Stand still beside the shock
Choose the local axis to point from upstream to downstream. Let the incoming and outgoing gas speeds be positive along it. Write mass density in kilograms per cubic metre and pressure in pascals. A tiny planar patch is sufficient when its thickness is much smaller than the curvature radius.
Follow a fixed area of this stationary front. In a short time, upstream gas fills a cylinder whose length is speed times time. Multiplying its volume by density gives the mass crossing the area. A steady shock cannot accumulate an ever-growing pile of mass inside that narrow transition, so the incoming and outgoing mass fluxes agree.
Call their common value . Define compression . Mass conservation immediately gives . Compression therefore requires a slower downstream flow in this frame. This does not say that shocked ambient gas moves inward in the remnant-centre frame.
Keep pressure in the momentum balance
Moving gas transports momentum, and pressure also pushes on the boundary of a control volume. Their sum is the momentum flux. Take the upstream gas to be cold enough that its pressure is negligible compared with its ram pressure. This is the strong-shock assumption used here.
Substitute into the momentum balance. The downstream pressure is . The gas has slowed, but pressure now carries the missing part of the momentum flux.
Pressure alone has not fixed the compression. We still need to track energy. Take a nonradiative, nonmagnetic ideal gas with no energetic-particle feedback. Its internal energy per unit mass is pressure divided by density and by one less than the adiabatic index. Transporting this gas also requires pressure work; internal energy plus that work is the enthalpy.
The adiabatic index is the ratio of heat capacities at constant pressure and at constant volume. For a monatomic gas whose thermal energy occupies the three translational directions, this ratio is .
The energy equation selects the compression
Divide the energy flux by the common mass flux. What remains is kinetic energy plus enthalpy per unit mass. Neglect upstream enthalpy, insert the downstream pressure from momentum conservation and divide by the incoming kinetic energy.
The resulting quadratic has a root , which means no shock, and the compressive root shown below. A monatomic gas has , yielding . The outgoing speed is one quarter of the incoming speed, and the downstream pressure is three quarters of the incoming ram pressure.
These values are linked consequences of the same three balances. Choosing a fourfold density jump and an unrelated downstream speed would lose mass conservation. A radiative shock can compress much more; magnetic stresses or a dynamically important energetic-particle population also change this ideal-gas problem.
Reading: Chevalier (1982), Self-similar solutions for the interaction of stellar ejecta with an external medium, ApJ258,790–797 · Sandbox Physics independent supernova-remnant Python science reader
Calculate a conditional temperature
To translate pressure into temperature, write . Here is Boltzmann’s constant, is the proton mass and is the mean mass per thermal particle, counting both ions and electrons. For fully ionized hydrogen, one proton and one electron share each hydrogen atom’s mass, so .
For a worked example, use and an equilibrated fully ionized hydrogen gas. The jump gives , corresponding to . The factor of four in density is essential in this calculation.
A collisionless astrophysical shock need not give electrons and ions the same temperature immediately. Its mean pressure jump can be well defined while the electron temperature relevant to an X-ray spectrum differs from this single-temperature estimate. Predicting that spectrum also needs ionization and emission physics.
Reading: Sandbox Physics independent supernova-remnant Python science reader
Return to the expanding remnant
A forward shock moving outward at into stationary ambient gas has . Its downstream gas moves outward at in the remnant-centre frame. Subtracting the shock speed recovers the downstream shock-frame speed with the appropriate inward-pointing local axis.
The reverse shock meets ejecta that are already expanding. Their unshocked speed at the reverse radius is . The upstream relative speed is this ejecta speed minus the reverse-shock speed. Using the forward shock’s speed there would heat the wrong incoming flow. This difference leads naturally to the two-shell problem in the next chapter.