Visual walkthrough — Speed of sound — a = √(γRT) — derivation

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Step 1 — What even is a sound wave?

WHAT. Picture a long tube of still air. At one end you give the air a tiny shove. That shove is a thin region where the air is squeezed slightly tighter than its neighbours — a little extra pressure. That squeezed region does not stay put: it travels down the tube. The speed it travels at is what we call $a$ (for acoustic).

WHY start here. Before any math, we must agree on the object of study: not the air moving bodily down the tube (it barely moves), but the disturbance — the pattern of "slightly squeezed" — racing along. Think of a stadium wave: the people stay in their seats, only the pattern travels.

PICTURE. The amber band is the thin squeezed region (the wave). Untouched air is ahead of it; already-relaxed air is behind. The band moves right at speed $a$ while individual molecules barely budge.

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Step 2 — Ride the wave (freeze the picture)

WHAT. We jump onto the wave and travel along with it. In our new frame the squeezed band sits still, and instead the gas comes rushing at us from the right at speed $a$, gets processed through the band, and leaves on the left slightly changed.

WHY. In the ground frame everything is moving and changing in time — a nightmare to analyse. Riding the wave makes the flow steady: at every fixed point the numbers stop changing with time. Steady flow is where the simple bookkeeping laws (mass in = mass out) work cleanly. This is the single most important trick in the whole derivation.

PICTURE. Same wave as Step 1, but now the wave is nailed to the centre of the frame. Incoming gas: speed $a$, pressure $p$, density $\rho$. Outgoing gas (after crossing the band): speed $a-dV$, pressure $p+dp$, density $\rho +d\rho$. The little changes are exaggerated so you can see them.

Related idea: Stagnation properties and energy equation uses the same "ride with the flow" mindset.

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Step 3 — Mass cannot pile up (continuity)

WHAT. Count the mass. In steady flow, whatever mass streams into the thin band each second must stream out the other side each second. Nothing accumulates inside a zero-thickness band.

WHY. If more mass entered than left, the band would get heavier and heavier forever — impossible. So "mass rate in = mass rate out." Mass rate through an area $A$ is $\rho \times (\text{speed})\times A$; the $A$ is the same on both sides, so it cancels.

$\underset{\text{in: density}}{\underbrace{\rho }}\underset{\text{in: speed}}{\underbrace{a}}=\underset{\text{out: density}}{\underbrace{(\rho +d\rho )}}\underset{\text{out: speed}}{\underbrace{(a-dV)}}$

PICTURE. Two flux arrows of equal "thickness" — the left one built from $\rho$ and $a$, the right one from $\rho +d\rho$ and $a-dV$. Equal mass rate means the arrows carry the same stuff-per-second.

Multiply out the right side and kill the tiny×tiny term $d\rho dV$ (negligibly small):

\rho a = \rho a - \rho\,dV + a\,d\rho \;\Rightarrow\; \boxed{\;\rho\,dV = a\,d\rho\;}\tag{1}

Reading equation (1): the term $\rho dV$ (density times the tiny slow-down) exactly balances $ad\rho$ (sound speed times the tiny density bump). Speed change and density change are locked together.

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Step 4 — Newton's law for the slab (momentum)

WHAT. Now push, not mass. The pressure on the left face of the band and the pressure on the right face are slightly different. That pressure difference is a net force, and a net force must change the momentum of the gas passing through.

WHY. Newton's second law in flow language: $\text{(net force)}=(\text{mass per second})\times (\text{change in velocity})$. The mass per second is $\dot{m}=\rho aA$ (from Step 3). The velocity change of the gas is $(a-dV)-a=-dV$.

$\underset{\text{push from left}}{\underbrace{pA}}-\underset{\text{push from right}}{\underbrace{(p+dp)A}}=\underset{\dot{m}}{\underbrace{\rho aA}}\underset{-dV}{\underbrace{[(a-dV)-a]}}$

PICTURE. The band as a slab with two opposing pressure arrows on its faces; the right arrow is a touch longer ($p+dp$). The imbalance points left and is exactly what decelerates the through-flow by $dV$.

