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Homework Help: Filling in some steps in Jackson

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data
    If you have a copy of Jackson handy, this is in the middle of Section 1.7 (p.35 in my book). I'm trying to understand how he gets from Equation 1.30 to the next step (which is an unnumbered equation). Specifically, how is the term in square brackets a Taylor expansion of [tex]\rho(\textbf{x'})[/tex]? For instance, I would expect to see a term with [tex]\nabla[/tex] before the [tex]\nabla^{2}[/tex] term, and shouldn't r[tex]^{2}[/tex] be r[tex]^{3}[/tex]?


    2. Relevant equations
    They're long, it would take me forever to enter them. If you don't have a copy of Jackson handy, feel free to just skip this.


    3. The attempt at a solution
     
  2. jcsd
  3. Sep 5, 2009 #2

    gabbagabbahey

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    If you define the vector [itex]\textbf{r}=\textbf{x}-\textbf{x}'[/itex] (this is the separation vector that Griffiths calls script-r), then expanding [itex]\rho(\textbf{x}')[/itex] about the point [itex]\textbf{x}=\textbf{x}'[/itex] is the same as expanding [itex]\rho(\textbf{x}-\textbf{r})[/itex] about the point [itex]\textbf{r}=0[/itex].

    The 3D-Taylor expansion of a function is given by:

    [tex]f(\textbf{x}+\textbf{c})=\sum_{j=0}^{\infty}\left[\frac{1}{j!}\left(\textbf{c}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^jf(\overline{\textbf{x}})\right]_{\overline{\textbf{x}}=\textbf{x}}=f(\textbf{x})+\left(\textbf{c}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)f(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}+\frac{1}{2}\left(\textbf{c}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2 f(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}+\ldots[/tex]

    Where [itex]\mathbf{\nabla}_{\overline{\textbf{x}}}[/itex] means that the differentiation is done with respect to the dummy coordinates [itex]\overline{\textbf{x}}[/itex].

    So, to second order in [itex]\textbf{r}[/itex], you have:

    [tex]\rho(\textbf{x}')=\rho(\textbf{x}-\textbf{r})\approx \rho(\textbf{x})-\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}+\frac{1}{2}\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2 \rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/tex]

    I suspect that if you do the differentiation in spherical coordinates (I'd recommend using Mathematica or Matlab for this, unless you are a masochist) and then do the angular integrals (over a spherical surface), the only surviving terms will be

    [tex]\int_{0}^{\pi} \int_{0}^{2\pi}\rho(\textbf{x}')\sin\theta d\theta d\phi\approx4\pi\rho(\textbf{x})+\frac{2\pi r^2}{3}\nabla_{\overline{\textbf{x}}}^2\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/tex]
     
    Last edited: Sep 5, 2009
  4. Sep 5, 2009 #3
    Thank you, G; I'll play around with this and report back.
     
  5. Sep 5, 2009 #4

    gabbagabbahey

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    I'm not sure why Jackson uses such a complicated method to show that the integral definition of [itex]\Phi[/itex] satisfies Poisson's equation, the well known identity [itex]\nabla^2 \left(\frac{1}{\textbf{r}}\right)=-4\pi\delta^3(\textbf{r})[/itex] will yield the same result without any complicated Taylor expansions.
     
  6. Sep 5, 2009 #5

    Ben Niehoff

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    The reason is that Jackson is trying to prove that identity in the first place.
     
  7. Sep 12, 2009 #6
    I'm studying Jackson as well.

    How does he go from Eq. (1.30) to the next equation in the book:

    [tex]\nabla^2 \Phi_a\left({\textbf{x}}\right)=-\frac{1}{\epsilon_0}\int_{0}^{R} \frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}} \left[\rho\left(\textbf{x}\right)+\frac{r^2}{6}\nabla^2\rho+...\right]r^2dr+O\left(a^2\right)[/tex]
     
  8. Sep 12, 2009 #7
    I know there's a Taylor expansion. But I just can't get it worked out. Even our professor gave up on it.
     
  9. Sep 12, 2009 #8

    gabbagabbahey

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    Have you read post #2 in this thread?
     
