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Homework Help: Nabla Operator Question

  1. Jun 13, 2009 #1
    Does my solution look correct to you guys?

    1. The problem statement, all variables and given/known data

    [tex]\nabla \varphi (r)[/tex]

    [tex]\varphi (r) = \frac{1}{4\pi\epsilon_{0}}\frac{1}{r}[/tex]

    with: [tex]r = \sqrt{x^{2}+y^{2}+z^{2}}[/tex]

    2. Relevant equations


    3. The attempt at a solution

  2. jcsd
  3. Jun 13, 2009 #2


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    You used the divergence not the gradient. Remember taking the gradient of a scalar field "creates" a vector.
  4. Jun 13, 2009 #3
    Is this better? I can't figure out how to simplify it further.

    Last edited: Jun 13, 2009
  5. Jun 13, 2009 #4


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    That is correct, but you can simplify it much further. What is [itex]x^2+y^2+z^2[/itex]? What is [itex]x \hat{e_1}+y\hat{e_2}+z \hat{e_3}[/itex]?

    Hint: [itex]\vec{r}=?[/itex]
  6. Jun 13, 2009 #5
    It is the derivative w.r.t. r times r-hat, because if you change r by dr, the step length in the r-hat direction is dr.

    By definition, you have for a differentiable function f(x1,x2,...) that:

    df = nabla f dot ds (1)

    where ds is the displacement vector.

    You also have that:

    df = df/dx1 dx1 + df/dx2 dx2 + df/dx3 dx3 + ... (2)

    In case of Cartesian coordinates, ds = (dx1, dx2,...), but in general this is not the case.
  7. Jun 13, 2009 #6
    [tex]r^{2}[/tex] = [tex]x^{2}+y^{2}+z^{2}[/tex]?

    [itex]- \hat{r} = -x \hat{e_1}-y\hat{e_2}-z \hat{e_3}[/itex]?

    So final answer:
    [tex]\frac{-\hat{r}}{r^{5/2}}[/tex] ?
    Last edited: Jun 13, 2009
  8. Jun 13, 2009 #7


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    No, [itex]|\hat{r}|=1[/itex] yet we know that [itex]|x \hat{e_1}+y\hat{e_2}+z \hat{e_3}| \neq 1[/itex]. Again, do you know what [itex]\vec{r}[/itex] looks like in Cartesian coordinates?
  9. Jun 13, 2009 #8
    I'm thinking hard but I'm not sure..

    I'm thinking something along the lines of.. the x-component of [tex]\hat{r}[/tex].. the y-component. or something? >_<
  10. Jun 13, 2009 #9


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    You're getting close. We are given that [itex]|\vec{r}|=\sqrt{x^2+y^2+z^2}[/itex]=r. So we want to find s an expression for [itex]\vec{r}[/itex] whose length is r and points in the [itex]\hat{r}[/itex] direction.
  11. Jun 14, 2009 #10
    I don't understand the differences between:

    [tex]\vec{r},\hat{r}[/tex] and [tex]r[/tex]!!

  12. Jun 14, 2009 #11


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    A vector has a direction and a magnitude. A unit vector is a vector with magnitude 1. For example [itex]\hat{e_1}[/itex] is the unit vector pointing in the x-direction with [itex]|\hat{e_1}|=1[/itex]. These are used in Cartesian coordinates and I am sure you're familiar with them. In your problem however we are working with spherical coordinates. the unit vectors for spherical coordinates are [itex]\hat{r},\hat{\theta}, \hat{\phi}[/itex]. So whereas you express a vector in Cartesian coordinates as [itex]a \hat{e_1}+b\hat{e_2}+c \hat{e_3}[/itex] in spherical coordinates you would express it as [itex]\alpha \hat{r}+\beta \hat{\theta}+\gamma \hat{\phi}[/itex].

    In spherical coordinates the length of a vector [itex]\vec{r}[/itex] is given by [itex]|\vec{r}|=\sqrt{x^2+y^2+z^2}=r[/itex]. Therefore r is the magnitude of [itex]\vec{r}[/itex]. A vector also has a direction and the direction of [itex]\vec{r}[/itex] is given by[itex]\hat{r}[/itex].

    It is given in the problem statement that [itex]|\vec{r}|=\sqrt{x^2+y^2+z^2}[/itex] and we have a vector,[itex]x \hat{e_1}+y\hat{e_2}+z \hat{e_3}[/itex] in Cartesian coordinate which we would like to write as a vector in spherical coordinates. If you draw a Cartesian coordinate system and draw a radial vector on the axes, then how can you express this radial vector in terms of x,y,z? Compare it to the vector we're trying to write in spherical coordinates, also compare the length of [itex] x \hat{e_1}+y\hat{e_2}+z \hat{e_3}[/itex] with the length of [itex]\vec{r}[/itex].

    Lastly you also need to write [itex](x^2+y^2+z^2)^{-3/2}[/itex] in terms of r correctly, it is not [itex]r^{-5/2}[/itex].
    Last edited: Jun 14, 2009
  13. Jun 14, 2009 #12


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    This is almost there... if you fix your confusion in how to notate each kind of vector, you only have to fix a minor arithmetical error
  14. Jun 14, 2009 #13


    so it's the negative r unit vector pointing in the r direction.. over r cubed.. right? =(

    I seem to have a huge knowledge hole about vectors
  15. Jun 14, 2009 #14


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    It is the correct answer, but how did you get to it?

    No it is the unit vector pointing in the [itex]-\hat{r}[/itex] direction multiplied by the scalar r. That means it is a vector pointing radially towards the origin of the coordinate system and has a length r.
    Last edited: Jun 14, 2009
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