Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations: Visual walkthrough

Step 1 — Two buckets of energy in a moving gas

WHAT. Picture a little parcel of air flying to the right. It owns energy in two separate buckets:

a thermal bucket — the invisible jiggling of its molecules, which we feel as temperature $T$ ;
a motion bucket — the whole parcel sailing along at speed $V$ (metres per second).

WHY start here. Temperature is literally stored microscopic energy. If we stop the parcel, the motion bucket has to empty somewhere — and the only place it can go (no heat allowed to escape) is the thermal bucket. So temperature must rise. To turn that story into numbers we need an energy accounting rule.

PICTURE. The blue arrow is the bulk motion (bucket 2); the orange fuzz is molecular jiggling (bucket 1). Watch the total stay fixed while the split changes.

Step 2 — The energy ledger: enthalpy + motion = constant

WHAT. Follow the parcel down a pipe. We write down a rule that says its total energy never changes. That rule is the steady-flow energy equation:

$thermal energy per kg (enthalpy) h + motion energy per kg \frac{V ^{2}}{2} = the total, a constant h_{0}$

WHY enthalpy $h$ , not plain heat? In a flowing pipe the gas also does push-work on the gas ahead of it. The quantity that already includes that push-work is enthalpy $h$ (that is why the parent's Mistake #3 warns: use $c_{p}$ , not $c_{v}$ ). So the honest "thermal bucket" for a flow is $h$ , not internal energy.

Why is the right side a constant? No heat crosses the pipe wall (adiabatic) and no shaft does work on the gas. With those two switches off, the First Law collapses to "sum of the two buckets doesn't change." Call that unchanging sum $h_{0}$ — the stagnation enthalpy, the value you'd have if all motion were converted to thermal ( $V = 0$ ).

PICTURE. As the pipe narrows the parcel speeds up (motion bucket fills) and cools (thermal bucket drains) — but the two heights always add to the same dashed line $h_{0}$ .

Step 3 — Trade enthalpy for temperature

WHAT. We want temperature, but the ledger speaks in enthalpy. For an ideal gas with constant specific heat there is a simple exchange rate:

$h = c_{p} T \Rightarrow thermal, in K units c_{p} T + \frac{V ^{2}}{2} = c_{p} T_{0}$

WHY this swap. $c_{p}$ (specific heat at constant pressure) is exactly the "how many joules per kg to raise one kelvin" number. It is the currency converter between energy (joules/kg) and temperature (kelvin). It must be $c_{p}$ (not $c_{v}$ ) because we are converting enthalpy.

Divide through by $c_{p} T$ and rearrange:

$T_{0} = T + \frac{V ^{2}}{2 c _{p}}$

Read it: the stagnation temperature $T_{0}$ is the static temperature $T$ plus a bonus $\frac{V ^{2}}{2 c _{p}}$ paid entirely by the motion bucket.

PICTURE. A thermometer in the still, stopped gas reads higher ( $T_{0}$ ) than one drifting with the flow ( $T$ ); the gap is the emptied motion bucket.

Step 4 — Rewrite the bonus using Mach number

WHAT. The clumsy $\frac{V ^{2}}{2 c _{p}}$ becomes clean if we measure speed in Mach number. We need two known facts:

$a^{2} = γ R T (sound speed), c_{p} = \frac{γ R}{γ - 1} (ideal-gas identity) .$

WHY these two. The first converts $T$ into a speed we can compare $V$ against. The second rewrites $c_{p}$ in terms of $γ$ and $R$ , which will magically cancel against $a^{2} = γ R T$ .

Substitute into $T_{0} / T = 1 + \frac{V ^{2}}{2 c _{p} T}$ :

$\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 \cdot \frac{γ R}{γ - 1} \cdot T} = 1 + \frac{γ - 1}{2} = V^{2} / a^{2} = M^{2} \frac{V ^{2}}{γ R T}$

The under-brace is the whole trick: $γ R T$ is $a^{2}$ , so $V^{2} / (γ R T) = M^{2}$ .

PICTURE. A rising curve: $T_{0} / T$ vs $M$ , gently curving upward like a parabola because of the $M^{2}$ .

Step 5 — From temperature to pressure needs smoothness

WHAT. Temperature only cared about energy. Pressure cares about how you stop the gas. Stop it smoothly (no rubbing) and you recover the most pressure; stop it roughly and some is lost forever. The smooth ideal is isentropic — reversible and adiabatic.

