# Why is $r$ a function of $\theta$? [closed]

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:

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

• How does that prove r is not a function of $\theta$? r is a function of time also. – user234190 Nov 30 '19 at 11:54
• 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. – Madhuchhanda Mandal Nov 30 '19 at 11:55
• On the contrary it can be said, if r is a function of theta then your claim is never true. – Madhuchhanda Mandal Nov 30 '19 at 12:00
• 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. – Madhuchhanda Mandal Nov 30 '19 at 12:02
• 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. – user4552 Nov 30 '19 at 14:46

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.

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.

• r= sin(theta) , 0<theta<2pi represents a circle. Not sure if your argument is valid. – Madhuchhanda Mandal Nov 30 '19 at 13:29

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.

• Your last paragraph ignores the fact that $\theta$ is invertible (unless it's constant). – WillO Nov 30 '19 at 16:11
• I am not ignoring that fact at all. Look at the quoted text. It implies that $\theta$ is constant – Dale Nov 30 '19 at 16:51
• 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. – WillO Nov 30 '19 at 16:54
• 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 – Dale Nov 30 '19 at 17:04