Shock compression ratio

The derivation of the shock compression factor.

$$\rho v = const$$

$$\rho v^2 + P = const$$

$$\rho v(\frac{1}{2}v^2 + \frac{\gamma}{\gamma-1}\frac{P}{\rho}) = const$$

Let 1(2) to indicates the shock upstream (downstream) quantity.

The compression factor is defined by 

$$r = \frac{\rho_2}{\rho_1} = \frac{v_1}{v_2} .$$

$$P_2 = \rho_1 v_1^2 + P_1 - \rho_2 v_2^2$$

$$\frac{P_2}{\rho_2} = \frac{P_1}{\rho_2} + \frac{\rho_1 v_1^2}{\rho_2} - v_2^2 = \frac{P_1}{\rho_1 r} + \frac{v_1^2}{r} - (\frac{v_1}{r})^2$$

$$\frac{1}{2}v_1^2+\frac{\gamma}{\gamma-1}\frac{P_1}{\rho_1} = \frac{1}{2}v_2^2+ \frac{\gamma}{\gamma-1}(  \frac{P_1}{\rho_1 r} + \frac{v_1^2}{r} - (\frac{v_1}{r})^2)$$

multiply by $\frac{2}{v_1^2}$,

$$r^2(1+\frac{1}{\gamma-1}\frac{2}{M^2}) -\frac{2}{\gamma-1}(\frac{1} {M_1^2}+\gamma)r+\frac{2\gamma}{\gamma-1}-1 \,*$$

It is easy to verify that $r_1=1$ is one of the solution. Thus, we can write the equation as 

$$(r-1)(r-r_2) = r^2-(r_2-1)r+r_2$$

Comparing with equation *, we can get

$$r_2 = (\frac{2\gamma}{\gamma-1}-1)/(1+\frac{2}{(\gamma-1)M_1^2}) = \frac{(\gamma +1)M_1^2}{(\gamma-1)M_1^2+2}$$

cosmic rays modified shock

Three regions:

(0) far upstream, the presence of accelerated particles can be neglected

(1) in the shock precursor just upstream of the subshock

(2) behind the shock

$$\rho_0 v_0 = \rho_1 v_1 = \rho_2 v_2$$

$v_0=V_{s}$ is the velocity of the gas with respect to the shock. $V_s$ is the shock velocity.

Momentum flux conservation

$$P_0 + \rho_0 V_s^2 = P_1 + \rho_1 v_1^2 = P_2 +\rho_2 v_2^2$$

$P=P_{th}+P_{cr}$ is the pressure.

Mach number for the shock structure far upstream

$$M_0^2 \equiv \frac{1}{\gamma_g} \frac{\rho_0 V_s^2}{P_0}$$

Compressive ratio across the different regions 0, 1, and 2.

$$\xi_1 = \frac{\rho_1}{\rho_0}, \qquad \xi_2 = \frac{\rho_2}{\rho_1}, \qquad \xi_{12}=\xi_1 \xi_2 = \frac{\rho_2}{\rho_0}$$

Assuming the thermal pressure in the precursor in only due to adiabatic heating in the precursor, $P_{1,th}=P_0\xi_1^{\gamma_g}$ and across the sub-shock the pressure associated with the accelerated particles does not change.

The cosmic ray terms cancel each other when considering the momentum flux from region 1 to 2, $P_{1,cr}=P_{2,cr}$.

$$M_1^2 = \frac{\rho_1 v_1^2}{\gamma_g P_1} = \frac{\rho_0 V_s^2/\xi_1}{\gamma_g P_0 \xi_1^{\gamma_g}} = M_0^2\xi^{-(\gamma_g+1)}$$

$$\frac{P_2}{\rho_0 V_s^2} = \frac{P_{2,cr} + P_{2,th}}{\rho_0 V_s^2} = \frac{1}{\gamma_g M_0^2} + 1 - \frac{1}{\xi_{12}}$$

$$\frac{P_{2,th}}{\rho_1 v_1^2} = \frac{\xi_1^{\gamma_g+1}}{\gamma_g M_0^2} + 1 - \frac{1}{\xi_2}$$

The compression ratio of the sub-shock is given by the standard shock relation:

$$\xi_2 = \frac{(\gamma_g+1)M_1^2}{(\gamma_g-1)M_1^2 +2}$$

$$\xi_{12} = M_0^{\frac{2}{\gamma_g+1}} [\frac{\xi_2^{\gamma_g}(\gamma_g+1)-\xi_2^{\gamma_g+1}(\gamma_g-1) }{2}]^{\frac{1}{\gamma_g+1}}$$

$$\frac{P_{th1}}{P_{w1}} = \frac{p_{th1}}{p_{th0}} \frac{P_{th0}}{P_{w1}}$$

$$\frac{P_{th0}}{P_{w1}} = \frac{P_{th0}}{\rho_0 u_0^2} / \frac{P_{w1}}{\rho_0 u_0^2}$$

I will update later...  

Burlaga, L. F., Ness, N. F., Gurnett, D. A., & Kurth, W. S. 2013, ApJL, 778, L3

*** IMPORTANT  WAVE PROPAGATION AT OBLIQUE SHOCKS: HOW DID TYCHO GET ITS STRIPES?

Vainio & Schlickeiser (1999, hereafter VS99)

Reference:

THE RELATION BETWEEN POST-SHOCK TEMPERATURE, COSMIC-RAY PRESSURE,AND COSMIC-RAY ESCAPE FOR NON-RELATIVISTIC SHOCKS