# A Vec norm in polar coord. differs from norm in cartesian coor

#### Emil_M

I am really confused about coordinate transformations right now, specifically, from cartesian to polar coordinates.

A vector in cartesian coordinates is given by $x=x^i \partial_i$ with $\partial_x, \partial_y \in T_p \mathcal{M}$ of some manifold $\mathcal{M}$ and and $x^i$ being some coefficients.

Assuming an euclidean metric $[\delta_{ij}]= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$, the norm of the vector $x$ is given by $$\sqrt{\delta(x,x)}=\sqrt{\delta_{ij}x^i x^j}=\sqrt{x^2+y^2}=:r$$.

Now, we make a transformation into polar coordinates:

$x=r \cos{\phi}$
$y=r \sin{\phi}$

Which gives us the vector $x=y^i \partial_i$ in terms of the basis vectors $\partial_r, \partial_\phi$.

The euclidean metric in polar coordinates looks as follows:
$[\hat{\delta_{ij}}]= \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$

Computing the norm in the polar coordinates yields:

$$\sqrt{\delta(x,x)}=\sqrt{\hat{\delta_{ij}}y^i y^j}=\sqrt{r^2+r^2\phi^2}.$$

As $\delta(x,x)$ is a tensor relation, though, it should be coordinate independent. Furthermore, the inner product is invariant over coordinate transformations. So why is the norm different?

To add to the confusion, I tried transforming the coefficients from cartesian to polar coordinates:

$$x'^i=\frac{\partial f^i}{\partial x^j} x^j \;\;\; (1)$$

With $\frac{\partial f^i}{\partial x^j}=\left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\frac{\sin{\phi}}{r} & \frac{\cos{\phi}}{r} \end{array} \right)$ being the matrix of partial derivatives.

Applying $(1)$ for the first component $r$ yields:

$$x'^1= \frac{\partial f^1}{\partial x^j} x^j= \cos{\phi}\cdot r \cos{\phi} + \sin{\phi}\cdot r \sin{\phi}=r$$

So far so good. For the second component $\phi$, however, we get:

$$x'^2= \frac{\partial f^2}{\partial x^j} x^j=-\frac{\sin{\phi}}{r} \cdot r \cos{\phi}+\frac{\cos{\phi}}{r} \cdot r \sin{\phi}=0$$

So what's going on here? Why does any general vector $x=(x,y)^T$ have a zero second component in polar coordinates?

Help!

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#### Orodruin

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The position vector is $r\partial_r$ in polar coordinates, not $r\partial_r + \theta\partial_\theta$. The coordinate transformations are the coordinate transformations, they are generally not the components of a vector.

#### Emil_M

I'm sorry, I don't understand. A basis for $\mathbb{R}^2$ needs two basis vectors. And any vector in $\mathbb{R}^2$ is then a linear combination of said basis vectors, right? Are the basis vectors of polar coordinates not given by $\partial_r$ and $\partial_\phi$? If so, how can a single basis vector form a basis for a two-dimensional vector space?

#### Orodruin

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I'm sorry, I don't understand. A basis for $\mathbb{R}^2$ needs two basis vectors. And any vector in $\mathbb{R}^2$ is then a linear combination of said basis vectors, right? Are the basis vectors of polar coordinates not given by $\partial_r$ and $\partial_\phi$? If so, how can a single basis vector form a basis for a two-dimensional vector space?
Yes, it has two basis vectors, but the $\phi$-component of the position vector is zero, not $\phi$.

#### Orodruin

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with $\partial_x, \partial_y \in T^*_p \mathcal{M}$ of some manifold $\mathcal{M}$ and and $x^i$ being some coefficients.
I think a part of your confusion is that you are confusing the components of some general vector with the actual coordinates. The coordinates are not the components of a vector. Furthermore, the partial derivatives are not in $T_p^*M$, they are in $T_pM$ (they are a set of basis vectors in $T_p^M$, the components multiplying them are scalar fields!)

Why does any general vector x=(x,y)Tx=(x,y)Tx=(x,y)^T have a zero second component in polar coordinates?
That is not a general vector, it is your position vector because you chose it to have the coordinates as components in Cartesian coordinates.

#### kent davidge

The euclidean metric in polar coordinates looks as follows:
$[\hat{\delta_{ij}}]= \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$
Here you gave the metric obtained by means of the cartesian coordinate system! So you should also use the cartesian components in
$$\sqrt{\delta(x,x)}$$
which reads $\sqrt{r^2 \cos^2 \phi + r^2 \sin^2 \phi} = r$

As to why the $\phi$-component of the position vector is zero, it simply because the position vector is defined in this way! It is the position of a given particle, which by the way you calculated when you started out with the vector in cartesian coordinates.

