In many questions and answers about high speed aerodynamics here it is said that as speed approaches the speed of sound, there is also a decrease in air density.

Can someone explain why?.

  • 3
    $\begingroup$ Can you add a link to at least one such question? $\endgroup$
    – Jim
    Aug 26, 2022 at 14:38

1 Answer 1


We start with the Laplace equation for the speed of sound: $$a = \sqrt{\frac{\gamma \cdot p}{\rho}}$$ and differentiate and square it: $$a^2 = \frac{\delta p}{\delta\rho}$$ Now we can write: $$\delta p = \delta\rho \cdot a^2$$ Next to be replaced is the pressure differential $\delta p$, using the equation for the conservation of momentum in its differentiated form: $$\delta p = -\rho \cdot v \cdot\delta v$$ $$\rightarrow\:-\frac{\delta v}{v} = \frac{\delta\rho}{\rho} \cdot \left(\frac{a}{v}\right)^2$$ which can be written as $$-\frac{\delta v}{v} \cdot Ma^2 = \frac{\delta\rho}{\rho}$$ Interpretation: At small Mach numbers, changes in speed cause negligible changes in density, but as Mach approaches unity, both are of similar magnitude. With $Ma>>1$, changes in density will become dominant. A speed increase is always coupled to a decrease in density.

  • $\begingroup$ wonderful derivation! -NN $\endgroup$ Aug 27, 2022 at 18:19

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