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Gradient problem

  1. Jan 4, 2015 #1
    1. The problem statement, all variables and given/known data
    image024.gif

    Solving part (c) which should be
    [itex]\overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r[/itex]

    2. Relevant equations


    Let [itex]\nabla=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}[/itex]

    and [itex] \overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}[/itex]
    [itex] r = \mid r\mid=\sqrt{x^{2}+y^{2}+z^{2}}[/itex]

    3. The attempt at a solution

    Consider left side of the inequality.

    Now [itex]\nabla.\overrightarrow{r}=
    (\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}
    ).\left(x\hat{i}+y\hat{j}+z\hat{k}\right)=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3[/itex]

    L.H.S. [itex]=\overrightarrow{r}.(\nabla.\overrightarrow{r})[/itex]

    L.H.S. [itex]= \left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right).3=3\overrightarrow{r}[/itex]


    Now consider right side of the inequality.

    [itex]\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}\right)[/itex]

    [itex]\left(r\nabla\right) =\hat{i}\frac{\partial}{\partial x}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{j}\frac{\partial}{\partial y}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}+\hat{k}\frac{\partial}{\partial z}\left(x^{2}+y^{2}+z^{2}\right)^{\frac{1}{2}}[/itex]

    [itex]\left(r\nabla\right)=\hat{i}\frac{2x}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}\frac{2y}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}\frac{2z}{2}\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

    [itex]\left(r\nabla\right)=
    \hat{i}x\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{j}y\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}+\hat{k}z\left(x^{2}+y^{2}+z^{2}\right)^{-\frac{1}{2}}[/itex]

    [itex]\left(r\nabla\right)=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}[/itex]

    R.H.S. [itex]=r\left(r\nabla\right)=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^{2}+y^{2}+z^{2}}}=\overrightarrow{r}[/itex]

    Hence L.H.S. [itex]\neq[/itex] R.H.S.
     
  2. jcsd
  3. Jan 4, 2015 #2

    vela

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    That still can't be right. See below.

    Your last line is incorrect. You found ##\nabla\cdot\vec{r} = 3##, so you should have ##\vec{r}\cdot(\nabla\cdot\vec{r}) = \vec{r}\cdot 3##, which doesn't make sense because you can't dot a vector with a scalar.


    You can't change the order of ##r## and ##\nabla##. The expression on the RHS is equal to ##\nabla r##, not ##r \nabla##.

    What you actually calculated is ##\nabla r = \hat{r}##. It's not for nothing, however, as you need this to correctly evaluate the righthand side.

    The RHS you started with was ##r\nabla r##. At the end, you incorrectly say it's ##r(r\nabla)##.
     
  4. Jan 5, 2015 #3
    Thx Vela! I'll give it another try.

    There is a misprint in the problem statement.
    image024.gif
    I think (c) part should be [itex] \overrightarrow{r}.(\nabla.\overrightarrow{r)}\neq\left(r\nabla\right)r [/itex]
    Are you OK with it?
     
  5. Jan 5, 2015 #4

    vela

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    No.
     
  6. Jan 5, 2015 #5

    haruspex

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    In the photocopied text, all occurrences of 'r' are in bold, implying vectors.
     
  7. Jan 5, 2015 #6

    BvU

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    In your chapter 1 in the Chow book, under "Formulas involving ##\nabla## " , you find ##\nabla\cdot {\bf r} = 3## and ##({\bf A}\cdot \nabla){\bf r} = {\bf A}## for A any differentiable vector field function. Substituting ##{\bf A} = {\bf r}## gives ##({\bf r}\cdot \nabla){\bf r} = {\bf r}##.

    ##{\bf r} \cdot (\nabla\cdot {\bf r}) ## would be a vector dotted into a scalar. Doesn't make sense.

    And in ##({\bf r} \nabla){\bf r} = {\bf r}## the ##({\bf r} \nabla){\bf r}## (without the ##\cdot## ) could be a matrix for all we know.

    As in 1.18(b), mr Chow seems to be a bit sloppy, or at least inconsistent with his notation. Best I can make of it is that he wants you to prove that
    ##{\bf r} \; (\nabla\cdot {\bf r}) \ne ({\bf r} \cdot \nabla)\;{\bf r} ##, which by now should be a piece of cake for you :)
     
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