- #1

Hamiltonian

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- Homework Statement
- For ##Q=\mathbb{R}^2## and ##\mathcal{L}=\frac{1}{2}(\dot{x}^2+\dot{y}^2)## the EL equations are ODEs whose solutions are straight lines travelled at constant speed. Rewrite ##\mathcal{L}## in polar coordinates ##r,\theta## and write down the corresponding Euler-Lagrange equations, thus deriving the ODEs for straight lines in polar coordinates.

- Relevant Equations
- The Euler Lagrange Equations:

##\frac{d}{dt}\left( \frac{\partial\mathcal{L}}{\partial\dot{q}_i}\right)=\frac{\partial\mathcal{L}}{\partial q_i}##

In polar coordinates, ##x=rcos(\theta)## and ##y=rsin(\theta)## and their respective time derivatives are

$$\dot{x}=\dot{r}cos(\theta) - r\dot{\theta}sin(\theta)$$

$$\dot{y}= \dot{r}sin(\theta)+r\dot{\theta}cos(\theta)$$

so the lagrangian becomes after a little simplifying,

$$\mathcal{L}=\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2)$$

The Euler-Lagrange equations are:

$$\frac{d}{dt}\left( \frac{\partial\mathcal{L}}{\partial \dot{r}}\right)=\frac{\partial\mathcal{L}}{\partial r}$$

$$\frac{d}{dt}\left( \frac{\partial\mathcal{L}}{\partial \dot{\theta}}\right)=\frac{\partial\mathcal{L}}{\partial \theta}$$

First the EL-equation in terms of ##r## we get,

##\frac{\partial\mathcal{L}}{\partial\dot{r}}=\dot{r}##

##\frac{\partial\mathcal{L}}{\partial r}=r\dot{\theta}^2##

putting it all together we get, $$\ddot{r}=r\dot{\theta}^2$$

The EL-equation in terms of ##\theta## we get,

##\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=r^2\dot{\theta}##

##\frac{\partial\mathcal{L}}{\partial\theta}=0##

putting it all together we have,

##\frac{d(r^2\dot{\theta})}{dt}=0##

which can either simply be written as ##r^2\dot{\theta}=c## where ##c## is some constant or we can differentiate w.r.t ##t## and get ##2r\dot{r}\dot{\theta}+r^2\ddot{\theta}=0##

So I essentially have to equations, one from each EL-equation,

$$\ddot{r}=r\dot{\theta}^2$$ $$r^2\dot{\theta}=c$$

are these ODEs for straight lines in polar coordinates? I don't know what they mean by that or what ODEs for straight lines in polar coordinates should look like.

$$\dot{x}=\dot{r}cos(\theta) - r\dot{\theta}sin(\theta)$$

$$\dot{y}= \dot{r}sin(\theta)+r\dot{\theta}cos(\theta)$$

so the lagrangian becomes after a little simplifying,

$$\mathcal{L}=\frac{1}{2}(\dot{r}^2+r^2\dot{\theta}^2)$$

The Euler-Lagrange equations are:

$$\frac{d}{dt}\left( \frac{\partial\mathcal{L}}{\partial \dot{r}}\right)=\frac{\partial\mathcal{L}}{\partial r}$$

$$\frac{d}{dt}\left( \frac{\partial\mathcal{L}}{\partial \dot{\theta}}\right)=\frac{\partial\mathcal{L}}{\partial \theta}$$

First the EL-equation in terms of ##r## we get,

##\frac{\partial\mathcal{L}}{\partial\dot{r}}=\dot{r}##

##\frac{\partial\mathcal{L}}{\partial r}=r\dot{\theta}^2##

putting it all together we get, $$\ddot{r}=r\dot{\theta}^2$$

The EL-equation in terms of ##\theta## we get,

##\frac{\partial\mathcal{L}}{\partial\dot{\theta}}=r^2\dot{\theta}##

##\frac{\partial\mathcal{L}}{\partial\theta}=0##

putting it all together we have,

##\frac{d(r^2\dot{\theta})}{dt}=0##

which can either simply be written as ##r^2\dot{\theta}=c## where ##c## is some constant or we can differentiate w.r.t ##t## and get ##2r\dot{r}\dot{\theta}+r^2\ddot{\theta}=0##

So I essentially have to equations, one from each EL-equation,

$$\ddot{r}=r\dot{\theta}^2$$ $$r^2\dot{\theta}=c$$

are these ODEs for straight lines in polar coordinates? I don't know what they mean by that or what ODEs for straight lines in polar coordinates should look like.