Physics Asked on September 26, 2021
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
$$
begin{aligned}mathbf B(mathbf y)&=frac{mu_0}{4pi}int_Omegamathbf J(mathbf x)timesnabla_{mathbf x}frac{1}{|mathbf x-mathbf y|}d^3mathbf x
&=frac{mu_0}{4pi}int_Omegaleft[frac{1}{|mathbf x-mathbf y|}nabla_{mathbf x}timesmathbf J(mathbf x)-nabla_{mathbf x}timesleft(frac{mathbf J(mathbf x)}{|mathbf x-mathbf y|}right)right]d^3mathbf x
&=frac{mu_0}{4pi}int_Omegafrac{1}{|mathbf x-mathbf y|}nabla_{mathbf x}timesmathbf J(mathbf x)d^3mathbf x-frac{mu_0}{4pi}int_{partialOmega}mathbf n(mathbf x)timesleft(frac{mathbf J(mathbf x)}{|mathbf x-mathbf y|}right)d^2 S(mathbf x)
end{aligned}$$
where $partialOmega$ 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 $partialOmega$ (this can be achieved by slightly enlarging $Omega$ if $mathbf J(mathbf x)$ is not zero at $partialOmega$) we can then drop the second term on the last line. Now we simply have
$$
begin{aligned}mathbf B(mathbf y)&=frac{mu_0}{4pi}int_Omegafrac{1}{|mathbf x-mathbf y|}nabla_{mathbf x}timesmathbf J(mathbf x)d^3mathbf x
end{aligned}.$$
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?
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.
Answered by Xiaodong Qi on September 26, 2021
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: $$underbrace{nabla {times} vec{B} = mu_0 vec{J}}~text{ } ~(+ mu_0varepsilon_0 frac{partial vec{E}}{partial t})$$
(In other words, currents attract/repel: $F_{1,2}= mu_0 frac{vec{I_1}cdotvec{I_2}L}{2pi r}$)
*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:
$$underbrace{nabla {times} vec{B} = mu_0varepsilon_0 frac{partial vec{E}}{partial t} }~text{ } ~(+ mu_0 vec{J})$$
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:
Answered by Al Brown on September 26, 2021
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