newtonian mechanics - Understanding the definition of tangent basis

This question could sound silly but I though a lot about it and I'm not new to physics.

Let's say I have a plane on which I use polar coordinates, it means a point $P$ can be indicated by its coordinates $(r, \theta)$. Then we need a basis in order to write the vectors as tuples of numbers, the tangent basis for this coordinate system is: $(\frac {\partial P}{\partial r},\frac {\partial P}{\partial \theta})$.

What is a derivative of $P$? I know $P$ is a point of the plane that is represented by its coordinates $(r, \theta)$. I don't have a mathematical form of $P$ with a dependence on $r$ and $\theta$ that I can differentiate.

Answer

The basis you're looking for is not $({\partial P\over\partial r},{\partial P\over\partial \theta})$; it is $({\partial\over\partial r},{\partial\over\partial\theta})$.

Tangent vectors specify directions in which you can take derivatives, so you can identify a tangent vector with the operator that takes a derivative in that direction. For the tangent vector ${\partial/\partial r}$, the operator can be described roughly as "take the directional derivative in the $r$ direction", or slightly less roughly as "take the derivative in the one and only direction in which the derivative of $r$ is $1$ and the derivative of $\theta$ is $0$". Similarly (with $r$ and $\theta$ reversed) for $\partial/\partial\theta$.

When we apply $\partial/\partial r$ (or $\partial /\partial\theta$) to a function $f$, we call the result $\partial f/\partial r$ (or $\partial f/\partial\theta)$.

The above is the main idea; what follows is a little more involved and might or might not be more than you want right now. Maybe you'll want to come back and reread it from time to time.

I. A tangent vector $T$ at $P$ is (by definition!) an operator that takes differentiable functions defined near $P$ and turns them into scalars. It is required to satisfy several conditions:

First, it should be linear, so $T(f+g)=Tf+Tg$ and $T(\alpha f)=\alpha Tf$ (where $f$ and $g$ are any functions and $\alpha$ is any scalar).

Next, if $f$ and $g$ agree in a neighborhood of $P$, then $T(f)$ should equal $T(g)$.

Next, if $f$ is any constant function, then $T(f)$ should be zero.

Next, if $f$ is a product of two differentiable functions that both vanish at $P$, then $T(f)$ should be zero.

II. Start with any coordinate system defined near $P$ --- say $(x,y)$. Then it is possible to prove that there is exactly one tangent vector $T$ such that $T(x)=1$ and $T(y)=0$. We call that tangent vector ${\partial\over\partial x}$. Likewise there's just one tangent vector $U$ such that $U(y)=1$ and $U(x)=0$. We call that tangent vector ${\partial\over\partial y}$.

Or start with a different coordinate system, like $(r,\theta)$. Look for the one and only tangent vector that takes $r$ to $1$ and $\theta$ to $0$. That tangent vector is called $\partial\over\partial r$. The one and only tangent vector that takes $\theta$ to $1$ and $r$ to $0$ is called ${\partial\over\partial \theta}$.

(Dangerous Curve: The coordinate $r$ can be part of more than one coordinate system. The tangent vector $\partial/\partial r$ will be different depending on what coordinate system you're starting with. So if your coordinate system is $(r,\theta)$, then ${\partial/\partial r}$ is a tangent vector that takes $\theta$ to zero; if your coordinate system is $(r,y)$ then ${\partial/\partial r}$ is a tangent vector that takes $y$ to zero, and despite having the same name, these are not the same tangent vector!)

Of course you probably want to think about tangent vectors geometrically, which is fine, but there's a one-to-one correspondence between your geometric picture of a tangent vector and the algebraic definition of a tangent vector as an operator --- and it pays to learn to go back and forth between the two.