1

Consider a point in cartesian and polar coordinates, $P = \{x,y\} = \{r,\theta\}$, such that $x = r \cos\theta$ and $y = r\sin\theta$, where $r$ is a radial distance that is not necessarily unity. I have an arbitrary vector (e.g. a velocity vector) at point $P$, and I know it's cartesian coordinates, $\vec{v} = a^i e_i = a^0 e_0 + a^1 e_1$, where $e_i = \{\hat{x}, \hat{y}\}$, and I want to find the components in a bases that is the polar tangent vectors, $e'_i = \{\hat{r},\hat{\theta}\}$, i.e. $\vec{v} = a^{\prime i} e'_i = a^{\prime 0} e'_0 + a^{\prime 1} e_1$.

(Ultimately I need to do this exercise in 3D spherical coordinates, and do a couple of coordinate rotations in between, but I think this case encompasses my [current] confusion).

Attempt:

I can define a transformation between coordinates based on, $$e'_j = e_i \, {S^i}_j = e_i \frac{\partial e'_j}{\partial e_i},$$

i.e. a covariant transformation. For example, $e'_0 = \hat{r} = \frac{\partial}{\partial r} = \frac{\partial}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial}{\partial y} \frac{\partial y}{\partial r}.$

Thus the tensor that transforms the bases $\{\hat{x},\hat{y}\} \rightarrow \{\hat{r},\hat{\theta}\}$, can be calculated as

$${S^i}_j = \pmatrix{\frac{\partial e'_0}{\partial e_0} & \frac{\partial e'_1}{\partial e_0} \\ \frac{\partial e'_0}{\partial e_1} & \frac{\partial e'_1}{\partial e_1}} = \pmatrix{\frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y}} = \pmatrix{\cos\theta & -\frac{1}{r} \sin\theta \\ \sin\theta & \frac{1}{r} \cos\theta},$$

and the values of $\{r,\theta\}$ here are, again, fixed based on the point $P$. Using this, we immediately get, $e'_i = \{\hat{r}, \hat{\theta} \} = \{\hat{x}\cos\theta + \hat{y}\sin\theta, \,\, - (\hat{x}/r)\sin\theta + (\hat{y}/r) \cos\theta \}$, which matches what I get from geometry. So this seems good.

The components of the vectors transform contravariantly, and so the transformation should be given by, ${T^i}_j = \partial e_j \, / \, \partial e'_i$ (NOTE: is this still true, in non-orthonormal bases?). Going through the same procedure, I find the components of the tensor that transforms the coordinates (transposes because these are 'columns') $\{a_0,a_1\}^T \rightarrow \{a'_0, a'_1\}^T$ as,

$${T^i}_j = \pmatrix{\frac{\partial e_0}{\partial e'_0} & \frac{\partial e_1}{\partial e'_0} \\ \frac{\partial e_0}{\partial e'_1} & \frac{\partial e_1}{\partial e'_1}} = \pmatrix{\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}} = \pmatrix{\cos\theta & \sin\theta \\ -r \sin\theta & r \cos\theta}.$$

Now if we try this, we immediately get a problem: $a^{\prime i} = \{x\cos\theta + y\sin\theta, \,\, -rx\sin\theta + ry\cos\theta \}^T$. The $0$th component seems fine, but the $1$th has the wrong dimensions: length squared instead of dimensionless! So it seems like this $T$ tensor should actually be transposed... but then it no longer seems to represent a contravariant transformation.

Am I missing a transpose somewhere? Or am I assuming an orthonormality somewhere where it doesn't exist? Rows-vs-columns, and left-vs-right matrix multiplication has always confused me, so I'm primarily following the indices and their positions, but I've still tried to be consistent with rows and columns. One "reason" to add a transpose, however, could be that while the inverse matrix should transform the components (as apposed to bases), the components are 'columns' while the bases 'rows', so perhaps that adds a transpose? But where does this come from in index notation? Any help or pointers (even on improper terminology / syntax) is much appreciated!

