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Derive grad T in spherical coordinates

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data

    ##x=r\sin\theta\cos\phi,\,\,\,\,\,y=r\sin\theta\sin\phi,\,\,\,\,\,z=r\cos\theta##

    ##\hat{x}=\sin\theta\cos\phi\,\hat{r}+\cos\theta\cos\phi\,\hat{\theta}-\sin\phi\,\hat{\phi}##
    ##\hat{y}=\sin\theta\sin\phi\,\hat{r}+\cos\theta\sin\phi\,\hat{\theta}+\cos\phi\,\hat{\phi}##
    ##\hat{z}=\cos\theta\,\hat{r}-\sin\theta\,\hat{\theta}##

    2. Relevant equations
    By chain rule,

    ##\frac{\partial T}{\partial x}=\frac{\partial T}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial T}{\partial\theta}\frac{\partial\theta}{\partial x}+\frac{\partial T}{\partial\phi}\frac{\partial\phi}{\partial x}##
    ##\frac{\partial T}{\partial y}=\frac{\partial T}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial T}{\partial\theta}\frac{\partial\theta}{\partial y}+\frac{\partial T}{\partial\phi}\frac{\partial\phi}{\partial y}##
    ##\frac{\partial T}{\partial z}=\frac{\partial T}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial T}{\partial\theta}\frac{\partial\theta}{\partial z}+\frac{\partial T}{\partial\phi}\frac{\partial\phi}{\partial z}##

    ##\nabla T=\frac{\partial T}{\partial x}\hat{x}+\frac{\partial T}{\partial y}\hat{y}+\frac{\partial T}{\partial z}\hat{z}##

    3. The attempt at a solution

    ##\nabla T=(\frac{\partial T}{\partial r}\frac{1}{\sin\theta\cos\phi}+\frac{\partial T}{\partial\theta}\frac{1}{r\cos\theta\cos\phi}+\frac{\partial T}{\partial\phi}\frac{-1}{r\sin\theta\sin\phi})\hat{x}+(\frac{\partial T}{\partial r}\frac{1}{\sin\theta\sin\phi}+\frac{\partial T}{\partial\theta}\frac{1}{r\cos\theta\cos\phi}+\frac{\partial T}{\partial\phi}\frac{1}{r\sin\theta\sin\phi})\hat{y}##
    ##+(\frac{\partial T}{\partial r}\frac{1}{\cos\theta}+\frac{\partial T}{\partial\theta}\frac{-1}{r\sin\theta}+\frac{\partial T}{\partial\phi}0)\hat{z}##

    Just by looking at the coefficient of ##\hat{r}##, we get

    ##\frac{\partial T}{\partial r}+\frac{\partial T}{\partial\theta}\frac{\tan\theta}{r}-\frac{\partial T}{\partial\phi}\frac{1}{r\tan\theta}+\frac{\partial T}{\partial r}+\frac{\partial T}{\partial\theta}\frac{\tan\theta}{r}+\frac{\partial T}{\partial\phi}\frac{\tan\theta}{r}+\frac{\partial T}{\partial r}-\frac{\partial T}{\partial\theta}\frac{1}{r\tan\theta}##,

    which is clearly not correct, since

    ##\nabla T=\frac{\partial T}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial T}{\partial\theta}\hat{\theta}+\frac{1}{r\sin\theta}\frac{\partial T}{\partial\phi}\hat{\phi}##,

    the coefficient of ##\hat{r}## should be just ##\frac{\partial T}{\partial r}##.
     
    Last edited: Apr 19, 2015
  2. jcsd
  3. Apr 19, 2015 #2

    vela

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    You should rethink some of those derivatives. For example, you have ##r = \sqrt{x^2+y^2+z^2}##, so
    $$\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2+y^2+z^2}} = \frac{r \sin\theta \cos\phi}{r} = \sin\theta\cos\phi.$$ You somehow got the reciprocal.
     
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