I'm studying introductory vector calculus and need to confirm/clarify my concepts. The definition of the derivative of a vector (for example in $\mathbb{R}^2$) if the unit vectors are constant throughout the 2D space is in terms of its components: if we have $r(t)=(x(t), y(t))$ in the standard Cartesian basis, then


Now if we move to polar coordinates $\rho, \phi$, then the unit basis vectors $e_{\rho},e_{\phi}$ will change direction depending on the location in 2D space. To define the derivative in this case, the book that I'm studying gives the following quick method: we see that $r=\rho e_{\rho}$ (where $\rho$ is the distance of the vector's endpoint from the origin), which means $$\frac{dr}{dt}=\frac{d\rho}{dt}e_{\rho}+\rho\frac{de_{\rho}}{dt}$$ enter image description here

So far, so good: $\frac{d\rho}{dt}$ can be calculated since we can express $\rho$ in terms of $x(t)$ and $y(t)$, and differentiate that expression w.r.t. $t$. In this specific case, we can also express $e_{\rho}=(\cos\phi)e_x + (\sin\phi)e_y$. It turns out that $$\frac{de_{\rho}}{dt}=\frac{d\phi}{dt}e_{\phi}$$ because of the specific way $e_{\rho}$ and $e_{\phi}$ are defined in terms of $e_x$ and $e_y$.

Expressing the same vector $r$ in a general curvilinear coordinate system $u,v$, enter image description here

To even start differentiating $r$, we need to find the components of $r$ in the new system. I'm assuming the way to identify $r$ is to identify it as the intersection of two coordinate curves $u=c_1$ and $v=c_2$ - in this case, $u=5$ and $v=4$. Is my understanding correct? Is this the way to identify the components of a vector in a curvilinear system?

So if we have some differentiable functions $f,g$ such that $u=f(x,y)$ and $v=g(x,y)$ and $r=ue_u+ve_v$, then $$\frac{dr}{dt}=\frac{du}{dt}e_u+u\frac{de_u}{dt}+\frac{dv}{dt}e_v+v\frac{de_v}{dt}$$

$\frac{du}{dt}$ can be identified as $\frac{df(x(t),y(t))}{dt}$ and can be evaluated. How does one, in general, express basis vectors $e_u$ and $e_v$ in terms of $e_x$ and $e_y$? And even if we do manage to define curvilinear basis vectors in terms of $e_x,e_y$, it's not necessary that we'll get a nice expression for $\frac{de_u}{dt}$ and $\frac{de_v}{dt}$ in terms of $e_u$ and $e_v$. How do we get the curvilinear components of $\frac{dr}{dt}$ in that case?


It's not clear what you mean by "differentiation" in curvilinear coordinates. But you can set all this up with parameterizations. If you want to describe a point in $\mathbb R^2$, "the target", you can do it by using the natural coordinates, mapping them via functions from $\mathbb R^2$, "the source". That is, define

$\phi:\mathbb R^2\to \mathbb R^2$ by $\phi(x,y)=(\phi_1(x,y),\phi_2(x,y)).$

For example, if we take $\phi_1(x,y)=\sqrt{x^2+y^2}$ and $\phi_2(x,y)=\tan^{-1}\left(\frac{y}{x}\right)$, we get polar coordinates. Of course, we have to be careful: if $x=0$, this parameterization won't work, but if we use another branch of arctan, we will get a different parameterization, one that works.

As another example, take $\phi(x,y)=(x,y),$ the identity. This will give us the usual coordinates on $\mathbb R^2.$

Notice that in the second example, there is no problem if we move the points around; the same $\phi$ will work for all of them. Whereas, in the first case, there may be some "bad points", for which $\phi$ won't work.

Now let $(u,v):=(\phi_1(x,y),\phi_2(x,y))$ for conveience, and notice that if we fix a point $(u_0,v_0)$, and if we move out from $(u_0,v_0)$ to some nearby $(u,v)$, the points will generally traverse a complicated path. So it makes sense to affix a $\textit{whole coordinate system at}\ (u_0,v_0)$, and then as the point moves, we will get coordinate systems that move along with it as we go from $(u_0,v_0)$ to $(u,v)$ along the path specified by $\phi.$ You can visualize this process by thinking of pasting tangents at each point on the curvy lines you drew above. These are the (local) coordinate systems.

Notice also that if $\phi(x,y)=(x,y)$, the identity, the coordinate systems all look the same as we go from $(x_0,y_0)$ to $(x,y)$. At some intermediate point $(x_1,y_1)$ the $x$ component "points in the $x$ direction" and the $y$ component "points in the $y$ direction".

But when $\phi$ is more complicated, we see that the tangents are changing as we move along, so we need to know how to calculate them, because it is the tangents that will give us our (local) coordinate systems. But this is easy! We already know how to find tangents for vector-valued functions that map open subsets of $\mathbb R$ into $\mathbb R^2.$

More precisely, if we fix $v_0$ and let $u_0$ move to a nearby value $u_0+\Delta u$, we get a $\textit{curve},\ \vec r$ in $\mathbb R^2, $ as $(u,v_0)$ traces out a path in $\mathbb R^2$ and its tangent at $(u_0,v_0)$ is $\frac{\partial \vec r}{\partial u}$, evaluated at $(u_0,v_0).$ Now, define $\vec e_1:=\frac{\partial \vec r}{\partial u}/\left|\frac{\partial \vec r}{\partial u}\right|$, where the denominator is just the scaling factor that normalizes $\frac{\partial \vec r}{\partial u}$.

