Visual walkthrough — Prandtl-Meyer expansion waves — isentropic, supersonic turning

Before we start, three words we will use constantly, in plain language:

Everything below lives in the supersonic world $M>1$, because only there do the wafer-thin waves we are about to draw exist.

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Step 1 — One wall bend = one ripple

WHAT. Look at the figure. A supersonic stream (thick black arrow) runs along a flat wall. At one point the wall kinks away from the flow by a tiny angle we call $d\theta$ (Greek letter "theta", our name for turn angle; the little $d$ in front means "an infinitesimally small amount of"). A single, wafer-thin Mach wave (the red line) springs from that kink.

WHY. We refuse to bite off the whole turn at once. A big turn is hard; a microscopic turn is easy because across one thin wave almost nothing changes. We will handle the microscopic turn exactly, then add up millions of them (that is what integration will do in Step 7).

PICTURE. The red Mach wave sits at the Mach angle $\mu$ (Greek "mu") measured from the flow direction. From the Mach-wave picture, $\mu =\arcsin \left(\frac{1}{M}\right).$ Reading it term by term: $\arcsin$ asks "which angle has this sine?"; $\frac{1}{M}$ is the ratio $a/V$ (signal speed over flow speed). A faster flow (bigger $M$) gives a smaller $\mu$ — the ripple lies flatter. Hold that thought; it is why a fan opens up later.

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Step 2 — Freeze the velocity into a triangle

WHAT. Across the red wave the flow speed changes from $V$ to $V+dV$ and the direction swings by $d\theta$. We split each velocity into two pieces: one along the wave (call it $V_t$, the tangential part) and one across it (the normal part). The figure shows both the "before" (black) and "after" (red) velocity arrows resolved onto the wave.

WHY. A Mach wave is the weakest cousin of an oblique shock. Its one rule is: the tangential velocity component is unchanged — only the normal component is nudged. If one component is frozen, the geometry collapses to plain right-triangle trigonometry, which we can solve with no calculus at all.

PICTURE. In the figure, $V_t$ (the projection onto the red line) is identical before and after — the dashed guide lines land on the same length. All the change lives in the small normal wedge that produces the turn $d\theta$.

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Step 3 — Why a tiny wedge angle is $d\theta$

WHAT. Before we extract the turn, settle one thing cleanly: for a razor-thin wedge, the angle and its tangent are the same number. The figure overlays, for a small angle, three things that look identical: the straight tangent side, the curved arc, and the straight chord.

WHY. Measure the angle $d\theta$ in radians — the length of the arc on a unit circle. The tangent of that angle is the length of the tangent segment. As the wedge shrinks, arc, chord and tangent all collapse onto the same length, so $\tan (d\theta )\approx \sin (d\theta )\approx d\theta (\text{for tiny }d\theta ,\text{ in radians}).$

Term by term:

• $d\theta$ — the arc length on a unit circle, i.e. the angle itself in radians.
• $\tan (d\theta )$ — the tangent segment's length; for a big angle it bulges away from the arc, but for a tiny angle the gap vanishes.

PICTURE. Watch the red tangent line in the figure hug the black arc ever more tightly as the angle drops from $40^{\circ }$ to $5^{\circ }$. That collapse is the entire licence to write the wedge angle as $d\theta$ with no correction term — legitimate only because Step 1 chose an infinitesimal bend.

Recall Why may we replace

$\tan (d\theta )$ by $d\theta$? Because the bend is infinitesimal ::: for a tiny angle in radians, $\tan (d\theta )\approx \sin (d\theta )\approx d\theta$.

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Step 4 — Read $d\theta$ off the triangle

WHAT. Using "tangential component unchanged" plus the small-angle fact from Step 3, the before/after triangle gives a clean link between how much the speed changes and how much the direction changes: $d\theta =\sqrt{M^2-1}\frac{dV}{V}.$

Term by term:

• $d\theta$ — the tiny turn (what we want to accumulate).
• $\frac{dV}{V}$ — the fractional speed change: not "how many m/s" but "what percent." This is the natural currency because the trig ratio only cares about proportions.
• $\sqrt{M^2-1}$ — the conversion factor. It equals $\cot \mu =1/\tan \mu$, the "run over rise" of the red wave. Notice it needs $M>1$, otherwise the square root of a negative number appears and there is no real wave. Supersonic-only, baked right into the algebra.

WHY these tools. We used $\tan \mu =\frac{1}{\sqrt{M^2-1}}$ (from the same right triangle whose hypotenuse is $M$ and whose short side is $1$, because $\sin \mu =1/M$). $\tan$ is exactly "opposite over adjacent" on that triangle — the ratio that measures the wave's steepness, which is precisely what turns speed change into direction change.

