Why Is This Basis Non-Coordinate?

In summary, the author is confused by Schutz's notion of non-coordinate basis. He is seeking clarification on what it means.
  • #1
desic
4
0
I'm self-studying Schutz Geometric methods of mathematical physics, having problems with ex. 2.1. page 44.

r=cos(theta)x+sin(theta)y
theta=-sin(theta)x+cos(theta)y

show this is non-coordinate basis, i.e. show commutator non-zero.

I try to apply his formula 2.7, assuming

V1=cos(theta), V2=sin(theta)
W1=-sin(theta), W2=cos(theta)
x(r)=r cos(theta)
y(r)=r sin(theta)
x(theta)=cos(theta)
y(theta)=sin(theta)

These parametrics I got from integrating back from the components of r and theta

I believe the component of x should be (sin(theta))/r, however I get (sin(theta) - r sin(theta))/r.

would appreciate any help
 
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  • #2
I'm sorry I can't help you but I'm intrigued by this notion of "non-coordinate basis". Does the author give a definition? (And if so can you post it here please o:))
 
  • #3
I'm as confuzed as quasar, but I noticed you defined x and y in terms of r and theta... if they're two separate functions, you're going to confuse the heck out of yourself. If they're not, then you've confused the heck out of yourself. Either way, better labelling is better
 
  • #4
Well according to Schutz, "any linearly indepedent set of vector fields can serve as a basis, and one can easily show that not all of them are derivable from coordinate systems". The basic test is whether the fields commute. So if:
l and m are independent parameters that generate integral curves over a space
components of vector V are dx/dl, applied to basis d/dx partial derivative for each coordinate x
likewise W=dx/dm
then if commutator of vectors [V,W]=VW-WV does not equal 0, then the parameters l and m form a non-coordinate basis.

Geometrically, if you travel from P along V curve (delta l = e) to point R, then along W curve (delta m = e) to point A, and travel from P along W curve (delta m = e) to Q, then along V curve (delta l = e) to B, then A is not necessarily the same point as B. The distance from A to be is e squared times the commutator [V,W].

As to the original problem, yes I remain confused. I'm only guessing that is how you derive the parametric equations, and in my calculations I try to keep the processing of parameters separate. It seems to get a near result, and a non-zero result, but not the right result.
 
  • #5
2nd last para of my previous note should read (that's what happens when you type too fast):

Geometrically, if you travel from P along V curve (delta l = e) to point R, then along W curve (delta m = e) to point A, and travel from P along W curve (delta m = e) to Q, then along V curve (delta l = e) to B, then A is not necessarily the same point as B. The vector from A to B is e squared times the commutator [V,W] (which is a vector on basis of partial derivatives of x coordinates).
 
  • #6
Dear DESIC, I´m having same problem with Ex 2.1 p. 44 of Schutz Geom. Methods of Math. Physics. Have you come any closer to resolving your querry? The answer given by Schutz is
[r,theta]= -theta/r. Isn´t the magnitude of r just unity? Bendon
 
  • #7
first of all whether you have a coordinate basis or not depends on what coordinates you use on local patches of the manifold..
and by that i mean that in r,θ coordinates [tex]\frac{\partial}{\partial r}[/tex]
and [tex]\frac{\partial}{\partial \theta}[/tex]
are a coordinate basis indeed...
but if you re-express these basis vectors in x,y coordinate language the first becomes
[tex]\frac{\x}{\sqrt[x^2+y^2]}\frac{\partial}{\partial x} + \frac{\y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial y}[/tex]
and etc for the second...
so in x,y basis they are non coordinate vectors and their commutator if you do the math is not zero
[tex]\frac{\x}{\sqrt{x^2+y^2}} [/tex]
 
  • #8
:rolleyes:sry i messed up my latex
 

Related to Why Is This Basis Non-Coordinate?

1. What is a noncoordinate basis vector field?

A noncoordinate basis vector field is a set of vector fields that are not expressed in terms of a coordinate system, but rather in terms of a different set of basis vectors. These basis vectors do not necessarily align with the coordinate axes and may vary throughout the space.

2. How are noncoordinate basis vector fields different from coordinate basis vector fields?

Noncoordinate basis vector fields are different from coordinate basis vector fields in that they are not tied to a specific coordinate system. Coordinate basis vector fields are defined by the coordinate axes, while noncoordinate basis vector fields may vary in direction and magnitude throughout the space.

3. What are some examples of noncoordinate basis vector fields?

Some examples of noncoordinate basis vector fields include polar basis vectors in two-dimensional space, spherical basis vectors in three-dimensional space, and cylindrical basis vectors in cylindrical coordinates.

4. Why are noncoordinate basis vector fields useful?

Noncoordinate basis vector fields are useful because they allow for more flexibility in describing vector fields in complex or non-standard coordinate systems. They also allow for a more natural representation of certain physical phenomena such as rotation or translation.

5. How are noncoordinate basis vector fields used in scientific research?

Noncoordinate basis vector fields are used in scientific research in various fields such as physics, engineering, and mathematics. They are particularly useful in solving problems that involve non-standard coordinate systems or complex vector fields, such as in fluid dynamics, electromagnetism, and quantum mechanics.

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