Coordinate systems in ##\mathbb{R}^2##

[kent davidge]

I want to show some of my current understanding/findings involving vector spaces. The reason is two fold: to ask whether my current understanding is ok and to give context for a specific question in the end.

The set $\{(x,0), (0,y) \}$, with $x,y \in \mathbb{R}$, spans $\mathbb{R}^2$. For example, $\{(1,0), (0,1) \}$ is the canonical basis.

It's clear that a vector will remain the same under a change of basis. For instance, consider a vector having components $a,b$ in the basis $\{(2,0),(0,3) \}$, i.e. $a(2,0) + b(0,3)$. It naturally will be $2a(1,0) + 3b(0,1)$ in the canonical basis, but it's the same vector in $\mathbb{R}^2$.

Now consider $x = r \cos \theta, y = r \sin \theta$, with $r, \theta \in \mathbb{R}$, the so called polar coordinates. It also spans $\mathbb{R}^2$. By the above reasoning regarding the basis, for each pair $(r, \theta)$ (not both equal to zero) we have a different basis for $\mathbb{R}^2$.

The problem I see is that people do not typically writes a vector like $a(r \cos \theta, 0) + b(0, r \sin \theta)$. So this make me suspicious that my way of doing it is not correct.

Also, following the presented reasoning, if we want the measures of distances (yea, I should have said metric) to be the same in the two given basis, I found that we need to do the following.

Evaluate $dx$ and $dy$ at $\theta = 0$ to get $\{(dr, 0), (0, r d \theta) \}$. Then $dx^2 + dy^2 = dr^2 + r^2 d \theta^2$. The question that arises is why this would happen exactly if $\theta = 0$.

 

[BvU]

To start with the last question: can you write down the coordinate transform between the cartesian and polar bases (i.e. $r = f(x,y), \ \ \theta = g(x,y)\$? And the inverse transform (yes you can and you did: $x = r\cos\theta,\ y = r\sin\theta$).

Differentiate the latter to see what the general $dx^2+dy^2$ expression looks like. There are no mixed terms because the $r$-axis and the $\theta$-axis are perpendicular.

And one typcally writes a vector in polar coordinates as $\ r\hat r + \theta \hat \theta\$, in short $\ (r,\theta)$ .

Practice transform and inverse transform with the cartesian $(1,1)$ and $(-1,1)$ (perpendicular), and then with e.g. $(1,1)$ and $(1,2)$.

 

[kent davidge]

Differentiate the latter to see what the general $dx^2+dy^2$ expression looks like

it's equal to $dr^2 + r^2 d \theta^2$. So the way of obtaining it as I gave above was just a coincidence?

And one typcally writes a vector in polar coordinates as $\ r\hat r + \theta \hat \theta\$, in short $\ (r,\theta)$

Can we identify $\hat r$ with $(r \cos \theta,0)$ and $\hat \theta$ with $(0, r \sin \theta)$ in the context of post #1?

 

[BvU]

the way of obtaining it as I gave above was just a coincidence

It is not a coincidence:
"The question that arises is why this would happen exactly if $\theta = 0$" is moot. It happens at any $\theta$.

The fact that no mix term ($...dr\;d\theta$) appears in $ds^2$ is due to the fact that $\hat r$ and $\hat \theta$ are perpendicular.

Can we identify $\hat r$ with $(r \cos \theta,0)$ and $\hat \theta =(0, r \sin \theta)$ in the context of post #1?

No. Make a drawing and see for yourself that $\hat r$ does not point in the direction $(r\cos\theta,0)$ is pointing. Can you give us the correct "identification" ? It's not complicated and it is a meaningful step towards writing the base transform as a matrix [edit]comment not useful

Furthermore: In general the hat indicates a unit vector, not a base vector.

 

[Mark44]

I want to show some of my current understanding/findings involving vector spaces. The reason is two fold: to ask whether my current understanding is ok and to give context for a specific question in the end.

The set $\{(x,0), (0,y) \}$, with $x,y \in \mathbb{R}$, spans $\mathbb{R}^2$. For example, $\{(1,0), (0,1) \}$ is the canonical basis.

If xy = 0, then your vectors don't span $\mathbb R^2$, and hence aren't a basis for this space.

It's clear that a vector will remain the same under a change of basis. For instance, consider a vector having components $a,b$ in the basis $\{(2,0),(0,3) \}$, i.e. $a(2,0) + b(0,3)$. It naturally will be $2a(1,0) + 3b(0,1)$ in the canonical basis, but it's the same vector in $\mathbb{R}^2$.

Now consider $x = r \cos \theta, y = r \sin \theta$, with $r, \theta \in \mathbb{R}$, the so called polar coordinates. It also spans $\mathbb{R}^2$. By the above reasoning regarding the basis, for each pair $(r, \theta)$ (not both equal to zero) we have a different basis for $\mathbb{R}^2$.

The problem I see is that people do not typically writes a vector like $a(r \cos \theta, 0) + b(0, r \sin \theta)$. So this make me suspicious that my way of doing it is not correct.

Also, following the presented reasoning, if we want the measures of distances (yea, I should have said metric) to be the same in the two given basis, I found that we need to do the following.

Evaluate $dx$ and $dy$ at $\theta = 0$ to get $\{(dr, 0), (0, r d \theta) \}$.
Then $dx^2 + dy^2 = dr^2 + r^2 d \theta^2$. The question that arises is why this would happen exactly if $\theta = 0$.

This notation is ambiguous. For example, $dx^2$ could be interpreted as the differential of $x^2$, but I believe you mean the square of the differential, or $(dx)^2$. Same for the other terms.

 

[Mark44]

There are no mixed terms because the $r$-axis and the $\theta$-axis are perpendicular.

There really aren't any axes in a polar coordinate system. The x- and y-axes in Cartesian coordinates correspond to the rays $\theta = 0$ and $\theta = \frac \pi 2$.

 

[BvU]

hello Mark,

I agree. Did I use the word axis ? I did ! But, like a genuine ass, I only did it once: in #4 I write that $\hat r$ and $\hat\theta$ are perpendicular

But drawing the direction of the unit vectors $\hat r$ and $\hat\theta$ at an arbitrary point is a useful exercise

 

[kent davidge]

"The question that arises is why this would happen exactly if $\theta = 0$" is moot. It happens at any $\theta$

Here you do not appear to be getting what I meant. By following the procedure described at the end of the opening post, you will get the two metrics equal only if you choose a $\theta = k \pi / 2, k \in \mathbb{N}$, not just any $\theta$.

If xy = 0, then your vectors don't span $\mathbb{R}^2$, and hence aren't a basis for this space.

That's it. I would add this in the opening post, but choose not to do so for the post to be smaler.

No. Make a drawing and see for yourself that $\hat r$ does not point in the direction $(r\cos\theta,0)$ is pointing

There really aren't any axes in a polar coordinate system. The x- and y-axes in Cartesian coordinates correspond to the rays $\theta = 0$ and $\theta = \frac \pi 2$.

What should I take out of these remarks?

Can you give us the correct "identification" ?

I don't know what would be the correct identification.

 

[BvU]

Here you do not appear to be getting what I meant. By following the procedure described at the end of the opening post, you will get the two metrics equal only if you choose a $\theta = k \pi / 2, k \in \mathbb{N}$, not just any $\theta$.

Is that so ? Perhaps you could show me a counterexample where $ds^2 = dx^2 + dy^2 \ne dr^2 + r^2 d\theta^2$ for some $\theta\ne k\pi$ ?

I don't know what would be the correct identification.

Can we identify $\hat r$ with $(r \cos \theta,0)$ and $\hat \theta$ with $(0, r \sin \theta)$ in the context of post #1?

I have explained that $\hat r\ne (r \cos \theta,0)$ and, consequently $\hat \theta$ can not be $(0, r \sin \theta)$. I asked you to provide -- in cartesian coordinates -- a unit vector in the direction of $\vec r$ for the point with polar coordinates $(r,\theta)$, and then idem $\hat \theta$

 

[kent davidge]

Is that so ? Perhaps you could show me a counterexample where $ds^2 = dx^2 + dy^2 \ne dr^2 + r^2 d\theta^2$ for some $\theta\ne k\pi$ ?

That doesn't work when using what I described at the end of the opening post. But it does work though if we evaluate $dx$ and square it, and also $dy$, and add them up.

I asked you to provide -- in cartesian coordinates -- a unit vector in the direction of $\vec r$ for the point with polar coordinates $(r,\theta)$, and then idem $\hat \theta$

Woult it be $(x e_x + y e_y) / (e_x \cdot e_y)$, where $\cdot$ is the inner product, $e_x,e_y$ are the cartesian basis vectors.

 

[BvU]

That doesn't work when using what I described at the end of the opening post. But it does work though if we evaluate dxdxdx and square it, and also dydydy, and add them up.

Please show. I can't comment this way.

Would it be

$e_x\cdot e_y = 0$, so: no.
Can you post a drawing ?

