Equations of motion of relativistic perfect fluid subject to Bjorken’s boost-invariant conditions are analysed. General relations between the gradient of temperature and the shape of stream lines are derived. The case of pure transverse motion (vanishing radial velocity) is studied in some detail. It is shown that the stable solution exists only for a very restricted class of the equations of state of the fluid

Dyrek, A.

Florkowski, Wojciech

Acta Physica Polonica B

English

Jagiellonian Univeristy Repository

Vol. B15 (1984) ACTA PHYSJCA POLONICA No 7BOOST-INVARIANT MOTION OF RELATIVISTIC PERFECT FLUID*♦ This research has been supported in part by the M. Skłodowska-Curie Foundation, Grant No F7-071-P.** Address: Instytut Fizyki, Uniwersytet Jagielloński, Reymonta 4, 30-059 Kraków, Poland.(653)By A. Dyrek and W. FlorkowskiInstitute of Physics, JageIIonian University, Cracow**(Received January 16, 1984)Equations of motion of relativistic perfect fluid subject to Bjorken,s boost-invariant conditions are analysed. Generał relations between the gradient of temperaturę and the shape of stream lines are derived. The case of pure transverse motion (vanishing radial velocity) is studied in some detail. It is shown that the stable solution exists only for a very restricted class of the equations of state of the fluid.PACS numbers: 47.75.+f, 12.40.-y/. IntroductionIn this paper we study the equations of motion of the relativistic perfect fluid satisfying the Bjorken condition [1] of boost-invariance. It was argued in [1] that these equations describe the evoiution of quark-gluon plasma possibly created in the central region of the collision of heavy ions at very high energies.Some particular cases of such boost-invariant motion of relativistic fluid were already considered by several authors. Bjorken [1] derived the solution depending only on z and t and independent on x, y variables. Baym et al. [2] considered radial motion of the fluid with rotational symmetry around the collision (z) axis and found several solutions corresponding to Riemann initial conditions. In Ref. [3] solutions for static initial condi­tions were presented. Finally, in Ref. [4] the purely transverse motion (vanishing radial velocity) again with rotational symmetry around the collision axis was considered. It was proven that if equation of state is that of an ideał gas, no solution exists.In this paper we continue investigations along the lines suggested in Refs. [1-4], We study the rotationally-symmetric motions but do not restrict the fluid velocity other- wise. An interesting relation is found between the gradient of temperaturę and the shape of flux lines. Namely, for the temperaturę decreasing with inereasing distance, far from 654the centre, the fluid expands along the radial lines. We also discuss the purely transverse motion and find a class of equations of state of the fluid which admit a stable solution. Some comments are given on the problem of generał equations of motion in case of weak dependence on azimuthal angle.Evolution of the perfect fluid is governed by the conservation laws for energy and momentum∂μTμv = 0 (1.1)with the stress tensor given byTμv = (ε+p)uμuv+pgμ'l. (1.2)Here ε is the energy density, p the pressure, g"v the metric tensor, uμ is the four-velocityuμ = γ(l,υ), y = (l-t>2)~ιz2. (1.3)Taking the four-divergence of Eq. (1.2) we obtainuv[Mμ∂μ(ε+p)+(ε +p)<3μuμ]+(ε+p)uμ∂μuv -∂yp = 0. (1.4)Multiplication by uv gives u'13με+(ε+p)δμMμ = 0 (1.5)thus (1.4) can be written asuvuμ∂μp-∂vp + (ε+p)uμ∂μuv = 0. (1.6)The last equations are rełations between ε, p and v. It is convenient for further discus- sion to consider thermodynamic variables: entropy and temperaturę rather than ε and p. This change of variables can be performed using the rełations (they are valid