Ohmic Power of Ideal Pulsars Andrei Gruzinov CCPP, Physics Department, New York University, 4 Washington Place, New York, NY 10003 1 1 0 2 n ABSTRACT a J Ideal axisymmetric pulsar magnetosphere is calculated from the standard stationary force- 1 free equation but with a new boundary condition at the equator. The new solution predicts 3 Ohmic heating. About 50% of the Poynting power is dissipated in the equatorial current layer outside the light cylinder,with about10%dissipatedbetween1 and1.5 lightcylinder radii. The ] E Ohmic heat presumably goes into radiation, pair production, and accelerationof charges – in an H unknown proportion. . h p 1. Introduction - o r We have shown that ideal pulsars calculated in st the force-freelimit ofStrong-FieldElectrodynam- 1 1 a ics (SFE) dissipate a large fraction of the Poynt- 0.5 0 [ 0 ing flux in the singular current layer outside the -1 -0.5 1 lightcylinder(Gruzinov2011). Thisresult–finite -1 -2 v damping in an ideal system – is not really that 0 1 2 3 0 0.5 1 4 unusual. Burgers equation, for instance, with vis- 4 3 1 8 cosity+0,dissipatesfiniteenergyininfinitelythin 0.8 5 shocks. 0.6 0.4 1. The standard axisymmetric pulsar magneto- 2 0.2 0 sphere features a nearly head-on1 collision of 0 0 0.5 1 1 Poynting fluxes right outside the light cylinder. 1 It is to be expected, although merely by common : 1 v sense,thatsuchacollisionshouldbeaccompanied i by damping. X Here we show that our SFE solution also ob- r 0 a tains from the standard force-free magnetosphere 0 1 2 3 equationofScharlemant&Wagoner(1973),ifone usesthe“correct”boundaryconditionattheequa- torial current layer. Fig. 1.— Everything is in pulsar units. rs =0.25. Lower Left: Isolines of ψ, integer multiples of Weproposethatthe“correct”boundarycondi- 0.1ψ0, ψ0 = 1.44. Upper Left: Poynting flux L tionatthesingularcurrentlayer(whichnowexists and the field invariant I ≡ B2 −E2 at equator only outside the light cylinder) is vs r. Multiple curves for I are different discrete 2 2 approximations – a rough rendering of numerical B −E =0. (1) accuracy. Upper Right: F. LowerRight: Poynting This condition is Lorentz invariant, comes up in flux L through the light cylinder on the field lines with ψ(1,z)<ψ. 1154.6◦,Gruzinov(2005) 1 2 the SFE simulations , and has a clear physical dipole near the surface inside the star. The star meaning (at equator, the field becomes electric- is assumed to be a perfect conductor. Ω is the like in order to drive large current). angular velocity of the star. Wecannotbesurethatourproposalworks,un- We get til one justifies the full SFE, or just eq.(1), micro- φ=ψ, A=A(ψ), (5) scopically. But conversionof 50% of the Poynting where A is an arbitrary function of ψ, and we flux into the Ohmic power (radiation, electron- also get the “Grad-Shafranov-like”pulsar magne- positron pairs) occurring close to the light cylin- tosphere equation for ψ der must have consequences for the pulsar phe- nomenology, and needs to be studied. 2 2 (1−r )∆ψ− ψr+F(ψ)=0. (6) In§2wederivethepulsarmagnetosphereequa- r tion and explain how Contopoulos, Kazanas & Here ∆≡∇2, and F ≡AA′, where the prime de- Fendt(1999)solveit. In§3weputtogetherallthe notestheψ-derivative. Thepulsarmagnetosphere equationswhichareneededtocalculatethepulsar equation (6) is solved outside the star magnetosphere. In §4 we describe the numerical solution and the corresponding physics results. 2 2 2 r +z >rs, (7) 2. Ideal pulsar magnetosphere with the boundary condition at the surface of the star Goldreich & Julian (1969) proposed that neu- r2 2 2 2 tdriotinonstar magnetospheres obey the force-free con- ψ = rs3, r +z =rs. (8) ρE+j×B=0. (2) The pulsar magnetosphere equation (6) con- Surprisingly, it turns out that this simple equa- tains F – an arbitrary function of ψ, and it is tions allows a full calculation of the pulsar mag- not clear how one should solve it. This was ex- netosphere (Scharlemant & Wagoner 1973, Con- plained and done by Contopoulos, Kazanas & topoulos, Kazanas & Fendt 1999, Gruzinov 2005, Fendt (1999) (CKF). Spitkovsky 2006). Thepulsarmagnetosphereequationiselliptical For the stationary axisymmetric case, the cal- bothinsideandoutsidethelightcylinder,andcan culation is as follows. Using axisymmetry and