Axisymmetric dynamo action produced by differential rotation, with anisotropic electrical conductivity and anisotropic magnetic permeability
UUnder consideration for publication in J. Plasma Phys. Axisymmetric dynamo actionproduced by differential rotation,with anisotropic electrical conductivityand anisotropic magnetic permeability
Franck Plunian † , and Thierry Alboussi`ere Universit´e Grenoble Alpes, Universit´e Savoie Mont Blanc, CNRS, IRD, IFSTTAR, ISTerre,38000 Grenoble, France Univ. Lyon, Univ. Lyon 1, ENSL, CNRS, LGL-TPE, F-69622, Villeurbanne, France(Received xx; revised xx; accepted xx)
The effect on dynamo action of an anisotropic electrical conductivity conjugated to ananisotropic magnetic permeability is considered. Not only is the dynamo fully axisym-metric, but it requires only a simple differential rotation, which twice challenges the well-established dynamo theory. Stability analysis is conducted entirely analytically, leading toan explicit expression of the dynamo threshold. The results show a competition betweenthe anisotropy of electrical conductivity and that of magnetic permeability, the dynamoeffect becoming impossible if the two anisotropies are identical. For isotropic electricalconductivity, Cowling’s neutral point argument does imply the absence of an azimuthalcomponent of current density, but does not prevent the dynamo effect as long as themagnetic permeability is anisotropic.
Key words:
Magnetohydrodynamics, Dynamo effect, Anisotropy
1. Introduction
The dynamo effect is a magnetic instability produced by the displacement of anelectrically conducting medium, without the aid of a magnet or a remanent magneticfield. A part of the kinetic energy of the moving medium is thus transferred into magneticenergy. This process is the most likely candidate to explain the ubiquity of magnetic fieldsobserved in astrophysical objects (Rincon 2019). The increasing resolution of numericalsimulations of dynamo equations makes it possible to reproduce, ever better, the magneticfeatures measured in natural objects (Schaeffer et al. et al. et al. et al. et al. † Email address for correspondence: [email protected] a r X i v : . [ phy s i c s . p l a s m - ph ] F e b F. Plunian, T. Alboussi`ere dynamo effect (Miralles et al. et al. et al. et al. et al. et al. et al. et al.
2. Conductivity and permeability anisotropy
We consider a material such that the electrical conductivity and magnetic perme-ability are denoted σ (cid:107) and µ (cid:107) in a given direction q , and σ ⊥ and µ ⊥ in the directionsperpendicular to q .Writing Ohm’s law, J = σ (cid:107) E in the direction of q and J = σ ⊥ E in the directionsperpendicular to q , leads to the following conductivity tensor:[ σ ij ] = σ ⊥ δ ij + ( σ (cid:107) − σ ⊥ ) q i q j . (2.1) ynamo action with anisotropic conductivity and permeability (cid:3045)(cid:3087) 𝛼 (cid:3053) Figure 1.
Left: The inner-cylinder of radius R rotates as a solid-body within an outer cylinderat rest. The radius R (cid:48) of the outer cylinder is taken as infinite. Right: The curved lines areperpendicular to q and describe logarithmic spirals. They correspond to the directions alongwhich σ = σ ⊥ and µ = µ ⊥ . Inverting (2.1) leads to the resistivity tensor: (Ruderman & Ruzmaikin 1984)[ σ ij ] − = 1 σ ⊥ δ ij + ( 1 σ (cid:107) − σ ⊥ ) q i q j . (2.2)Similarly, a magnetic permeability tensor can be defined as[ µ ij ] = µ ⊥ δ ij + ( µ (cid:107) − µ ⊥ ) q i q j , (2.3)with the inverse tensor [ µ ij ] − = 1 µ ⊥ δ ij + ( 1 µ (cid:107) − µ ⊥ ) q i q j . (2.4)We choose q as a unit vector in the horizontal plane: q = c e r + s e θ , (2.5)where ( e r , e θ , e z ) is a cylindrical coordinate system, with c = cos α and s = sin α , α being a prescribed angle.In figure 1 the curved lines are perpendicular to q and describe logarithmic spirals.They correspond to the directions along which σ = σ ⊥ and µ = µ ⊥ . We consider thesolid-body rotation U of a cylinder of radius R embedded in an infinite medium at rest.Both regions are made of the same material, with therefore identical conductivity tensorsand identical permeability tensors. F. Plunian, T. Alboussi`ere