DilithiumMatrix
  • 796
  • 1
    My 2 cents: completely forget everything about matrices and transposes for such calculations (if at the end you want to store all the information in a matrix, fine, but for computing the coordinate transformations, I suggest you avoid them), because it is a completely arbitrary rule for organizing a collection of numbers and is very often misused, and very easy to make errors (both numerical and conceptual). The number of transposes needed depends on your conventions etc. I think if you transpose your $T^i_j$, you should get the right answer (but I haven't checked the details). – peek-a-boo Jan 04 '20 at 18:43
  • 1
    Take a look at this answer of mine https://math.stackexchange.com/questions/3492505/taking-the-dot-product-in-polar-coordinates-using-the-metric-tensor/3492817#3492817 (If you don't know much about manifolds and tangent spaces, that's still fine, I think the ideas should be ok enough to understand.) Although the question being asked is different, I do address how to transform the components of a vector, and leter in the comments, I talk a little bit about the effect orthonormality has. – peek-a-boo Jan 04 '20 at 19:14
  • @peek-a-boo already very helpful, thank you! Something I now realize I was wrong about was actually $\hat{\theta} \neq \partial / \partial \theta = r \hat{\theta}$. So, $\hat{\theta}$ is a basis-vector, the unit-vector "in the $\theta$ direction". But I'm confused about what $\partial / \partial \theta \equiv \partial_\theta$ is then. Is this "the $\theta$ tangent-vector"? Is there ever a time when you would choose to express something as $A^r \partial_r + A^\theta \partial_\theta$ instead of $A^r \hat{r} + A^\theta \hat{\theta}$? – DilithiumMatrix Jan 04 '20 at 22:06
  • 1
    What is $\partial/\partial \theta$? The crude answer is that it is just a tangent vector to (the manifold) $\Bbb{R}^2$. To give you atleast a slightly satisfying answer, I would have to describe a bit of manifolds and tangent spaces, and for this it is worth being slightly general, and go directly to the $n$ dimensional case. – peek-a-boo Jan 04 '20 at 22:11
  • 1
    Strictly speaking you should imagine two copies of the plane. The first is your $r ,\theta$ "space of parameters" and the second is the "actual" plane $\Bbb{R}^2$ which describes actual points. I hope this distinction is clear: a rectangle $[r_1, r_2] \times [\theta_1, \theta_2]$ in the "parameter space" is a real rectangle. However, if you map these set of points using polar coordinates $(r, \theta) \mapsto \alpha(r,\theta) := (r \cos \theta, r \sin \theta)$, then the image of this rectangle under this map will be an annulus (the region between two circles of radii $r_1, r_2$). – peek-a-boo Jan 04 '20 at 22:34
  • 1
    So, $\alpha$ is the function which maps points in the "parameter space" to points in "the real world". I hope this is clear: if you tell me $r=2, \theta = \pi$, then these are just a pair of numbers, and the tuple $(2, \pi)$ is just an element in the "parameter space". It is only its image point under the map $\alpha$, namely $\alpha(2, \pi) = (2 \cos \pi, 2 \sin \pi)$ which has the physical meaning of being the point in the plane which is a distance $2$ from the origin and an angle $\pi$ from the positive $x$ axis. – peek-a-boo Jan 04 '20 at 22:39
  • 1
    It is this function $\alpha$ (or sometimes by convention its inverse$ \alpha^{-1}$) which formally is what is meant by the term "polar coordinate system". Note that even though the space of parameters and the "real world" are both subsets of $\Bbb{R}^2$, think of them as completely distinct copies of $\Bbb{R}^2$, which play different roles. Now we can finally talk about the vectors. Formally, for each $(r, \theta) \in (0, \infty) \times (0, 2 \pi)$ in the parameter space, we define $\dfrac{\partial}{\partial r}\bigg|{\alpha(r, \theta)} = \dfrac{d}{dt} \bigg|{t=0} \alpha((r, \theta) + t e_1)$ – peek-a-boo Jan 04 '20 at 22:42
  • 1
    In other words, $\partial/ \partial r$ is that vector obtained by first moving along the $e_1 = (1,0)$ direction in the parameter space of ($r, \theta$) (we move along the $e_1$ direction, because the way I described things, I made $r$ the first coordinate of the parameter space an $\theta$ the second component). Now, we map this straight line in the parameter space via the polar coordinate map $\alpha$ to get the curve $t \mapsto \alpha((r,\theta) + t e_1) = \alpha(r+t, \theta) = ((r+t) \cos \theta, (r+t) \sin \theta)$. – peek-a-boo Jan 04 '20 at 22:49
  • 1
    This is now a curve in the "real world",and you're finding it's velocity at time $t=0$ by differentiating at $t=0$. Similarly, we define $\dfrac{\partial}{\partial \theta} \bigg|{(r, \theta)} = \dfrac{d}{dt} \alpha\left( (r, \theta) + t e_2\right)$. i.e we first move along the $e_2$ (i.e $\theta$) direction in the parameter space in a straight line. Then $\alpha$ maps this line in parameter space into the real world to give us a curve. We then differentiate this curve at $t=0$ to get the velocity tangent vector $\dfrac{\partial}{\partial \theta} \bigg|{(r, \theta)}$. – peek-a-boo Jan 04 '20 at 22:52
  • 1
