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Given a current density distribution $\mathbf J(\mathbf x)$ inside a closed bounded region $\Omega$, the magnetic field at any point $\mathbf y$ outside of $\Omega$ can be expressed as $$ \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\mathbf J(\mathbf x)\times\nabla_{\mathbf x}\frac{1}{|\mathbf x-\mathbf y|}d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\left[\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)-\nabla_{\mathbf x}\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)\right]d^3\mathbf x\\ &=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x-\frac{\mu_0}{4\pi}\int_{\partial\Omega}\mathbf n(\mathbf x)\times\left(\frac{\mathbf J(\mathbf x)}{|\mathbf x-\mathbf y|}\right)d^2 S(\mathbf x) \end{aligned}$$ where $\partial\Omega$ is the boundary of $\Omega$, $n(\mathbf x)$ is the unit normal of $\partial \Omega$ and $S(\mathbf x)$ is the area of the surface element. Now, if the current density $\mathbf J(\mathbf x)$ is zero at the boundary $\partial\Omega$ (this can be achieved by slightly enlarging $\Omega$ if $\mathbf J(\mathbf x)$ is not zero at $\partial\Omega$) we can then drop the second term on the last line. Now we simply have $$ \begin{aligned}\mathbf B(\mathbf y)&=\frac{\mu_0}{4\pi}\int_\Omega\frac{1}{|\mathbf x-\mathbf y|}\nabla_{\mathbf x}\times\mathbf J(\mathbf x)d^3\mathbf x \end{aligned}.$$

If the current density $\mathbf J(\mathbf x)$ is continuous and differentiable, the above conclusion should be correct. However, $\mathbf J(\mathbf x)$ might not be continuous in $\Omega$, e.g., infinite thin coils inside $\Omega$ carrying electrical current. Is the above derivation correct for $\mathbf J(\mathbf x)$ containing delta functions? What kind of singularities in $\mathbf J(\mathbf x)$ is permitted?

Qmechanic
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Jasper
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  • I think that the above is always true, simply because the the definition of the derivative of a distribution (such as a delta-function or a step function, which is how we describe the current configurations you're talking about) is done via a similar formula. The Mathworld article on distributions (aka "generalized functions") might be worth a look on your part. – Michael Seifert Oct 06 '15 at 23:47
  • Thanks for bringing the reference. As you said, it is correct even if $\mathbf J$ contains delta functions, since it can be verified that $\int_\Omega f(\mathbf x)\nabla\delta(\mathbf x-\mathbf x_0)d^3\mathbf x=-\nabla f(\mathbf x)|_{\mathbf x=\mathbf x_0}$ for $\mathbf x_0$ in the interior of $\Omega$ for any differentiable function $f(\mathbf x)$. – Jasper Oct 07 '15 at 18:32

2 Answers2

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It seems you might be assigning causality to Maxwell that isn’t there, and/or believing that the magnetic field contribution from curls around electric field come from or include currents, as only one term? Not sure.

Let’s start by clarifying a few basics about what Maxwell says and what is causal. Then mention and link to Jefimenko.

The four Maxwell Equations (five relationships) are best understood as:

1.Electric charges cause electric fields that converge/diverge at the charge: $$\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} $$

(In other words, charges attract/repel: $F_{1,2}= k_e \frac{q_1q_2 }{r^2}$)

2.Currents cause magnetic fields that curl around the current: $$\underbrace{\nabla {\times} \vec{B} = \mu_0 \vec{J}}~\text{ } ~(+ \mu_0\varepsilon_0 \frac{\partial \vec{E}}{\partial t})$$

(In other words, currents attract/repel: $F_{1,2}= \mu_0 \frac{\vec{I_1}\cdot\vec{I_2}L}{2\pi r}$)

*By considering propagation time, Jefimenko (below) derives all of electromagnetism from just the above two.

3.Magnetic field lines are always closed, with no sources or drains of field: $$\nabla \cdot \vec{B} =0 $$

(Quite straightforward until here.)

4.The electric field curls around changes in the magnetic field: $$\nabla \times \vec{E} = \frac{-\partial \vec{B}}{\partial t}$$

This is a consequence, but during Maxwell is considered an additional relationship.

5.The magnetic field curls around changes in the electric field:

$$\underbrace{\nabla {\times} \vec{B} = \mu_0\varepsilon_0 \frac{\partial \vec{E}}{\partial t} }~\text{ } ~(+ \mu_0 \vec{J})$$

This is a consequence, but during Maxwell is considered an additional relationship.

From Jefimenko's equations we know that 4., 5. are not causal - not from the curl to the derivative nor vice versa. The terms are due to current variations affecting each field individually. Fields are tools. If fields (Maxwell) are used, the terms in 2 and 5 must both be included, even if they come from one object.

Time-varying application of this, Jefimenko Equations:

https://en.m.wikipedia.org/wiki/Jefimenko%27s_equations

Al Brown
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Interesting observation. As you have stated, the second equation only valid when the boundary contains all the current distribution inside. But is this what you are asking? You should open this question for objections as well.