provided chemical ρotential vanishes — this will be assumed throughout all the paper)dε = Tds, dp = sdT, ε+p = Ts, (1.7)where s is the entropy density.One obtains, after some manipulationsaμ(sM'1) = o, (i.8)uμ∂μ(Tuv) = ovT. (1.9)Only 3 components of the last equations are independent, sińce ιΛzμ ≈ 1. Therefore we have 4 equations for 5 unknowns: vx, vy, vx, T, s. This system should be supplemented by equation of states=f(T). (1.10)In the next Section we rewrite Eqs. (1.8) and (1.9) using the Bjorken conditionvz = z∣t (1.11) 655and the requirement that all other quantities depend only onτ = ft2-z2 and transverse coordinates. In that Section we consider relations between the gradient of temperaturę and the shape of stream lines. The case of purely transverse motion (vortex) is discussed in Section 3. Our conclusions are listed in the last Section. Finally, in the appendices details of calculations are given.2. Generał equations of motion of boost-invariant fluidUsing the Bjorken ansatz (1.11) for vx we are left with two-dimensional problem, in the piane perpendicular to the collision axis. If we represent the two-dimensional velocity vector v by its length v ≡ ∣υ∣ and the angle a between v and the radial versor ⅛ (see Fig. 1) we havevR = v cos α, vτ = v sin a, (2.1)where vR and vτ are radial and transversal components of the velocity.Fig. 1. Definition of kinematic variablesWithout any loss of generality, we can restrict our considerations to z = 0, where v- = 0 and τ = t.Using standard methods of differential geometry we then obtain from Eqs. (1.8) and (1.9) the following three conditions:(a) Equation for entropy∂t(rtsγ')+8r(rtsγv cos a) + ∂φ(tsγv sin a) = 0.(b) Equations for temperaturę and velocity∂t(rTγv) + r cos α 5r(Ty) + sin a bφ{Tf) = 0, da,∖ + 77) (2.2)(2-3)(2.4)(2.5)cos a , (υ sin asin a t>,T---------- 5_T = Tγ⅛ ---------r \ rd „ ,where the differential operator —— is denned as dtd u sin a— ≡ ∂t+υ cos a σrH---------- ∂φdt r656and ∂t, ∂r, ∂φ denote partial derivatives with respect to the corresponding variables. The operator (2.5) represents the change of any quantity in the co-moving frame, i.e. in the frame where the local velocity of the fluid vanishes.In case of fuli rotational symmetry T, s, v and a do not depend on φ (smali non-sym- metric perturbations are discussed in Appendix 3). Consequently, Eqs. (2,2)-(2.4) reduce to∂l(rtsγ) + ∂r(rtsγv cos a) = 0,o,(Tγr) + cos a ∂r(Tγ) = 0,sin a δrT = Tγ2v da \(2.6)(2.7)(2-8)+For a = 0 we recover the Eqs. of Ref. [2] and for a = π∕2 we recover the Eqs. of Ref. [4], Eq. (2.8) has an interesting consequence. Considering the motion of a smali fragment of the fluid we may treat φ and a as functions of t. Let us observe that the quantityv sin α vτ dφ _r r dtis the angular velocity of the fragment.Fig. 2. The shape of the stream lines for (a) temperaturę decreasing with increasing distance from the centre; (b) temperaturę increasing with increasing distance from the centre; (c) constant temperaturę657Furthermore, one sees that the quantityv sin a da d---------- 1—— = — (φ+a+const) r-------- dt dtis the rotational speed of the velocity vector, i.e. the ratę of change of the angle between the velocity vector and a certain radial linę. Consequently, the behavior of the angle α+φ+const is closely related to the sign of crT. To be specific, let us assume that ω > 0, i.e. α∈(0,π). Then, in the co-moving frame:for 8rT < 0 — this angle decreases with increasing t;for ∂τT > 0 — this angle increases with decreasing t;for ∂rT = 0 — this angle is constant and the stream linę is a straight one.These relations are illustrated in Fig. 2. One sees that for the most interesting case of temperaturę decreasing with increasing r (orT < 0), the rotational speed is negative. In the co-moving frame a must be a monotonie function of time, decreasing with increasing t, to balance the inerease of φ (ω > 0).Fig. 3. Generał view of the stream lines and central static regionFig. 