Insight into polar coordinates (Newtonian mechanics)?

[Leo Liu]

I am learning to use polar coordinates to describe the motions of particles. Now I know how to use polar coordinates to solve problems and the derivations of many equations. However, the big picture of polar coordinates remains unclear to me. Would you mind sharing your insight with me so that I can better grasp the related concepts, and hopefully be equipped to see the big picture of it? Thanks.

Equations:
$$\vec r = r\hat r (\theta) $$
$$\vec v = \dot r \hat r + r\hat \theta \dot \theta$$
$$\vec a = \hat r (\ddot r - r \dot \theta^2)+ \hat \theta (r \ddot \theta + 2\dot r \dot \theta)$$

 

[anuttarasammyak]

The relations on unit vectors
$$d\hat{\mathbf{r}}=\hat{\phi}\ d\phi$$
$$d\hat{\phi}=\hat{\mathbf{r}}\ d\phi$$
and the differential rule give the formula you wrote. These relations tell not radial $dr$ but only angle $d\phi$ of position displacement should change the unit vectors.

EDIT sign error as pointed out in post #4 correction
$$d\hat{\phi}=-\hat{\mathbf{r}}\ d\phi$$

 

[etotheipi]

If you have a coordinate system whose coordinates $y^a$ are related to the coordinates $x^i$ in a Cartesian system, then the tangent vector basis is (ignoring issues of normalisation)$$\vec{E}_a = \frac{\partial \vec{x}}{\partial y^a} = \frac{\partial x^i}{\partial y^a}\vec{e}_i$$For instance if $x = r\cos{\theta}$ and $y = r\sin{\theta}$, then$$\hat{r} = \frac{\partial x}{\partial r}\hat{x} + \frac{\partial y}{\partial r}\hat{y} = \cos{(\theta)} \hat{x} + \sin{(\theta)} \hat{y}$$Everything else in a dynamics context works in pretty much the same way as before, you just need to have the correct forms for $\vec{v}$ and $\vec{a}$. For instance for planar circular motion you might decompose the forces on a particle into the tangent vector basis and write$$F_r = ma_r = m(\dot{r} - r\dot{\theta}^2)$$but you also have the constraint that $\dot{r} = 0$.

 

[Leo Liu]

The relations on unit vectors
$$d\hat{\mathbf{r}}=\hat{\phi}\ d\phi$$
$$d\hat{\phi}=\hat{\mathbf{r}}\ d\phi$$
and the differential rule give the formula you wrote. These relations tell not radial $dr$ but only angle $d\phi$ of position displacement should change the unit vectors.

For the second equation, I think there should be a negative sign on the right side because $$d \phi = -\cos \phi d\phi \: \hat i - \sin \phi d\phi \: \hat j \implies d\phi=-(\cos \phi \hat i + \sin \phi \hat j)d \phi$$

 

[mpresic3]

The big picture is this. The most useful application of polar coordinates to physics is examining the central force problem. Fore example when you study Keplers problem of planetary motion. A nice undergrad exercise and lecture is showing the paths or orbits in an inverse square field are conic sections, hyperbola (path), parabola(path), ellipse (orbit), or circle (orbit). Because the force law is radial (i.e. central) , plane polar coordinates is a natural way to approach the problem. Cartesian coordinates would be much harder.

The position, velocity, and acceleration, you listed for plane polar coordinates look good to me. I do not find any error or any reference to phi, which is usually used for azimuth?

 

[etotheipi]

To follow up on @mpresic3's comment, the total energy of an orbiting body in polar coordinates is$$E = \frac{1}{2}mv^2 + U(r) = \frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) + U(r)$$However the angular momentum in polar coordinates is $$L = mr^2 \dot{\theta}$$which means that, with $h := \frac{L}{m},$$$r^2 \dot{\theta}^2 = \frac{L^2}{m^2 r^2} = \frac{h^2}{r^2}$$ $$E = \frac{1}{2}m \left(\dot{r}^2 + \frac{h^2}{r^2} \right) + U(r) = \frac{1}{2}m\dot{r}^2 + \left[\frac{mh^2}{2r^2} + U(r) \right]$$ We define the effective potential $$U_{\text{eff}}(r) := \frac{mh^2}{2r^2} + U(r)$$so that$$E = U_{\text{eff}}(r) + \frac{1}{2}m\dot{r}^2$$and we have reduced our problem to essentially a one dimensional case! This is quite a bit easier to work with, for instance$$F_{\text{eff}}(r) = -\frac{dU_{\text{eff}}(r)}{dr} = m\ddot{r}$$
this is useful for e.g. checking if an orbit is stable.

