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Sandbox Physics
Binary mass transfer and common envelopes

AC29 · L2

Why exchanging mass can widen an orbit

Sandbox Physics ·

A lighter donor feeding a heavier companion can widen its orbit. A heavier donor can initially make the orbit shrink. The deciding information is how mass is redistributed and how much angular momentum any escaping material carries.

Add the angular momentum carried by both stars

Start with a circular point-mass binary. Temporarily neglect spin, angular momentum stored in a disc, and other external torques. The stars orbit at different distances from their common center of mass; the lighter one travels farther from it. Adding their orbital angular momenta gives a quantity that depends on both the mass distribution and separation.

M=Md+Ma,Jo=MdMaGaMM=M_d+M_a,\qquad J_o=M_dM_a\sqrt{Ga\over M}
Total mass is MM and separation is aa. Donor and accretor subscripts retain the previous chapter’s definitions.
dln⁡a=2 dln⁡Jo−2 dln⁡Md−2 dln⁡Ma+dln⁡Md\ln a=2\,d\ln J_o-2\,d\ln M_d-2\,d\ln M_a+d\ln M
Logarithmic differentiation turns the circular relation into a response equation. Given the mass and angular-momentum ledgers, the separation follows.

These assumptions are physical choices. A disc or a stellar spin reservoir can exchange angular momentum with the orbit. If that exchange matters, orbital angular momentum cannot simply be held fixed. Declaring it negligible gives the conservative calculation below a clear, limited purpose.

Reading: Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1

The conservative minimum lies at equal masses

Let x>0x>0 be mass removed from the donor and entirely kept by the accretor. Then Md=Md0−xM_d=M_{d0}-x and Ma=Ma0+xM_a=M_{a0}+x. With both total mass and orbital angular momentum fixed, the response integrates directly.

aa0=(Md0Ma0MdMa)2,dln⁡adx=2Md−2Ma{a\over a_0}=\left({M_{d0}M_{a0}\over M_dM_a}\right)^2,\qquad {d\ln a\over dx}={2\over M_d}-{2\over M_a}
A heavier donor gives a negative derivative; a lighter donor gives a positive one. Equal masses mark the turnover.

For an initial 3Modot3M_odot donor and 2Modot2M_odot accretor, transferring 0.5Modot0.5M_odot makes both masses 2.5Modot2.5M_odot. Their product is largest there, so the separation is smallest.

amin⁡a0=(3×22.5×2.5)2=0.9216,Pmin⁡P0=0.92163/2=0.884736{a_{\min}\over a_0}=\left({3\times2\over2.5\times2.5}\right)^2=0.9216,\qquad {P_{\min}\over P_0}=0.9216^{3/2}=0.884736
Kepler’s law supplies the period ratio at fixed total mass. This analytic example removes one sixth of the initial donor, beyond the interactive local 8% range. It is not a full stellar evolutionary history.
Computed conservative separation curves versus removed fraction for initial donor-to-accretor mass ratios 0.4, 1.5 and 3
Conservative circular orbits, with mass and orbital angular momentum retained. Each curve directly evaluates the mass ledger; this is imposed removal, not a stellar evolutionary timescale.

The figure uses removed mass as a fraction of the initial donor mass. Different initial mass ratios give different trends even though every curve keeps all matter in the system. When a lighter donor feeds a heavier accretor, their mass product decreases; preserving the same orbital angular momentum requires a wider orbit.

Reading: Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1

Give escaped matter its own ledger entry

Suppose the accretor retains only a fraction bb. The donor loses xx, the accretor gains bxbx, and matter outside the binary gains (1−b)x(1-b)x. The third entry remains necessary even though that matter no longer belongs to the binary.

dJo=−jesc(1−b) dx,jesc=γJoMdJ_o=-j_{\rm esc}(1-b)\,dx,\qquad j_{\rm esc}=\gamma{J_o\over M}
This prescription defines escaped specific angular momentum as a multiple of orbital angular momentum per unit total mass.
JoJo0=(MM0)γ,aa0=(MM0)2γ+1(Md0Ma0MdMa)2{J_o\over J_{o0}}=\left({M\over M_0}\right)^\gamma,\qquad {a\over a_0}=\left({M\over M_0}\right)^{2\gamma+1}\left({M_{d0}M_{a0}\over M_dM_a}\right)^2
Both retention bb and this specifically defined gammagamma are held constant here. The orbit and escaped matter together carry the initial angular momentum.

The same retained fraction can give different separations. Matter escaping with a larger specific angular momentum reduces the remaining orbital angular momentum more strongly. That effect competes with the response to redistributed masses. The orbit additionally depends on the specific angular momentum of that lost matter.

Reading: Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1

The launch site supplies a different lever arm

Isotropic loss still carries the orbital motion of its source star. A donor wind differs from material that first reaches the accretor and is then isotropically re-emitted. Using each star’s distance from the center of mass gives the following factors in the same specific-angular-momentum convention.

γdonor=MaMd=1q,γreemission=MdMa=q\gamma_{\rm donor}={M_a\over M_d}={1\over q},\qquad\gamma_{\rm reemission}={M_d\over M_a}=q
These factors change with the mass ratio. Keeping their initial numerical values constant would describe another loss prescription.

For a fully unretained slow donor wind, b=0b=0. Substituting the donor lever arm into the response equation gives dlna=−dlnMdln a=-dln M, hence the limit below.

aM=a0M0aM=a_0M_0
This is the slowly varying isotropic-wind limit with an approximately circular orbit. Impulsive loss can change eccentricity and does not obey the same adjustment rule.

MESA’s fixed jdot_ml_check example combines donor wind, accretor re-emission and ring loss. Its parameters must be read together with the specific-angular-momentum definitions. A ring radius or ring lever-arm factor is not automatically the gamma used here. Only the mathematical relationship is evaluated in this independent ODE comparison. Running the stellar structure solver and its full MESA evolution is a separate calculation.

Reading: Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1 · MESA 25.12.1, jdot_ml_check; commit 4a68bfbce0f6930c3f861d0611e914f02569a899

Closing two ledgers does not conserve orbital energy

Eorb=−GMdMa2aE_{\rm orb}=-{GM_dM_a\over2a}
Evaluate this diagnostic from the resulting masses and separation. It is not the total energy of the stars, gas, heat and radiation.

Even conservative transfer generally changes orbital energy while preserving total mass and orbital angular momentum. A real binary must account for accretion, stellar spin, heat and radiation in the larger energy problem. This chapter converts a declared mass/angular-momentum prescription into a circular orbit; it has not closed the gas energy equation.

Removed mass is also not an elapsed-time coordinate. A transfer rate needs an additional observation or a physical transfer prescription. The later Beta Lyrae tutorial supplies a historical period trend for a conditional rate calculation. It never infers that rate from how fast the trajectory animation plays.

Reading: Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1 · Ivanova et al. (2013), Common envelope evolution; arXiv:1209.4302v2

Exercises and explanations

  1. How much conservative transfer makes an initial 3Modot3M_odot plus 2Modot2M_odot pair closest, and what is its minimum separation ratio?

    Read the explanation

    At fixed total mass the mass product is largest at equality. Solving 3−x=2+x3-x=2+x gives x=0.5Modotx=0.5M_odot, then amin/a0=0.9216a_{min}/a_0=0.9216. This does not establish whether physical transfer can continue that far.

  2. The accretor retains half of 0.02Modot0.02M_odot removed from the donor. Write all three mass entries.

    Read the explanation

    The donor change is −0.02Modot-0.02M_odot, the accretor change +0.01Modot+0.01M_odot, and escaped matter gains 0.01Modot0.01M_odot. They sum to zero. Escaped angular momentum requires its own additional entry.

  3. Why can the slow-wind relation not simply be applied to an instantaneous loss of the same mass?

    Read the explanation

    The slow relation assumes an orbit that can adjust while remaining nearly circular. Instantaneous loss leaves the old position and velocity initially unchanged, and can produce eccentricity or unbind the pair. An equal total mass loss does not define an equal dynamical process.

Try the prediction

Send the same lost mass through different channels

  1. Set initial mass ratio to three and removed fraction to 8%. In the expanded controls set retention to zero.
  2. Compare donor wind with prescribed specific angular momentum at factor two. Why can the same lost mass widen one orbit and shrink the other?
  3. Export both records and check mass and escaped angular momentum separately. Trace the orbital-energy change into the larger unresolved stellar and gas energy problem.

Sources and calculation scope

  1. Soberman, Phinney & van den Heuvel (1997), astro-ph/9703016v1

    Sections 2–3, angular-momentum loss channels, response exponents and AppendixA. Translate the paper’s donor/accretor names before using its mass-ratio convention.

  2. MESA 25.12.1, jdot_ml_check; commit 4a68bfbce0f6930c3f861d0611e914f02569a899

    The native inlist and run_binary_extras.f90 give an analytic mixed-loss reference. Our ODE checks the mathematics; we did not compile or pass the MESA evolutionary test. Native code is LGPL3-or-later and is not copied into this Lab.

  3. Ivanova et al. (2013), Common envelope evolution; arXiv:1209.4302v2

    Energy formalism, sources/sinks, timescales and core/remnant boundary discussions. The scalar endpoint does not predict the flow or an ejection morphology.

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