Gold Member
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.

Related Classical Physics News on Phys.org
Gold Member
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##.

Gold Member
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.

dextercioby
Homework Helper
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.

#### Attachments

• 18 KB Views: 412
vanhees71
Gold Member
2019 Award
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:
• dextercioby and MathematicalPhysicist
Gold Member
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.

vanhees71
Gold Member
2019 Award
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?

Gold Member
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.

vanhees71
Gold Member
2019 Award
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.

Gold Member
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. :-)

• vanhees71
Gold Member
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.

vanhees71
Gold Member
2019 Award
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
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \frac{1}{r} = -\int_{0}^{2 \pi} \mathrm{d} \varphi \int_{0}^{\pi} \mathrm{d} \vartheta \sin \vartheta=-4 \pi.$$

• MathematicalPhysicist
Gold Member
@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.

vanhees71
Gold Member
2019 Award
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.

Gold Member
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.

Gold Member
Gold Member
That's what I wrote, isn't it?
You wrote ##\Delta (1/r) = 0 ## for ##\vec{x}\ne 0 ##, it's not the same, as in ##r = | \vec{x}-\vec{x}'|##.

vanhees71
Gold Member
2019 Award
I see. That's why I prefer to write out all variables, i.e., the equation means
$$\Delta_x \frac{1}{|\vec{x}-\vec{x}'|}=0 \quad \text{for} \quad \vec{x} \neq \vec{x}'.$$

vanhees71
Gold Member
2019 Award
I have the solution to problem 1.5 of Jackson's third edition here:
https://www.scribd.com/doc/31752818/Classical-Electrodynamics-3rd-Ed-J-D-Jackson-Solutions-214-Pg

(in the above link it's the solution to problem 1.4 on pages 5-6.

My question is why there isn't a minus sign in the total orbital electronic charge Q, i.e why is it Q=q and not Q=-q?

The minus is carried out from the current distribution obtained by using the Poisson equation.
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).

Gold Member
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##.
Ok, can you give a look at post number 16?

Gold Member
I am bit puzzled by the solution to problem (1.9) from Jackson's 3rd edition.
In the solution manual here:
https://www.scribd.com/doc/31752818/Classical-Electrodynamics-3rd-Ed-J-D-Jackson-Solutions-214-Pg
the solution to problem (1.9) is in (1.8) on pages 8-10; what I don't understand is the calculation being done on page 10, the answer to (ii)Parallel cylinders.

in the second line, it's written that: ##Q=C\Delta V ## and ##dQ/dx = \Delta V d/dx \{ \bigg[4\ln (x/a) \bigg]^{-1} \} = \frac{-1}{4x\ln(x/a)}##.
This cannot be true, at least I don't see why is that right?

By calculating the derivative on the RHS I get the following:
##dQ/dx = \Delta V \frac{-1}{4\ln^2(x/a) x} = \frac{-Q}{x\ln(x/a)}##
where I plugged in ##\Delta V = Q[4\ln(x/a)]##.
Am I wrong? is the solution manual wrong?

vanhees71
Gold Member
2019 Award
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.
Introduce spherical coordinates with ##\hat{n}## as the polar axis. Further use ##\hat{\rho} \cdot \vec{\nabla}=\partial_{\rho}##. Then you get for your final integral
$$\int_{S_2} \mathrm{d}^2\Omega \int_0^{\rho} \mathrm{d}\rho \partial_{\rho} \frac{1}{\sqrt{\rho^2+1+2 \rho \cos \vartheta}} = \int_{S_2} \mathrm{d}^2\Omega \left (\frac{1}{\sqrt{\rho^2+1+2 \rho \cos \vartheta}} \right)_{\rho=0}^{\rho \rightarrow \infty}=-4 \pi.$$

• MathematicalPhysicist and dextercioby
Gold Member
I found solutions someone prepared for chapter one; and I am a little bit perplexed as to the solution to problem 1.23b on page 41; how did he infer the charges of ##Q_c## and ##Q_i##? I understand that the input to the potentials are the ones which were obtained in 1.23a.

But I don't understand how did he infer that it's ##\epsilon_0 (5\Phi -\Phi_1-2\Phi_2-2\Phi_3)##.