3. Induction equation
In the magnetohydrodynamic approximation, the Maxwell equations and Ohm’s lawtake the form H = [ µ ij ] − B , (3.1) J = ∇ × H , (3.2) ∂ t B = −∇ × E (3.3) ∇ · B = 0 , (3.4) J = [ σ ij ]( E + U × B ) , (3.5)where H , B , J , E and U are the magnetic field, the induction field, the current density,the electric field and the velocity field. The induction equation then takes the form ∂ t B = ∇ × ( U × B ) − ∇ × (cid:0) [ σ ij ] − ∇ × (cid:0) [ µ ij ] − B (cid:1)(cid:1) . (3.6)Renormalizing the distance, electrical conductivity, magnetic permeability and time byrespectively R, µ ⊥ , σ ⊥ and µ ⊥ σ ⊥ R , the dimensionless form of the induction equation isidentical to (3.6), but with[ σ ij ] − = δ ij + σq i q j , σ = σ ⊥ σ (cid:107) − , (3.7)[ µ ij ] − = δ ij + µq i q j , µ = µ ⊥ µ (cid:107) − U = (cid:40) rΩ e θ , r < , r > , (3.9)where Ω is the dimensionless angular velocity of the inner cylinder. We note that ( σ, µ ) ∈ [ − , + ∞ [ , with σ = 0 and µ = 0 corresponding to respectively isotropic conductivityand isotropic permeability.Provided the velocity is stationary and z -independent, an axisymmetric magneticinduction can be searched in the form B ( r, z, t ) = ˜ B e θ + ∇ × (cid:16) ˜ A e θ (cid:17) , (3.10)with ( ˜ A, ˜ B ) = ( A ( r ) , B ( r )) exp( γt + i kz ), where γ is the instability growth rate and k thevertical wavenumber of the corresponding eigenmode, and where A and B depend onlyon the radial coordinate r . Thus the magnetic induction takes the form B = (cid:18) − i kA, B, r ∂ r ( rA ) (cid:19) exp( γt + i kz ) , (3.11)with dynamo action corresponding to (cid:60){ γ } > ∇ × ( U × B ) = 0 in each region r < r > γA + (1 + σs ) D k ( A ) + µc k A = i csk ( σ − µ ) B, (3.12) γB + (1 + µs ) D k ( B ) + σc k B = − i csk ( σ − µ ) D k ( A ) , (3.13)where D ν ( X ) = ν X − ∂ r (cid:0) r ∂ r ( rX ) (cid:1) . The derivation of (3.12) and (3.13) is givenin appendix B. For σ = µ = 0, corresponding to isotropy of both conductivity andpermeability, (3.12) and (3.13) are diffusion equations, leading to a free decaying solution ynamo action with anisotropic conductivity and permeability µ = 0, (3.12) and (3.13) are identicalto the equations derived in Plunian & Alboussi`ere (2020).
4. Dynamo threshold
General form of the solutions
Looking for non-oscillating solutions, the dynamo threshold then corresponds to γ = 0.Thus, taking γ = 0 in (3.12) and (3.13), it can be shown (Appendix C) that (cid:0) D k µ ◦ D k σ (cid:1) ( A ) = (cid:0) D k σ ◦ D k µ (cid:1) ( B ) = 0 , (4.1)where k σ = k (cid:18) σ σs (cid:19) / , k µ = k (cid:18) µ µs (cid:19) / . (4.2)We note that the two operators D k σ and D k µ are commutative. Therefore in (4.1) wecan apply the two operators in the order we want, D k µ ◦ D k σ or D k σ ◦ D k µ , to both A and B . The set of functions X ( r ), satisfying the fourth-order differential equation (cid:0) D k µ ◦ D k σ (cid:1) ( X ) = 0, is a vector space of dimension 4. Now, we know that, whatever ν , the solutions of D ν ( X ) = 0 are a linear combination of I ( νr ) and K ( νr ), where I and K are modified Bessel functions of first and second kind, of order 1. Therefore thesolutions of (4.1) are a linear combination of I ( k σ r ), K ( k σ r ), I ( k µ r ) and K ( k µ r ).Looking for A in the form A = α σ I ( k σ r ) + β σ K ( k σ r ) + α µ I ( k µ r ) + β µ K ( k µ r ) , (4.3)and specifying that A must be finite at r = 0 and that lim r →∞ A = 0, leads to A = (cid:40) r < , α σ I ( k σ r ) + α µ I ( k µ r ) r > , β σ K ( k σ r ) + β µ K ( k µ r ) , (4.4)where α σ , β σ , α µ and β µ are free parameters that will be constrained by additionalboundary conditions at r = 1. Replacing (4.4) in (3.12) for γ = 0 leads to the followingexpression for BB = r < , i cks (cid:18) α σ I ( k σ r ) + µs µs α µ I ( k µ r ) (cid:19) r > , i cks (cid:18) β σ K ( k σ r ) + µs µs β µ K ( k µ r ) (cid:19) , (4.5)the derivation of which being given in Appendix D.4.2. Boundary conditions at r = 1From the Maxwell equations and Green-Ostrogradski and Stokes theorems, the radialcomponent of B and the tangential components of H = [ µ ij ] − B must be continuousat r = 1. Taking the expression of B and H given in (3.11) and (B 1), these continuityconditions can be written in terms of A and B as A ( r = 1 − ) = A ( r = 1 + ) , (4.6) B ( r = 1 − ) = B ( r = 1 + ) , (4.7) ∂ r A ( r = 1 − ) = ∂ r A ( r = 1 + ) . (4.8) F. Plunian, T. Alboussi`ere