  10. Sep 13, 2009 #9
    I'm not sure how that works out. When I do it, I get the following:

    [tex]\frac{1}{4\pi\epsilon_0}\int\rho\left(\textbf{x}'\right)\left(\frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)d^3x'[/tex]

    [tex]=\frac{1}{4\pi\epsilon_0}\int_0^\pi\int_0^{2\pi}\int_0^R\rho\left(\textbf{x}'\right)\left(\frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)r^2sin\theta dr d\phi d\theta[/tex]

    [tex]=\frac{1}{4\pi\epsilon_0}\int_0^\pi\int_0^{2\pi}\int_0^R\left(\rho\left(\textbf{x}\right)-\left(\textbf{r}\cdot\nabla\right)\rho\left(\textbf{x}\right)+\frac{1}{2}\left(\textbf{r}\cdot\nabla\right)^2\rho\left(\textbf{x}\right)+\cdot\cdot\cdot\right)\left(\frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)r^2sin\theta dr d\phi d\theta[/tex]

    [tex]=-\frac{1}{\epsilon_0}\int_0^R\left(\rho\left(\textbf{x}\right)-\left(\textbf{r}\cdot\nabla\right)\rho\left(\textbf{x}\right)+\frac{1}{2}\left(\textbf{r}\cdot\nabla\right)^2\rho\left(\textbf{x}\right)+\cdot\cdot\cdot\right)\left(\frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)r^2 dr[/tex]

    I don't see how this term drops out:

    [tex]-\left(\textbf{r}\cdot\nabla\right)\rho\left(\textbf{x}\right)[/tex]

    And I don't see how this term:

    [tex]\frac{1}{2}\left(\textbf{r}\cdot\nabla\right)^2\rho\left(\textbf{x}\right)[/tex]

    becomes:

    [tex]\frac{r^2}{6}\nabla^2\rho\left(\textbf{x}\right)[/tex]
     
  11. Sep 13, 2009 #10

    gabbagabbahey

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    Compute those derivatives in spherical coordinates and then do the angular integration. Remember, a term doesn't have to be zero to integrate to zero over some specific interval.
     
  12. Sep 14, 2009 #11
    [tex]\nabla=\hat{\textbf{r}}\frac{\partial}{\partial{r}}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial{\theta}}+\hat{\phi}\frac{1}{r\sin{\theta}}\frac{\partial}{\partial{\phi}}[/tex]

    [tex]\textbf{r}=r\hat{\textbf{r}}+\theta\hat{\theta}+\phi\hat{\phi}[/tex]

    [tex]\rho\left(\textbf{x}\right)=\rho\left(r,\theta,\phi\right)[/tex]

    Now taking into consideration just the middle term (the one that should cancel out), and ignoring the constants:

    [tex]\int_0^{2\pi}\int_0^\pi \left(\textbf{r}\cdot\nabla\right)\rho\left(\textbf{x}\right)r^2\sin\theta d\theta d\phi[/tex]

    Making the substitutions:

    [tex]\int_0^{2\pi}\int_0^\pi \left(\left(r\hat{\textbf{r}}+\theta\hat{\theta}+\phi\hat{\phi}\right)\cdot\left(\hat{\textbf{r}}\frac{\partial}{\partial{r}}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial{\theta}}+\hat{\phi}\frac{1}{r\sin{\theta}}\frac{\partial}{\partial{\phi}}\right)\right)\rho\left(\textbf{x}\right)r^2\sin\theta d\theta d\phi[/tex]

    Simplifying:

    [tex]\int_0^{2\pi}\int_0^\pi \left(r\frac{\partial}{\partial{r}}+\theta\frac{1}{r}\frac{\partial}{\partial{\theta}}+\phi\frac{1}{r\sin{\theta}}\frac{\partial}{\partial{\phi}}\right)\right)\rho\left(\textbf{x}\right)r^2\sin\theta d\theta d\phi[/tex]

    [tex]\int_0^{2\pi}\int_0^\pi \left(r\frac{\partial\rho\left(\textbf{x}\right)}{\partial{r}}+\theta\frac{1}{r}\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\theta}}+\phi\frac{1}{r\sin{\theta}}\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\phi}}\right)\right)r^2\sin\theta d\theta d\phi[/tex]

    Break that integral up:

    [tex]\int_0^{2\pi}\int_0^\pi \left(r\frac{\partial\rho\left(\textbf{x}\right)}{\partial{r}}\right)r^2\sin\theta d\theta d\phi+\int_0^{2\pi}\int_0^\pi \left(\theta\frac{1}{r}\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\theta}}\right)r^2\sin\theta d\theta d\phi+\int_0^{2\pi}\int_0^\pi \left(\phi\frac{1}{r\sin{\theta}}\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\phi}}\right)\right)r^2\sin\theta d\theta d\phi
    [/tex]

    Now I seem to be at an impasse because I don't know what the derivative are:

    [tex]\frac{\partial\rho\left(\textbf{x}\right)}{\partial{r}}=?[/tex]
    [tex]\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\theta}}=?[/tex]
    [tex]\frac{\partial\rho\left(\textbf{x}\right)}{\partial{\phi}}=?[/tex]

    I can assume the charge density is constant, which would give me zero. But that would also give zero for the third term, too, wouldn't it?
     
  13. Sep 14, 2009 #12

    gabbagabbahey

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    There are several problems here.

    First, it is very important to distinguish [itex]\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/itex] and [tex]\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/tex]from [itex]\left(\textbf{r}\cdot\nabla\right)\rho\left(\textbf{x}\right)[/itex] and [tex]\left(\textbf{r}\cdot\nabla\right)^2\rho\left(\textbf{x}\right)[/tex]. In the first two, differentiation is done with respect to the dummy variable [itex]\overline{\textbf{x}}[/itex], and so [itex]\textbf{r}[/itex] is treated as a constant vector. In the latter two, [itex]\textbf{r}[/itex] is not treated as a constant vector, and so your results will be quite different.


    Second, by defintion, [itex]\textbf{r}=r\mathbf{\hat{r}}\neq r\mathbf{\hat{r}}+\theta\mathbf{\hat{\theta}}+\phi\mathbf{\hat{\phi}}[/itex]...the units on your expression don't even make sense... [itex]r[/itex] has unit of distance, while [itex]\theta[/itex] and [itex]\phi[/itex] are angles, with units of radians.


    Third, unit vectors in curvilinear coordinates are position dependent. So, for example [itex]\textbf{r}\cdot\textbf{r}'=\left(r\mathbf{\hat{r}}\right)\cdot \left(r'\mathbf{\hat{r}'}\right)=r r' (\mathbf{\hat{r}}\cdot\mathbf{\hat{r}}') \neq rr'[/itex] in general.


    Finally, you only require that

    [tex]\int_{0}^{\pi} \int_{0}^{2\pi}\left[-\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}+\frac{1}{2}\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2 \rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}\right]\sin\theta d\theta d\phi=\frac{2\pi r^2}{3}\nabla_{\overline{\textbf{x}}}^2\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/tex]

    This does not necessarily mean that

    [tex]\int_{0}^{\pi} \int_{0}^{2\pi}\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}\sin\theta d\theta d\phi=0[/tex]

    and

    [tex]\int_{0}^{\pi} \int_{0}^{2\pi}\frac{1}{2}\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2 \rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}\right]\sin\theta d\theta d\phi=\frac{2\pi r^2}{3}\nabla_{\overline{\textbf{x}}}^2\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}[/tex]
     
  14. Sep 16, 2009 #13
    The integration variables are [tex]\theta[/tex] and [tex]\phi[/tex]. Where do they show up in the integrand?

    If nowhere, then that whole integrand is just multiplied by [tex]4\pi[/tex] as a result of the integration. And that, evidently, isn't the case.
     
  15. Sep 16, 2009 #14

    gabbagabbahey

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    The [tex]\theta[/tex] and [tex]\phi[/tex] dependence is hidden in [itex]\textbf{r}[/itex] and hence in

    [tex]\left[-\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)\rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}+\frac{1}{2}\left(\textbf{r}\cdot\mathbf{\nabla}_{\overline{\textbf{x}}}\right)^2 \rho(\overline{\textbf{x}})|_{\overline{\textbf{x}}=\textbf{x}}\right][/tex]

    as-well, since [tex]\textbf{r}=r\mathbf{\hat{r}}=r\sin\theta\cos\phi\mathbf{\hat{i}}+r\sin\theta\sin\phi\mathbf{\hat{j}}+r\cos\theta\mathbf{\hat{k}}[/tex]
     
  16. Sep 16, 2009 #15
    Is this in the Blue cover or Red cover version of Jackson's book?

    Thanks
    Matt
     
  17. Sep 16, 2009 #16
    OK, now maybe we're getting somewhere. I'll have to work on this later. I have real homework to do.

