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I Coordinate transformation - Rotation

  1. Oct 11, 2017 #1
    ort1.jpg
    ort2.jpg
    How author derives these old basis unit vectors in terms of new basis vectors ? Please don't explain in two words.

    [itex] \hat{e}_x = cos(\varphi)\hat{e}'_x - sin(\varphi)\hat{e}'_y[/itex]
    [itex] \hat{e}_y = sin(\varphi)\hat{e}'_x + cos(\varphi)\hat{e}'_y[/itex]​
     
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  3. Oct 11, 2017 #2

    Orodruin

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    How about one word? Trigonometry.

    It is just a matter of projecting the old basis on the new. Since the new basis is orthonormal, it holds that ##\vec e_i = (\vec e_i\cdot\vec e’_j)\vec e’_j##.
     
  4. Oct 11, 2017 #3
    We obtain:

    [itex] \vec{e}_i = (\vec{e}_i \cdot \vec{e}'_j)\vec{e}'_j [/itex]
    [itex] \vec{e}_i = | \vec{e}_i | | \vec{e}'_j |cos(\frac{\pi}{2}+\varphi) \vec{e}'_j [/itex]
    [itex] \vec{e}_i = -sin(\varphi)\vec{e}'_j [/itex]

    Right ? But we didn't get transformation formula.
     
    Last edited: Oct 11, 2017
  5. Oct 11, 2017 #4

    Orodruin

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    No, you have not done it correctly. You cannot have different free indices on the different sides of the equation. Also note that repeated indices must be summed over.
     
  6. Oct 11, 2017 #5
    [itex] \vec{e}_1 = | \vec{e}_1 | | \vec{e}'_1 |cos(\varphi) \vec{e}'_1 + | \vec{e}_1 | | \vec{e}'_2 |cos(\frac{\pi}{2}+\varphi) \vec{e}'_2[/itex]
    [itex] \vec{e}_1 = cos(\varphi) \vec{e}'_1 -sin(\varphi)\vec{e}'_2 [/itex]

    [itex] \vec{e}_2 = | \vec{e}_2 | | \vec{e}'_1 |cos(\frac{\pi}{2}-\varphi) \vec{e}'_1 + | \vec{e}_2 | | \vec{e}'_2 |cos(\varphi) \vec{e}'_2[/itex]
    [itex] \vec{e}_2 = sin(\varphi) \vec{e}'_1 +cos(\varphi)\vec{e}'_2 [/itex]

    Now am I right ?

    But I want to understand in projection language (trigonometry) how author derives these formulas, but I didn't find explanation of unit vector basis transformation using trigonometry. I tried but didn't figured out how. Point me right direction.
     
  7. Oct 11, 2017 #6

    Orodruin

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    The old basis vector is a unit vector that forms the hypothenuse of a right triangle who's other sides are its projections onto the new coordinate directions. The rest is just applying the definitions of sine and cosine in terms of the sides of a right triangle.
     
  8. Oct 11, 2017 #7

    Orodruin

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    The other way around is to define a rotation as the transformation that relates two orthonormal bases. Looking at the two-dimensional case, you would have
    $$
    \vec e_x = a \vec e'_x + b \vec e'_y \quad \mbox{and} \quad \vec e_y = c \vec e'_x + d \vec e'_y.
    $$
    From here you can apply the orthonormality conditions of both systems in asserting that ##\vec e_x^2 = \vec e'^2_x = \vec e_y^2 = \vec e'^2_y = 1## and ##\vec e_x \cdot \vec e_y = \vec e'_x \cdot \vec e'_y = 0##. This will give you a set of relations among the coefficients ##a##, ##b##, ##c##, and ##d## (please derive these relations yourself). You should get three independent relations and since you have four parameters, you will be left with one free parameter (the angle of rotation). You will be able to make the interpretation of the angle of rotation by looking at the inner product between ##\vec e_x## and ##\vec e'_x##.
     
  9. Oct 11, 2017 #8
    What you said is evident from the picture, but how ? I drew it many times but cannot get sum of projections that is a old basis unit vector.
     
  10. Oct 11, 2017 #9

    Orodruin

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    I don't understand what you are trying to say here. It is evident from the picture but it is not evident?
     
  11. Oct 11, 2017 #10
    How to project [itex]\hat{e}_x[/itex] on [itex]\hat{e}'_x[/itex] and [itex]\hat{e}'_y[/itex] ?

    basis.jpg
     
    Last edited: Oct 11, 2017
  12. Oct 11, 2017 #11

    Svein

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    Rotate your figure until [itex]\hat{e}'_x[/itex] is horizontal. Then the projections will be obvious.
     
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