- #1
Von Neumann
- 101
- 4
In polar coordinates we have [itex] \vec{r} = r \hat{r} [/itex] [itex]\Rightarrow \vec{v} = \frac{d}{dt}({r \hat{r}}) = \dot{r}\hat{r} + r \frac{d \hat{r}}{dt} [/itex].
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 [itex] \hat{r} [/itex].) Then he goes on to say it's a good guess that the other term is the component in the tangential [itex] \left( \hat{\theta} \right) [/itex] direction. He proves this is so in 3 ways; namely by proving [itex] \frac{d\hat{r}}{dt} [/itex] is in the [itex] \hat{\theta} [/itex] direction). The first two ways I understand - It's the third one I'm stuck on.
He starts by drawing two position vectors [itex] \vec{r} [/itex] and [itex] \vec{r} + \Delta \vec{r} [/itex] at the respective times [itex] t [/itex] and [itex] t + \Delta t [/itex], along with their respective unit vectors [itex] \hat{r}_{1} [/itex] , [itex] \hat{r}_{2} [/itex] , [itex] \hat{\theta}_{1} [/itex] , and [itex] \hat{\theta}_{2} [/itex]. From the geometry, we see that [itex] \Delta \hat{r} = \hat{\theta}_{1}sin\Delta\theta - \hat{r}_{1}(1-cos\Delta\theta) [/itex], where [itex] \Delta\theta [/itex] is the angle between the two position vectors).
From this we see that [itex] \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)[/itex]. Almost there, just need to take the limit of this quantity as [itex] \Delta t [/itex] tends to 0. So we need to evaluate [itex] \frac{d \hat{r}}{dt} = \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\hat{r}}{\Delta t}[/itex].
He make the following argument which concludes the proof:
"In the limit [itex] \Delta t \to 0 [/itex], [itex]\Delta\theta [/itex] approaches zero, but [itex]\Delta\theta/\Delta t[/itex] approaches the limit [itex]d\theta/dt[/itex]. Therefore, [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]. The term in [itex] \hat{r} [/itex] entirely vanishes in the limit and we are left with [itex] \frac{d \hat{r}}{dt}= \dot{\theta} \hat{\theta} [/itex]."
I understand that [itex] \Delta\theta/\Delta t[/itex] approaches [itex]d\theta/dt[/itex] as [itex] \Delta t \to 0 [/itex], but I'm lost after that. How does one come to the conclusion that [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]? Then, how does this lead us to the conclusion that [itex] \hat{r} [/itex] 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.
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 [itex] \hat{r} [/itex].) Then he goes on to say it's a good guess that the other term is the component in the tangential [itex] \left( \hat{\theta} \right) [/itex] direction. He proves this is so in 3 ways; namely by proving [itex] \frac{d\hat{r}}{dt} [/itex] is in the [itex] \hat{\theta} [/itex] direction). The first two ways I understand - It's the third one I'm stuck on.
He starts by drawing two position vectors [itex] \vec{r} [/itex] and [itex] \vec{r} + \Delta \vec{r} [/itex] at the respective times [itex] t [/itex] and [itex] t + \Delta t [/itex], along with their respective unit vectors [itex] \hat{r}_{1} [/itex] , [itex] \hat{r}_{2} [/itex] , [itex] \hat{\theta}_{1} [/itex] , and [itex] \hat{\theta}_{2} [/itex]. From the geometry, we see that [itex] \Delta \hat{r} = \hat{\theta}_{1}sin\Delta\theta - \hat{r}_{1}(1-cos\Delta\theta) [/itex], where [itex] \Delta\theta [/itex] is the angle between the two position vectors).
From this we see that [itex] \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)[/itex]. Almost there, just need to take the limit of this quantity as [itex] \Delta t [/itex] tends to 0. So we need to evaluate [itex] \frac{d \hat{r}}{dt} = \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\hat{r}}{\Delta t}[/itex].
He make the following argument which concludes the proof:
"In the limit [itex] \Delta t \to 0 [/itex], [itex]\Delta\theta [/itex] approaches zero, but [itex]\Delta\theta/\Delta t[/itex] approaches the limit [itex]d\theta/dt[/itex]. Therefore, [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]. The term in [itex] \hat{r} [/itex] entirely vanishes in the limit and we are left with [itex] \frac{d \hat{r}}{dt}= \dot{\theta} \hat{\theta} [/itex]."
I understand that [itex] \Delta\theta/\Delta t[/itex] approaches [itex]d\theta/dt[/itex] as [itex] \Delta t \to 0 [/itex], but I'm lost after that. How does one come to the conclusion that [itex] \displaystyle \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t}(\Delta\theta)^n [/itex] for [itex] n>0 [/itex]? Then, how does this lead us to the conclusion that [itex] \hat{r} [/itex] 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.