#### Orodruin

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Here you gave the metric in terms of the cartesian coordinate system!
No he did not. There is no $r^2$ in the metric in a Cartesian coordinate system.

#### kent davidge

No he did not. There is no $r^2$ in the metric in a Cartesian coordinate system.
I meant the given metric is obtainable through $$\frac{\partial x}{\partial x'} \frac{\partial x}{\partial x'}g$$ where $x, x'$ are the cartesian and polar coordinates, respectively, and $g$ the metric in cartesian coordinates.

#### Orodruin

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First of all, he should not be putting the Cartesian components into $\hat \delta$ as your post seems to suggest. He is correct in wanting to put the components of the polar coordinate system into the metric expressed in the polar coordinates. The problem is, as I have already stated, that he has mixed up coordinates with vector components. Nothing else.

Second, the expression
$$\frac{\partial x}{\partial x'} \frac{\partial x}{\partial x'}g$$
really makes no sense unless you add in the appropriate indices.

#### Emil_M

I think a part of your confusion is that you are confusing the components of some general vector with the actual coordinates. The coordinates are not the components of a vector. Furthermore, the partial derivatives are not in $T_p^*M$, they are in $T_pM$ (they are a set of basis vectors in $T_p^M$, the components multiplying them are scalar fields!)
Woops, you're right, that was a typo.

That is not a general vector, it is your position vector because you chose it to have the coordinates as components in Cartesian coordinates.
I'm not sure I understand this. I just chose some local coordinate representation of the vector $x$, that does not make it any less general, does it?

Thank you for taking the time to help me, btw! I appreciate it.

#### Orodruin

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I'm not sure I understand this. I just chose some local coordinate representation of the vector $x$, that does not make it any less general, does it?
No, you did not. You chose a very specific vector, the position vector, which in Cartesian coordinates $x$ and $y$ is expressed as $x\partial_x + y\partial_y$. This vector will in general not be expressed as $\chi^a\partial_a$ in an arbitrary coordinate system with coordinates $\chi^a$. As you have already shown with the computations in your first post, the components of the position vector in polar coordinates are $r$ and $0$, respectively.

A general vector field would be of the form $V = V^1\partial_x + V^2 \partial_y$, where the $V^i(x,y)$ are general functions of the coordinates. If you apply the transformation rules to this general vector you will find that
• It generally has non-zero components in both directions.
• Its magnitude is the same regardless of the coordinate system.

#### Dale

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Are the basis vectors of polar coordinates not given by ∂r∂r\partial_r and ∂ϕ∂ϕ\partial_\phi?
You have to be careful here. Coordinates do not have basis vectors. The tangent space is what has basis vectors. Each point in the manifold has a different tangent space. In a flat manifold the distinction can be blurred, but I think it is a good distinction to keep in mind

#### Orodruin

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You have to be careful here. Coordinates do not have basis vectors.
There is a basis of the tangent space associated to any coordinate system, i.e., the holonomic basis. Of course, $\partial_i$ belongs to different tangent spaces at each point of the manifold, but that is a different matter.

#### Dale

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There is a basis of the tangent space associated to any coordinate system, i.e., the holonomic basis.
Yes. As you say, it is a basis of the tangent space. It is derived from the coordinates, but it is not a basis of the coordinates because the coordinates are not a vector space in general.

#### Orodruin

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Yes. As you say, it is a basis of the tangent space. It is derived from the coordinates, but it is not a basis of the coordinates because the coordinates are not a vector space in general.
I think nomenclature is also somewhat to blame here as the holonomic basis is sometimes referred to as a "coordinate basis". That just means a holonomic basis, not a basis for the coordinates.

