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A Thread about Jackson's Classical Electrodynamics 3rd edition

  1. Apr 6, 2017 #1

    MathematicalPhysicist

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    In this thread I gather my questions concerning derivations in the textbook of equations and of solutions to the exercises.

    I hope every student and professional will benefit from this thread.

    I'll start in the next post.
     
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  3. Apr 6, 2017 #2

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    On page 35, how do you get the Taylor expansion with the laplacian from the usual multivariable Taylor expansion?

    I.e of ##\rho(\vec{x'})=\rho(\vec{x})+r^2/6 \nabla^2 \rho(\vec{x})+\ldots##, and why there's no term with first order derivatives? where did it disappear?
    where ##|\vec{x}-\vec{x'}|=r##.
     
  4. Apr 6, 2017 #3

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  5. Apr 6, 2017 #4

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    Ok, I understand why the quadratic term appears the way it does, but not why the linear term with the first order derivatives disappear over there.
     
  6. Apr 6, 2017 #5

    dextercioby

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    But it's written there why. The Taylor expansion includes the linear term, but it's under an integral over R3. It's a triple improper integral which has an antisymmetric integrand under a symmetric domain. It vanishes.
     

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  7. Apr 6, 2017 #6

    vanhees71

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    Admittedly the calculation is a bit confusing, because of the shorthand notation used by Jackson. Calculated is the volume integral
    $$\Delta \phi_a(\vec{x})=-\frac{1}{4 \pi \epsilon_0} \int_{K_R(\vec{x})} \mathrm{d}^3 \vec{x}' \frac{3 a^2}{[(\vec{x}'-\vec{x})]+a^2]^{5/2}} \rho(\vec{x}'),$$
    where ##K_R(\vec{x})## is the sphere of radius ##R## around ##\vec{x}##. Substitute
    $$\vec{x}'=\vec{x}+\vec{r}.$$
    Then ##\vec{r}## runs over the sphere of radius ##R##, ##K_R##, around the origin, and we have
    $$\Delta \phi_a(\vec{x}) =-\frac{1}{4 \pi \epsilon_0} \int_{K_R} \mathrm{d}^3 \vec{r} \frac{3a^2}{(r^2+a^2)^{5/2}} \rho(\vec{x}+\vec{r}).$$
    Now you expand the charge distribution around ##\vec{r}=0##:
    $$\rho(\vec{x}+\vec{r})=\sum_{n=0}^{\infty} \frac{r_{j_1} \cdots r_{j_n}}{n!} \partial_{j_1} \cdots \partial_{j_n} \rho(\vec{x}).$$
    Now the contribution the integral for ##n=0## obviously is
    $$\Delta \phi_a(\vec{x})=-\frac{1}{4 \pi \epsilon_0} \rho(\vec{x}) 4 \pi \int_0^{R} \mathrm{d} r \frac{3a^2 r^2}{(r^2+a^2)^{5/2}}=-\frac{1}{\epsilon_0} \rho(\vec{x}) \frac{R^3}{(R^2+a^2)^{3/2}} \rightarrow -\frac{1}{\epsilon_0} \rho(\vec{x}) \quad \text{for} \quad a \rightarrow 0.$$
    For ##n=1## you get ##0## because of symmetry, because the integrand is proportional to the integral
    $$\vec{\nabla} \rho \int_{K_R} \mathrm{d}^3 \vec{r} \vec{r} \frac{3a^2}{\sqrt{a^2+r^2}}=0.$$
    For ##n \geq 2## I cannot verify generally Jackson's result, but all you need is to show that the integral is 0 for ##a \rightarrow 0##. Now introduce again spherical coordinates, and you see that all these integrals are proportional to
    $$3 a^2 \int_0^R \mathrm{d} r \frac{r^{n+2}}{(r^2+a^2)^{5/2}}=3a^2 I_n.$$
    [corrected in view of the hint in #7]
    For ##n=2## we get
    $$I_2=\int_0^R \mathrm{d} r \frac{r^{4}}{(r^2+a^2)^{5/2}}=\mathrm{arsinh} \left ( \frac{R}{a} \right )-\frac{4R^3+3ra^3}{4(R^2+a^2)^{3/2}}.$$
    Obviously one has
    $$\lim_{a \rightarrow 0} a^2 I_2=0.$$
    For ##n \geq 3## one can take the limit ##a \rightarrow 0## under the integral
    $$\lim_{a \rightarrow 0} I_n=\int_0^R \mathrm{d} r r^{n-3}=\frac{1}{n-2} R^{n-2}.$$
    Thus for ##a \rightarrow 0## also all contributions for ##n\geq 3## vanish, and that was what we wanted to prove.
     
