# Standard basis vectors in polar coordinates?

• Aug 17th 2012, 05:25 PM
gralla55
Standard basis vectors in polar coordinates?
I always thought that a set of "standard basis vectors", meant a set of orthonormal vectors such that a given point in the coordinate system is just a linear combination of those vectors. In Cartesian coordinates, the point (2,1,1) for instance, is just the linear combination 2i + 1j + 1k.

However, my vector calculus book defines the standard basis coordinates for polar coordinates as the set of unit vectors where one vector goes in the r-direction and the other in the theta-direction. I understand that this set will be an orthonormal basis, but is it correct to label them as "standard basis vectors" ? The polar coordinates of the given point, does NOT equal the corresponding linear combination of the alleged "standard basis vectors". In fact, it will just equal a scaled version of the vector in the r-direction.

Any help clarifying this would be greatly appreciated! =))
• Aug 17th 2012, 06:33 PM
GJA
Re: Standard basis vectors in polar coordinates?
Hi, gralla55.

I don't know if I understand things perfectly, but I will respond to what I think you mean; you can judge for yourself if what I've said is of any use or not.

First, there might be some confusion because the point (2,1,1) is a point in 3-dimensional space and polar coordinates are used to describe points in 2-dimensional space (i.e. the plane). If we wanted to use something besides Cartesian coordinates to describe the point (2,1,1), we could use something like cylindrical or spherical coordinates.

Next, when giving the coordinates of a point in two different coordinate systems we will (generally) not have the same numbers for the individual coordinates, but both sets of coordinates will give the location of the same point.

I think an example would do us some good.

Say we are considering the point (0,1) in Cartesian coordinates in the plane (this is the point that sits one unit above the origin). The polar coordinates of this point would be $(1,\pi/2)$ (let me know if it's unclear why this is so). We notice that neither coordinates "match up" (so to speak), but both describe the same point.

Perhaps what you're after is the conversion that allows us to convert a point from polar coordinates to Cartesian coordinates. These are:

$x=r\cos\theta$

$y=r\sin\theta.$

Notice that if we plug $r=1$ and $\theta=\pi/2$ in the above equations we end up with our original Cartesian coordinates (0,1).

Not sure if this is what you're after, but I thought I'd take a stab in the dark. If there's still some confusion, let me know and we can revist the topic :)

Good luck!
• Aug 17th 2012, 08:15 PM
Deveno
Re: Standard basis vectors in polar coordinates?
let's take a slighty less trivial example:

consider the vector i + j in 2-dimensional space. often this vector is represented by its coordinates as: (1,1).

in this convention: we identify i with the point (1,0) (the unit vector in the i direction, the direction of the x-axis) and j with the point (0,1), so that:

i + j = (1,0) + (0,1) = (1,1).

of course, we can also represent this as a pair (r,θ), where:

r = √(x2+y2)

θ = arctan(y/x) *

* with some exceptions and fiddling to get θ in the range we want: usually either 0 ≤ θ < 2π, or -π < θ ≤ π. usually you have to define quadrant by quadrant, and make special rules for points on the y-axis.

it would be nice if we could represent this pair as a linear combination of some unit vectors r and θ.

well, the direction a vector is pointing in, depends on the angle in a critical way. so the "unit radial vector" is going to depend on θ.

we DEFINE: r = cos(θ)i + sin(θ)j.

and θ = -sin(θ)i + cos(θ)j. this second vector is called "the unit transverse vector".

but i promised an example. so back to our point (1,1). to express in polar coordinates, we need to find r and θ.

r = √(12+12) = √(1+1) = √2

θ = arctan(y/x) = arctan(1/1) = arctan(1) = π/4 (no "fiddling" is required in the first quadrant).

so in "polar coordinates" this is (√2,π/4).

now let's determine our two "polar basis vectors":

r = cos(π/4)i + sin(π/4)j = (√2/2)(i + j). note this points in the direction of i+j, but is "scaled down to unit length".

