Chapter 4 SHOCK WAVES IN ASTROPHYSICS 1. Physics of the shock waves 6.1. What are the shock waves? Shock waves are found in many high energy astrophysical phenomena and play a key role in many different astrophysical environments like star formation in the spiral arms of galaxies, the high velocity outflows from young stars, extra-galactic radio sources and compact galactic nuclei, and in the interaction of the solar wind with the magnetic field of the Earth 1 . It is a general property of perturbations in a gas that they are propagated away from their source at the speed of sound in the medium. It is therefore obvious that if a disturbance is propagated at a velocity greater than the speed of sound, the disturbance cannot behave like a sound wave at all. There is a discontinuity between the regions behind and ahead of the disturbance and the latter region can have no prior knowledge of its imminent arrival-the sound waves which would transmit the information are propagated at a speed less than that of the disturbance. These discontinuities are called shock waves. They commonly arise in situations such as a) explosions or b) where gases flow past obstacles at supersonic velocity or, equivalently, c) objects move supersonically through a gas. The basic phenomenon is the flow of a gas at a supersonic velocity relative to the local velocity of sound. 1 6.2. The basic properties of plane shock waves We assume that there is an abrupt discontinuity between the two regions of fluid flow. In the undisturbed region ahead of the shock wave, the gas is at rest with pressure p1 , density 1 and temperature T1 . Behind the shock wave, the gas moves supersonically and its pressure, density and temperature are p 2 , 2 and T2 , respectively. It is more convenient to analyze the shock waves in a reference frame moving at velocity U in which the shock wave is stationary. Then, the undisturbed gas flows towards the discontinuity at velocity v1 U and, when it passes through it, its velocity becomes v 2 in the moving reference frame. The behavior of the gas passing through the shock wave is described by a set of conservation relations. First, mass is conserved on passing through the discontinuity and hence 2 1 v1 2 v 2 . (1) Second, the energy flux, i.e., the energy passing per unit time through unit area parallel to v1 is continuous. It is one of the standard results of fluid dynamics that 1 the energy flux through a surface normal to the vector v is v v 2 w , where w is the 2 enthalpy per unit mass, w m pV , where m is the internal energy per unit mass and V 1 is the specific volume, i.e. the volume per unit mass. We consider only plane shock wave, which are perpendicular to v1 and v 2 . Therefore the conservation of energy flux implies 1 2 1 2 1 v1 v12 w1 2 v 2 v 22 w2 . (2) Notice that it is the enthalpy per unit mass and not the energy per unit mass which appears in this relation. The reason for this is that, in addition to internal energy, work is done on any element of the fluid by the pressure forces in the fluid and this energy is available for doing useful work. Another way of looking at this relation is in terms of Bernoulli’s equation of fluid mechanics in which the quantity 1 2 1 p v w v 2 m is conserved along streamlines, which is the case for normal 2 2 flow through the shock wave. Finally, the momentum flux through the shock wave should be continuous. For the perpendicular shocks considered here, the momentum flux is given by p v 2 and hence p1 1 v12 p2 2 v 22 . (3) Notice that the pressure p , being the force per unit area, contributes to the momentum flux of the gas. Equations (1)-(3) are the three conservation equations which are often referred to as the shock equations. 3 For simplicity we shall study the case of shock waves in a perfect gas for which the enthalpy is w pV / 1 , where is the ratio of the specific heats. One can obtain many important results for this case. First of all, we define the mass flux per unit area as j 1 v1 2 v 2 . Then, from Eq. (3), which describes the conservation of the momentum, we immediately find j2 p2 p1 . V1 V2 (4a) In addition, we obtain an expression for the velocity difference: v1 v 2 j V1 V2 p2 p1 V1 V2 . (4b) The next step is to find the ratio V2 /V1 as a function of p1 and p 2 for a perfect gas. We begin with the equation of conservation of the energy flux Eq. (2) and substitute as follows: 1 2 1 1 1 v1 w1 v 22 w2 , j 2V12 w1 j 2V22 w2 . 2 2 2 2 Using Eq. (4b), this expression reduces to w1 w2 1 V1 V2 p 2 p1 0 . 2 We can now substitute the perfect gas expression w pV / 1 to obtain V2 1 p1 1 p2 , V1 1 p1 1 p2 (5) 4 which gives the relation between the pressures and specific volumes on either side of the shock. We can immediately find the relation between T2 and T1 from the perfect gas law, p1V1 p 2V2 , T1 T2 T2 p2V2 p2 1 p1 1 p2 . T1 p1V1 p1 1 p1 1 p2 Also using Eq. (5) we can eliminate V2 from the expression for the flux density j , obtaining j2 1 p1 1 p2 . 2V1 From this equation we can find the velocities of the gas in front of and behind the shock, V 1 p1 1 p 2 V . (6) v j V 1 1 p1 1 p2 , v 22 j 2V22 2 2 1 p1 1 p 2 2 2 2 1 2 2 1 It is most convenient to write these results in terms of the Mach number M of the shock wave, which is defined as M 1 v1 / c1 , where c1 p1 / 1 is the velocity of sound for the undisturbed gas. Thus M 12 v12 . p1V1 From Eqs. (6) we obtain the pressure ratios as p 2 2M 12 1 , 1 p1 5 and the density ratio 2 v1 1 p1 1 p 2 1 . 1 v 2 1 p1 1 p 2 1 2 2 M1 Finally we find the temperature ratio T2 2M 12 1 2 1M 12 . T1 12 M 12 It is useful to look at these ratios in the limit of very strong shocks, M 12 1 . Then p 2 2M 12 2 1 T2 2 1M 12 , , . p1 1 1 1 T1 12 These results show that in the limit of very strong shocks, the temperature and pressure can become arbitrarily large but the density ratio attains the finite value 1 . For example, a monatomic gas has 4 / 3 and hence 2 / 1 4 in the limit of 1 very strong shocks. These results demonstrate how efficiently strong shock waves can heat gas to very high temperatures and this is found to be the case in supernova explosions and supernova remnants. 6 What exactly is happening in the shock front? It is apparent that the undisturbed gas is both heated and accelerated when it passes through the shock front, and this is mediated by the atomic or molecular viscosity of the gas. It can be shown that the acceleration and heating of the gas takes place over a physical scale of the order of the mean free path of the atoms, molecules or ions of the gas. Thus, the shock front is expected to be very narrow and the heating very strong over this short distance. 6.3. The supersonic piston A common situation in high energy astrophysics is one in which an object is driven supersonically into a gas, or equivalently, a supersonic gas flows past a stationary object. A useful illustrative example is that of a piston driven supersonically into tube containing stationary gas. A shock wave forms ahead of the piston and the gas behind the shock moves at the velocity of the piston v p U . In the frame of reference of the shock front, which moves at some yet unknown velocity v s , the velocity of inflow of the stationary gas is 7 v1 v s and the gas behind the shock moves at velocity v 2 . As yet we do not know v1 and v 2 but we know that their difference v1 v 2 U . Using Eq. (4b) we find v1 v 2 U p2 p1 V1 V2 . Substituting for V2 and squaring we find that the above equation can be written in terms of the pressure ratio p2 / p1 : p2 p2 U 2 1U 2 2 1 1 0. 2 p1V1 2 p1V1 p1 p1 2 We can now write p1V1 c12 , where c1 is the velocity of the sound in the undisturbed medium, and solve for p2 / p1 : p2 1U 2 U 12 U 2 1 1 p1 c1 4c12 16c12 1/ 2 . The velocity v1 v s follows from Eq. (6); V1 c12 p2 v 1 p1 1 p2 1 1 . 2 2 p1 2 1 Substituting for p2 / p1 gives 12 U 2 vs U c12 4 16 1 8 1/ 2 . This is a very neat result since it determines the thickness of the layer of shocked gas ahead of the piston for any supersonic velocity U . Let us look at the case of a very strong shock wave U c1 . Then vs 1 2 U. Thus the ratio of the position of the shock front to the position of the piston is v s / U 1 / 2 . For a monatomic perfect gas 5 / 3 and hence v s / U 4 / 3 . Thus, all the gas which was originally in the tube between x 0 and the position of the shock wave is squeezed into a smaller distance v s U t . It follows that the density increase over the undisturbed gas is 2 / 1 v s / v s U 1 / 1 . This simple calculation gives some feel for what is observed when supersonically moving gas encounters an obstacle or is ejected into a stationary gas. Ahead of the obstacle there is a shocked region which runs ahead of the advancing piston. This is what is expected to occur when a supernova ejects a sphere of hot gas into the interstellar medium. It also shows that there is a stand-off distance of a shock front from a blunt object placed in the flow. 9 2. Fundamentals of gamma-ray astrophysics 6.4. Basic concepts Astronomy is the attempt to gather information from cosmic objects through the detection of particles (photons, leptons, hadrons, neutrinos etc.), emitted or affected by them. Astrophysics is the attempt to use our knowledge of the physical laws such as we know them on the Earth to propose theories that can represent adequately the physical events behind the astronomical observations 2 . Since astronomers try to detect particles, it is convenient to start by introducing as a basic concept the number of particles incident per unit of surface area per unit of solid angle per unit of time arriving at a given, unspecified detector. We will call this basic quantity the intensity of the particles, and we will denote it by I . In general we will use a subscript to indicate the type of particles, e.g. I and I p denote the intensities of gamma rays and protons, respectively. Once the intensity is introduced, we can define the particle flux as F I cos d , where the angle is determined by the direction of motion of the particles with respect to the normal to the area, and the integration is performed over the solid angle. For isotropic radiation the flux is F I and the number of particles per unit volume is N 4 I. v In most cases v c is a good approximation because we deal with relativistic particles. 10 Normally, we will have particles with different energies, so it is useful to introduce a particle energy distribution N (E ) such that N N ( E )dE . 0 The number of particles with energies greater than E is obtained just by integrating from E . In a similar way, the integrated flux density is F E F ( E )dE . E The luminosity of a source located at a distance d that radiates isotropically is given by L E 4d 2 F ( E )dE , E where d is the distance to the source. The energy density of the particles is w Emax EN ( E )dE . Emin The energy flux is obtained from this expression just by multiplying by c / 4 , if we deal with relativistic particles or photons. Let us consider now that a flux of particles of type a , with velocity v a , interacts with some target formed by particles of type b within a volume dV . The number of particle interactions of a given type, dN i , occurring in a time dt in the volume dV will be proportional to the number of particles b in the volume dV and 11 to the number on incident particles that traverse the cross section dA of that volume in the time dt : dN i d i nb0 dV na v a dt . In this expression n b0 and n a are the densities of target and incident particles in a coordinate system with the target at rest. The differential cross section d i characterizes the number of reactions of type i occurring per unit of time in unit volume for a unit flux density of incident particles and unit density of the target. It is measured in units of area, the standard unit being the mb (i.e. 10 3 Barn, 1 Barn= 10 24 cm 2 ). The total cross section for a given interaction i is the sum over all possible momenta of the resulting particles after the interaction. Both i and d i are relativistic invariants. The total cross section tot is obtained by summing the cross sections of all possible processes that occur upon the interaction of particles a and b , i.e. tot i . i The relative probability of a given reaction channel is given simply by p i / tot . In the case of gamma-ray emission, if the generation of the gamma rays is due to the interaction of particles of type i with a given intensity I i Ei , r with a target of density nr , we can write the intensity of the radiation from the resulting gamma-ray source as I E nr Ei , E I i Ei , r dEi dr , l E 12 where l defines the direction along the line of sight (i.e. l r / r ). The emissivity of the gamma-ray source is defined as q E , r nr Ei , E I i Ei , r dEi , E in such a way that I E q E , r dr . l 6.5. Synchrotron radiation A relativistic particle moving in a magnetic field will emit photons within an angle ~ mc 2 / E of its direction of motion. In a magnetic field B an electron moves along a helical path with an angular frequency B given by eB me c 2 B me c E 13 The radiation spectrum of the electron is given by P( E ) 3e 3 E B 2 Ec me c K d , 5/3 E / Ec where E h is the energy of the radiation, B B sin , is the pitch angle and K 5 / 3 is the modified Bessel function of the second kind. The characteristic energy of the photons is given by 2 3h eB E . Ec 4 me c me c 2 The maximum of P (E ) occurs at Emax 1.9 10 11 B / Gauss E / GeV GeV. 2 We see that only for extremely energetic particles and strong magnetic fields we can get gamma-ray photons from synchrotron radiation. The total energy rate loss by synchrotron radiation of an electron moving in a field B can be obtained by integrating the radiation spectrum. The result is 2 e2 dE e c 2 dt syn 3 me c 2 2 2 B , where E e / me c 2 is the Lorentz factor of the particle. Introducing the Thomson cross section T 8e 4 / 3me2 c 4 0.665 10 24 cm 2 and averaging over an isotropic pitch angle distribution, the expression for the energy losses can be set in the following convenient form 14 4 dE e T cwmag 2 , dt syn 3 where wmag B 2 / 8 is the magnetic energy density and B is measured in Gauss. Let’s assume that we have a homogeneous and isotropic power-law electron distribution in a random magnetic field, given by N e E e dE e K e E e p dE e . Then the resulting spectrum is given by E 3e 3 K e p I E B d r E dE e e E Ec me c 2 l with 3h eB Ee Ec 4 me c me c 2 K d , 5/3 E / Ec 2 . By introducing a new variable of integration E / Ec 4me cE / 3heB Ee / me c 4me c 1 dE e me c 2 2 3heB 2 2 , 4me c Ee 2 me c 3heB 1/ 2 E1 / 2 1 / 2 , 1/ 2 E1 / 2 3 / 2 d , and by making the assumption that the end points of the electron energy spectrum do not contribute (this means that the lower limit on the E e integration may be replaced by zero), we obtain e3 I E a p me c 2 3e 4m 3 c 5 e p 1 / 2 B ( p 1) / 2 K e LE( p 1) / 2 , where L dr is the characteristic size of the emitting region and a ( p ) a p (0) , where l 15 a p ( x) p 1 2 x K 5 / 3 dd . It can be shown that a ( p ) is given by 3 p 1 3 p 19 p 5 2 ( p 1) / 2 3 12 12 4 . a( p) p7 8 p 1 4 a ( p ) is 0.147, 0.103, 0.0852 and 0.0742 for p 1.5, 2, 2.5 and 3, respectively. The emission is a power law with index ( p 1) / 2 . 6.6. Inverse Compton interactions The scattering of relativistic electrons on soft photons can produce gamma rays. The intensity of the radiation from this process when the soft photon field has a density n ph E ph , r is I IC E l E E , E , E I E , r n E 0 e ph e e ph ph , r dEe dE ph dr . We can introduce a parameter Ee E ph / me c 2 such that for 1 the scattering is classical. In such a limit the cross section can be approximated by the Thomson cross scattering T and the average energy of the emerging photons will be 16 2 E 4 E ph 2 , 3 where E ph is the average energy of the target photons. The energy losses for an electron in a photon field of energy density w ph when 1 can be approximated as dE e T cw ph 2 2 10 14 w ph 2 eV/s. dt IC We see that at this energies the ratio of synchrotron to IC cooling times is simply t IC / t syn wmag / w ph . If the incident electron spectrum is a power law and the photon field can be approximated by a mono-energetic distribution, then we obtain 1 I E n ph L T me c 2 2 IC 1 p 4 E ph 3 ( p 1) / 2 K e E( p 1) / 2 . Here, L is the typical source dimension, and E ph and n ph are average values for the photon energy and photon number density in the source. If the photon field is thermal radiation, then n ph E E ph ph me c exp E / me c 2 2 2 2 3 c 1 , ph / kT 1 where c 2me c h is the Compton wavelength of the electron. By assuming again a 1 homogeneous and isotropic power-law form N e E e K e E e p , with K e a constant, we have I e E e c K e E e p . 4 Therefore we obtain for I IC 17 electron distribution of the r02 ( p 5) / 2 I E LK e kT F p E( p 1) / 2 , 2 3 2 4 c IC where r0 e 2 / me c 2 is the classical electron radius, T is the temperature and p 5 p 5 2 p 3 p 2 4 p 11 2 2 , F ( p) p 32 p 1 p 5 where is the Riemann function. For p 1.5, 2 and 2.5, F ( p ) 3.91, 5.25 and 7.57, respectively. Notes 1 The physics of shock waves is discussed in many books. Classical references are L.D. Landau and E.M. Lifshitz, Fluid mechanics, Oxford, England: Pergamon Press, 1987 and Ya. B. Zel′dovich and Yu. P. Raizer, Physics of shock waves and high-temperature hydrodynamic phenomena, New York: Academic Press, 1966-1967. The presentation in this notes mainly follow M. S. Longair, High energy astrophysics, Cambridge: Cambridge University Press, 1992. 2 The presentation of the astrophysics of the gamma-rays is based on G. E. Romero and K. S. Cheng, Fundamentals of gamma-ray astrophysics, in K. S. Cheng and G. E. Romero, editors, Kluwer Academic Publishers, Dordrecht, Boston, London, 2004. 18