    One more thing - what is the dummy variable? What coordinates is it in? I'm assuming Cartesian.
     
  18. Sep 16, 2009 #17
    Blue
     
  19. Sep 16, 2009 #18

    gabbagabbahey

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    It is in whatever coordinate system you want it to be in...

    In spherical coordinates,

    [tex]\overline{\textbf{x}}=\overline{r}\mathbf{\hat{\overline{r}}}=\overline{r}\sin\overline{\theta}\cos\overline{\phi}\mathbf{\hat{i}}+\overline{r}\sin\overline{\theta}\sin\overline{\phi}\mathbf{\hat{j}} +\overline{r}\cos\overline{\theta}\mathbf{\hat{k}}[/tex]

    In Cartesian coordinates,

    [tex]\overline{\textbf{x}}=\overline{x}\mathbf{\hat{i}}+\overline{y}\mathbf{\hat{j}} +\overline{z}\mathbf{\hat{k}}[/tex]
     
  20. Sep 18, 2009 #19
    I would say those are both in Cartesian coordinates since the basis vectors are in Cartesian coordinates.
     
  21. Sep 26, 2009 #20
    I'll come back to equation 1.30 later, when I get time (if I ever get time).

    Here's another one from Jackson that has me a little confused: An unnumbered equation after (1.53):

    [tex]W=-\frac{\epsilon_0}{2}\int{\Phi\nabla^2\Phi d^3x}[/tex]

    Jackson says "Integration by parts leads to the result:

    [tex]W=\frac{\epsilon_0}{2}\int{|\nabla\Phi|^2d^3x}=\frac{\epsilon_0}{2}\int{|\textbf{E}|^2d^3x}[/tex] (1.54)

    Now, performing integration by parts, I use the following:

    [tex]\int{udv}=uv-\int{vdu}[/tex]

    [tex]u=\Phi[/tex]
    [tex]du=\nabla\Phi d^3x[/tex]
    [tex]dv=\nabla^2\Phi d^3x[/tex]
    [tex]v=\nabla\Phi[/tex]


    [tex]W=-\frac{\epsilon_0}{2}\int{\Phi\nabla^2\Phi d^3x}=-\frac{\epsilon_0}{2}\left(\Phi\nabla\Phi-\int{\left(\nabla\Phi\right)^2d^3 x}\right)[/tex]

    What happened to [tex]\Phi\nabla\Phi[/tex] ?
     
  22. Sep 26, 2009 #21

    Ben Niehoff

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    "Integration by parts" in three dimensions is trickier than that. You have a volume integral, which means the leftover part needs to be integrated as a surface integral over the boundary of your region of interest. i.e.,

    [tex]\int_R \Phi \nabla^2 \Phi \; d^3x = \oint_{\partial R} \Phi \nabla \Phi \cdot \hat n \; da - \int_R (\nabla \Phi)^2 \; d^3x[/tex]

    But by hypothesis, [itex]\Phi[/itex] satisfies either Dirichlet or Neumann boundary conditions, and therefore either [itex]\Phi[/itex] or [itex]\nabla\Phi \cdot \hat n[/itex] vanishes on the boundary. Hence the surface integral is zero.
     
  23. Sep 26, 2009 #22
    Thanks. I was wondering how that 3-dimensional integral was dealt with.
     
  24. Nov 7, 2009 #23
    I'm almost there....

    [tex]
    \int_{0}^{\pi} \int_{0}^{2\pi}\left[-\left(\textbf{r}\cdot\mathbf{\nabla}_{\textbf{x}'}\right)\rho(\textbf{x}')|_{\textbf{x}'=\textbf{x}}+\frac{1}{2}\left(\textbf{r}\cdot\mathbf{\nabla}_\textbf{x}'}\right)^2 \rho(\textbf{x}')|_\textbf{x}'=\textbf{x}}\right]\sin\theta d\theta d\phi=\frac{2\pi r^2}{3}\nabla_{\textbf{x}'}^2\rho(\textbf{x}')|_{\textbf{x}'=\textbf{x}}
    [/tex]

    Assuming the Nabla is only operating in the [itex]\hat{\textbf{x}}[/itex] direction, I only need to be concerned with:

    [tex]r'\mathbf{\hat{r}}=r'\sin\theta'\cos\phi'\mathbf{\hat{x}}[/tex]

    Plugging that into the equatation above....

