6.2 Alternative Patterns
6.2.1 Large Rossby Number
When the Rossby number is large, the wind moves in an essentially ballistic manner.
As we have argued in discussing the Rhines scale, the flow may be considered to be quasi-two-dimensional, as we take the wind to move along isobars. Note that in this limit the rotation of the star is largely irrelevant. As a result, the system is axisymmetric about the line connecting the pulsar and the companion. In analyzing this case, then, we eschew our standard conventions for ˆz and instead take ˆz to lie along this line. The angular coordinates θ andφ are then redefined accordingly, so vθ is now the wind speed from one side of the star to the other, while vφ measures the wind speed along the symmetric direction.
Suppose that the star has some temperature differential ∆T between the day and night sides12. Then in moving around the star, the wind acts as a heat engine. The specific power it moves is given by
ε0 =cpv·∇T. (6.7)
Now the rate at which the wind may extract work from this process is just the heat engine efficiency multiplied by the specific power. Using the endoreversible heat engine efficiency, a common approximation used in place of the maximal Carnot
12This is equivalent to speaking about the amplitude of the first nontrivial spherical harmonic.
6. GLOBAL WIND PATTERNS 84 efficiency for real-world systems, we find that
W˙ =
1−
s
1−∆T T
(cpv·∇T)≈ ∆T
2T (cpv·∇T), (6.8) and so approximating the gradient yields
W˙ = cpvθ∆T2
2πRT = v2svθ 2πR
∆T T
!2
kB cvµ
!
, (6.9)
where the final term on the right is 3/2 for an ideal gas and of order unity generally.
The intuitive picture here is that the wind moves to the hot side of the star, picks up heat, and then moves to the cold side to release it. It then moves back to the hot side to warm up again, and the process repeats.
There are a variety of structures that the global flow could take on. One would simply be to set vθ to some uniform nonzero value. Given that there is nothing driving the flow in the ˆφ direction, and no spontaneous symmetry breaking, we may then setvφ= 0 up to turbulent corrections. This has the disadvantage of causing a net mass flux across the star. This could be remedied by setting vθ to some value which varies as a periodic function of φ, spontaneously breaking the axisymmetry.
There is, however, no physical process giving the φ scale for this symmetry breaking.
Additionally, this solution leads to a singularity in the continuity equation near the poles, which is somewhat harder to remove.
There are two natural ways to correct the problems uncovered in the previous examples. The first would be to make use of Hadley cells. These preserve axisymmetry and require no additional length scale. They respect the isobaric nature of the flow up to corrections of order hs13, but have the advantage that there is no longer a singularity in the continuity equation. The other possibility is to assume once more circumferential transport. The axis along which the the transport aligns would be set by some combination of the weak residual effects of rotation and the magnetic anisotropy in the underlying microscopic viscosity, both of which will tend to weakly align it with the star’s rotation axis. The question is then of whether or not all of the gas moves with the same handedness around the star. There is no physical process which breaks the symmetry here, so if there are zones with alternating handedness we expect the scale of alternation to be R. In either case, we refer to the relevant speed as vθ.
13That is, they have deviations of orderhs.
6. GLOBAL WIND PATTERNS 85 Using the above results, we may equate ˙W totaled over a spherical shell of thickness l with the specific power lost to viscous drag, giving
4πR2lvθvs2∆T2
2πRT2 =vθ24πR2νvl−1+ 2πRlv2νhR−1 kB cvµ
!
(6.10)
= 2πlνhvθ2 1 + 2
R l
2 νv νh
! kB cvµ
!
(6.11)
∴Rv2s∆T2
πT2 =νhvθ 1 + 2
R l
2 νv νh
! kB cvµ
!
. (6.12)
Note that because ˙W is always a small fraction of ε, we do not need to worry about including heat produced by viscous effects in the calculation of heat transport.
We must now consider the radiative and convective cases separately. In the radiative case,
νh =vθR (6.13)
and
νv =vθ2 α+νh
glℵ(∇ad− ∇). (6.14)
As a result, we may write vs2∆T2
πT2 =vθ2 1 + 2
R l
2
1 + α vh
v2θ gℵl
!! kB cvµ
!
(6.15)
=vθ2 1 + 2
R l
2
1 + α νh
vθ2 vs2
! γ ℵ2
!
kB cvµ
!
(6.16)
=vθ2 1 + 2
R l
2
1 + k
ρcpvθR
! v2θ v2s
! γ ℵ2
!
kB cvµ
!
