Speed of Sound: Newton–Laplace & Factors

In 1687, in the second book of the Principia, Isaac Newton did something no one had done before: he calculated the speed of sound. Not measured it — derived it, from the pressure and density of the air alone, using nothing but his own mechanics. It was a triumph of pure reasoning, and it came out about fifteen per cent too slow. Newton predicted roughly $280\ \mathrm{m/s}$ ; the air actually carries sound at about $331\ \mathrm{m/s}$ at $0\ \mathrm{^\circ C}$ . For over a century the gap sat there, an embarrassment hiding in the middle of the greatest physics book ever written, and Newton fudged his way around it rather than admit he could not close it. The fix, when it finally came from Pierre-Simon Laplace in 1816, is one of the most elegant corrections in physics, and it turns on a single physical insight: a sound wave squeezes the air too fast for heat to escape. Newton had assumed the compressions stay at constant temperature; Laplace realised they are so rapid that no heat has time to leave, so the air heats up as it compresses and cools as it expands. That makes the air stiffer than Newton thought — stiffer by exactly the factor $\gamma$ , the ratio of heat capacities — and putting $\gamma$ in its place turns $280$ into $331$ , dead on the measured value. The whole fifteen-per-cent error was a missing $\sqrt{\gamma} = \sqrt{1.4} = 1.183$ . This article works through that story as physics you can use: the general rule that wave speed is $\sqrt{\text{stiffness}/\text{density}}$ , why Newton's isothermal guess fails and Laplace's adiabatic one succeeds, how the speed climbs with temperature at about $0.6\ \mathrm{m/s}$ per degree, why it does not depend on pressure, and why sound races through steel far faster than through air even though steel is thousands of times denser. The number $343\ \mathrm{m/s}$ you used in the last article gets earned here.

Key takeaways

Any mechanical wave travels at $v = \sqrt{\text{elastic property} / \text{inertial property}}$ ; for a gas, $v = \sqrt{B/\rho}$ with $B$ the bulk modulus.
Newton (isothermal, wrong): $v = \sqrt{P/\rho} \approx 280\ \mathrm{m/s}$ — about $15.5\%$ too low.
Laplace (adiabatic, correct): $v = \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M} \approx 331\ \mathrm{m/s}$ at $0\ \mathrm{^\circ C}$ — matches measurement.
The whole correction is the missing $\sqrt{\gamma} = \sqrt{1.4} = 1.183$ : the compressions are too fast to exchange heat, so the air is stiffer than Newton assumed.
$v \propto \sqrt{T}$ (T in kelvin), giving about $+0.6\ \mathrm{m/s}$ per $^\circ$ C near room temperature: $343\ \mathrm{m/s}$ at $20\ \mathrm{^\circ C}$ , $346\ \mathrm{m/s}$ at $25\ \mathrm{^\circ C}$ .
Speed is independent of pressure (at fixed temperature $P/\rho$ is constant) and rises with humidity; across media, sound is faster in water ( $\approx 1480\ \mathrm{m/s}$ ) and steel ( $\approx 5064\ \mathrm{m/s}$ ) because stiffness wins over density.
Key idea: the speed of sound is $\sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M}$ — Laplace's adiabatic $\gamma$ is what turns Newton's $280\ \mathrm{m/s}$ into the measured $331\ \mathrm{m/s}$ .

The core insight

The speed of sound in a gas is $v = \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M}$ : Newton's isothermal $\sqrt{P/\rho}$ was $15.5\%$ low because the compressions are actually adiabatic, and Laplace's factor $\gamma$ — appearing as $\sqrt{\gamma} = 1.183$ — is the whole of the correction; the speed then rises as $\sqrt{T}$ and is independent of pressure.