    There are still a few subtleties here, but I'll skip them for now. This idea is very general. For example, in $3D$ spherical coordinates, we have the function $\beta$ which maps points in spherical parameter space to the "real world" via $\beta(r, \theta, \phi) = (r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta)$, where $(r, \theta, \phi) \in (0, \infty) \times (0, \pi) \times (0, 2 \pi)$. Once again, we define very similarly, $\dfrac{\partial}{\partial \phi} \bigg|{\beta(r, \theta, \phi)} = \dfrac{d}{dt} \bigg|{t=0} \beta((r, \theta, \phi) + t e_3)$ – peek-a-boo Jan 04 '20 at 22:57
  • 1
    hopefully this gives atleast some geometric idea of how to think of these tangent vectors: tangent vectors are simply the velocity vectors to curves in the manifold. Note that there are several equivalent ways of defining tangent spaces; but I highly recommend that you watch the lecture series by Frederic Schuller which I referenced also in the comments to the other answer. (I now realize it might have been better to write this all up as an answer simply because of the length of everything, and I already notice some typos which I now can't fix :( ) – peek-a-boo Jan 04 '20 at 23:01
  • 1
    to answer your last question. Strictly speaking, there is no difference which basis you use. An arrow is an arrow, and it doesn't care about any basis/coordinate systems. However, one thing I realised is that it is easier (for me) to write things as $A = A_r \partial_r + A_{\theta} \partial_{\theta}$, as opposed to $A = a_r \hat{r} + a_{\theta} \hat{\theta}$, because then I can directly write down the transofrmation law: $A_x = \dfrac{\partial x}{\partial r} A_r + \dfrac{\partial x}{\partial \theta} A_{\theta}$ This is NOT true for the little $a$'s because you have to wrry abt normalization. – peek-a-boo Jan 04 '20 at 23:17
  • 1
    It is easy in the sense that I just blindly apply that formula (because I've used it so many times it's part of my memory, so if I deviate from it without thinking, I'll make all sorts of mistakes). But if you wanted, then using $a_r = A_r$ and $A_{\theta} = \dfrac{a_{\theta}}{r}$, we have: $A_x = \dfrac{\partial x}{\partial r} a_r + \dfrac{\partial x}{\partial \theta} \dfrac{a_{\theta}}{r}$. So, at some point, if you know the basic ideas then it's just a matter of preference/how you learn things... and this is how I think about these coefficient transformation behaviour. – peek-a-boo Jan 04 '20 at 23:21
  • @peek-a-boo I think there may be an additional subtlety. It seems as though $\partial_\theta$ but not $\hat{\theta}$ allows your coordinate to be an angle instead of a distance. i.e. $a_\theta$ has units of distance, but $A_\theta$ is unitless. For example, if out vector $A$ is supposed to represent a velocity, then in polar coordinates, I would expect the '$\theta$' component to have units of inverse-seconds, and thus I would need to use the $\partial_\theta$ frame, instead of the $\hat{\theta}$ frame. But I don't understand why! What makes $\partial_\theta$ correspond to the angle?! – DilithiumMatrix Jan 05 '20 at 21:29
  • The issue of units is always confusing, and tbh I haven't thought too much about it (because it never bothered me). I think you're right that $\partial_{\theta}$ and $\hat{\theta}$ differ in units by a factor of $[r] = \text{length}$. But to me, the components of a vector by themselves do not have any meaning. So, while you're worried about $A_{\theta}$ and $a_{\theta}$ having different units, my claim is that the product $A_{\theta} \partial_{\theta}$ and $a_{\theta}\hat{\theta}$ have the same units (in fact $A_{\theta}, a_{\theta}$ were defined so that these quantities are equal) – peek-a-boo Jan 06 '20 at 00:54
  • Again, while $A_r$ is dimensionless, the product $A_r \partial_r$ has the same units as $A_{\theta} \partial_{\theta}$ (which is why it makes sense to add them I guess). Also, take a look on physics SE; there are several questions regarding units of objects like the metric tensor and other tensorial objects, and they all seem to echo the fact that the units of the components of a tensor depends heavily on the coordinate system (/basis) with respect to which you're taking the components of the vector. As for why $\partial_{\theta}$ corresponds to angle, well I suggest you draw a picture – peek-a-boo Jan 06 '20 at 00:58
  • For example, draw two copies of the plane. In the first one, draw a line in the $r, \theta$ "parameter space" corresponding to $r_0=1$, and let $\theta$ increase from $\theta_0$. i.e consider the curve $c(t) = (1, \theta_0 + t \theta)$. Now, draw the image of this straight line (in the second copy of the plane) under the polar coordinate map $\alpha$ I defined above (i.e sketch the composite curve $\alpha \circ c$). The image will be part of the arc of a unit circle. Then, draw the vector $(\alpha \circ c)'(0).$ You'll find that this vector will be tangential to the circle, and point towards – peek-a-boo Jan 06 '20 at 01:03
  • increasing values of $\theta$. Recap: we kept $r$ fixed and varied $\theta$ in the parameter space. We then used the polar coordinate function $\alpha$ to map this vertical line in parameter space "into the real world", where the image will be part of the arc of unit circle. The direction of the velocity vector is exactly what we defined as $\partial_{\theta}$ and it points in the direction of change in angle. – peek-a-boo Jan 06 '20 at 01:06

