# Derivation of Ampere's Law in Jackson

The derivation of Ampere's Law in Jackson E&M from the Biot Savart law is for the most part fairly traditional, using the $\nabla\times(\nabla\times A)$ identity on the vector potential:

$$\nabla\times\mathbf{B}=\nabla\times\nabla\times\frac{\mu_0}{4\pi}\int\left(\frac{\mathbf{J(x')}}{\mathbf{|x-x'|}}\right)\,d^3x$$

and exploiting some identities (like the Laplacian/Dirac identity derived earlier) to simplify the problem.

However, as he is wont to do, he leaves out most of the steps in the computation, and there is an especially egregious omission from 5.20 to 5.21 that I am having trouble understanding.

He simplifies the first expression to something more tractable, namely:

$$\nabla\times\mathbf{B}=-\frac{\mu_0}{4\pi}\,\nabla\int\mathbf{J(x')}\cdot\nabla'\left(\frac{1}{|\mathbf{x}-\mathbf{x'}|}\right)\,d^3x'+\mu_0\mathbf{J(x)}$$

And then, in a rather cryptic statement, says "integration by parts yields:"

$$\nabla\times\mathbf{B}=-\frac{\mu_0}{4\pi}\,\nabla\int\frac{\nabla'\cdot\mathbf{J(x')}}{|\mathbf{x}-\mathbf{x'}|}\,d^3x'+\mu_0\mathbf{J(x)}$$

I've worked through most of the work, but there are still a few steps that escape me.

By integration by parts, he almost certainly means applying the identity $\mathbf{A}\cdot\nabla \psi = \nabla\cdot(\psi\mathbf{A})-\psi(\nabla\cdot\mathbf{A})$, which reduces the problem to

$$\nabla\times\mathbf{B}=\nabla\int\frac{\nabla'\cdot\mathbf{J(x')}}{|\mathbf{x-x'}|}\,d^3x'-\nabla\int\nabla'\cdot\left(\frac{\mathbf{J(x')}}{|\mathbf{x-x'}|}\right)\,d^3x'+\ ...$$

For steady state magnetic phenomena, the first integrand vanishes (since $\nabla\cdot\mathbf{J}=0$), but the second remains.

$$\nabla\times\mathbf{B}=-\nabla\int\nabla'\cdot\left(\frac{\mathbf{J(x')}}{|\mathbf{x-x'}|}\right)\,d^3x'+\ ...$$

I have no idea how to show that this term vanishes. I'd use the divergence theorem to transform the volume integral to a surface integral, but the integrand is not $C^1$ with the singularity at zero, so I'm not sure I can. Jackson does something clever with an $a$-potential

$$\lim_{a\rightarrow 0}\left(\nabla\cdot\left(\frac{1}{(x^2+a^2)^{1/2}}\right)\right)$$

in a somewhat similar electrostatic case earlier, but I don't think it's applicable here.

• Griffiths has a lot more details and explanation of this proof than Jackson does. Jul 31, 2016 at 2:16
• I also have Griffith next to me, and it's much more lucid. I am still concerned about applying the Divergence Theorem to singular forms. Griffith does the same thing in deriving Gauss' Law, but I am not well enough versed in differential forms to dispute or confirm it. Jackson seems to go to great lengths to avoid this problem by constructing elaborate limits of the potential around $r=0$, so I think there must be some concern for non $C^1$ forms. Jul 31, 2016 at 2:21
• @JacobAustin How do you interpret the $A=\frac{\mu_0}{4\pi}\int\frac{J(x')}{|x-x'|}d^3x'$ where the integrand blows up at $x'=x$? I am not answering or giving hint to your question. This is also something I don't understand. See physics.stackexchange.com/q/249559 . But I think it's related to your question. Jul 31, 2016 at 3:10

