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Question regarding tensors derive acceleration in polar form

  1. Jun 21, 2005 #1

    learningphysics

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    I'm having trouble with this question. It's from Rindler's Introduction to Special Relativity which I'm going through myself. I'm just starting to learn about tensors.

    <<<<i) A vector [tex]A^i[/tex] has components [tex]\dot{x}, \dot{y}[/tex] in rectangular Cartesian coordinates; what are its components in polar coordinates?>>>>

    This part I believe I know. The components are [tex]\dot{r}, r\dot{\theta}[/tex]. The first component is the [tex]a_r[/tex] component and the second is the [tex]a_{\theta}[/tex] component.

    <<<<ii) A vector [tex]B^i[/tex] has components [tex]\ddot{x}, \ddot{y}[/tex] in rectangular Cartesian coordinates; prove, directly from A.3 that its components in polar coordinates are [tex]\ddot{r}-r{(\ddot{\theta})}^2, \ddot{\theta}+2\dot{r}\dot{\theta}/r[/tex]>>>>

    This is what A.3 says:
    <<<<An object having components [tex]A^{ij....n}[/tex] in the [tex]x^i[/tex] system of coordinates and [tex]A^{i'j'...n'}[/tex] in the [tex]x^{i'}[/tex] system is said to behave as a contravariant tensor under the transformation [tex]\{x^i\}->\{x^{i'}\}[/tex] if [tex]A^{i'j'....n'}=A^{ij....n}{p_i}^{i'}{p_j}^{j'}....{p_n}^{n'}[/tex]>>>>

    I'm not sure how this is to be done. The [tex]a_{\theta}[/tex] coordinate in part ii) seems to be divided by r. I don't know if this is a mistake in the book or there is some reason for it.

    How do I use the definition of contravariant tensors to derive the formula for acceleration in polar coordinates? I really have no clue. I can derive the formula just using derivatives, but I don't see how to use tensors to derive it.

    Thanks a bunch for your help!
     
  2. jcsd
  3. Jun 21, 2005 #2

    dextercioby

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    [tex] A' ^{i}=\frac{\partial x' ^{i}}{\partial x^{j}} A^{j} [/tex] (1)

    For

    [tex] A^{j}=\left(\dot{x},\dot{y}\right) [/tex] (2)

    So

    [tex] A'^{1}=\frac{\partial\rho}{\partial x} \dot{x}+\frac{\partial\rho}{\partial y} \dot{y} =\cos\phi \ \left(\dot{\rho}\cos\phi-\rho\dot{\phi}\sin\phi\right)+\sin\phi \ \left(\dot{\rho}\sin\phi+\rho\dot{\phi}\cos\phi\right)=\dot{\rho} [/tex]

    This is for the first comp of the transformed velocity.

    U do the other "3" (one for velocity & 2 for acceleration).

    Daniel.
     
  4. Jun 21, 2005 #3

    learningphysics

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    Cool! Thanks a bunch dexter! :biggrin:
     
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