Bondi flow revisited

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by  Satadal Datta

Bondi flow revisited
Satadal Datta
Harish-Chandra research Institute, Chhatnag Road, Jhunsi, Allahabad-211019, INDIA
Email: [email protected]
January 7, 2016
6
1
0 Abstract
2
n Newtonian spherically symmetric transonic accretion has been studied by including the mass of the
a accretingmatter,whileconsideringthegrowthoftheaccretoritselftobenegligiblysmall. Anoveliterative
J method has been introduced to accomplish that task. It has been demonstrated that the inclusion of the

6 massofthefluidchangesthecriticalpropertiesoftheflowaswellasthetopologicalphaseportraitsofthe
stationary integral solution.

]
R
S
.
h
p
1 Introduction 2 Governing Equations
-
o
r
t Spherically symmetric Bondi [1] flow studies the As we are considering steady flow, the Bernoulli’s
s dynamicsoftheinfallingtestfluids. Inthepresent equation reads as [4]1
a
[ work, we introduce a novel iterative method to
study the effect of the inclusion of the mass of u2 (cid:90) dp
1 + +V(r)=constant (1)
the accreting material. We did not consider the 2 ρ
v
7 growth of the accretor as a consequence of the ac- where fluid velocity is u, pressure is p and V(r) is
0 cretion and hence the direct effect of the inclu-
thegravitationalpotentialduetotheaccretorand
3 sion of the self gravity has not been studied. For
the infalling fluid itself. V(r) satisfying Poisson’s
1
usual astrophysical accretion, the aforementioned
0 equation is as below,
approximation, that the accretion rate as well as
.
1 thetimescaletostudytheproblemisnotsolarge GM 4πG(cid:90) r (cid:90) ∞
0 that the mass of the accretor will change – is a V(r)=− − ρr2dr−4πG ρ(r)rdr
6 valid approximation. r r R(cid:63) r
1 (2)
: We consider steady state accretion of one tem- whereR istheradiusoftheaccretor. Forbrevity
v (cid:63)
perature fluid onto a non spinning stationary (the
i we are considering adiabatic flow. So our barat-
X observer is in the co-moving frame) accretor. We
ropic equation is
r consideramoreappropriateeffectivegravitational
a
potentialincomparisontowhathadbeenassumed p=Kργ (3)
by Chia [2].
We found that the inclusion of fluid mass where K is a constant. And sound speed follows
alters the location of the critical point of the the relation,
flow as well as the values of the values of the
(cid:18) (cid:19)
∂p γp
accretion variables measured at the critical point. c2 = = (4)
We also find that the Mach number vs radial s ∂ρ ρ
s
distance profile, the usual topology of the phase
where s = entropy. Now, we have the energy of
portrait of the Bondi flow, takes a different form
fluid element per unit mass is E. Now, at infinity
for such massive accretion. We characterize our
flow profile using a realistic set of astrophysically 1Inthispaper,thebasicfluidequationsaretakenfrom
relevant parameters. thiscitedbook.
1
fluid velocity is zero so according to Bernoulli’s As a consequence we have the following:
equation, The critical values x , y and z correspond to the
c c c
Mach number to be unity, i.e; the transonic curve
u2 c2 c2
+ s +V(r)=E = s∞ (5) corresponds to maximum accretion rate.
2 (γ−1) (γ−1)
We have from definition
Physically, we are taking our infinity to be such
λ =x2y z
a large distance such that the potential terms are c c c c
vanishing there. We have discussed this issue in Using equation (9), we get,
details later in section 6. Now we define certain
dimensionless variables as Mach number = m=yczc−(γ−21) =1
r u ρ ————————————————-
x= ,y = ,z =
GM c ρ Now, from equation (8) and (10), we get,
c2 s∞ ∞
s∞
wherec isthesoundspeedatinfinityandρ is x 1+αI
s∞ ∞ c = 1c (11)
the density of fluid at infinity. From equation (5), x 1+α(γ−1)I
b 2c
introducing the dimension less variables, we get z
( c)(γ−1) =(1+α(γ−1)I ) (12)
y2 z(γ−1) 1 (cid:18)I (cid:19) zb 2c
+ − −α 1 +I =0 (6)
2 (γ−1) x x 2 Using equation (9), we get
where α = 4πρ∞G3M2c−s∞6, I1 = (cid:82)xzx2dx and λλcb =(1+αI1c)2(1+α(γ−1)I2c)2(5(γ−−3γ1)) (13)
x∗
I2 = (cid:82)∞zxdx with x∗= (GRM(cid:63) ) wsphheerreicxabl,uzsub,alλBboanrdeitahcecrcertitioicnalwviathluoeusticnocnassideeor-f
x c2s∞
ing fluid mass. These values are respectively as
Thus the mass accretion rate which is obtained
follows:
by integrating the continuity equation,
(5−3γ)
M˙ =4πr2ρu=4π(GM)2c−s∞3ρ∞x2yz xb = 4
1
Wedefineanotherdimensionlessvariableλ=x2yz z(γ−1) =
and from above equation mass accretion rate is b 2xb
proportional to λ. Now, we can rewrite equation λ2 =z(γ+1)x4
b b b
(6) as folows,
λ2 (zγ−1−1) 4 Equation of motion of the
+ x4−(1+αI )x3−αx4I =0 (7)
2z2 (γ−1) 1 2
infalling fluid
3 Calculations of Critical
To find the behaviour of the Mach number vs
Point of the flow radial distance profile for self gravitating steady
state Bondi accretion, we have to know the equa-
Now, our aim is to maximize accretion rate. We tion of motion of the infalling fluid. According to
willfindx,y,zforwhichλismaximized. So,from Euler’s momentum conservation equation:
e∂∂qλzu|a(txi=oxnc,y(7=)ycs,ze=ttzicn)g=0∂∂λ,xw|e(xg=extc,rye=sypce,zc=tizvce)=ly 0 and ρDD(cid:126)ut =−(cid:79)(cid:126)p−ρ(cid:79)(cid:126)V
4(z(γ−1)−1)x V is the external potential, in our case V is
c(γ−1) c −3(1+αI1c)−4αxcI2c =0 (8) the gravitational potential. So considering steady
flow, we have
and
du 1dp GM 4πG(cid:90) r
λ2 =zγ+1x4 (9) u =− − − ρr2dr (14)
c c c dr ρdr r2 r2
Using equation (9), from equation (7), we get R(cid:63)
Again considering spherically symmetric steady
zc(γ−1)xc+(zc(γ−1)−1)xc−1−α(I +αx I )=0 flow, we have from continuity equation, accretion
2 (γ−1) 1c c 2c rate,
(10) M˙ =4πr2ρu=constant (15)
2
Considering steady flow, using equation (3), (4), of α is small. Now we will find the values of the
(14), we have dimensionless quantities from equations (11), (12)
and (13) by method of iteration. To find the crit-
du (2c2s − GM − 4πG(cid:82)r ρr2dr) ical parameters from the above equations we have
= r r2 r2 R(cid:63) (16)
dr (u− c2s) to find the integrals I1c and I2c. So, as our first
u iteration we will put the value of z as the usual z
Now, using equation (15), (3), (4), we have ofsphericallysymmetricBondiflow. xc appearsin
thedefiniteintegralsI andI asupperlimitand
1c 2c
(cid:18) (cid:19)
dcs = cs(1−γ) 2 + 1du (17) lower limit respectively and in this case as our 1st
dr 2 r udr iteration, we will replace x by x (= (5−3γ)/4).
c b
Thusasour1stiteration,wecanfindthevaluesof
If we can solve these equations we can have the
the integrals. And then we can compute the criti-
Mach number vs radial distance profile except
cal parameters from equation (11), (12) and (13).
at the transonic point because at transonic point
Again, as our 2nd iteration, we can put these new
from equation (16) takes 0/0 form comes. So, ap-
values as the limits of the integrals described be-
plying L’Hospital’s rule there we have a quadratic
fore and so on and thus we can improve our preci-
equation as
sion in each iteration. As we are considering weak
Aq2+Bq+C =0 gravitational field due to fluid mass, we have de-
scribed only the 1st iteration here. In the 2nd
where, iteration the terms of order α2 will appear (obvi-
ous from equation (11), (12) and (13)). We are
A=(1+γ) neglecting the order α2 and the terms of higher
4c (γ−1) order in α. So, in the 1st iteration, the values of
B = sc
rc I1c and I2c are as below.
2c2 (2γ−1) 2GM 8πG (cid:90) xb
C = sc − +4πGρ − I I = z (x)x2dx (19)
r2 r3 c r3 1c 1c Bondi
c c c x∗
du (cid:90) ∞
q = |
dr critical point I2c = zBondi(x)xdx (20)
xb
where z (x) is z for the usual Bondi flow with-
Bondi
c is the sound speed at critical point. r is the out considering fluid mass. Now, again we have a
sc c
radial distance of critical point from the centre of problem that there is no proper analytic form for
the accretor. So, zBondi(x), so we will find the integrals using the
√ numerical values of zBondi(x) at different x .
−B± B2−4AC
q = (18)
2A
6 The limits of the integrals
Still, the problem is unsolvable because to find
thevaluesofthedimensionlessvariablesatcritical x is the upper limit and lower limit of the in-
b
point (which will serve as the initial conditions to tegrals I and I respectively. The lower limit
1c 2c
solvedifferentialequations(16)and(17)),weneed appearing in the integral I is x*. Now, in the
1c
densityprofileoftheinfallingfluidinthefirstplace integral, we can put x* using values of radius of
and we have no idea about the density profile. So some known stars and we can put the mass of the
in the next section, we develop a methodology to stars in equations. Alternatively, we can find the
solve the differential equations (16) and (17). radius of a star when the mass of a star is given
from any empirical relation. We have used the
following emperical relations[3] between mass and
5 Method of iteration
radius of a Main Sequence star.
In equation (11), (12), (13) and (16) ρ appears to R (cid:18) M (cid:19)0.945
(cid:63) =1.06 M <1.66M (21)
beavariableaboutwhichwehavenoinformation. R M (cid:12)
(cid:12) (cid:12)
So,wewillconsiderthatevenifweincludemassof
and
the medium it does not change the critical values
x , y , z and λ abruptly. According to equa- R (cid:18) M (cid:19)0.555
c c c c (cid:63) =1.33 M >1.66M (22)
tions (11), (12) and (13), in this case, the value R M (cid:12)
(cid:12) (cid:12)
3
whereR , R aretheradiusofthestarandofthe (cid:90) x∞
(cid:63) (cid:12) I = z (x)xdx (28)
2c Bondi
Sun and M, M are the mass of the star and of
(cid:12) xb
the Sun respectively. Thus the lower limit of the
Now, we can numerically calculate the values I
1c
integral I is fixed.
1c and I and put them in equation (11), (12) and
2c
Now, ∞ appears in the upper limit of the integral
(13) to find the changes in the critical parameters
I . By ∞ we mean some large distance from the
2c due to inclusion of fluid mass.
accretorwheregravityisweak. Let’scallthislarge
distance to be r and corresponding value of the
∞
dimensionlesssparametertobex . Themass,the 7 Physical Importance of Q1
∞
energy and γ of the fluid are given. At infinity as
and Q2
the velocity of the fluid is zero, so we have
c2 GM 4πG(cid:90) r∞ Atinfinitywecanwritetheenergyofinfallingfluid
s∞ − − ρr2dr =E (23)
(γ−1) r r as
∞ ∞ R(cid:63)
Now, we will compare the magnitude of the 1st c2
E = s∞ +EQ
term in potential with energy and we will set r∞ (γ−1)
tosomevaluesuchthattheterm GM isnegligibly
small compared to energy of the inr∞falling fluid. where Q(= Q1+Q2) gives total precision of the
Quantitatively, we set a very small quantity Q1 problem. Soinsummery, whenE,M, γ andQare
according to our desired precision as below. given, we at first set Q1 to find appropriate r∞ of
the problem and setting Q2 accordingly gives us
(GM)
Q1= r∞ (24) density of the infalling matter at infinity. If we
E closely look at the expressions of Q1 and Q2, we
Now, we can easily find r . For the 2nd term of see that the ratio
∞
the potential, we will put the density of the fluid
Q2 Mass of the entire medium
as the density of fluid in case of usual non self- s= = (29)
Q1 Mass of the accretor
gravitatingsphericallysymmetricBondiflow(The
method of iteration discussed in section 5). There Where mass of the medium means the calculated
we need the density of the fluid at infinity. So, we mass of the medium considering density profile
will choose the density of fluid at infinity to be of as that for usual Bondi flow, i.e; the mass of
suchasmallvaluesothattheterm 4πG(cid:82)r∞ρr2dr the medium calculated in the 1st iteration. In-
r∞ R(cid:63)
isnegligiblecomparedtotheenergyEofthefluid. tuitively,wecansaythatgreaterthevalueofratio
Quantitatively, we set a very small quantity Q2 Q2/Q1 greater the effect of inclusion of fluid mass
according to our desired precision as below. or greater the departure from the usual Bondi ac-
4πG(cid:82)r∞ρr2dr cretion.
Q2= r∞ R(cid:63)
E
[4πGρ∞(GM)3I1∞] 8 Behaviour of the critical pa-
= r∞c6s∞
E rameters
Q2Er (E(Y −1))3
ρ∞ = 4πG∞(GM)3I (25) We can now find the critical parameters. We have
1∞ the mass M of the accretor, E or the temperature
where we have used the method of iteration and
of the fluid at infinity, γ of the fluid as inputs
as the 2nd and 3rd term of the equation (23) are
and also we have fixed Q to be 0.02. Choosing s
taken to be negligibly small compared to the en-
gives us r and density of fluid at infinity. Our
∞
ergy E of the infalling fluid we have used the fol-
parameter space is γ (4/3 to 5/3), E (0 to 1 in
lowing to find out cs∞. units of c2). Now for neutral hydrogen-interstellar
c2 medium (H1 region)[5] temperature varies nearly
E = s∞ (26) from 10◦K to 1000◦k and corresponding energy
(γ−1)
per unit mass varies nearly of the order of 10−14
Andalsoaccordingtothemethodofiterationand to 10−12 in units of c2 . Here we have plotted
using the value of r from equation (24) we have,
∞ percentage change in the critical values due to
(cid:90) xb inclusion of fluid mass with the γ-E parameter
I = z (x)x2dx (27)
1c Bondi space. Temperature of the fluid at infinity is
x∗
4
2.4 on (-Dx/x )x100-E (in log scale) plane.
2.2 b
2 Fig. 2 and Fig. 3 implies that when s and E
s=3 s=3 1.8
2.5 1.6 are given and also total precision Q is given, the
s=1s=1 1.4
1.2 shift in transonic surface does not depend on E
ff s=1s/3=1/3 01.8 and that’s why only a single curve appear in Fig.
1 0.6
2, otherwise there would be several curves for
((--DDxx//xxbb))xx110000%%
0.00000010x01000-01120 different Es. We can generate the surface for s=1
0.0000000000008
1.3 1.35 1.4 γ 1.45 1.5 1.55 1.6 1.605.00000200x.010000-001003000020000005 E(c2 unit) jEaullsatExbisfy.ortTraahnpesalsrahttiiicfotunlianorfcγrtihtaiencadcluQprv.oeiInnitntxhFbeigiFs.isga2.ma3elotfnhoger
Fig. 1: Behaviour of percentage change in xb verticallinessignifythesamething,i.e; thepoints
where f = -(Dx/xb)×100 and Dx = (xc−xb) for a single vertical line gives the percentage
change in x for different γs. A single vertical line
b
corresponds to a single energy. Actually fixing Q
and s of the problem give Q1 and Q1 in return
taken to vary nearly 50◦K to 1000◦K. We find the
givesr foragivenE.ThenfromQ2 deriving ρ
∞ ∞
shift in transonic surface due to inclusion of fluid
somehowwipeoutthedependenceonE.Inclusion
mass for different s.
of fluid mass decreases the radius of sonic surface
More the total mass of the medium increases
The planes corresponding to different s
greater the shift in x . Relatively small γs give
b
seem parallel to γ-E plane because the variations
significant changes. We find the changes in other
in x for different s are incomparable. The whole
b critical parameters.
picture becomes clear when we plot for single s.
1.504
8
(Dx/xb)x100% 7
1.502 6
7
5
s=3s=3 4
1.5 5 ss==11 3
3 s=1/3 2
Dx/xb)x100%f 11..449968 s=1 ((DDzz//gzzgbb))xx110000%% 0s.=s01=0/310/0301x01000-01020 110
( 0.0000000000008
1.494 1.3 1.35 1.4 γ 1.45 1.5 1.55 1.6 0.0000200x0.100000-0102030x00100020000005 E(c2 unit)
1.65
1.492 Fig. 4: Behaviour of percentage change in z
b
where g = (Dz/z )×100 and Dz = (z −z )
1.49 b c b
1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65
γ
5.5
Fig. 2: Projection of the surface of Fig. 1 for s=1
(Dz/zb)x100%
on (-Dx/x )x100-γ plane.
b 5
1.504 4.5
1. 510.52 Dz/zb)x100%g 3 .45 s=1
(
-(Dx/xb)x100%f 11..449968 2 .35
1.494
2
1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65
γ
1.492
Fig. 5: Projection of the surface of Fig. 4 for
1.49 E=5e-15(c^2) 2.05e-13(c^2) s=1 on (Dz/zb)x100-γ plane.
0.0000000000000 0.0000000000000 0.0000000000001 0.0000000000010
E(c2) in logscale
Fig. 3: Projection of the surface of Fig. 1 for s=1 For brevity, we have not plotted E-(Dz/z )x100.
b
5
Similarly,percentagechangeinz doesnotdepend negligible compared to that for the other critical
b
on E and z is significantly increased for γ close parameters. One can safely say that percentage
b
to 4/3. Fig. 4 implies that if we increase s we change in x merely depends on γ and actually
b
increase the change in z which is intuitively it’s a function of only one variable, s. Again s can
b
obvious. Another important point is that if we not be arbitrarily large because in that case the
compare the percentage changes in z with that of change in mass of the accretor will be significant.
b
x we see that inclusion fluid mass effect z more Inclusion of fluid mass changes z much more
b b b
than x . significantly than the other critical parameters.
b
We find how do the critical parameters depend
onthemassratioofthefluidmediumandthestar.
s=3 44.5 2.6
s=3 3.5
4 3 2.4
2.5
2 2.2
2.5 1.5
ss==11 1 2
0.5
1 ss==1/13/3 0 1.8
((DDλλ//λλbb))xx11h 00100.3%% 1.35 1.4 γ 1.45 1.5 1.55 1.6 1.605.00000002.00x0100000-01003000.20000000000000.50000000000100x8E010(0c0-201 u02n1i0t) (-Dx/xb)x100%f 111 ...1246 gamma=1.41
0.8
Fig. 6: Behaviour of percentage change in 0.6
λb where h =(Dλ/λb)×100 and Dλ=(λc−λb) 0.4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
The surfaces seem to converge towards γ=1.65. s
3
datapoints
1.5+0.633842*log(x)
3
2.5
2.5
2
%
λλD/b)x100% h 1 .25 (Dx/xb)x100f 1 .15
(
1 s=1
0.5
0.5
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
s
0 1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65 Fig. 8: Dependence of percentage change in xb
γ
on s and data points are fitted. We choose upper
Fig. 7: Projection of the surface of Fig. 6 for s=1
limit of s to be 5.
on (Dλ/λ )x100-γ plane.
b
As percentage change in x is only function
b
Similarly, percentage change in λ does not
b of s, we have fitted the data points to find the
depend on E and it is significantly increased for γ
nature of dependence on s. We have obtained an
close to 4/3. Observing Fig. 6 implies that if we
empirical relation as
increase s we increase the change in λ which is
b
intuitively obvious. Dx
f =(− ×100)=a+b ln(s) (30)
Increasing s increases the shift in the critical
x
b
parameters.
Another important observation is that the per- In our case, the positive constants are as a = 1.5
centage change in x with respect to γ is not only and b = 0.633842. The fact that a to be 1.5 is
b
smallerthantheothercriticalparametersbutalso also obvious from Fig. 2.
the variation in percentage change of x with γ is
b
6
8 is obvious from Fig. 6 and 4, g (γ) is also
λ,z
monotonically decreasing function of γ.
7
gamma=1.33 Now we find find α[2] dependence of the percent-
age change in critical parameters.
6
gamma=1.41
Dz/zb)x100%g 45 gamma=1.5 4 .45
(
3.5
s=3
3
3
%
12 λλD/b)x100h 2 .25 s=1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 (
s 1.5
Fig. 9: Dependence of percentage change in z on
b
1 s=1/3
s.
0.5
4
0
0 2e-05 4e-05 6e-05 8e-05 0.0001 0.00012
3.5 α
Fig. 11: Dependence of percentage change in λ
gamma=1.33 b
3 on α.
2.5
λλ(D/b)x100%h 1 .25 gamma=1.41 Ffsoaormrxebbnraeavntuidtrye,z.b wbIneeccraheuaassveientghneαoytianprcelroetmatseoedrseptoehrrecleensstasamgoeef
change in the critical parameters. Again as per-
1 gamma=1.5 centage change in x is a slowly varying function
b
of γ, it is also a slowly varying function of α. We
0.5
have fitted the above plot as below,
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
s 3
Fig. 10: Dependence of percentage change in λb fidttaintga pcouirnvtes
on s.
2.5
The changes in critical parameters increase
2
with s. Fig. 9 and Fig 10. shows also that why % s=1
d=o1t.h6e5.suWrfeacseeseftohraztbthaendseλpbarcaotniovnersgbeettowweaenrdtshγe D/Db)x100λλh 1.5
(
curves of Fig. 9 and 10 for γ = 1.33 and 1.41 is a=2.4388e+12
1
b=-6.0685e+08
greater than the separation between the curves
for γ =1.41 and 1.5.
0.5 c=69243.3
Equation (30) implies that the test fluid ap- d=-0.576482
proximation is good for s (cid:54) exp(-a/b). This
0
1e-05 2e-05 3e-05 4e-05 5e-05 6e-05 7e-05 8e-05 9e-05 0.0001 0.00011
is the region of s where Bondi’s assumptions α
and his results are applicable. In our case, Fig. 12: The data points are fitted to the curve
that region corresponds to s (cid:54) 0.094 and it’s aα3+bα2+cα+d
important to mention that a and b depends on
Q. Similarly one can fit the same for λ and z We have found empirically a cubic polynomial,
b b
as well and one will find some best fitting curves Dλ
as (f (γ)+g (γ)ln(s)) respectively. From Fig. h=( )×100=aα3+bα2+cα+d (31)
λ,z λ,z λ
b
7 and 5 we can conclude that f (γ) are mono-
λ,z
The coefficients a, b, c, d depend on s, E and
tonically decreasing functions of γ. As the curves
Q. Similarly percentage change in z and x have
for other s have same nature as s=1, g (γ) can b b
λ,z
cubic dependence on α and in the same way, we
not be monotonically increasing function and as
canfitthemintocubicpolynomials(notshownfor
the surfaces are not parallel to each other which
brevity).
7
9 Ambiguity of the velocity 10 Mach number vs radial
gradient distance profile
Now, we can calculate du/dr at transonic surface
Now, we have found the initial conditions to find
from equation (18). We have obtained the critical
variation of Mach number with radial distance
parameters, we will now just put them to find q.
from the centre of the accretor. To do this, we
Important point is to note that in I , appearing
1c have to numerically solve equation (16) and (17)
in the term C of the quadratic equation of q, we
simultaneously. Mach number(=u/c ) can then
s
will use the new value of x which is found by
c be found. We have now x , y , z and q as initial
c c c 2
1st iteration. Now, according to equation (18),
conditions and we will use the numerical values
we will obtain two values of q. As inclusion of
of density for spherically symmetric Bondi flow
fluid mass does not changes the flow abruptly so
without including fluid mass effect. So we find
we will choose the value of q which is closer to
Mach number vs radial distance profile as below.
thevalueofcritical(du/dr)whenfluidmasseffect
isn’t considered. So, we plot the behaviour of the
derivatives and have compared those values with 1.5 fluid mass considered
fluid mass neglected
the spherical Bondi flow without including fluid 1.4
mass. We are calling the solutions of q as below,
1.3
√ 1.2
−B+ B2−4AC
qq12 == −B−√22BAA2−4AC Mach No.Mach Number 01 ..191
0.8
and we are calling 0.7
0.6
5e+13 1e+14 1.5e+14 2e+14 2.5e+14 3e+14
du Radial Distance(km)
q = |
dr Bondi
Fig. 14: Mach number vs radial distance profile
where mass of the Main-Sequence star is 1 solar
where du| is the value of du/dr at tran-
dr Bondi mass and γ of the fluid is 1.41, temperature of
sonic surface without inclusion of gravity due to
the infalling fluid at infinity is about 50◦K and
medium.
density of the fluid at infinity is of the order
10−29kg/m3.
