Earth magnetism-Researching News

How Earth’s Magnetic Field Influences Flows in the Planet’s Core

(This article has been published long ago, just re-posted in this blog for further reference. Anyhow, comments are welcome.)

October 31, 2024• Physics 17, 142

 

A “Little Earth Experiment” inside a giant magnet sheds light on so-far-unexplained flow patterns in Earth’s interior.

 

Sketch of outer-core flows derived from data from the Swarm mission (plotted looking down on the North Pole). The black lines indicate flows averaged over two decades, showing patterns that violate the Taylor-Proudman theorem by crossing a boundary defined by the solid-core radius. The red and blue colors indicate azimuthal-flow components.

Earth’s inner core is a hot, solid ball—about 20% of Earth’s radius—made of an iron alloy. The planet’s outer core, beneath the rocky mantle, is a colder, liquid metal. Geophysics models explain that, since the movement of a liquid metal induces electrical currents, and currents induce a magnetic field, convection and rotation produce our planet’s magnetic fields. But these models typically neglect an important contribution: how Earth’s magnetic field influences the very flows that generate it. Alban Pothérat of Coventry University, UK, and collaborators have now developed a theory that accounts for such feedback and vetted it using a lab-based “Little Earth Experiment” [1]. Their results inform a model pinpointing processes that might explain the discrepancies between theoretical predictions and satellite observations of Earth, opening new perspectives on the study of geophysical flows.

Understanding flows in planetary interiors is a long-standing challenge. “If you don’t account for the fact that the magnetic field itself changes the flow, then you won’t get the right flow,” says Pothérat. Indeed, both satellite data from the European Space Agency’s Swarm mission and state-of-the-art numerical simulations indicate certain circulating core flows where liquid produced at the boundary of the inner core is fed into the outer core, moves upward toward the poles, and from there finally flows back inward (Fig. 1).

These observations can’t be explained by the established theory for rotating fluids, which assumes that magnetic-field-induced forces on the flow can be neglected, as they are dominated by rotation-induced Coriolis forces. If rotation is fast enough, goes the theory, flows of liquid are two dimensional and lie in the plane normal to the rotation axis. For planetary interiors, this imposes a constraint known as the Taylor-Proudman theorem: Fluid cannot flow across a boundary, called the tangent cylinder, defined by a radial distance equal to the solid-core radius. The flows documented on Earth, however, violate this condition. To explain the discrepancy, what’s needed is an experiment that captures convection, rotation, and magnetism all at once, says Pothérat.

Sketch of the setup used by Pothérat and co-workers. A liquid-containing hemisphere, a heating element, and a water tank represent Earth’s outer core, inner core, and mantle, respectively. The structure sits on a rotating table in a 10-tesla magnetic field, while a laser injected from the top allows the researchers to map the flow field.

Pothérat and his colleagues have realized such an experiment, using it to vet an extension of the established model associated with the Taylor-Proudman constraint. Previous Earth-mimicking setups used rotating tanks filled with a highly conductive but opaque liquid metal, which prevented the visualization of liquid flows. The team’s “Little Earth Experiment” (Fig. 2) instead used a low-conductivity, transparent sulphuric acid solution, which allowed the acquisition of maps that visualize flow structures in a magnetorotating fluid. “That’s the big originality,” says Pothérat. Since their liquid was much less conducting than a liquid metal, the team needed a large external magnetic field to create magnetic forces sufficiently strong to mimic planet-like conditions. They achieved such conditions using a 10-tesla magnetic field produced by the giant magnet of the Grenoble High Magnetic Field Laboratory in France. “The real feat of our experiment is fitting a rotating tank inside this enormous magnet,” says Pothérat.

The Little Earth Experiment involves a rotating hemispheric dome sitting on a flat, rotating table. The transparent-glass hemisphere was filled with a conductive fluid, which represented the outer core of Earth. At the hemisphere’s flat bottom, a cylindrical heating element protruding into the liquid played the role of Earth’s inner solid core, driving convection. A water-filled, cylindrical plastic tank sat on top of the hemisphere, providing a cooling effect that mimicked that due to Earth’s mantle. By lacing the conductive fluid with hundreds of thousands of micron-diameter hollow glass particles, which scattered incoming laser light, the researchers could track the particles’ positions and measure the fluid’s velocity at multiple points as the tank rotated.

