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Nabla problem -- Gradient

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate [tex]\nabla e^{i\vec{k}\cdot \vec{r}}[/tex]

    2. Relevant equations
    [tex]\nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r} [/tex]


    3. The attempt at a solution
    I have a problem. I know result
    [tex]=\nabla e^{i\vec{k}\cdot \vec{r}}=i\vec{k} e^{i\vec{k}\cdot \vec{r}}[/tex]
     
    Last edited: Nov 3, 2014
  2. jcsd
  3. Nov 3, 2014 #2

    RUber

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    This problem is much clearer when broken up componentwise:
    ##e^{i \vec k \cdot \vec r } = e^{i (k_x x + k_y y + k_z z) } ##
    Where ##\vec k = k_x \hat x + k_y \hat y + k_z \hat z ## and ##\vec r = x \hat x + y \hat y + z \hat z##
     
  4. Nov 4, 2014 #3
    Yes but for that way we need a lot of time. Perhaps
    [tex]\nabla=\sum_w\vec{e}_w\frac{\partial}{\partial x_w}[/tex]
    [tex]e^{i \vec{k}\cdot \vec{r}}=e^{i\sum_q k_q x_q} [/tex]
    but I get a problem with this sums. Maybe
    [tex]\sum_w\vec{e}_w\frac{\partial}{\partial x_w}e^{i\sum_q k_q x_q}=[/tex]
    [tex]=\sum_w\vec{e}_we^{i\sum_q k_q x_q}(\frac{\partial}{\partial x_w}i\sum_q k_q x_q)=[/tex]
    [tex]=e^{i\vec{k}\cdot \vec{r}}i \sum_w \vec{e}_w k_{w}=[/tex]
    [tex]=i\vec{k}e^{i\vec{k}\cdot \vec{r}}[/tex]

    but again I did not use theorem
    [tex]\nabla f(r)=\frac{df}{dr}\nabla r[/tex]
     
  5. Nov 4, 2014 #4

    RUber

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    let ##f(\vec r) = e^{ik_x \hat x +ik_y \hat y+ik_z \hat z}## and ##\vec r = ik_x \hat x +ik_y \hat y+ik_z \hat z = i \vec k \cdot \vec r##
    then use the theorem.
     
  6. Nov 4, 2014 #5
    I think that "relevant equation",
    [itex]\nabla f(r)=\frac{df}{dr}\nabla r=\frac{df}{dr}\frac{\vec{r}}{r}[/itex],
    is not that relevant here:
    [itex]e^{i\vec{k}\cdot \vec{r}}\neq f(r)[/itex] since [itex]r = (x^2 + y^2 + z^2)^{1/2}[/itex].

    Rather,
    [itex]\nabla f(g(\vec r))=\frac{df}{dg}\nabla g(\vec r)[/itex]
    with
    [itex] f(g) = e^g [/itex] and [itex] g(\vec r) = i \vec k \cdot \vec r[/itex].
     
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