Why Does the Area Element Vanish at the Origin in Polar Coordinates?

[davidge]

Consider the mapping $f: \mathbb{D}^1 \longrightarrow [0,1] \times (0,2 \pi]$ where $\mathbb{D}^1$ is the unit disk. This is the familiar polar coordinate system.

The area element is $dx \wedge dy$ in $\mathbb{D}^1$ and $r dr \wedge d\theta$ in $[0,1] \times (0,2 \pi]$*. Now at $r = 0$ this vanishes. But $dx \wedge dy = r dr \wedge d\theta$ and that means $dx \wedge dy = 0$ as well.

We know that the polar system is singular at $r = 0$ because there cannot be a bijection: for every $\theta \in (0,2 \pi]$ we have the point $(0,0)$ in $\mathbb{D}^1$.

But what is the meaning of the vanishing area element $dx \wedge dy$?

*I know this is two dimensional and the area element as defined in vector calculus as the vector product of two vector requires a third vector, which means a third dimension. But I think this is not relevant here.

 

[fresh_42]

Consider the mapping $f: \mathbb{D}^1 \longrightarrow [0,1] \times (0,2 \pi]$ where $\mathbb{D}^1$ is the unit disk. This is the familiar polar coordinate system.

The area element is $dx \wedge dy$ in $\mathbb{D}^1$ and $r dr \wedge d\theta$ in $[0,1] \times (0,2 \pi]$*. Now at $r = 0$ this vanishes. But $dx \wedge dy = r dr \wedge d\theta$ and that means $dx \wedge dy = 0$ as well.

We know that the polar system is singular at $r = 0$ because there cannot be a bijection: for every $\theta \in (0,2 \pi]$ we have the point $(0,0)$ in $\mathbb{D}^1$.

But what is the meaning of the vanishing area element $dx \wedge dy$?

*I know this is two dimensional and the area element as defined in vector calculus as the vector product of two vector requires a third vector, which means a third dimension. But I think this is not relevant here.

It means that you have allowed the RHS to be multiplied by zero and the LHS not. The equations doesn't hold for $r=0\,$.

 

[davidge]

The equations doesn't hold for $r=0\,$.

Does it mean we have to remove the point $(0,0)$ from the disk?

 

[fresh_42]

Does it mean we have to remove the point $(0,0)$ from the disk?

In order to do what? I consider $(dx,dy)$ as Cartesian coordinates. If we give them coefficients, zero shouldn't be a problem. To me it is as if you asked, whether the origin of the plane should be removed, because the standard basis vectors aren't zero, whereas the radius can be zero.

 

[davidge]

To me it is as if you asked, whether the origin of the plane should be removed, because the standard basis vectors aren't zero, whereas the radius can be zero.

Yes, this is what I was asking you

I consider $(dx,dy)$ as Cartesian coordinates.

Me too

In order to do what?

In order not to have a inconsistent equation when $r = 0$.

 

[fresh_42]

Well, the relation $(0,0) \longleftrightarrow \{0\} \times (0,2\pi]$ is obviously not one-to-one which causes problems. To remove a point is problematic, as it directly affects the functions defined on $\mathbb{D}^1$. I'd rather change the coordinate system depending on the task.

 

[davidge]

Well, the relation $(0,0) \longleftrightarrow \{0\} \times (0,2\pi]$ is obviously not one-to-one which causes problems. To remove a point is problematic, as it directly affects the functions defined on $\mathbb{D}^1$. I'd rather change the coordinate system depending on the task.

Any examples of such coordinate systems?
This is not a specific problem, I was just thinking about what I mentioned in post #1.

 

[fresh_42]

I assume in most cases it doesn't make a difference. If so, I'd simply use the Cartesian version, but that's only because I'm no big fan of polar coordinates. It would be interesting to have an example, where it makes a difference to see, what exactly shouldn't work.

 

[WWGD]

Basically, your coordinate change map is not defined for r=0:

We have:
$x=rcos\theta$
$y=rsin\theta$
But this holds for $r \neq 0$ only , as you said . Then the same is true for $dx \wedge dy = (\frac {\partial}{\partial \theta}d\theta+ \frac{\partial}{\partial r}dr )( rcos\theta+ ysin\theta)$. The wedge is not defined at 0 because the coordinate expression for r=0 is not defined.
EDIT: the one way I can see $dx \wedge dy$ turned into 0 is if the determinant of the Jacobian of a change of coordinate is 0. This is what $f(x)dx\wedge dy$ measures; how much f(x) stretches/shrinks the area element under a trasnformation.

 

[davidge]

EDIT: the one way I can see $dx \wedge dy$ turned into 0 is if the determinant of the Jacobain of a change of coordinate is 0

Which is not the case we are considering, because for this case we as you said, $r=0$ is not on the domain, isn't?

 

[WWGD]

Which is not the case we are considering, because for this case we as you said, $r=0$ is not on the domain, isn't?

