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

3.1.3 · D2 · Physics › Compressible Flow & Aerodynamics › Speed of sound — a = √(γRT) — derivation

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Step 1 — Sound wave hoti kya hai?

KYA HAI. Ek lamba tube socho jisme still air hai. Ek end pe tum air ko ek chhoti si dhakka dete ho. Woh dhakka ek patla sa region hota hai jahan air apne neighbours se thoda zyada squeeze hoti hai — thoda extra pressure. Woh squeezed region apni jagah nahi rehta: woh tube mein aage travel karta hai. Jis speed se woh travel karta hai, use hum $a$ kehte hain (acoustic ke liye).

YAHAAN SE KYUN SHURU KAREIN. Kisi bhi math se pehle, hum study ke object pe agree karte hain: air ka poora tube mein chalte rehna nahi (woh barely hilti hai), balki disturbance — "thoda squeezed" ka pattern — racing along. Stadium wave socho: log apni seats mein rehte hain, sirf pattern travel karta hai.

PICTURE. Amber band woh patla squeezed region hai (wave). Untouched air uske aage hai; already-relaxed air peeche hai. Band right mein speed $a$ se move karta hai jabki individual molecules barely hilte hain.

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Step 2 — Wave pe sawari karo (picture ko freeze karo)

KYA HAI. Hum wave pe jump karte hain aur uske saath travel karte hain. Humare naye frame mein squeezed band still baithta hai, aur instead gas right se hamare taraf speed $a$ se rush karti hai, band se process hoti hai, aur left pe thodi changed nikal jaati hai.

KYUN. Ground frame mein sab kuch time ke saath move aur change ho raha hai — analyse karna nightmare hai. Wave pe sawari karne se flow steady ho jaata hai: har fixed point pe numbers time ke saath change karna band kar dete hain. Steady flow wahan hai jahan simple bookkeeping laws (mass in = mass out) cleanly kaam karte hain. Yeh poori derivation ka sabse important trick hai.

PICTURE. Step 1 jaisi wave, lekin ab wave frame ke centre mein nailed hai. Incoming gas: speed $a$, pressure $p$, density $\rho$. Outgoing gas (band cross karne ke baad): speed $a-dV$, pressure $p+dp$, density $\rho +d\rho$. Chhote changes exaggerate kiye gaye hain taaki tum dekh sako.

Related idea: Stagnation properties and energy equation isi "flow ke saath ride karo" mindset use karta hai.

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Step 3 — Mass pile up nahi ho sakti (continuity)

KYA HAI. Mass count karo. Steady flow mein, jo bhi mass har second patli band mein stream in hoti hai woh dusri side se stream out zaroor honi chahiye. Zero-thickness band ke andar kuch bhi accumulate nahi hota.

KYUN. Agar zyada mass enter karti exit se, toh band hamesha ke liye heavier hoti jaati — impossible. Toh "mass rate in = mass rate out." Area $A$ se mass rate $\rho \times (\text{speed})\times A$ hai; $A$ dono sides pe same hai, toh cancel ho jaata hai.

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

PICTURE. Do flux arrows equal "thickness" ke — left wala $\rho$ aur $a$ se bana, right wala $\rho +d\rho$ aur $a-dV$ se. Equal mass rate matlab dono arrows same stuff-per-second carry karte hain.

Right side multiply karo aur tiny×tiny term $d\rho dV$ ko kill karo (negligibly small):

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

Equation (1) padhna: term $\rho dV$ (density times tiny slow-down) exactly $ad\rho$ (sound speed times tiny density bump) ko balance karta hai. Speed change aur density change lock hain ek saath.

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Step 4 — Slab ke liye Newton ka law (momentum)

KYA HAI. Ab push karo, mass nahi. Band ke left face pe pressure aur right face pe pressure thoda alag hain. Woh pressure difference ek net force hai, aur net force ko band se guzarne wali gas ka momentum change karna chahiye.

KYUN. Newton's second law flow language mein: $\text{(net force)}=(\text{mass per second})\times (\text{change in velocity})$. Mass per second hai $\dot{m}=\rho aA$ (Step 3 se). Gas ka velocity change hai $(a-dV)-a=-dV$.

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

PICTURE. Band ek slab ki tarah apne faces pe do opposing pressure arrows ke saath; right arrow thoda lamba hai ($p+dp$). Imbalance left point karta hai aur exactly through-flow ko $dV$ se decelerate karta hai.

Area $A$ cancel ho jaata hai aur signs tidy hote hain:

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

Equation (2) padhna: tiny pressure rise $dp$ exactly $\rho a$ per unit of slow-down $dV$ cost karta hai. Zyada push karo → gas ko aur slow karo.

