frac{dx}{dt}\right) - \frac{k}{2}x^2$ is the //Lagrangian//.

Starting from this action, we can derive the equation of motion  by using the [[equations:euler_lagrange_equations|Euler-Lagrange equation]].

$$ \frac{\partial {\cal L}}{\partial x}-\frac{d}{dt}(\frac{\partial  {\cal L}}{\partial dot{x}})= 0   $$

$$ \Rightarrow  m \frac{d^2}{dt^2} x -kx= 0$$

<--

--> Pendulum#

The Lagrangian for the pendulum can be written
\[{\cal L=T-U}=\frac{mr^2\dot{\theta}^2}{2}+mgr\cos\theta\]
The equation of motion is
\[\frac{\partial {\cal L}}{\partial \theta}-\frac{d}{dt}(\frac{\partial  {\cal L}}{\partial dot{\theta}})=
-mgr\sin\theta-mr^2\frac{d\dot\theta}{dt}=0\]
giving:
\[\ddot{\theta}+\frac{g}{r}\sin\theta=0\]

**Small Amplitude Approximation**

If  $\theta <<1$ then $\sin\theta\approx\theta$
and
\[\frac{d^2\theta}{dt^2}+\frac{g}{r}\theta\]
This solution describes simple harmonic motion
\[\theta=A\cos(\omega t-\phi)\]
where 
\[\omega=\sqrt{\frac{g}{r}}\]
and the constants $A$ and $\phi$ are determined from 
the initial condition
E.g. if $\theta=0,\;t=0$ then $\phi=\pm \frac{\pi}{2}$.
If $d\theta/dt=0,\;t=0$ then $\phi=0,$ or $\pi$
(depending on whether the pendulum is moving to the left or 
right initially)
The period is
\[\tau=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{r}{g}}\]

**Finite Amplitude**

If the amplitude  is not small we have
to solve the  nonlinear equation
\[\frac{d^2\theta}{dt^2}+\frac{g}{r}\sin \theta=0\]
(nonlinear because sin $\theta$ is a nonlinear
function of the dependent variable $\theta$.)
 
<--

-->The Atwood Machine#
A frictionless pulley with two masses, $m_1$ and $m_2$, hanging from it.

We have
\begin{align*}
 K &= \frac{1}{2}(m_1+m_2)(\frac{d}{dt}(\ell-x))^2  = 
\frac{1}{2}(m_1+m_2)\dot{x}^2 \\
 V &= -m_1 g x - m_2 g(\ell-x) \\
 \text{so }
 L &= K-V = \frac{1}{2}(m_1+m_2)\dot{x}^2 + m_1gx+m_2g(\ell-x)
\end{align*}
The configuration space is $Q=(0,\ell)$, and $x\in(0,\ell)$.  Moreover $TQ=(0,\ell)\times\mathbb{R}\ni(x,\dot{x})$.
As usual $L:TQ\rightarrow\mathbb{R}$.  Note that solutions of the Euler-Lagrange equations will only be defined for \emph{some} time $t\in\mathbb{R}$, as eventually the solutions reaches the ``edge'' of $Q$.

The momentum is:
\[
 p = \frac{\partial L}{\partial\dot{x}} = (m_1+m_2)\dot{x}
\]
and the force is:
\[
 F = \frac{\partial L}{\partial x} = (m_1-m_2)g
\]
The Euler-Lagrange equations say
\begin{align*}
 \dot{p} &= F \\
 (m_1+m_2)\ddot{x} &= (m_1-m_2)g \\
 \ddot{x} &= \frac{m_1-m_2}{m_1+m_2}g
\end{align*}
So this is like a falling object in a downwards gravitational acceleration $a=\left(\frac{m_1-m_2}{m_1+m_2}\right)g$.

We integrate the expression for $\ddot{x}$ twice to obtain the general solution to the motion $x(t)$.  Note that $\ddot{x}=0$ when $m_1=m_2$, and $\ddot{x}=g$ if $m_2=0$.
<--

<tabbox Abstract> 
Lagrangian mechanics can be formulated geometrically using [[advanced_tools:fiber_bundles|fibre bundles]].

The Lagrangian function is defined on the tangent bundle $T(C)$ of the configuration space $C$.

In contrast, the Hamiltonian function is defined on the cotangent bundle $T^\star(C)$, which is also called phase space.

The map from $T^\star(C) \leftrightarrow T(C)$ is called [[advanced_tools:legendre_transformation|Legendre transformation]].

----

* See page 471 in Road to Reality by R. Penrose, page 167 in Geometric Methods of Mathematical Physics by B. Schutz and http://philsci-archive.pitt.edu/2362/1/Part1ButterfForBub.pdf.

</tabbox>

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