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A coordinate system is often thought of as labelling points in space as if a perpendicular were to be dropped from that point to each of the coordinate axes, and the values on each of the axes read of and recorded in a tuple $(x,y,z)$. Consequently, the coordinates are interpreted as signed perpendicular distances to the coordinate axes.

However, I haven't been able to find any references which make clear the connection between vector spaces and coordinate systems. Given a vector space $V$ equipped with a set of basis vectors, I'm inclined to say that a coordinate system is imposed onto the vector space such that the axes are in the direction of the basis vectors, and that the coordinates are the scalar components of the vector to any given point?

So, is it only possible to construct coordinate systems within a vector space (that is, can coordinate systems exist without a vector space?). And if so, how can we express this mathematically?

Thank you!

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I liked the discussion in "A Geometric Approach to Differential Forms" by David Bachman, and "Tensors, Differential Forms And Variational Principles" Lovelock & Rund.

Firstly, vector spaces do not need coordinate systems. Vector spaces can be defined in abstract terms very well. A great book here is Halmos "Finite-dimensional vector spaces". Axler's book (suggested above) is also great.

In physics we usually do want to talk about spaces, and in particular topological spaces, and in particular about differentiable manifolds. Lets say you have such a manifold $M$ and a scalar function $F$ defined on this $M$. The function can be seen as a map from manifold to the space of real numbers $F: M\to \mathbb{R}$, i.e. for each point $p\in M$ on the manifold the function has a real value.

Next, working with abstract points on the manifold is cumbersome, so one usually defines a map from real numbers, or Cartesian product of several real number spaces to the manifold, i.e. $\varphi:\mathbb{R}\times\dots\mathbb{R}\to M$. Such that for each $p\in M$ there is a unique tuple of real numbers $\{x^i\}_{i=1 \dots N}$, such that $\varphi\left(x^1,\,x^2\,\dots\right)=p$. This is the coordinate system. We can now define $f=F\circ\varphi: \mathbb{R}\times\dots\mathbb{R}\to\mathbb{R}$.

Next, we usually want to know how much $F$ changes when we move from $p_1=\varphi\left(x^1\dots\right)$ to $p_2=\varphi\left(x^1+\delta x^1\dots\right)$. This can be expressed as $df=\sum_i \delta x^i \frac{\partial f}{\partial x^i}=\delta x^i \partial_i f$. Now, we can note that there is a similarity between vector spaces and partial derivatives, both can be added, multiplied by real numbers etc... The analogy is so good that you can define a tangent vector space at point $p\in M$. This tangent vector space, denoted by $T_p M$, contains all linear combinations of first-order partial derivatives at $p$, i.e. $T_p M=\{\partial_1, \partial_1 + \partial_2, \partial_1 -3 \partial_2\dots\}$. That's the vector space you were after. In particular a vector $v\in T_p M$, $v=v^i\partial_i$ is a differential operator that can be applied to any function on $M$ to give $v.f=v^i \partial_i f$, where $v^i$ are (real ) numbers.

Note that this vector space is only defined at a single point of the manifold. The collection of tangent spaces at all points is called the tangent bundle. There are many more things to cover there. Sternberg discusses it in "Group Theory and Physics".

So what is the practical utility of such long definition? Well, defining your vector basis through derivatives can be quite elegant. For example Cartesian basis vectors in 3d can be shown to be given by $\mathbf{\hat{x}}=\boldsymbol{\nabla}x,\,\mathbf{\hat{y}}=\boldsymbol{\nabla}y,\,\mathbf{\hat{z}}=\boldsymbol{\nabla}z$. Similarly, we can defined non-normalized basis for the spherical coordinates as $\boldsymbol{e}_r = \boldsymbol{\nabla}r,\, \boldsymbol{e}_\theta = \boldsymbol{\nabla}\theta, \boldsymbol{e}_\phi = \boldsymbol{\nabla}\phi$. So for any function $f=f\left(r,\,\theta,\,\phi\right)$, by definition,

$\boldsymbol{\nabla}f=\boldsymbol{e}_r\partial_r f + \boldsymbol{e}_\theta\partial_\theta f+ \boldsymbol{e}_\phi\partial_\phi f$,

but equivalently

$\boldsymbol{\nabla}f=\mathbf{\hat{x}}\partial_x f + \mathbf{\hat{y}}\partial_y f + \mathbf{\hat{z}}\partial_z f$.

What if $f=\theta$? Then:

$\boldsymbol{e}_\theta = \mathbf{\hat{x}}\partial_x \theta + \mathbf{\hat{y}}\partial_y \theta + \mathbf{\hat{z}}\partial_z \theta$

So now you know the decomposition of one of the spherical basis vectors into the Cartesian basis from calculus - no need for those pesky diagrams! For example $\tan\theta=\sqrt{x^2+y^2}/z$, so

$\partial_x \tan\theta=\frac{1}{\cos^2\theta}\partial_x \theta = \frac{x}{z\sqrt{x^2+y^2}}=\frac{\cos\phi}{r\cos\theta}$, so:

$\partial_x \theta = \frac{\cos\phi\cos\theta}{r}$

etc.