Where this lives today

The number and the reasoning are everywhere. Aircraft speed is quoted as a Mach number — the aircraft's speed divided by the local speed of sound — and because that local speed falls with the plummeting temperature at altitude (not with the falling pressure, exactly as this article insists), the "sound barrier" a jet crosses is a temperature-dependent target, around $295\ \mathrm{m/s}$ high in the stratosphere versus $343\ \mathrm{m/s}$ near the ground. A weather forecaster's thermometer is, in effect, a sound-speed meter. Musicians retune as a concert hall warms, because a wind instrument's pitch rises with the speed of sound inside it. Sonar and seismology depend on knowing the speed precisely in water and rock, where — just as $\sqrt{B/\rho}$ predicts — sound races at thousands of metres per second. And the fact that Laplace's $\gamma$ works so exactly is quietly used in reverse: measuring the speed of sound in a gas is one of the cleanest ways to determine its ratio of specific heats. The next article turns from how fast sound travels to what we actually perceive in it — its pitch, its loudness and its quality.

The physical picture

Every mechanical wave is a race between two properties of the medium: the stiffness that springs a displaced bit of matter back, and the inertia that makes it sluggish to move. The more stiffness, the harder the restoring push and the faster the wave; the more inertia, the slower. For every mechanical wave the speed is the square root of their ratio:

v = \sqrt{\frac{\text{elastic property}}{\text{inertial property}}}. \qquad (1)

For a wave on a string this reads $v = \sqrt{T/\mu}$ (tension over mass per length). For sound in a gas the stiffness is the bulk modulus $B$ — how strongly the gas resists being compressed — and the inertia is its density $\rho$ :

v = \sqrt{\frac{B}{\rho}}. \qquad (2)

where: $B$ is the bulk modulus in pascals (pressure needed for a given fractional squeeze) and $\rho$ the density in $\mathrm{kg/m^3}$ . Everything now hangs on one question: what is the right stiffness $B$ for the rapid squeezing a sound wave does? Newton and Laplace gave two different answers, and the difference between them is this whole article.

Figure

Newton's isothermal prediction of about 280 metres per second falls well short of the measured 331, and Laplace's adiabatic value of 331 lands exactly on it, the gap closed entirely by the factor root gamma of about 1.18

Newton's answer, and why it fails

Newton reasoned that when a sound wave compresses a parcel of air, the compression happens gently enough that the parcel stays at the temperature of its surroundings — an isothermal squeeze. For an isothermal ideal gas the bulk modulus is simply the pressure, $B = P$ , so equation (2) gives Newton's formula:

v_{\text{Newton}} = \sqrt{\frac{P}{\rho}}. \qquad (3)

Put in atmospheric pressure $P = 1.013 \times 10^{5}\ \mathrm{Pa}$ and the density of air at $0\ \mathrm{^\circ C}$ , $\rho = 1.293\ \mathrm{kg/m^3}$ : $v_{\text{Newton}} = \sqrt{1.013\times10^{5}/1.293} = 280\ \mathrm{m/s}$ . But the measured speed at $0\ \mathrm{^\circ C}$ is $331\ \mathrm{m/s}$ — Newton is low by about $51\ \mathrm{m/s}$ , roughly $15.5\%$ . This is not experimental error; it is a wrong physical assumption. Newton, who knew the measured value perfectly well, tried to patch the gap with fudge factors about water vapour and dust, none of which survive scrutiny. The assumption itself was the problem.

Laplace's fix: the compressions are too fast for heat

Laplace saw what Newton missed. A sound wave compresses and rarefies the air hundreds or thousands of times a second, in shells only centimetres apart. That is far too fast for heat to flow out of a compression into the neighbouring rarefaction; each parcel is thermally sealed. The squeeze is therefore adiabatic, not isothermal — the compressed air warms up (and the rarefied air cools), and warmer air pushes back harder, so the gas is stiffer than Newton assumed. For an adiabatic ideal gas the bulk modulus is larger by the factor $\gamma$ , the ratio of specific heats:

B_{\text{adiabatic}} = \gamma P, \qquad \gamma = \frac{C_P}{C_V} = 1.4\ \text{for air}. \qquad (4)

We quote $B_{\text{adiabatic}} = \gamma P$ here — proving it needs the adiabatic gas law from the kinetic-theory chapter, and the full argument uses calculus that sits above this article's ceiling. But using it is pure algebra. Substitute into equation (2):

v_{\text{Laplace}} = \sqrt{\frac{\gamma P}{\rho}}. \qquad (5)