The key measurements were maps of the fluid’s velocity at two heights—one near the solid core and one at latitudes close to the top of the hemisphere—obtained for electromagnetic and rotational forces representative of Earth-like conditions. Fitting such data, the researchers calculated that at least 10% of the liquid flowed in a circulating pattern similar to that occurring in Earth’s core: From the solid core the liquid flowed toward the top of the dome and then toward the equatorial regions. “We can finally see what the flow looks like,” says Pothérat.

The researchers established that the polar component of the magnetic field drives flow between polar and equatorial regions, concluding that this field-induced flow must be taken into account for explaining convective flows that can’t be described by rotation alone. Modifying the conventional theory by adding a magnetic field pointed in the polar direction allowed the researchers to predict exactly when and how much liquid flowed across. The magnetic field fully explains why the flow crosses the tangent-cylinder boundary, says Pothérat.

Hao Cao, a geophysics researcher at the University of California, Los Angeles’s Department of Earth, Planetary, and Space Sciences, calls the work “impressive” and says that the experimental verification of flow regimes “illustrates the critical role of magnetic fields in shaping fluid dynamics in Earth’s and planetary cores.” He adds a cautionary note regarding its direct application to real planetary cores, pointing out that “the fluid dynamics in this experiment has minimal impact on the magnetic field itself.” That’s not the situation expected in planetary cores, he says.

–Rachel Berkowitz

Rachel Berkowitz is a Corresponding Editor for Physics Magazine based in Vancouver, Canada.

References

1. A. Pothérat et al., “Magnetic Taylor-Proudman constraint explains flows into the tangent cylinder,” Phys. Rev. Lett. 133, 184101 (2024).

 

CHAPTER III-MOON'S TOTAL CHARGE

MOON's TOTAL CHARGE ON E.M

Every previous solution is solved with many assumptions. Anyhow, the results are meaningful; they do confirm that the semi-moons in every day of Moon are approximately the reverting points when the Moon’s hemisphere effect changes from negative to positive or vice versa. Hence the two semi-moons divide the Moon’s orbit into 2 different, opposite halves; the effect on one half is going against or contrarian the current E.M, while the other’s effect is supporting or backing the current E.M.

The radiation from the Sun is changing and the Moon charge therefore is changing accordingly; so there must be a sequent change in Moon’s effect on E.M, and this chapter is a discussion about the change in circumstance when Moon’s charge is changing and the total charge ‘’Qa+Qb’’ is no more a ‘’0’’.

We leave the influence of permeability of the media between Moon and Earth centre for another problem; just consider the Moon’s influence on Earth’s surface.

Let’s set and solve the following problems for Moon’s effects.

2/-The problem of Moon’s total charge for Houston:

Houston of U.S chosen as observer in this problem is rather close to Earth's equator and quite in the middle of Van Allen belt and plasma sphere; therefore, it is very hard to detect the Moon’s effect. Nonetheless, we keep working with that location as observer then sort out the issue, and following is the problem:

U.S observer location (30N, 95W), USA on 02^nd May 2017, Moon’s total charge Qm and other parameters:

-Illumination: 49.1%≈50%, this is to make sure that each of the two hemispheres is capable to eliminate the other, so the hemispheric effect can be null.

-Meridian passing: 19.31 hrs.

-Moon rise/set (excerpt from Moon’s time table):

-Distance: 377652 km from point P to Moon.

-Moon’s average radius: R=1738.1 km.

Question:

With Moon’s total charge Qm≠0, consider the amplitude of magnetism that directly induced by Moon on Houston at 19.00 hrs LT.

2/1-To understand more about the question (analysis):

-The meridian passing is on 19.31 hrs at 75.7⁰.

-From the Moon’s timetable: Moon rise/set is 12.38/(next day)02.20 hrs. or 12.38/26.20  (or Moon’s on sky within 822 minutes).

