Shocks Without Collisions: How the Solar Wind Thermalises in 300 Kilometres — Epoche C2
The problem A shock wave is a thin surface across which a supersonic flow is abruptly slowed, compressed and heated. In air, the heating is unmysterious: molecules collide, the ordered kinetic energy of the flow is redistributed among random motions, and the transition is a few mean free paths thick. Because this is the only mechanism most of us are taught, it is natural to conclude that a plasma in which particles almost never collide cannot support a shock at all. Yet Earth's bow shock stands permanently in the solar wind, and it thermalises the flow across a layer a few hundred kilometres thick. This essay sets out the problem quantitatively, then the mechanism that resolves it. Units are Gaussian CGS throughout, as in the space-plasma literature cited: magnetic fields in gauss, lengths in centimetres, energies in erg and eV. Take representative solar-wind parameters at 1 AU and hold them for the whole argument: proton number density $n_p = 5$ cm$^{-3}$, proton temperature $T_p = 10^5$ K, magnetic field strength $B = 5\times10^{-5}$ G (that is 5 nT), bulk speed $u_1 = 4\times10^{7}$ cm s$^{-1}$, and Coulomb logarithm $\ln\Lambda = 25$. Sizing the problem The Coulomb collision frequency for protons follows from the standard Rutherford-scattering result, in which small-angle deflections accumulate: $\nu_{pp} = 4.8\times10^{-8}\,n_p \ln\Lambda\, T_p^{-3/2}$ s$^{-1}$, with $n_p$ in cm$^{-3}$ and $T_p$ in eV. Here $T_p = 10^5\ \mathrm{K} = 8.62$ eV, so $T_p^{3/2} = 25.3$, and $\nu_{pp} = 4.8\times10^{-8} \times 5 \times 25 / 25.3 = 2.4\times10^{-7}$ s$^{-1}$. The collision time is therefore $\tau_{pp} = 4.2\times10^{6}$ s, or about seven weeks. The proton thermal speed is $v_{th,p} = \sqrt{2k_BT_p/m_p} = 4.1\times10^{6}$ cm s$^{-1}$, giving a mean free path $$\lambda_{mfp} = v_{th,p}\,\tau_{pp} = 4.1\times10^{6} \times 4.2\times10^{6} = 1.7\times10^{13}\ \mathrm{cm} = 1.15\ \mathrm{AU},$$ since 1 AU $= 1.496\times10^{13}$ cm. A proton in the solar wind travels roughly the Earth–Sun distance between collisions. Now the thickness of the shock. The natural scale is the ion inertial length $d_i = c/\omega_{pi}$, where $\omega_{pi} = \sqrt{4\pi n_p e^2/m_p}$ is the ion plasma frequency — the frequency at which displaced ions oscillate against the electrons. With $e = 4.80\times10^{-10}$ esu and $m_p = 1.67\times10^{-24}$ g this gives $\omega_{pi} = 2.9\times10^{3}$ s$^{-1}$ and $d_i = 3\times10^{10}/2.9\times10^{3} = 1.0\times10^{7}$ cm, about 100 km. Spacecraft crossings put the magnetic ramp at a few $d_i$; take $3d_i \approx 3.1\times10^{7}$ cm. The ratio of the two lengths is $1.7\times10^{13}/3.1\times10^{7} = 5.6\times10^{5}$, which is between five and six orders of magnitude. Collisions are not merely inefficient here. They are irrelevant. What survives the loss of collisions, and what does not It is worth separating two things that are usually taught together. The Rankine–Hugoniot conditions — conservation of mass, momentum and energy flux across the layer — do not mention collisions at all. They therefore still apply, and they still fix the jump. For an adiabatic index $\gamma = 5/3$ the compression ratio is $r = (\gamma+1)M^2/[(\gamma-1)M^2 + 2]$ with $M$ the magnetosonic Mach number; at these parameters $M \approx 6.5$ and $r \approx 3.7$, close to the strong-shock limit $(\gamma+1)/(\gamma-1) = 4$., and the Alfvén Mach number is $M_A = u_1/v_A$ with $v_A = B/\sqrt{4\pi n_p m_p} = 4.9\times10^{6}$ cm s$^{-1}$. This gives $M_A = 4\times10^{7}/4.9\times10^{6} = 8.2$, and hence $r = 3.83$, close to the strong-shock limit $(\gamma+1)/(\gamma-1) = 4$. The downstream temperature follows from the same conditions: $k_BT_2 = \tfrac{3}{16}m_pu_1^2 = 0.1875 \times 1.67\times10^{-24} \times 1.6\times10^{15} = 5.0\times10^{-10}$ erg, that is 313 eV, or $3.6\times10^{6}$ K — a factor of 36 above the upstream $10^5$ K. What the conservation laws do not supply is a mechanism. They say the entropy must rise; they are silent about what raises it. That is the gap the microphysics must fill. The mechanism: reflect a few ions, let them make the waves Edmiston and Kennel showed in 1984 that there is a first critical Mach number, between 1 and 2.76 depending on the field angle and plasma beta, above which no amount of resistive or viscous dissipation can satisfy the jump conditions: the downstream flow simply cannot be made slow enough. With $M_A = 8.2$ the bow shock is far above this, and something other than wave damping must remove the excess flow energy. The shock's answer is to reject part of the incoming population. A fraction of incoming ions — roughly 20% at this Mach number for a quasi-perpendicular geometry — is turned back by the combination of the magnetic ramp and the cross-shock electrostatic potential. The scale of that potential is a fraction of the incoming ion energy: with $\tfrac{1}{2}m_pu_1^2 = 1.34\times10^{-9}$ erg $= 835$ eV, a typical potential drop of 10–50% corresponds to 80–420 V, which is what spacecraft measure. The reflected ions do not simply leave. They gyrate in the upstream field, and their turning distance is $0.68\,u_1/\Omega_{ci}$ with $\Omega_{ci} = eB/m_pc = 0.48$ rad s$^{-1}$, giving $0.68 \times 4\times10^{7}/0.48 = 5.7\times10^{7}$ cm, about 570 km. This is the shock foot, and it is directly observed. The step that does the real work is what happens next. Two ion populations now stream through each other: the incoming beam and the reflected beam. Counter-streaming beams are unstable — this is the free energy that Sagdeev identified in 1966 — and the resulting electromagnetic waves grow on the ion gyro-timescale, of order $\Omega_{ci}^{-1} = 2$ s, rather than the collisional timescale of $4.2\times10^{6}$ s. That is a difference of $4.2\times10^{6}/2 \approx 2\times10^{6}$, six orders of magnitude, and it is exactly what is needed to close the gap computed above. The waves then scatter the particles' pitch angles, which randomises the directed motion. Dissipation in