1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook