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