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Covariant and Contravariant components in Oblique System

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    9QuWu.png
    In the oblique coordinate system K' defined in class the position vector r′ can be written as:
    [itex]r'=a\hat{e'}_{1}+b\hat{e'}_{2}[/itex]

    Are a and b the covariant (perpendicular) or contravariant (parallel) components of r′? Why? Give an explanation based on vectors’ properties and another based on tensors’ properties.

    2. Relevant equations
    [itex]\hat{e'}_1=\hat{e}_1[/itex]
    [itex]\hat{e'}_2=e_1cos(\alpha )+e_2sin(\alpha )[/itex]

    [itex]{x'}_{1\perp}={x}_1[/itex]
    [itex]{x'}_{2\perp}=x_1cos(\alpha )+x_2sin(\alpha )[/itex]

    3. The attempt at a solution
    My best effort is that a and b are the covariant components of r' since they are the perpendicular projects of r' onto the [itex]\hat{e'}_1[/itex] and [itex]\hat{e'}_2[/itex] basis vectors, so they're essentially equivalent to [itex]{x'}_{1\perp}[/itex] and [itex]{x'}_{2\perp}[/itex]. So I think this would be my vector solution for the problem, but I don't know exactly how to represent it. As for the an explanation based on the tensor's properties, I don't even know where to start...

    Thanks in advance.
     
    Last edited: Sep 8, 2012
  2. jcsd
  3. Sep 8, 2012 #2
    If you used the perpendicular projections onto the axes with the basis vectors given, would you recover the point [itex]r[/itex]? It doesn't look like that to me. You'd end up way out above and to the right.
     
  4. Sep 9, 2012 #3
    Ah, I see. I have been struggling with this problem for the past couple days and think I have developed a rough proof:
    a and b are contravariant components of r' since they both give the magnitude of the vectors that add to r'. We found that:
    [itex]\hat{e'}_{1}=\hat{e}_{1}[/itex]
    [itex]\hat{e'}_{2}=\hat{e}_{1}cos(\alpha)+\hat{e}_{2}sin(\alpha)[/itex]
    And
    [itex]x'_{1\parallel}=x_{1}-cot(\alpha)x_{2}[/itex]
    [itex]x'_{2\parallel}=x_{2}csc(\alpha)[/itex]

    Making the substitutions into [itex]r'=a\hat{e'}_{1}+b\hat{e'}_{2}[/itex]:
    [itex]r'=(x_{1}-cot(\alpha)x_{2}) \hat{e}_{1}+(x_{2}csc({\alpha}))(\hat{e}_{1}cos( \alpha)+\hat{e}_{2}sin(\alpha))[/itex]
    We get
    [itex]r'={x}_{1}\hat{e}_{1}+{x}_{2}\hat{e}_{2}[/itex]

    So I think this is good for the vector explanation. Any advice for the tensor representation?
     
    Last edited: Sep 9, 2012
  5. Sep 9, 2012 #4
    Yeah, they should be contravariant components, but to be honest, when the question says "give an explanation based on the tensors' properties," I'm not even clear on which tensors are being talked about.
     
  6. Sep 9, 2012 #5
    Maybe I'll have to present that to the professor. I thought that maybe she just wanted the same proof, but using Einstein notation for the matrices that make the transformation.
     
  7. Sep 10, 2012 #6
    e1' and e2' are unit vectors along the coordinate directions of the primed coordinate system. e'1 is the partial derivative of the arbitrary position vector r with respect to x1' and e'2 is the partial derivative of the arbitrary position vector r with respect to x2', where I have used a more suggestive notation here for "parallel" x primes. Thus,

    r = x1'e1' + x2'e2'

    Therefore, a = x1' and b = x2'. These are the contravariant components.
     
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