1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Noncoordinate basis vector fields

  1. Oct 28, 2006 #1
    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
     
  2. jcsd
  3. Oct 28, 2006 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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:))
     
  4. Oct 28, 2006 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
     
  5. Oct 28, 2006 #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.
     
  6. Oct 28, 2006 #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).
     
  7. Jul 19, 2010 #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? Best regards, Bendon
     
  8. Dec 19, 2010 #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]
     
  9. Dec 19, 2010 #8
    :uhh:sry i messed up my latex
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Noncoordinate basis vector fields
  1. Electric field vector (Replies: 17)

Loading...