Wave drag is often defined as the component of drag that arises due to the presence of shockwaves. Shocks result in an entropy rise and a corresponding increase in drag. However, when using Supersonic Potential Flow Theory (Ackeret Theory), the flow is assumed to be fully isentropic (i.e. no shockwaves) and it can be shown (see Bertin, Houghton, Anderson for reference) that the wave drag is non-zero for such flows. So, what is the mechanism behind why the wave drag arises in such a flow without shocks?

More Detail (from Aerodynamics for Engineers by Bertin) enter image description here

  • $\begingroup$ Weak shock can be described by isentropic assumption fairly well. $\endgroup$
    – JZYL
    Commented Feb 5, 2020 at 22:56
  • $\begingroup$ Yes, but if wave drag is caused by the increase in entropy across a shock, then how does the isentropic assumption lead to a non-zero wave drag? $\endgroup$
    – Nick Hill
    Commented Feb 6, 2020 at 20:46
  • $\begingroup$ Maybe isentropic does not imply that there are no shockwaves, just that the increase of entropy across them is ignored. $\endgroup$
    – Orbit
    Commented Mar 25, 2020 at 10:54

1 Answer 1


The drag can be computed from Navier-Stokes momentum equations as:

$$D=-\oint_{S_\infty}p\hat{i}\cdot d\vec{S} - \oint_{S_\infty}\rho (\hat{i}\cdot\vec{u})(\vec{u}\cdot d\vec{S})$$

Usually, all boundaries other than the rear boundary (Trefftz plane) would have the same flow properties as the free-stream, and they fall out of the integral. We are left with:

$$D=\int_{S_T}{\rho u(V_\infty-u)dS}$$

As you correctly pointed out, the momentum defect of a potential flow should be exactly zero, which means that drag should also be zero.

However, for oblique shock in small-disturbance potential flow, the oblique shock (or rather, each constant potential wave characteristic) extends to infinity, carrying energy with it. If you want to calculate the far-field drag, you would need to do the integral around the entire far-field. This is different from energy dissipation in viscous boundary layer or subsonic normal shock.

Oblique shock

Picture ref. https://nikander.github.io/compflow/Anderson/Chapter4/print/

  • $\begingroup$ While I see that the constant potential wave characteristic carries energy to infinity I am still confused by why drag is produced. The Mach waves which are occuring are infinitesimally weak so they should have no dissipation at all. So, why is it that when I integrate (even if I include the entire far field) that I get a finite energy loss (or drag)? After all, the flow field is still consisting of isentropic waves right? $\endgroup$
    – Nick Hill
    Commented Mar 24, 2020 at 13:36
  • 2
    $\begingroup$ @NickHill In the potential flow, the oblique shock is just a wave. It carries energy away from it and the net effect is drag. Take induced drag for example, there is no entropy produced nor momentum defect, but the drag is still there. $\endgroup$
    – JZYL
    Commented Mar 24, 2020 at 13:49

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