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Step 5 — Mass aur momentum fuse karo → $a^2=dp/d\rho$

KYA HAI. Humare paas do facts hain, (1) aur (2), dono mein mysterious $dV$ hai. $dV$ eliminate karo aur dekho $a$ ke baare mein kya bachta hai.

KYUN. $dV$ ek bookkeeping helper tha, koi directly measurable cheez nahi. Use eliminate karne se measurable cheezoin ke beech relation milta hai: pressure change, density change, aur wave speed.

(1) se: $dV=\frac{ad\rho }{\rho }$. Ise (2) mein daalo:

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

PICTURE. Ek "gears mesh" diagram: mass gear (eqn 1) aur momentum gear (eqn 2) ek saath turn karte hain aur $dV$ neeche drop ho jaata hai, clean shaft $a^2=dp/d\rho$ bacha ke.

Equation (3) padhna: $a^2$ exactly stiffness over inertia hai. $dp/d\rho$ poochta hai "diye gaye extra density pe mujhe kitna extra pressure milta hai?" Ek gas jo squeeze hone pe hard push back karti hai (chhote $d\rho$ ke liye bada $dp$) sound fast carry karti hai. Dhyan do: abhi tak koi temperature nahi, koi $\gamma$ nahi — yeh sirf mass aur momentum se aaya.

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Step 6 — KAUN SA squeeze? Heat-has-no-time insight

KYA HAI. $dp/d\rho$ ko number mein turn karne ke liye hum yeh jaanna chahte hain ki squeeze ke dauran pressure aur density kaise related hain. Do candidates:

• Isothermal (constant temperature): heat freely flow karta hai toh $T$ fixed rehta hai.
• Adiabatic (no heat exchange): squeeze itni fast hoti hai ki heat escape nahi kar sakti.

ADIABATIC KYUN. Sound wave air ke har parcel ko ek second mein hundreds of times squeeze aur release karti hai. Gas mein heat conduction sluggish hoti hai — ek cycle ke ek flick mein temperature level out karne ke liye bahut zyada slow. Toh har parcel squeeze hone pe heat up hota hai aur stretch hone pe cool hota hai, koi heat bahar nahi jaati. Yahi adiabatic ki definition hai. (Dekho Adiabatic vs isothermal processes.)

PICTURE. Do side-by-side parcels squeeze ho rahe hain. Left = isothermal: chhote heat arrows leak ho rahe hain, $T$ flat. Right = adiabatic: ek sealed box, koi arrows nahi, $T$ squeeze ke saath spike karta hai. Ek clock "too fast!" stamp karta hai sealed wale ke upar.

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Step 7 — Adiabatic gas law $\gamma$ ko answer mein daalta hai

KYA HAI. Ek ideal gas jo adiabatically compress hoti hai, uske liye pressure aur density follow karte hain

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

jahan $C$ ek constant hai aur $\gamma$ (Greek "gamma") heat-capacity ratio hai.

YEH SHAPE KYUN. Isentropic relations p ∝ ρ^γ se: $p{V}^{\gamma }=$ const, aur kyunki density mass-over-volume hai, $\rho \propto 1/V$, toh $p\propto {\rho }^{\gamma }$. Exponent $\gamma >1$ curve ko steeper banata hai isothermal straight-ish line se — steeper matlab stiffer matlab faster sound.

PICTURE. Ek $p$-versus-$\rho$ graph. Isothermal curve $p\propto \rho$ (gentle, cyan) aur adiabatic curve $p\propto {\rho }^{1.4}$ (steep, amber). Slope $dp/d\rho$ tangent line hai — adiabatic curve pe steeper, toh bada $a$.

$p=C{\rho }^{\gamma }$ differentiate karo (amber curve ka slope):

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

toh equation (3) ke saath combine karte hue:

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

Equation (4) padhna: bare stiffness $p/\rho$ (jo Newton ke paas tha) factor $\gamma$ se boost hoti hai — amber curve ki extra steepness.

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Step 8 — Ideal gas law ise sirf temperature tak collapse karta hai

KYA HAI. Ideal gas law kehta hai $p=\rho RT$, jahan $R$ specific gas constant hai (dekho Ideal gas law and specific gas constant). Rearrange karte hain: $p/\rho =RT$.

KYUN. Yeh last substitution hai. Yeh reveal karta hai ki ratio $p/\rho$ koi independent cheez nahi — yeh simply $RT$ hai. Individual pressure aur density ke baare mein sab kuch wash out ho jaata hai.

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

Final formula symbol by symbol padhna:

• $\gamma$ — adiabatic "no-heat-escapes" boost (Step 6–7).
• $R$ — specific gas constant $R_u/M$; halke molecules (chhota molar mass $M$) bada $R$ dete hain aur faster sound.
• $T$ — absolute temperature kelvin mein; garam gas = zippier molecules = faster sound.

