Why is the electric field time independent in the induction equation of the magnetohydrodynamics?

Question

Why does the induction equation of the magnetohydrodynamics assume the electric field $E$ is time $t$ independent or $\frac{\partial E}{\partial t}=0$?

Note: I am querying about the general case described in the Section Mathematical statement of the Wikipedia page linked to earlier, not the special case of infinite conductivity describe in the subsequent section Perfectly-conducting limit.

Specifically, Maxwell's equation stipulates $$\nabla \times \vec B = \mu_0\Big(\vec J + \varepsilon_0 \frac{\partial \vec E} {\partial t} \Big).$$ The $\frac{\partial \vec E} {\partial t}$ term would produce a $\frac{\partial^2 \vec B} {\partial t^2}$ term giving a wave component which is currently missing in the final PDE for $\vec B$.

The derivation in the first paragraph has $$\nabla \times \vec B = \mu_0 \vec J.$$ Where did the $\frac{\partial E}{\partial t}$ term go?

Answer 1

The Wikipedia article you link makes it explicit that the time-varying electric field is small compared to the current density $\mathbf{J}$,

The displacement current can be neglected in a plasma as it is negligible compared to the current carried by the free charges

So if your question is as simple as that, then the answer is found in the very article you linked, just under the equations you've referenced.

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That said, the derivation of the induction equation should be starting with Faraday's law, $$\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}\tag{1}$$ combined with the Lorentz force, $$\mathbf{F}=q\mathbf{E}+q\boldsymbol{\beta}\times\mathbf{B}\tag{2}$$ where $\boldsymbol{\beta}=\mathbf{v}/c$. Then, under the assumption of infinite conductivity, results in $$\mathbf{E}=-\boldsymbol{\beta}\times\mathbf{B}\tag{3}$$ It is by using Equation (3) and placing it into Equation (1) that we obtain the MHD induction equation, $$\frac{\partial\mathbf{B}}{\partial t}+\nabla\times\left(\boldsymbol{\beta}\times\mathbf{B}\right)=0$$

Extensions to non-ideal scenarios can be obtained through various modifications to the Lorentz force law or through various electric field equations (e.g., Ohm's law in which $\mathbf{J}\sim\mathbf{E}+\boldsymbol{\beta}\times\mathbf{B}$ from which you obtain the diffusive MHD model with a term like $\nabla^2\mathbf{B}$)

Answer 2

MHD theory does not assume $\partial_\mathrm{t}\mathbf{E}=0$! It assumes $v/c\ll1$.

Let me first write down the Maxwell-Faraday equation in SI and CGS systems respectively $$\nabla\times\mathbf{E}=-\partial_\mathrm{t}\mathbf{B}\quad(\mathrm{SI}),\quad \nabla\times\mathbf{E}=\partial_\mathrm{t}\mathbf{B}/c\quad (\mathrm{CGS}).$$ Denoting by $L$ the spatial size of the plasma system and by $T$ the temporal scale of the process, we have \begin{align}E/L\sim B/T\quad(\mathrm{SI}),\quad E/L\sim B/cT\quad (\mathrm{CGS})\end{align} or \begin{align}E\sim vB\quad(\mathrm{SI}),\quad E\sim (v/c)B\quad (\mathrm{CGS})\tag{1}\label{eq1}\end{align} where $E,v,B$ are the magnitudes of the characteristic electric field, velocity, and magnetic field of the phenomenon that concerns us. In case of relativistic phenomenon, such as electromagnetic waves, whose characteristic velocity is close to the speed of light in vacuum ($v\sim c$), the relation becomes $$E\sim cB\quad(\mathrm{SI}),\quad E\sim B\quad (\mathrm{CGS}).$$ Magnetohydrodynamics (MHD, not relativistic magnetohydrodynamics), concerns the plasma phenomena in which velocity is much less than the speed of light in vacuum $v/c\ll1$. This statement leads to an immediate consequence, that electromangetic waves in plasmas are not the part MHD theory shall consider, since their velocities are close to the speed of light in vacuum. For the Maxwell-Ampere equation $$\nabla\times\mathbf{B}=\mu_0\mathbf{j}+\partial_\mathrm{t}\mathbf{E}/c^2\quad(\mathrm{SI}),\quad \nabla\times\mathbf{B}=4\pi\mathbf{j}/c+\partial_\mathrm{t}\mathbf{E}/c\quad (\mathrm{CGS}).\tag{2}\label{eqjflw}$$ we can write the following relations for the three terms $$\frac{B}{L}\sim \mu_0 j +\frac{1}{c^2}\frac{E}{T} \quad(\mathrm{SI}),\quad\frac{B}{L}\sim \frac{j}{c} +\frac{1}{c}\frac{E}{T} \quad(\mathrm{CGS})$$ Using the relations \eqref{eq1} we arrive at $$B\sim \mu_0 jL +\frac{v^2}{c^2}B \quad(\mathrm{SI}),\quad B\sim jL/c +\frac{v^2}{c^2}B \quad(\mathrm{CGS})$$ It is now obvious that, for phonomena of $v/c\ll1$, the term of electric displacement in \eqref{eqjflw} must be much less than the curl of magnetic induction. To maintain the equality in \eqref{eqjflw}, the electric current, whose determination depends not only on the Maxwell equations but also on the plasma model, should be close to the curl of magnetic induction. Thus the term of temporal derivative of electric field can be omitted.

In fact, we can put it inversely. It is the omitting of this term, among other manipulations (approximations), that makes MHD theory, which is designed for non-relativitic phenomena in plasmas.

Answer 3

You are not assuming that the electric field is stationary. Indeed, from Maxwell-Faraday’s law, you recover a time dependence. You are just neglecting the displacement current (as indicated by the wikipedia article). Physically, you are cutting part of the feedback loop between the EM fields, so you are neglecting radiation.

Pedestrianly, this amounts to setting the speed of light to infinity. More realistically, it’s rather about saying that the typical velocity of your fluid is small compared to the speed of light. Note that this is not about the actual velocity field of the fluid. Any combination of the parameters which gives a velocity works just as well. For example at large distance with fixed frequency this approximation would necessarily break down.

Hope this helps.