Right, we are not really transforming the space, we are expressing it in different terms, i.e., changing coordinate systems, so no squashing ( nor stretching) is happenning. EDIT: By ( many) definition(s) , actually , a change of variable is supposed to be a diffeomorphism, meaning the ais invertible,and therefore its determinant is non-zero. Which kind of makes sense, that you would want that for a coordinate change.

 

[Orodruin]

Am I the only one bothered by the fact that the set $[0,1]\times [0,2\pi)$ is not open and so the described mapping is not a chart? Polar coordinates can never give you a chart at $r=0$ and so it makes no sense to try to interpret coordinate expressions for tensors in polar coordinates at the origin.

 

[WWGD]

Am I the only one bothered by the fact that the set $[0,1]\times [0,2\pi)$ is not open and so the described mapping is not a chart? Polar coordinates can never give you a chart at $r=0$ and so it makes no sense to try to interpret coordinate expressions for tensors in polar coordinates at the origin.

But I think the map is just intended to be a (local) change of coordinates. Is this equivalent to $[0,1] \times [0,2\pi)$ being a chart?EDIT: I can see how a manifold is a way of assigning coordinates, but I had never seen things that way. Still, I agree, the coordinate change is a local diffeomorphism so it should preserve the topology of $\mathbb D^1$.

 

[Orodruin]

But I think the map is just intended to be a (local) change of coordinates. Is this equivalent to $[0,1] \times [0,2\pi)$ being a chart?

A chart is a homeomorphism from an open subset of the manifold to an open subset of $\mathbb R^n$. It does not need to cover the entire manifold. Remove $r=0$, $r=1$ and $\theta = 0$ and you have a chart that covers almost the entire disc.

 

[davidge]

@Orodruin I've been considering $[0,1] \times [0,2 \pi)$ throughout this thread because that set covers the entire disk and I think it's the most familiar way of mapping points on the disk, although it's not homeomorphic to $\mathbb{R}^2$.

it makes no sense to try to interpret coordinate expressions for tensors in polar coordinates at the origin.

Is the area element and all other quantities defined only when we have charts? Is it what you mean here?

 

[Orodruin]

@Orodruin I've been considering $[0,1] \times [0,2 \pi)$ throughout this thread because that set covers the entire disk and I think it's the most familiar way of mapping points on the disk, although it's not homeomorphic to $\mathbb{R}^2$.

Is the area element and all other quantities defined only when we have charts? Is it what you mean here?

If you want to use the coordinate expressions at a point, you must use a chart that is valid at that point. Writing $r\, dr\wedge d\theta$ makes no sense at $r = 0$ because your coordinate system does not define a chart there. There are some charts that cover all of the disc, your coordinate system with the given set is not one of them because it is not a chart. If you use any chart that covers the origin (such as Cartesian coordinates), you will find that the area element is non-zero there.

 

[davidge]

If you want to use the coordinate expressions at a point, you must use a chart that is valid at that point. Writing $r\, dr\wedge d\theta$ makes no sense at $r = 0$ because your coordinate system does not define a chart there. There are some charts that cover all of the disc, your coordinate system with the given set is not one of them because it is not a chart. If you use any chart that covers the origin (such as Cartesian coordinates), you will find that the area element is non-zero there.

I got it. Thanks.

 

[lavinia]

Consider the mapping $f: \mathbb{D}^1 \longrightarrow [0,1] \times (0,2 \pi]$ where $\mathbb{D}^1$ is the unit disk. This is the familiar polar coordinate system.

This mapping cannot be continuous.

In the other direction the map is continuous but singular.

 

[davidge]

In the other direction the map is continuous but singular.

Where it is singular?

 

[lavinia]

Where it is singular?

The map in the other direction maps the entire half open interval ${0}×(0,2π]$ onto the center of the unit disk.

 

[davidge]

The map in the other direction maps the entire half open interval ${0}×(0,2π]$ onto the center of the unit disk.

Oh, yea. Thanks.

 

[WWGD]

This mapping cannot be continuous.

In the other direction the map is continuous but singular.

Are you referring to the coordinate change map $x=rcos\theta, y=rsin\theta$?

 

[lavinia]

Are you referring to the coordinate change map $x=rcos\theta, y=rsin\theta$?

The map from the disk to $[0,1]×(0,2π]$ cannot be continuous. The reverse is singular.

 

[WWGD]

The map from the disk to $[0,1]×(0,2π]$ cannot be continuous. The reverse is singular.

I meant, by "The map" , are you referring to the change of variable map $x=rcos\theta, y =rsin\theta$?

 

[lavinia]

I meant, by "The map" , are you referring to the change of variable map $x=rcos\theta, y =rsin\theta$?

The OP described the inverse of polar coordinates as a surjective map $D^1→[0,1]×(0,2π]$. If this map were continuous - as you also pointed out - then $[0,1]×(0,2π]$ would be compact - which it isn't. So no surjective map can be continuous.

The reverse map from $[0,1]×(0,2π]→D^1$ can be defined as $(α,β) →(α$cos $(β),α$ sin$(β))$ but this map is singular.