# Velocity Vector in Polar Coordinates (Kleppner p.30)

1. Jun 6, 2013

### Von Neumann

In polar coordinates we have $\vec{r} = r \hat{r}$ $\Rightarrow \vec{v} = \frac{d}{dt}({r \hat{r}}) = \dot{r}\hat{r} + r \frac{d \hat{r}}{dt}$.

In the book Introduction to Mechanics, K & K says the right term is the component of velocity directed radially outward. (Surely a typo, as the left term is the velocity component associated with the direction $\hat{r}$.) Then he goes on to say it's a good guess that the other term is the component in the tangential $\left( \hat{\theta} \right)$ direction. He proves this is so in 3 ways; namely by proving $\frac{d\hat{r}}{dt}$ is in the $\hat{\theta}$ direction). The first two ways I understand - It's the third one I'm stuck on.

He starts by drawing two position vectors $\vec{r}$ and $\vec{r} + \Delta \vec{r}$ at the respective times $t$ and $t + \Delta t$, along with their respective unit vectors $\hat{r}_{1}$ , $\hat{r}_{2}$ , $\hat{\theta}_{1}$ , and $\hat{\theta}_{2}$. From the geometry, we see that $\Delta \hat{r} = \hat{\theta}_{1}sin\Delta\theta - \hat{r}_{1}(1-cos\Delta\theta)$, where $\Delta\theta$ is the angle between the two position vectors).

From this we see that $\frac{\Delta\hat{r}}{\Delta t} = \hat{\theta}_{1}\frac{sin\Delta\theta}{\Delta t} - \hat{r}_{1}\frac{(1-cos\Delta\theta)}{\Delta t} = \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right)$. Almost there, just need to take the limit of this quantity as $\Delta t$ tends to 0. So we need to evaluate $\frac{d \hat{r}}{dt} = \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\hat{r}}{\Delta t}$.

He make the following argument which concludes the proof:

"In the limit $\Delta t \to 0$, $\Delta\theta$ approaches zero, but $\Delta\theta/\Delta t$ approaches the limit $d\theta/dt$. Therefore, $\displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n$ for $n>0$. The term in $\hat{r}$ entirely vanishes in the limit and we are left with $\frac{d \hat{r}}{dt}= \dot{\theta} \hat{\theta}$."

I understand that $\Delta\theta/\Delta t$ approaches $d\theta/dt$ as $\Delta t \to 0$, but I'm lost after that. How does one come to the conclusion that $\displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n$ for $n>0$? Then, how does this lead us to the conclusion that $\hat{r}$ entirely vanishes in the limit?

I've been trying my hardest to work through this text, but I tend to get snagged for quite some time on explanations like the above. Usually I can fill in the missing steps myself. I feel as though I cannot thoroughly penetrate this textbook as I have others.

Thanks in advance for any guidance.

2. Jun 6, 2013

### TSny

Maybe you can make use of the product law for limits if you think of $\frac{\Delta \theta}{\Delta t}$ and $(\Delta \theta)^n$ as functions of $\Delta t$.

3. Jun 6, 2013

### Von Neumann

Well, since $\displaystyle \lim_{x \to a} f(x)g(x)= \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$ then $\Rightarrow \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot \lim_{\Delta t \to 0} (\Delta\theta)^n = \lim_{\Delta t \to 0}\frac{\Delta\theta}{\Delta t} \cdot 0 = 0$.

How does this help though?

4. Jun 6, 2013

### TSny

I hope I'm not misunderstanding your original question. I am assuming you want to show that $\displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n$ equals zero. If so, isn't that what you have essentially shown?

5. Jun 6, 2013

### Von Neumann

The goal is to prove that $\frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta}$.

I do now understand why the previous limit holds. However, why is it relevant to proving that $\frac{d \hat{r}}{dt} = \dot{\theta}\hat{\theta}$?

Since a limit of products is identical to the product of the separated limits, any limit containing a factor of $(\Delta\theta)^n$ will result in 0. Is this why $\hat{r}$ vanishes? We can factor out a $(\theta)^2$ from the second term of $\frac{\Delta\hat{r}}{\Delta t}$. And, since $n=2,>0$, then the limit is 0. Right?

But this reasoning suggests that we can factor out a $(\theta)$ from the first term of $\frac{\Delta\hat{r}}{\Delta t}$. And since $n=1,>0$ then the limit is 0 for this one as well. Which is certainly not correct, since $\hat{\theta}$ is not supposed to vanish.

EDIT: Hold on! I figured it out, I just need to type out all the details.

Last edited: Jun 6, 2013
6. Jun 6, 2013

OK. Good.

7. Jun 6, 2013

### Von Neumann

I think I got it:

\begin{align*} \frac{d\hat{r}}{dt} &= \displaystyle \lim_{\Delta t \to 0} \left( \frac{\Delta\hat{r}}{\Delta t} \right) \\ &= \lim_{\Delta t \to 0} \left [ \hat{\theta}_{1} \left( \frac{\Delta\theta - \frac{1}{6}(\Delta\theta)^3+\cdots}{\Delta t} \right) - \hat{r}_{1} \left( \frac{\frac{1}{2}(\Delta\theta)^2 - \frac{1}{24}(\Delta\theta)^4+\cdots}{\Delta t} \right) \right] \\ & = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left( \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} - \frac{1}{6} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^3}{\Delta t} + \cdots \right) - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left( \frac{1}{2} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^2}{\Delta t} - \frac{1}{24} \lim_{\Delta t \to 0} \frac{(\Delta\theta)^4}{\Delta t} + \cdots \right) \\ & = \lim_{\Delta t \to 0} \hat{\theta}_{1} \cdot \left[ \dot{\theta} - \frac{1}{6} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^2 \right) + \cdots \right] - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot \left[ \frac{1}{2} \lim_{\Delta t \to 0} \left ( \frac{\Delta\theta}{\Delta t} \cdot \Delta\theta \right) - \frac{1}{24} \lim_{\Delta t \to 0} \left( \frac{\Delta\theta}{\Delta t} \cdot (\Delta\theta)^3 \right) + \cdots \right] \\ & = \dot{\theta} \cdot \lim_{\Delta t \to 0} \hat{\theta}_{1} - \lim_{\Delta t \to 0} \hat{r}_{1} \cdot 0 \ \left( \text{since} \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n \ \text{for} \ n>0 \right) \\ &= \dot{\theta} \hat{\theta} \end{align*}

Correct?

8. Jun 7, 2013

### TSny

Yes, that looks OK. In going from the second to third line you don't really need to take the limits of $\hat{\theta}_1$ and $\hat{r}_1$ since they are fixed vectors that don't depend on $\Delta t$. But, what you did is fine.

9. Jun 7, 2013

### Von Neumann

I meant it implicitly in going from the 5th to the 6th line. Thanks TSny!