### 2.3 Exact solution of the Riemann problem in SRHD

Let us first consider the one-dimensional special relativistic flow of a perfect fluid in the absence of a gravitational field. The Riemann problem then consists of computing the breakup of a discontinuity, which initially separates two arbitrary constant states (left) and (right) in the fluid (see Figure 1 with and ). For classical hydrodynamics the solution can be found, e.g., in [62]. In the case of SRHD, the Riemann problem was considered by Martí and Müller [180], who derived an exact solution for the case of pure normal flow generalizing previous results for zero initial velocities [276]. More recently, Pons, Martí and Müller [235] have obtained the general solution in the case of non-zero tangential speeds.

The solution to this problem is self-similar, because it only depends on the two constant states defining the discontinuity and , where , and on the ratio , where and are the initial location of the discontinuity and the time of breakup, respectively. Both in relativistic and classical hydrodynamics the discontinuity decays into two elementary nonlinear waves (shocks or rarefactions) which move in opposite directions towards the initial left and right states. Between these waves two new constant states and (note that and in Figure 1) appear, which are separated from each other by a contact discontinuity moving with the fluid. Accordingly, the time evolution of a Riemann problem can be represented as

where and denote a simple wave (shock or rarefaction) and a contact discontinuity, respectively. The arrows ( / ) indicate the direction (left / right) from which fluid elements enter the corresponding wave.

As in the Newtonian case, the compressive character of shock waves (density and pressure rise across the shock) allows us to discriminate between shocks () and rarefaction waves ():

where p is the pressure, and subscripts a and b denote quantities ahead and behind the wave. For the Riemann problem and for and , respectively. Thus, the possible types of decay of an initial discontinuity can be reduced to

Across the contact discontinuity the density exhibits a jump, whereas pressure and normal velocity are continuous (see Figure 1). As in the classical case, the self-similar character of the flow through rarefaction waves and the Rankine–Hugoniot conditions across shocks provide the relations to link the intermediate states () with the corresponding initial states . They also allow one to express the normal fluid flow velocity in the intermediate states ( for the case of an initial discontinuity normal to the axis) as a function of the pressure in these states.

The solution of the Riemann problem consists in finding the intermediate states and , as well as the positions of the waves separating the four states (which only depend on , , , and ). The functions and allow one to determine the functions and , respectively. The pressure and the flow velocity in the intermediate states are then given by the condition

where .

In the case of relativistic hydrodynamics, the major difference to classical hydrodynamics stems from the role of tangential velocities. While in the classical case the decay of the initial discontinuity does not depend on the tangential velocity (which is constant across shock waves and rarefactions), in relativistic calculations the components of the flow velocity are coupled by the presence of the Lorentz factor in the equations. In addition, the specific enthalpy also couples with the tangential velocities, which becomes important in the thermodynamically ultrarelativistic regime.

The functions are defined by

where () denotes the family of all states which can be connected by a rarefaction (shock) with a given state ahead of the wave.

The fact that one Riemann invariant is constant across any rarefaction wave provides the relation needed to derive the function . In differential form, the function reads

where is the absolute value of the tangential velocity, and
and where
the () sign corresponding to (). In the previous expressions, stands for the local sound speed.

Considering that in a Riemann problem the state ahead of the rarefaction wave is known, the integration of Equation (19) allows one to connect the states ahead () and behind the rarefaction wave. Moreover, using Equation (21), the EOS, and the following relation obtained from the constraint , that holds across the rarefaction wave,

the ODE can be integrated, the solution being only a function of p. Let us point out that the integration of Equation (19) is along an adiabat of the EOS.

In the limit of zero tangential velocities, , the function g does not contribute. In this limit and in the case of an ideal gas EOS one has

(where is the adiabatic exponent of the EOS) recovering expression (30) in [180]. The equation can be then integrated to give [180]
with
the () sign of corresponding to (). In the above equation, is the sound speed of the state , and is given by

The family of all states , which can be connected through a shock with a given state ahead of the wave, is determined by the shock jump conditions. One obtains

where the () sign corresponds to (). and denote the shock velocities for shocks propagating to the right and left, respectively. They are given by
Tangential velocities in the initial states are hidden within the flow Lorentz factor . On the other hand, stands for the modulus of the mass flux across the shock front,
where the enthalpy of the state behind the shock can be obtained from the Taub adiabat,
In the general case, the above nonlinear equation must be solved together with the EOS to obtain the post-shock enthalpy as a function of p. In the case of ideal gas EOS with constant adiabatic index, the post-shock density can be easily eliminated, and the post-shock enthalpy is the (unique) positive root of the quadratic equation [180]

Finally, the tangential velocities in the post-shock states can be obtained from [235]

Figure 2 shows the solution of a particular mildly relativistic Riemann problem for different values of the tangential velocity. The crossing point of any two lines in the upper panel gives the pressure and the normal velocity in the intermediate states. The range of possible solutions in the ()-plane is marked by the shaded region. While the pressure in the intermediate state can take any value between and , the normal flow velocity can be arbitrarily close to zero in the case of an extremely relativistic tangential flow. The values of the tangential velocity in the states and are obtained from the value of the corresponding functions at in the lower panel of Figure 2. The influence of initial left and right tangential velocities on the solution of a Riemann problem is enhanced in highly relativistic problems. We have computed the solution of one such problem (see Section 6.2.2 below, Problem 2) for different combinations of and . The initial data are  = 103,  = 1,  = 0;  = 10–2,  = 1,  = 0, and the 9 possible combinations of  = 0, 0.9, 0.99. The results are given in Figure 3 and Table 1, and a complete discussion can be found in [235].

Table 1: Solution of the relativistic Riemann problem at t = 0.4 with initial data  = 103,  = 1.0,  = 0.0,  = 10–2,  = 1.0, and  = 0.0 for 9 different combinations of tangential velocities in the left () and right () initial state. An ideal EOS with  = 5/3 was assumed. The various quantities in the table are: the density in the intermediate state left () and right () of the contact discontinuity, the pressure in the intermediate state (), the flow speed in the intermediate state (), the speed of the shock wave (), and the velocities of the head () and tail () of the rarefaction wave.
 0.00 0.00 9.16 × 10–2 1.04 × 10+1 1.86 × 10+1 0.960 0.987 –0.816 +0.668 0.00 0.90 1.51 × 10 –1 1.46 × 10+1 4.28 × 10+1 0.913 0.973 –0.816 +0.379 0.00 0.99 2.89 × 10 –1 4.36 × 10+1 1.27 × 10+2 0.767 0.927 –0.816 –0.132 0.90 0.00 5.83 × 10–3 3.44 × 10+0 1.89 × 10–1 0.328 0.452 –0.525 +0.308 0.90 0.90 1.49 × 10–2 4.46 × 10+0 9.04 × 10–1 0.319 0.445 –0.525 +0.282 0.90 0.99 5.72 × 10–2 7.83 × 10+0 8.48 × 10+0 0.292 0.484 –0.525 +0.197 0.99 0.00 1.99 × 10–3 1.91 × 10+0 3.16 × 10–2 0.099 0.208 –0.196 +0.096 0.99 0.90 3.80 × 10–3 2.90 × 10+0 9.27 × 10–2 0.098 0.153 –0.196 +0.094 0.99 0.99 1.29 × 10–2 4.29 × 10+0 7.06 × 10–1 0.095 0.140 –0.196 +0.085

Finally, let us note that the procedure to obtain the pressure in the intermediate states is valid for general EOS. Once has been obtained, the remaining state quantities and the complete Riemann solution,

can be easily derived. In Section 9.4, we provide two FORTRAN programs called RIEMANN (Section 9.4.1) and RIEMANN-VT (Section 9.4.2), which allow one to compute the exact solution of an arbitrary special relativistic Riemann problem for an ideal gas EOS with constant adiabatic index, both with zero and non-zero tangential speeds using the algorithm discussed above.

Solving a Riemann problem involves the solution of an algebraic equation for the pressure (Equation (17)). Moreover, the functional form of this equation depends on the wave pattern under consideration (see expressions (16). In a recent paper [241], Rezzolla and Zanotti have presented a procedure, suitable for implementation into an exact Riemann solver in one dimension, which removes the ambiguity arising from the wave pattern. That method exploits the fact that the expression for the relative velocity between the two initial states is a (monotonic) function of the unknown pressure, , which determines the wave pattern. Hence, comparing the value of the (special relativistic) relative velocity between the initial left and right states with the values of the limiting relative velocities for the occurrence of the wave patterns (16), one can determine a priori which of the three wave patterns will actually result (see Figure 4). In [242] the authors extend the above procedure to multi-dimensional flows.