#### stevendaryl

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I'm sorry, I don't understand. A basis for $\mathbb{R}^2$ needs two basis vectors. And any vector in $\mathbb{R}^2$ is then a linear combination of said basis vectors, right? Are the basis vectors of polar coordinates not given by $\partial_r$ and $\partial_\phi$? If so, how can a single basis vector form a basis for a two-dimensional vector space?
Cartesian coordinates lead to bad habits that only show up when you start using non-Cartesian coordinates. One of them is talking about "position vectors". A location is not a vector. People think of the point with coordinates $(x,y)$ this way:
1. Start at the origin ($x=0, y=0$).
2. Travel $x$ units in the x-direction.
3. Travel $y$ units in the y-direction.
4. Then you have arrived at the point $(x,y)$
Non-Cartesian coordinates don't work that way. The point with coordinates $(r, \theta)$ cannot be described as:
1. Start at the origin.
2. Travel $r$ units in the r-direction.
3. Travel $\theta$ units in the $\theta$-direction
One reason this description makes no sense is because "the r-direction" or "the $\theta$-direction" is not a unique direction. What direction is meant by "the r-direction" depends on your location.

So $(x,y)$ can be associated with a vector in a way that $(r, \theta)$ cannot be.

#### Orodruin

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One of them is talking about "position vectors". A location is not a vector.
This is a problem of affine space, not of Cartesian coordinates. Given an affine space, you can select an origin and by the definition of an affine space, there is a bijection from the tangent space to the affine space. In this sense, it is perfectly fine to talk about a position vector in an affine space as long as you have defined an origin. The position vector takes a vector value at each point and is therefore actually a vector field. It is perfectly well defined to express this vector field in any coordinates you want to. The problem appears when one starts to assume that the coordinates are the components of this field in all coordinate systems, which is simply not true for the reasons you stated.

So (x,y)(x,y)(x,y) can be associated with a vector in a way that (r,θ)(r,θ)(r, \theta) cannot be.
The position vector can perfectly well be related to the $r$ and $\theta$ coordinates in an affine space as long as you have chosen your origin. If the origin is chosen to be the point $r = 0$ (as is customary), the position vector field is given by $r\partial_r$.

#### Dale

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Hmm, when is a manifold an affine space?

#### stevendaryl

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The position vector can perfectly well be related to the $r$ and $\theta$ coordinates in an affine space as long as you have chosen your origin. If the origin is chosen to be the point $r = 0$ (as is customary), the position vector field is given by $r\partial_r$.
Yes, but the original confusion was that in the case of Cartesian coordinates, the point $(x,y)$ means "start at the origin, and go x units in the x-direction and then y units in the y-direction". The point $(r,\theta)$ cannot be interpreted that way. For one thing, "the r-direction" is undefined at the origin. You can associate a position $(r,\theta)$ with a vector at the origin, but that vector is not one with components $(r,\theta)$.

#### Orodruin

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Hmm, when is a manifold an affine space?
When it is homeomorphic to $\mathbb R^n$, see also https://en.wikipedia.org/wiki/Affine_manifold (an affine manifold is not necessarily an affine space, it does not need to be all of $\mathbb R^n$, just a subset).

Yes, but the original confusion was that in the case of Cartesian coordinates, the point $(x,y)$ means "start at the origin, and go x units in the x-direction and then y units in the y-direction". The point $(r,\theta)$ cannot be interpreted that way. For one thing, "the r-direction" is undefined at the origin. You can associate a position $(r,\theta)$ with a vector at the origin, but that vector is not one with components $(r,\theta)$.
Yes, which is why I find it an inherently bad idea to represent vectors as (a,b,c) without any reference to the basis used and prefer to write out the basis explicitly.

#### vanhees71

Gold Member
I am really confused about coordinate transformations right now, specifically, from cartesian to polar coordinates.

A vector in cartesian coordinates is given by $x=x^i \partial_i$ with $\partial_x, \partial_y \in T_p \mathcal{M}$ of some manifold $\mathcal{M}$ and and $x^i$ being some coefficients.

Assuming an euclidean metric $[\delta_{ij}]= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$, the norm of the vector $x$ is given by $$\sqrt{\delta(x,x)}=\sqrt{\delta_{ij}x^i x^j}=\sqrt{x^2+y^2}=:r$$.

Now, we make a transformation into polar coordinates:

$x=r \cos{\phi}$
$y=r \sin{\phi}$

Which gives us the vector $x=y^i \partial_i$ in terms of the basis vectors $\partial_r, \partial_\phi$.

The euclidean metric in polar coordinates looks as follows:
$[\hat{\delta_{ij}}]= \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)$

Computing the norm in the polar coordinates yields:

$$\sqrt{\delta(x,x)}=\sqrt{\hat{\delta_{ij}}y^i y^j}=\sqrt{r^2+r^2\phi^2}.$$

As $\delta(x,x)$ is a tensor relation, though, it should be coordinate independent. Furthermore, the inner product is invariant over coordinate transformations. So why is the norm different?