Taking A and B given in (4.4) and (4.5) and replacing them in (4.6), (4.7) and (4.8)leads to (Appendix E) α σ I ( k σ ) − β σ K ( k σ ) = 0 (4.9) α µ I ( k µ ) − β µ K ( k µ ) = 0 (4.10) α σ k σ I ( k σ ) + α µ k µ I ( k µ ) + β σ k σ K ( k σ ) + β µ k µ K ( k µ ) = 0 , (4.11)where I and K are modified Bessel functions of first and second kind, of order 0. Itis convenient to introduce the parameters λ σ = α σ I ( k σ )i k/s and λ µ = α µ I ( k µ )i k/s .Then, using (4.9) and (4.10), we can rewrite A and B in the following form:i kA = r < , s (cid:18) λ σ I ( k σ r ) I ( k σ ) + λ µ I ( k µ r ) I ( k µ ) (cid:19) r > , s (cid:18) λ σ K ( k σ r ) K ( k σ ) + λ µ K ( k µ r ) K ( k µ ) (cid:19) , (4.12) B = r < , c (cid:18) λ σ I ( k σ r ) I ( k σ ) + µs µs λ µ I ( k µ r ) I ( k µ ) (cid:19) r > , c (cid:18) λ σ K ( k σ r ) K ( k σ ) + µs µs λ µ K ( k µ r ) K ( k µ ) (cid:19) . (4.13)The continuity of ∂ r A at r = 1, given by (4.11), then leads to the following identitybetween λ σ and λ µ λ σ Γ ( k σ ) + λ µ Γ ( k µ ) = 0 , (4.14)with Γ ( x ) = x (cid:18) I ( x ) I ( x ) + K ( x ) K ( x ) (cid:19) ≡ ( I ( x ) K ( x )) − , (4.15)the last equality coming from the Wronskian relation I m ( x ) K m +1 ( x ) + I m +1 ( x ) K m ( x ) = 1 /x. (4.16)In figure 2 the eigenmodes i kA and B are plotted versus r for λ σ = Γ ( k µ ) and λ µ = − Γ ( k σ ) such that (4.14) is satisfied.Finally, the tangential components E θ and E z of the electric field E = − U × B + [ σ ij ] − J (4.17)have to be continuous at r = 1. The expression of the current density J , which is derivedin Appendix F, is given by J r = − i ckλ σ (cid:40) r < , I ( k σ r ) /I ( k σ ) r > , K ( k σ r ) /K ( k σ ) , (4.18) J θ = − σsc σs J r , (4.19) J z = ck σ λ σ (cid:40) r < , I ( k σ r ) /I ( k σ ) r > , − K ( k σ r ) /K ( k σ ) , (4.20)where the coefficient exp(i kz ) has been dropped for convenience.From (3.7), (3.9) and (4.19), we find that E θ = 0, which is in agreement with ynamo action with anisotropic conductivity and permeability Figure 2.
Eigenmodes i kA (dashed lines) and B (solid lines) versus r , for σ = 10 , k = 1 . , α = 0 . π , λ σ = Γ ( k µ ), λ µ = − Γ ( k σ ) and for µ ∈ {− . , , . } . axisymmetric solutions. Indeed, Maxwell equation (3.3) taken at the threshold implies ∇ × E = 0. Applying the Stokes theorem to the integral of ∇ × E on a disc of radius r ,and assuming axisymmetry, then leads to E θ ( r ) = 0.The continuity of E z implies the following identity:( J z + rΩ B r )( r = 1 − ) = J z ( r = 1 + ) . (4.21)Replacing (4.12) and (4.20) in (4.21), and using (4.14), leads to the dynamo threshold Ω c = cs ( I ( k σ ) K ( k σ ) − I ( k µ ) K ( k µ )) − . (4.22)
5. Analysis of the results
Dispersion relation
A striking consequence of (4.22) is that Ω c ( σ, µ ) is antisymmetric, satisfying Ω c ( σ, µ ) = − Ω c ( µ, σ ) . (5.1)In addition for identical anisotropies of conductivity and permeability, µ = σ , thethreshold is infinite, leading to the impossibility of an axisymmetric dynamo,lim | σ − µ |→ | Ω c ( σ, µ ) | → + ∞ . (5.2)This is illustrated in figure 3, in which the curves of a few isovalues of Ω c are plottedversus k σ and k µ . In particular, having both σ (cid:29) µ (cid:29) k σ ≈ k µ ≈ k/s .The antisymmetry property (5.1) of Ω c ( σ, µ ) can also be derived directly from the set F. Plunian, T. Alboussi`ere
Figure 3.
Isovalues of Ω c ∈ {± , ± , ± , ± , ± } in the ( k σ , k µ ) map, for α = 0 . π .The diagonal k σ = k µ corresponds to Ω c → ±∞ . Figure 4.
