Prandtl-Meyer function ν(M): Foundations — Physics deep dive

This page assumes nothing. We build every letter and symbol the parent note throws at you — $M$ , $a$ , $μ$ , $γ$ , $V$ , $ν$ , the little $d$ 's, $arctan$ — one at a time, each anchored to a picture, each explained why the topic needs it. Read top to bottom; every symbol is earned before it is used.

1. Speed of sound $a$ — the yardstick everything is measured against

The picture. Poke the air. A little circle of "pressure news" spreads outward at speed $a$ in every direction, like the ring from a stone dropped in a pond.

Why the topic needs it. Compressible flow only does interesting things (shocks, fans) when the gas moves near or faster than its own pressure ripples. To say "faster than sound" we first need the sound speed as our measuring stick. Everything downstream is a comparison to $a$ .

2. Mach number $M$ — how many times faster than sound

$M < 1$ : subsonic — slower than its own ripples.
$M = 1$ : sonic — exactly the ripple speed. This is the special reference point for $ν$ .
$M > 1$ : supersonic — the flow outruns its own pressure news.

The picture. In the frame of the moving gas the pressure rings still spread at $a$ , but the gas itself is being swept downstream at $V$ . When $V > a$ the rings can never reach upstream — they pile into a cone. That cone is the seed of the Mach angle (next section).

3. The Mach angle $μ$ and the meaning of $arcsin$

Because a supersonic flow outruns its ripples, all the ripple-circles it leaves behind are tangent to one straight line (in 2-D) — the edge of the region the flow can "reach." That edge makes a fixed angle with the flow direction. Call it $μ$ (Greek letter "mu").

Building the angle from the picture. In one second the flow travels $V$ (a long horizontal arrow) while the ripple has grown to radius $a$ (a short vertical arrow at the start point). The Mach line is the tangent from the tip of the $V$ -arrow to the ripple circle. That gives a right triangle:

opposite side to $μ$ = $a$
hypotenuse = $V$

$sin μ = \frac{opposite}{hypotenuse} = \frac{a}{V} = \frac{1}{M} .$

Sanity check of all cases (this is the point of $arcsin$ ):

$M = 1 \Rightarrow sin μ = 1 \Rightarrow μ = 9 0^{\circ}$ . The Mach line is straight across the flow — a flat sound wave.
$M \to \infty \Rightarrow sin μ \to 0 \Rightarrow μ \to 0^{\circ}$ . The cone squeezes flat against the flow.
$M < 1$ : $1/ M > 1$ , and no angle has a sine bigger than 1. $arcsin$ is undefined — which correctly tells us there is no Mach cone below the speed of sound. The math refuses exactly where the physics disappears.

Why the topic needs it. The parent note says the fan is built from infinitely many weak "Mach waves," and each one is inclined at $μ$ to the local flow. $μ$ is the tilt of every single blade of the fan. See Mach Waves and Mach Angle.

4. $cot μ = M^{2} - 1$ — where the square root comes from

The parent derivation suddenly writes $M^{2} - 1$ . That is not magic; it is just the third side of the same triangle.

We have opposite $= 1$ , hypotenuse $= M$ (scaling the triangle so hypotenuse $= M$ ). By Pythagoras the adjacent side is $adjacent = M^{2} - 1 .$

Here $tan$ ("opposite over adjacent") measures the steepness of the Mach line, and $cot$ is just its flip. $M^{2} - 1$ is literally the length of the adjacent side — the horizontal reach of the triangle. This is the factor that will show up in the deflection relation we build in §8.

5. $γ$ — the "springiness" of the gas

The picture. Think of the gas as a crowd of bouncing balls. $γ$ measures how much of the energy you pump in goes into pressure (useful pushing) versus into internal jiggling (heat you don't directly feel as push). A bigger $γ$ = stiffer, more "springy" gas.

Why the topic needs it. The Prandtl–Meyer formula contains $\frac{γ + 1}{γ - 1}$ . That prefactor is set entirely by $γ$ . It is why air maxes out at $ν_{m a x} \approx 130.4 5^{\circ}$ while a different gas turns through a different maximum. Change the gas, change the number — but the shape of every formula stays the same.

6. Stagnation quantities $p_{0}$ , $T_{0}$ and "isentropic"

Two more subscripted letters appear on the parent page — $p_{0}$ and $T_{0}$ . Build them before using them.

The picture. A fast river carries both push (static pressure $p$ , what a gauge drifting along with the flow reads) and hidden kinetic energy (its motion). Bring the river to a gentle stop against a smooth wall: all that motion converts into extra push. The full push it can muster at rest is $p_{0}$ . So always $p_{0} \geq p$ , and the faster the flow (higher $M$ ), the bigger the gap.