In Jackson's textbook, paragraph $\S 5.3$ starts with equation (5.14) given here as (001) $$\mathbf{B}\left(\mathbf{x}\right)=\dfrac{\mu_{o}}{4\pi}\int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\times}\dfrac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3}d^{3}x' \tag{001}$$ Then using the relation just above (1.15) in this same textbook, easily proved and given here as (002) $$\dfrac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3}= - \boldsymbol{\nabla} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) \tag{002}$$ equation (001) is expressed as $$\mathbf{B}\left(\mathbf{x}\right)= - \dfrac{\mu_{o}}{4\pi}\int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\times}\boldsymbol{\nabla} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right)d^{3}x' \tag{003}$$ If in the vector formula $$\boldsymbol{\nabla}\boldsymbol{\times} \left(\psi \mathbf{a} \right)=\boldsymbol{\nabla}\psi \boldsymbol{\times}\mathbf{a}+\psi\left(\boldsymbol{\nabla}\boldsymbol{\times} \mathbf{a} \right) \tag{004}$$ we replace $$\psi=\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|} , \quad \mathbf{a}=\mathbf{J}\left(\mathbf{x}'\right) \tag{005}$$ then $$\boldsymbol{\nabla}\boldsymbol{\times} \left( \dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|} \mathbf{J}\left(\mathbf{x}'\right) \right)=\boldsymbol{\nabla} \left( \dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|} \right) \boldsymbol{\times} \mathbf{J}\left(\mathbf{x}'\right) + \dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|} \underbrace{\left [\boldsymbol{\nabla}\boldsymbol{\times} \mathbf{J}\left(\mathbf{x}'\right) \right]}_{=\mathbf{0}} \tag{006}$$ So $$\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\times}\boldsymbol{\nabla} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right)= - \boldsymbol{\nabla}\boldsymbol{\times} \left( \dfrac{\mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right|} \right) \tag{007}$$ Replacing this expression under the integral in (003) $$\mathbf{B}\left(\mathbf{x}\right)= \dfrac{\mu_{o}}{4\pi}\int\:\boldsymbol{\nabla}\boldsymbol{\times} \left( \dfrac{\mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right|} \right)d^{3}x' \tag{008}$$ The curl $\:\boldsymbol{\nabla}\boldsymbol{\times}\:$ concerns differentation with respect to $\: \mathbf{x}\:$ so it's exported out of the integral since the integration variable is $\: \mathbf{x}'\:$ $$\mathbf{B}\left(\mathbf{x}\right)= \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla}\boldsymbol{\times} \int\:\dfrac{\mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right|} d^{3}x' \tag{009}$$ identical to equation (5.16) in textbook.
From above equation we have $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla}\boldsymbol{\times} \boldsymbol{\nabla}\boldsymbol{\times} \int\:\dfrac{\mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right|} d^{3}x' \tag{010}$$ identical to equation (5.18) in textbook. Using the following formula for any vector field $$\boldsymbol{\nabla} \boldsymbol{\times} \left( \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{A}\right) =\boldsymbol{\nabla}\left(\nabla \boldsymbol{\cdot}\mathbf{A}\right)- \nabla^{2}\mathbf{A} \tag{011}$$ equation (010) yields $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla} \int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\cdot}\boldsymbol{\nabla} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) d^{3}x' - \dfrac{\mu_{o}}{4\pi} \int\:\mathbf{J}\left(\mathbf{x}'\right)\nabla^{2} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) d^{3}x' \tag{012}$$ equation (5.19) in textbook.

Note that all differential operators like

$\:\boldsymbol{\nabla}=\text{grand},\boldsymbol{\nabla}\boldsymbol{\times}=\text{curl},\boldsymbol{\nabla}\boldsymbol{\cdot}=\text{div} \:$