PICTURE. The figure draws the $\mu$-triangle: hypotenuse of length $M$, opposite side $1$, adjacent side $\sqrt{M^2-1}$ (red). That red side is the factor sitting in the formula.

Recall Where does

$\sqrt{M^2-1}$ come from? From $\sin \mu =1/M$, the adjacent side is $\sqrt{M^2-1}$ ::: so $\cot \mu =\sqrt{M^2-1}$, exactly the factor multiplying $dV/V$.

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Step 5 — Swap "speed change" for "Mach change"

WHAT. Speed $V$ is awkward to tabulate; the Mach number $M$ is the physicist's ruler. Since $V=Ma$, take the natural logarithm (written $\ln$; throughout this page "log" always means $\ln$, the logarithm to base $e$) and differentiate: $\ln V=\ln M+\ln a⟹\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}.$

WHY. The natural log turns a product into a sum, and its derivative is the friendly $d(\ln x)=dx/x$ — a fractional change. So "the fractional change of a product = the sum of the fractional changes." That is the cleanest way to see that a change in $V$ has two sources: the flow can go faster ($dM$), and the local speed of sound can change ($da$) because the gas cools.

Term by term:

• $\frac{dM}{M}$ — fractional change of Mach number (the thing we want).
• $\frac{da}{a}$ — fractional change of the local speed of sound. This is not zero: as the gas expands it cools, and cooler gas carries sound more slowly.

PICTURE. The bar chart in the figure shows one $dV/V$ bar splitting into a red $dM/M$ piece and a black $da/a$ piece. We must not forget the $da/a$ piece — that is the classic slip.

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Step 6 — Kill $da/a$ using the isentropic energy law

WHAT — the cancellation, unpacked line by line. We now do three little pieces of algebra and glue them together. The figure shows the same three arrows feeding into one clean result.

Piece A — link $a$ to $T$. For a perfect gas $a=\sqrt{\gamma RT}$, so $a\propto \sqrt{T}$. Taking $\ln$ and differentiating (same trick as Step 5): $\frac{da}{a}=\frac{1}{2}\frac{dT}{T}.$ (Half, because $a$ depends on the square-root of $T$.)

Piece B — link $T$ to $M$. The isentropic energy relation, using the constant $T_0$, says $\frac{T_0}{T}=1+\frac{\gamma -1}{2}{M}^2$. Rearranged, $T=T_0/\left(1+\frac{\gamma -1}{2}{M}^2\right)$. Since $T_0$ is a constant, taking $\ln$ and differentiating kills it and leaves $\frac{dT}{T}=-\frac{(\gamma -1)MdM}{1+\frac{\gamma -1}{2}{M}^2}.$ (The minus sign is the physics: speeding up, i.e. $dM>0$, cools the gas, $dT<0$.)

Piece C — combine A and B. Substitute B into A ($da/a=\frac{1}{2}dT/T$), so the factor $(\gamma -1)$ becomes $\frac{\gamma -1}{2}$: $\frac{da}{a}=-\frac{\frac{\gamma -1}{2}MdM}{1+\frac{\gamma -1}{2}{M}^2}.$

The cancellation. Now feed this into Step 5's $\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}$. Write $\frac{dM}{M}=\frac{dM}{M}\cdot 1$ and combine over the common denominator $D=1+\frac{\gamma -1}{2}{M}^2$:

