✎ Lagrangian Mechanics [The Physics Travel Guide]

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

<note tip>
Explanations in this section should contain no formulas, but instead colloquial things like you would hear them during a coffee break or at a cocktail party.
</note>
  
<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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