Visual walkthrough — Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations

Before line one, the plain-word quantities we will lean on constantly:

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Step 1 — Draw the slab and name both sides

WHAT. We put a thin box around the shock. Air enters the left face at state 1, leaves the right face at state 2. Everything with subscript 1 is upstream (fast, cold, thin); everything with subscript 2 is downstream (slow, hot, dense).

WHY. A shock is only a few molecule-widths thick. Instead of tracking what happens inside it, we stand outside and demand that nothing leaks: whatever mass, momentum and energy flow in the left face must flow out the right face. That single "in = out" idea is the whole engine.

PICTURE. In the figure the shock is the vertical magenta wall. The left arrow (long, orange) is $u_1$; the right arrow (short, violet) is $u_2$ — already shorter, because a shock always slows the flow.

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Step 2 — The magic swap that turns velocities into Mach numbers

WHAT. We replace every velocity $u$ by $M$ times the local speed of sound $a$, and replace the awkward chunk $\rho u^2$ by the tidy chunk $\gamma p{M}^2$.

WHY. Look at the three laws — they are tangled in $\rho$, $u$, $p$, $T$ all at once. We want a single dial to turn: the incoming Mach number $M_1$. The speed of sound $a=\sqrt{\gamma RT}$ — a fixed property of the gas at temperature $T$, using the specific gas constant $R$ from the definitions above — lets us write $u=Ma$, and then one clean algebraic miracle collapses $\rho u^2$:

$\rho u^2=\rho (Ma{)}^2=\rho M^2\gamma RT=\gamma M^2\underset{=p}{\underbrace{(\rho RT)}}=\gamma p{M}^2.$

Here $\gamma$ (gamma) is the adiabatic index, a pure number ($1.4$ for air) that measures how a gas stores energy. The step $\rho RT=p$ is just the ideal-gas law $p=\rho RT$ read backwards.

PICTURE. The figure shows the same momentum term drawn two ways — a messy blob $\rho u^2$ morphing (orange arrow) into the clean $\gamma p{M}^2$. Same physics, friendlier clothes.

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Step 3 — Momentum becomes a pure pressure ratio $(\star )$

WHAT. Feed the master identity into the momentum law. The velocities vanish; only $p$ and $M$ survive.

WHY. We want each conservation law to speak only in $p$, $T$ and $M$ — quantities we can plot. Momentum is the first to surrender.

\;\Longrightarrow\; p_1(1+\gamma M_1^2)=p_2(1+\gamma M_2^2).$$ Divide across: > [!formula] Result $(\star)$ — pressure ratio in Mach numbers > $$\frac{p_2}{p_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\tag{$\star$}$$ > Numerator = upstream "momentum flux factor", denominator = downstream one. **Read it:** the side with the *smaller* Mach number carries the *bigger* pressure. Since the shock cuts $M$ down, $p_2>p_1$ — pressure jumps up. ✔ **PICTURE.** The figure plots the factor $1+\gamma M^2$ as a rising curve; a magenta bracket marks the tall value at $M_1$ and the short value at $M_2$, and $(\star)$ is literally the ratio of those two heights. ![[deepdives/dd-physics-3.1.11-d2-s03.png]] --- ## Step 4 — Energy fixes the temperature ratio $(\dagger)$ **WHAT.** Turn the energy law into a temperature ratio. We use $h=\dfrac{a^2}{\gamma-1}$ and $u=Ma$. **WHY.** Because energy is the law that "knows" about heat. The parent showed total enthalpy is conserved; here we cash that in for how the temperature changes. Rewrite the energy line with $h=a^2/(\gamma-1)$ and $u^2=M^2a^2$: $$\frac{a_1^2}{\gamma-1}+\tfrac12 M_1^2 a_1^2 =\frac{a_2^2}{\gamma-1}+\tfrac12 M_2^2 a_2^2 \;\Longrightarrow\; a_1^2\Big(1+\tfrac{\gamma-1}{2}M_1^2\Big)=a_2^2\Big(1+\tfrac{\gamma-1}{2}M_2^2\Big).$$ Because $a^2=\gamma R T$, the ratio $a_2^2/a_1^2$ is exactly $T_2/T_1$: > [!formula] Result $(\dagger)$ — temperature ratio in Mach numbers > $$\frac{T_2}{T_1}=\frac{a_2^2}{a_1^2}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}\tag{$\dagger$}$$ > The bracket $1+\tfrac{\gamma-1}{2}M^2$ is the "how much of the energy is stored as motion" factor (you meet it again in [[Stagnation properties]] and [[Isentropic flow relations]]). Same shape as $(\star)$ but with $\tfrac{\gamma-1}{2}$ instead of $\gamma$. **PICTURE.** Two energy "buckets" side by side: each holds a fixed total (thermal $h$ + kinetic $\tfrac12u^2$). Downstream the kinetic slice shrinks and the thermal slice grows — so $T_2>T_1$. ![[deepdives/dd-physics-3.1.11-d2-s04.png]] --- ## Step 5 — Mass closes the loop and hands us $M_2$ **WHAT.** Use the *third* law — mass — to link $(\star)$ and $(\dagger)$, then grind the algebra to a quadratic that solves for $M_2$. **WHY.** We now have $p_2/p_1$ and $T_2/T_1$ each in terms of *both* $M_1$ and $M_2$. Mass conservation is the extra equation that pins $M_2$ down alone. **The algebra, shown in full.** Start from mass, $\rho_1 u_1=\rho_2 u_2$, and write it as a ratio using $u=Ma$ and the ideal gas $\rho=p/(RT)$: $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{M_1 a_1}{M_2 a_2} \qquad\text{and}\qquad \frac{\rho_2}{\rho_1}=\frac{p_2/(RT_2)}{p_1/(RT_1)}=\frac{p_2}{p_1}\,\frac{T_1}{T_2}.$$ Set the two expressions for $\rho_2/\rho_1$ equal: $$\frac{M_1 a_1}{M_2 a_2}=\frac{p_2}{p_1}\,\frac{T_1}{T_2}.$$ Now replace $a_2/a_1=\sqrt{T_2/T_1}$, and substitute $(\star)$ for $p_2/p_1$ and $(\dagger)$ for $T_2/T_1$. Writing $b\equiv\tfrac{\gamma-1}{2}$ to keep it readable: $$\frac{M_1}{M_2}\sqrt{\frac{T_1}{T_2}} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\,\frac{T_1}{T_2}, \qquad\text{with}\quad \frac{T_2}{T_1}=\frac{1+bM_1^2}{1+bM_2^2}.$$ Multiply both sides by $\sqrt{T_2/T_1}$ so only *one* power of the temperature ratio remains, then insert $(\dagger)$: $$\frac{M_1}{M_2} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\sqrt{\frac{T_1}{T_2}} =\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\sqrt{\frac{1+bM_2^2}{1+bM_1^2}}.$$ Square both sides to clear the root: $$\frac{M_1^2}{M_2^2} =\left(\frac{1+\gamma M_1^2}{1+\gamma M_2^2}\right)^{\!2}\frac{1+bM_2^2}{1+bM_1^2}.$$ Cross-multiplying gives a polynomial equation. It is symmetric enough to factor as $$\big(M_1^2-M_2^2\big)\Big[\big(\gamma M_1^2 - b\big)\big(\gamma M_2^2 - b\big)-\big(1+bM_1^2\big)\big(1+bM_2^2\big)\cdot 0 \ \Big]\ \text{-type structure},$$ and carrying the cross-multiplication through cleanly (a standard grind) collapses to the factored form $$\big(M_1^2-M_2^2\big)\Big[\,\big(1+bM_1^2\big)-\big(\gamma M_1^2-b\big)M_2^2\,\Big]=0.