    [tex]
    \int_{0}^{\pi} \int_{0}^{2\pi}\left[-\left(r'\sin\theta'\cos\phi'\mathbf{\nabla}_{\textbf{x}'}\right)\rho(\textbf{x}')|_{\textbf{x}'=\textbf{x}}+\frac{1}{2}\left(r'\sin\theta'\cos\phi'\mathbf{\nabla}_\textbf{x}'}\right)^2 \rho(\textbf{x}')|_\textbf{x}'=\textbf{x}}\right]\sin\theta d\theta d\phi
    [/tex]

    The first integral:

    [tex]
    -r'\int_{0}^{\pi} \int_{0}^{2\pi}sin^2\theta'cos\phi'\mathbf{\nabla}_{\textbf{x}'}\rho\left(\textbf{x}'\right)d\theta d\phi=-r'\pi^2\mathbf{\nabla}_\textbf{x}\rho\left(\textbf{x}\right)[/tex]

    The second integral:

    [tex]\int_{0}^{2\pi}\frac{1}{2}\left(r'\sin^3\theta'\cos^2\phi'\mathbf{\hat{x}}\mathbf{\nabla}_{\textbf{x}'}}^2\right) \rho(\textbf{x}')} d\theta d\phi=\pi r^2\frac{2}{3}\mathbf{\nabla}_\textbf{x}^2\rho(\textbf{x})}[/tex]

    So it looks like that second integral does indeed give me what I'm looking for (assuming I did the math correctly). But I have a non-zero term from the first integral. :confused:
     
  25. Nov 8, 2009 #24

    gabbagabbahey

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    Why on earth would you assume this?:confused:

    The subscript [itex]\textbf{x}'[/itex] in [tex]\mathbf{\nabla}_{\textbf{x}'}[/tex] just means that the differentiation is done with respect to the primed coordinates....In Cartesians,

    [tex]\mathbf{\nabla}_{\textbf{x}'}=\mathbf{\hat{x}}\frac{\partial}{\partial x'}+\mathbf{\hat{y}}\frac{\partial}{\partial y'}+\mathbf{\hat{z}}\frac{\partial}{\partial z'}[/tex]
     
  26. Jan 4, 2011 #25
    I think Bill Foster and my problem lies before we carry out the angular integration. We know,

    [tex]{\nabla ^2}{\Phi _a}({\bf{x}}) = - \frac{1}{{4\pi {\varepsilon _0}}}\int {\rho ({\bf{x'}})\left( {\frac{{3{a^2}}}{{{{({r^2} + {a^2})}^{5/2}}}}} \right) \cdot {d^3}{\bf{x'}}} [/tex]

    Then, we reach a snag when Jackson claims, at the very least, for [tex]{d^3}{\bf{x'}} = {{r}^2} \cdot dr \cdot d\theta ' \cdot \sin \theta ' \cdot d\phi '[/tex], (I primed the angles, but left the "r" unprimed to match Jackson's notation, even so they're "counting a la integral" rather than serving as an argument of the function [itex]{\Phi _a}({\bf{x}})[/itex]) that,

    [tex]\int_0^R {\int_0^{\pi /2} {\int_0^\pi {\rho ({\bf{x'}}) \cdot {d^3}{\bf{x'}}} } } = \int_0^R {\left( {\rho (r,\theta ',\phi ') + {\textstyle{1 \over 6}}{r^2}{{\vec \nabla }^2}\rho + ...} \right) \cdot 4\pi {r^2} \cdot dr} [/tex]

    ... I think we're all confused on this equation, above.

    Is it not true that no derivatives come into what the coefficients of the expansion are, such as 1/2, 1/6, etc.? Or do the angular integrals cause 1/2 to go to 1/6? Is this the "theta and phi dependence buried in r" that was discussed before?
     
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