Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 38C / Solution

For a physical adiabatic ideal gas one has $\gamma >1$; the following algebra also holds for positive $\gamma \ne 1$. Compression work implies $p=\rho W^{\prime }(\rho )-W$, obtained by differentiating the total internal energy $VW(M/V)$ at fixed mass. With the stated pressure law this integrates toThe term proportional to mass is an arbitrary energy reference; set $C=0$. Mass flux is $\rho u$, momentum flux is $p+\rho u^2$, and dividing the energy flux by mass flux givesfor steady flow.

Work in the shock's rest frame. If the shock speed in the laboratory is $D$, upstream speed relative to it is $D$ and downstream speed is $D/r$, where $r={\rho }_1/{\rho }_0$. Conservation of momentum gives $p_1-p_0={\rho }_0{D}^2(1-1/r)$. Conservation of specific total enthalpy givesEliminating $D^2$ and using $p_1/p_0=1+\beta$ yieldsIts weak-shock expansion is $1+\beta /\gamma -(\gamma -1){\beta }^2/(2{\gamma }^2)+O({\beta }^3)$, exactly the expansion of $(1+\beta {)}^{1/\gamma }$ through second order. Thus the adiabatic pressure-density relation agrees to order ${\beta }^2$; entropy production first affects this comparison at higher order. The same isentropic constant is not exactly preserved across a finite shock. The wording allows $\gamma =1$, but the printed enthalpy formula is then undefined. The compression calculation instead gives $W=(p_0/{\rho }_0)\rho \log (\rho /{\rho }_0)+C\rho$, and the specific enthalpy is $(p_0/{\rho }_0)[1+\log (\rho /{\rho }_0)]+C$. Thus the requested formula containing $1/(\gamma -1)$ requires $\gamma \ne 1$; its physical ideal-gas application uses $\gamma >1$.