Converting Velocity Formula: Polar to Cartesian

[n3pix]

I have a little question about converting Velocity formula that is derived as,

$\vec{V}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\hat{x}+\frac{dy}{dt}\hat{y}+\frac{dz}{dt}\hat{z}$

in Cartesian Coordinate Systems $(x, y, z)$. I want to convert this into Polar Coordinate System $(r, \theta)$.

$\vec{r}=r(t)\hat{r}(t)$

$\frac{\vec{r}}{dt}=\frac{d}{dt}[r(t)\hat{r}(t)]=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}$

There are 2 methods to get velocity vector formula that is defined on Polar Coordinate Systems.

First method is like this,

$\hat{r}=\hat{x}\cos{\theta}+\hat{y}\sin{\theta}$

$\hat{\theta}=-\hat{x}\sin{\theta}+\hat{y}\cos{\theta}$

$\frac{d\hat{r}}{dt}=-\sin{\theta}\dot{\theta}\hat{x}+\cos{\theta}\dot{\theta}\hat{y}$

$\frac{d\hat{r}}{dt}=\dot{\theta}(-\sin{\theta}\hat{x}+\cos{\theta}\hat{y})$

$\frac{d\hat{r}}{dt}=\dot{\theta}(\hat{\theta})=\dot{\theta}\hat{\theta}$

$\frac{d\vec{r}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}=\frac{dr}{dt}\hat{r}+r\dot{\theta}\hat{\theta}$

This is the first method. The second method is more complex than the first one as my opinion.

Let me show you the second method,

As $\hat{r}$ and $\hat{\theta}$ unit vectors are defined on a circle, to understand them and their derivation over time we have to examine a particle's motion along a circle. In circular motion, at $\Delta t$ time the $\hat{r}$ unit vector becomes $\hat{r}+\Delta \hat{r}$. When $\Delta t\to 0$, $\Delta \hat{r}$ and $\hat{\theta}$ are in the same direction (Fig. 2.20a, 2.20b). If $\Delta \theta \to 0$ next to $\Delta t$, the magnitude of $\Delta \hat{r}$,

$\left|\Delta \hat{r}\right|=\left|\hat{r}\right|\Delta \theta=\Delta \theta$

Because $\left|\hat{r}\right|=1$. So,

$\Delta \hat{r}=\Delta \theta\hat{\theta}$

[Q-1: Here, how this equation is possible to equal? Because $\Delta \theta$ is a radian or degree and what does mean it's multiplication with $\hat{\theta}$ unit vector? A figure, schema, graph or example would be fine!]

[Q-2: Also, when we multiply the both sides of $\left|\Delta \hat{r}\right|=\left|\hat{r}\right|\Delta \theta=\Delta \theta$ with $\hat{\theta}$ how we get $\Delta \hat{r}=\Delta \theta\hat{\theta}$?]

$\frac{\Delta\hat{r}}{\Delta t}=\frac{\Delta \theta}{\Delta t}\hat{\theta}$

[Q-3: Here, we divide both sides with $\Delta t$ but on the right side we have $\frac{\Delta \theta}{\Delta t}\hat{\theta}$ instead of $\frac{\Delta \theta\hat{\theta}}{\Delta t}$, because the $\hat{\theta}$ is changing by time and it can be rewrited as a function of time (t) $\hat{\theta}(t)$. We haven't touched $\hat{\theta}$ on the equation althogh it's changing by time. Why?]

In $\Delta t\to 0$ limit case, we get the following result

$\frac{d\hat{r}}{dt}=\frac{d\theta}{dt}\hat{\theta}$

With the smilar processes, we get $\frac{d\hat{\theta}}{dt}=-\frac{d\theta}{dt}\hat{r}$. This process is related with the given figure below (Fig. 2.20c).

And lastly, if we put the $\frac{d\hat{r}}{dt}$ and $\frac{d\hat{\theta}}{dt}$ into the velocity vector formula, we get

$\vec{V}=\frac{d\vec{r}}{dt}=\frac{d}{dt}(r(t)\hat{r}(t))=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}$

$\vec{V}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$

That's it!

Thanks!

 

[n3pix]

Any idea? :(

I'm working on this questions, more specificly on the second way for 3-4 days after the school :(

 

[PeroK]

Any idea? :(

I'm working on this questions, more specificly on the second way for 3-4 days after the school :(

I like the first method.

Note that another way to look at it is:

$\Delta \vec{r} \approx \Delta r \hat{r} + r \Delta \theta \hat{\theta}$

 

[n3pix]

I like the first method.