1
0 q1
q2=q 1.02
-1 flufluidid m maasss sc onnesgildeecrteedd
1.015 1
115du/dr(/sec)x105 --32 1 1.0.0015
No. 1
-4 Mach mber 0.995
Nu
-5 s=1 Mach 0.99
0.985
-6
1.3 1.35 1.4 1.45 1.5 1.55 1.6 1.65
γ 0.98
Fig. 13: q , q and q are plotted against γ for
1 2 0.975
1.4e+14 1.42e+14 1.44e+14 1.46e+14 1.48e+14 1.5e+14
comparison Radial Distance(km)
From the above figure, we see that q is much
2
closertoqthanq . Asweareconsideringthatthe Fig. 15: Close view near transonic surface.
1
inclusion of fluid mass does not alter the critical
values abruptly, so we will choose q over q . So, inclusion of fluid mass make the infalling
2 1
8
fluid to cross sound speed nearer to the accretor. Usingthismethodology,wehavecomparedthere-
sults with the usual spherically symmetric Bondi
flow and we have found that inclusion of fluid
mass changes the nature of flow of the infalling
11 Summary and Conclusions fluid. Mass accretion rate is increased and radius
of the transonic surface is decreased when fluid
mass comes into picture.
In this paper, we have studied the non relativis-
tic spherically symmetric accretion including fluid
mass. We haven’t considered the growth of the 12 Acknowledgements
accretor during the process of accretion. We have
consideredsteadyflowofinfallingfluidandtheac- The author is thankful to his Ph.D. supervisor for
cretor to be a candidate of Main-sequence stars. introducinghimtothisproblemandforhishelpin
We have designed a methodology to find the modifying the calculations and the overall presen-
changes in spherically symmetric accretion when tation. He is also thankful to his colleague Dhruv
fluid mass is taken into account. We have set Q, Pathak for useful discussions.
i.e; the precision of the problem and then we have
considered several accretion systems by our input
References
parameters. Wemayhaveingeneralseveralinputs
as: massMoftheaccretor,γ ofthefluid,energyE
of the fluid, density of the fluid at infinity ρ , the [1] Bondi, H., 1952. MNRAS 112, 195.
∞
mass ratio of the medium and the accretor s and
[2] Chia,T.T,1978.Mon..astr.Soc.,185,561.
the extent of the medium r etc. Now according
∞
to our methodology after fixing Q when we take s [3] Demircan Osman and Kaharaman Go¨ksel.,
into account that in return fixes the extent of the 1991Ap and SS.181..313D.
medium r according to the input energy E. ρ
∞ ∞
is then fixed in accordance with γ and E of the [4] ClarkeC.J.andCarswellR.F.,2007,
fluid. In a summary when we have given the 5 in- ‘PrincipleofAstrophysicalFluidDynamics’,
put parameters M, Q, s, γ and E of the problem CAMBRIDGE UNIVERSITY PRESS.
r and ρ are fixed. r and ρ are no longer
∞ ∞ ∞ ∞ [5] Spitzer Lymann.,Jr., 2004, ‘Physical Pro-
independent input parameters. Actually, these 5
cesses in the Interstellar Medium’, WILEY-
input parameters contain the informations about
VCH Verlag GmbH and Co. KGaA, Wein-
ρ andr . Onecanstartinotherwayroundlike
∞ ∞ heim.
by taking input parameters, r and ρ first and
∞ ∞
then setting Q of the problem. In that case, our 5
independentparameterswouldbeM,Q,γ,r and
∞
ρ . Then one can find the changes in the critical
∞
parameters by varying r and ρ . That will be
∞ ∞
anotherwayoflookingatthesameproblem. Sim-
ilarly, one can have other several ways of looking
at the same problem.
Now if an arbitrary accretion system is given in a
way that some input parameters are known then
at first we have to check that the known input pa-
rameters are sufficient to find any solution or not
and if the given input parameters are sufficient to
analysethesystemthenwehavetofindwhatisthe
precision Q and if Q is close to 1 then the system
isoutsideofourformalism. Otherwise,wewilltry
to find s of the problem and if s is too small or if
sliesinsidetheregionwheretestfluidapproxima-
tion is valid (as discussed in section IX) then one
cansafelyproceedwithoutincludingfluidmassef-
fectotherwiseonehavetoproceedbyfollowingthe
procedure discussed in the above sections.
9