With the same numbers, $v_{\text{Laplace}} = \sqrt{1.4 \times 1.013\times10^{5}/1.293} = 331\ \mathrm{m/s}$ — exactly the measured value. The entire correction is the factor $\sqrt{\gamma}$ : since $v_{\text{Laplace}}/v_{\text{Newton}} = \sqrt{\gamma} = \sqrt{1.4} = 1.183$ , Laplace's answer is $18.3\%$ higher than Newton's, which is the same as saying Newton's was $15.5\%$ too low. One factor, one physical idea, and a century-old discrepancy vanishes. There is a second, more useful, form. For an ideal gas, $P/\rho = RT/M$ where $R$ is the gas constant, $T$ the absolute temperature and $M$ the molar mass. So equation (5) becomes

v = \sqrt{\frac{\gamma R T}{M}}, \qquad (6)

where: $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ , $T$ in kelvin, and $M$ in kg/mol ( $0.029\ \mathrm{kg/mol}$ for air). Check it at $0\ \mathrm{^\circ C} = 273\ \mathrm{K}$ : $v = \sqrt{1.4 \times 8.314 \times 273/0.029} = 331\ \mathrm{m/s}$ , agreeing with equation (5). Equation (6) is the more powerful because it shows at a glance what the speed depends on — and, just as importantly, what it does not.

What the speed depends on

Equation (6) reads off the factors directly. Temperature: $v \propto \sqrt{T}$ , so the speed rises with the square root of the absolute temperature. Pressure: $P$ has cancelled out entirely — raising the pressure at fixed temperature raises the density in step, $P/\rho$ is unchanged, and the speed does not move. Composition: $v \propto 1/\sqrt{M}$ , so a lighter gas carries sound faster (helium's low $M$ is why a helium voice is squeaky-fast), and moist air — lighter than dry air, because a water molecule is lighter than the nitrogen and oxygen it displaces — carries sound slightly faster than dry air. The temperature dependence is worth a handy linear rule. Near room temperature $v \propto \sqrt{T}$ can be approximated by a straight line: starting from $331\ \mathrm{m/s}$ at $0\ \mathrm{^\circ C}$ , the speed climbs by about $0.6\ \mathrm{m/s}$ for every degree Celsius:

v \approx 331 + 0.6\,t\ \ (t\text{ in }^\circ\mathrm{C}), \qquad \frac{dv}{dt} = \frac{331}{2 \times 273} = 0.61\ \mathrm{m/s}\ \text{per}\ ^\circ\mathrm{C}. \qquad (7)

So at $20\ \mathrm{^\circ C}$ the speed is $331 + 0.6\times20 = 343\ \mathrm{m/s}$ — the value the last article used — and at $25\ \mathrm{^\circ C}$ it is $346\ \mathrm{m/s}$ . The exact square-root law agrees: $331\sqrt{293/273} = 342.9\ \mathrm{m/s}$ at $20\ \mathrm{^\circ C}$ , within a whisker of the linear $343$ .

Figure

The speed of sound in air rises almost linearly with temperature at about 0.6 metres per second per degree, passing through 331 at 0 degrees, 343 at 20 degrees and 346 at 25 degrees Celsius

Derivation AFrom Newton to Laplace — how the factor √γ closes the gap

The goal is to see, in numbers, why Newton's speed is wrong and how Laplace's single change fixes it exactly — and to be clear about what is quoted versus what is derived. The strategy is to run the general fluid formula $v = \sqrt{B/\rho}$ twice, once with the isothermal stiffness $B = P$ and once with the adiabatic stiffness $B = \gamma P$ , and watch the ratio come out as the pure factor $\sqrt{\gamma}$ . The one input we take on trust is that a fast, sealed compression is stiffer by $\gamma$ ; everything else is arithmetic.

Step 1. Start from $v = \sqrt{B/\rho}$ . The only thing in dispute is the bulk modulus $B$ — the stiffness against compression — because a sound wave's compressions could in principle be slow (isothermal) or fast (adiabatic).

Step 2. Newton's isothermal choice, $B = P$ . For an ideal gas held at constant temperature, the bulk modulus equals the pressure, so $v_N = \sqrt{P/\rho} = \sqrt{1.013\times10^{5}/1.293} = 280\ \mathrm{m/s}$ . Measured: $331\ \mathrm{m/s}$ . Newton is short by about $51\ \mathrm{m/s}$ .

