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I thought that basis vectors were of magnitude one and located at the origin and were each linearly independent, so how in things like polar coordinates can the basis vectors be moving?

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    $\begingroup$ Vectors are not located anywhere, for starters. $\endgroup$ Oct 9, 2021 at 23:10
  • $\begingroup$ I thought the unit vectors had magnitude one and were located a the origin of the reference point? $\endgroup$
    – CatsOnAir
    Oct 9, 2021 at 23:17
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    $\begingroup$ @CatsOnAir No. If I say go "west", does "west" have an origin point? You have an origin point, yes, but west is just west and that doesn't change just because your location changes. $\endgroup$
    – DKNguyen
    Oct 10, 2021 at 0:35
  • $\begingroup$ Are you interested in manifolds and tangent spaces, or is your concern more about "position vectors" used to describe a point in space? $\endgroup$
    – JEB
    Oct 10, 2021 at 2:08
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    $\begingroup$ But "go 1 kilometer west" is the vector here, and it applies equally well to you as to me, even though we are not in the same place. $\endgroup$ Oct 10, 2021 at 16:11

7 Answers 7

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The only requirements for basis vectors is that they must be linearly independent and they must span the vector space. There is no requirement that they be normalized nor that they be orthogonal to each other.

Now, when you speak of basis vectors moving in polar coordinates what you mean is that you have some manifold with a polar coordinate chart on it. That manifold is not a vector space, but at each point in the manifold we can construct a vector space called the tangent space. The tangent space at each point is a separate vector space from the tangent space at any other point.

In each tangent space, we can construct a set of basis vectors from the coordinates called a holonomic basis. These vectors are linearly independent and span the tangent space at the given point in the manifold. So in that sense they are each at the origin of a different vector space.

The idea of the basis vectors moving is based on a connection between neighboring tangent spaces that allows us to transport a vector from one tangent space to a neighboring space. In this sense, the holonomic basis vector associated with a given coordinate at one point in the manifold is not the same as the corresponding basis vector from another point transported to the first point.

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    $\begingroup$ While there is no requirement that basis vectors be normalized or orthogonal, calculations are often simplified if they are. So it is common to chose basis vectors that are. $\endgroup$
    – mmesser314
    Oct 10, 2021 at 4:45
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The short answer is: Because there is a separate vector space with a separate basis at every point of space and you can choose a basis for each of theses vector spaces independently.

To understand this properly you need the theory of manifolds (as explained in Dale's answer).

It is just an "accident" of Euclidean space that vectors don't change when you transport them around. We can identify all the vector spaces at all points and choose a single basis for them in Cartesian coordinates because of this accident.

Acutally, you can also work in polar coordinates for the position, and still use the global Cartesian basis vectors to express your velocity/electric field/.... It is just very convenient to match the basis vectors to your coordinates, as it simplifies calculations.

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The situation that you describe is a common one and relies on there being just one vector space. Moreover, here the basis vectors are normalised so they have unit magnitude. However, in general a basis vector belongs to a basis and this is a collection of linearly independent and spanning set. These vectors need not be normalised and so may not have unit magnitude.

There need not be just one vector field. More precisely, over a manifold, at each point there is a tangent space and bundling them all together gives the tangent bundle over that manifold. We can pick a basis of tangent vectors at each point. And hence we get a field of tangent vectors, or a tangent field for short, over the manifold.

This picture is static. However, by choosing a curve on the manifold and sending a particle moving along this curve we see we have a moving choice of basis along the curve. This is the basis of Cartan's method of the moving frame where frame is just a synonym for basis.

As an example: take a plane, this is a manifold. One curve we can take is the circle. And a common choice of tangent vectors in the plane which is on the circle is the obvious one, the unit radial vectors and the unit tangent vectors along the circle. We can call this the polar basis. Then sending a particle along the circle gives the moving polar basis that you mention.

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When vector are present in cartesian coordinates, the basis vectors look useless. After all, a vector (4,2) = 4(1,0) + 2(0,1) doesn't add new information.

In polar coordinates, each point has a set of basis vectors, and any vector acting in the point is a linear combination of them.

