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Change of basis

  1. Jul 22, 2005 #1
    Say, i have a vector a, defined in a coordinate system, x-y-z and i rotate the axes by an angle theta around the z-axis, so i have my z-component invariant in this change of basis. Can someone show me why,

    [tex] a_x' = a_x cos\theta + a_y sin\theta [/tex] and

    [tex] a_x' = -a_x sin\theta + a_y cos\theta [/tex]

    i can't seem to find any place where they show how this is done.
     
  2. jcsd
  3. Jul 22, 2005 #2

    robphy

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    Draw a vector and two sets of orthogonal axes.
    For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.
    Do a little trigononometry.
    I'll post a URL with a picture, if I can find one.

    Ok here's one:
    http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html [Broken]
    (although the vector is not drawn from the origin to the little box).

    (Check the subscript on the LHS of your second equation.)
     
    Last edited by a moderator: May 2, 2017
  4. Jul 22, 2005 #3

    LeonhardEuler

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    The new x-axis is a line tilted at some angle, [itex]\theta[/tex], to the horizontal. Basically, if you want [itex]a'_y[/tex], you want the distance between a point and this line. The formula for this is
    [tex]|a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}}[/tex]
    where the equation of the line is [itex]y=mx+b[/tex]. Since the line passes through the origin, b=0. [itex]tan(\theta)=m[/tex], so:
    [tex]|a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}}[/tex]
    [tex]= \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|}[/tex]

    [tex]= |tan(\theta)a_x-a_y|\times|cos(\theta)|[/tex]

    [tex]=|sin(\theta)a_x-cos(\theta)a_y|[/tex]

    Since, for [itex]\theta=0[/tex], we should have [itex]a'_y=a_y[/tex], this means the signs will only work out properly if:
    [tex]a'_y=-sin(\theta)a_x+cos(\theta)a_y[/tex]
    The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.
     
    Last edited: Jul 22, 2005
  5. Jul 22, 2005 #4

    robphy

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    Here's a vector-algebraic proof.

    For the first set of axes, we have unit vectors [tex]\hat x[/tex] and [tex]\hat y[/tex].
    For the second set, which is rotated counterclockwise by angle [tex]\theta[/tex], we have [tex]\hat x'[/tex] and [tex]\hat y'[/tex].

    Note that [tex]A_x= \vec A\cdot \hat x[/tex] and so on.

    Now, compute [tex]A_{x'}= \vec A\cdot \hat x'[/tex] where one writes
    [tex] \vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y[/tex].
    So, let's plug it in:
    [tex]\begin{align*}
    A_{x'}
    &= \vec A\cdot \hat x' \\
    &= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x' \\
    &=(A_x) \hat x\cdot \hat x' + (A_y) \hat y\cdot \hat x' \\
    &=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\
    &=(A_x) \cos\theta + (A_y) \sin\theta \\
    \end{align}
    [/tex]
    Do the same for [tex]A_{y'}[/tex], and note that [tex] \hat x\cdot \hat y' =\cos(90^\circ+\theta)= -\sin\theta [/tex] and [tex] \hat y\cdot \hat y' =\cos\theta[/tex].
     
    Last edited: Jul 22, 2005
  6. Jul 22, 2005 #5
    The original vector had an x-component and a y-component. I find it easiest to just follow those two components separately.

    When the original vector gets rotated the original x-component, let’s call that X, will not be only in the direction of the x-axis anymore, the component of the rotated “original x-component“ that will still be in the direction of the x-axis is:
    [tex]X\cos(\theta)[/tex]
    but it will now also have a component in the direction of the y-axis, which will be:
    [tex]-X\sin(\theta)[/tex]

    The same happens with the original y component. The component of the rotated “original y component“ that will still be in the direction of the y-axis is:
    [tex]Y\cos(\theta)[/tex]
    and the component in the direction of the x-axis will be:
    [tex]Y\sin(\theta)[/tex]

    To get the x-component of the new vector add the parts in the direction of the x-axis:
    [tex]X\cos(\theta) + Y\sin(\theta)[/tex]
    To get the y-component of the new vector add the parts in the direction of the y-axis:
    [tex]-X\sin(\theta) + Y\cos(\theta)[/tex]
     
  7. Jul 22, 2005 #6
    hey thanks alot for the help. Yup the subscript on the LHS should be y. Here's what i have done, a little algebra,

    [tex] a_x' cos \theta = a_x[/tex]

    [tex] a_x' sin \theta= a_y [/tex]

    what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for [tex] a_x'[/tex]

    for [tex] a_y' [/tex] i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

    Is that right?

    edit: i'll look at the rest of the methods too. : )
     
    Last edited: Jul 22, 2005
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