    Last edited: Apr 6, 2017
  8. Apr 6, 2017 #7

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    Ah, ok I see, but then we should have:
    $$\nabla\rho(\vec{x}) \cdot \int d^3\vec{r} \vec{r} \frac{3a^2}{(a^2+r^2)^{5/2}} = \bigg| \nabla \rho \bigg| \int d^3r r\cos\theta 3a^2 / (a^2+r^2)^{5/2}$$

    The integral on the theta angle gets vanished; as for your calculation of the last integral, the integrand should be ##r^{n-3}##.

    The answer from MSE gives me an answer as to why ##r^2\nabla^2 \rho/6## appears there, he just separates between a traceless part which gets vanished by symmetry and the part that stays in this expression.
    Thanks.
     
  9. Apr 6, 2017 #8

    vanhees71

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    You are right about the error in my power of ##r##. I corrected my proof from before.

    For the linear term you have to keep ##\vec{\nabla} \rho(\vec{x})## in front. Why should you be allowed to write the modulus there?
     
  10. Apr 6, 2017 #9

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    Just one more thing that I want to clear out, @vanhees71 and @dextercioby , in the link that I linked to on post #3 from MSE.

    I just want to see it clearly as to why the traceless part gets vanished in the spherical integral, it appears from the next link: http://mathinsight.org/taylors_theorem_multivariable_introduction that we have from the traceless part:
    ## \vec{r}^T(H-1/3 tr(H)I) \vec{r}##; I mean if I write it for the 3 dimensional case I get:
    ##(x,y,z) (0 , H_{xy} , H_{xz} ; H_{xy} , 0 , H_{yz} ; H_{zx}, H_{yz} , 0) (x,y,z)^T = 2xy H_{xy}+2xzH_{xz}+2yzH_{yz}##
    where the ##H##'s are the entries of the Hessian and thus are constants with respect to ##r##, these are derivatives of #\rho# evaluated at the point ##\vec{x}##.

    So the mixed integrals have some form of ##\sin \phi \cos \phi## or ##\cos \theta \sin^2 \theta ##, which all vanish.

    Is this the symmetry the post in MSE refers to?

    Thanks.
     
  11. Apr 6, 2017 #10

    vanhees71

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    The charge distribution can be anything, and thus you need to full Taylor expansion. You cannot leave out terms, but in integrating over a sphere some terms vanish.
     
  12. Apr 6, 2017 #11

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    The zeros in the matrix that I wrote are for the traceless matrix, so for this matrix the mixed terms get cancelled, since these Hessian entries are constants with respect to ##r##, they are derivatives of ##\rho## evaluated at ##\vec{x}## (we did the Taylor expansion around ##\vec{x}## and ##r= \vec{x'}-\vec{x}##).

    Just saying symmetry without being explicit is just not my type of answer I am looking for. :-)
     
  13. Apr 6, 2017 #12

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    On the same page, he writes that: ##\nabla^2 (1/r)=0 ## for ##r\ne 0## and its volume integral is ##-4\pi##.

    The first identity is simple, and I did it; and indeed it's zero.

    My problem is with the integral, I used Green's first identity (I might as well use Green's theorem), to get:
    $$ \int_V \nabla^2(1/r) d^3r = \int_S \hat{n}\cdot \nabla(1/r) da$$.

    ##\nabla(1/r)=(x,y,z)/r^3##, my problem is with how to handle the integration limit of the RHS and the normal vector, I thought it was: ##\hat{n} = (x,y,z)/r##, but then how to choose ##da## and ##S##? I understand that ##V## is the volume of the sphere.

