Deriving ODEs for straight lines in polar coordinates for a given Lagrangian

  • #1
Hamiltonian
296
190
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.
 
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  • #2
Hamiltonian said:
In polar coordinates, ##x=rcos(\theta)## and ##y=rsin(\theta)##
and from that it follows that ##x^2+y^2=r^2##. Does this relation look like a straight line?

You need a constraint.
 
Last edited:
  • #3
kuruman said:
Does this relation look like a straight line?
I am a little confused as to what you mean by ##x^2+y^2=r^2## looks like a straight line, in cartesian coordinates that's a circle with radius ##r##.
I looked at the link you provided and it talks about how the equation for a straight line in polar coordinates is,
$$r=\frac{1}{sin\theta-cos\theta}$$ as if you substitute ##x=rcos\theta## & ##y=rsin\theta## you get the equation of a straight line in cartesian coordinates. But I still don't understand how the two ODEs I got ##\ddot{r}=r\dot{\theta}^2## and ##r^2\dot{\theta}=c## represent straight lines in polar coordinates.
Does it mean that the solutions to the ODEs must be straight lines? or that ##r(\theta)=\frac{1}{sin\theta-cos\theta}## must solve the ODEs but ##r## and ##\theta## are functions of time and not each other.
 
  • #4
kuruman said:
and from that it follows that ##x^2+y^2=r^2##. Does this relation look like a straight line?

You need a constraint.
Huh? This is irrelevant. It is just a coordinate transformation and ##r## is a new coordinate. What the coordinate level surfaces look like is of little importance here.

The approach of the OP is correct.
 
  • #5
Hamiltonian said:
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}##

So I essentially have to equations, one from each EL-equation,
r¨=rθ˙2 r2θ˙=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.
These are the differential equations that a straight line in Euclidean space must satisfy when written in polar coordinates. There is nothing magical to it.

You could also derive them by simply making the appropriate coordinate transformation of the Cartesian coordinate equations ##\ddot x = \ddot y = 0##.
 
  • #6
Orodruin said:
These are the differential equations that a straight line in Euclidean space must satisfy when written in polar coordinates. There is nothing magical to it.
Yeah, that's what the question says, "thus deriving the ODEs for straight lines in polar coordinates". I do not understand why every straight line must satisfy these ODEs when written in polar coordinates is it because the path that minimizes the action for this particular lagrangian must be straight lines, as there is no force acting on the particle and hence it must by Netwons first law move in a straight line?
or is it an obvious mathematical fact that a straight line satisfies these ODEs, if it is I don't see it.

My understanding of what it means for a straight line written in polar coordinates to satisfy an ODE might be wrong?
I thought what it means is that the functions ##r(t)## and ##\theta (t)## that satisfy both the ODEs represent a time-parameterized straight line.
 
  • #7
Hamiltonian said:
I do not understand why every straight line must satisfy these ODEs when written in polar coordinates is it because the path that minimizes the action for this particular lagrangian must be straight lines, as there is no force acting on the particle and hence it must by Netwons first law move in a straight line?
The action is the action that gives straight lines in Cartesian coordinates when extremised. Changing coordinates will not change this so if you change coordinates and then apply the EL equations, those same straight lines must satisfy the differential equations you get out because they are still the same straight lines.

Forces have little to do with things here as there is no actual physics involved. (Even though you can make a parallel to the Lagrangian of a freely moving particle.)

Hamiltonian said:
or is it an obvious mathematical fact that a straight line satisfies these ODEs, if it is I don't see it.
You just derived it! I don’t see how you can really see it any more than actually proving it to be the case. Perhaps I am not getting what you mean by ”see it”? What exactly are you looking for?

Hamiltonian said:
My understanding of what it means for a straight line written in polar coordinates to satisfy an ODE might be wrong?
I thought what it means is that the functions r(t) and θ(t) that satisfy both the ODEs represent a time-parameterized straight line.
That is exactly what it means.
 
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  • #8
Orodruin said:
You just derived it! I don’t see how you can really see it any more than actually proving it to be the case. Perhaps I am not getting what you mean by ”see it”? What exactly are you looking for?
I was originally thinking that once we derive the two ODEs from the EL equation we must somehow prove that the solutions to these ODEs will be straight lines, by solving the ODEs.
But I think I understand now that is exactly what I have proved(without solving them ofcourse), the question states "EL equations are ODEs whose solutions are straight lines travelled at constant speed" and all I did was write the EL equations for polar coordinates instead of cartesian coordinates and hence the pair of ODEs I derived must still be for straight lines in polar coordinates.
Thank you!
 
  • #9
Hamiltonian said:
I am a little confused as to what you mean by ##x^2+y^2=r^2## looks like a straight line, in cartesian coordinates that's a circle with radius ##r##.
I looked at the link you provided and it talks about how the equation for a straight line in polar coordinates is,
$$r=\frac{1}{sin\theta-cos\theta}$$ as if you substitute ##x=rcos\theta## & ##y=rsin\theta## you get the equation of a straight line in cartesian coordinates. But I still don't understand how the two ODEs I got ##\ddot{r}=r\dot{\theta}^2## and ##r^2\dot{\theta}=c## represent straight lines in polar coordinates.
Does it mean that the solutions to the ODEs must be straight lines? or that ##r(\theta)=\frac{1}{sin\theta-cos\theta}## must solve the ODEs but ##r## and ##\theta## are functions of time and not each other.
Orodruin said:
Huh? This is irrelevant. It is just a coordinate transformation and ##r## is a new coordinate. What the coordinate level surfaces look like is of little importance here.

The approach of the OP is correct.
Sorry. I misread the question.
 

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