Curves of the dynamo threshold Ω c versus k , for µ = 0, α = 0 . π and σ ∈ {− . , − . , − . , − . , , , , + ∞} . ynamo action with anisotropic conductivity and permeability Figure 5.
Curves of the dynamo threshold Ω c versus k , for σ = 10 , α = 0 . π and µ ∈ {− . , − . , − . , , , , } . of equations (3.12-3.13) taken for γ = 0, the boundary conditions (4.6-4.8) and (4.21),without deriving explicitly the expressions of A and B . This is shown in Appendix G.Alternatively, changing α to − α in (4.22) also changes Ω c to − Ω c . This can be alsoderived directly from (3.12-3.13), taken for γ = 0, the boundary conditions (4.6-4.8), and(4.21), by changing A to − A (or B to − B ).We check that for an isotropic permeability, µ = 0, Ω c is the same as that given inPlunian & Alboussi`ere (2020). In figure 4 the curves of the dynamo threshold Ω c versus k are plotted for µ = 0 (isotropic magnetic permeability), α = 0 . π and different valuesof σ . The negative values of σ correspond to an electrical conductivity that is the highestin the direction parallel to q .In figure 5 the curves of the dynamo threshold Ω c versus k are plotted for σ = 10 , α = 0 . π and different values of µ . For µ = 0, the minimum value of | Ω c | is obtainedfor k = 1 . α = 0 . π , and is equal to min k,α | Ω c | = 14 .
61 (Plunian & Alboussi`ere2020). For positive values of µ , | Ω c | increases with µ , showing the detrimental effect ofhaving both a high σ and a high µ . For negative values of µ , | Ω c | decreases with | µ | ,showing that the dynamo effect is favoured if the permeability is higher in the directionparallel to q .For s = 0 (radial q ) or c = 0 (azimuthal q ), the dynamo is impossible. This is obviousfor s = 0 as the threshold given by (4.22) is infinite. For c = 0, (4.13) implies that B = 0.In addition, as s = 1, (4.2) implies that k σ = k µ = k . Then, from (4.12) and (4.14), wefind that A = 0. 5.2. Current density
Concerning the current density J , given at the threshold by (4 . . σ , and not on µ . In other words, taking an anisotropic magnetic0 F. Plunian, T. Alboussi`ere permeability µ (cid:54) = 0 does not change the geometry of the current density with respect tothe isotropic case µ = 0.For an isotropic conductivity σ = 0, we find that J θ = 0. This corresponds to theneutral point argument of Cowling (1934), after which a toroidal current density cannotbe produced if axisymmetry is assumed. However, and although such a neutral pointargument is satisfied for σ = 0, this does not exclude the possibility of dynamo actionfor an anisotropic magnetic permeability µ (cid:54) = 0.From (4.19), we note that the projection in the ( r , θ ) plane of the current density J describes spiralling trajectories. In the limit σ → + ∞ , we find that J · q = 0.5.3. Magnetic induction
From the expression of B given in (3.11), and applying (4.12), (4.13) and (E 5), leadsto the following expressions for the magnetic induction components B r = − s r < , λ σ I ( k σ r ) I ( k σ ) + λ µ I ( k µ r ) I ( k µ ) r > , λ σ K ( k σ r ) K ( k σ ) + λ µ K ( k µ r ) K ( k µ ) , (5.3) B θ = c r < , λ σ I ( k σ r ) I ( k σ ) + λ µ (cid:18) µs µs (cid:19) I ( k µ r ) I ( k µ ) r > , λ σ K ( k σ r ) K ( k σ ) + λ µ (cid:18) µs µs (cid:19) K ( k µ r ) K ( k µ ) , (5.4) B z = − i sk r < , λ σ k σ I ( k σ r ) I ( k σ ) + λ µ k µ I ( k µ r ) I ( k µ ) r > , − λ σ k σ K ( k σ r ) K ( k σ ) − λ µ k µ K ( k µ r ) K ( k µ ) , (5.5)where, again, the coefficient exp(i kz ) has been dropped for convenience. In contrast to J ,the induction field depends not only on σ , but also on µ , implying the following remarks.In the case of identical anisotropic conductivity and permeability, σ = µ , as mentionedearlier the dynamo is impossible. From (4.2) and (4.14) we have k σ = k µ and λ σ + λ µ = 0,implying that B r = B z = 0. In that case the induction field B is then purely toroidal.This is in agreement with the antidynamo theorem of Kaiser et al. (1994), after whichan invisible dynamo, with a purely toroidal magnetic field, is impossible.In the limit µ → ∞ , from (5.3) and (5.4) we have c B r = − s B θ , implying that B · q =0. The projection in the ( r , θ ) plane of the induction field B thus describes spirallingtrajectories perpendicular to q .In figure 6 the current lines of B and J are plotted in the horizontal plane for differentvalues of ( σ, µ ). In figures 6a, 6b and 6c, σ = 10 and µ ∈ {− .