It is easy to understand for a point in the earth surface. It is defined by 2 coordinates: longitude and latitude. For each pair of that coordinates, we can set 2 vectors: One of them pointing North and other East for example. Now it is possible to define a velocity vector with a magnitude and a direction, based on these basis vectors for a neighborhood of that location.

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enter image description here

you know the components ($~v_x~,v_y~$) of a vector at point 1

$$\vec v=\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}=v_x\,\underbrace{\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}}_{\vec e_x}+v_y\,\underbrace{\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}}_{\vec e_y}$$

where $~\vec e_x~,\vec e_y~$ are the basis vectors, they are linear independent with $~\vec e_x\cdot e_x=1~,~\vec e_y\cdot e_y=1~,\vec e_x\perp\vec e_y~$

note that the components are $$v_x=\vec v\,\cdot \vec e_x\\ v_y=\vec v\,\cdot \vec e_y$$

and the magnitude $~v=\sqrt{v_x^2+v_y^2}$

If you move the vector $~\vec v~$ to point 2 and you choose new basis vectors for convenience $~\vec e_t~,\vec e_n~$ hence

$$\vec v=\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}=v_t\,\vec e_t+v_n\,\vec e_n$$

the basis vectors fulfilled the requirement form $~\vec e_x~$ and $~\vec e_y~$ basis vectors.

the new components are

$$v_t=\vec v\,\cdot \vec e_t\\ v_n=\vec v\,\cdot \vec e_n$$

and the magnitude is unchanged $~v=\sqrt{v_x^2+v_y^2}$

at point 3 you obtain another components of the vector $\vec v~$ , again the magnitude is unchanged.

so nothing is wrong with this calculation ,choosing the basis vectors is only depending on what you want to achieve

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Part 1 - Addressing your confusion about basis vectors

I thought that basis vectors were of magnitude one

Consider what it means to express the same point two different bases.

Suppose you have the canonical (standard) basis $B = \{\hat I, \hat J\}$. Take some arbitrary pair of space-spanning (so, linearly independent) vectors $\vec i$ and $\vec j$ (not necessarily orthogonal, and not necessarily of unit length) and have them form a different basis $B' = \{\vec i, \vec j\}$.


Aside: Commonly (in a slight abuse of notation), you'd write an ordered basis as $B = \{e_1, e_2\}$, with the subscripts defining the ordering (so that you know which coordinate is associated with which basis vector). Since I'm using $I, J$ and $i, j$, there are no subscripts, but the usual ordering is assumed (dictionary ordering, $i \to j$).


Now, suppose you have some vector $\vec r$. Expressed in the standard basis, it's a linear combination of the basis vectors in $B$:

$$\vec{r} = x\hat I + y\hat J\space, \space\text{coordinates:}\space(x, y)$$

and in terms of the primed basis, the same vector can be expressed as a linear combination of $\vec i$ & $\vec j$, it's just that the coordinates will be different. So, in terms of $B'$, it's:

$$\vec{r} = u\vec i + v\vec j\space, \space\text{coordinates:}\space(u, v)$$

It's basically saying "take $u$ steps in the $\vec i$ direction, and take $v$ steps in the $\vec j$ direction". Note that this means that the $(u, v)$ coordinates of $\vec i$ and $\vec j$ in the primed basis are $(1, 0)$ and $(0, 1)$, respectively. So, expressed in their own basis, $\vec i$ & $\vec j$ are both of unit length, and orthogonal. Basically, working in terms of (u, v) coordinates is the same as working in the standard basis, it's just that there's this larger context, a coordinate-independent space of points, and the concrete point associated to each coordinate pair is different depending on which basis you choose.

enter image description here

The neat thing is that this lets you convert between the two. Suppose that

$$\vec{i} = (2, 0.5)_B = 2\hat I + 0.5\hat J \\ \vec{j} = (0.5, 1)_B = 0.5\hat I + 1\hat J $$

(the subscript $(x, y)_B$ indicating the coordinate frame), and that you have $\vec r = (u, v)_{B\space '} = (1, 2)_{B\space '}$, and you want to know the corresponding $(x, y)_B$ representation. Well, just scale $\vec i$ & $\vec j$ by $u$ & $v$, respectively:

$$\vec{r} = 1\vec i + 2\vec j \\ = 1(2\hat I + 0.5\hat J) + 2(0.5\hat I + 1\hat J) \\ = 3\hat I + 2.5\hat J $$

So you get that the corresponding $(x, y)_B$ is $(3, 2.5)_B$.