WHY a new ingredient. Two parcels can share the same $T_{0}$ yet reach different $P_{0}$ if one was slowed with friction. To pin $P_{0}$ down we must demand the clean, reversible stop. The isentropic relations link temperature to pressure and density:

$\frac{P _{0}}{P} = (\frac{T _{0}}{T})^{\frac{γ}{γ - 1}}, \frac{ρ _{0}}{ρ} = (\frac{T _{0}}{T})^{\frac{1}{γ - 1}}$

Where these come from. Isentropic means $P v^{γ} = const$ (a stiff spring law); combine with the ideal-gas law $P v = R T$ to eliminate volume $v$ , and out drop those exponents.

PICTURE. Two ramps from static to stagnation: the smooth green ramp climbs all the way to full $P_{0}$ ; the jagged red ramp (friction) tops out lower — same $T_{0}$ , smaller $P_{0}$ .

Step 6 — Assemble the two boxed pressure/density laws

WHAT. Drop Result 1 into the isentropic exponents:

WHY the exponents order this way. Squeeze a gas and pressure jumps fastest, density next, temperature least — hence "Temp is Tame, Press is Power." A consistency check that must hold (ideal gas law): $\frac{P _{0}}{P} = \frac{ρ _{0}}{ρ} \cdot \frac{T _{0}}{T}$ , since $3.5 = 2.5 + 1$ . ✓

PICTURE. Three curves rising from $M = 0$ : temperature (gentle), density (steeper), pressure (steepest) — the exponent ranking made visible.

Step 7 — Edge case A: $M \to 0$ collapses to Bernoulli

WHAT. Set the speed tiny. Expanding the pressure bracket for small $M^{2}$ :

$\frac{P _{0}}{P} \approx 1 + \frac{γ}{2} M^{2} + \frac{γ}{8} M^{4} + \dots \Rightarrow P_{0} - P \approx \frac{1}{2} ρ V^{2}$

WHY it must. At crawl speed the gas barely compresses, so the compressible law has to fold into the familiar Bernoulli result $P_{0} = P + \frac{1}{2} ρ V^{2}$ . It does. The leftover $M^{4}$ piece is the compressibility correction — negligible below $M \approx 0.3$ , essential above it.

PICTURE. The exact pressure curve and the Bernoulli line hug each other near $M = 0$ , then peel apart as $M$ grows — showing exactly where "incompressible" breaks.

Step 8 — Edge case B: irreversibility drops $P_{0}$ (the shock trap)

WHAT. Send the gas through a normal shock (or any friction). The stop is now rough, entropy rises, and $P_{0}$ falls — while $T_{0}$ does not budge.

WHY the asymmetry. $T_{0}$ rides on energy conservation, which a shock respects (still adiabatic). $P_{0}$ rides on reversibility, which a shock violates. So:

$T_{0}$ : conserved across a shock. ✓
$P_{0}$ : decreases across a shock. ✗

This is the parent's Mistake #1 made visual — never assume $P_{0}$ survives every adiabatic process.

PICTURE. Across the red shock line: the $T_{0}$ bar stays level, the $P_{0}$ bar drops by the loss $Δ P_{0}$ .

The one-picture summary

Everything above on a single map: the shared bracket feeds three exponents; the smooth road keeps full $P_{0}$ , the rough road (shock/friction) loses some.

Recall Feynman retelling — the whole walkthrough in plain words

A moving gas carries energy two ways: warmth and rush. Stop it gently and the rush turns into warmth, so the stopped gas is hotter — that hotter reading is the total temperature $T_{0}$ , and it only asks that no heat leak out. Because temperature is really just stored energy, one clean energy ledger ( $h + \frac{1}{2} V^{2} =$ constant) gives $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ , where $M$ is speed measured in sound-speeds. Pressure is fussier: it depends on how you stop the gas. Stop it smoothly and you get the full total pressure $P_{0}$ ; stop it roughly (a shock, friction) and some squeeze is lost forever even though the warmth is untouched. That smoothness rule (isentropic) turns the same bracket into pressure and density laws, just with bigger exponents — pressure biggest, density next, temperature smallest. Crawl slowly and it all melts back into plain Bernoulli. That's the entire story: one bracket, three exponents, one warning about rough stops.

Recall Term-by-term self-test

Which quantity survives a shock, $T_{0}$ or $P_{0}$ ? ::: $T_{0}$ (adiabatic); $P_{0}$ drops (irreversible). What is the shared bracket in all three formulas? ::: $1 + \frac{γ - 1}{2} M^{2}$ . Why $c_{p}$ not $c_{v}$ in $T_{0} = T + V^{2} /2 c_{p}$ ? ::: Because we converted enthalpy $h = c_{p} T$ , not internal energy. What does the $M^{4}$ term represent in the Bernoulli expansion? ::: The compressibility correction. Exponents on $P_{0} / P$ , $ρ_{0} / ρ$ , $T_{0} / T$ for air? ::: $3.5, 2.5, 1$ .