= Solution
Integrate any conservation equation from part (a) through a vanishingly thin control volume moving with shock speed $s$. Its singular terms require $[F]=s[Q]$. The stationary shock has $s=0$, so all five fluxes are continuous. Substitution gives the <Rankine-Hugoniot conditions for a perfect gas>:
$$
\boxed{[\rho v]=0,\quad[\rho v^2+p]=0,\quad
[\rho vu_y]=[\rho vu_z]=0,\quad
[\rho v(u^2/2+w)]=0.}
$$
Let the common mass flux be $j=\rho_1v_1=\rho_2v_2>0$. Dividing the transverse conditions by $j$ proves $u_{y2}=u_{y1}$ and $u_{z2}=u_{z1}$. Dividing the energy condition by $j$ shows that $u^2/2+w$ is unchanged. Thus tangential kinetic energies cancel when solving for the normal shock compression. No global equality of the <entropy> parameter $p/\rho^\gamma$ across the shock has been assumed: physical dissipative shocks increase that parameter.
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