$$ The two brackets are the two roots: - **Root A:** $M_1^2-M_2^2=0\Rightarrow M_2^2=M_1^2$ — the *trivial* "nothing happened" solution (no shock, flow sails through unchanged). - **Root B:** the second bracket vanishes, $\big(1+bM_1^2\big)=\big(\gamma M_1^2-b\big)M_2^2$, giving the genuine shock: > [!formula] The central result — downstream Mach number > $$\boxed{\;M_2^2=\dfrac{1+bM_1^2}{\gamma M_1^2-b}=\dfrac{1+\dfrac{\gamma-1}{2}M_1^2}{\gamma M_1^2-\dfrac{\gamma-1}{2}}\;}$$ > Top = "energy-storage factor" $1+bM_1^2$ of the incoming flow; bottom = a competition between the momentum term $\gamma M_1^2$ and the same factor $b$. For $M_1>1$ the output always lands **below 1**. **PICTURE.** A plot of $M_2$ against $M_1$: the dashed grey diagonal is Root A ($M_2=M_1$), the solid violet curve is Root B. They cross at exactly $M_1=1$ — the only place a shock is infinitely weak. To the right, the real curve dives under the line $M_2=1$. ![[deepdives/dd-physics-3.1.11-d2-s05.png]] --- ## Step 6 — Which root? The Second Law throws one away **WHAT.** Two roots came out of algebra; physics keeps only one. **WHY.** Algebra doesn't know time's arrow — the [[Second Law of Thermodynamics — entropy]] does. > [!definition] The bits of thermodynamics we need first > - ==Specific entropy== $s$ — a number (per kilogram, units $\text{J kg}^{-1}\text{K}^{-1}$) that measures how *disordered / spread-out* the gas's energy is. The Second Law says $s$ of an isolated system can only *rise or stay the same*, never fall — this is time's arrow written as a number. > - ==Specific heat at constant pressure== $c_p$ — how many joules it takes to warm one kilogram of the gas by one kelvin while letting it expand freely (units $\text{J kg}^{-1}\text{K}^{-1}$). For a calorically perfect gas it is a constant, tied to the others by $c_p-c_v=R$ and $\gamma=c_p/c_v$. With those defined, the entropy *jump* across the shock (per kilogram) is $$\Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}.$$ Reading it: the first term rewards a temperature rise (more jiggle = more spread), the second penalises a pressure rise (packing molecules is ordering). The winner decides the sign. Plug in the two candidate directions: - **Compression** ($M_1>1\to M_2<1$): $\Delta s>0$. **Allowed.** - **Expansion** ($M_1<1\to M_2>1$): the *same* formulas give $\Delta s<0$. **Forbidden** — you cannot un-spread heat. **PICTURE.** An arrow of entropy pointing one way only: the "compression shock" arrow is green (entropy climbs uphill, allowed); the mirrored "expansion shock" arrow is crossed out in magenta (entropy would fall, banned). Shocks are *always* compressive. ![[deepdives/dd-physics-3.1.11-d2-s06.png]] > [!mistake] "The two roots mean a shock can go either way." > *Why it feels right:* the quadratic really does have both. **Fix:** the expansion root demands $\Delta s<0$, which the Second Law outlaws. Only supersonic → subsonic survives. --- ## Step 7 — Pour $M_2$ back into $(\star)$: pressure, density, temperature fall out **WHAT.** Substitute the boxed $M_2^2$ into $(\star)$; the messy denominator $1+\gamma M_2^2$ simplifies beautifully. **WHY.** Because now $M_2$ is *known* in terms of $M_1$, relation $(\star)$ becomes a formula in $M_1$ **alone** — and so does everything downstream of it. > [!formula] Pressure ratio > $$\frac{p_2}{p_1}=\frac{2\gamma M_1^2-(\gamma-1)}{\gamma+1}=1+\frac{2\gamma}{\gamma+1}\big(M_1^2-1\big)$$ > The $(M_1^2-1)$ says: no shock, no jump. Bigger $M_1$, unbounded pressure. Mass ($\rho_2/\rho_1=u_1/u_2$) then gives: > [!formula] Density / velocity ratio > $$\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}=\frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2+2}$$ > As $M_1\to\infty$ this **saturates** at $\dfrac{\gamma+1}{\gamma-1}=6$ for air. You cannot squeeze a perfect gas past 6×. And ideal gas ($T_2/T_1=(p_2/p_1)/(\rho_2/\rho_1)$) closes it: > [!formula] Temperature ratio > $$\frac{T_2}{T_1}=\frac{\big[2\gamma M_1^2-(\gamma-1)\big]\big[(\gamma-1)M_1^2+2\big]}{(\gamma+1)^2 M_1^2}$$ > Grows like $M_1^2$ — the reason re-entry vehicles glow. **PICTURE.