2 Answers2

1

First some differential topology of surfaces (like $\mathbb R^2$) that helps how to handle the coordinates change techniques to get how the change of components behave.

If you have two parametrization $$\Phi:\Omega\hookrightarrow\Sigma \qquad{\rm and}\qquad \Psi:\Gamma\hookrightarrow\Sigma$$ of a surface $\Sigma$, the jacobians $J\Phi, J\Psi$ of them allows to assign the tangents frames at $p\in\Sigma$ via $$J\Phi(a):\mathbb R^2\to T_p\Sigma, $$ where $a\in\Omega$, $\Phi(a)=p$ and $T_p\Sigma$ is the tangent space at $p$, so $$\partial_0=J\Phi(a)e_0\quad{\rm and}\quad \partial_1=J\Phi(a)e_1.$$ Also $$\tilde \partial_0=J\Psi(b)e_0\quad{\rm and}\quad \tilde \partial_1=J\Psi(b)e_1,$$ with the other parametrization such that $\Psi(b)=p$. Now, one can get a map $\lambda:\Omega\to\Gamma$ which complies $$ \Phi=\Psi\circ\lambda\quad{\rm and}\quad \lambda(a)=b $$ So, by the chain's rule $J\Phi=J\Psi\cdot J\lambda$ and $$J\Phi(a)=J\Psi(b)\cdot J\lambda(a),$$ and $$J\Phi(a)e_0=J\Psi(b)\cdot J\lambda(a)e_0 \quad{\rm and}\quad J\Phi(a)e_1=J\Psi(b)\cdot J\lambda(a)e_1.$$ But, if $$ J\lambda(a)=\left(\begin{array}{cc} \lambda^0{}_0&\lambda^0{}_1\\ \lambda^1{}_0&\lambda^1{}_1\end{array}\right), $$ then \begin{eqnarray*} \partial_0&=&J\Psi(b)(\lambda^0{}_0e_0+\lambda^1{}_0e_1)\\ &=&\lambda^0{}_0J\Psi(b)e_0+\lambda^1{}_0J\Psi(b)e_1\\ &=&\lambda^0{}_0\ \tilde\partial_0+\lambda^1{}_0\ \tilde\partial_1 \end{eqnarray*} and similarly $$\partial_1=\lambda^0{}_1\ \tilde\partial_0+ \lambda^1{}_1\ \tilde\partial_1.$$ For an arbitrary tangent vector in the first coordinates $\vec v=a^0\partial_0+a^1\partial_1$, the new components that would have, will be $$\vec v=(a^0\lambda^0{}_0+a^1\lambda^0{}_1)\tilde\partial_0+ (a^0\lambda^1{}_0+a^1\lambda^1{}_1)\tilde\partial_1, $$ which correspond with the operation $$ \left(\begin{array}{c}a^0\\a^1\end{array}\right) \to \left(\begin{array}{cc} \lambda^0{}_0&\lambda^0{}_1\\ \lambda^1{}_0&\lambda^1{}_1 \end{array}\right) \left(\begin{array}{c}a^0\\a^1\end{array}\right). $$