99; 0; 5 } . The currentlines of J are identical because, as previously seen, J does not depend on µ . From 6ato 6c, increasing µ has the effect of distorting the B current lines in the outer cylinder,such that the current lines of B reach the same curvature as the current lines of J ,which eventually is detrimental to dynamo action. In figures 6d, 6e and 6f, µ = 10 and σ ∈ {− .
99; 0; 10 } . From 6d to 6f, increasing σ has the effect of distorting the J currentlines in both inner and outer cylinders, such that the current lines of J reach the samecurvature direction as the current lines of B , which ultimately is again detrimental tothe dynamo action. We note that the J current lines in figures 6a, 6b and 6c and the B ynamo action with anisotropic conductivity and permeability Figure 6.
Current lines of B (solid lines) and J (dashed lines) in the horizontal plane for α = 0 . π , k = 1 .
1, and for (a) ( σ, µ ) = (10 , − . σ, µ ) = (10 , σ, µ ) = (10 , σ, µ ) = ( − . , ), (e) ( σ, µ ) = (0 , ), (f) ( σ, µ ) = (10 , ). current lines in figures 6d, 6e and 6f are identical. This is because σ (cid:29) J · q ≈
0, and µ (cid:29) B · q ≈
6. Dynamo mechanism
The set of equations (3.12-3.13) can be rewritten in terms of B r and B θ as γ B r = cs ( σ − µ ) k B θ − (1 + σs ) D ˜ k σµ ( B r ) (6.1) γ B θ = cs ( σ − µ ) D k ( B r ) − (1 + µs ) D ˜ k µσ ( B θ ) , (6.2)with ˜ k σµ = k (cid:18) µc σs (cid:19) / , ˜ k µσ = k (cid:18) σc µs (cid:19) / . (6.3)On the right hand side of each equation (6.1) and (6.2), the first term is a source term forthe dynamo effect, while the second term is a decay term. In (6.1), resp. (6.2), the term2 F. Plunian, T. Alboussi`ere cs ( σ − µ ) k B θ , resp. cs ( σ − µ ) D k ( B r ), corresponds to the generation of B r from B θ ,resp. B θ from B r . The differential rotation between the inner and outer cylinders alsoparticipates in the generation of B θ from B r , through the boundary condition (4.21). Thelatter is, however, not sufficient in itself. Therefore, it is clear why increasing the value of | σ − µ | helps for the dynamo effect, and why the dynamo is impossible for σ = µ . Dynamoaction thus occurs through differential rotation conjugated to anisotropic diffusion.From the point of view of basic Maxwell and Ohm equations, and in the case of anisotropic conductivity ( σ = 0) and anisotropic magnetic permeability ( µ (cid:54) = 0), dynamoaction can be understood in the following way. Suppose there exists an axisymmetricmagnetic induction disturbance with a non-zero radial component B r at some height z along the shear zone ( r = 1). Ohm’s law (3.5) then drives two opposite currents inthe axial direction e z within the rotor and stator. In a medium of isotropic electricalconductivity, this current forms closed loops in the meridian planes, as can be seenin Fig. 6(e). From Amp`ere’s law (3.2), this generates an azimuthal magnetic field H θ .Finally from (3.1), a radial component of the induction vector B r is generated from H θ because of the anisotropic magnetic permeability. Depending on the orientation of theanisotropic permeability tensor and the direction of the solid rotation, the generated B r can either reinforce (dynamo action is possible) or oppose the initial seed of radialmagnetic induction (no dynamo).
7. Conclusions
For an anisotropic electrical conductivity ( σ ⊥ (cid:54) = σ (cid:107) ) conjugated to an anisotropic mag-netic permeability ( µ ⊥ (cid:54) = µ (cid:107) ), we could think that maximizing the ratio ( σ ⊥ µ ⊥ ) / ( σ (cid:107) µ (cid:107) )could help for the dynamo action. This would correspond to minimizing magneticdiffusivity in the perpendicular direction relative to that in the parallel direction. Thisis not true for two reasons. First, contrary to the isotropic case, defining an anisotropicmagnetic diffusivity is meaningless, because the electrical conductivity and magneticpermeability are now tensors. Second, it has been shown that taking σ ⊥ /σ (cid:107) (cid:29) µ ⊥ /µ (cid:107) (cid:29) J and magneticinduction B in the same direction q ⊥ . In contrast, having σ ⊥ /σ (cid:107) (cid:29) µ ⊥ /µ (cid:107) = 1,or σ ⊥ /σ (cid:107) = 1 and µ ⊥ /µ (cid:107) (cid:29) | Ω c | = 14 .