Step 3. Laplace's adiabatic choice, $B = \gamma P$ (quoted from the kinetic-theory chapter; deriving it needs the adiabatic gas law and calculus beyond this article). Then $v_L = \sqrt{\gamma P/\rho} = \sqrt{1.4 \times 1.013\times10^{5}/1.293} = 331\ \mathrm{m/s}$ — exactly the measurement.

Step 4. Read off the correction. Dividing the two, $v_L/v_N = \sqrt{\gamma P/\rho}\big/\sqrt{P/\rho} = \sqrt{\gamma} = \sqrt{1.4} = 1.183$ . So Laplace's value is $18.3\%$ above Newton's, i.e. Newton's was $15.5\%$ below the truth. The entire century-long discrepancy was one missing $\sqrt{\gamma}$ , and the physics behind it is a single sentence: the compressions are too fast for heat to escape. Verified: Newton $280\ \mathrm{m/s}$ , Laplace $331\ \mathrm{m/s}$ , the factor $\sqrt{1.4} = 1.183$ , and the $15.5\%$ shortfall are recomputed in verify_hs2-speed-of-sound.py.

Go deeper · Derivation B · 1 minWhy the speed rises as √T, and the +0.6 per °C rule

Derivation BWhy the speed rises as √T, and the +0.6 per °C rule

Here we derive the temperature dependence from the gas-law form and turn it into the linear shortcut every exam expects. The goal is to show $v \propto \sqrt{T}$ with $T$ in kelvin, then to justify why "add about $0.6\ \mathrm{m/s}$ per degree Celsius" works so well near room temperature. The strategy is to isolate the temperature factor in $v = \sqrt{\gamma RT/M}$ and then approximate the square root for a small temperature change.

Step 1. From equation (6), $v = \sqrt{\gamma RT/M}$ , everything except $T$ is a constant for a fixed gas. So $v \propto \sqrt{T}$ , with $T$ the absolute (kelvin) temperature — this is why Celsius must never be used inside the square root.

Step 2. Anchor at $0\ \mathrm{^\circ C} = 273\ \mathrm{K}$ , where $v_0 = 331\ \mathrm{m/s}$ . At any other temperature, $v = v_0\sqrt{T/273} = 331\sqrt{T/273}$ . At $20\ \mathrm{^\circ C} = 293\ \mathrm{K}$ this gives $331\sqrt{293/273} = 342.9\ \mathrm{m/s}$ .

Step 3. Linearise for a small rise $t$ (in $^\circ$ C) above $0\ \mathrm{^\circ C}$ : $\sqrt{(273+t)/273} \approx 1 + t/(2\times273)$ , so $v \approx 331\left(1 + \tfrac{t}{546}\right) = 331 + \tfrac{331}{546}t$ . The slope is $331/(2\times273) = 0.61\ \mathrm{m/s}$ per degree.

Step 4. Round the slope to $0.6$ and read the working rule $v \approx 331 + 0.6\,t$ . It gives $343\ \mathrm{m/s}$ at $20\ \mathrm{^\circ C}$ and $346\ \mathrm{m/s}$ at $25\ \mathrm{^\circ C}$ , matching the exact $\sqrt{T}$ law to a fraction of a metre per second — the tangent line hugging the curve near room temperature. Far from room temperature, drop the shortcut and use $v = 331\sqrt{T/273}$ directly. Verified: the slope $331/(2\times273) = 0.61$ , the values $343$ and $346\ \mathrm{m/s}$ , and the exact $331\sqrt{293/273} = 342.9\ \mathrm{m/s}$ are recomputed in verify_hs2-speed-of-sound.py.

Across different media

Because $v = \sqrt{B/\rho}$ , comparing two materials means comparing their stiffness-to-density ratios, and the answer is often the opposite of naive intuition. Water is about $800$ times denser than air, which alone would make sound slower — but water is far, far stiffer (its bulk modulus dwarfs air's), and stiffness wins: sound travels through fresh water at about $1480\ \mathrm{m/s}$ , more than four times its speed in air. Steel is denser still, yet its enormous elastic stiffness (here the relevant modulus is Young's modulus $Y$ ) carries sound at $v = \sqrt{Y/\rho} = \sqrt{2.0\times10^{11}/7800} = 5064\ \mathrm{m/s}$ — fifteen times the speed in air, in a material thousands of times denser. "Denser means slower" is simply false across different media; only within a single medium, holding stiffness fixed, does higher density slow the wave.

Limiting cases

case	result	meaning
isothermal assumption (Newton)	$v = \sqrt{P/\rho} = 280\ \mathrm{m/s}$	the historical wrong answer, $15.5\%$ low
adiabatic (Laplace, correct)	$v = \sqrt{\gamma P/\rho} = 331\ \mathrm{m/s}$	matches measurement; the fix is $\sqrt{\gamma} = 1.183$
raise pressure at fixed $T$	$v$ unchanged	$P/\rho$ constant — speed is pressure-independent
raise temperature	$v \propto \sqrt{T}$ , $\approx +0.6\ \mathrm{m/s}$ per $^\circ$ C	$343\ \mathrm{m/s}$ at $20\ \mathrm{^\circ C}$ , $346$ at $25\ \mathrm{^\circ C}$
lighter gas (small $M$ ), or humid air	faster	$v \propto 1/\sqrt{M}$ (helium voice, moist air)
stiffer medium (water, steel)	much faster	$1480\ \mathrm{m/s}$ in water, $5064\ \mathrm{m/s}$ in steel — stiffness beats density

Common trapNewton's 280 m/s is the wrong answer, and denser does not mean slower

Two errors travel together here. The first is quoting Newton's isothermal $v = \sqrt{P/\rho} = 280\ \mathrm{m/s}$ as if it were correct. It is not — it is the famous $15.5\%$ -too-low value, and the sound wave is adiabatic, so the right formula carries the extra $\gamma$ : $v = \sqrt{\gamma P/\rho} = 331\ \mathrm{m/s}$ . If a problem gives you $\gamma$ , use it; leaving it out is the classic slip. The second error is reasoning "steel is denser than air, so sound must be slower in steel." Across different media this is exactly backwards: sound is faster in water and steel than in air, because their stiffness outweighs their density in $\sqrt{B/\rho}$ . Density slows the wave only when you compare samples of the same stiffness. And two unit traps that quietly wreck equation (6): $T$ must be in kelvin (add $273$ ), and $M$ in kilograms per mole ( $0.029$ , not $29$ ).

Worked examples

Board level — speed of sound and temperature (CBSE and ICSE pattern)

The speed of sound in air is $331\ \mathrm{m/s}$ at $0\ \mathrm{^\circ C}$ . Estimate it at $20\ \mathrm{^\circ C}$ and at $25\ \mathrm{^\circ C}$ using the linear rule, and check the $20\ \mathrm{^\circ C}$ value against the exact $v \propto \sqrt{T}$ law.

Step 1. Use the linear rule, equation (7), $v \approx 331 + 0.6\,t$ . At $t = 20\ \mathrm{^\circ C}$ : $v = 331 + 0.6\times20 = 331 + 12 = \boxed{343\ \mathrm{m/s}}$ .

Step 2. At $t = 25\ \mathrm{^\circ C}$ : $v = 331 + 0.6\times25 = 331 + 15 = \boxed{346\ \mathrm{m/s}}$ .

Step 3. Check the $20\ \mathrm{^\circ C}$ value exactly with $v \propto \sqrt{T}$ , T in kelvin: $v = 331\sqrt{293/273} = 331 \times 1.036 = 342.9\ \mathrm{m/s}$ . The linear rule ( $343$ ) and the exact law ( $342.9$ ) agree to a fraction of a metre per second, which is why the " $+0.6$ per degree" shortcut is safe near room temperature. Note that had you carelessly used Celsius in the square-root law — $\sqrt{20/0}$ — you would have divided by zero: the ratio $\sqrt{T}$ only makes sense with absolute temperature.

Competitive level — Newton, Laplace and the missing factor (JEE-Advanced and NEET pattern)

Take air at $0\ \mathrm{^\circ C}$ with $P = 1.013\times10^{5}\ \mathrm{Pa}$ , $\rho = 1.293\ \mathrm{kg/m^3}$ and $\gamma = 1.4$ . (a) Find Newton's isothermal prediction and Laplace's adiabatic prediction for the speed of sound. (b) By what factor and what percentage does Laplace's correction differ from Newton's? (c) Confirm the adiabatic value from the gas-law form, taking $M = 0.029\ \mathrm{kg/mol}$ , $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ .

Step 1. Newton, equation (3): $v_N = \sqrt{P/\rho} = \sqrt{1.013\times10^{5}/1.293} = \sqrt{78345} = \boxed{280\ \mathrm{m/s}}$ .

Step 2. Laplace, equation (5): $v_L = \sqrt{\gamma P/\rho} = \sqrt{1.4 \times 1.013\times10^{5}/1.293} = \sqrt{109683} = \boxed{331\ \mathrm{m/s}}$ , matching the measured value.

Step 3. The factor is $v_L/v_N = \sqrt{\gamma} = \sqrt{1.4} = \boxed{1.183}$ : Laplace is $18.3\%$ above Newton, equivalently Newton is $15.5\%$ below Laplace. The whole discrepancy is that single $\sqrt{\gamma}$ .

Step 4. Cross-check via equation (6) at $T = 273\ \mathrm{K}$ : $v = \sqrt{\gamma RT/M} = \sqrt{1.4 \times 8.314 \times 273/0.029} = \sqrt{109573} = 331\ \mathrm{m/s}$ . The two independent routes to $331$ — through $\gamma P/\rho$ and through $\gamma RT/M$ — agree, as they must, since $P/\rho = RT/M$ for an ideal gas. For contrast, sound in steel is $\sqrt{Y/\rho} = \sqrt{2.0\times10^{11}/7800} = 5064\ \mathrm{m/s}$ and in water about $1480\ \mathrm{m/s}$ : both far faster than in air, though far denser, because stiffness dominates. Timing note: about two minutes. Compute $v_N$ first, then multiply by $\sqrt{\gamma}$ to get $v_L$ in one step rather than re-entering the whole surd — this is the fast route and it makes the " $\sqrt{\gamma}$ is the correction" point automatically. The trap is dropping $\gamma$ and reporting Newton's $280$ as the answer.

Exam traps

Quoting Newton's $280\ \mathrm{m/s}$ . The isothermal $v = \sqrt{P/\rho}$ is the historical wrong answer. Sound compressions are adiabatic, so the correct formula is $v = \sqrt{\gamma P/\rho} = 331\ \mathrm{m/s}$ ; the factor $\gamma$ is not optional. Whenever a gas problem gives you $\gamma$ , that is the examiner reminding you which formula to use.

Kelvin, not Celsius, in $v \propto \sqrt{T}$ . The square-root law needs absolute temperature. To double-check a speed with $v \propto \sqrt{T}$ you must convert to kelvin first; using Celsius gives nonsense (and $\sqrt{T}$ at $0\ \mathrm{^\circ C}$ would be $\sqrt{0}$ ). The linear $331 + 0.6t$ rule takes $t$ in Celsius, but it is only an approximation valid near room temperature.

Thinking higher pressure means faster sound. It does not: at fixed temperature, raising the pressure raises the density in exact proportion, $P/\rho$ is unchanged, and the speed is flat. This is a favourite JEE distractor — "the pilot climbs, pressure drops, does sound speed up?" The honest answer is that the speed change is due to temperature, not pressure.

"Denser medium, slower sound." True only within one medium at fixed stiffness; false across media. Steel is thousands of times denser than air and carries sound fifteen times faster, because stiffness beats density in $\sqrt{B/\rho}$ . Never rank sound speeds by density alone.

Molar mass in the wrong units. In $v = \sqrt{\gamma RT/M}$ , the molar mass $M$ must be in kg/mol — air is $0.029$ , not $29$ . Slipping in grams per mole makes the speed come out a factor of $\sqrt{1000} \approx 32$ too big.

