Practive Derivations for MT 1 GSI: Goni Halevi SOLUTIONS

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Noel Derick Jefferson
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1 Prctive Derivtions for MT GSI: Goni Hlevi SOLUTIONS Note: These solutions re perhps excessively detiled, but I wnted to ddress nd explin every step nd every pproximtion in cse you forgot where some of them cme from Also, I use the cronym OOM to men order-of-mgnitude, in cse tht s confusing Tidl Accelertion & Height of Tidl Bulge Tidl Accelertion: The tidl ccelertion is the difference between grvittionl ccelertion on one side of the object nd tht on the other In this derivtion, I ll use M s the mss of the centrl, mssive object, R s the rdius of the orbiting object, nd s the semi-mjor xis Therefore, tidl = ner fr = GM 2 GM ) + R) 2 = GM [ ] R/) 2 [ = GM 2 + R ) 2 ] Now, we Tylor expnd with the ssumption tht the rdius of the orbiting body is much less thn its distnce wy from the centrl body Mthemticlly, R << R/ << The pproprite Tylor expnsion is the first order binomil one for smll x, In this cse, x = R/ nd α = 2, so + x) α + αx ) tidl GM 2 [ 2R ] Dropping the 2 becuse it s OOM equl to, the tidl ccelertion is tidl GMR 3 Height of Tidl Bulge: Here, we envision the bulge s piece of shell with height h nd set the tidl ccelertion on the body equl to the grvittionl ccelertion the bulge feels due to the body it belongs to, which we will give mss m nd mss density ρ The mss of complete shell of thickness h on the orbiting body is m shell = SA)hρ = 4πR 2 hρ where SA is the surfce re of the body nd we hve mde the implicit ssumption tht the entire shell is t distnce of R wy from the center where R + h) R becuse h << R Let s mke the OOM pproximtion tht the bulge hs /4π) the mss of n entire shell with thickness h, so the mss of the bulge lone is, to OOM, m bulge R 2 hρ The ccelertion on the bulge due to the grvity of the body it belongs to is then g,bulge,m Gm bulge R 2 = Ghρ

2 But the density of the body is just ρ m/r 3, to OOM, so g,bulge,m Ghm R 3 Setting this equl to the tidl ccelertion nd solving for h, g,bulge,m tidl Ghm R 3 GMR 3 h MR4 m 3 h M ) 3 R R m where, to cler up the nottion, M nd m re the mss of the centrl body nd the orbiting body respectively, R is the rdius of the orbiting body, nd is the semi-mjor xis 2 Tidl Disruption Rdius The condition for tidl disruption is tht, to OOM, the height of tidl bulge on body is equl to tht body s rdius This occurs t tidl disruption rdius denoted by td, which is simple to determine once you hve h, s derived bove h R ) 3 M R R R m td ) 3 M m td R ) /3 M td = R m where, gin, M nd m re the mss of the centrl body nd the orbiting body respectively, R is the rdius of the orbiting body, nd is the semi-mjor xis In this cse, td is the distnce between the bodies when m is tidlly disrupted 3 Event Horizon of Blck Hole This one is quick nd simple, s long s you know wht the EH physiclly entils The EH is where light, nd thus, nothing t ll, cn not exit the blck hole, so the escpe velocity must be greter thn the speed of light At EH, v esc = c The escpe velocity for body with mss M nd rdius R is found by considering test prticle m test ) t the edge of the body nd setting its totl energy equl to zero E test = T test + U test = 0 2 m testvesc 2 GMm test R 2GM v esc = R 2 The nottion gets bit confusing here, since is used to denote both ccelertion, like in the first derivtion, nd semi-orbitl ccess, like here Be creful with this = 0 2

3 For blck hole, which doesn t hve trditionl rdius, this velocity equls c t the event horizon R = EH ) 2GM 2 = c EH 4 Roche Lobe Rdius EH = 2GM c 2 Eugene uses Hill Sphere rdius to denote the sme thing Nottion: We will give the two bodies msses of M nd m, where M is the more mssive one duh), nd distnce between them of r We will cll the rdius of m s Roche lobe r m Let s consider test prticle with mss m t est The msses re rrnged such tht M >> m >> m test The ide is tht ech mss, M nd m, hs it s own turf nd M s turf is lrger thn m s becuse of their respective msses This turf is the Roche lobe, which isn t ctully sphericl but we cn pretend right now At the Roche rdius of m, denoted r m, the ccelertion due to the pull of m on test prticle should equl the differentil ccelertion between the pull of M on the test prticle t r m nd the pull on the test prticle if it were t the loction of m Equting ccelertions such tht m test cncels out, g,m,rm g,m,rm g,m,r Gm r 2 m = GM r r m ) 2 GM r 2 2) The first term on the right hnd side ccelertion due to M t r m ) needs to be Tylor expnded, GM r r m ) 2 = GM r 2 r m /r) 2 = GM r 2 r ) 2 m r Using binomil expnsion Eqution ) with x = r m /r << nd α = 2, GM r 2 + 2r ) m r Plugging this bck into Eqution 2, Gm r 2 m GM r 2 + 2r ) m GM r r 2 Solving for r m, we find Gm r 2 m r m = GM r 2 2r m r m ) /3 r 2M Agin, to clrify nottion, r m is the Roche lobe rdius of m, which is in binry with the more mssive str M nd the distnce between M nd m is r between them Note tht if we hd done this rigorously insted of to OOM, the 2 would be 3 3

