Vectors and Polar Coordinates
A breezy walkthrough of cross products, the BAC-CAB rule, coordinate transformation matrices, and polar coords — all the vector stuff that actually felt new in university physics.
Vectors. We’ve been doing these since high school, so… there shouldn’t be anything scary here, right?!!!
The first thing that felt actually new in university vector-land was probably the cross product, right??
And right after we learned the cross product, out of nowhere~~~ the professor goes “memorize this one, kids ^^” — and that thing was the BAC-CAB rule.
$$\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) \quad = \quad \overrightarrow{B}(\overrightarrow{A} \cdot \overrightarrow{C}) \quad - \quad \overrightarrow{C}(\overrightarrow{A} \cdot \overrightarrow{B})$$First thing to notice: a vector triple product gives you… a vector!!!!
Why?????? Well, if you’ve actually learned the cross product, you already know why…
And we tend to just memorize BAC-CAB as a formula, but like — why does it come out looking exactly like that?!?!?!
I think trying to derive it yourself is a genuinely good move. You get why it lands in that shape.
It’s not hard. Seriously, give it a shot on your own!
OK, next up — the coordinate transformation matrix.
The coordinate transformation matrix, which was basically bread-and-butter in linear algebra…!
A coordinate transformation matrix is just a linear map (a function), and the easy way to analyze a linear map is to look at where it sends the basis vectors of your existing coordinate system.
For the full story on this, go check out ‘Linear Algebra I Studied.’ For those of you who just came here for classical mechanics, I’ll do it textbook-style…
First, I’m going to express the same vector using the basis vectors of two different coordinate systems.
Then we’ll find the relationship between those two systems.
If we write vector A in the basis of the Oxyz (black) system,
$$\overrightarrow{A} \quad = \quad A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k}$$it’ll look like this, yeah??
Now — off to Ox’y’z’!!!!
Let’s go find $A_{x'}$. If we dot vector A with the unit vector $\hat{i}'$ along the x’ axis (a.k.a. projecting it),
$$A_{x'}$$this should drop out, right??
$$\left| A_{x'} \right| \quad = \quad \overrightarrow{A} \cdot \hat{i}' \quad = \quad \overrightarrow{A} \quad = \quad \left( A_{x}\hat{i}+A_{y}\hat{j}+A_{z}\hat{k} \right) \hat{i}' \quad = \quad A_{x}\hat{i}\cdot\hat{i}'+A_{y}\hat{j}\cdot\hat{i}'+A_{z}\hat{k}\cdot\hat{i}'$$Same idea for the other two:
$$\left| A_{y'} \right| \quad = \quad A_{x}\hat{i}\cdot\hat{j}'+A_{y}\hat{j}\cdot\hat{j}'+A_{z}\hat{k}\cdot\hat{j}'$$$$\left| A_{z'} \right| \quad = \quad A_{x}\hat{i}\cdot\hat{k}'+A_{y}\hat{j}\cdot\hat{k}'+A_{z}\hat{k}\cdot\hat{k}'$$Writing those three equations in matrix form — honestly, a 10th grader could do this. Matrices show up in 10th grade, right???
Anyway, stuffing them into a matrix:
$$\begin{pmatrix} \left| A_{x'} \right| \\ \left| A_{y'} \right| \\ \left| A_{z'} \right| \end{pmatrix} \quad = \begin{pmatrix} ii' & ij' & ik' \\ ij' & jj' & kj' \\ ik' & jk' & kk' \end{pmatrix} \begin{pmatrix} \left| A_{x} \right| \\ \left| A_{y} \right| \\ \left| A_{z} \right| \end{pmatrix}$$…yeah, pretty easy, right???? One more little thing:
By the exact same logic, the rotation matrix that spins your coordinate system by $\theta$~
$$\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$$the reason this is what you get is exactly
$$\begin{pmatrix} ii' & ji' \\ ij' & jj' \end{pmatrix}$$and you’ll be able to see straight through to why.
Honestly it’s so easy I’m starting to wonder why I’m even writing this up, heh heh.
Polar coordinates
From here it gets a little spicier, haha. Because you don’t do this in high school! heh heh heh.
Cartesian coordinates (the rectangular coordinate system) are, like, just~~ nice, haha.
But there are times when using them makes things absurdly complicated.
And ditching Cartesian can make those times dramatically easier!!
(I’m a physics kid, so let me give a physics example. In quantum mechanics, the hydrogen atom is a Schrödinger equation problem — but if you write that equation in Cartesian form?
Cooked. Not just cooked, basically unsolvable. But the moment you look at it in polar coordinates?
Solvable!!!!!!! (Still not easy though, haha. Sorry about that.))
So now — we have to abandon the x-axis and y-axis we’ve been treating like a religion. …(cries)
Before we really get into it, quick reminder: “coordinates” are originally what you get when you take the “coefficients in front of the basis vectors” and shove them into an “ordered pair.”
It’s less like “(2,3) means 2 on x and 3 on y,” and more like —
$$2\hat{i}+3\hat{j}$$this is what’s really going on, and that’s why we write it as the ordered pair (2,3). That’s the more honest way to say it.
