5
$\begingroup$

This question already has an answer here:

The standard theory for the earth's magnetic field is that the planet's core is a geodynamo.

I know moving electric charges can induce a magnetic field, but I also know that opposite charges produce opposite fields, and unless I missed something, normal matter has no net charge on the macro level.

In Maxwell's equations, $\mathrm{\textbf{J}}=0$ means $\mathrm{\textbf{B}}=0$, correct?

So if the Earth's core is electrically neutral, how is geodynamism possible?

$\endgroup$

marked as duplicate by John Rennie electromagnetism Jan 17 at 6:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See popularmechanics.com/science/a12075/4277476 $\endgroup$ – David White Jan 15 at 19:25
  • $\begingroup$ That article has no explanation. $\endgroup$ – spraff Jan 15 at 21:42
  • $\begingroup$ This is an active area of research, but it is obviously possible to generate a magnetic field with an electrically neutral liquid metal, such as sodium or iron. $\endgroup$ – David White Jan 15 at 22:38
  • 1
    $\begingroup$ Quick clarification: $\mathbf J=0$ does not generally imply that $\mathbf B=0$ because of the displacement current term $\epsilon_0\mu_0 \frac{\partial}{\partial t} \mathbf E$ in Ampere's law. Even if we discard this term (as we usually do in MHD, the typical setting for looking at geodynamos), $\nabla \times \mathbf B =0$ and $\nabla \cdot \mathbf B=0$ do not imply that $\mathbf B=0$. However - if we additionally impose the reasonable physical requirement that $|\mathbf B|\rightarrow 0$ faster than $1/r$, then the Helmholtz Decomposition Theorem implies that $\mathbf B=0$. $\endgroup$ – J. Murray Jan 17 at 5:05
4
$\begingroup$

The focus on “neutral medium” is a red herring.

A copper wire is neutral. Move it through a magnetic field, and an EMF is induced. Since copper conducts, that in turn drives a current. And, properly arranged (i.e. a “dynamo” configuration), that current can create more magnetic field and make it all stronger.

The energy is coming from the motion of the material. So long as the material conducts, a current can flow even though it’s neutral overall.

It doesn’t even matter whether the current is negative electrons going one way or positive ions going the other.

$\endgroup$
2
$\begingroup$

Maxwell's 4th equation in steady state, states $$\rm \vec\nabla \times \vec B= \mu_0 \vec J,$$ so, so far you're right, if $\rm J=0$, then $\rm B=0$. However why do you think that $\rm J$ is zero? The current density is defined over the charge momentum densities, and for a partially or fully ionized fluid consisting only of ions and electrons this is $$\rm \vec J = \sum_{species} \vec J_s = \vec J_{e} + \vec J_{p} = n_e q_e \vec{v}_e +n_i q_i \vec{v}_i = n_e q_e(\vec{v}_i - \vec{v}_e)$$ where in the last equality one only uses the ion to electron number ratio and the fact that the electron charge is opposite to that of the ions. Thus, even in the ion-rest-frame (equal to the lab frame in good approximation, due to their slower movement/high inertia) there is a net current, when the net charge $$\rm Q = \sum_{\rm species} Q_s = n_e q_e + n_i q_i = 0$$ is zero.

So the answer to your question doesn't even have anything to do with the Maxwell's equations.
Zero net charge does NOT imply zero current.

$\endgroup$
  • $\begingroup$ In a nutshell, you're saying electron velocity is not equal to proton velocity in the earth's core? $\endgroup$ – spraff Jan 16 at 11:53
  • $\begingroup$ @spraff: Exactly! In no electrically conductive medium are they equal. Why would they? Current IS the relative motion of the two (or more) species against each other. $\endgroup$ – AtmosphericPrisonEscape Jan 16 at 12:25
  • $\begingroup$ In that case what is causing the net motion of electrons relative to protons? $\endgroup$ – spraff Jan 16 at 15:58
  • $\begingroup$ @spraff: That's a whole different can of worms. Particularly, the acceleration of electrons doesn't have to be represented by the static Maxwell-equations. That's where the research comes in. But you have your answer for "So if the Earth's core is electrically neutral, how is geodynamism possible?" $\endgroup$ – AtmosphericPrisonEscape Jan 16 at 16:22
  • $\begingroup$ @spraff: In a bike dynamo the story is simple: You have a static magnet, and you turn a wire through it mechanically. Now the electrons couple easily to the magnetic field, while ions are caught in the metal lattice. The EMF then forces the electrons to move in a time-dependent manner, which creates a source term for $J$. The geodynamo in the end must do something similar, create an initial movement from the EMF which is then self-consistently kept over coupling with the Maxwell-equations. $\endgroup$ – AtmosphericPrisonEscape Jan 16 at 16:32
1
$\begingroup$

When an electrically conductive medium is set into rotation, it will produce a magnetic field even though it is electrically neutral overall. Geodynamism occurs inside the earth because the electrically conductive liquid ore is in convective motion, because of the temperature difference between the center of the earth and the surface. The dynamics of this process as it takes place inside the earth is complex and involves the earth's rotation, the convective motion of liquid iron, and heat transfer between the core and the outer layers of the earth.