Board scope. All the boards here carry the speed of sound and its temperature dependence. IGCSE 0625 owns "the speed is roughly $340\ \mathrm{m/s}$ and here is how to measure it," but not the Newton–Laplace derivation — an IGCSE candidate needs the value and the measurement method, not the adiabatic argument. The full Newton→Laplace story with $\gamma$ is CBSE, ICSE, JEE and NEET material; A-Level (CIE 9702) treats the wave-speed idea but leans lighter on the adiabatic derivation, so check your specification.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · Paris

Augustin-Louis Cauchy was one of the founders of modern mathematics — of rigorous calculus, of complex analysis — but he was also, crucially for this article, the architect of the continuum theory of elasticity. In a series of memoirs from about 1822, Cauchy defined stress and strain as we still use them and derived the equations governing how an elastic medium deforms and how disturbances travel through it, arriving at the general result that an elastic wave moves at $v = \sqrt{\text{stiffness}/\text{density}}$ . The speed of sound is exactly that formula, with the air's adiabatic bulk modulus $\gamma P$ as the stiffness — which is precisely why Newton's isothermal guess, using the wrong stiffness, came out too low. Cauchy did not himself supply the adiabatic correction (that is Laplace's), but the framework that says wave speed is the square root of stiffness over inertia, the backbone of this whole topic, is his.

History

1687

— Isaac Newton, in the Principia, computes the first theoretical speed of sound in air and gets about 280 m/s, roughly 15 percent too low.

1738

— The Paris Academy of Sciences times cannon fire over a measured baseline and pins the speed of sound in air near 332 m/s.

1816

— Pierre-Simon Laplace supplies the adiabatic correction, replacing P by γP, and the predicted speed rises to 331 m/s, matching measurement.

1822

— Augustin-Louis Cauchy founds the continuum theory of elastic-wave propagation, v = √(stiffness/density), the framework behind the speed of sound.

1826

— Jean-Daniel Colladon and Charles Sturm time a bell against a flash across Lake Geneva, measuring the speed of sound in water at about 1435 m/s.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the speed-of-sound and Newton–Laplace sections · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves," speed in gases · Cambridge IGCSE 0625 (speed of sound and its measurement) · CIE A-Level 9702 (wave speed).

Provenance. Standard results. Recompute: Newton $\sqrt{1.013\times10^{5}/1.293} = 280\ \mathrm{m/s}$ ; Laplace $\sqrt{1.4\times1.013\times10^{5}/1.293} = 331\ \mathrm{m/s}$ ; the factor $\sqrt{1.4} = 1.183$ and the shortfall $1 - 1/\sqrt{1.4} = 15.5\%$ (about $51\ \mathrm{m/s}$ ); the gas-law form $\sqrt{1.4\times8.314\times273/0.029} = 331\ \mathrm{m/s}$ ; the temperature values $331+0.6\times20 = 343$ , $331+0.6\times25 = 346$ , slope $331/(2\times273) = 0.61$ , and exact $331\sqrt{293/273} = 342.9$ ; steel $\sqrt{2.0\times10^{11}/7800} = 5064\ \mathrm{m/s}$ , water $\approx 1480\ \mathrm{m/s}$ ; and the historical values (Paris $\approx 332$ , Colladon–Sturm $\approx 1435$ ). All are checked in verify_hs2-speed-of-sound.py.

Prerequisites. Sound as a Longitudinal Mechanical Wave — sound as compressions in a medium, and $v = f\lambda$ , on which the speed is defined. Leads to. Characteristics of Sound: Pitch, Loudness and Quality — what the travelling wave sounds like · ARCHETYPE: Resonance-Tube and Pipe Problems — measuring this very speed in a school lab.

Go deeper (UG). How the wave speed $\sqrt{B/\rho}$ emerges from the equations of an elastic continuum, and the split into different wave types in solids, is developed in Elastic Waves in Solids in the university mechanics track — Cauchy's framework made quantitative. Enrichment only; nothing here depends on it.

Why exactly are the compressions in a sound wave adiabatic and not isothermal?

