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Gradient transforms under axes rotation

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]f[/itex] is a function of two variables: y, z. I want to show that the gradient:

    [tex]\nabla f=\frac{\partial f}{\partial y}\hat y + \frac{\partial f}{\partial z}\hat z[/tex]
    Transforms as a vector under rotation of axes.
    2. Relevant equations
    The rotation of axes:
    [tex]A \left\{\begin{array}{l}\bar {y} = y \cdot \cos \phi + z \cdot \sin \phi \\ \bar {z} = -y \cdot \sin + z \cdot \cos \end{array}\right.[/tex]
    3. The attempt at a solution
    I use the chain rule:
    [tex]\frac{\partial f}{\partial \bar {y}}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar {y}}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \bar {y}}[/tex]
    And the similar for [itex]\frac{\partial f}{\partial \bar {z}}[/itex]. In order to find [itex]\frac{\partial y}{\partial \bar {y}}[/itex] and [itex]\frac{\partial z}{\partial \bar {z}}[/itex] i solve A for y and z and get:

    [tex]\left\{\begin{matrix} y= \bar {y}\cdot \cos \phi - \bar {z}\cdot \sin \phi \\ z= \bar {y}\cdot \sin \phi +\bar {z}\cdot \cos \phi\end{matrix}\right.[/tex]

    Now i get:
    [tex]\left\{\begin{array}{l}\frac{\partial f}{\partial \bar {y}}=\frac{\partial f}{\partial y}(-\sin \phi)+\frac{\partial f}{\partial z}\cos \phi \\ \frac{\partial f}{\partial \bar {z}}=\frac{\partial f}{\partial y}(-\cos \phi)+\frac{\partial f}{\partial z}(-\sin \phi)\end{array}\right.[/tex]
    Not as i expected, in the form of A.
    It is closer to switching between the axes, but that's all i understand from this result.
     
  2. jcsd
  3. Feb 9, 2012 #2

    vela

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    How did you get ##\partial y/\partial \bar{y} = -\sin \phi##? That's what your final results imply. Similar questions for the other derivatives.
     
  4. Feb 14, 2012 #3
    For [itex]\partial y/\partial \bar{y}[/itex] I differentiated the formula i derived earlier:
    [tex]y= \bar {y}\cdot \cos \phi - \bar {z}\cdot \sin \phi[/tex]
     
  5. Feb 14, 2012 #4

    vela

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    Which doesn't give you ##\partial y/\partial \bar{y} = -\sin \phi##. That's why I'm asking where did you get that from.
     
  6. Feb 15, 2012 #5
    You mean i have a mistake in differentiating [itex]y= \bar {y}\cdot \cos \phi - \bar {z}\cdot \sin \phi[/itex]? how come?
    The derivative of [itex]\cos[/itex] is [itex]-\sin[/itex], and i ignore the second member [itex] \bar {z}\cdot \sin \phi[/itex] because it doesn't contain [itex]\bar {y}[/itex].
    Or you ask how did i get: [itex]y= \bar {y}\cdot \cos \phi - \bar {z}\cdot \sin \phi[/itex]?
     
  7. Feb 15, 2012 #6

    HallsofIvy

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    [itex]-sin(\phi)[/itex] is the derivative of [itex]cos(\phi)[/itex] with respect to [itex]\phi[/itex]. You are not differentiating with respect to [itex]\phi[/itex]. [itex]\phi[/itex] is a constant.
     
  8. Feb 15, 2012 #7
    Thanks!!!
     
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