61, for k = 1 . α = 0 . π .As an application let us consider an experimental demonstration of the dynamoeffect based on such conductivity and permeability spiral anisotropy, with differentialrotation between two cylinders, as sketched in figure 1. An anisotropic conductivity,resp. permeability, can be manufactured by alternating thin layers of two materials withdifferent conductivities, resp. permeabilities. Although the resulting medium is no longeraxisymmetric, our model is still a good approximation of such an experiment. To realizethe first case σ ⊥ /σ (cid:107) (cid:29) µ ⊥ /µ (cid:107) = 1, we can alternate spiral layers of a high electricalconductivity material, e.g. copper, and a material which is electrically insulating, e.g.epoxy resin, both having a relative magnetic permeability equal to unity. To realize thesecond case σ ⊥ /σ (cid:107) = 1 and µ ⊥ /µ (cid:107) (cid:29) µ -metal (permalloy), and a material with a relative magneticpermeability equal to unity, e.g. stainless steel, both having approximately the sameelectrical conductivity. The current lines of B and J of these two cases are illustratedin figure 6b and 6e. For the second case, a crucial issue will be to guarantee a goodelectrical contact between both materials, µ -metal and stainless steel. Indeed, if this is ynamo action with anisotropic conductivity and permeability σ ⊥ /σ (cid:107) (cid:29) µ ⊥ /µ (cid:107) (cid:29) σ ⊥ /σ (cid:107) = 1, and as illustrated infigure 6e, the azimuthal current density is null, J θ = 0 , which is in agreement with theneutral point argument of Cowling. As J = ∇ × H , this implies that the circulation ofthe poloidal component of H on a closed current line is zero (Cowling 1934). However, asshown in (B 1), in the case of an anisotropic magnetic permeability this does not implythat the poloidal component of B is zero. Therefore, although the neutral point argumentof Cowling still holds, it does not imply the impossibility of a dynamo effect. Acknowledgements
We thank R. Deguen for fruitful discussions on Earth’s inner core anisotropy.
Appendix A. Derivation of ∇ × ( U × B ) = 0 Assuming axisymmetry ( ∂ θ = 0), the curl of the cross product of U = rΩ e θ and B = ( B r , B θ , B z ) is given by ∇ × ( U × B ) = ( ∂ z ( rΩB z ) + ∂ r ( rΩB r )) e θ . Assuming that Ω is constant in space and using the solenoidality of B , ∇· B = 0, leads to ∇× ( U × B ) = 0. Appendix B. Derivation of (3.12) and (3.13)
The product of [ µ ij ] − = µc µcs µcs µs
00 0 1 given by (3.8), and the inductionfield B = − i kAB r ∂ r ( rA ) exp( γt + i kz ) given by (3.11), leads to the magnetic field H = [ µ ij ] − B = − i k (1 + µc ) A + µcsB − i kµcsA + (1 + µs ) B r ∂ r ( rA ) , (B 1)where, from now, the exponential term is dropped for convenience. Assuming axisym-metry, the curl of H takes the form ∇ × H = − i kH θ i kH r − ∂ r H z r ∂ r ( rH θ ) , leading to the currentdensity J = ∇ × H = − k µcsA − i k (1 + µs ) BD k ( A ) + µc k A + i µcskB − i µcsk r ∂ r ( rA ) + (1 + µs ) r ∂ r ( rB ) , (B 2)where D ν ( X ) = ν X − ∂ r (cid:0) r ∂ r ( rX ) (cid:1) . The product of [ σ ij ] − = σc σcs σcs σs
00 0 1 given by (3.7), and J given by (B 2), leads to[ σ ij ] − J = − k µcsA + σcsD k ( A ) − i k (1 + σc + µs ) Bµc k A + (1 + σs ) D k ( A ) − i csk ( σ − µ ) B − i µcsk r ∂ r ( rA ) + (1 + µs ) r ∂ r ( rB ) . (B 3)4 F. Plunian, T. Alboussi`ere
