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Vector and Basis Transformations

  1. Sep 15, 2011 #1
    Hi, I'm working through Schutz's intro to GR on my own, and I'm trying to do problems as I go to make sure it sinks in. I've encountered a bump in chapter 5, though. I don't think this is a tough problem at all, I think it's just throwing me off because x and y are coordinates as well as variables, if you know what I mean.

    1. The problem statement, all variables and given/known data
    Hopefully my latex tags work!
    Let [tex] f= x^2 +y^2 + 2xy [/tex]
    [tex] V \rightarrow (x^2 +3y, y^2+3x) [/tex]
    [tex] W \rightarrow (1,1) [/tex]

    So, among other things, I have to express f as a function of r and theta, and find the components of V and W in the r, theta basis. I thought I knew what I was doing, but the new components of my V vector are pretty ugly, which always makes me suspicious.
    2. Relevant equations

    [tex] x^2 + y^2 = r^2 [/tex]
    [tex] x = rcos\theta[/tex]
    [tex] y=rsin\theta[/tex]
    [tex] V^{\alpha '} = \Lambda^{\alpha '}_{\beta} V^{\alpha} [/tex]
    3. The attempt at a solution

    Well, I'm pretty sure f is just [tex] f(r, \theta) = r^2 + 2r^2\cos\theta [/tex],
    I'd rather not write out the matrix equation, but using the equation above for the transformation of vector components, and writing out V in r and theta as [tex] V \rightarrow (r^2\cos^2\theta +3r\sin\theta, r^2\sin^2\theta +3r\cos\theta) [/tex], and I've calculated the transformation matrix between cartesian and polar to have components from left to right, top to bottom, of [tex] \cos\theta, \sin\theta, -\sin\theta /r, \cos\theta/r [/tex], I get pretty ugly vector components for the new coordinate system (complete with things that don't simplify much, like cos^3 + sin^3, etc).
    Does it look like I'm doing it right? Completely wrong matrix?
  2. jcsd
  3. Sep 15, 2011 #2


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    You left out a factor of sin theta in the second term.
    Looks okay to me so far. What are you getting?
    Last edited: Sep 15, 2011
  4. Sep 15, 2011 #3
    You are right, I did leave out sin theta. That was happily just a typo/transcription error, though.

    I think I do know what [tex] \hat{r}, \hat{theta}[/tex] are. They're the unit basis vectors in the r, theta coordinate system, corresponding to the radial direction and the angular direction, they're orthogonal, etc.
    But it's also part of my confusion. The question specifically asks for the components of the vector. I know when you transform the basis vectors, the transformation is
    [tex] e_{\alpha'} = \Lambda^{\beta}_{\alpha'} e_{beta} [/tex]
    Where the unprimed labels are cartesian, and the primed in this case is the polar. That gives

    [tex] e_r = \cos\theta e_x + \sin\theta e_y [/tex]
    [tex] e_{\theta} = -r\sin\theta e_x +r\cos\theta e_y [/tex]

    So I guess I don't understand how to combine these. The question specifically asks for the "V and W components on the polar basis, as functions of r and theta."

    I calculated the components (which are ugly, which lead me here), but I never transformed the basis vectors. I suppose it's possible that the correct answer is a combination of the components I calculated and the coefficients that come from transforming the basis vectors, but I don't think so. I'm just not clear on this process, I guess.
  5. Sep 15, 2011 #4
    I guess I should put in my ugly vector: The answer I had got for the components of V is

    [tex] r( r(\cos^3\theta + \sin^3\theta) + 3\sin\theta\cos\theta, \frac{3(\cos^2\theta-\sin^2\theta)}{r} +\sin^2\theta\cos\theta - \cos^2\theta\sin\theta) [/tex]

    Thanks for the help, also!
  6. Sep 15, 2011 #5
    Okay, to think out loud for a moment:

    I do think that the components of the vector are just those that I displayed above. The component transformation should not depend on the basis transformation, I think. It's really just that the answer I got was so ugly that I thought I must be wrong.
  7. Sep 15, 2011 #6


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    The [itex]3\sin\theta\cos\theta[/itex] in the r-component is off by a factor of 2, but otherwise looks good.
  8. Sep 15, 2011 #7
    You are right, that should have been 6, not 3.
    Thanks for the help!
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