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I am reading "Key Points of Mechanics" by Haruo Yoshida.

He wrote $r$ was a function of $\theta$. (please see below.)

But if $\theta(t_0)=\theta(t_1)$ and $r(t_0) \neq r(t_1)$ for $t_0 \neq t_1$, $r$ is not a function of $\theta$.

Please explain why $r$ is a function of $\theta$.

The equations of motion in a polar coordinate system are the following:

$m (\ddot{r}-r\dot{\theta}^2) = f_r$,
$m(r\ddot{\theta}+2\dot{r}\dot{\theta})=f_{\theta}$.

If $f_r := -\frac{GMm}{r^2}, f_{\theta}:=0$, then
$\ddot{r}-r\dot{\theta}^2 = -\frac{GM}{r^2}$,
$r\ddot{\theta}+2\dot{r}\dot{\theta}=0$.

$\frac{d}{dt}(r^2\dot{\theta}) = 2r\dot{r}\dot{\theta}+r^2\ddot{\theta} = r(r\ddot{\theta}+2\dot{r}\dot{\theta})=0$.

So, $r^2\dot{\theta}$ is constant.
Let $h := r^2\dot{\theta}$.

$\ddot{r}-r\dot{\theta}^2 = \ddot{r}-r(\frac{h}{r^2})^2 = \ddot{r}-\frac{h^2}{r^3} = -\frac{GM}{r^2}$.

$r$ is a function of $\theta$ and the following equation holds:

$\frac{d}{d\theta}(\frac{1}{r^2}\frac{dr}{d\theta})=\frac{1}{r}-\frac{GM}{h^2}$.

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  • $\begingroup$ How does that prove r is not a function of $\theta$? r is a function of time also. $\endgroup$ – user234190 Nov 30 '19 at 11:54
  • $\begingroup$ What's the problem? Your claim is right. But how does it affect the fact that r can be a function of theta? If your claim is right, r cannot be a function of theta. But your claim is not guaranteed to hold for every r. $\endgroup$ – Madhuchhanda Mandal Nov 30 '19 at 11:55
  • $\begingroup$ On the contrary it can be said, if r is a function of theta then your claim is never true. $\endgroup$ – Madhuchhanda Mandal Nov 30 '19 at 12:00
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    $\begingroup$ Also, do note that theta is not equal to theta+2pi . Hence spiral motions can be perfectly represented by r as a function of theta. $\endgroup$ – Madhuchhanda Mandal Nov 30 '19 at 12:02
  • $\begingroup$ Please edit the question to give some context so we know what this is about. Also, please edit the title so people can understand what is being asked about. $\endgroup$ – Ben Crowell Nov 30 '19 at 14:46
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You've got $r^2\dot{\theta}$ constant. Unless that constant is zero, this means that $\dot{\theta}$ never changes sign, which means that $\theta$ is either always increasing or always decreasing as a function of $t$. Therefore you never have to worry about the possibility that $\theta(t_0)=\theta(t_1)$ with $t_0\neq t_1$.

The only exception is where the constant value of $\dot{\theta}$ is zero, in which case the motion is all back and forth along a single radius.

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Typically you have three variables, $r, \theta, t$. You are correct that r cannot be a function of $\theta $ if you limit the range of $\theta$ to for example $[0,2\pi]$. If you do not project to this interval then you can take $t$ as the independent variable and have $r$ depend on $t$ via $\theta $. This is a fairly standard approach to derive Kepler orbits.

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    $\begingroup$ r= sin(theta) , 0<theta<2pi represents a circle. Not sure if your argument is valid. $\endgroup$ – Madhuchhanda Mandal Nov 30 '19 at 13:29
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Please explain why 𝑟 is a function of 𝜃

Unfortunately, there is not much content to the statement. Basically, both $\theta$ and $r$ are functions of $t$. And if $\theta$ is an invertable function then $t=\theta^{-1}(\theta(t))$ so $r=r(t)=r(\theta^{-1}(\theta(t)))$.

Now, we can write $R=(r \circ \theta^{-1})$ so then $r=R(\theta)$ shows that $r$ is a function of $\theta$

But if 𝜃(𝑡0)=𝜃(𝑡1) and 𝑟(𝑡0)≠𝑟(𝑡1) for 𝑡0≠𝑡1, 𝑟 is not a function of 𝜃

Yes, you are correct. In this case $\theta$ is not invertable so $\theta^{-1}(\theta(t))=t$ does not exist and the above approach falls apart.

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  • $\begingroup$ Your last paragraph ignores the fact that $\theta$ is invertible (unless it's constant). $\endgroup$ – WillO Nov 30 '19 at 16:11
  • $\begingroup$ I am not ignoring that fact at all. Look at the quoted text. It implies that $\theta$ is constant $\endgroup$ – Dale Nov 30 '19 at 16:51
  • $\begingroup$ No, the assumption in the quoted text does not imply that $\theta$ is a constant. You also need the equation that says $r^2\dot{\theta}$ is constant. $\endgroup$ – WillO Nov 30 '19 at 16:54
  • $\begingroup$ Sure, conservation of angular momentum together with the quoted assumption imply that $\theta$ is constant. Even without conservation of angular momentum the quoted assumption alone means that $\theta$ is not invertable. My answer is correct as is and I am not ignoring anything. Not sure why you think I am. The assumption is a perfectly valid assumption, I simply chose to answer it as a valid assumption instead of rejecting the assumption as you did in your answer. I don’t have a problem with your approach, not sure why you have a problem with mine $\endgroup$ – Dale Nov 30 '19 at 17:04

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