concern differentiation with respect to $\:\mathbf{x}\:$ and so they can be inserted freely under the integrals with respect to $\:\mathbf{x}'\:$.
Now the following equations are valid (unnumbered in textbook) $$\boldsymbol{\nabla} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right)= - \boldsymbol{\nabla}' \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) \tag{013}$$ essentially identical to (002) and
$$\nabla^{2} \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right)= - 4\pi \delta\left(\mathbf{x}-\mathbf{x}'\right) \tag{014}$$ Replacing these two expressions under the 1rst and 2nd integral respectively in the rhs of (012) we have $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= - \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla} \int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\cdot}\boldsymbol{\nabla}' \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) d^{3}x' + \mu_{o} \underbrace{\int\:\mathbf{J}\left(\mathbf{x}'\right)\delta\left(\mathbf{x}-\mathbf{x}'\right) d^{3}x'}_{=\mathbf{J}\left(\mathbf{x}\right)} \tag{015}$$ or $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= - \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla} \int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\cdot}\boldsymbol{\nabla}' \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) d^{3}x' + \mu_{o} \mathbf{J}\left(\mathbf{x}\right) \tag{016}$$ equation (5.20) in textbook. Integration by parts of the integral in the rhs of (016) yields $$\int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\cdot}\boldsymbol{\nabla}' \left(\dfrac{1}{\left|\mathbf{x}-\mathbf{x}'\right|}\right) d^{3}x' = - \int\:\dfrac{ \boldsymbol{\nabla}' \boldsymbol{\cdot} \mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right| } d^{3}x' \tag{017}$$ and (016) gives $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= \dfrac{\mu_{o}}{4\pi}\boldsymbol{\nabla} \int\:\int\:\dfrac{ \boldsymbol{\nabla}' \boldsymbol{\cdot} \mathbf{J}\left(\mathbf{x}'\right)}{\left|\mathbf{x}-\mathbf{x}'\right| } d^{3}x' + \mu_{o} \mathbf{J}\left(\mathbf{x}\right) \tag{018}$$ equation (5.21) in textbook. For steady-state magnetic phenomena $\:\boldsymbol{\nabla}' \boldsymbol{\cdot} \mathbf{J}\left(\mathbf{x}'\right)=0\:$, so that we obtain $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{B}= \mu_{o} \mathbf{J}\left(\mathbf{x}\right) \tag{019}$$ equation (5.22) in textbook.

The main question is how equation (017) is valid integrating by parts. This is proved in the ADDENDUM .