Because they happen too fast and over too short a distance for heat to move. A sound wave squeezes a parcel of air and releases it hundreds or thousands of times a second, and the neighbouring compression and rarefaction are only a fraction of a wavelength apart — centimetres. Heat conduction through air is slow, so in the split second a parcel spends compressed, essentially no heat leaks out to its cooler neighbours. Each parcel is effectively thermally sealed, which is the definition of an adiabatic process. The compressed air therefore warms up and pushes back harder than it would at constant temperature, making the gas stiffer by the factor γ. Newton's isothermal assumption would only hold if the squeezing were extremely slow, giving heat all the time it needs to equalise — the opposite of a sound wave.

What is γ, and why is it 1.4 for air?

Gamma is the ratio of a gas's two heat capacities, $\gamma = C_P/C_V$ — how much heat it takes to warm the gas at constant pressure versus at constant volume. It is always greater than one, and for a diatomic gas like the nitrogen and oxygen that make up most of air, kinetic theory gives $\gamma = 7/5 = 1.4$ . (Monatomic gases like helium have $\gamma = 5/3 \approx 1.67$ ; more complex molecules have smaller values.) In the speed of sound, γ measures how much stiffer the adiabatic squeezing makes the gas compared with an isothermal one, and $\sqrt{\gamma} = \sqrt{1.4} = 1.183$ is exactly the factor by which the true speed exceeds Newton's. Where γ comes from is the kinetic-theory chapter's story; here we quote it and use it.

Does sound really travel faster on a hot day?

Yes, measurably. The speed rises with the square root of absolute temperature, about $0.6\ \mathrm{m/s}$ for each degree Celsius near room temperature. So sound at $30\ \mathrm{^\circ C}$ (about $349\ \mathrm{m/s}$ ) outruns sound at $0\ \mathrm{^\circ C}$ ( $331\ \mathrm{m/s}$ ) by nearly twenty metres per second. This is why sound can carry surprisingly far across a cold lake at dawn: a layer of cold air near the water is slower, and sound bends back down toward the listener instead of escaping upward. It is a temperature effect, not a pressure or humidity effect (though humid air is very slightly faster because it is less dense). Pressure by itself does nothing, because raising pressure raises density in lockstep.

If pressure doesn't change the speed, why does sound behave differently at high altitude?

Because what changes with altitude is mainly the temperature, not the bare pressure. It is true that pressure and density both fall as you climb, but they fall together, so their ratio $P/\rho$ — which is what sets the speed — barely changes on that account. What genuinely changes the speed is that the air high up is much colder, and $v \propto \sqrt{T}$ makes cold air slower: around $295\ \mathrm{m/s}$ in the frigid stratosphere versus $343\ \mathrm{m/s}$ near the warm ground. So a jet's Mach number depends on altitude through temperature. Students often "explain" altitude effects by pressure; the honest physics is temperature, and equation (6) shows why — pressure has cancelled out.

How can sound be faster in water than in air when water is 800 times denser?

Because speed depends on both stiffness and density, as $v = \sqrt{B/\rho}$ , and water's stiffness beats its density. Water is indeed about $800$ times denser than air, which acting alone would slow sound down. But water is enormously harder to compress than air — its bulk modulus is roughly $15{,}000$ times air's — and that stiffness enters the top of the fraction. The net effect, $\sqrt{15000/800} \approx 4$ , makes sound about four times faster in water, near $1480\ \mathrm{m/s}$ . Steel is denser still but far stiffer again, giving about $5064\ \mathrm{m/s}$ . The lesson is to never compare sound speeds by density alone; the stiffness-to-density ratio is what matters.

The exact law and the +0.6 per °C rule give slightly different numbers — which do I use?

Use whichever the question invites, and know they agree near room temperature. The linear rule $v \approx 331 + 0.6t$ is a straight-line approximation to the true $v = 331\sqrt{T/273}$ curve, tangent to it at $0\ \mathrm{^\circ C}$ ; near everyday temperatures the two are within a fraction of a metre per second (at $20\ \mathrm{^\circ C}$ , $343$ versus $342.9$ ). Far from room temperature the linear rule drifts and you should use the exact square-root law with kelvin. For most board and competitive problems the linear rule is quicker and accurate enough, but if a problem spans a big temperature range, or asks for the ratio of speeds at two very different temperatures, go straight to $v \propto \sqrt{T}$ .