Note that another way to look at it is:

$\Delta \vec{r} \approx \Delta r \hat{r} + r \Delta \theta \hat{\theta}$

Thank you for your answer! You made me happy, I was checking here every 5 minutes to see something new :P

I'm sorry, I don't understand how you made that derivation of $\Delta \vec{r}$.

Also, I like the first method as well. I think I'm a bit perfectionist person, I can't skip everything easily something make me focus on them instead of keep going. I ask myself what could I lose by skipping this method and I don't have an exact answer of course. I think like I could face against this method or these things in the future. What do you think about this?

Thanks again!

 

[vanhees71]

I don't know, where your problem is. You did it already right with your first method in #1. I don't know, what the 2nd method should show. Note that vectors are invariant objects. The only thing that changes are the components when you change the basis!

 

[vanhees71]

I cannot open the pdf file in #6.

Nevertheless, I'm still puzzled where your problem really is. So let's do the concepts again. You start with Cartesian coordinates of the Euclidean plane, i.e., there are two perpendicular unit vectors $\vec{e}_j$. The position vector is given as a function of time by
$$\vec{r}(t)=x_j(t) \vec{e}_j.$$
Here and in the following the (Cartesian) Einstein summation convention is used, i.e., you have to sum over equal indices from 1 to 2.

Now we introduce polar coordinates. They are related to the Cartesian ones by
$$\vec{r}=r (\cos \theta \vec{e}_1+\sin \theta \vec{e}_2).$$
Now you want to use at any point a basis adapted to the generalized coordinates $q_1=r$ and $q_2=\theta$.

To that end you think of the plane to be covered by coordinate lines, i.e., by $r$-lines where $\theta$ is hold constant and $r$ runs over its range, i.e., $r \in \mathbb{R}_{>0}$ and $\theta$-lines where $r$ is held fixed and $\theta$ is runing over its range, i.e., $\theta \in [0,2\pi)$.

Now in a first step you introduce the holonomous basis, which simply is given by the tangent vectors to the $r$ lines and the $\theta$ lines, given by the corresponding partial derivatives wrt. the generalized coordiantes,
$$\vec{b}_r=\partial_r \vec{r}=(\cos \theta \vec{e}_1 + \sin \theta \vec{e}_2), \quad \vec{b}_{\theta}=\partial_{\theta} \vec{r}=r (-\sin \theta \vec{e}_1 + \cos \theta \vec{e}_2).$$
Now you realize that these vectors are always perpendicular to each other, i.e., $\vec{b}_r \cdot \vec{b}_{\theta}=0$.

It's now convenient in such a case to normalize this vector and have a Cartesian basis at any point in the plane that consists of the perpendicular normalized vectors. To that end you calculate
$$g_r=|\vec{b}_r|=1, \quad g_{\theta}=|\vec{b}_{\theta}|=r$$
and then introduce
$$\vec{e}_r=\frac{1}{g_r} \vec{b}_r=\vec{b}_r=(\cos \theta \vec{e}_1+\sin \theta \vec{e}_2), \quad \vec{e}_{\theta}=\frac{1}{g_{\theta}} \vec{b}_{\theta}=(-\sin \theta \vec{e}_1+\cos \theta \vec{e}_2).$$
Now if you have an arbitrary vector field you can decompose it in its components with respect to this basis, but this basis depends on where you are in space. Thus when taking derivatives you have to take into account this dependence of the basis on the position. For that you need the partial derivatives of the basis vectors wrt. the generalized coordinates. These you can easily calculate using the Cartesian components given above. You get
$$\partial_r \vec{e}_r=0, \quad \partial_{\theta} \vec{e}_{r}=\vec{e}_{\theta}, \quad \partial_r \vec{e}_{\theta}=0, \quad \partial_{\theta} \vec{e}_{\theta}=-\vec{e}_r.$$

Now for your example of the velocity of the point mass. In the polar coordinates you have
$$\vec{r}=r \vec{e}_r.$$
Now to get the velocity you have to take the time derivative of this vector, including the basis vectors. For this you use the chain rule for functions depending on the independent variables $r$ and $\theta$:
$$\dot{\vec{r}}=\dot{r} \vec{e}_r + r \dot{\vec{e}}_r = \dot{r} \vec{e}_r +r (\dot{r} \partial_r \vec{e}_r + \dot{\theta} \partial_{\theta} \vec{e}_r) = \dot{r} \vec{e}_r + r \dot{\theta} \vec{e}_{\theta}.$$
That's all you need!

 

[Ibix]

I cannot open the pdf file in #6.

Despite the file extension, it was a jpg of a design for an over-unity perpetual motion machine, posted here by a different user and apparently now deleted.