4 5 Accretion Luminosity Consider test mss M t distnce wy from n ccreting object of mss M nd rdius It is flling towrd the object, losing grvittionl potentil energy tht is converted into photons The finl energy of the test mss is entirely due to grvittionl potentil, since it is t rest once it hs fllen in, nd thus hs no kinetic energy The initil energy is both kinetic nd grvittionl Thus, the chnge in energy of M between when it begins flling in, t distnce, nd when it reches the surfce of M, is E = E f E 0 GM M = ) GM M + 2 M GM ) = GM M + GM M 2 Here we re going to mke n pproximtion using <<, which just mens tht the mss flling in origintes t distnce much frther thn the rdius of the object it flls into With this ssumption in plce, we cn sy tht the second term is much smller thn the first, since its denomintor is much lrger Thus, we will eliminte it completely remember, OOM) E GM M Since this is the energy lost by M hence the negtive sign) nd energy needs to be conserved, we cn sy tht the energy emitted in photons is the bsolute vlue of this, E photons = E = GM M The luminosity is defined s being energy per unit time, so the totl luminosity is given by L totl = E t = GM M t We re going to cll the quntity M/ t) Ṁ, which is just nottionl difference Physiclly, Ṁ represents the rte of mss ccretion In terms of this quntity, we hve L totl = GM Ṁ We cn repet this but for finl stte tht is insted one where it is still in orbit, just bout to hit the str s surfce, nd hs finl energy of insted In this cse, the chnge in energy is The luminosity of the disk itself is then just E f = T f + U f = GM M 2 E GM M 2 L disk = E t L disk = GM Ṁ 2 The luminosity of the disk represents hlf of the totl luminosity of the object, where the other hlf comes from the surfce boundry lyer 4

5 6 Temperture of n Annulus Let s determine the temperture of n nnulus of the disk by considering test mss M tht strs t rdius + from the str with mss M nd is then t distnce of, where << Assuming both kinetic nd potentil energy ie M is orbiting M s it flls in), the energy difference between when it is t + nd just t is given by E = E fter E before = GM M 2 = GM M 2 [ ) + GM M 2 + ) + ) ] Since / <<, we use my fvorite Tylor expnsion see Eqution ), nd get E = GM M 2 Now, we re going to determine the flux from this prt of the disk, which is defined s energy per unit time per unit re Mthemticlly, F = E t A 3) The relevnt re, A, is the differentil re for ring t rdius with thickness, formul tht you should definitely know: A = 2π Thus, plugging into Eqution 3 for E nd A, we find F = GM 4π 3 M t We recognize the second multiplictive term s Ṁ nd cn mkes the ssumption of stedy flow so tht Ṁ is essentilly constnt to substitute it in We lso consider tht the disk is ctully ccreting from both sides, so its effective re is not A, but 2 A Since we put A in the denomintor of F, this mounts to dividing our expression for flux by 2 Thus, F = GM Ṁ 8π 3 is the ctul flux If we ssume the nnulus is emitting flux like blckbody not ctully true, but we ll dd corrective constnt lter), we cn use F σt 4, where σ is the Stefn-Boltzmnn constnt nd T is the temperture Applying this, T GM Ṁ 8π 3 σt 4 ) /4 GM Ṁ 3/4 8πσ The pproximtion cme in when we ssumed blckbody emission, so our nswer is not exct To be rigorous, we include fctor of 3 inside the prentheses, which is importnt to mention only if you re not doing OOM things Thus, the finl result for the nnulus s temperture is ) /4 3GM Ṁ T = 3/4 8πσ where gin, for clrity, Ṁ is the rte of mss ccretion for n object with mss M, is the distnce to the nnulus whose temperture T we re clculting, nd σ is the Stefn-Boltzmnn constnt ) 5

6 7 Eddington Limit The Eddington limit is the mximum luminosity for stedy, sphericlly symmetric object of mss M To derive it, we re going to imgine n electriclly neutrl box t some distnce from the object, filled with electrons nd protons in equl number The box will experience force due to pressure rdition nd force due to grvity, which we equte to find the Eddington limit Let s define photon momentum of p = E c The momentum flux is defined to be momentum per unit re per unit time, so F p = E cat = E t ca = L ca = L 4π 2 c where we hve used the fct tht luminosity is energy per unit time nd tht the re should be the surfce re of n imginry sphere extending from M to the box, t distnce wy Note tht F p hs units of pressure, so we cn think of it s rdition pressure The force due to this rdition pressure is the pressure times the re of electrons 2 tht it hits The number of electrons is denoted N e nd the cross-sectionl re for n electron is σ T hompson σ T Thus, the force due to rdition pressure is F rd = F p N e σ T The box lso experiences grvittionl force, which is rigorously given in terms of the mss of n electron m e, mss of proton m p, nd number of ech, N e nd N p respectively s F g = MGm en e + m p N p ) 2 However, becuse we know tht electrons re much less mssive thn protons, m e << m p nd we cn keep only the second term without losing much Therefore, F g MGm pn p 2 is the pproximte force due to grvity Equting the two forces cting on the box rdition pressure nd grvity), we hve LN e σ T 4π 2 c = MGm pn p 2 nd we cn cncel s well s N e nd N p the box is neutrl, so they re equl), which is good becuse these re ll rbitrry Solving for L, we hve L edd = 4πGMm pc σ T which vries only with the mss of he str M nd includes constnts: G, c, the mss of proton m p, nd the Thompson cross-section of n electron σ T A few notes bout the Eddington Limit: It mkes the ssumption of Hydrogen plsm It lso doesn t work for explosions, since they ren t sted IT cn help you determine the mximum msses of strs becuse the luminosity fusion, L fusion M 3 while L edd M Thus, there is mss such tht luminosity due to fusion nd L edd re equl, nd no str cn be bove this mss 2 We ignore the protons here becuse the Thompson cross-section of proton is much less thn tht of n electron; their effective re is negligible 6

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl