|May16-12, 06:26 AM||#18|
trying to understand contravariance
Here is another example for you to consider.
Assume that you have a flat compact disc (CD), and that the CD lies within the horizontal x-y plane of a rectangular cartesian coordinate system x-y-z. The axis of the CD coincides with the z-axis of the cartesian coordinate system, and the CD is rotating as a rigid body about its axis (relative to the x-y-z coordinate system) with a constant angular velocity of ω. Each material particle of the CD travels in a perfect circle about the z-axis. Assume that there is also a cylindrical polar coordinate system (r-θ-z) present, that coincides with the x-y-z coordinate system.
As reckoned from the r-θ-z coordinate system, the velocities of the various material particles comprising the CD are given by:
V = (ωr) iθ = ω (r iθ) = ω eθ
where eθ is the coordinate basis vector in the θ direction:
eθ = r iθ
According to the equations above, even though the velocity vector V varies with r and θ, its contravariant component in the r-direction is zero, and its contravariant component in the θ-direction is a constant, and equal to ω:
Vr = 0
Vθ = ω
Note that the velocity V is a vector tangent to the circles (trajectories) around which the material particles are traveling.
Now use your transformation formula to show that, in cartesian coordinates, the contravariant components of the velocity V are given by:
Vx = -ωy
Vy = +ωx
V = -ωy ix + ωx iy
|May17-12, 11:30 AM||#19|
all right, i think i got it now
thank you everyone!!
|contravariance, covariance, riemann curvature|
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