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--- theories:classical_mechanics:lagrangian [2018/04/12 16:54]
bogumilvidovic [Concrete]
+++ theories:classical_mechanics:lagrangian [2018/10/11 14:12] (current)
jakobadmin [Abstract]
@@ Line 1: / Line 1: @@
 ====== Lagrangian Mechanics ======
+//see also [[formalisms:lagrangian_formalism]] //
 <tabbox Intuitive>
-<note tip>
+In Lagrangian mechanics we derive how a particle will evolve using the idea that ‘total amount that happened’ from
-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.
+one moment to another as a particle traces out a path is minimal.
-</note>
+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\defeq K(t)-V(q(t))
+ L\equiv K(t)-V(q(t))
 \end{equation}
 called the __Lagrangian__. In addition, for any trajectory
-$q \colon [t_0,t_1]\rightarrow\mathbb{R}^n$ with $q(t_0)=a$, $q(t_1)=b$, we
+$q$ with $q(t_0)=a$, $q(t_1)=b$, we
 we define the corresponding __action__
 \begin{equation}
- S(q)\defeq \int_{t_0}^{t_1}L(t)\,dt.
+ S(q)\equiv \int_{t_0}^{t_1}L(t)\,dt.
 \end{equation}
+{{ :theories:classical_mechanics:lagrangianpfad.png?nolink&400|}}
 The basic idea is now that nature causes
 particles to follow the trajectories with the //least// amount of action.
-Using this approach, we can derive [[theories:newtonian_mechanics|Newtonian mechanics]]. Alternatively, we can start with Newtonian mechanics and derive Lagrangian mechanics.
+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:classical_mechanics:newtonian|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 $m_2=0$.
+<--
+----
+**Reading Recommendations**
+  * The best book on Lagrangian mechanics is The Lazy Universe by Coopersmith
+  * Many problems with solutions are collected in [[https://archive.org/details/SchaumsTheoryAndProblemsOfTheoreticalMechanics|Schaum's Outline of Theory and Problems of Theoretical Mechanics]] by Murray R Spiegel
 <tabbox Abstract>
 Lagrangian mechanics can be formulated geometrically using [[advanced_tools:fiber_bundles|fibre bundles]].
@@ Line 36: / Line 223: @@
 ----
+  * [[https://core.ac.uk/download/pdf/4887416.pdf|Lectures on Mechanics]] by Marsden
   * 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.

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