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SBU PHY 521 - Binary Stars

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1Binary StarsConsider a binary composed of two stars of masses M1and M2. Wedefine M = M1+ M2and µ = M1M2/M. If a1and a2are the meandistances of the stars from the center of mass, then M1a1= M2a2. Themean separation of the stars is a = a1+ a2. If the orbit is elliptical witheccentricity e , then the separation at periastron is a(1 − e) and at apastronit is a(1 + e). The total energy and angular momentum of the binary areE = −12GM1M2aJ = µqGaM1 − e2= µΩa2p1 − e2.Kepler’s Law isΩ2=2πP2=GMa3.The projected orbital velocity of star 1 isv1= Ωa1sin i.The quantityf1(M1, M2, i) =(M2sin i)3M2=v31GΩis known as the mass function since it depends only on observables v1, P .If Doppler shifts f r om star 2 are measured, thenf2(M1, M2, i) =(M1sin i)3M2=v32GΩcan also be found. ThenM2M1=v1v2independent of i. If the binary is eclipsing, the angle i can be determinedand the masses individually determined as well.Mass Transfer and Roche LobesThe total potential of a binary is−Φ =GM1r1+GM2r2+12d2Ω2,where r1and r2are the distances to stars 1 and 2 and d is the distance tothe rotation axis. Restricting ourselves to the orbital plane, with the originat the center of mass,−Φ (x, y) =GM1q(x − a1)2+ y2+GM2q(x + a2)2+ y2+12x2+ y2GMa3.2In dimensionless coordinates ¯x = x/a, ¯y = y/a, m1= M1/M, m2= M2/M:−Φ (¯x, ¯y) =GMam1q(¯x − m2)2+ ¯y2+m2q(¯x + m1)2+ ¯y2+12¯x2+ ¯y2.Contours of constant Φ are shown in the figure. There a re deep minima atthe stellar centers, and maxima at five so-called Lagrangian points. The L1point between the stars is significant because if a star expands and reachesthe potential surface passing through it, mass can be transferred to itscompanion.The equipotential surface that passes through L1is called the Rochelobe, and its size depends upon the mass ratio of the binary. Kopal (1959)gives for the radius RRwith nearly the same volume as the Roche lobe:RRa= 0.46M1M1/3. (1)A better fit is by Eggleton:RR/a = 0.49".6 +M1M2−2/3ln 1 +M1M21/3!#−1. (2)3Assume the binary is circular. Thena =MJ2GM21(M − M1)2, da =2aM12M1− MM − M1dM1, (3)if dM = dJ = 0. This shows that if M2< M1, transferring mass from M1to M2results in a shrinkage of the orbit. Am episode of conservative masstransfer in a binary results inafinal= ainitial M1,initialM2,initialM1,finalM2,final!2.Eq. (3) implies that in terms of the mass ratio of the binary, q = M2/M1,da =2aqq − 11 + qdq, (4)or a ∝ (1 + q)4q−2. Expressing Eq. (1) in terms of q, then taking thederivative and combining with Eq. (4),dRL= RLdaa−RL3dq1 + q= RL2qq − 11 + q−13 (1 + q)dq.This implies that the Roche lobe size reaches its minimum value whenq = 6/5, or M1= 5M/11.On the other hand, suppose that mass is lost from one star in the formof a wind and is not accreted onto the companion. Then we might expectthatafinal= ainitialMM − ∆M,and mass loss will cause an increase in a binary’s separation.Now consider mass transfer when star 1 fills its Roche lobe. Stable masstransfer occurs when the change in ra dius of star 1 after transferring anincrement of mass through the inner Lagrangian point is not offset by acorresponding change in the Roche radius, triggered by the new mass ratioof the binary. This requires that the logarithmic change of radius with massfor star 1 satisfiesd ln Rd ln M1≡ α ≥d ln RRd ln M1=d ln ad ln M1+13= 22M1− MM − M1+13.4In an equal mass binary, the first term vanishes. Generally, we can expectthat this condition is generally satisfied. It is not, however, for a star witha convective envelope, for which γ = 5/3 and R ∝ M−1/3.In some situations, mass transfer will be driven by losses of orbital angu-lar momentum. The primary sources o f angular momentum loss are mag-netic braking and g r avitational radiation. We have˙aa= 2˙JJ− 21 −M2M1˙M2M2,where the donor star is taken to be 2, so that˙M2< 0. Using the simpleRoche lobe formula,˙RRRR= 2˙JJ− 21 −M2M1˙M2M2+13˙M2M2.Assume that˙R2/R2= α(˙M2/M2, where α = −1/3 for a non-relativisticdegenerate, or convective, star, and α = 1 for a main sequence star. Forstable mass transfer, R2should remain equal to RR. Then we have˙JJ=56+α2−M2M1˙M2M2.Since bo th sides of this equation must be negative, we findM2M1≤56+α2.When α = −1/3(1), M2/M1≤ 2/3(4/3). G r avitational radiation leads to˙JJ= −32G35c5M1M2(M1+ M2)a4s−1. (5)Explosive Mass LossAnother case o f mass transfer occurs after a supernova explosion, buthere the mass loss is sudden and catastrophic and the companion does notaccept the mass. If too much mass is lost from the system, the binary will bedisrupted. The survival of the binary depends on the am ount of mass loss,the phase of the binary (i.e., is it near periastron (a more stable situation)or apastron (a less stable situation)), and the magnitude a nd direction ofany “kick” imparted to the supernova remnant star. If the explosion is5perfectly symmetric, the kick velocity is zero. However , if the explosion isasymmetric, the kick velocity is determined by momentum conservation. A1% asymmetry in the neutrinos released in a supernova could impart a kickvelocity of approximately∆V =∆EMnsc≃ 360 kms−1assuming a neutron star mass of Mns= 1.4 M⊙and ∆E = 3 × 1051erg.This, in ma ny cases, exceeds the relative orbital velocities of the stars priorto the explosion. As we will see, a kick in the direction of orbital motion ofthe remnant destabilizes the system, but a reverse kick can stabilize it.Following J. Hills, Ap. J. 267 (1983) 322, the total energy of the binaryprior to the explosion isEo= −GMo1M22ao= −GMo1M2r+12µoV2owhere the subscript o refers to the initial system. r is the instantaneousseparation o f the two stars (equal to aoif the orbit is circular). At per i-astron, r = ao(1 − eo) and at apastron, r = ao(1 + eo). µo= Mo1M2/Mois the reduced mass and Mois the initial total mass. Vois the initial rel-ative velocities of the two stars. We can furthermore define a parameterV2c= GMo/aoas the relative velocity in the circular orbit case.Immediately after the explosion, if the orbit is still bo und, we haveE = −GM1M22a= −GM1M2r+12µV2by comparison, where µ = M1M2/M. Defining ∆M = Mo1−M1= Mo−M,it is straightforward to demonstrate thataao=1 − ∆M/Mo1 − (2ao/r) (∆M/Mo) +V2o− V2/V2c.In order to remain bound, it is therefore


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SBU PHY 521 - Binary Stars

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