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Metric tensor and gradient in spherical polar coordinates

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data

    Let ##x##, ##y##, and ##z## be the usual cartesian coordinates in ##\mathbb{R}^{3}## and let ##u^{1} = r##, ##u^{2} = \theta## (colatitude), and ##u^{3} = \phi## be spherical coordinates.

    1. Compute the metric tensor components for the spherical coordinates ##g_{r\theta}:=g_{12}=\big\langle\frac{\partial}{\partial r}\ ,\ \frac{\partial}{\partial \theta}\big\rangle\ \text{etc.}##
    2. Compute the coefficients ##(\nabla\ f)^{j}## in ##\nabla\ f = (\nabla\ f)^{r}\frac{\partial}{\partial r}+(\nabla\ f)^{\theta}\frac{\partial}{\partial \theta}+(\nabla\ f)^{\phi}\frac{\partial}{\partial \phi}##.
    3. Verify that ##\frac{\partial}{\partial r}##, ##\frac{\partial}{\partial\theta}##, and ##\frac{\partial}{\partial\phi}## are orthogonal, but that not all are unit vectors. Define the unit vectors ##{\bf{e}}'_{j}=\frac{(\partial / \partial u^{j})}{||\partial / \partial u^{j}||}## and write ##\nabla\ f## in terms of this orthonormal set ##\nabla\ f = (\nabla\ f)'^{r}{\bf{e}}'_{r}+(\nabla\ f)'^{\theta}{\bf{e}}'_{\theta}+(\nabla\ f)'^{\phi}{\bf{e}}'_{\phi}##.
    2. Relevant equations

    3. The attempt at a solution

    ## g_{r\theta}:=g_{12}=\big\langle\frac{\partial}{\partial r}\ ,\ \frac{\partial}{\partial \theta}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial r}\frac{\partial x}{\partial \theta}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial \theta}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial \theta}\\
    =(\text{sin}\ \theta\ \text{cos}\ \phi)(r\ \text{cos}\ \theta)(\text{cos}\ \theta)+(\text{sin}\ \theta\ \text{sin}\ \phi)(r\ \text{cos}\ \theta)(\text{sin}\ \phi)+(\text{cos}\ \theta)(-r\ \text{sin}\ \theta)\\
    =0##

    Am I correct so far?
     
    Last edited: Sep 23, 2016
  2. jcsd
  3. Sep 23, 2016 #2

    Orodruin

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    Yes
     
  4. Sep 23, 2016 #3
    Ok, so now let me do the remaining calculations in parts 1 and 2.

    ## g_{r\phi}:=g_{13}=\big\langle\frac{\partial}{\partial r}\ ,\ \frac{\partial}{\partial \phi}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial r}\frac{\partial x}{\partial \phi}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial \phi}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial \phi}\\
    =(\text{sin}\ \theta\ \text{cos}\ \phi)(-r\ \text{sin}\ \theta\ \text{sin}\ \phi)+(\text{sin}\ \theta\ \text{sin}\ \phi)(r\ \text{sin}\ \theta\ \text{cos}\ \phi)+(\text{cos}\ \theta)(0)\\
    =0
    ##

    ##
    g_{\theta\phi}:=g_{23}=\big\langle\frac{\partial}{\partial \theta}\ ,\ \frac{\partial}{\partial \phi}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \phi}+\frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \phi}+\frac{\partial z}{\partial \theta}\frac{\partial z}{\partial \phi}\\
    =(r\ \text{cos}\ \theta\ \text{cos}\ \phi)(-r\ \text{sin}\ \theta\text{sin}\ \phi)+(r\ \text{cos}\ \theta\ \text{sin}\ \phi)(r\ \text{sin}\ \theta\ \text{cos}\ \phi)+(-r\ \text{sin}\ \theta)(0)\\
    =0
    ##




    2

    ## ({\nabla f })^{r}=g_{rr}=:=g_{11}=\big\langle\frac{\partial}{\partial r}\ ,\ \frac{\partial}{\partial r}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial r}\frac{\partial x}{\partial r}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial r}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial r}\\
    =(\text{sin}\ \theta\ \text{cos}\ \phi)^{2}+(\text{sin}\ \theta\ \text{sin}\ \phi)^{2}+(\text{cos}\ \theta)^{2}\\
    =1
    ##
    ## ({\nabla f })^{\theta}=g_{\theta\theta}=:=g_{22}=\big\langle\frac{\partial}{\partial \theta}\ ,\ \frac{\partial}{\partial \theta}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \theta}+\frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \theta}+\frac{\partial z}{\partial \theta}\frac{\partial z}{\partial \theta}\\
    =(r\ \text{cos}\ \theta\ \text{cos}\ \phi)^{2}+(r\ \text{cos}\ \theta\ \text{sin}\ \phi)^{2}+(-r\ \text{sin}\ \theta)^{2}\\
    =r^{2}
    ##
    ## ({\nabla f })^{\phi}=g_{\phi\phi}=:=g_{22}=\big\langle\frac{\partial}{\partial \phi}\ ,\ \frac{\partial}{\partial \phi}\big\rangle\\
    = \big\langle\frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\big\rangle\\
    = \frac{\partial x}{\partial \phi}\frac{\partial x}{\partial \phi}+\frac{\partial y}{\partial \phi}\frac{\partial y}{\partial \phi}+\frac{\partial z}{\partial \phi}\frac{\partial z}{\partial \phi}\\
    =(-r\ \text{sin}\ \theta\ \text{sin}\ \phi)^{2}+(r\ \text{sin}\ \theta\ \text{cos}\ \phi)^{2}+(0)^{2}\\
    =r^{2}\text{sin}^{2}\theta
    ##

    Are they correct?
     
  5. Sep 23, 2016 #4

    Orodruin

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    What is ##f##? Also, normally, the gradient is a covariant vector. Your notation is therefore not very clear to me.
     
  6. Sep 23, 2016 #5
    I mention the definition below:

    If ##M^{n}## is a (psuedo)-Riemannian manifold and ##f## is a differentiable function, the gradient vector ##\nabla f## is the contravariant vector associated to the covector ##df## such that ##df({\bf{w}})=\langle\nabla f, {\bf{w}}\rangle## for some vector ##\bf{w}##.

    Now, ##\displaystyle{df = \frac{\partial f}{\partial x^{j}}dx^{j}}## and ##\nabla f## is the contravariant vector associated to the covector ##df## so that ##\displaystyle{(\nabla f)^{i} = g^{ij}\frac{\partial f}{\partial x^{j}}}##.
     
  7. Sep 23, 2016 #6

    Orodruin

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    You still have not defined what ##f## is. Based on your problem statement, it is not necessarily the coordinate functions
     
  8. Sep 23, 2016 #7
    Thanks for pointing that out. I now understand my mistake in my proposed solution.

    I think the function ##f## has to be kept untouched. Using the definition ##\displaystyle{(\nabla f)^{i} = g^{ij}\frac{\partial f}{\partial x^{j}}}##,

    ## ({\nabla f})^{r}=g^{rr}\frac{\partial f}{\partial r}=(g_{rr})^{-1}\frac{\partial f}{\partial r}=\big\langle\frac{\partial}{\partial r}\ ,\ \frac{\partial}{\partial r}\big\rangle ^{-1}\frac{\partial f}{\partial r}\\
    = \big\langle\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}\big\rangle ^{-1}\frac{\partial f}{\partial r}\\
    = (\frac{\partial x}{\partial r}\frac{\partial x}{\partial r}+\frac{\partial y}{\partial r}\frac{\partial y}{\partial r}+\frac{\partial z}{\partial r}\frac{\partial z}{\partial r}) ^{-1}\frac{\partial f}{\partial r}\\
    =[(\text{sin}\ \theta\ \text{cos}\ \phi)^{2}+(\text{sin}\ \theta\ \text{sin}\ \phi)^{2}+(\text{cos}\ \theta)^{2}] ^{-1}\frac{\partial f}{\partial r}\\
    =\frac{\partial f}{\partial r}
    ##

    ## ({\nabla f })^{\theta}=g^{\theta\theta}=(g_{\theta\theta}) ^{-1}\frac{\partial f}{\partial \theta}=\big\langle\frac{\partial}{\partial \theta}\ ,\ \frac{\partial}{\partial \theta}\big\rangle ^{-1}\frac{\partial f}{\partial \theta}\\
    = \big\langle\frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}\big\rangle ^{-1}\frac{\partial f}{\partial \theta}\\
    = (\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \theta}+\frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \theta}+\frac{\partial z}{\partial \theta}\frac{\partial z}{\partial \theta}) ^{-1}\frac{\partial f}{\partial \theta}\\
    =[(r\ \text{cos}\ \theta\ \text{cos}\ \phi)^{2}+(r\ \text{cos}\ \theta\ \text{sin}\ \phi)^{2}+(-r\ \text{sin}\ \theta)^{2}] ^{-1}\frac{\partial f}{\partial \theta}\\
    =\frac{1}{r^{2}}\frac{\partial f}{\partial \theta}
    ##
    ## ({\nabla f })^{\phi}=g^{\phi\phi}=(g_{\phi\phi}) ^{-1}\frac{\partial f}{\partial\phi}=\big\langle\frac{\partial}{\partial \phi}\ ,\ \frac{\partial}{\partial \phi}\big\rangle ^{-1}\frac{\partial f}{\partial\phi}\\
    = \big\langle\frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\ ,\ \frac{\partial x}{\partial \phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial \phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial \phi}\frac{\partial}{\partial z}\big\rangle ^{-1}\frac{\partial f}{\partial\phi}\\
    = (\frac{\partial x}{\partial \phi}\frac{\partial x}{\partial \phi}+\frac{\partial y}{\partial \phi}\frac{\partial y}{\partial \phi}+\frac{\partial z}{\partial \phi}\frac{\partial z}{\partial \phi}) ^{-1}\frac{\partial f}{\partial\phi}\\
    =[(-r\ \text{sin}\ \theta\ \text{sin}\ \phi)^{2}+(r\ \text{sin}\ \theta\ \text{cos}\ \phi)^{2}+(0)^{2}] ^{-1}\frac{\partial f}{\partial\phi}\\
    =\frac{1}{r^{2}\text{sin}^{2}\theta}\frac{\partial f}{\partial\phi}##
     
  9. Sep 23, 2016 #8

    Orodruin

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    Note that it is not always true that ##g^{ij} = (g_{ij})^{-1}##. This is only true if ##\partial_r## is orthogonal to the other basis vectors. However, this is the case here.
     
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