The area $A$ cancels and the signs tidy up:

-dp\,A = \rho a A(-dV) \;\Rightarrow\; \boxed{\;dp = \rho a\,dV\;}\tag{2}

Reading equation (2): a tiny pressure rise $dp$ costs exactly $\rho a$ per unit of slow-down $dV$. Push harder → slow the gas more.

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Step 5 — Fuse mass and momentum → $a^2=dp/d\rho$

WHAT. We have two facts, (1) and (2), both containing the mysterious $dV$. Eliminate $dV$ and see what is left about $a$.

WHY. $dV$ was a bookkeeping helper, not something we can measure directly. Eliminating it leaves a relation between measurable things: pressure change, density change, and the wave speed.

From (1): $dV=\frac{ad\rho }{\rho }$. Slot it into (2):

dp = \rho a \cdot \frac{a\,d\rho}{\rho} = a^2\,d\rho \;\Rightarrow\; \boxed{\;a^2 = \dfrac{dp}{d\rho}\;}\tag{3}

PICTURE. A "gears mesh" diagram: the mass gear (eqn 1) and momentum gear (eqn 2) turn together and $dV$ drops out the bottom, leaving the clean shaft $a^2=dp/d\rho$.

Reading equation (3): $a^2$ is stiffness over inertia made exact. $dp/d\rho$ asks "how much extra pressure do I get for a given extra density?" A gas that pushes back hard when squeezed (big $dp$ for small $d\rho$) carries sound fast. Notice: no temperature yet, no $\gamma$ yet — this came purely from mass and momentum.

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Step 6 — WHICH squeeze? The heat-has-no-time insight

WHAT. To turn $dp/d\rho$ into a number we must know how pressure and density are related during the squeeze. Two candidates:

• Isothermal (constant temperature): heat flows freely so $T$ stays fixed.
• Adiabatic (no heat exchange): the squeeze happens so fast heat cannot escape.

WHY adiabatic. A sound wave squeezes and releases each parcel of air hundreds of times a second. Heat conduction through gas is sluggish — far too slow to level out the temperature within one flick of a cycle. So each parcel heats up when squeezed and cools when stretched, with no heat leaving. That is the definition of adiabatic. (See Adiabatic vs isothermal processes.)

PICTURE. Two side-by-side parcels being squeezed. Left = isothermal: little heat arrows leak out, $T$ flat. Right = adiabatic: a sealed box, no arrows, $T$ spikes with the squeeze. A clock stamps "too fast!" over the sealed one.

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Step 7 — The adiabatic gas law puts $\gamma$ into the answer

WHAT. For an ideal gas compressed adiabatically, pressure and density obey

$p=C{\rho }^{\gamma }$

where $C$ is a constant and $\gamma$ (Greek "gamma") is the heat-capacity ratio.

WHY this shape. From Isentropic relations p ∝ ρ^γ: $p{V}^{\gamma }=$ const, and since density is mass-over-volume, $\rho \propto 1/V$, so $p\propto {\rho }^{\gamma }$. The exponent $\gamma >1$ makes the curve steeper than the isothermal straight-ish line — steeper means stiffer means faster sound.

PICTURE. A $p$-versus-$\rho$ graph. The isothermal curve $p\propto \rho$ (gentle, cyan) and the adiabatic curve $p\propto {\rho }^{1.4}$ (steep, amber). The slope $dp/d\rho$ is the tangent line — steeper on the adiabatic curve, so bigger $a$.

Differentiate $p=C{\rho }^{\gamma }$ (slope of the amber curve):

$\frac{dp}{d\rho }=C\gamma {\rho }^{\gamma -1}=\gamma \cdot \frac{C{\rho }^{\gamma }}{\rho }=\gamma \frac{p}{\rho }$

so combining with equation (3):

\boxed{\;a^2=\gamma\,\frac{p}{\rho}\;}\tag{4}

Reading equation (4): the bare stiffness $p/\rho$ (what Newton had) is boosted by the factor $\gamma$ — the amber curve's extra steepness.

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Step 8 — Ideal gas law collapses it to temperature only

WHAT. The ideal gas law says $p=\rho RT$, where $R$ is the specific gas constant (see Ideal gas law and specific gas constant). Rearranged: $p/\rho =RT$.

