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Given I know how the charge distribution $\rho(\vec{x}, t)$, I can define a vector $\vec{j}$ that is supposed to show the flow of this charge distribution. To do that, $\vec{j}$ only has to suffice \begin{align} \partial_i j^i + \partial_t \rho = 0 \end{align}

Thing is, given another divergence-free vectorfield $\vec{c}$, I could add it to $\vec{j}$ without changing above equation. So I ask: Is there an ambiguity in the definition of $\vec{j}$?

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  • $\begingroup$ Imagine if there was only one solution to that equation. Then there would be only one possible current density that could be found in the universe, and it wouldn't depend on the physical structure the current is flowing in, or the sources or fields nearby. The current density in an AWG 000 wire connected to mains would be the same as in an AWG 30 wire connected to a 9 V batter and the same as in a region of vacuum with no energy sources nearby. $\endgroup$ – The Photon Oct 1 '18 at 21:46
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No, there is no ambiguity in the definition of $\vec j.$ It is not defined by that equation. It only satisfies it.

The definition of $\vec j$ is that $\int_D \vec{j} \cdot d\vec{S}$ should equal the net current passing any surface $D$.

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  • $\begingroup$ You might want to share what the definition of $\vec{j}$ is. $\endgroup$ – Quantumwhisp Oct 2 '18 at 5:37
  • $\begingroup$ @Quantumwhisp. I've now added a definition. It might not be the best definition, but I don't think that the definition on Wikipedia is better. $\endgroup$ – md2perpe Oct 17 '18 at 14:36
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Charge density $\rho$ known for every point x in the whole space does not uniquely define current density $\mathbf j$. $\mathbf j$ is largely independent of $\rho$, one can have non-zero $\mathbf j$ of any value even if $\rho=0$ (current inside wires).

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