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Homework Help: Gradient question

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    This is not a hw problem, just a question
    [tex]\nabla[/tex](A.B) = (B.[tex]\nabla[/tex]) A +(A.[tex]\nabla[/tex])B+Bx([tex]\nabla[/tex]xA)+Ax([tex]\nabla[/tex]xB)

    A,B are vectors
    2. Relevant equations

    3. The attempt at a solution

    I can't make sense of the first 2 terms on the right hand side - is (B.[tex]\nabla[/tex])
    just div of B?

    Also, how do I solve, [tex]\nabla[/tex](e-xr-2 [tex]\widehat{r}[/tex])
    Can I treat the unit vector [tex]\widehat{r}[/tex] as constant?
  2. jcsd
  3. Mar 22, 2010 #2


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    No, it's more complicated than that. In Cartesian Coordinates, the Del operator is

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


    [tex]\textbf{B}\cdot\mathbf{\nabla}=B_x\frac{\partial}{\partial x}+B_y\frac{\partial}{\partial y}+B_x\frac{\partial}{\partial z}[/tex]

    And so,

    [tex](\textbf{B}\cdot\mathbf{\nabla})\textbf{A}=B_x\frac{\partial \textbf{A}}{\partial x}+B_y\frac{\partial \textbf{A}}{\partial y}+B_x\frac{\partial \textbf{A}}{\partial z}[/tex]

    The gradient of a vector is a second rank tensor. Is this really what you are trying to calculate? What is the original problem?
  4. Mar 23, 2010 #3
    The original problem is :

    Derive the Vector identity:

    (Ji.[tex]\nabla'[/tex])[tex]\nabla'[/tex](e-[tex]\gamma[/tex]r/r) =

    [[tex]\gamma[/tex]2((Ji.[tex]\hat{r}[/tex])[tex]\hat{r}[/tex]+3/r([tex]\gamma[/tex]+1/r)(Ji.[tex]\hat{r}[/tex])[tex]\hat{r}[/tex] - Ji/r([tex]\gamma[/tex]+1/r)]e-[tex]\gamma[/tex]r/r

    I solved
    [tex]\nabla'[/tex] (e-[tex]\gamma[/tex]r/r) and got

    ([tex]\gamma[/tex]+1/r) (e-[tex]\gamma[/tex]r/r2) [tex]\hat{r}[/tex]
    Last edited: Mar 23, 2010
  5. Mar 23, 2010 #4


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    I'm having a difficult time reading your expression...

    [tex](\textbf{J}\cdot\mathbf{\nabla})\left(\mathbf{\nabla}\frac{e^{-\gamma r}}{r}\right)=\left[\gamma^2(\textbf{J}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}+\frac{3}{r}\left(\gamma+\frac{1}{r}\right)(\textbf{J}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}}-\frac{1}{r}\left(\gamma+\frac{1}{r}\right)\textbf{J}\right]\frac{e^{-\gamma r}}{r}[/tex]

    ^^^ Is this what you meant? I assume that [itex]\textbf{r}[/itex] is just the usual position vector, [itex]r[/itex] is its magnitude and [tex]\hat{\mathbf{r}}[/itex] is its direction? If so, why does your expression have primes next to the nablas?

    Again, I can only assume that you mean

    [tex]\mathbf{\nabla}\left(\frac{e^{-\gamma r}}{r}\right)=-\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}[/tex]

    If so, then yes, that's correct.

    Are you still having difficulty carrying out the derivative [tex](\textbf{J}\cdot\mathbf{\nabla})[/tex] on this expression?
  6. Mar 23, 2010 #5
    The expressions do have primes next to nabla. The nabla operates on primed coordinates. Source is at the primed coordinate and effect is at the unprimed coordinates.

    Yes, the difficulty i was having was with the J.nabla
    Also, say if you try to take the gradient of
    [tex]-\left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}
    how would you deal with direction vector r^
  7. Mar 23, 2010 #6


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    Surely this means that [itex]\textbf{r}[/itex] isn't the position vector, but rather the separation vector between the position vectors of the field and source points;



    First, the negative sign shouldn't be there if my above assumption is correct.

    Second, you aren't taking the gradient of that (if you did, you would end up with a second rank tensor, not a vector). To take the partial derivative of something like [tex]f(r)\hat{\mathbf{r}}[/tex], you will have to use the product and chain rules. For example,

    [tex]\begin{aligned}\frac{\partial}{\partial x'} \left[f(r)\hat{\mathbf{r}}\right] &= \frac{\partial f}{\partial x'}\hat{\mathbf{r}}+f(r)\frac{\partial \hat{\mathbf{r}}}{\partial x'} \\ &= \left(\frac{\partial f}{\partial r}\right)\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}+f(r)\frac{\partial}{\partial x'}\left[\frac{(x-x')\hat{\mathbf{x}}+(y-y')\hat{\mathbf{y}}+(z-z')\hat{\mathbf{z}}}{r}\right] \\ &= f'(r)\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}+f(r)\left[\frac{-1}{r}\left(\frac{\partial r}{\partial x'}\right)\hat{\mathbf{r}}-\frac{1}{r}\hat{\mathbf{x}}\right]\end{aligned}[/tex]
  8. Mar 23, 2010 #7
    Yes, you are right

    and also no -ve sign in the expression
    \left(\gamma+\frac{1}{r}\right)\frac{e^{-\gamma r}}{r^2}\hat{\mathbf{r}}


    Thanks for explaining the second rank tensor expression. I thought of the product rule, but then didn't pursue it coz I couldn't figure out how to evaluate d(r hat)/dt
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