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.

  • $\begingroup$ By way of clarifying what's troibling you: Euclid routinely talks about points in the plane without invoking coordinates. Do you find that baffling? $\endgroup$ – WillO Jul 4 at 3:47
  • $\begingroup$ Aha! I see that @G.Smith beat me to it on the invocation of Euclid. $\endgroup$ – WillO Jul 4 at 3:49
  • $\begingroup$ No, coordinate systems are immaterial to physical observations. If you have a reference frame where you see different "physics", then something is wrong with the theory (at least all this is valid in inertial frames). $\endgroup$ – user2820579 Jul 4 at 4:56
  • $\begingroup$ By the way, you can look at all the equations of physics are given in covariant form. They don't depend on specific coordinates. E.g. the quintessential $\vec{F} = m\vec{a}$: it never says that you have to use spherical or rectangular coordinates and either solution must give the same physical results. $\endgroup$ – user2820579 Jul 4 at 4:59
  • $\begingroup$ i edited the question because I think you misunderstood me $\endgroup$ – SimoBartz Jul 4 at 12:26

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.


There is nothing a priori about coordinate systems. They have no physical significance and are invented by humans, not by nature.

I recommend that you read Misner, Thorne, and Wheeler’s discussion in Gravitation about how coordinates are like telephone numbers assigned simply to keep track of which events in spacetime are close to which other events. (This was back in the 1970s when two houses with numerically close telephone numbers were geographically close to each other.)

  • $\begingroup$ I agree on that, I mean that when we start a description of a problem we assume the points of the space as the coordinates on some coordinate system. And this the a priori coordinate system because a post is the same as its coordinates for us. is it? $\endgroup$ – SimoBartz Jul 3 at 20:53
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    $\begingroup$ Points, lines, planes, spheres, etc. have a geometric meaning and “existence” completely independent of the coordinates and equations that might get assigned to represent them. Euclid did not need coordinates to reason about Euclidean geometry. Descartes introduced coordinates about 2000 years later! Coordinates can be useful but they are not fundamental. $\endgroup$ – G. Smith Jul 3 at 21:09
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    $\begingroup$ Take the gradient. $\endgroup$ – G. Smith Jul 3 at 21:24
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    $\begingroup$ This discussion is too long. Post a new question. $\endgroup$ – G. Smith Jul 3 at 21:28
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    $\begingroup$ G.Smith we can continue in chat $\endgroup$ – SimoBartz Jul 3 at 21:35

Points and Vectors can be thought of as fundamental objects and operated on without a coordinate system. One only needs a coordinate system to measure things. This approach most commonly focuses on vectors, and represents points as displacements without loss of generality. Accordingly, I'll do the same and focus on Vectors.

There are many properties of vectors which can be described geometrically, rather than with numbers. For example, the dot product $\mathbf{a}\cdot \mathbf{b}$ is defined to be $|\mathbf{a}||\mathbf{b}|sin\theta$, where theta is the angle between them. This is invariant under all coordinate systems you might measure the vector in. You are welcome to think of it as $\mathbf{a}\cdot \mathbf{b} = \sqrt{(a_1 b_1 + a_2 b_2 + \ldots + a_nb_n)}$, which makes sense in a given coordinate system, but you don't have to have the coordinate system.

One property of Vectors is that in an N dimensional space, N linearly independent vectors form a basis. If you have a basis, $\mathbf{b_1} \mathbf{b_2} \ldots \mathbf{b_n}$ you can write any arbitrary vector $v$ as $c_1\mathbf{b_1} + c_2\mathbf{b_2} + \ldots + c_n\mathbf{b_n}$. The set of $c$ values that arise from this is independent of any coordinate system (it is dependent on the particular basis vectors you chose).

Where coordinate systems come into play is when one wishes to write a vector as $<c_1, c_2, \ldots, c_n>$, and wish to define the vector operations as algebraic operations on these components. The difference between this and the basis example above is that, with a coordinate system, we assume some basis is more important than others, and begin to define our vectors with respect to that basis. That's merely a choice.

Even without such a basis, the vector identities hold true. In a Euclidean space, $|\mathbf{b}-\mathbf{a}|+|\mathbf{c}-\mathbf{b}| \ge |\mathbf{c}-\mathbf{b}|$, regardless of whether you describe these vectors in terms of their coordinates or not.

This thinking then extends to curvilinear coordinate systems, which permit fancier cases like polar coordinates where the bases are not always vectors, but curves. This requires a whole host of extra complexity (such as understaning covariant and contravariant bases), but regardless, it all works without specifying a coordinate system!