 

[kent davidge]

Please show

I explained that in post #1. I don't see value in repeating it over here.

$e_x\cdot e_y = 0$, so: no.
Can you post a drawing ?

 

[BvU]

Almost agree: you want to take the square root of the denominator.

Now substitute $x_0= r\cos\theta$ and $y_0=r\sin\theta$ and you get $\hat r = (\cos\theta,\sin\theta)$ . Analogously, what is $\hat \theta$ ?

Re

don't see value in repeating it over here

No comment, except: I still don't see what it is that makes you say

why this would happen exactly if $\theta = 0$ .

because it happens at all $\theta$.

 

[kent davidge]

Oh cool. But normalizing the vector requires an inner product defined, right? So this wouldn't be the most general approach. Is'nt there a way of working with polar coordinates without taking inner products?

 

[BvU]

Dot product is easy in polar coordinates: $$\vec a\cdot\vec b=|a||b|\cos(\theta_a-\theta_b)$$

Is'nt there a way

Depends on what you want to do ...
Normalizing a single vector is even easier; $\hat r = {\vec r\over r}$

 

[kent davidge]

Dot product is easy in polar coordinates: $$\vec a\cdot\vec b=|a||b|\cos(\theta_a-\theta_b)$$
Depends on what you want to do ...
Normalizing a single vector is even easier; $\hat r = {\vec r\over r}$

Yes, but taking a inner product requires a inner product being defined on the vector space, i.e. an additional step.

Returning to your post #13

Now substitute $x_0= r\cos\theta$ and $y_0=r\sin\theta$ and you get $\hat r = (\cos\theta,\sin\theta)$ . Analogously, what is $\hat \theta$ ?

It's a little bit strange that our general vector is not just $a(\cos \theta,0) + b(0, \sin \theta)$, because I'm not seeing how would we get an additional vector $\hat \theta$.

 

[BvU]

In your picture, $\hat r$ points in the direction of the red vector $(x_0, y_0)$. In polar coordinates, the same vector is described as $r, \theta$ with $r^2 = x_0^2, y_0^2$ and $\tan\theta ={y_0\over x_0}$.

The direction of $\hat \theta$ is found by increasing $\theta$ with a small amount $d\theta$... And the unit vector $\hat \theta$ is $\vec {d\theta}\over {|\vec {d\theta}|}$ !

 

[kent davidge]

In your picture, $\hat r$ points in the direction of the red vector $(x_0, y_0)$. In polar coordinates, the same vector is described as $r, \theta$ with $r^2 = x_0^2, y_0^2$ and $\tan\theta ={y_0\over x_0}$.

The direction of $\hat \theta$ is found by increasing $\theta$ with a small amount $d\theta$... And the unit vector $\hat \theta$ is $\vec {d\theta}\over {|\vec {d\theta}|}$ !

Oh, ok. Got it now.

How would the components of a general vector transform between the polar and cartesian coordiantes in this case? One way is by requiring that they transform in a way such that the inner product is invariant, as one does in the context of manifolds. But is there other possibilites for how the components transform?

 

[Mark44]

In your picture, $\hat r$ points in the direction of the red vector $(x_0, y_0)$. In polar coordinates, the same vector is described as $r, \theta$ with $r^2 = x_0^2, y_0^2$ and $\tan\theta ={y_0\over x_0}$.

Don't you mean $r^2 = x_0^2 + y_0^2$?

 

[romsofia]

I think this would be a great step for you to jump into differential forms!

I'd recommend reading this book:https://www.amazon.com/dp/B00OKUG1UO/?tag=pfamazon01-20 (but he has a more condensed free version which you can find here): http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/start

More specifically in regards to your post about line elements, this page will be the most helpful (but not sure how much you'll get without checking out the first couple of pages): http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/polar and http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/polar2

Hopefully this helps clear up your question about why we can consider these two basis equal.

 

[kent davidge]

Thank you @romsofia for the material. I had already learned about that. I figured a clear way of knowing what the basis vectors are.

Denoting them as $\{e_r, e_\theta \}$, as they form a basis for $\mathbb{R}^2$ they must be orthogonal. The line element is $ds^2 = dx e_x \cdot dx e_x + dy e_y \cdot dy e_y = dr e_r \cdot dr e_r + d\theta e_\theta \cdot d\theta e_\theta$. But we know what the first equality is, it's $dr^2 + r^2 d \theta^2$. Collecting this out we may conclude that $$e_r \cdot e_r = 1 \\ e_\theta \cdot e_\theta = r^2 \\ e_r \cdot e_\theta = 0$$ From this and the drawing showing what $e_r$ is, we get $e_r = (\cos\theta, \sin\theta)$ and $e_\theta = (-r \sin\theta, r \cos\theta)$.