Now, using polar coordinates means we’re changing the basis vectors,
which means the coefficients change too, which means the numbers we jam inside ( , ) are going to be different.
So let’s first take a rough look at how Cartesian basis vectors and polar basis vectors differ.
Polar coordinates have no fixed basis.
Meaning: the basis keeps changing on you…
For us, brainwashed by the x,y,z coordinate system since forever, this probably doesn’t hit right away.
Let’s nail it with one picture.
The basis vectors $\hat{i},\hat{j}$ describing vectors A, B, C are the same across the board…!
Now look at polar.
the direction of the vectors representing this, and
$$\hat{\theta}$$the direction of the vectors representing this — they’re different for each vector.
In other words, when you describe this vector vs. that vector, the basis itself is different.
OK.
Feels like I’m making things murkier, so why did I even bring this up???
Because when we look at velocity and acceleration in polar coordinates, if I hadn’t said this, things would get confusing fast.
If you’re not confused, feel free to ignore all of it!!!!
Anyway — to describe velocity, acceleration, etc. in polar coordinates,
we’re going to take the time derivative of the position vector once, and then again right after.
The position vector in polar coordinates is
$$\overrightarrow{r} = r\hat{r}$$(here $r$ is just a number, and $\hat{r}$ is the direction vector.)
Now I’ll differentiate with respect to time.
$$\frac{d\overrightarrow{r}}{dt} = \frac{dr}{dt}\hat{r} \quad + \quad r\frac{d\hat{r}}{dt}$$Done!, except — in Cartesian the derivative of the basis vector was 0… but now, as we just saw above,
the story just changed!! heh. Why?? Because the basis vectors are not constant anymore!!!!
So from here on we have to wrestle with
$$\frac{d\hat{r}}{dt}$$this guy!!!
If the unit vector pointing in the r direction just kept pointing in the r direction, it wouldn’t change at all —
so if a change in that unit vector does happen,
it’s because the thing changed in the $\theta$ direction!!!!
So the direction of the change in
$$\hat{r}$$the direction of the change of $\hat{r}$ points along
$$\hat{\theta}$$and the magnitude is
$$\Delta \hat{r} \quad = \quad \hat{\theta}\Delta\theta$$divide both sides by $\Delta t$:
$$\therefore \frac{d\hat{r}}{dt} = \dot{\theta}\hat{\theta}$$and that’s what $d\hat{r}/dt$ is.
Same deal for
$$\hat{\theta}$$the unit vector in the $\theta$ direction — its change points along
$$-\hat{r}$$!!!!
Why the minus???
You kinda feel it here, right??
When $\theta$ increases (in the + direction), the direction of that change points in the $-\hat{r}$ direction!!
Still not clicking? OK:
If “before the change” is red and “after the change” is blue,
the “change in the vector” is that thick red one, right??
So what I’m saying is: the vector representing the change in $\hat{\theta}$ points straight toward the center.
$$\therefore \frac{d\hat{\theta}}{dt} = \quad -\dot{\theta}\hat{r}$$OK, now let me just barrel through the time derivative of the position vector:
$$\overrightarrow{v} \quad = \quad \frac{d\overrightarrow{r}}{dt} \quad = \quad \frac{dr}{dt}\hat{r} \quad + \quad r\frac{d\hat{r}}{dt} \quad = \quad \dot{r}\hat{r} \quad + \quad r(\dot{\theta}\hat{\theta})$$Done~~~
Now let’s tack on acceleration and keep rolling!
$$\overrightarrow{a} \quad = \quad \frac{d}{dt}\overrightarrow{v} \quad = \quad \frac{d}{dt}(\quad \dot{r}\hat{r} \quad + \quad r(\dot{\theta}\hat{\theta})) \quad \\ \quad = \quad \ddot{r}\hat{r} \quad + \quad \dot{r}\frac{d\hat{r}}{dt} \quad + \quad \dot{r}\dot{\theta}\hat{\theta} \quad + \quad r\ddot{\theta}\hat{\theta} \quad + \quad r\dot{\theta}\frac{d\hat{\theta}}{dt}\\ \quad = \quad \ddot{r}\hat{r} \quad + \quad \dot{r}(\dot{\theta}\hat{\theta}) \quad + \quad \dot{r}\dot{\theta}\hat{\theta} \quad + \quad r\ddot{\theta}\hat{\theta} \quad + \quad r\dot{\theta}(-\dot{\theta}\hat{r})\\ \quad = \quad (\ddot{r}-r\dot{\theta}^{2})\hat{r} \quad + \quad (2\dot{r}\dot{\theta} \quad + \quad r\ddot{\theta})\hat{\theta}$$Please, please don’t just memorize this as a formula~
Just be able to derive it. That’s enough.
OK so — we get polar coordinates now, but wait, earlier we were doing coordinate transformations, right??
“Hey man, shouldn’t you also give us the matrix that transforms from Cartesian to polar???”
— fair pushback,
but even though mechanics textbooks don’t really show it, you can get it super easily using the Jacobian!!!
Go google “Jacobian”!! heh heh heh heh.
There are probably plenty of great posts out there already, but if anyone actually asks, I’ll write one up on the spot!!
Alright, bye bye~
Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.