4. Distribution of temperaturę corresponding to picture in Fig. 3658Since α is ρositive, it has a lower boundary and the increase of φ must be limited and far away from the centre it makes the radial motion predominating.For such a shape of motion there is an area the stream lines cannot enter: the interior of a certain circle (see Fig. 3). Hence, we have a static hot core surrounded by the expanding fluid. However, sińce v = 0, Eqs. (2.7) and (2.8) imply 8rT = 0 inside the core. Thus, the temperaturę as a function of r must be given by a bell-shaped function (see Fig. 4), or temperaturę increases with increasing r somewhere.3. Boost-invariant vortexFor purely transveιsal motion (vR = 0, a = π∕2) Eqs. (2.6)-(2.8) simplify intoĄ(tsy) = 0,∂t(Tγv) = 0, (3.1)(3-2)(3.3)∂rT = Tγ2υ2rThese Eqs were studied in Ref. [4] where it was shown that their solutions are inconsistent with the equation of state for perfect fluid, i.e.s = const • T3. (3.4)In Appendix 1 we show that the only equations of state consistent with Eqs. (3.1)-(3.3) are of the forms = const ∙ T ∙ exρ (cT2'), (3.5)and exp (cT2)√T2 + k2 k ,where c and k are constants.Let us first discuss the solution (3.5). As derived in Appendix 1, temperaturę and velocity of the fluid in this case are given byandv = — r(3-7)(3.8)r0(t) is an increasing positive function of time. It is plotted in Fig. 6 for several values of parameters c and k.As seen from Eq. (3.7) for r < r0(t), T2 becomes negative and thus the solution does not have a physical meaning. We also observe that at r = r0(t), v = 1. For r > r0(t) the solu­tions (3.7) and (3.8) are plotted in Fig. 5.As seen from Fig. 5 the temperaturę is an increasing function of the distance from the centre, starting from T = 0 at r = r0(t). Thus this solution describes an expanding infiniteτ = k∣4-— 4,√ r20(t) r2659Fig. 5. The stable solution for the vortex: (a) temperaturę; (b) velocityFig. 6. The radius of a circle of indefinable temperaturę as a function of time660ring of plasma with temperaturę increasing from inner to outer region. The ring cools down with increasing time, the limiting temperaturę at r → ∞ being k∣r0(t). It is also worth to notice that this solution satisfies the standard condition for nonrelativistic vortices, namely that its circulation is independent of r. However, it does depend on time. This is easily seen from Eq. (3.8). It is proven in Appendix 2 that this solution is stable against smali perturbations introducing radial component of velocity.The corresponding results for the solution (3.6) areυ =kT = 1 (3-9)(3.10)l+21n-⅞btwhere b = const and m = --.. —-.2ck2It is shown in Appendix 2 that this solution is unstable against smali perturbations. We thus shall not discuss it any further.Fig. 7. The unstable solution for the vortex: (a) temperaturę; (b) velocity6614. ConclusionsWe have investigated solutions of the hydrodynamical equations describing space-time evolution of the quark-gluon plasma with cylindrical symmetry along the collision axis.The generał relations between the gradient of temperaturę and the shape of stream lines show that temperaturę decreases with di stance from the centre only when far from collison region the fluid expands outside along the radial lines.Exact solutions of pure transversal motion (vortex) are given. It is interesting that in this case the form of the equation of state of the fluid is determined. This solution desc- ribes an expanding ring of rotating plasma. The temperaturę increases with increasing distance from the center and decreases with increasing time. It seems, however, unlikely that such a solution can be realizable in naturę.The authors