    Thanks in advance!
     
  14. Apr 7, 2017 #13

    vanhees71

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    First of all since
    $$\Delta \frac{1}{r}=0 \quad \text{for} \quad \vec{x} \neq 0,$$
    it's clear that the integral over any volume containing the origin in its interior is the same as the integral over any sphere with the origin as center (think about why!). So we can use ##V=B_{a}(0)## and then the surface integral is
    $$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \frac{1}{r} = -\int_{\partial V}\mathrm{d}^2 \vec{f} \cdot \frac{\vec{r}}{r^3}.$$
    Now in the usual spherical coordinates this reads
    $$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \frac{1}{r} = -\int_{0}^{2 \pi} \mathrm{d} \varphi \int_{0}^{\pi} \mathrm{d} \vartheta \sin \vartheta=-4 \pi.$$
     
  15. Apr 7, 2017 #14

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    @vanhees71 first thing it should be ##\Delta(1/r)=0## for ##r\ne 0##.

    As for your remark I believe we can separate the integral for when ##r=0## and when ##r\ne 0## for the latter the integral will vanish because of the above calculation of the integrand.
    OK, I see it now.
     
  16. Apr 7, 2017 #15

    vanhees71

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    That's what I wrote, isn't it?

    The other part is correct: Formally the argument goes as follows: Since the origin by assumption is within the interior of the arbitrary volume ##V##. Thus there's an entire ball ##B_a(0)## contained in the interior of ##V##. Now you can apply Gauss's Integral law to the volume ##V'=V \setminus B_a(0)## (the volume ##V## with the ball taken out), which reads
    $$0=\int_{V'} \mathrm{d}^3 \vec{x} \Delta \frac{1}{r}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \frac{1}{r} - \int_{\partial B_a(0)} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \frac{1}{r}.$$
    The reason for the minus in front of the 2nd surface integral is that the standard orientation for the surface-element vectors is out of the sphere, but when integrating over ##V'## it must point into the sphere.
     
  17. Apr 8, 2017 #16

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    I have another question, on page 42 he writes that the dimensionless integral: ##\int \frac{\vec{\rho}\cdot (\vec{\rho}+\hat{n}}{\rho^3 |\vec{\rho}+\hat{n}|^3}d^3\rho##; it's written that ##\frac{(\vec{\rho}+\hat{n})}{|\vec{\rho}+\hat{n}|^3} =- \nabla_\rho (1/|\vec{\rho}+\hat{n}|)##; and from that the dimensionless integral can easily be shown to have value ##4\pi##.

    I am not sure how to show this, we have: ##\int \frac{\vec{\rho}\cdot (\vec{\rho}+\hat{n}}{\rho^3 |\vec{\rho}+\hat{n}|^3} = -\int d\Omega \int d\rho \hat{\rho}\cdot\nabla_\rho(1/|\vec{\rho}+\hat{n}|)##.

    How to proceed with the last expression?

    Thanks.
     
  18. Apr 10, 2017 #17

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  19. Apr 10, 2017 #18

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    You wrote ##\Delta (1/r) = 0 ## for ##\vec{x}\ne 0 ##, it's not the same, as in ##r = | \vec{x}-\vec{x}'|##.
     
  20. Apr 10, 2017 #19

    vanhees71

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    I see. That's why I prefer to write out all variables, i.e., the equation means
    $$\Delta_x \frac{1}{|\vec{x}-\vec{x}'|}=0 \quad \text{for} \quad \vec{x} \neq \vec{x}'.$$
     
  21. Apr 10, 2017 #20

    vanhees71

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    In my 3rd edition of Jackson's book it's about spheres with given charge distribution (conducting and going like ##\rho \propto r^n## for ##0 <r<R## and total charge ##Q##). This I'd solve with Poisson's equation in spherical coordinates. It's a simple ordinary differential equation since the electrostatic potential only depends on ##r##.

    In the next problem it's about the ground state of the hydrogen atom. Here the total charge is of course ##q=-e<0## (i.e., the charge of one electron).
     
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