-The distance from A or B to a general P (on longitude of P) can be viewed as the same and depicted in the following figure:

  

-The given P is on land surface with eye height ‘’0’’, the horizon viewed from P is a plane with the straight line from Sunrise (A) to Sunset (B).

-Note that AP = PB, by Pythagorean theory AP=(AO²-OP²)^1/2 (Figure 4/IV), where OP=Re, the value of Earth average radius despite ‘’O’’ is not coinciding on Oe (Figure 5/IV).

-A circle with centre (O) is designed to meet AP=(AO²-OP²)^1/2 and distance from P to Moon 377652 km. This centre (O) may not be coinciding on the terrestrial geometric centre (Oe).

 2/2-Solution:

2/2/1-Step 1: to re-think about the real things.

Like any other electric current or moving charge, the Moon definitely contributes magnetism. Moon’s total charge has been assumed as neutral with 2 opposite hemispheres in previous problem.

But in reality the total charge of any object under solar wind can’t be neutral at all the time; it is actually changing, and is given a value Qm (positive) in this problem.

The questions are how magnificent is the magnetism induced by the lunar total charge? Which wise (north or south) does it direct to?

2/2/2-Step 2: Assumptions.

- P (30N, 95W) is assumed as the stance for observer with eye height ‘’0’’.

- Distance from P to A is the same as P to B or any position on the longitude of P to those 2 positions of Moon.

-The Moon is assumed as on ‘’no real move’’ throughout the time of calculation. By this condition, the Moon’s real move is not considered. Instead, we consider the Moon’s relative move or 12(+) hr. trace around the Earth.

-Moon’s total charge (Qm=Qa+Qb and Qm≠0) stays unchanged throughout this calculation.

-Distance: 377652 km is from P to the nearest position of the Moon. The distance from P to charge centre of Moon can be approximately added with ½ of Moon’s average radius 1/2*1738.1 km.

The distance d=378521 km is assumed as distance from observer to Moon’s charge centre.

3-Step 3: Solution with reliable accuracy.

-Recall Biot-Savart expression for a single charge. We add some more lines on the basic figure 12b/v to facilitate the solution.

-In the above figure:  The relative Moon’s trace (not the Moons’ orbit) in 12 hrs is demonstrated (from Moonrise till Moon set) in 2D coordinates, the larger half of the circle (from A to B through under part) is not viewed. Within this duration, the Earth completes almost a half of cycle around its own axis. We can imagine that a certain observer standing somewhere near either North or South Pole, with a restricted view she/he can observes the Moon rising like depicted in the figure (while the Moon after Moonset is not seen, and the Earth makes no move).

The problem becomes pure geometric on 2D coordinates; and the figure requires some additional assumptions and quantities:

+Moon is at a general position (M) on sky, its distance to P is ‘’r’’ which varies against time and reaches the given value of d at meridian passing;

+Moon is rising at A and setting at B;

+Moon is expecting to pass meridian at Z over P; where angle APM=angle APZ =90⁰ while Moon’s altitude reaches its max of that day 75.7⁰.

+Vector  ''v'' is perpendicular to the radius that drawn through Moon (OM);

+PD is perpendicular to OM and parallel to;

+The angles are named after each point and numbered: M1, M2 and O1,O2; or P1,P2,P3,P4;

+Parameters: d=378521*10³m (minimum distance from the given position P to Moon’s centre).

+The Moon’s angle (P2 or angle APM) is taken by a professional equipment (Sextant) (at the given time); (the ‘’altitude’’ of Moon or any object on sky is understood as the angle between the line from observer to the object and horizontal plane at observer. Therefore, a large gap between the two aforementioned concepts (altitude and P2) is found, and no chance for a mixing up or confusion). Nonetheless, we can’t deny that the Moon’s altitude depends on P2 and reaches max almost at the middle of Moon’s trace between A and B, therefore we may keep considering altitude in relation with Moon’s angle P2. Furthermore, by the laws of 3D trigonometric, we can calculate P2 from altitude or vice versa, but no discussion for that issue is encouraged in this book.