. (6.17)
Now note that k
ρcpvθR = F
ρcpvθR|∂rT| = F P
ρcpvθR∇T|∂rP| = F hs
ρcpT vθR∇ = F vθgRρ∇
kB µcp
!
. (6.18) The last term and ∇are both of order unity. The flux is generally within two orders of magnitude of 1012erg/cm2,R is within an order of magnitude of 1010cm, g is close to 104cm/s2, so this term may be written roughly as v−1θ ρ−110−2g/cm2/s. In the limit of fast winds, we expect this to be small, and hence may neglect the vertical shear, while for slower winds or lower densities we may neglect the horizontal shear.
6. GLOBAL WIND PATTERNS 86 Regardless, solving for vθ yields
vθ =
v u u u u u u t
s
1 + 8vs2∆TπT22
c
vµ kB
R l
2
1 + ρck
pvθR
vθ2 vs2
γ
ℵ2
−1 4Rl21 + ρck
pvθR
v2θ v2s
γ
ℵ2
. (6.19)
This may be simplified by dropping factors of ℵ, γ, cvµ/kB, all of which are quite close to unity, and by assuming ∆T /T to be small. Doing so yields
vθ =vs∆T
πT . (6.20)
The heat transported is therefore ε0 =vθcp∆T
πR = 1
πRvscpT ∆T T
!2
≈ vs3 πR
∆T T
!2
. (6.21)
On the other hand, in the convective case
νv =lmax (vc, vθ), (6.22) while
νh = max (lvc, Rvθ). (6.23)
As R > l, their ratio is 1 for lvc< Rvθ, R/lfor vθ > vc, and Rvθ/lvc in between. As Rl, then, the term (R/l)2νh/νv is always dominant over unity, so we may write
Rv2s ∆T2
2πT2 =νvvθ
R l
2 kB cvµ
!
. (6.24)
Once more we will drop the rightmost term, for it should be very close to unity. This done, we may substitute in the expression for νv and find
v2s ∆T T
!2
l R
!
= 2πvθmax (vc, vθ). (6.25) To solve this, we first assume vc> vθ. This yields
vθ = v2s 2πvc
∆T T
!2
l R
!
. (6.26)
6. GLOBAL WIND PATTERNS 87 If this exceeds vc then we next take vc< vθ and find
vθ =vs
∆T T
s l
2πR. (6.27)
Ifvc> vθ then ε0 =vθcp∆T
πR = v2scpT 2πRvc
l R
! ∆T T
!3
≈ vs4 2πRvc
l R
! ∆T T
!3
, (6.28)
while in the other case ε0 =vθcp∆T
πR = vscpT πR
∆T T
!2s
l
2πR ≈2l−1vs3 l 2πR
!3/2
∆T T
!2
. (6.29) The similar structure of all of the heat transport equations indicates that we may simplify, and write them each as
ε0 =ξl−1v3s, (6.30)
where ξ is a dimensionless quantity given in the radiative case as ξ= 2 l
2πR
! ∆T T
!2
, (6.31)
in the convective vc > vθ case as ξ = 2πvs
vc l 2πR
!2
∆T T
!3
, (6.32)
and in the convective vc < vθ case as ξ= 2 l
2πR
!3/2
∆T T
!2
. (6.33)
From the form of ξ we may gain some intuition about the system. To begin, note that ξ/l depends onl only in the convective case. This is because in the convective case the nature of the turbulence which resists moving heat circumferentially depends onl, whereas in the radiative case this dependence is not there, for the turbulence there depends only on the Richardson viscosity scale. Additionally,l/2πRis typically of order 10−3, while vs/vc is typically of order 103. If ∆T /T is smaller than unity,
6. GLOBAL WIND PATTERNS 88 we see that the most efficient transport comes when the turbulence is suppressed by entropic stratification, as in the radiative case.
Note that the above expressions were derived assuming ∆T /T is somewhat smaller than unity. The functional forms become somewhat more complicated as the temperature difference increases, so we will keep in mind that there could be deviations from the above behavior and use them primarily as guidelines for intuition and estimation. Having said that, note thatl/Ris generally on the order of 10−3, and vs/vc is generally not more than 103, so all of the cases considered thus far indicate that vθ approaches vs as ∆T approaches T. This behavior is expected regardless of the underlying turbulent model, so the fact that it occurs in all of the cases indicates that we are not missing substantial qualitative physics. Furthermore, when vθ > vs, we simply substitute vθ =vs to get the correct physics, for winds generally cannot travel much above the sound speed without incurring tremendous losses. In this case,
ε0 =vθcp∆T
πR ≈ v3s∆T
πRT . (6.34)