Now for your case the surface is $\mathbb R^2$, $\Phi=1\!\!1$ the identity maps and $\Psi$ is $$ \left(\begin{array}{c} r\\\theta\end{array}\right)\mapsto \left(\begin{array}{c}r\cos\theta\\r\sin\theta\end{array}\right), $$ Hence for $\lambda$ we have $$ r=\sqrt{x^2+y^2}\quad {\rm and}\quad \theta=\arctan\frac{y}{x}, $$ and its derivative is $$ \left(\begin{array}{cc}\dfrac{x}{\sqrt{x^2+y^2}} &\dfrac{y}{\sqrt{x^2+y^2}} \\ -\dfrac{y}{\sqrt{x^2+y^2}}& \dfrac{x}{\sqrt{x^2+y^2}}\end{array}\right), $$ which happens to be the inverse of the derivative $J\Psi$ but expressed in orthonormal coordinates.

Then the base change with respect to this is $$e_0=\frac{x}{\sqrt{x^2+y^2}}\ e'_0-\frac{y}{\sqrt{x^2+y^2}}\ e'_1 \quad {\rm and}\quad e_1=\dfrac{y}{\sqrt{x^2+y^2}}\ e'_0+\dfrac{x}{\sqrt{x^2+y^2}}\ e'_1, $$ or $$e_0=\cos\theta\ e'_0-\sin\theta\ e'_1 \quad {\rm and}\quad e_1=\sin\theta\ e'_0+\cos\theta\ e'_1, $$ in polar terms.

Now insert those into your linear combination $\vec v=a^0e_0+a^1e_1$ to get $$\vec v=(a^0\cos\theta+a^1\sin\theta)e'_0 +(-a^0\sin\theta+a^1\cos\theta)e'_1. $$ You could see how the multiplication $$\left(\begin{array}{cc} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{array}\right) \left(\begin{array}{c}a^0\\a^1\end{array}\right), $$ matches how the new components of $\vec v$ are get.

janmarqz
  • 10,538
  • wow! let me add that the vectorizations like the matrix borne arranged data about geometric and analytic processes have an extreme super utility at the time of organizing information and pave the way to a better understanding of how algebra geometry and analysis are intertwined. Thanks! – janmarqz Jan 06 '20 at 01:26
0

There were two key problems here, both "easily" discovered with some insights, advice, and tutelage from @peek-a-boo.

1) I added an accidental inverse in each Jacobian element. I associated the Jacobian matrix elements, $\partial e'_j / \partial e_i$ with (e.g. for $j=0, i=1$) $\partial r / \partial y$, which is wrong. $e'_j \neq x^{\prime j}$, instead, $e'_j = \partial / \partial x^{\prime j}$, and also $x^{\prime 0} = r$ and $x^1 = y$ are coordinates associated with those basis vectors. i.e coordinate - basis-vector combos are $x^i e_i$ or in the prime (polar) basis $x^{\prime i} e'_i$. Thus ${S^i}_j = \partial e'_j / \partial e_i = \partial x^i / \partial x^{\prime j}$.