$\endgroup$
  • 1
    $\begingroup$ I think the intended question is how does a neutral liquid in motion generate current, and if the liquid is not neutral then why is that so? $\endgroup$ – hyportnex Jan 11 at 10:39
  • $\begingroup$ will edit. -Niels $\endgroup$ – niels nielsen Jan 11 at 19:38
  • $\begingroup$ Yeah, how can a a neutral moving material generate current? It sounds like a symmetry violation. $\endgroup$ – spraff Jan 12 at 17:11
  • $\begingroup$ wikipedia has a good article on how this works. $\endgroup$ – niels nielsen Jan 12 at 19:14
  • $\begingroup$ @nielsnielsen Do you mean this? If so I am confused because it seems that $\mathrm{\textbf{J}}=0$ $\endgroup$ – spraff Jan 15 at 21:49
0
$\begingroup$

The dynamo on my bicycle is also electrically neutral. Magnetism is caused by current - or spin - not by charge.

$\endgroup$
  • $\begingroup$ This is a helpful comparison, but it doesn't really give much detailed insight. It's true that modern dynamos are self-exciting, but to me the fact that that works is as surprising as the fact the earth's dynamo works. It's explaining one surprise with another equal surprise. $\endgroup$ – Ben Crowell Jan 11 at 21:22
  • $\begingroup$ @Ben Crowell The OP is not asking for detailed insight. My point is that neutrality does not imply J=0. So it is back to the drawing board for the OP. $\endgroup$ – my2cts Jan 16 at 12:52
  • $\begingroup$ If I was the OP I would ask for detailed insight. I would probably shrug and accept it if I didn't get any. But why not ask? The worst that can happen is that smart informed people say no, they won't share. $\endgroup$ – J Thomas Jan 17 at 5:04
-1
$\begingroup$

If we have a neutral medium containing on average an equal amount of positive and negative charges $N>>1$, then the fluctuation of the total charge is $\sqrt{N}e$. Thus, a neutral medium is formed charge. The correct description of the formula for the electric current is described AtmosphericPrisonEscape . The conductor with a current on average is electrically neutral. But in order to generate an electric current, a seed electric field strength is needed. This seed intensity is formed due to charge fluctuations, which is described in the formulas I have proposed.

I think this explains the inversion of the Earth’s magnetic field. When the fluctuation of the Earth’s charge reaches a large value, sufficient to change the polarity of the magnetic field, the Earth’s magnetic field is inverted. If the Earth is partly a dielectric, and partly a conductor, then persistent charge fluctuations can form in the Earth.

We note the coincidence of the values of average time between the inversions of the magnetic field $T\approx 10^6 year$ and the characteristic time of fluctuations is equal $T=\frac{N_{av}r_g}{c}=2*10^{13}s$, $N_{av}$ the Avogadro number, $r_g$ the gravitational radius. The frequency of fluctuations, equals on the Earth $\frac{c}{r_g}$, on the Sun is orders of magnitude higher due to the high temperature and thermonuclear reactions and the characteristic time of fluctuations is 9-12 years. The frequency of fluctuations of thermonuclear reactions is much higher in the Sun than the frequency of fluctuations in the Earth. The gravitational radius is replaced by the core radius. The exponential multiplier reduces this frequency. As a result, such a characteristic time of inversion at the Sun is formed. Note that the characteristic time of inversion of the Earth’s magnetic field is not constant when the next fluctuation occurs. The process of formation of fluctuations and its development is long and can last from 200 years to several thousand years. In this case, the magnetic field of the Earth will weaken and the Earth will be unprotected. The last time the inversion of the Earth’s magnetic field occurred 780 thousand years ago. But the period of inversion is not constant and the calculated period is approximate. Thus, no periodicity in the change of poles was detected, and this process is considered stochastic. There are time intervals of tens of millions of years when the inversion did not occur.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.