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In matrix form:

$$ \begin{bmatrix} 3 \\ 2.5 \\ \end{bmatrix} = \begin{bmatrix} 2 & 0.5 \\ 0.5 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} $$

Part 2 - Addressing your confusion about local frames

I thought that basis vectors were [...] located at the origin [...], so how in things like polar coordinates can the basis vectors be moving?

Mathematically, vectors don't have distinct origin points - they are just a direction and a magnitude (geometric vectors, anyway). You can visualize this by thinking of all vectors as emanating from the origin of a Euclidean coordinate system; the endpoint then corresponds to the coordinate representation in the standard basis.

In other words, even things like velocities, accelerations, forces - all these by themselves don't have an intrinsic origin point. Each is, really, just a direction and a magnitude. This makes the math work out. But in physics, you often want to attach an extra bit of "metadata" to such vectors, to better describe the bigger picture (beyond just the "mathematical tooling" of linear algebra), so you often translate the vector in space so that it starts at some sensible-looking point on an object.

Or, think about force fields; to each point in space, a vector (force) is attached. The vectors are often represented as emanating from the associated point. But, another way to think about this is that you have a function that takes a point in space, and produces a vector. So, imagine some application that provides UI that lets you select a point as input, and that has a separate window to show the output vector emanating from the origin; as you move the input point around, the output vector changes magnitude and direction. You can get back to the usual picture by mentally translating the vector to the input point.

You can do the same thing with coordinate frames. A problem may be difficult to solve or express in the standard basis, and easier to deal with in some other (primed) basis. The new basis vectors do emanate from the origin, but the primed basis itself is often associated with some point in space - perhaps it's only useful at the immediate vicinity of that point (good enough for calculus!).

So you can think of it as of a local frame associated with that point.

As you change the "input" point, the associated basis changes - the basis vectors change direction and magnitude. In essence, you have a different local basis for each point in space. But because directions and magnitudes vary smoothly, you can think of the process as of a local basis "moving", "rotating" and more generally, "changing shape".

enter image description here

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  • $\begingroup$ so if I'm understanding correctly every point in space has basis vectors to describe any vector at that point ( so if there is a force on object B and object B is at some point the force has a basis from that point that describe its direction and magnitude. So in physics when we are describing a position from a point that's why the basis vectors seem to be only at the origin, because in the case of position we are using magnitude to represent distance when in fact we could choose any point to base our position on and the basis vectors would move with that choice. (pt one) $\endgroup$
    – CatsOnAir
    Oct 11, 2021 at 20:38
  • $\begingroup$ However in polar coordinates we make our set our bases vectors around a point lets say P that is moving with respect to the Cartesian coordinates but not our new bases (kind of like how you said its a different frame of reference) and then we can do a change of biases to get the velocity vector in terms of polar coordinates into a velocity in terms of our Cartesian cords. (pt two) $\endgroup$
    – CatsOnAir
    Oct 11, 2021 at 20:38
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"Coordinates" is a more general term that "basis vectors". Basis vectors apply only to vector spaces, while coordinates apply to any manifold (and, if one uses the term loosely enough, pretty much any space). Basis vectors provide a coordinate system by simply taking their coefficients, but not all coordinate systems correspond to a set of basis vectors. Also, while basis vectors do have to be independent, they do not have to have magnitude $1$, and in fact vector spaces do not have to have any norm at all to have basis vectors, and if a space doesn't have a norm, then "magnitude $1$" is meaningless. Additionally, vectors aren't really "located" at the origin. The are often represented as displacements from the origin to another point in affine space, but that's not quite the same thing as being "at" the origin.

Polar coordinates are not based on basis vectors, although we can use them to define local basis vectors for each point.

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