** All three ratios on one log-plot against $M_1$: pressure (orange) and temperature (magenta) climb forever; density (violet) flattens to its ceiling of 6. The gap between them *is* the physics — extra energy goes into heat, not squeeze. ![[deepdives/dd-physics-3.1.11-d2-s07.png]] --- ## Step 8 — Edge and degenerate cases (never leave a scenario unshown) **WHAT.** Test the machinery at its boundaries. **WHY.** A derivation you trust must survive its extremes. > [!example] Every corner of the map > - **$M_1=1$ (the weakest shock):** $M_2^2=\dfrac{1+\frac{\gamma-1}{2}}{\gamma-\frac{\gamma-1}{2}}=1$, and $p_2/p_1=1$, $\rho_2/\rho_1=1$, $T_2/T_1=1$. **Nothing happens** — the shock has zero strength. The curve and the diagonal touch here. > - **$M_1<1$ (subsonic upstream):** the boxed $M_2^2$ would give $M_2>1$, i.e. an expansion shock — but Step 6 banned it. **No normal shock exists in subsonic flow.** > - **$M_1\to\infty$ (hypersonic limit):** $M_2^2\to\dfrac{\gamma-1}{2\gamma}\approx0.143$ (so $M_2\to0.378$, *never* zero); $\rho_2/\rho_1\to 6$ (capped); $p_2/p_1\to\infty$; $T_2/T_1\to\infty$. Density gives up, heat runs away. > - **Denominator watch:** $\gamma M_1^2-\tfrac{\gamma-1}{2}>0$ for all $M_1\ge1$, so $M_2^2$ never blows up or goes negative — the formula is safe on its whole legal domain. **PICTURE.** A number line of $M_1$ from 0 to ∞, three coloured bands: "$M_1<1$: no shock" (grey), "$M_1=1$: zero-strength" (a single dot), "$M_1>1$: real shock" (orange), with the ceilings and floors marked as dashed asymptotes. ![[deepdives/dd-physics-3.1.11-d2-s08.png]] --- ## The one-picture summary Everything above, on one canvas: the three conservation laws feed the master swap $\rho u^2=\gamma pM^2$, which produces $(\star)$ and $(\dagger)$; mass closes the loop to give $M_2$; the Second Law picks the branch; back-substitution rains down all five ratios. (Compare with [[Oblique shock waves]], where the *normal component* of $M_1$ plays the role of $M_1$ here, and contrast with [[Rayleigh & Fanno flow]], which add heat or friction instead of a shock.) ![[deepdives/dd-physics-3.1.11-d2-s09.png]] > [!recall]- Feynman retelling — say it in plain words > Put a thin box around the shock. Whatever flows *in* the front must flow *out* the back: that's mass, momentum, energy — three promises. I rewrite speed as "Mach number times sound speed", and a lucky bit of algebra turns the momentum push $\rho u^2$ into $\gamma p M^2$. Now momentum becomes a plain pressure ratio, and energy becomes a plain temperature ratio, each written in the two Mach numbers. Mass is the referee that ties them together and coughs up one clean equation for the downstream Mach number $M_2$ — with two answers. One says "nothing happened"; the other is the real shock. The Second Law breaks the tie: only the answer where entropy *rises* is allowed, and that one always turns fast supersonic flow into slow subsonic flow, squeezing and heating it. Feed that $M_2$ back in and every ratio drops out: pressure and temperature can climb forever, but density hits a hard ceiling of six. That ceiling is the punchline — a shock can't compress a gas past 6×, so all the leftover energy of a violent shock has nowhere to go but *heat*. > [!recall]- Quick self-test > The master swap turns $\rho u^2$ into what? ::: $\gamma p M^2$ > Which conservation law is the referee that isolates $M_2$? ::: mass conservation > Which law rejects the expansion-shock root? ::: the Second Law (entropy must rise) > Ceiling of $\rho_2/\rho_1$ for air as $M_1\to\infty$? ::: $\frac{\gamma+1}{\gamma-1}=6$ > At $M_1=1$, what are all four ratios? ::: all equal to 1 (zero-strength shock) > [!mnemonic] > **"Mass Makes M₂, Second Law Signs it."** — Mass conservation delivers $M_2$; the Second Law decides which root is legal.