θ = -sin(π/4)i + cos(π/4)j = (√2/2)(j - i).

what do we see?

i+j = rr. what happened to θ?

to understand more fully what is going on here, imagine something travelling around a central point of a circle, in a plane. we could fix the origin of our plane at the travelling object, and impose two fixed perpendicular axes, and resolve the "horizontal component" (as something times i) and the "vertical component" (as something times j) of the objects's velocity. this is the "rectangular view". if we picked "some other origin", we just "translate" by the vector difference.

but, we could alternatively fix our origin as the center of the circle the object is orbiting. then r is "the unit vector in the direction the object is moving", and θ is "the unit vector of the angle the center is AT". in other words r and θ are useful in describing circular MOTION (or angular forces).

let us imagine that we have the vector with its "tail" at (√2/2,-√2/2) and its "head" at (1 + √2/2, 1 - √2/2). this is just the same "vector" as (1,1) but we've shifted the origin from (0,0) to (√2/2,-√2/2). now r reflects that our vector has length √2 in the direction of i+j, and θ reflects that (0,0) is at angle 3π/4 (counter-clockwise) away from (√2/2,-√2/2).

*******

so what is going on here? we are used to thinking of coordinates in space as "absolute", the origin is fixed, and the relationships are rigid. but in fact, the choice of a coordinate system is somewhat arbitrary, the space itself doesn't care. we PICK a point to be the origin, and base our calculations on that.

and sometimes, the value of a vector-valued function, might depend on "where you are". for example: you're on a mountain, and the way a marble will roll depends on if you're in a valley, or on one side of the peak. well, for "each point" of a vector space, we can create a COPY of that vector space that is in every respect like the original, except we have "moved the origin to where we are". such a copy is called a "tangent space", and although the RULES for calculating in a tangent space are exactly the same as in our original space, the "coordinates" (the actual numbers) are different.

polar coordinate systems take advantage of a situation where there is radial symmetry...like when we are travelling in a circle. if we aren't travelling in a circle...they are inconvenient, and often confusing. rectangular coordinates are (in a sense), more "uniform", there's no special point to pick as the origin that makes anything any easier, or harder (we add x's to x's, y's to y's, no matter where we are).

perhaps, if you noticed above, it turns out that θ = d(r)/dθ. if we think of r as our "radial" velocity, this means that an object in a circular orbit around a central point is "accelerating towards the center" (this is called "centripetal force"). this corresponds with our usual notion of thinking of "acceleration" as the derivative of "velocity".
• Aug 18th 2012, 03:15 PM
gralla55
Re: Standard basis vectors in polar coordinates?
Thank you for your very detailed reply! I now understand the usefulness of such vectors, but I'm still not sure if I get why they are labeled "standard basis" vectors. As you say, the coordinates of polar coordinates does NOT equal the corresponding coefficients of these vectors. Rather, you use them to measure relative velocity from your starting point. I always thought the standard basis for polar and spherical coordinate systems would include some weird-looking bent vectors or something, haha. Perhaps I've misunderstood the definition of a "standard basis"? Thank you so much again!
• Aug 18th 2012, 05:08 PM
Deveno
Re: Standard basis vectors in polar coordinates?
well, one can think of a basis in two different, but related ways:

1) a linearly independent set that gives coordinates meaning

2) "axes" for our space....a frame of reference

the first view sees space as "static", and if we use rectangular coordinates (1) and (2) work out to be the same thing. the second view is "the view from the moving object", often θ is a function of time. in 3-dimensions you get: the tangent vector t (to the path you're on, corresponding to the derivative of position, or velocity), the normal vector n (this is the transverse vector in 2 dimensions, corresponding to the acceleration, and is the derivative of t with respect to the parameter (usually time)), and the "bi-normal" or "outward-normal" b to the osculating plane defined by t and n, which is the cross-product of t and n.

this is a "moving coordinate system" its origin is always at the "moving point" (object/particle/etc.). see here:

Frenet

the "polar basis coordinate system" is not quite the same thing as "polar coordinates", and to make matters worse, the same "variable" names are used. r, θ act just like i and j, just "moved to where the action is". they're clearly *related* to polar coordinates, because we need to know the polar coordinates to actually calculate them. they're tools for looking at "curved" motion, and most of linear algebra is concerned with "straight stuff".

but disregarding how r and θ are defined, the important fact is: they ARE both unit vectors, and they ARE perpendicular. so they form an orthonormal basis. you can think of r as a "local x-axis", and θ as a "local y-axis", that varies over time (as the angle changes).

an analogy: if you wanted to describe vectors in our solar system, you'd probably choose the sun as the origin, and pick two arbitrary axes to represent "the plane of planetary orbit". but if you want to launch a rocket from earth, we're "the fixed origin" and our coordinate system changes (relative to the other one) as we orbit the sun.