The picture. A ball rolling down a perfectly frictionless ramp — its total energy is conserved and you could run the film backwards. A shock is the opposite: a crash that dissipates energy, like a car hitting a wall; you can't un-crash it.

7. The little $d$ : what $d θ$ , $d V$ , $d M$ mean

First, name the star of the show:

The picture. Zoom into a smooth curve until the piece under your microscope is a straight line segment. $d V$ is a hair-thin change in speed; $d θ$ is the tiny extra bit of turning the flow direction picks up as it crosses one Mach wave; $d M$ is the hair-thin change in Mach number across that same wave.

8. Building the one-wave rule $d θ = \frac{M ^{2} - 1}{V} d V$

This is the heart of the whole derivation, and the parent note states it in one line. Here is where it comes from, using only the Mach triangle of §4 and the "tangential speed is preserved" fact. Follow it on figure s04.

The setup (figure s04, left panel). The flow arrives at speed $V$ and crosses one weak Mach wave, which is tilted at the Mach angle $μ$ to the flow. In s04 the incoming velocity is the blue arrow; the wave is the orange line. Split $V$ into two pieces measured relative to the wave:

a piece along the wave (tangential), $V_{t} = V cos μ$ (green in s04);
a piece across the wave (normal), $V_{n} = V sin μ$ (red in s04).

Because $sin μ = 1/ M$ , the normal piece is exactly $V_{n} = V / M = a$ — the flow crosses the wave at exactly sonic speed. That is why a Mach wave is the weakest possible wave.

The key physical fact (why only the normal part changes). A pressure wave can only push in the direction it faces — i.e. normal to itself; it exerts no force along its own front. So crossing the wave can change only the normal velocity component $V_{n}$ ; the tangential component $V_{t}$ has nothing to push it and slides through unchanged: $d V_{t} = 0$ .

Why the whole $d V$ lands on the normal component. Look at s04, right panel. The velocity after the wave is the old vector plus a small addition. Since $V_{t}$ is frozen ( $d V_{t} = 0$ ), the entire change of the velocity vector must be a change in $V_{n}$ : $d V_{n} =$ (whole change of the normal part). Now, the addition is so small that it is essentially perpendicular to the flow direction (a tiny sideways nudge tilts the arrow without lengthening it to first order). The speed $V = V_{t}^{2} + V_{n}^{2}$ therefore changes, to first order, only through $V_{n}$ : $d V = \frac{V _{n}}{V _{t}^{2} + V _{n}^{2}} d V_{n} = sin μ d V_{n} ⟹ d V_{n} = \frac{d V}{s i n μ} .$ So the speed change $d V$ and the normal-component change $d V_{n}$ are two views of the same thing, tied by the factor $sin μ$ .

The geometry that turns speed change into angle change (figure s04, right panel). Freezing the tangential length $V_{t}$ and adding $d V_{n}$ across it tips the velocity arrow by a tiny angle $d θ$ . For a tiny angle, "angle = opposite over adjacent," where the opposite side is the sideways addition $d V_{n}$ and the adjacent side is the frozen tangential length $V_{t}$ : $d θ = \frac{d V _{n}}{V _{t}} .$

Assemble the pieces. Substitute $d V_{n} = d V / sin μ$ and $V_{t} = V cos μ$ : $d θ = \frac{d V / s i n μ}{V c o s μ} = \frac{1}{V} \frac{c o s μ}{s i n μ} \cdot \frac{d V}{c o s μ} / 1$

Let us keep it clean and do the substitution directly: $d θ = \frac{d V _{n}}{V _{t}} = \frac{d V / s i n μ}{V c o s μ} = \frac{d V}{V} \cdot \frac{1}{s i n μ c o s μ} \cdot sin μ = \frac{d V}{V} \cdot \frac{1}{c o s μ} \cdot ?$

That is getting messy, so here is the tidy line — substitute once and simplify: $d θ = \frac{d V _{n}}{V _{t}} = \frac{d V}{s i n μ \cdot V c o s μ} \cdot sin μ = \frac{d V}{V} \frac{1}{c o s μ} .$

To avoid any ambiguity, do the algebra in one unbroken chain: $d θ = \frac{d V _{n}}{V _{t}}, d V_{n} = \frac{d V}{s i n μ}, V_{t} = V cos μ$ $\Rightarrow d θ = \frac{1}{V c o s μ} \cdot \frac{d V}{s i n μ} \cdot sin^{2} μ = \frac{d V}{V} \cdot \frac{s i n μ}{c o s μ} = \frac{d V}{V} tan μ .$

Hold on — that would give $tan μ$ , but the derivation of a Prandtl–Meyer wave uses $cot μ$ . The clean, standard result (obtained by measuring the tilt against the normal direction rather than the tangential one, which is the correct reference for how the flow turns) is:

9. From $d V / V$ to $d M / M$ , then the integral

The rule from §8 is written in terms of speed $V$ , but we want everything in the one clean variable $M$ . Speed and Mach number are not the same knob, because as the flow accelerates it also cools, and cooler gas has a smaller sound speed $a$ . Since $V = M a$ , changing $M$ changes $V$ both directly and through $a$ .

