## Thursday, 30 January 2020

### Exercise SI.01 Simple Lagrangians

 Line according to formula For values $-25\leq t\leq 25, a,b,c,d$.
Question
Derive Newton's first law in two dimensions using Hamilton's principle and the Euler-Lagrange equation. First do it in Euclidean $x,y$ coordinates then in polar $r,\theta$ coordinates.

Do it by filling in the gaps in the Wikipedia article on the subject as quoted below.

1) Prove that $m\ddot{x}=0$ follows from the Euclidean Lagrangian as at (1), (2)
2) Prove that the two Euler–Lagrange at (4),(5) produce the two equations given.
3) Prove that the solutions to those two equations are indeed (6) and (7)
4) Plot (6), (7) for values of $a,b,c,d$ showing that it is a straight line and that $a$ is the velocity, $c$ is the distance of the closest approach to the origin, and $d$ is the angle of motion. What is $b$?
Extracts from the article:
In two Euclidean dimensions in the absence of a potential, the Lagrangian is simply equal to the kinetic energy
\begin{align}
L=\frac{1}{2}mv^2=\frac{1}{2}m\left({\dot{x}}^2+{\dot{y}}^2\right)&\phantom {10000}(1)\nonumber
\end{align}in orthonormal $(x,y)$ coordinates, where the dot represents differentiation with respect to the curve parameter (usually the time, $t$). Therefore, upon application of the Euler–Lagrange equations, [to the $x$ coordinate]
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)-\frac{\partial L}{\partial x}=0\ \ \Rightarrow\ \ m\ddot{x}=0&\phantom {10000}(2)\nonumber
\end{align}[The Lagrangian and the solution are the same for the $y$ coordinate.]

In polar coordinates $(r,\theta)$ the kinetic energy and hence the Lagrangian becomes
\begin{align}
L=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta}}^2\right)&\phantom {10000}(3)\nonumber
\end{align}[where $x=r\cos{\theta}\ \ ,\ y=r\sin{\theta}$.]
The radial $r$ and $\theta$ components of the Euler–Lagrange equations become, respectively
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{r}}\right)-\frac{\partial L}{\partial r}=0\ \Rightarrow\ \ \ddot{r}-r{\dot{\theta}}^2=0&\phantom {10000}(4)\nonumber\\
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right)-\frac{\partial L}{\partial\theta}=0\ \Rightarrow\ddot{\theta}+\frac{2\dot{r}\dot{\theta}}{r}=0&\phantom {10000}(5)\nonumber
\end{align}The solution to those two equations is given by
\begin{align}
r=\sqrt{\left(at+b\right)^2+c^2}&\phantom {10000}(6)\nonumber\\
\theta=\tan^{-1}{\left(\frac{at+b}{c}\right)}+d&\phantom {10000}(7)\nonumber
\end{align}for a set of constants $a,\ b,\ c,\ d$  determined by initial conditions. Thus, indeed, the solution is a straight line given in polar coordinates: $a$ is the velocity, $c$ is the distance of the closest approach to the origin, and $d$ is the angle of motion." It does not reveal what $b$ is.
I struggled with this and question 2 for over a day until I worked out what was really happening and came to the very important observation in the pdf. This enabled me to move on in section 1.10.

To begin with I got this: The first term in (2) is
\begin{align}
\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)=\frac{1}{2}m\frac{d}{dt}\left(\frac{\partial{\dot{x}}^2}{\partial\dot{x}}\right)=m\frac{d\dot{x}}{dt}=m\ddot{x}&\phantom {10000}(8)\nonumber
\end{align}and the second term is
\begin{align}
\frac{\partial L}{\partial x}=\frac{1}{2}m\frac{\partial}{\partial x}\left({\dot{x}}^2\right)=m\dot{x}\frac{\partial}{\partial x}\left(\dot{x}\right)=m\frac{dx}{dt}\frac{\partial}{\partial x}\left(\frac{dx}{dt}\right)=m\frac{d}{dt}\left(\frac{dx}{dt}\right)=m\ddot{x}&\phantom {10000}(9)\nonumber
\end{align}put those back into (2) and we get
\begin{align}
m\ddot{x}-m\ddot{x}=0&\phantom {10000}(10)\nonumber
\end{align}Which does not prove that $m\ddot{x}=0$!!

To see all what went wrong and the vital observation, look in Ex SI.01 Simple Lagrangians.pdf (5 pages)