After some work, you can recover the normalized spherical basis $\mathbf{\hat{r}}=\boldsymbol{e}_r,\,\boldsymbol{\hat{\theta}}=r\boldsymbol{e}_\theta\,\boldsymbol{\hat{\phi}}=r\sin\theta\boldsymbol{e}_\phi$

So what? Well how about taking curl?

$\boldsymbol{\nabla}\times\boldsymbol{\hat{\theta}}=\boldsymbol{\nabla}\times r.\boldsymbol{\nabla}\theta=\boldsymbol{\nabla}r\times\boldsymbol{\nabla}\theta=\boldsymbol{\hat{r}} \times \boldsymbol{\hat{\theta}}/r$

Much easier than trying to work from Cartesian coordinates

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  • $\begingroup$ Thank you, this is an amazing resource! Plus a lot of other suggested reading to look into! $\endgroup$ – James Wirth Feb 22 at 15:23
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Can coordinate systems exist without a vector space?

Yes. For example, there is a coordinate system on the surface of the Earth called latitude and longitude. But the surface of a sphere is not a vector space.

How can we express this mathematically?

In the case of a spherical surface, one common way is with a polar angle $\theta$ and an azimuthal angke $\phi$.

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  • $\begingroup$ Though wouldn't such a spherical coordinate system come equipped with basis vectors $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$? How would these basis vectors exist without defining a vector space? $\endgroup$ – James Wirth Feb 21 at 19:08
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    $\begingroup$ You are thinking about a 3D Euclidean space in which a spherical surface is embedded. That is a vector space. The space of the surface itself is not a vector space. Mathematicians often consider spaces without ever thinking about whether or not they are embedded or embeddable in a higher-dimensional space. $\endgroup$ – G. Smith Feb 21 at 19:10
  • $\begingroup$ Ah, I understand. So is the concept of a coordinate system completely separate to vector spaces? Because the basis vectors are often aligned with our coordinate axes, and we sometimes speak of a Cartesian coordinate system as consisting of basis vectors $\hat{x}$, $\hat{y}$ and $\hat{z}$. $\endgroup$ – James Wirth Feb 21 at 19:15
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    $\begingroup$ @JamesWirth, Not sure what you're asking with "...completely separate...?" But, (A) You can define a vector space without ever mentioning the idea of coordinates, and (B) having defined a vector space, you can then pick any minimal set of vectors that completely span it, and use that set as the basis for a coordinate system. Make of that what you will, I guess. $\endgroup$ – Solomon Slow Feb 21 at 19:36
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    $\begingroup$ @JamesWirth, I did not say that "axes need to be aligned..." In fact, I did not say "axes" at all. But using basis vectors (as per my answer) is a simple, formal way to completely specify a coordinate system that works for any vector space. If you have some other method of doing it, and it solves some problem for you, then great! But if your method maps all of the linear operations on vectors onto corresponding linear operations on coordinate tuples, then there's a pretty good chance that your system of assigning coordinates actually is isomorphic to defining them relative to basis vectors. $\endgroup$ – Solomon Slow Feb 21 at 19:58
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I'm inclined to say that a coordinate system is imposed onto the vector space such that the axes are in the direction of the basis vectors, and that the coordinates are the scalar components of the vector to any given point?

Basically, yes. But more formally speaking, If you choose a basis set, $B_1, B_2, ..., B_n$ for some vector space, then for any $V$ that is a member of the space you can find unique scalars $a_1, a_2,...,a_n$ such that $V=a_1B_1+a_2B_2+\dots+a_nB_n$.

The $a_1, a_2,...,a_n$ are the coordinates of $V$.

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  • $\begingroup$ I upvoted because you put it really nicely. Though how do we apply this definition if we're using something like spherical coordinates: then the scalar components of the three basis vectors would not be the coordinates? $\endgroup$ – James Wirth Feb 21 at 19:54
  • $\begingroup$ @JamesWirth, See G. Smith's answer. The mere fact that you can assign coordinates (e.g., lattitude and longitude) to all of the points in some abstract space (e.g., the surface of a sphere) does not mean that you can meaningfully apply the laws of linear algebra to those points (i.e., you can't necessarily manipulate them as vectors.) I highly recommend this book if you want to grok vectors and vector spaces. $\endgroup$ – Solomon Slow Feb 21 at 20:05
  • $\begingroup$ Awesome, thanks for the suggestion - I'll go and check it out! Many thanks for putting up with me, if I'm coming across as pedantic it's just because it's quite a confusing topic for the uninitiated! $\endgroup$ – James Wirth Feb 21 at 20:10

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