thereforebe solvedif someconditionsaregivenat stationarity,in cylindricalcoordinates(r,θ,z), we all boundaries and if F is known. We first pick a represent the fields by the three scalars φ, ψ, and trial F. A, which depend on r and z but not on θ: Then inside the light cylinder, we have all the boundaryconditions: (i)weknowψ atthesurface 1 E=−∇φ, B= (−ψz,A,ψr), (3) of the star, (ii) ψ = 0 at z = ±∞, (iii) at r = r 1 the boundary condition is given by the pulsar where the subscripts denote the partial deriva- magnetosphere equation itself, ψr = F/2. So we tives. can find ψ inside the light cylinder, say by the We plug (3) into (2) and use ρ = ∇· E and relaxation method using the variation principle. j = ∇×B. We also use the boundary conditions Outside the light cylinder, CKF postulate the at the surface of the star – the continuity of the boundary condition at the equator, normal component of the magnetic field and the tangentialcomponent of the electric field. We use ψ(r,0)=ψ(1,0), r >1. (9) the pulsar units Then, with some boundary conditions at infinity µ=Ω=c=1, (4) (we use ψ =0 at z =±∞ and ψr =0 at r =∞), andwiththesameboundaryconditionatthelight whereµisthemagneticdipolemomentofthestar. cylinder, ψr = F/2 at r = 1, one can solve the It is assumed that the magnetic field is a pure pulsar magnetosphere equation outside the light cylinder too. 2SeeFig.5ofGruzinov(2008). 2 ForagenericF, this proceduregivesa solution We then manually chose g(r), so as to make I(r) with ψ(1 − 0,z) 6= ψ(1 + 0,z). But one might as close to zero as we can at all r > 1. It turns hope that there is a (unique?) F which gives a out, that smooth solution. CKF use a feedback procedure 0.48 g(r)=0.52+ , (18) – numerical adjustment of F leading to a smooth r light cylinder crossing. nullifies I(r) to about the numerical accuracy 3. Inourcase,theboundaryconditionattheequa- The resulting magnetosphere, the function tor outside the light cylinder is F(ψ), the Poynting flux at different radii, and the invariant I(r) are shown in Fig.1. 2 2 2 (r −1)(∇ψ) =A , z =±0, r >1. (10) One gets the spin-down power (the Poynting At the light cylinder, r=1, this gives flux at the stellar surface = the Poynting flux through the light cylinder): A(ψ0)=0, ψ0 ≡ψ(1,0), (11) µ2Ω4 meaning that there is no singular return current Lsd ≈0.9 c3 . (19) on the field line ψ = ψ0. The only singularity is The Ohmic power (the Poynting flux on the field the equatorial current layer. lines that cross the equatorial current layer) is 3. Pulsar magnetosphere equation LOhm ≈0.5Lsd. (20) Insummary,wesolvethefollowingpulsarmag- TheOhmicpowerbetween1and1.5lightcylinder netosphere equation radii is 2 2 ′ LOhm1.5 ≈0.1Lsd. (21) (1−r )∆ψ− ψr +AA =0, (12) r REFERENCES r2 2 2 2 ψ = r3, r +z =rs, (13) Contopoulos, I., Kazanas, D. , Fendt, C., 1999, s Ap.J. 511, 351 2 2 2 (r −1)(∇ψ) =A , z =±0, r >1. (14) Goldreich, P., Julian, W. H., 1969,Ap.J. 157,869 The function A(ψ) is (uniquely?) determined by Gruzinov, A., 2005,Phys. Rev. Lett., 94, 021101 the continuity atthe lightcylinder. The Poynting power is L=R dψA. Gruzinov, A., 2008,JCAP, 11, 002 Gruzinov, A., 2011,arXiv:1101.3100 4. Numerical Solution and Results Scharlemant, E. T., Wagoner, R. V., 1973, Ap.J. Ournumericalprocedureis asfollows. We pick 182, 951 a trial function Spitkovsky, A., 2006, ApJ, 648 L51 ′ g(r)>0, g (r)<0, r >1, g(1)=1. (15) Insteadoftheboundarycondition(14),weimpose 3This numerical procedure, though formally correct, is methodologically inappropriate. In fact, we have ψ(r,0)=ψ0g(r), r >1, ψ0 ≡ψ(1,0). (16) “cheated”. The function g(r) has been read off the SFE time-dependent simulation (we then confirm that various WeapplytheCKFrelaxationmethodtocalculate otherprofilesg(r)don’tnullifyI(r)totheaccuracyshown F(ψ)(ignoringthesingularreturncurrent). Then inFig.1.). What one reallywants isaCKF-typefeedback loop, which would adjust g(r) so as to nullify I(r). Or, wecalculatethefieldinvariantattheequatorout- maybe,onecanenforcethecorrectboundarycondition(14) side the light cylinder throughouttheF relaxation. Wewereunabletodevelopa numericalschemewhichwouldsolvetheproblem(12-14) 1 2 2 2 allbyitself. I(r)= (A +(∇ψ) )−(∇ψ) , z =±0, r >1. r2 This2-columnpreprintwaspreparedwiththeAASLATEX (17) macrosv5.2. 3