Solving Dynamical Systems with Polar Coordinates

  • Context: Graduate 
  • Thread starter Thread starter Juggler123
  • Start date Start date
  • Tags Tags
    Polar
Click For Summary

Discussion Overview

The discussion revolves around solving dynamical systems using polar coordinates, specifically focusing on deriving formulas for the angular velocity \(\dot{\theta}\) and the radial velocity \(\dot{r}\). Participants explore various mathematical approaches and expressions related to these derivatives.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant mentions using the formula \(r\dot{r} = x\dot{x} + y\dot{y}\) to solve for \(\dot{r}\) and seeks a similar formula for \(\dot{\theta}\).
  • Another participant suggests starting from \(\tan \theta = y/x\) and hints at a relationship involving \(\sec^2 \theta\) for deriving \(\dot{\theta}\).
  • A participant expresses difficulty in formulating a nice expression for \(\dot{\theta}\) despite working from the tangent relationship.
  • One participant provides an expression for \(\dot{\theta}\) as \(\theta' = (xy' - yx')/r\) but later questions its correctness based on dimensional analysis.
  • Another participant proposes differentiating \(r^2 = x^2 + y^2\) to derive \(\dot{r}\) and \(\ddot{r}\) correctly.
  • A participant shares their exploration of deriving equations related to Kepler's laws and expresses uncertainty about their previous calculations.
  • One participant concludes with a refined expression for \(\dot{\theta}\) as \(\theta' = (xy' - yx')/r^2\) after correcting earlier mistakes.

Areas of Agreement / Disagreement

Participants express varying degrees of confidence in their derived formulas, with some acknowledging potential mistakes. There is no consensus on the correctness of the expressions for \(\dot{\theta}\) and \(\dot{r}\), and multiple competing views remain regarding the derivations.

Contextual Notes

Some participants indicate that their derivations may contain errors, particularly in dimensional analysis and the relationships between variables. The discussion includes attempts to connect the mathematical expressions to physical laws, such as Newton's laws and Kepler's laws, but these connections remain unresolved.

Juggler123
Messages
80
Reaction score
0
I've been solving some dynamical systems and have had to convert some of the ode's to polar co-ordinates, I've been using the formula

r\dot{r} =x\dot{x} y\dot{y}

to easily solve for \dot{r}

I'm just wondering if there's an easy to use formula to find \dot{\theta}

I've been trying to formulate one but with no real luck!

Any suggestions?
 
Physics news on Phys.org
Hi Juggler123! :smile:

(you missed out a + :wink:)

tanθ = y/x, so … sec2θ dθ = … ? :smile:
 
Ooops yeah missed that +!

I've worked on from tanθ = y/x but can't end up a nice formula for theta dot. Just wondering if one exists!?
 
Show us how far you've got. :smile:
 
tex function doesn't seem to be working for me which is annoying...

theta-dot*r^2=x^2*d/dt(y/x)

which I think gives

theta-dot*r^2=x^2*[y-dot*x^(-1) - y*x-dot^(-2)]

not particularly nice though..
 
Looks ok to me …

r2θ' = xy' - yx' :wink:
 
\frac{48}{2}(9+3)
 
There are about 4 pages of this stuff in my english notebook... (my physics notebook was full)

This is what I got:

θ' = (xy' - yx')/r
r' = (xx' + yy')/r

of course its possible that I made a mistake. most of my work from then on dealt with trying to derive formulas for θ'' and r''.

I got r'' = (x''x + y''y + (θ')² )/r

oh, and I did all this because I was trying to figure out like Newton's Laws and trying to get a differential equation that would end up with like r = r0/( 1 + e cos θ ) to validate kepler's 1st law, and also get a parametric equation for r and θ in terms of time, but I wasn't able to since I haven't taken math that high in school yet, and an upperclassman told me that there was a better way, so geometric differentiation I guess, because I know that Newton did a lot of that.
 
Welcome to PF!

HI DarthPickley! Welcome to PF! :smile:
DarthPickley said:
θ' = (xy' - yx')/r

No, that can't be right, the dimensional analysis is wrong …

the LHS is T-1, but the top line of the RHS is L2T-1, and the bottom line is only L

(similarly your r'' must be wrong …

the best way of getting r'' would be to differentiate r2 = x2 + y2 twice :wink:)
 
  • #10
I thought I had probably made a mistake.

but anyways I just learned about Feynman's lost lecture so I have another way of understanding the Kepler stuff, but anyways,
I think that here, for my derivations, I will redo things:

having a vector v such that |v| = r, and then you have v + dv, with |v+dv| = r+dr. to get what dr is, you have to project dv onto v, so you have dv ∙ v / |v| = (x*dx + y*dy)/r. to find dr / dt, you get (x*dx/dt + y*dy/dt)/r = ( x*x' + y*y' ) / r.
to get r * sin dθ [##], which is the angular component of dv, you have to find the part of dv that is perpendicular to v. since the vector perpendicular to v would be < -y , x >, you can project dv onto this vector to get r * sin dθ. so
r * sin dθ = dv ∙ <-y,x> / r = ( -y*dx + x*dy ) / r.
sin dθ = (x*dy - y*dx) / r²
also, for very small dθ, sin dθ --> dθ. (they are about equal)
so, dθ/dt = (x*dy - y*dx )/r² / dt = (x*dy/dt - y*dx/dt)/r² = ( x*y' - y*x' ) / r²

Thus, the REAL final answer is θ' = (xy'-yx')/r²

[##] -- this is where I messed up before (probably). I think I said something silly like that θ is the same as the tangential component. Thanks for pointing this out, I puzzled over it awhile before realizing.

also, this is just the same thing that tiny-tim gave.
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
3
Views
3K