Myself, I ran into this sort of fun when writing software for a frame conversion utility. Its very hard to develop a notation for a vector which is easily processed by a computer and doesn't rely on a coordinate system. I had to define a "standard" coordinate system for each of my frames (which happened to be a normal cartesian coordinate system), and state that all vectors were rendered into components using that coordinate system for framing operations. In the peer review, it was very hard to get people to distinguish between "ECEF," which is a coordinate system, and "The Earth Fixed frame," which was a frame. The coupling between them was so tight it was hard to see why they had to be separated.

  • $\begingroup$ I edited the question because I think you misunderstood it, I'll be happy to know your opinion about it $\endgroup$ – SimoBartz Jul 4 at 12:26
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    $\begingroup$ @SimoBartz I think what I wrote still applies. For example, you may think that physics requires being able to do $\frac{\partial P}{\partial r}$, but in fact it can be done with other clever tools such as total derivatives which do not require a basis. They act much like a generalized version of your displacement vector. Concepts such as differential forms permit one to do calculus with such derivatives. $\endgroup$ – Cort Ammon Jul 4 at 18:42

The derivative of $P$ may look confusing in the way that you have presented it, but really you are misunderstanding what $P$ is. A point can be described in terms of $(r,\theta)$, just as a point can be described in cartesian coordinates as $(x,y)$ but both of these refer only to the point itself NOT to the function in which it contained.

The point is not a function and therefore there is no such thing as $\frac{\partial P}{\partial r}$ or $\frac{\partial P}{\partial\theta}$.

If you ask how much does the point $P$ change as $r$ changes $\frac{\partial P}{\partial r}$ the answer is it doesn't, points do not change. $\frac{\partial P}{\partial r} = 0$. Similarly $\frac{\partial P}{\partial\theta} = 0$.

What you can ask, however, is how much do the x and y components of the point individually change with respect to a change in $\theta$ or $r$:

We can find: ($\frac{\partial X}{\partial\theta}$, $\frac{\partial Y}{\partial\theta}$) , ($\frac{\partial X}{\partial r}$, $\frac{\partial Y}{\partial r}$)

to do this you need to define your X and Y coordinates of a point as functions of r &/or $\theta$. This can be done fairly simply using the following definitions

cos($\theta$) = $\frac{x}{r}$

sin($\theta$) = $\frac{y}{r}$

r = $\sqrt{x^2 + y^2}$

For example, lets say we have a simple polar function:

$r = cos(\theta)$

a substitution reveals

$\sqrt{x^2 + y^2}= cos(\theta)$

$ x^2 = cos(\theta)^2 - y^2 $

$ x = \sqrt{cos(\theta)^2 - y^2} $

but sin($\theta$) = $\frac{y}{r}$ so $y = rsin(\theta)$

*but for this problem we have defined r in terms of theta, so $y = cos(\theta)sin(\theta)$

Now with another substitution we can define x soley in terms of r and $\theta$

$ x = \sqrt{cos(\theta)^2 - cos(\theta)sin(\theta)} $

From here you could solve for $\frac{\partial X}{\partial\theta}$, or using our initial polar equation you could rewrite in terms of r and solve for $\frac{\partial X}{\partial r}$

The conversion from polar to cartesian is pretty annoying sometimes, but its the only way to do it. And if you wanted to find how fast the point was moving in terms of displacement with respect to $\theta$ you could do a pythagorean sum of the X & Y partial derivatives. THis is as close as youre gonna get to a "$\frac{\partial P}{\partial\theta}$", same goes for $\frac{\partial P}{\partial r}$

$\frac{\partial P}{\partial\theta}$ = $\sqrt{\frac{\partial X}{\partial\theta}^2+ \frac{\partial Y}{\partial\theta}^2}$

  • $\begingroup$ If this is not the issue, you should consider rewriting or reposting your question as the question is not clear in the way it is currently formatted. $\endgroup$ – definitelynotbs Jul 13 at 18:33
  • $\begingroup$ this is exactly the issue, and I'm still confused about it $\endgroup$ – SimoBartz Jul 14 at 8:31
  • $\begingroup$ "The derivative of $P$ may look confusing in the way that you have presented it" so how would you present it? $\endgroup$ – SimoBartz Jul 14 at 8:36
  • $\begingroup$ @SimoBartz I think I figured out what you were asking-- how quickly the point is changing with respect to theta or r? If this is your question I edited my original post and answered it. For some reason your original post confused me a little but now I see I must have been misreading it. But I hope this helps! $\endgroup$ – definitelynotbs Jul 15 at 2:19

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