Similarly, we define $\vec e_2:=\frac{\partial \vec r}{\partial v}/\left|\frac{\partial \vec r}{\partial v}\right|.$

Now that we have the machinery in place, we can do an example:

$u=\sqrt{x^2+y^2}:=r$ and $v=\tan^{-1}\left(\frac{y}{x}\right)$. Of course, we usually write $u=r$ and $v=\theta.$ Then, $\vec r(r,\theta)=(r\cos\theta,r\sin\theta)$ and so

$\frac{\partial \vec r}{\partial r}=(\cos\theta,\sin\theta)$ and $\frac{\partial \vec r}{\partial \theta}=(-r\sin\theta,r\cos\theta).$ Therefore,

$\vec e_1=(\cos\theta,\sin\theta)/1=(\cos\theta,\sin\theta)$ and $\vec e_2=\frac{1}{r}(-r\sin\theta,r\cos\theta).$

Now that we have the basis vectors (in the tangent space), as functions of $r$ and $\theta$, we can differentiate them in the usual way.

If you repeat this calculation for $\phi(x,y)=(x,y)$, you will get $\vec e_1=(1,0)$ and $\vec e_2=(0,1)$, and obtain the familiar result that, in standard coordinates, the tangent spaces at each point in $\mathbb R^2$ all point in the same direction, which is why we get away with treating them as constants when we differentiate vectors, expressed in that coordinate system.

edit: you mentioned covariant and contravariant components in the comments. This picture may help to see what is going on

enter image description here

The point $p=(u_0,v_0)$ is fixed in the lower left-hand corner. $e_1$ and $e_2$ are the tangent vectors to the curves $\vec r(u,v_0)$ and $\vec r(u_0,v)$ at $p$.

The covariant components are the orthogonal projections onto the coordinate axes defined by $e_1$ and $e_2$ while the contravariant components are the coordinates of $M$ in the $\{e_1,e_2\}$ basis at $p$.

Everything is going on at $p$. If we change it, then presumably, the $x_i,\ x^i$ and $e_i$ change as well. And the calculation we did above tells us exactly how they change as $p$ changes.

  • $\begingroup$ Perfect answer! One thing then that I'd still like to confirm: from what you said, the basis vectors in curvilinear coordinate systems are defined locally, so it makes sense to express the vector differential $d\mathbf{r}$ in terms of the local basis vectors $\mathbf{e}_u,\mathbf{e}_v$. By "local" I mean the vectors defined in the tangent space of the origin point of $d\mathbf{r}$ [cont'd]... $\endgroup$ – Shirish Kulhari Dec 27 '19 at 16:09
  • $\begingroup$ [cont'd]... Does that mean it's not possible to express the non-differential vector $\mathbf{r}$ (as you see in the 2nd figure of the question) in terms of basis vectors in curvilinear coordinates? i.e. Does that mean the expression $\mathbf{r}=u\mathbf{e}_u+v\mathbf{e}_v$ makes no sense? $\endgroup$ – Shirish Kulhari Dec 27 '19 at 16:09
  • $\begingroup$ $\bf r$ is OK. It's a function that maps from $\mathbb R^2$ to $\mathbb R^2$, if I am reading it correctly. The point is that in curvilinear coordinates, we use the $tangents$ to measure what happens as the point moves along. This amounts to placing coordinate systems defined by the partial derivatives, at each point and seeing how they "move" as we move the point. I look at it like changing the camera angle as the point moves along. This does not happen in standard coordinates because the "camera angle" does not change. Mathematically: the position and tangent vectors are the same. $\endgroup$ – Matematleta Dec 27 '19 at 16:27
  • $\begingroup$ I see, but what I meant was: suppose you just have a vector in curvilinear coordinates (ref. i.stack.imgur.com/wHM0D.png ). I can express this nicely in canonical coordinates as, say $\mathbf{a}=a_x\hat i + a_y\hat j$. What I'm struggling with is making the transition to curvilinear. How can we write $\mathbf{a}$ in this same form for curvilinear? Can we even write $\mathbf{a} = a_u\mathbf{e}_u+a_v\mathbf{e}_v$ for some $a_u$ and $a_v$? And if so, what's the "recipe" for finding their values? Your answer gives me the big picture, but could I have concrete details? $\endgroup$ – Shirish Kulhari Dec 27 '19 at 16:49
  • $\begingroup$ Yes. In a nbhd of $(u_0,v_0)$ you have $\vec r(u,v)=a(u,v)\vec e_1(u,v)+b(u,v)\vec e_2(u,v)$. I do not use the subscripts $u$ and $v$ so as not to confuse them with partial derivatives. What's going on here is that $\phi$ is locally invertible, so you use the inverse function theorem on it. Think of two copies of $\mathbb R^2$ and a diffeomorphism $\phi$ that maps a small open set $(x_0,x_1)\in U\subseteq \mathbb R^2$ to a small open set $(u_0,v_0)\in V\subseteq \mathbb R^2$. $\endgroup$ – Matematleta Dec 27 '19 at 17:09

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