The following is a figure that depicts approximately the altitude of Moon and the Moon’s angle P2:

 Figure 9/V-Moon’s altitude

(At Moon’s meridian passing, the angle APM is obviously 90⁰ or ½ π while Moon’s altitude keeps varying (angle MPH is not always to be 90⁰ even if we measure it from Earth equator)).

Again, we refer to the basic expression of Biot-Savart:

With Biot-Savart expression, this problem is the normal geometric one in astrophysics, we can demonstrate every important parameter on 2D coordinates. The sensitive issue in this expression is the product of vector ''v''  and unit vector ''r'', a vector directs from this page toward reader, and its  non-directional value is:

v*Sin(M1)

The non-directional value of B is:

b/1- To seek for (r ):

+Consider the triangle MDP (Vector ''v'' and PD are extended for a better view with M’P’//MP):

 Cos(O2)=Cos(07.95⁰)=0.990389

+In the triangle MPO:

In the triangle ODP:

OD=Re*Cos(O2)= 6356.8*0.990389=6295.704795 km

And Sin (M1) = Sin(81.91⁰+07.95⁰)=0.99999

MD = MO-OD or MD=(Re+d)-Re*Cos(O2)=6356.8+378521-6356.8*0.990389=

=378582.0952048 Km

Thus, r=MD/Sin(M1)= 378582.0952048/0.99999 =378585.88106 km=

r=378585.88106*10³m.

b/2-Angle M1:

M1=P2+P3 (Vector v is parallel with PD, the rule for alternate angles is applied)

P2 is Sextant reading (or astronomy almanac), and the P3=O2. Thus:

Angle M1=P2+O2, with given P2=81.91⁰ and P3=07.95⁰   

 Angle M1=81.91⁰+07.95⁰=89.86⁰ = 89051’’

Final calculation:

-Replace the followings in the expression and calculate value of B(tc):

Radius r=378585.881*10³m

Speed v=21230.8834 m/s

Angle M1=89⁰51”, Sin(M1)=0.99997≈1.

As assumption of this problem, q=Qm Coulomb, every length is in meter (m), the unit of B is Tesla (T).

We have:

B(tc)=10^-7*21230.8834*378585.881^-2*10^-6*Qm=

=1.481287*10^-20*Qm Tesla

(The previous approximate calculation gives result: B(tc)=1.439480*10^-20*Qm)

Because every of (r,v,M1) is depending on observer’s position, therefore the above result cannot be used for everywhere. Nonetheless it is to confirm that the Moon exerts its static electric force on the

Earth and especially induces a magnetic field on it, the value can be calculated.

For instant, a reading on a Tesla meter is indicating that the Moon’s contribution at Houston (South U.S.A) reaches its max +5*10^-7 T (normally the meter reading about 2-3 hrs before meridian passing can indicate the average magnetic level); we can find Moon’s total charge as following:

B(tc)= 1.481287*10^-20*Qm=5*10^-7 Tesla

Qm=B(tc)/ 1.481287*10^-20=5*10^-7/ 1.481287*10^-20 =

=3.375443*10¹³ Coulombs.

Everywhere inside the loop, the magnetic vector is by right-hand law.

(Problem: (the following problem is designated for 60 N to make sure that Moon signal is detected without being totally barred by plasma sphere): 3.375443*10¹³ Coulombs is a value of Moon’s charge that we detected when semi-moon (illumination 50%) is passing meridian of a position ‘’P’’ (60N, 0.0). If in 36 hrs later, when Moon’s illumination of 40% we detect Moon’s charge to be ‘’0’’ at a certain position P’ (60N, 180.0). Calculate the Moon’s total charge components Qa & Qb? 

This theoretic problem is definitely inspired by many science students but not included in this book. After this problem, we find out more about the roles of 2 hemispheres of Moon on E.M).

Figure 13/V-Earth’s magnetic field amplitude (in mG)

The above is a real graph of magnetic amplitude observed by our correspondent at latitude 70 N on Atlantic, about new moon. The records are bold for 10 minutes before and after the Moon’s meridian passing.