A question about mechanics and generalized coordinates.

[Lajka]

Hello,

I wasn't quite sure where to make this topic, so I hope I didn't do wrong by putting it here.

The question I'm having is somewhat difficult to describe and I guess it's more of a mathematical question really, but since I'm learning mechanics now and came up with it, I thought it would be best to ask it here.

So what I'm interested in is what really makes generalized coordinates... coordinates?
Let me explain this.

The way I figure coordinates is that they are just a numbers, a coefficients if you will, which multiply the basis vectors in order to produce a needed vector as a superposition of these. So, the first thing I need is a basis, a set of linearly independent vectors which are able to span the whole vector space.

Since I learn mechanics, it's only natural to observe R³ space, and a standard basis for that is, of course, (1,0,0), (0,1,0) and (0,0,1) vectors, or [tex]\vec{i}[/tex], [tex]\vec{j}[/tex] and [tex]\vec{k}[/tex] vectors.

So, if I have my coordinates [tex](x,y,z)[/tex], the only thing I need to do is to multiply my unit vectors with them, add them, and voila, I have my desired vector.

My problem here is because I can't think of any other coordinates that satisfy all these conditions.
For example, I'm well aware of spherical coordinates [tex](r,\varphi,\theta)[/tex] and I'm well aware that these are equivalent to a former triple [tex](x,y,z)[/tex] in a sense that they can fully determine any position, and that they also have three orthogonal unit vectors, but here's the thing...

First of all, you don't even use all three of these unit vectors, you just use one of them to determine the position, and not one of them is even constant! I don't have problem with this from a physical point of view, but I do have from a mathematical point of view. Like I said before, if I have new coordinates, I need a new basis as well, and although these unit vectors are orthogonal, they are not constant, and that just doesn't make sense if they are to constitute a basis.

Further more, if these new coordinates are coordinates on their own right, they should be able to stand alone for themselves, but that's not the case! If I want to determine a position in spherical system, i write [tex]\vec{r}=r*\vec{r_{0}}[/tex], right? But to calculate [tex]\vec{r_{0}}[/tex], I need my [tex]\varphi[/tex], [tex]\theta[/tex] AND [tex]\vec{i}[/tex] and [tex]\vec{j}[/tex] unit vectors from before!
WHY on Earth would I need these OLD unit vectors when I should be able to calculate everything without them? Remember, I have chosen a new basis and new coordinates now! Old coordinates, old basis, begone!

What I'm trying to say is that, for me, the process is simple: I have coordinates, I have constant basis and I calculate a needed vector. That's the definition that I've learned from math when I was learning vector spaces, and as far as I can tell, the only basis I've seen that satisfies these rigorous conditions is the triple [tex](\vec{i},\vec{j},\vec{k})[/tex].
All other coordinates (and their unit vectors consequently) depend on that triple in some way, so how can one look at them as a set of independent coordinates in their own right?

When I see in my book from mechanics something like "generalized coordinates are defined as [tex](q_{1},,q_{2},q_{3})[/tex], and that can be [tex](x,y,z)[/tex] or [tex](r,\varphi,\theta)[/tex] or...", I can't help but wonder if that's really fair, because I only see one example of 'true' coordinates with fully independent, constant basis.
So, what am I missing here?

P.S. I should say that I'm also aware that [tex](x,y,z) \mapsto (r,\varphi,\theta)[/tex] is a map and I know what a Jacobian is and so forth, I've learned all that from a multivariable calculus, but I don't know how and if any of that can be used here for some reasoning. After all, I used that for integrating functions, this something different, these are vector spaces.

 

[Dale]

Hi Lajka

This is a pretty detailed discussion that is usually just glossed over, and there are a lot of hairy details that I really won't be able to cover in this format, but I will do what I can.

The way I figure coordinates is that they are just a numbers, a coefficients if you will, which multiply the basis vectors in order to produce a needed vector as a superposition of these. So, the first thing I need is a basis, a set of linearly independent vectors which are able to span the whole vector space.

A coordinate system is a map from points on a manifold to points in R(N). In general, curvilinear coordinates do not form a vector space, as it does not generally make sense to perform the defining operations of scalar multiplication and vector addition.

Coordinate systems (at least those used in physics) do form metric spaces. So you can find distances and angles in terms of coordinates by using the metric.

For example, I'm well aware of spherical coordinates [tex](r,\varphi,\theta)[/tex] and I'm well aware that these are equivalent to a former triple [tex](x,y,z)[/tex] in a sense that they can fully determine any position, and that they also have three orthogonal unit vectors, but here's the thing...