2) I incorrectly associated the coordinate-induced tangent vectors, e.g., $\partial_{x}$ with the basis vectors, e.g., $\hat{x}$. While this happens to be true in the cartesian coordinates, i.e. $\partial_x = \hat{x}$ and $\partial_y = \hat{y}$, it is not generally true, for example (importantly), $\partial_\theta = r \hat{\theta} \neq \hat{\theta}$.

DilithiumMatrix
  • 796
  • 1
    btw regarding (2), the terminology you're using is incorrect. Note that $\partial_x, \partial_{\theta}, \hat{x}, \hat{r}...$ are all called tangent vectors. $\hat{x}, \hat{r}, \hat{\theta}$ are simply the normalized versions of $\partial_x, \partial_r, \partial_{\theta}$ (it just so happens that the first two are already normalized). Next, it's an abuse of language to call $\hat{x}$ alone a basis vector. A basis by definition is a collection of vectors, for example in this 2D case, ${\partial_x, \partial_y}$ is one basis, called the cartesian basis ${\partial_r, \partial_{\theta}}$ is – peek-a-boo Jan 05 '20 at 16:58
  • 1
    another basis, called the polar coordinates-induced basis. These two bases happen to be orthogonal, in the sense that given the "standard" metric, we have $g(\partial_r, \partial_{\theta}) = g(\partial_x, \partial_y) = 0$. Another example of a basis is $B = {\partial_x, \partial_r}$ or even $B' = {\partial_y, \partial_{\theta}}$ (basically pick any two distinct vectors from $\partial_x \partial_y, \partial_r, \partial_{\theta}$). However, $B$ and $B'$ are not an orthogonal basis. (recall that a basis of a vector space is a linearly independent set which spans the whole space) – peek-a-boo Jan 05 '20 at 17:02
  • 1
    Given any coordinate system, we can construct tangent vectors (by taking the velocity vectors to curves as I desribed briefly in my comments above). We call these the coordinate-induced tangent vectors. For example, we call $\partial_x, \partial_y$ the cartesian-coordinate induced tangent vectors, and $\partial_r, \partial_{\theta}$ the polar coordinate-induced tangent vectors). If we want, we could introduce parabolic coordinates $\xi(\cdot), \eta(\cdot)$ and talk about parabolic coordinates-induced tangent vectors $\partial_{\xi}, \partial_{\eta}$ (all the vectors are taken at ONE point p). – peek-a-boo Jan 05 '20 at 17:07
  • 1
    So, a more correct way of saying (2) is that you incorrectly assumed that the coordinate-induced tangent vectors are actually unit vectors. – peek-a-boo Jan 05 '20 at 17:08
  • @peek-a-boo (minor point): so even though ${\hat{x}, \hat{y}}$ are basis vectors, it's inappropriate to call $\hat{x}$ a basis vector? – DilithiumMatrix Jan 05 '20 at 17:28
  • 1
    From a super formal persepective I think it is incorrect...but now I'm picking at straws. Almost everyone (even an extremely pedantic individual like me) refers to $\hat{x}$ as a basis vector... BUT the reason is because in this context, one often means $B = {\hat{x}, \hat{y}}$ is the basis in question, and for convenience/ease-of-language, you simply refer to $\hat{x}$ as a basis vector (which is a part of the basis $B$). So, as long as you know the reason why you're calling a certain object by a certain name, then you can say whatever you want. – peek-a-boo Jan 05 '20 at 17:35
  • 1
    If in the entire discussion there is only one basis $B$ involved, then of course it is perfectly meaningful and unambiguous to refer to elements of $B$ as basis vectors. But in our discussion, we have had several bases $B, B'$ (and others which I didn't give names) So, it is always more proper (to avoid any confusion) to say "... is an element of the basis ..." or "... is a basis vector belonging to the basis ...". At the end of the day, math is about ideas, and you need to communicate them with others. People often speak informally because they know what the other intends to say – peek-a-boo Jan 05 '20 at 17:38
  • 1
    but if needed, a mathematician will always be able to tone the level of precision/detail up or down as required so as to have an unambiguous communication of ideas with the intended audience (and of course, before you can communicate ideas with someone else, those ideas need to be clear to you)... so I just try to point out some of these more subtle points regarding language/terminology. – peek-a-boo Jan 05 '20 at 17:41