 

[JD_PM]

Given the energy and angular momentum equations

$$E = \frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) + U(r) \tag 1$$

$$\dot{\theta}= \frac{L}{mr^2} \tag 2$$

We eliminate $\dot \theta$ (by plugging $(2)$ into $(1)$) to get

$$E = \frac 1 2 m \dot{r}^2 + \frac{L^2}{2 r^2 m} + U(r)= \frac 1 2 m \dot{r}^2 + U_{\text{eff}}(r) \tag 3$$

As etotheipi explained.

$(3)$ is sometimes called the radial motion equation, which given initial conditions, leads to determine the change in the polar coordinate $r$ with respect to time. The change of the polar coordinate $\theta$ with respect to time is determined by $(2)$.

In practise one does not seek to solve $(3)$ analytically (I encourage you to try to get $r(t)$ given the attractive inverse squared field $\vec F = \frac{-k}{r^2} \hat r$ and see what happens ). Instead we try to simplify the problem and only get an equation that describes the path taken by the particle (which means that we do not seek to know where the particle is on the its path at a particular time but only to get an equation describing its path). Let's show explicitly how.

We start off by noticing that any particle moving under a central force field does so in a plane. This fact suggests we should use a 2D coordinate system: polar coordinates. Next we use Newton's equations in polar coordinates (it is indeed a good exercise to show it; you only need to use the formulas you provided in #1)

$$m \ddot r - mr \dot \theta^2 = F(r) \tag{4.1}$$

$$mr \ddot \theta +2m \dot r \dot \theta \tag{4.2} = 0$$

What we want is to get rid of the polar-coordinate derivatives with respect to time ( i.e. $\ddot r$ and $\dot \theta$) in Newton's equations in order to get these in terms of a new coordinate whose derivatives are not going to be with respect to time. It turns out that there's a beautiful change of variables which makes this possible (I'd say it was Bernoulli's trick)

$$u:= \frac{1}{r} \tag 5$$

Next we get $\dot r$ in terms of the new coordinate

$$\dot r = \frac{d}{dt} \Big( \frac{1}{u} \Big) = \frac{-1}{u^2}\frac{du}{dt} = \frac{-1}{u^2}\frac{du}{d\theta} \dot \theta = \frac{-L}{m} \frac{du}{d \theta} \tag 6$$

Where I've used the chain rule and the angular momentum equation in terms of the new coordinate.

To get $\ddot r$ we simply differentiate $(6)$ to get

$$\ddot r = \frac{-L}{m} \frac{d}{dt} \Big( \frac{du}{d \theta} \Big) = \frac{-L}{m} \frac{d}{d \theta} \Big( \frac{du}{dt} \Big) = \frac{-L}{m} \frac{d}{d \theta} \Big( \frac{du}{d\theta} \dot \theta \Big) = -\Big(\frac{Lu}{m} \Big)^2 \frac{d^2 u}{d \theta^2} \tag7$$

Thus, having $r \dot \theta^2 = \frac{L^2 u^3}{m^2}$, $(4.1)$ becomes

$$-\Big(\frac{Lu}{m} \Big)^2 \frac{d^2 u}{d \theta^2} - \frac{L^2 u^3}{m} = F(1/u) $$

Rearranging a bit we get

$$\frac{d^2 u}{d \theta^2} +mu = -m^2 \frac{F(1/u)}{L^2 u^2} \tag 8$$

$(8)$ is known as the path equation.

Source: you can find all I wrote and more in chapter 7 of Gregory's Classical Mechanics book.

 

[JD_PM]

What shocked me about $(8)$ when I first learned about it was that it is linear only when $F(r) \sim \frac{1}{r^2}$ and when $F(r) \sim \frac{1}{r^3}$. The former is the most important case in the theory of fields.

Mmm I've always wondered whether this is just coincidence...

 

[etotheipi]

Very neat. And that orbit equation also nicely paves the way for Bertrand's theorem, which I won't dare try and derive but will instead leave a link to a derivation here