=\frac{dM}{M}\cdot\frac{D-\frac{\gamma-1}{2}M^2}{D}.$$ Look at the numerator: $D-\tfrac{\gamma-1}{2}M^2 = \left(1+\tfrac{\gamma-1}{2}M^2\right)-\tfrac{\gamma-1}{2}M^2 = 1$. The $\tfrac{\gamma-1}{2}M^2$ pieces **cancel exactly**, leaving just $1$: $$\boxed{\;\frac{dV}{V}=\frac{1}{1+\frac{\gamma-1}{2}M^2}\,\frac{dM}{M}.\;}$$ Term by term of the survivor: - $\gamma$ (Greek "gamma") — the gas's heat-capacity ratio; $1.4$ for air. - $R$ — cancels out of every fractional relation, so it never reaches the final $\nu$ formula. - $1+\frac{\gamma-1}{2}M^2$ — the "temperature-drop" denominator. As $M$ climbs it grows, so each extra bit of Mach number buys **less** extra velocity: the flow saturates toward a ceiling speed. **WHY.** We needed one equation tying $a$ (or $T$) to $M$. Energy conservation for an adiabatic, isentropic stream (constant $T_0$) provides exactly that, and it is the *only* extra physics the whole derivation uses. **PICTURE.** The figure walks A → B → C into the boxed cancellation, then plots $1+\frac{\gamma-1}{2}M^2$ rising with $M$; the shrinking red gap between the $dV/V$ and $dM/M$ curves shows velocity flattening out at high $M$. --- ## Step 7 — Add up every ripple: integrate **WHAT.** Put Steps 4 and 6 together to get the turn caused by one thin wave, entirely in $M$: $$d\theta=\frac{\sqrt{M^2-1}}{1+\frac{\gamma-1}{2}M^2}\,\frac{dM}{M}.$$ Now stack infinitely many of these micro-turns from the sonic point $M=1$ up to any $M$. "Stacking infinitely many infinitesimals" is exactly what an integral $\int$ does. Define the total as the **Prandtl–Meyer function** $\nu(M)$: $$\nu(M)=\int_{1}^{M}\frac{\sqrt{m^2-1}}{1+\frac{\gamma-1}{2}m^2}\,\frac{dm}{m}.$$ Carrying out the integral (a standard substitution) gives the boxed result: > [!formula] Prandtl–Meyer function > $$\nu(M)=\sqrt{\frac{\gamma+1}{\gamma-1}}\;\arctan\!\sqrt{\frac{\gamma-1}{\gamma+1}\,(M^2-1)}\;-\;\arctan\!\sqrt{M^2-1},\qquad \nu(1)=0.$$ > Here $\nu$ comes out in **radians** (it is an integral of $d\theta$, an angle). Multiply by > $180/\pi$ to read it in degrees. ![[deepdives/dd-physics-3.1.16-d2-s07.png]] Term by term: - $\nu(M)$ — the **running total turn**, in radians, needed to accelerate from sonic ($M=1$) to $M$. - The lower limit $1$ is why $\nu(1)=0$: at exactly sonic you have not turned at all yet. - Two $\arctan$ terms appear because the integrand naturally splits into two pieces, each of the form "$\dfrac{dx}{1+x^2}$" — the fingerprint of $\arctan$. Why that fingerprint? Because if you set $x=\tan\phi$, then $dx=(1+x^2)\,d\phi$, so $\dfrac{dx}{1+x^2}=d\phi$ and the integral is just the angle $\phi=\arctan x$ (the inset in the figure shows this substitution collapsing the area under $1/(1+x^2)$ into a plain angle). $\arctan$ shows up precisely because our micro-turns were built from the tangent of $\mu$ back in Step 4. **WHY.** $\nu$ is a *bookkeeping* quantity. Because it is measured *from sonic*, the physics of any real corner only cares about the **difference** in $\nu$, so a corner of angle $\theta$ just adds: > [!formula] The working rule > $$\nu(M_2)=\nu(M_1)+\theta.$$ > Then all properties follow from the constant-$T_0,p_0$ [[Isentropic flow relations]]: > $$\frac{p_2}{p_1}=\left(\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\right)^{\!\frac{\gamma}{\gamma-1}},\qquad > \frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}.$$ **PICTURE.** The figure is the $\nu$-vs-$M$ curve: it starts at $(1,0)$, rises steeply, then bends toward a horizontal ceiling (Step 9). The red bracket marks how a $\theta$ of $10^\circ$ slides you along the curve from $M_1$ to $M_2$. --- ## Step 8 — Why the ripples spread into a fan **WHAT.** From Step 1, $\mu=\arcsin(1/M)$. Through the expansion $M$ **rises**, so $\mu$ **falls**. The first wave (at $M_1$) is steep; the last wave (at $M_2$) is shallow. Between them lies a continuous spray of waves — the **fan**. ![[deepdives/dd-physics-3.1.16-d2-s08.png]] **WHY.** This is the "diverging characteristics" claim from the parent, drawn out. Because every wave is infinitesimally weak and they *spread apart* (rather than pile up as in a compression), no wave ever catches another. Nothing steepens into a discontinuity, so the entropy rise stays $\Delta s\to 0$ — the process is ==isentropic==. Contrast an [[Oblique shock waves|oblique shock]], where converging characteristics collide into one lossy jump. **PICTURE.** The figure fans the red Mach lines from steep (leading edge, angle $\mu_1$) to shallow (trailing edge, angle $\mu_2$), with the flow arrows bending smoothly across. --- ## Step 9 — The edge case: turning into a vacuum **WHAT.