WHY. This is the last substitution. It reveals that the ratio $p/\rho$ is not an independent thing — it is simply $RT$. Everything about the individual pressure and density washes out.

$a^2=\gamma \frac{p}{\rho }=\gamma RT\Rightarrow \boxed{a=\sqrt{\gamma RT}}$

Reading the final formula symbol by symbol:

• $\gamma$ — the adiabatic "no-heat-escapes" boost (Step 6–7).
• $R$ — the specific gas constant $R_u/M$; lighter molecules (small molar mass $M$) give a bigger $R$ and faster sound.
• $T$ — absolute temperature in kelvin; hotter gas = zippier molecules = faster sound.

PICTURE. A "cancellation" panel: raise both $p$ and $\rho$ together (squeeze the gas), the two effects on $a$ pull opposite and annihilate, leaving a dial labelled $T$ as the only survivor controlling $a$.

The Mach number (Mach number and flow regimes) then compares flight speed to this local $a$; that is why compressibility only bites above $M\approx0.3$.

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Step 9 — Degenerate & limiting cases (never leave a gap)

WHAT & WHY. A good derivation must survive the extremes. Check them one by one.

• $T\to 0$ (absolute zero): $a=\sqrt{\gamma R\cdot 0}=0$. Molecules are motionless, so there is nothing to pass the "push" along — sound cannot propagate. Consistent.
• $T$ large (hot gas): $a$ grows as $\sqrt{T}$. Doubling temperature multiplies $a$ by $\sqrt{2}\approx 1.41$, not by $2$ — the square root matters.
• Isothermal (wrong) limit $\gamma \to 1$: the formula degrades to Newton's $\sqrt{RT}$. Setting $\gamma =1$ literally removes the heat-capacity boost — a built-in sanity dial.
• Heavy vs light gas ($M$ big vs small): since $R=R_u/M$, a heavy gas ($\mathrm{C}{\mathrm{O}}_2$, big $M$) has small $R$ → slow sound; a light gas (helium, small $M$) has big $R$ → fast sound. Same $T$, different $a$.
• Finite (not infinitesimal) pressure jump: if the "pulse" is large it steepens into a shock that travels faster than $a$; our derivation assumed $dp,d\rho$ infinitesimal, which is exactly why $a$ is the small-signal speed.

PICTURE. Four mini-panels: the $a$-vs-$T$ square-root curve pinned to the origin; a $\gamma =1$ vs $\gamma =1.4$ comparison; a helium-vs-air-vs-CO₂ bar of speeds; and a small pulse smoothly travelling versus a steep shock outrunning it.

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The one-picture summary

WHAT. Every step folded into a single flow-chart-with-pictures: ride the wave → mass gives (1) → momentum gives (2) → fuse to $a^2=dp/d\rho$ → choose adiabatic → $p\propto {\rho }^{\gamma }$ gives $\gamma p/\rho$ → ideal gas gives $\gamma RT$.

Recall Feynman retelling — the whole walk in plain words

A sound is a little squeeze racing through the air. To study it, hop onto the squeeze so it looks frozen and the air streams past you steadily. Now do two counts. First, mass: the air can't pile up inside the squeeze, so whatever flows in flows out — this ties the tiny slow-down of the air to the tiny bump in its density. Second, push: the squeeze has a slightly higher pressure on one face, and that little imbalance is a force; Newton says force slows the passing air — this ties the pressure bump to the same slow-down. Put the two together, the slow-down cancels, and out pops a beautiful fact: the sound speed squared equals how much extra pressure you get per extra density — stiffness over heaviness. Now the key twist: the squeeze happens so fast that heat has no time to leak away, so the air heats up as it's compressed. That "sealed-in heat" makes the gas springier than you'd naively think, and the springiness boost is a number called gamma. Plug the springy (adiabatic) law in, then use the ideal-gas law, and pressure-over-density turns into just $R$ times temperature. Everything about the specific pressure and density cancels, and you're left with $a=\sqrt{\gamma RT}$: the speed of sound rides entirely on temperature, because temperature is really just how fast the molecules are already zooming — and sound can't travel faster than the molecules carrying it.