PICTURE. Ek "cancellation" panel: $p$ aur $\rho$ dono ek saath raise karo (gas squeeze karo), $a$ pe dono effects opposite direction mein pull karke annihilate ho jaate hain, $T$ labelled ek dial bacha rehta hai jo akela $a$ control karta hai.

Mach number (Mach number and flow regimes) phir flight speed ko is local $a$ se compare karta hai; isliye compressibility sirf $M\approx0.3$ ke upar bite karti hai.

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Step 9 — Degenerate & limiting cases (koi gap mat chhoddo)

KYA HAI & KYUN. Ek achi derivation extremes mein survive karni chahiye. Unhe ek-ek karke check karo.

• $T\to 0$ (absolute zero): $a=\sqrt{\gamma R\cdot 0}=0$. Molecules motionless hain, toh "push" ko pass karne ke liye kuch nahi hai — sound propagate nahi ho sakti. Consistent.
• $T$ bada (hot gas): $a$ $\sqrt{T}$ ki tarah badhta hai. Temperature double karne se $a$ $\sqrt{2}\approx 1.41$ se multiply hota hai, $2$ se nahi — square root matter karta hai.
• Isothermal (galat) limit $\gamma \to 1$: formula Newton ke $\sqrt{RT}$ mein degrade ho jaata hai. $\gamma =1$ set karna literally heat-capacity boost remove karta hai — ek built-in sanity dial.
• Heavy vs light gas ($M$ bada vs chhota): kyunki $R=R_u/M$, ek heavy gas ($\mathrm{C}{\mathrm{O}}_2$, bada $M$) ka chhota $R$ hoga → slow sound; ek light gas (helium, chhota $M$) ka bada $R$ hoga → fast sound. Same $T$, different $a$.
• Finite (not infinitesimal) pressure jump: agar "pulse" bada ho toh woh ek shock mein steep ho jaata hai jo $a$ se faster travel karta hai; humari derivation ne assume kiya $dp,d\rho$ infinitesimal, exactly isliye $a$ small-signal speed hai.

PICTURE. Char mini-panels: $a$-vs-$T$ square-root curve origin pe pinned; $\gamma =1$ vs $\gamma =1.4$ comparison; helium-vs-air-vs-CO₂ speeds ka bar; aur ek chhota pulse smoothly travel kar raha hai versus ek steep shock usse outrun kar raha hai.

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Ek-picture summary

KYA HAI. Har step ek single flow-chart-with-pictures mein fold kiya: wave pe sawari karo → mass deta hai (1) → momentum deta hai (2) → fuse karo $a^2=dp/d\rho$ → adiabatic choose karo → $p\propto {\rho }^{\gamma }$ deta hai $\gamma p/\rho$ → ideal gas deta hai $\gamma RT$.

Recall Feynman retelling — poora walk plain words mein

Sound ek chhoti si squeeze hai jo air mein race karti hai. Use study karne ke liye, squeeze pe hop karo taaki woh frozen lage aur air steadily tumhare paas se stream kare. Ab do counts karo. Pehla, mass: air squeeze ke andar pile up nahi ho sakti, toh jo flow in hoti hai woh out hoti hai — yeh air ke tiny slow-down ko uski density mein tiny bump se tie karta hai. Doosra, push: squeeze ka ek face pe thoda zyada pressure hota hai, aur woh chhota imbalance ek force hai; Newton kehta hai force passing air ko slow karta hai — yeh pressure bump ko usi slow-down se tie karta hai. Dono ko saath rakho, slow-down cancel ho jaata hai, aur ek khoobsurat fact nikalta hai: sound speed squared equals har extra density pe extra pressure kitna milta hai — stiffness over heaviness. Ab key twist: squeeze itni fast hoti hai ki heat leak hone ka time nahi milta, toh air compress hone pe heat up hoti hai. Woh "sealed-in heat" gas ko zyada springy banati hai jo tumne naively socha hoga se, aur springiness boost ek number hai jise gamma kehte hain. Springy (adiabatic) law daalo, phir ideal-gas law use karo, aur pressure-over-density simply $R$ times temperature ban jaata hai. Specific pressure aur density ke baare mein sab kuch cancel ho jaata hai, aur tum $a=\sqrt{\gamma RT}$ ke saath bachte ho: speed of sound poori tarah temperature pe ride karti hai, kyunki temperature actually sirf yeh hai ki molecules already kitni fast zoom kar rahi hain — aur sound unse faster nahi travel kar sakti jo ise carry kar rahi hain.