** Push the corner harder and harder. As $M\to\infty$, both square-root arguments blow up and each $\arctan\to\tfrac{\pi}{2}$ (that is $90^\circ$). The formula settles on a hard ceiling. In radians it is $$\nu_{\max}=\frac{\pi}{2}\!\left(\sqrt{\frac{\gamma+1}{\gamma-1}}-1\right).$$ Converting to degrees (multiply by $180/\pi$, so the leading $\pi/2$ becomes $90^\circ$): $$\nu_{\max}\;\xrightarrow{\gamma=1.4}\;90^\circ\,(\sqrt6-1)\approx130.45^\circ.$$ ![[deepdives/dd-physics-3.1.16-d2-s09.png]] **WHY.** You cannot turn a sonic stream by more than about $130.5^\circ$. Beyond that the gas would need $M=\infty$ — infinite speed, zero pressure, zero temperature: an actual vacuum. The flow simply separates from the wall; no more turning is possible. This is the degenerate limit, and it also explains why over-expanded nozzles have a hard ceiling (see [[Nozzle design and overexpansion]]). **PICTURE.** The figure shows the $\nu$-curve of Step 7 flattening to its horizontal asymptote (red dashed line) at $\nu_{\max}$, with the last Mach wave lying nearly flat against the flow. > [!mistake] "Just keep bending the wall to turn the flow forever." > **Why it feels right:** Each extra degree of wall adds a degree of $\theta$. > **Fix:** $\nu$ has a ceiling $\nu_{\max}$. Once $\nu(M_1)+\theta$ would exceed it, the flow reaches > the vacuum limit and detaches — no further isentropic turning exists. --- ## The one-picture summary The final figure stitches the whole chain together on one canvas. A supersonic stream (black arrow) hits a convex corner, throws off a **fan** of red Mach waves that go from steep (leading edge) to shallow (trailing edge), and the labelled dials underneath record what happens across the fan: $M\uparrow$, $\mu\downarrow$, $p\downarrow$, $T\downarrow$, while the turning score climbs by exactly the wall's bend, $\nu(M_2)=\nu(M_1)+\theta$. Everything you derived across the nine steps — bend, frozen triangle, small-angle wedge, the $\sqrt{M^2-1}$ factor, the swap to $M$, the cancellation, the integral, the fan, and the vacuum ceiling — compressed into a single readable picture. ![[deepdives/dd-physics-3.1.16-d2-s10.png]] > [!recall]- Feynman retelling of the whole walkthrough > We wanted to know how much a fast air stream speeds up when a wall bends away from it. We couldn't > solve the big bend directly, so we bent the wall a *tiny* bit and watched one thin ripple (Steps 1–2). > Because the bend is tiny, the little wedge's angle *is* just $d\theta$ — its tangent, arc and chord > are the same length (Step 3). On that ripple one part of the velocity stays put, so plain > triangle-trig told us how much the direction swings per bit of speed change (Step 4). Speed is > clumsy, so we swapped it for Mach number, remembering that the air also *cools*, which slows its own > sound (Steps 5–6); the cooling term and the Mach term partly cancel, leaving one tidy denominator. > Then we glued millions of tiny bends together with an integral and christened the total $\nu$ — a > "turning score" measured from the sound barrier, counted in radians (Step 7). The waves fan out > because faster air makes flatter ripples (Step 8), and the whole thing has a hard limit near > $130.5^\circ$, where the air would have to become a vacuum (Step 9). To use it: add your wall's > bend to your starting score, look up the new Mach number, and read off the new pressure and > temperature. > [!mnemonic] The chain in six beats > **Bend → Wedge → Triangle → Swap V for M → Integrate → Add $\theta$.** --- ## Recall Why does the factor $\sqrt{M^2-1}$ force the flow to be supersonic? ::: For $M<1$ it is the root of a negative number — no real Mach wave exists; only $M>1$ supports waves. In Step 5, why must we keep the $da/a$ term? ::: Because the gas cools as it expands, so the local speed of sound changes too; ignoring it double-counts velocity as pure Mach change. In Step 6, what cancels to leave the tidy denominator? ::: Writing $dM/M$ over the common denominator $1+\frac{\gamma-1}{2}M^2$, the $\frac{\gamma-1}{2}M^2$ terms cancel, leaving numerator $1$. What is the total (stagnation) temperature $T_0$? ::: The temperature the gas would reach if brought losslessly to rest; it stays constant through an isentropic expansion. What lets us write the tiny wedge angle as $d\theta$ exactly? ::: The small-angle approximation $\tan(d\theta)\approx d\theta$, valid because the bend is infinitesimal (angle in radians). Why does $\arctan$ appear when we integrate? ::: The integrand splits into $dx/(1+x^2)$ pieces; substituting $x=\tan\phi$ turns each into $d\phi$, whose integral is $\arctan x$. What is $\nu_{\max}$ for air and what does it physically mean? ::: $\approx130.45^\circ$; the largest turn from sonic, reached only as $M\to\infty$ into a vacuum.