Taking the curl leads to ∇ × [ σ ij ] − J = − i kFG r ∂ r ( rF ) , with F = µc k A + (1 + σs ) D k ( A ) − i kcs ( σ − µ ) B, (B 4) G = i kcs ( σ − µ ) D k ( A ) + σk c B + (1 + µs ) D k ( B ) . (B 5)As ∇ × ( U × B ) = 0, the induction equation (3.6) is reduced to ∂ t B = −∇ × [ σ ij ] − J ,leading to γA = − F, (B 6) γB = − G, (B 7) γr ∂ r ( rA ) = − r ∂ r ( rF ) , (B 8)and then to (3.12) and (3.13). Appendix C. Derivation of the fourth-order differential equation(4.1) satisfied by A and B at the dynamo threshold Replacing γ = 0 in (3.12) and (3.13) leads to the following system(1 + σs ) D k ( A ) + µc k A = i csk ( σ − µ ) B, (C 1)(1 + µs ) D k ( B ) + σc k B = − i csk ( σ − µ ) D k ( A ) , (C 2)where, again, D ν ( X ) = ν X − ∂ r (cid:0) r ∂ r ( rX ) (cid:1) .It is straightforward to show that(1 + σs ) D k ( X ) = (1 + σs ) D k σ ( X ) − σc k X, (C 3)(1 + µs ) D k ( X ) = (1 + µs ) D k µ ( X ) − µc k X, (C 4)where k σ and k µ are defined in (4.2) and that we rewrite here for convenience k σ = k (cid:18) σ σs (cid:19) / , k µ = k (cid:18) µ µs (cid:19) / . Using (C 3) and (C 4) in (C 1) and (C 2) then leads to(1 + σs ) D k σ ( A ) = ck ( σ − µ )( ckA + i sB ) , (C 5)(1 + µs ) D k µ ( B ) = − ck ( σ − µ )( ckB + i sD k ( A )) . (C 6)Then to obtain (4.1) we need to demonstrate that D k µ ( ckA + i sB ) = 0 and D k σ ( ckB +i sD k ( A )) = 0. For that, we rewrite (C 1) and (C 2) as D k ( A ) = σ (cid:0) − s D k ( A ) + i cskB (cid:1) − µ (cid:0) c k A + i cskB (cid:1) , (C 7) D k ( B ) = − σ (cid:0) c k B + i cskD k ( A ) (cid:1) + µD k (cid:0) − s B + i cskA (cid:1) . (C 8)Multiplying (C 7) by ck , (C 8) by i s , and adding both quantities leads to(1 + µs ) D k ( ckA + i sB ) = − µc k ( ckA + i sB ) , (C 9)which, from (C 4) with X = ckA + i sB , is equivalent to D k µ ( ckA + i sB ) = 0 . (C 10)Applying (C 10) to (C 5) then leads to (cid:0) D k µ ◦ D k σ (cid:1) ( A ) = 0 . (C 11) ynamo action with anisotropic conductivity and permeability D k of (C 7) multiplied by i s on the one hand, and (C 8) multiplied by ck on theother hand, and adding both quantities leads to(1 + σs ) D k ( ckB + i sD k ( A )) = − σc k ( ckB + i sD k ( A )) , (C 12)which, from (C 3) with X = ckB + i sD k ( A ), is equivalent to D k σ ( ckB + i sD k ( A )) = 0 . (C 13)Applying (C 13) to (C 6) leads to (cid:0) D k σ ◦ D k µ (cid:1) ( B ) = 0 , (C 14)which, together with (C 11), corresponds to (4.1). Appendix D. Derivation of B , given in (4.5), at the dynamo threshold Starting from (4.4), which we rewrite here as A = (cid:40) r < , α σ I ( k σ r ) + α µ I ( k µ r ) r > , β σ K ( k σ r ) + β µ K ( k µ r ) , we will derive B from (3.12), which we write here for γ = 0 as B = (cid:0) (1 + σs ) D k ( A ) + µc k A (cid:1) / (i csk ( σ − µ )) . (D 1)Using the relations (C 3) and (C 4), and knowing that, whatever ν , D ν ( I ( k ν r )) = D ν ( K ( k ν r )) = 0 we find that D k ( A ) = r < , − σc k σs α σ I ( k σ r ) − µc k µs α µ I ( k µ r ) r > , − σc k σs β σ K ( k σ r ) − µc k µs β µ K ( k µ r ) . (D 2)Then, replacing in (D 1) the expressions of A and D k ( A ) given by (4.4) and (D 2) leadsto the following expression for B , which is also given in (4.5): B = r < , i cks (cid:18) α σ I ( k σ r ) + µs µs α µ I ( k µ r ) (cid:19) r > , i cks (cid:18) β σ K ( k σ r ) + µs µs β µ K ( k µ r ) (cid:19) . Appendix E. Derivation of the boundary conditions (4.9), (4.10) and(4.11) at the dynamo threshold
The continuity of A and B at r = 1, taken from their expressions (4.4) and (4.5), takesthe following form: α σ I ( k σ ) + α µ I ( k µ ) = β σ K ( k σ ) + β µ K ( k µ ) (E 1) α σ I ( k σ ) + µs µs α µ I ( k µ ) = β σ K ( k σ ) + µs µs β µ K ( k µ ) (E 2)leading to (4.9) and (4.10), which we rewrite here as α σ I ( k σ ) − β σ K ( k σ ) = 0 α µ I ( k µ ) − β µ K ( k µ ) = 0 . F. Plunian, T. Alboussi`ere
