Force-Free Black Hole Magnetospheres

Publications of the Astronomical Society of Japan

Okamoto, Isao¹ 

[1] National Astoronomical Observatory of Japan, 2-21-1 Osawa, Mitaka-shi, Tokyo 181-8588 Institute for Black Hole Mining, 114-6, Ochikawa, Hino-shi, Tokyo 191-0034

关键词: "black holes";    MHD;    "ISM: magnetic fields";    "stars: winds and outflows";   

DOI  :  10.1093/pasj/61.5.971

学科分类：天文学(综合)

来源: Oxford University Press

【 摘 要 】

Itisarguedthataforce-freedegenerateelectrodynamic(FFDE)magnetosphereofaKerrblackholewith0$$lt$$$$Omega_{mF}$$$$lt$$$${Omega_{mH}}$$consistsoftheouterclassicalandinnergeneral-relativisticdomains.ThisisdescribedbyasimpleDCdual-circuitmodel,withdissipativemembranesastwoloadsata“force-freeinfinitysurface”(S$$_{{mff}infty}$$)with$$omega$$$$=$$0andata“force-freehorizonsurface”(S$$_{mffH}$$)with$$omega$$$$=$$$${Omega_{mH}}$$,where$$omega$$,$${Omega_{mH}}$$,and$$Omega_{mF}$$aretheframe-dragging,thehorizonandthefieldlineangularfrequencies.Itisbeneathuppernullsurface,S$$_{mN}$$,at$$omega$$$$=$$$$Omega_{mF}$$betweenthetwodomainsthatdualunipolarbatteries(doubleEMF’s)existback-to-back,oppositelydirected,withapair-creationgapbetween.Thetotalenergyflux$${oldsymbol{S}}_E$$isalinearsumofthetwofluxes:thehole’soutwardspin-downenergyflux$${oldsymbol{S}}_{mSD}$$originatingatS$$_{mffH}$$andthePoyntingflux$$oldsymbol{S}_{mEM}$$emittedatS$$_{mN}$$inboththeoutwardandinwarddirections,with$${oldsymbol{S}}_E$$beingproportionalto$$Omega_{mF}$$,$${oldsymbol{S}}_{mSD}$$to$$omega$$and$$oldsymbol{S}_{mEM}$$to($$Omega_{mF}$$$$-$$$$omega$$)alongeachfieldline.Applyingaperturbationmethodforasplit-monopolarfieldwithaspin-parameter$$h$$$$ll$$1,theanalyticsolutionofthestreamequationisgiven,andthedoubleeigenvalueproblemduetothe`criticalitycondition’attheouter/innerfastsurfacesS$$_{moF}$$/S$$_{miF}$$andthe`boundarycondition’atS$$_{mN}$$issolvedtoyieldthefinaleigenvalue$$Omega_{mF}$$,intermsof$${Omega_{mH}}$$and$$f_{mH}$$$$=$$0.5676.TheratiooftheoutputpowerreachingS$$_{{mff}infty}$$tothedissipationonS$$_{mffH}$$is$$epsilon$$$$=$$1$$+$$(4$$/$$5)(1$$-$$$$f_{mH}$$)$$h^2$$.

【 授权许可】

Unknown