 

[kent davidge]

Also, from the above we may interpret our infinitesimal distance as a infinitesimal variation of each of the basis vectors, taken the inner product so that the distance is basis independent. Did I go too far on my reasoning?

 

[Mark44]

Denoting them as $\{e_r, e_\theta \}$, as they form a basis for $\mathbb{R}^2$ they must be orthogonal.

Just because two vectors form a basis for $\mathbb R^2$, they don't necessarily have to be orthogonal. For example, the vectors <1, 0> and <1, 1> are a basis for $\mathbb R^2$, but they aren't orthogonal.

 

[romsofia]

This thread popped into my head as I was reviewing somethings.

If we transform from polar to cartersian using $x=rcos\theta$ $y=rsin\theta$ we will see that it is defined everywhere except at the origin (r = 0) where it cannot be inverted. So we then can conclude that polar chart does not actually provide a full map of the Euclidean plane (R^2) so maybe that was your issue?

 

[Mark44]

This thread popped into my head as I was reviewing somethings.

If we transform from polar to cartersian using $x=rcos\theta$ $y=rsin\theta$ we will see that it is defined everywhere except at the origin (r = 0) where it cannot be inverted. So we then can conclude that polar chart does not actually provide a full map of the Euclidean plane (R^2) so maybe that was your issue?

If r = 0, then both x and y are 0. $\theta$ is arbitrary, though, so there is not a 1-1 correspondence between points in polar coordinates and points in the Cartesian plane.

 

[kent davidge]

So we then can conclude that polar chart does not actually provide a full map of the Euclidean plane (R^2) so maybe that was your issue?

Not really. I was more asking for confirmation of what I thought was correct, but I have learned a lot since I started this thread.

So we then can conclude that polar chart does not actually provide a full map of the Euclidean plane

@Mark44 latest comment makes me think that's not the case, because as he remembers when $r = 0$ we know that $x$ and $y$ must be zero as well. So $(0, \theta)$ covers $(0,0) \in \mathbb{R}^2$, although this will hold for any $\theta$ and thus we do not have a one-to-one map for this value of $r$. Or maybe I got this wrong?

 

[Mark44]

So $(0, \theta)$ covers $(0,0) \in \mathbb{R}^2$, although this will hold for any $\theta$ and thus we do not have a one-to-one map for this value of $r$. Or maybe I got this wrong?

No, you have it right, but it's not just at the origin. The Cartesian point (1, 1) can have many polar representations, including $(\sqrt 2, \pi/4), (\sqrt 2, 9\pi/4), (-\sqrt 2, 5\pi/4)$ and many others. Some of the representations can be eliminated if you insist that $r \ge 0$ and limit $\theta$ to the interval $[0, 2\pi)$, but that still leaves the multiple representations of the point at the pole.

 

[Stephen Tashi]

Let's consider the vocabulary used in the original post:

Now consider $x = r \cos \theta, y = r \sin \theta$, with $r, \theta \in \mathbb{R}$, the so called polar coordinates. It also spans $\mathbb{R}^2$.

That's correct, if you mean the set $\{x,y: x = r \cos(\theta), y = r \sin(\theta), r \in \mathbb{R}, \theta \in \mathbb{R} \}$ spans the vector space $\mathbb{R}^2$. In fact, that set is all of $\mathbb{R}^2$. ( "Span" has a technical definition. In general a set S can "span" a vector space V without being equal to all of V.)

By the above reasoning regarding the basis, for each pair $(r, \theta)$ (not both equal to zero) we have a different basis for $\mathbb{R}^2$.

What reasoning? I think you are reasoning about parameterizations of $\mathbb{R}^2$, not about sets that "span" $\mathbb{R}^2$. For example, the set of only two vectors $\{(0,1),(1,0)\}$ spans $\mathbb{R}^2$, but it is isn't a parameterization of $\mathbb{R}^2$.

The fact that a parameterization of a vector space involves two parameters (p,q) does not imply that the two parameters are basis vectors. A parameter is a variable. A basis vector is a single vector, not a "variable" vector.

That's the mathematical perspective.

In physics we do find the concept of a "variable basis". The general idea is that at an instant of time t1, we have one useful basis and at a different instant t2 we have a different useful basis. That's where the use of differentials like $r\ d\theta$ enters the picture.