thank Professors A. Białas and W. Czyż for suggesting this investigation and critical reading of the manuscript.APPENDIX 1Symmetric vortexIn the case of no radial motion the equations reduce to those shown in Section 3. Equations (3.1) and (3.2) can be writtentsγ = A(r), (Al.l)Tyv = B(r), (A1.2)where A(r) and B(r) are certain functions of r.We will show that the system of equations (Al.l), (A1.2), (3.3) and (1.10) is consistent only if special conditions for √4(r), B(r),f(T) are satisfied. Now we will fund these relations.Equations (Al.2) and (3.3) giveT2 = X(t) + Y(r), (Al.3)drwhere T(r) = 2 ∫ B2(r) — and X(t) is certain function of time. Let us assume that the func­tions X(t) and Y(r) can be inverted. Seeing that we can regard A and B as functions of Y and time as a function of X. We will use either X and Y or X+ Y and Y as independent variables in further considerations. Then, we transform Eq. (Al.l) and obtain[X+ Y+B2(Y)]t2(X)f2(^X+Y) = (X+ Y)A2(Y). (A1.4)We differentiate the above equation with respect to Y while X+ K is constant and then repeat twice differentiation with respect to X while Kis constant. After these manipulations the following equation is obtained[X+ r+B2(Y)]≠''W+2<∕>"(X) = 0, (Al.5)662where we introduced φ(X) = ln t(X) and prim denotes the derivative with respect to X. The variable Y enters only in the expressionY+B2(Y),thus Y+B2(Y) must be constant or φ(X) is a linear function. First we assumeY+B2(Y) = a = const. (Al.6)This condition leads to the form of B(r^) and Y(r).B(r) = -, γ(r)≈a-½τ (A1.7)r r(k is an arbitrary non-negative constant).Snbstituting (Al.7) into Eq. (Al.4) we obtain (after adjustment of the constants of integration) the exact form of the relation between entropy and temperaturę, i.e. the equation of states = const ∙ 7,exp (cΓ2) (A 1.8)(c is non-negative constant). We obtain also the implicit equation for X(f)exp(-cX(t))t = const —.√ X(t) + a(A1.9)One observes that X(t)+a is a monotonie function of time, exploding for time tending to zero and vanishing for large times like l∣t2.Since the function Y(r) becomes infinitely negative at r → 0 the expression (Al.3) for T2 is negative for r belo w some critical value r0(t). It means that the solution has no physical meaning for r ≤ r0. The radius ro(0 is given byfcr0(t)=7=, (Al.10)√X(t) + αck2ln t = ln r0------r + const. (Al. 11)The case of linear form of function φ(X) remains to be studied. One gets the solution by the method analogous to that used before. The corresponding equation of state is found to beTs = const ----- ■: exp {cT2), (Al.12)√T2 + ∕<2where c and k are ρositive constants.It may be verified easily that function B(r) is constant. The next appendix contains the argument which shows that B(r) given by a constant leads to instability, so we will not discuss this case anymore. However, it is interesting to notę that here also exists an expanding region of indefinable temperaturę (T2 < 0).663APPENDIX 2Stability of the solutions for a symmetric vortexLet us assume that v, at the piane z = 0, has transversal and radial componentsv = vTeT + vReR, (A2.1)wherevr Vγ. (A2.2)Because of cylindrical symmetry vτ, vR are functions of r and t. Substituting (A2.1) into Eqs. (1.8)-(1.9) we obtainγδ,(Tyυτ)+γ⅜δr(Tγι>τ) + Ty2 — = 0,r2 CT yδ,(Γy⅛)+7⅛or(Ty⅜)-Ty - +∂rT = 0, r(A2.3)(A2.4)15t(sy0+ — δr(rsγvtt) = 0.r (A2.5)γ can be expanded in powers of vR.. 23 ^RX0 = X¾)+y (⅜)∙y + ■■■, (A2.6)wherev = ∣p∣ = Vι>¾+⅛ (A2.7) Because vR < 1 we neglect ali the terms of higher than first order in vR. This way we get equations of similar form to Eqs. (A2.3)-(A2.5), but y(y) in all expressions is replaced by γ(uτ). In the next equations γ always means y(υτ).7¾Tyrτ)+7⅛aχ7>τ)+7Y — = 0,r(A2.8)Ty2v⅛ y∂t(Tyυf)- -^-r +∂rT = 0,r(A2.9)¾(syt)+ — d,(rsyvR) = 0.r(A2.10)Assuming that vT is a