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I am reading section 2.6 of Shercliff's book on MHD. He first establishes that (following his notation) \begin{align} \sum p &= q \delta \hspace{1em} \text{and} \tag{1} \\ \sum p(\mathbf u + \mathbf v) &= \mathbf j \delta \tag{2} \end{align} where $p$ is the charge of a free particle, $\delta$ is volume of a small representative region, $q$ free charge density, $\mathbf u$ is the particle's velocity w.r.t the fluid which is moving with a velocity $\mathbf v$, and $\mathbf j$ is the current density. The summations in above equations are over the representative volume in which macroscopic properties (all except $p$ and $\mathbf u$) remain constant.

He then introduces Ohm's law as a balance between electromagnetic and drag forces. \begin{align} \sum p \mathbf E + \sum p(\mathbf u + \mathbf v) \times \mathbf B = \sum k \mathbf u \tag{3} \end{align} where $k$ is some sort of drag coefficient. He goes forward stating that the RHS can be written as follows due to experiments \begin{align} \sum k \mathbf u = q \delta \frac{\mathbf j_c}{\sigma} \tag{4} \end{align} where $\mathbf j_c$ is the conduction current density and $\sigma$ is the conductivity.

Combining everything (putting (1), (2) and (4) in (3)), we have the Ohm's law as follows. \begin{align} \frac{\mathbf j_c}{\sigma} = \mathbf E + \frac{\mathbf j \times \mathbf B}{q} \tag{5} \end{align} This equation only relates $\mathbf j_c$ to $\mathbf j$, $\mathbf E$ and $\mathbf B$. To determine $\mathbf j$ for a given $\mathbf E$ and $\mathbf B$, we need one more relation between $\mathbf j_c$ and $\mathbf j$. How do we proceed?

In a later section, he says

In future therefore the conduction current is taken as being the total current, $\partial \mathbf P/\partial t$ and $q \mathbf v$ being negligible, so that \begin{align} \mathbf j = \sigma(\mathbf E + \mathbf v\times \mathbf B) \tag{6} \end{align}

Here $\partial \mathbf P/\partial t$ is the polarisation current.

Question

Is the following equation correct? $$ \mathbf j = \mathbf j_c + \partial \mathbf P/\partial t + q\mathbf v $$ If so, how is it derived, given (2) and (4). If not, what is that other relation between $\mathbf j_c$ and $\mathbf j$? And how does (5) simplify to (6) in light of these two relations between $\mathbf j_c$ and $\mathbf j$?

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Zxcvasdf
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  • From the text you typed it seems the author is considering a situation where only conduction current is present. The trouble with adding a polarization current is that the concept of polarization is natural only for collections of permanent neutral units such as air or glass, but less so for conductive materials such as metals where charges are mobile. In static solid metals, all current is thought as conduction current by convention. In liquid metal, there is a good reason to separate the convection term $\rho\mathbf v$. But I don't see how separating a polarization term would make any sense. – Ján Lalinský Aug 29 '19 at 20:31
  • Even if we don't bring polarisation current into picture, how does (5) become (6) for conducting liquids? This was my main query. – Zxcvasdf Aug 30 '19 at 01:23
  • You can derive it by postulating Ohm's law between current density and fields in the frame of the liquid element where it is at rest, then transforming back to lab frame, and neglecting some small terms. It is an approximate equation. – Ján Lalinský Aug 30 '19 at 12:30

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