Let the integral $$\mathrm{F}= \int\:\mathbf{A}\left(\mathbf{x}\right)\boldsymbol{\cdot}\boldsymbol{\nabla} \psi\left(\mathbf{x}\right) d^{3}x \tag{A-001}$$ where $\:\mathbf{A}\left(\mathbf{x}\right)=\left[A_{1}\left(\mathbf{x}\right),A_{2}\left(\mathbf{x}\right),A_{3}\left(\mathbf{x}\right) \right]\:$ and $\:\psi\left(\mathbf{x}\right)\:$ vector and scalar fields respectively of the vector variable $\:\mathbf{x}=(x_1,x_2,x_3)\:$ and the integration is over all space. $$\mathrm{F}=\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}\left(A_{1}\dfrac{\partial \psi}{\partial x_1}+A_{2}\dfrac{\partial \psi}{\partial x_2}+A_{3}\dfrac{\partial \psi}{\partial x_3}\right)dx_{1}dx_{2}dx_{3} \tag{A-002}$$
Now by an integration by parts $$\int\limits_{-\infty}^{+\infty}A_{\jmath}\dfrac{\partial \psi}{\partial x_{\jmath}}dx_{\jmath}=\biggl[A_{\jmath}\psi\biggr]_{x_{\jmath}=-\infty}^{x_{\jmath}=+\infty} - \int\limits_{-\infty}^{+\infty} \psi \dfrac{\partial A_{\jmath}}{\partial x_{\jmath}}dx_{\jmath}, \quad \jmath=1,2,3 \tag{A-003}$$ But in our case $$\psi\left(\mathbf{x}\right)=\dfrac{1}{\left|\mathbf{x}-\mathbf{x}_{0}\right|} \tag{A-004}$$ that is $$\lim_{x_{\jmath}\rightarrow\pm\infty}\psi\left(\mathbf{x}\right)=0 \tag{A-005}$$ so $$\int\limits_{-\infty}^{+\infty}A_{\jmath}\dfrac{\partial \psi}{\partial x_{\jmath}}dx_{\jmath}= - \int\limits_{-\infty}^{+\infty} \psi \dfrac{\partial A_{\jmath}}{\partial x_{\jmath}}dx_{\jmath}, \quad \jmath=1,2,3 \tag{A-006}$$ Formally by coordinate explicitly \begin{align} \int\limits_{-\infty}^{+\infty}A_{1}\dfrac{\partial \psi}{\partial x_{1}}dx_{1} & = - \int\limits_{-\infty}^{+\infty} \psi \dfrac{\partial A_{1}}{\partial x_{1}}dx_{1} \quad \Longrightarrow \nonumber\\ \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}A_{1}\dfrac{\partial \psi}{\partial x_1}dx_{1}dx_{2}dx_{3} & =- \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}\psi \dfrac{\partial A_{1}}{\partial x_{1}}dx_{1}dx_{2}dx_{3} \tag{A-006a}\\ \int\limits_{-\infty}^{+\infty}A_{2}\dfrac{\partial \psi}{\partial x_{2}}dx_{2} & = - \int\limits_{-\infty}^{+\infty} \psi \dfrac{\partial A_{2}}{\partial x_{2}}dx_{2} \quad \Longrightarrow \nonumber\\ \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}A_{2}\dfrac{\partial \psi}{\partial x_2}dx_{1}dx_{2}dx_{3} & =- \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}\psi \dfrac{\partial A_{2}}{\partial x_{2}}dx_{1}dx_{2}dx_{3} \tag{A-006b}\\ \int\limits_{-\infty}^{+\infty}A_{3}\dfrac{\partial \psi}{\partial x_{3}}dx_{3} & = - \int\limits_{-\infty}^{+\infty}\psi \dfrac{\partial A_{3}}{\partial x_{3}}dx_{3} \quad \Longrightarrow \nonumber\\ \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}A_{3}\dfrac{\partial \psi}{\partial x_3}dx_{1}dx_{2}dx_{3} & =- \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}\psi \dfrac{\partial A_{3}}{\partial x_{3}}dx_{1}dx_{2}dx_{3} \tag{A-006c} \end{align} and adding (A-006a),(A-006b) and (A-006c) we obtain \begin{align} \mathrm{F}&=- \int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{+\infty}\left(\psi\dfrac{\partial A_{1}}{\partial x_1}+\psi\dfrac{\partial A_{2}}{\partial x_2}+\psi\dfrac{\partial A_{3}}{\partial x_3}\right)dx_{1}dx_{2}dx_{3} \tag{A-007}\\ &=-\int\:\psi\left(\mathbf{x}\right)\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}\left(\mathbf{x}\right) d^{3}x \nonumber \end{align} so finally $$\int\:\mathbf{A}\left(\mathbf{x}\right)\boldsymbol{\cdot}\boldsymbol{\nabla} \psi\left(\mathbf{x}\right) d^{3}x=-\int\:\psi\left(\mathbf{x}\right)\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{A}\left(\mathbf{x}\right) d^{3}x \tag{A-008}$$
I think ((A-008) is valid in general under the condition (A-005) for $\:\psi\left(\mathbf{x}\right)\:$ and that $\:\mathbf{A}\left(\mathbf{x}\right)\:$ has finite values at $\:\pm\infty\:$.
• I wonder if this has solved the OP's question. You applied integration by parts in (A003), but $\psi$ is not continuous at $\textbf{x}_0$. Does integration by part hold in this case? Aug 1, 2016 at 18:38
You can punch a little hole of radius $\epsilon$ about the singularity at ${\bf x}= {\bf x}'$ and get a surface integral. The magnitude of the integrand is bounded by $|J({\bf x}|/\epsilon$ on the surface, and the surface area is $4\pi \epsilon^2$, so the contribution from the surface integral goes to zero as $\epsilon^2/\epsilon$ as $\epsilon\to 0$. The singularity has no effect therefore.