Binary Stars & Binary Evolution

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2 In dimensionless coordinates ¯x = x/a, ȳ = y/a, m 1 = M 1 /M, m 2 = M 2 /M: ⎡ ⎤ −Φ(¯x,ȳ) = GM ⎢ m 1 m ⎣ + √ 2 + a √(¯x 1 ( ¯x 2 + ȳ 2) ⎥ ⎦. − m 2 ) 2 + ȳ 2 (¯x + m 1 ) 2 + ȳ 2 2 Contours of constant Φ are shown in the figure. There are deep minima at the stellar centers, and maxima at five so-called Lagrangian points. The L 1 point between the stars is significant because if a star expands and reaches the potential surface passing through it, mass can be transferred to its companion. The equipotential surface that passes through L 1 is called the Roche lobe, and its size depends upon the mass ratio of the binary. Kopal (1959) gives for the radius R R with nearly the same volume as the Roche lobe: ( ) R 1/3 R M1 a = 0.46 . (1) M A better fit is by Eggleton: [ R R /a = 0.49 .6 + ( M1 M 2 ) −2/3 ln ( 1 + ( M1 M 2 ) 1/3 )] −1 . (2)
3 Assume the binary is circular. Then a = MJ 2 GM 2 1 (M − M 1) 2, ( ) ( ) 2a 2M1 da = − M dM M 1 M − M 1 , (3) 1 if dM = dJ = 0. This shows that if M 2 < M 1 , transferring mass from M 1 to M 2 results in a shrinkage of the orbit. Am episode of conservative mass transfer in a binary results in a final = a initial ( M1,initial M 2,initial M 1,final M 2,final ) 2 . Eq. (3) implies that in terms of the mass ratio of the binary, q = M 2 /M 1 , da = 2a ( ) q − 1 dq, (4) q 1 + q or a ∝ (1 + q) 4 q −2 . Expressing Eq. (1) in terms of q, then taking the derivative and combining with Eq. (4), da dR L = R L a − R [ ( ) ] L dq 2 q − 1 3 1 + q = R 1 L − dq. q 1 + q 3 (1 + q) This implies that the Roche lobe size reaches its minimum value when q = 6/5, or M 1 = 5M/11. On the other hand, suppose that mass is lost from one star in the form of a wind and is not accreted onto the companion. Then we might expect that M a final = a initial M − ∆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 mass transfer occurs when the change in radius of star 1 after transferring an increment of mass through the inner Lagrangian point is not offset by a corresponding change in the Roche radius, triggered by the new mass ratio of the binary. This requires that the logarithmic change of radius with mass for star 1 satisfies dln R dln M 1 ≡ α ≥ dln R R dln M 1 = dln a dlnM 1 + 1 3 = 2 ( 2M1 − M M − M 1 ) + 1 3 .
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Binary Stars & Binary Evolution

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