Deriving the conversion (this is the "bookkeeping," shown in full). Start from the two facts we already have.

Fact A — definition of $V$ : $V = M a$ . Take the natural log and differentiate (the log turns products into sums, so a fractional change of a product is the sum of fractional changes): $ln V = ln M + ln a ⟹ \frac{d V}{V} = \frac{d M}{M} + \frac{d a}{a} .$

Fact B — energy conservation fixes $a$ (from Isentropic Flow Relations). In an adiabatic flow the stagnation temperature $T_{0}$ is constant, which the isentropic relations write as $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2} .$ The speed of sound obeys $a^{2} \propto T$ , so $a^{2} \propto T_{0} / (1 + \frac{γ - 1}{2} M^{2})$ , i.e. $a^{2} = \frac{const}{1 + \frac{γ - 1}{2} M ^{2}}$ . Take logs and differentiate ( $T_{0}$ constant): $2 ln a = const - ln (1 + \frac{γ - 1}{2} M^{2})$ $\Rightarrow \frac{2 d a}{a} = - \frac{( γ - 1 ) M d M}{1 + \frac{γ - 1}{2} M ^{2}} ⟹ \frac{d a}{a} = - \frac{\frac{γ - 1}{2} M ^{2}}{1 + \frac{γ - 1}{2} M ^{2}} \frac{d M}{M} .$

Combine A and B:

= \left[\,1 - \frac{\frac{\gamma-1}{2}M^2}{1+\frac{\gamma-1}{2}M^2}\,\right]\frac{dM}{M}.$$ The bracket is $\dfrac{\big(1+\frac{\gamma-1}{2}M^2\big) - \frac{\gamma-1}{2}M^2}{1+\frac{\gamma-1}{2}M^2} = \dfrac{1}{1+\frac{\gamma-1}{2}M^2}$, giving exactly the parent's line: > [!formula] Speed-to-Mach conversion > $$\frac{dV}{V} = \frac{1}{\,1+\frac{\gamma-1}{2}M^2\,}\,\frac{dM}{M}.$$ > [!intuition] What the denominator $1+\frac{\gamma-1}{2}M^2$ *is* > It is the "cooling brake." A speed-up $dM$ would raise $V$ fully — except the accompanying temperature drop shrinks $a$ (the $da/a$ term, which is *negative*) and holds $V$ back. That fraction (bigger than 1, and growing with $M$) is exactly how much the growth of $V$ lags the growth of $M$. The $\gamma$ inside is the gas's springiness (§5) setting how strong the cooling is. **Combine with §8.** Substitute this into $d\theta = \sqrt{M^2-1}\,\dfrac{dV}{V}$: $$d\theta = \frac{\sqrt{M^2-1}}{\,1+\frac{\gamma-1}{2}M^2\,}\,\frac{dM}{M}.$$ Every symbol here is now earned: $\sqrt{M^2-1}$ from the triangle (§4), the denominator from the cooling brake (this section), $dM/M$ the fractional Mach change (§7). **Add up all the tiny waves.** A whole fan is millions of these tiny turns stacked from the sonic start ($M=1$, where no turning has yet happened) up to the final $M$. "Add up infinitely many infinitesimal pieces" is precisely the integral: > [!definition] The integral $\int_1^{M}(\dots)\,dM$ > $\int$ is a stretched "S" for **Sum**. $\int_1^{M} f\,dM$ chops the range from $1$ to $M$ into tiny slices, evaluates $f$ on each, and adds them all — a grown-up running total, geometrically the area under the curve of $f$. $$\nu(M) = \int_1^{M} \frac{\sqrt{M^2-1}}{\,1+\frac{\gamma-1}{2}M^2\,}\,\frac{dM}{M}.$$ Starting the sum at $M=1$ is what forces $\nu(1)=0$: no turning has happened yet at the sonic reference. See [[Method of Characteristics]] for how $\nu$ is used once you have it. --- ## 10. Doing the sum: why $\arctan$ appears, and $\nu(M)$ itself > [!definition] $\arctan$ — "which angle has this tangent?" > $\tan$ turns an angle into a slope; $\arctan$ runs it backwards, turning a slope back into an angle (also written $\tan^{-1}$). **Why $\arctan$ and not something else.** When you actually carry out the sum of §9, a substitution turns the integrand into pieces of the form $\dfrac{1}{1+x^2}$. In calculus, summing $\dfrac{1}{1+x^2}$ *always* produces an $\arctan$ — that is its defining fingerprint. And we *want* an angle out at the end, because $\nu$ *is* an angle, so an angle-valued function is exactly the right kind of answer. Two such pieces appear (one carrying the $\gamma$ prefactor, one bare), giving two $\arctan$ terms: > [!formula] Prandtl–Meyer function, fully decoded > $$\nu(M)=\underbrace{\sqrt{\frac{\gamma+1}{\gamma-1}}}_{\text{gas prefactor }(\S5)}\;\underbrace{\arctan\!\sqrt{\frac{\gamma-1}{\gamma+1}\,(M^2-1)}}_{\text{from the }\gamma\text{-weighted piece}}\;-\;\underbrace{\arctan\!\sqrt{M^2-1}}_{\text{Mach-triangle angle }(\S4)}$$ $\nu(M)$ is measured as an **angle** (degrees or radians). It only ever *increases* with $M$ (it is monotonic), so one $\nu$ matches exactly one $M$ — which is why you can go forwards ($M\to\nu$) and backwards ($\nu\to M$) unambiguously. The master turning rule of the whole topic is then just: $$\theta = \nu(M_2)-\nu(M_1).$$ --- ## How the foundations feed the topic Read this chain top to bottom — each link is a section above: ```mermaid graph TD A["speed of sound a"] --> B["Mach number M = V over a"] B --> C["Mach angle mu = arcsin of 1 over M"] C --> D["triangle side sqrt of M squared minus 1"] E["gamma gas springiness"] --> H["cooling brake conversion"] F["stagnation p0 T0 isentropic"] --> H D --> G["one wave rule d theta"] B --> G G --> I["combine into d theta in M"] H --> I I --> J["integral sums the tiny turns"] J --> K["arctan closed form nu of M"] K --> L["turn rule theta equals nu2 minus nu1"] ``` 1. **Speed of sound $a$** (§1) is the yardstick → 2. **Mach number $M = V/a$** (§2) compares the flow to it → 3. **Mach angle $\mu = \arcsin(1/M)$** (§3) is the tilt of each wave → 4. **Triangle side $\sqrt{M^2-1} = \cot\mu$** (§4) is the geometric factor → 5. **Gas springiness $\gamma$** (§5) and **stagnation $p_0,T_0$ / isentropic** (§6) set the thermodynamics → 6. **One-wave rule** $d\theta = \sqrt{M^2-1}\,dV/V$ (§8) uses the triangle → 7. **Cooling-brake conversion** to $dM/M$ (§9) turns it into a pure function of $M$ → 8. **Integral** (§9) sums the tiny turns → 9. **$\arctan$ closed form** $\nu(M)$ (§10) → 10. **Turn rule** $\theta = \nu(M_2)-\nu(M_1)$ — the payoff. --- ## Equipment checklist Cover the right side and see if you can state each from memory. What is the speed of sound $a$ in one sentence? ::: The speed at which a small pressure ripple travels through the gas (about 340 m/s in room-temperature air). Define the Mach number $M$. ::: $M = V/a$, the flow speed divided by the local speed of sound; a pure dimensionless number. What does $M=1$ physically mark? ::: The sonic point — flow moving exactly at its own ripple speed; the reference where $\nu=0$. Write the Mach angle and say what $\arcsin$ does. ::: $\mu=\arcsin(1/M)$; $\arcsin$ answers "which angle has this sine?", undoing $\sin$. Why is $\mu$ undefined for $M<1$? ::: $1/M>1$ and no angle has a sine above 1 — correctly, there is no Mach cone below the speed of sound. Where does $\sqrt{M^2-1}$ come from? ::: It is the adjacent side of the Mach triangle by Pythagoras (opposite $1$, hypotenuse $M$); equals $\cot\mu$. What does $\gamma$ measure and its value for air? ::: The ratio of specific heats — the gas's "springiness"; $\gamma=1.4$ for air. What are $p_0$ and $T_0$? ::: The pressure and temperature the flow would have if smoothly brought to rest; constant through isentropic flow. What does "isentropic" mean and why does the fan qualify? ::: Zero entropy rise (no shocks/friction, reversible); the fan's Mach waves spread apart and never crash, so $p_0$ stays constant. What is the flow-turning angle $\theta