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Preamble: Mathematically, the divergence of an eddy field is zero, thus for the magnetic field $$\nabla\cdot\nabla\times\boldsymbol B = \boldsymbol 0$$ and from the $\nabla\times\boldsymbol B$ Maxwell equation $$\nabla\cdot \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) = 0 \ .$$ The integral of the above over any volume is $0$. So is a closed surface integral (Gauss integration theorem) of the divergence's argument, i.e. $\forall A$ $$\oint_A \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S = 0$$

My thought experiment: As closed surface I choose a sphere $A$ centered at the origin. I split the sphere into a left part $L$ and right part $R$, both open surfaces with $A = L \cup R$. Surfaces $L$ and $R$ have the same orientation such that \begin{equation} \int_L \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S = \int_R \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S \end{equation} follows from $$\oint_{L \cup R} \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S = 0 \ .$$ Now I assume all current $\boldsymbol J$ flowing into the sphere is coming from the $L$-side and then stops inside (in a symmetric pattern around the origin), causing an accumulation of charge and a non-zero $\partial \boldsymbol E / \partial t$.

My problem: Due to the above, $\partial \boldsymbol E / \partial t$ has to be radially symmetric and so \begin{equation} \int_L \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S = \int_R \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S \end{equation}

EDIT
With my declaration of $L$ and $R$ orientation, this should be \begin{equation} \int_L \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S = -\int_R \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S \end{equation} which was my error and in hindsight renderes the rest of the question void!

/EDIT

are equal and what remains to fulfill the earlier flux equation is \begin{equation} \int_L \boldsymbol J \cdot d\boldsymbol S = \int_R \boldsymbol J \cdot d\boldsymbol S \ . \end{equation} But clearly we have \begin{align} \int_L \boldsymbol J \cdot d\boldsymbol S & \neq 0 \\ \int_R \boldsymbol J \cdot d\boldsymbol S & = 0 \ . \end{align}

So, where is my flaw? What is the missing piece to save the equations? Is this a naive consideration and I need to use the full set of Maxwell equations and consider an EM wave being emitted by deceleration of charge, thus yielding other sources of $\partial \boldsymbol E / \partial t$? Or include the cause of the stopping current in terms of an electric field?

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  • 1
    $\begingroup$ If $J$ is flowing, then I believe $\int_R \vec{J} \cdot d\vec{S} \neq 0$. If the charge is accumulanting it's got to get to the right side somehow. $\endgroup$ – QuantumBrick Oct 29 '14 at 14:05
  • $\begingroup$ Why does $\frac{\partial \vec{E}}{\partial t}$ have to be radially symmetric? $\endgroup$ – By Symmetry Oct 29 '14 at 14:08
  • $\begingroup$ @QuantimBrick If I assume all current to stop inside, there is no flux through the right side. $\endgroup$ – GDumphart Oct 29 '14 at 14:12
  • $\begingroup$ @GDumphart Yes there is. If the material is accumulating current (and it is, since you're injecting current in it from the left side), then, from the fact that this accumulation is symmetric, it needs to flow to the right side, too. $\endgroup$ – QuantumBrick Oct 29 '14 at 14:24
  • $\begingroup$ @BySymmetry I required the charges to stop in a symmetric around and close to the origin. Or theoretically exactly at the origin if you prefer. $\endgroup$ – GDumphart Oct 29 '14 at 14:28
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By your parition $A=L\cup R$, we have that \begin{gather}&\oint_A \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S = 0 &\implies\\ &\int_L \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S + \int_R \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S =0&\implies\\ &\int_L \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S = -\int_R \left( \boldsymbol J + \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} \right) \cdot d\boldsymbol S \end{gather} so the fluxes are not equal, they are opposite. So, although \begin{equation} \int_L \frac{\partial \boldsymbol E}{\partial t} d\boldsymbol S = \int_R \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S \end{equation} holds, this has no contradiction to \begin{equation} \int_L \boldsymbol J \cdot d\boldsymbol S \neq \int_R \boldsymbol J \cdot d\boldsymbol S \end{equation} We only have that \begin{gather}&\int_L \boldsymbol J \cdot d\boldsymbol S+\int_L \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} d\boldsymbol S=-\int_R \frac{\partial \boldsymbol E}{\partial t} \cdot d\boldsymbol S&\implies\\ &\int_L \boldsymbol J \cdot d\boldsymbol S=-\int_A \varepsilon_0\frac{\partial \boldsymbol E}{\partial t} d\boldsymbol S\end{gather}

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  • $\begingroup$ You skipped or misunderstood my declaration of $L$ and $R$ orientation when they're not joint (no worries, it's hard without a sketch). I took that scheme from here: users.ox.ac.uk/~math0391/EMlectures.pdf - very start of chapter 3, Eq (3.2). But you are totally right about me making a stupid sign boo boo, which solves my problem. Thank you for your answer! $\endgroup$ – GDumphart Oct 29 '14 at 16:58
  • $\begingroup$ I imagined that this could be the case, but I thought that there was a bigger chance to be a mistake here, than in the radially symmetric statement that the left and right fluxes were equal. $\endgroup$ – Mateus Sampaio Oct 29 '14 at 17:01
  • $\begingroup$ Rightfully so, I didn't emphasize my unnecessary assumption enough. $\endgroup$ – GDumphart Oct 29 '14 at 17:02

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