To write the continuity of ∂ r A at r = 1 we first need to calculate the expression of ∂ r A at any r . Using the following relations satisfied whatever ν : ∂ r ( I ( νr )) = νI ( νr ) − r I ( νr ) , (E 3) ∂ r ( K ( νr )) = − νK ( νr ) − r K ( νr ) , (E 4)the expression of ∂ r A is obtained by deriving (4.4): ∂ r A = r < , α σ (cid:18) k σ I ( k σ r ) − r I ( k σ r ) (cid:19) + α µ (cid:18) k µ I ( k µ r ) − r I ( k µ r ) (cid:19) r > , β σ (cid:18) − k σ K ( k σ r ) − r K ( k σ r ) (cid:19) + β µ (cid:18) − k µ K ( k µ r ) − r K ( k µ r ) (cid:19) . (E 5)Then, the continuity of ∂ r A at r = 1 leads to α σ ( k σ I ( k σ ) − I ( k σ )) + α µ ( k µ I ( k µ ) − I ( k µ )) = β σ ( − k σ K ( k σ ) − K ( k σ )) + β µ ( − k µ K ( k µ ) − K ( k µ )) . (E 6)Then, taking advantage of (4.9) and (4.10), (E 6) can be simplified to α σ k σ I ( k σ ) + α µ k µ I ( k µ ) + β σ k σ K ( k σ ) + β µ k µ K ( k µ ) = 0 , which is (4.11). Appendix F. Derivation of the current density J at the dynamothreshold
We rewrite the current density J which is given in (B 2) as J = ∇ × H = − i kφD k ( A ) + µc k A + i µcskB r ∂ r ( rφ ) , with φ = − i kµcsA + (1 + µs ) B . At the dynamo threshold A and B can be replaced bytheir expressions (4.4) and (4.5), leading to φ = r < , i cks α σ I ( k σ r ) r > , i cks β σ K ( k σ r ) . (F 1)Using the relations (E 3) and (E 4) leads to1 r ∂ r ( rφ ) = r < , i cks α σ k σ I ( k σ r ) r > , − i cks β σ k σ K ( k σ r ) . (F 2)Using (C 4), we find that D k ( A ) + µc k A + i µcskB = D k µ ( A ) + i µcsk µs φ. (F 3)Then from the expression of A given at the threshold by (4.4), we have D k µ ( A ) = (cid:40) r < , α σ D k µ ( I ( k σ r )) r > , β σ D k µ ( K ( k σ r )) . (F 4) ynamo action with anisotropic conductivity and permeability D k µ ( X ) = D k σ ( X ) + c k (cid:18) µ µs − σ σs (cid:19) X, (F 5)implying that D k µ ( A ) = r < , α σ c k (cid:18) µ µs − σ σs (cid:19) I ( k σ r ) r > , β σ c k (cid:18) µ µs − σ σs (cid:19) K ( k σ r ) , (F 6)where, again, we used the property that, whatever ν , D ν ( I ( k ν r )) = D ν ( K ( k ν r )) = 0.Therefore we find that D k µ ( A ) + i µcsk µs φ = r < , − σc k σs α σ I ( k σ r ) r > , − σc k σs β σ K ( k σ r ) . (F 7)Then the current density takes the following formfor r < , J = ck s α σ I ( k σ r ) − σc k σs α σ I ( k σ r ) i cks α σ k σ I ( k σ r ) , (F 8)for r > , J = ck s β σ K ( k σ r ) − σc k σs β σ K ( k σ r ) − i cks β σ k σ K ( k σ r ) . (F 9)Then, substituting α σ and β σ by their expressions in terms of λ σ , α σ = − i sλ σ / ( kI ( k σ ))and β σ = − i sλ σ / ( kK ( k σ )), leads to (4.18-4.20). Appendix G. Derivation of the antisymmetric relation (5.1)
Let us rewrite the set of equations (3.12-3.13) for γ = 0, the boundary conditions(4.6-4.8) and (4.21): (1 + σs ) D k ( A ) + µc k A = i csk ( σ − µ ) B, (G 1)(1 + µs ) D k ( B ) + σc k B = − i csk ( σ − µ ) D k ( A ) , (G 2)[ A ] r =1 + r =1 − = [ B ] r =1 + r =1 − = [ ∂ r A ] r =1 + r =1 − = 0 , (G 3)(1 + µs ) [ ∂ r B ] r =1 + r =1 − = − i kΩA ( r = 1 − ) , (G 4)where [ X ] r =1 + r =1 − = X ( r = 1 + ) − X ( r = 1 − ), and (G 4) being derived from (4.21) using(B 2). The system (G 1-G 4) is the complete system of equations leading to the dynamothreshold (4.22).Now, defining the new variables A (cid:48) and B (cid:48) as A (cid:48) = − σs µs A, (G 5) B (cid:48) = B − i k cs (cid:18) σs + µs σs (cid:19) A, (G 6)8 F. Plunian, T. Alboussi`ere and replacing them into (G 1-G 4) leads to(1 + µs ) D k ( A (cid:48) ) + σc k A (cid:48) = i csk ( µ − σ ) B (cid:48) , (G 7)(1 + σs ) D k ( B (cid:48) ) + µc k B (cid:48) = − i csk ( µ − σ ) D k ( A (cid:48) ) , (G 8)[ A (cid:48) ] r =1 + r =1 − = [ B (cid:48) ] r =1 + r =1 − = [ ∂ r A (cid:48) ] r =1 + r =1 − = 0 , (G 9)(1 + σs ) [ ∂ r B (cid:48) ] r =1 + r =1 − = i kΩA (cid:48) ( r = 1 − ) . (G 10)It shows that the new variables A (cid:48) and B (cid:48) obey to the same equations as A and B , inwhich σ and µ have been changed to µ and σ , and Ω to − Ω . REFERENCESAlboussi`ere, T., Drif, K. & Plunian, F.
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