To add to the confusion, I tried transforming the coefficients from cartesian to polar coordinates:

$$x'^i=\frac{\partial f^i}{\partial x^j} x^j \;\;\; (1)$$

With $\frac{\partial f^i}{\partial x^j}=\left( \begin{array}{cc} \cos{\phi} & \sin{\phi} \\ -\frac{\sin{\phi}}{r} & \frac{\cos{\phi}}{r} \end{array} \right)$ being the matrix of partial derivatives.

Applying $(1)$ for the first component $r$ yields:

$$x'^1= \frac{\partial f^1}{\partial x^j} x^j= \cos{\phi}\cdot r \cos{\phi} + \sin{\phi}\cdot r \sin{\phi}=r$$

So far so good. For the second component $\phi$, however, we get:

$$x'^2= \frac{\partial f^2}{\partial x^j} x^j=-\frac{\sin{\phi}}{r} \cdot r \cos{\phi}+\frac{\cos{\phi}}{r} \cdot r \sin{\phi}=0$$

So what's going on here? Why does any general vector $x=(x,y)^T$ have a zero second component in polar coordinates?

Help!
Don't overcomplicate things. It's an ingenious trick of the mathematicians to write $\partial_i$ for the holonomous basis vectors. You can just read it as the corresnponding differential operator, i.e., for arbitrary generalized coordinates you treat it as the partial derivative $\partial_i=\partial/\partial q^i$. Now we have
$$x^1=r \cos \varphi, \quad x^2=r \sin \varphi.$$
Then you get
$$\frac{\partial}{\partial x^i}=\frac{\partial q^j}{\partial x^i} \frac{\partial}{\partial q^i}.$$
Setting
$${T^i}_j=\frac{\partial q^j}{\partial x^i},$$
we have
$$\hat{T}=\hat{U}^{-1} \quad \text{with} \quad {U^i}_j=\frac{\partial x^i}{\partial q^j}.$$
So we first calculate $\hat{U}$:
$$\hat{U}=\begin{pmatrix}\cos \varphi & -r \sin \varphi \\ \sin \varphi & r \cos \varphi \end{pmatrix}.$$
Inverting gives
$$\hat{T}=\hat{U}^{-1} = \begin{pmatrix} \cos \varphi & \sin \varphi \\ -\sin \varphi/r & \cos \varphi/r \end{pmatrix}.$$
Thus you get
$$\frac{\partial}{\partial x^i} = {T^{i}}_j \frac{\partial}{\partial q_i} = (\partial_r,\partial_{\varphi}) \hat{T}=(\cos \varphi \partial_r-\sin \varphi/r \partial_{\varphi}, \sin \varphi \partial_r+\cos \varphi/r \partial_r)$$.
$$x^i \frac{\partial}{\partial x^i}=(\cos \varphi \partial_r-\sin \varphi/r \partial_{\varphi}, \sin \varphi \partial_r+\cos \varphi/r \partial_r) \begin{pmatrix}r \cos \varphi \\ r \sin \varphi \end{pmatrix}=r \partial_r.$$
This is of course already obvious without all this formalism, but it's a nice exercise. Note that $\vec{x}$ is not a "general vector" but a special vector, namely the position vector.

The norm transforms differently. This is also most easily seen in the Cartan notation. The metric coefficients in the two metrics are given by taking $\mathrm{d} q^i$ as the increments of the coordinates, i.e., you have
$$\mathrm{d} s^2=\delta_{ij} \mathrm{d} x^i \mathrm{d} x^j = \delta_{ij} \frac{\partial x^i}{\partial q^k} \frac{\partial x^j}{\partial q^l} \mathrm{d} q^j \mathrm{d} q^l.$$
The metric coefficients in cylinder coordinates are thus given by
$$g_{kl}=\delta_{ij} {U^{i}}_k {U^{j}}_l =[(\hat{U}^{T} \hat{U}]_{kl}.$$
Using the above result for $\hat{T}$ we get
$$\hat{g}\hat{T}^T \hat{T}=\mathrm{diag}(1,r^2).$$
For the position vector you get thus indeed
$$\delta_{ij} x^i x^j=(x^2+y^2)=r^2=(r,0) \hat{g} \begin{pmatrix} r \\ 0 \end{pmatrix}.$$
So everything is consistent, as it should be.

#### Emil_M

Thanks everybody for your amazing input!
I will think about all the information you have given me, and might return with follow-up questions :)

"Vec norm in polar coord. differs from norm in cartesian coor"

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