The three orthogonal unit vectors that you refer to are formed by the derivatives of the coordinates, and are called the coordinate basis. At each point in the manifold the coordinate basis can be used to define a "tangent space". This tangent space is a vector space, with vector addition and scalar multiplication, but it is a different tangent space at each point in the manifold.

If the metric is "flat" then it is possible to unambiguously relate vectors in one tangent space to vectors in another tangent space and thereby treat each of the tangent spaces as though they were all one and the same vector space.

Further more, if these new coordinates are coordinates on their own right, they should be able to stand alone for themselves, but that's not the case! If I want to determine a position in spherical system, i write [tex]\vec{r}=r*\vec{r_{0}}[/tex], right? But to calculate [tex]\vec{r_{0}}[/tex], I need my [tex]\varphi[/tex], [tex]\theta[/tex] AND [tex]\vec{i}[/tex] and [tex]\vec{j}[/tex] unit vectors from before!
WHY on Earth would I need these OLD unit vectors when I should be able to calculate everything without them? Remember, I have chosen a new basis and new coordinates now! Old coordinates, old basis, begone!

I don't know where you are getting the idea that you still need the Cartesian coordinates. Everything can be done in spherical coordinates.

 

[Lajka]

DaleSpam, thank you very much for your answer!

Since I'm an engineer, I'm not that well-versed in the theory of manifolds, so I'll guess I'll have to look into that. Granted, I don't have a lot of time for that at the moment, but I'm sure I'll figure something out. If you perhaps have any proposition apropos literature about this, that would be great too!

I don't know where you are getting the idea that you still need the Cartesian coordinates. Everything can be done in spherical coordinates.

Well, as far as I know, the only way to get [tex]\vec{r_{0}}[/tex] is to calculate it as
[tex]\vec{r_{0}} = cos(\theta)cos(\varphi)\vec{i} + cos(\theta)sin(\varphi)\vec{j} + sin(\theta)\vec{k}[/tex], right?

So, you are effectively still using [tex]\vec{i}, \vec{j}[/tex] and [tex]\vec{k}[/tex] unit vectors, which, it would seem to me, shouldn't be needed or used anymore, because they belong to Cartesian coordinates, and we're currenly using spherical ones. That's what I was referring to.


What I'm trying to say is this: someone gives you [tex](r,\varphi,\theta)[/tex], so what do you do? You have [tex]r[/tex], that's great, now you only need [tex]\vec{r_{0}}[/tex] and you're done. How to get [tex]\vec{r_{0}}[/tex]? Well, you calculate it as
[tex]\vec{r_{0}} = cos(\theta)cos(\varphi)\vec{i} + cos(\theta)sin(\varphi)\vec{j} + sin(\theta)\vec{k}[/tex],
and then you write [tex]\vec{r}=r*\vec{r_{0}}[/tex], and you're done, you have your vector. That's the only way, as far as I know, to get your desired [tex]\vec{r}[/tex] in spherical coordinates.

Was it a totally independent process? No, it wasn't, we had to use Cartesian unit vectors, and, if these coordinates were truly 'independent', that shouldn't be needed happen (I hope this makes some sense, English isn't my native language).

The way I see it, you should be able to use unit vectors and coordinates from one system of coordinates, and only that, to get any vector you want. That's the definition of basis vectors, and their coordinates respectively.

How is that possible for spherical coordinates? It clearly is for Cartesian ones, but I don't think it is for spherical ones, and I'm kinda baffled by that fact.

Do you think I'm wrong about this? If so, please, explain me why. Better yet, give me an example where you construct a vector in spherical system using only [tex](r,\varphi,\theta)[/tex] and [tex](\vec{r_{0}},\vec{\varphi_{0}},\vec{\theta_{0}})[/tex], i.e. only coordinates and unit vectors from that system particularly.

 

[Dale]

Well, as far as I know, the only way to get [tex]\vec{r_{0}}[/tex] is to calculate it as
[tex]\vec{r_{0}} = cos(\theta)cos(\varphi)\vec{i} + cos(\theta)sin(\varphi)\vec{j} + sin(\theta)\vec{k}[/tex], right?

This does not really accomplish anything. You have only expressed one named vector in terms of three other named vectors. But, how do you get i, j, and k? You could just as easily express i, j, and k in terms of r, theta and phi and make the same calculation in reverse.

The way I see it, you should be able to use unit vectors and coordinates from one system of coordinates, and only that, to get any vector you want. That's the definition of basis vectors, and their coordinates respectively.