solution of the above equations in the case vR = 0, we obtainTγvτ∂r(Tyυτ) + — = 0,r(A2.ll)∂,(7>r) = 0,δ,(rsyt⅛) = 0. (A2.12)(A2.13)664 In Appendix 1 we have shown that in the case of a symmetric vortex two possibilities exist for function B(r) = 7yrT:k(a) Tγvτ =r(b) Tγυτ = const.Eq. (A2.ll) shows that only the case (a) leads to stability. Taking into consideration (A2.ll) we obtain from (A2.12)and from (A2.13) ⅜ = <x(r)vτ, (A2.14)(A2.15)α(r) is arbitraiy function of r.Identity (a) implies an equation of state in the forms = const ∙ Texp (cT2), (A2.16)k2where T2 = X(t)+a-------. The solution of Eq. (A2.15) is α(r) = α0 expr2a(r) should be infinitesimal, therefore the solution is stable for all r > 0 only when c = 0 and α0 < 1.APPENDIX 3Non-azimuthally-symmetric perturbationThe generał equations of hydrodynamics - Eqs. (1.8) and (1.9) (with the condition of Bjorken) can be written in a form altemative to that of Eqs. (2.2)-(2.4), shown in Sec- tion 2. Taking υ as a pair of its components vR and υτ one obtains:(a) Equation for entropy(A3.1)where σ = ln s, v = ∣υ∣.(b) Equations for temperaturę and velocity(A3.2)(A3.3)665where O = ln T. The operator — is defined bydt— ≡ S,+vr∂γ+-∂. (A3.4)dt rWe assume the equation of state in the formσ = Ψ(0). (A3.5)It is much morę easy to obtain a solution when the unknowns do not depend on the azimuthal angle φ. However, if one knows such a solution, one can find an approximate result for the case with the functions slightly deformed. (The exact solution must be stable.) We will put all the dependence on the angle φ into these smali perturbations. So, we assume that the unknowns depend on φ but it is a weak dependence. The quantities should be periodical functions of the angle (period = 2π) and we can represent them with their Fourier series:θ(r, φ,t)≈Y θ<n∖r, t)einφ, σ(r, φ, t) = ∑ σ<">(r, t)ein*,n nv(r, <p, t) = ∑ υin∖r, t)e'nφ, va(r, φ,t) = £ ⅛0(r, t)ei"φ, (A3.6)n nwhere a = R, T and n — 0, +1, ±2, ...We assume that all the coefficients with their indices different from zero are infinites- imal and we may neglect their squares and products. Some coefficients are related. One observes that 40)4ł)+⅛0Hfc> = vwvm, k = 0, ± 1, ±2, ... (A3.7)sińce the Fourier expansion is unique and (t⅛)2 + (vτ)2 = (r)2. We can also write the ex- pansion of σ = Ψ(θ)∑σ<"V*  = σ = ψ(θ) = Ψ(0<o>) + Ψ'(0<o>) ∑ θ^einφ. (A3.8)n n≠0Consequently, σ0° and 0(B) must satisfy the following conditionsσ(-) = ψ'(0(o>)0<">, n ≠ o. (A3.9)For n = 0 we obtain equations equivalent to those obtained in case of rotational symmetry. For each n ≠ 0 we have three linear equations for 0(B), σw, ι⅛"∖ ⅛0 with two independent variables: r and t. They are[i -(√0*) 2]40⅛iθw+[i - (√0,)¾ew+ą4b)+40Mn>- 3(⅛0>)W°, - 2⅛0>are<0>+ar⅛0t×⅛>-240>(,⅛W0,+δrθ<0>+ = o, (A3.10)666 [i - (√oj)2]⅛o>¾0<b,+st⅛">+⅛0M0f ,,<°) ιn√0>lUi-(40))2-3(40))W0)+ 7- + }⅛>+[i-(√0>)2]+ 4°)0--2⅛Wo>)+Mo 4o = o,inO,n>r(A3.11)+ 4These equations are complicated, but tħey may simplify in some specific cases. E. g., for a vortex with smali radial and transversal perturbations we may neglect all the terms multiplied by Γr0* in the above system and substitute temperaturę, velocity and entropy of the symmetric pure vortex (see Appendix 1), for 0(O), ⅛0*,  a<0). Hence, we have three linear equations with known coefficients and some relations (A3.9) between the unknowns.REFERENCES[1] J. Bjorken, Phys. Rev. 027, 140 (1983).[2] G. Baym et al., Nuci. Phys. A407, 541 (1983).[3J A. Białas, W. Czyż, A. Kolawa, Acta Phys. Pol. B15, 229 (1984).[4] A. Białas, W. Czyż, Acta Phys. Pol. B15, 247 (1984).(A3.12)where

Boost-invariant motion of relativistic perfect fluid

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Boost-invariant motion of relativistic perfect fluid

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