Xiaodong Qi
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  • Thanks for your comments. I have updated my question to make a little clearer. It is easy to make the current at the boundary vanished, but my concern is that there might be requirements on the current inside the boundary. – Jasper Oct 06 '15 at 21:33
  • Can you give an example of singularity of current? The current is considered as a result of charge movement. As long as you can still talking about the continuous movements of charges, $\mathbf{J}$ may always be treated as continuous. Unless you are thinking of quantum processes when sudden jumps happen at the microscopic level and quantum electrodynamics should serve your purpose, otherwise I think you are safe. – Xiaodong Qi Oct 06 '15 at 21:44
  • Sure. Let's assume $\mathbf J(\mathbf x)$ is a segment of line current, e.g., $\mathbf J(\mathbf x)=I_c\int_{\tau_1}^{\tau_2}\delta(\mathbf x-\mathbf x^\prime(\tau))\frac{d\mathbf x^\prime(\tau)}{d\tau}d\tau$, where $\delta(\mathbf x-\mathbf x^\prime)$ is the Dirac delta function, $I_c$ is the amplitude of the current, $\mathbf x^\prime(\tau)\in c\in \mathbb R^3$ is the parametric representation of the line segment $c$ with parameter $\tau\in[\tau_1,\tau_2]$. – Jasper Oct 06 '15 at 21:50
  • In your case, the integral $\int_{t_1}^{t_2}\delta(x-x'(\tau))\frac{dx'}{d\tau}d\tau=\int_{x'(t_1)}^{x'(t_2)}\delta(x-x'(\tau))dx'=\mathrm{constant}$. Is that correct? – Xiaodong Qi Oct 06 '15 at 22:02
  • No, I don't think it is a constant. How do you perform a integral that has starting and ending points of vectors? What is the integral domain? – Jasper Oct 06 '15 at 22:40
  • The integral limit depends on how do you define the position as a function of time, but it should be given. – Xiaodong Qi Oct 06 '15 at 22:49
  • The integral limits as you write are 3D position vectors and they are known since they are the starting and ending points of the coil path. Here $\tau$ is not time; it is a parameter to access the points on the path. The integral $\int_{x'(t_1)}^{x'(t_2‌​)}\delta(x-x'(\tau))dx'$ you proposed is not a valid integral to me, since the integration domain is not clear. Is it a line, surface or volume integration? – Jasper Oct 07 '15 at 17:12
  • By definition of $\delta$ function integrals, that integral is either $1$ if $x\in[x'(\tau_1),x'(\tau_2)]$ (you can choose this region as a segment, area or volume) or $0$ otherwise. Sorry, the $t$ variable in my notation should be $\tau$ under your comment. Or, you may want to find other examples where the continuity condition fails. – Xiaodong Qi Oct 07 '15 at 20:40
  • Please give an example of "you can choose this region as a segment, area or volume." I think $x\in[x'(\tau_1),x'(\tau_2)]$ is an ill notation. Please go back to my original definition $\mathbf J(\mathbf x)=I_c\int_{\tau_1}^{\tau_2}\delta(\mathbf x-\mathbf x^\prime(\tau))\frac{d\mathbf x^\prime(\tau)}{d\tau}d\tau$ and think more carefully about its meaning. – Jasper Oct 07 '15 at 20:45
  • Maybe I shouldn't use that notation to fit in your meaning which I may not understand. You can simply make $\frac{dx'(\tau)}{d\tau}d\tau=dx'(\tau)$ and do the integral you defined on your own which, in the end, may be a continuous function or just a number as no other variable can be left over. Sorry, taking long comments here. – Xiaodong Qi Oct 07 '15 at 20:52
  • Note that $\mathbf x^\prime(\tau)$ is a well defined path. It is not a variable that can be used as a integration variable. However $\tau$ is. – Jasper Oct 07 '15 at 21:10
  • If $x'(\tau)$ is a path, you can still do a path integral at least in theory. That is not a problem at all. – Xiaodong Qi Oct 08 '15 at 03:24
  • You can try to write out your path integral to see what you get. Think this way. If $\mathbf x$ is not on the path $\mathbf x^\prime(\tau)$, $\mathbf J(\mathbf x)$ is zero. $\mathbf J(\mathbf x)$ is nonzero if $\mathbf x$ is on the path $\mathbf x^\prime(\tau)$. Then how possibly $\mathbf J(\mathbf x)$ is a constant? – Jasper Oct 08 '15 at 03:40
  • Are you trying to say that $\mathbf J(\mathbf x)$ is a constant only when $\mathbf x$ is restricted on the path $\mathbf x^\prime(\tau)$? Even if $\mathbf x$ is restricted on the path, $\mathbf J(\mathbf x)$ is still not a constant; its direction changes along the path and the direction at $\mathbf x$ is $\frac{d\mathbf x^\prime(\tau)}{d\tau}$ where $\mathbf x=\mathbf x^\prime(\tau)$. But its magnitude remains constant on the path. Is this last statement what you intended to mean? – Jasper Oct 08 '15 at 03:41
  • We shouldn't spend too much time on the integral without a third party plugging in. Can you simply give a reference where do you find this expression for $\mathbf{J}(\mathbf{X})$ in your example? Or, it is just simply not well defined? – Xiaodong Qi Oct 08 '15 at 03:43
  • I am quite sure it is well defined. You can ask your colleagues to confirm. – Jasper Oct 08 '15 at 03:46
  • In my view, yours derivations are correct for current with delta-functions. But you should tract integrals as principle values. It seem true. You can easily generalize vector-analysis for distributions and all the formulas will be true. – Artem Alexandrov Feb 17 '19 at 07:24