That is correct, and it is possible to do in any coordinate system including Cartesian and spherical coordinates. In order to calculate the unit vectors of any coordinate system you use the same formula. Specifically, the n-th unit vector is given by:

[tex]\hat{x_n}=\frac{\frac{d\vec{x}}{dx_n}}{\left|\frac{d\vec{x}}{dx_n}\right|}[/tex]

where

[tex]\vec{x}=(x_1,x_2,x_3)[/tex]

This is how r, theta, and phi are defined in spherical coordinates, it is also how i, j, and k are defined in Cartesian coordinates. Otherwise you are just using undefined symbols to represent some special set of named vectors.

 

[Lajka]

This does not really accomplish anything. You have only expressed one named vector in terms of three other named vectors. But, how do you get i, j, and k? You could just as easily express i, j, and k in terms of r, theta and phi and make the same calculation in reverse.

Well, I guess i kinda feel that [tex]\vec{i}[/tex], [tex]\vec{j}[/tex] and [tex]\vec{k}[/tex] vectors don't really need some formula for definition, because they look to me like they are almost inherently defined. How so? Well, they are (1,0,0), (0,1,0) and (0,0,1) respectively, when we use [tex](x,y,z)[/tex] coordinates.

I can't really do that with [tex](r,\varphi,\theta)[/tex] now, can I? (1,0,0) is not really [tex]\vec{r_{0}}[/tex], and so forth.

That is correct, and it is possible to do in any coordinate system including Cartesian and spherical coordinates. In order to calculate the unit vectors of any coordinate system you use the same formula. Specifically, the n-th unit vector is given by:
[tex]
\hat{x_n}=\frac{\frac{d\vec{x}}{dx_n}}{\left|\frac {d\vec{x}}{dx_n}\right|}
[/tex]
where
[tex]
\vec{x}=(x_1,x_2,x_3)
[/tex]
This is how r, theta, and phi are defined in spherical coordinates, it is also how i, j, and k are defined in Cartesian coordinates. Otherwise you are just using undefined symbols to represent some special set of named vectors.

Okay, I believe we're on something here, so just bear with me, please. See, I know this, and this is the problem:

Let's take [tex](x,y,z)[/tex]. So you say
[tex]
\hat{x_n}=\frac{\frac{d\vec{x}}{dx_n}}{\left|\frac{d\vec{x}}{dx_n}\right|} =
\frac{\frac{d(x,y,z)}{dx}}{\left|\frac{d(x,y,z)}{dx}\right|}
= (1,0,0) = \vec{i}
[/tex]

And the same thing for [tex]\vec{j}[/tex] and [tex]\vec{k}[/tex] vectors.

Great! Let's try this with spherical coordinates [tex](r,\varphi,\theta)[/tex] too now

[tex]
\hat{x_n}=\frac{\frac{d\vec{x}}{dx_n}}{\left|\frac {d\vec{x}}{dx_n}\right|} = \frac{\frac{d(r*\vec{r_{0}})}{dr}}{\left|\frac{d(r*\vec{r_{0}})}{dr}\right|} =
\frac{\vec{r_{0}}}{\left|\vec{r_{0}}\right|} = \vec{r_{0}}
[/tex]
(I used the fact that modulus of this unit vector is one, althought I haven't proved that anywhere)

Awesome! Let's try this one more time

[tex]
\hat{x_n}=\frac{\frac{d\vec{x}}{dx_n}}{\left|\frac {d\vec{x}}{dx_n}\right|} = \frac{\frac{d(r*\vec{r_{0}})}{d\varphi}}{\left|\frac{d(r*\vec{r_{0}})}{d\varphi}\right|} =
\frac{\frac{d(\vec{r_{0}})}{d\varphi}}{\left|\frac{d(\vec{r_{0}})}{d\varphi}\right|} =
...
[/tex]

So... what now?

In my opinion, I have to use
[tex]
\vec{r_{0}} = cos(\theta)cos(\varphi)\vec{i} + cos(\theta)sin(\varphi)\vec{j} + sin(\theta)\vec{k}
[/tex]

so I can see the dependancy [tex]\vec{r_{0}}[/tex] has of [tex]\varphi[/tex] to perform the derivation. Without this, I have no idea how to calculate the expression above and get my [tex]\vec{\varphi_{0}}[/tex].

So, again, I need [tex]\vec{i}[/tex], [tex]\vec{j}[/tex] and [tex]\vec{k}[/tex] vectors. I just do. And that's just wrong.

 

[Dale]

You may want to get a good textbook on manifolds and read up on them, or attend a class on differential geometry. Your questions are answerable, but this format is not very good for it. You do not need the Cartesian vectors to do math in other coordinate systems, in fact, for many manifolds it is not possible to define a Cartesian coordinate system, so you have no option to use them even if you wanted to.

 

[Lajka]

Okay, will do. I've no idea what textbook to use, but I'm sure I can google some.

Thank you again for all your help. :)