Thermal expansion — linear, area, volumetric: Visual walkthrough

1.7.7 · D2 · Physics › Thermodynamics › Thermal expansion — linear, area, volumetric

Hum sirf yeh assume karenge ki tum multiply kar sakte ho, aur tum jaante ho ki rectangle ka area "side times side" hota hai. Baaki sab kuch yahan build kiya gaya hai.

Step 1 — Atoms garm hone par zyada door kyun baithte hain

YAHAN KYA MATTER KARTA HAI. Is page par sab kuch sirf ek fact ke upar bookkeeping hai: ek single atom-to-atom gap garm hone par thoda sa badhta hai. Agar woh gap na badhta, kuch bhi expand nahi hota. Deep reason ke liye ki spring lopsided kyun hai, dekho Interatomic potential energy curve.

PICTURE. Figure mein, curve woh energy hai jo do atoms ke beech gap ke against store hoti hai. Bowl ka bottom woh jagah hai jahan ek thanda atom aaram karta hai (uski comfortable spacing). Ek garm atom ke paas zyada energy hoti hai, isliye woh bowl ki dono walls ke beech jhoolata hai — lekin left wall (squeeze) right wall (stretch) se zyada steep hai, isliye uske swing ka middle (uski average position, magenta dot) rightward drift karta hai, ek bade gap ki taraf.

Figure — Thermal expansion — linear, area, volumetric

Step 2 — Ek gap se poore rod tak

KYA ISLIYE. Growth fractional hai, fixed nahi. Das gaps jo har ek $1%$ badhte hain woh ek aisi chain banate hain jo $1%$ badhti hai — fraction ko parwah nahi ki kitne gaps hain. Isliye rod ki length aur ek atomic gap identical rule follow karte hain.

PICTURE. Top row: thanda rod, paanch gaps. Bottom row: garm rod, wahi paanch gaps har ek ek hi sliver (orange) se stretch hua. Total stretch $Δ L$ saare chhote slivers ka sum hai.

Step 3 — Growth factor ko ek baar naam do

KYA ISLIYE. Har shape ki har side usi same $k$ se multiply hoti hai (material isotropic hai — woh sabhi directions mein equally badhta hai). $(1 + α Δ T)$ ki jagah $k$ likhna hume symbols mein doobe bina geometry dekhne deta hai. Ant mein hum $k$ ko substitute karke expand karte hain.

PICTURE. $L_{0}$ length ka ek thanda segment $k L_{0}$ length ka ek garm segment ban jaata hai. Green bracket woh tiny extra bit hai, $α Δ T \cdot L_{0}$ .

Recall

$k$ sirf thoda sa 1 se bada kyun hai? Kyunki $α \approx 1 0^{- 5} K^{- 1}$ bahut chhota hai. $100 K$ rise ke liye, $α Δ T = 1 0^{- 3}$ , isliye $k = 1.001$ . Woh "barely 1 se bada hona" bilkul wahi reason hai jiske wajah se hum baad mein $α^{2}$ terms ko chhod sakte hain. Steel ( $α = 1.2 \times 1 0^{- 5}$ ) ko $100 K$ garm karne par $α Δ T$ kitna bada hoga? ::: $1.2 \times 1 0^{- 3}$ — lagbhag ek hajar mein ek part.

Step 4 — Area: do badhti sides ko multiply karo

$k^{2}$ kyun, $2 k$ nahi? Kyunki area do lengths ka product hai, sum nahi. Jab kisi product ke dono factors badhte hain, toh product unke growth factors ke product se badhta hai. Yahi poora reason hai ki area length se aage nikal jaata hai.

PICTURE. Thanda square magenta core hai. Heating ise rightward (ek orange strip) aur upward (ek doosri orange strip) wide karti hai, aur tiny violet corner fill karti hai jahan dono strips overlap hoti hain. Do strips = do $α Δ T$ contributions; corner = $α^{2} Δ T^{2}$ leftover.

$A = (k L_{0}) (k L_{0}) = k^{2} L_{0}^{2} = A_{0} k^{2}$

$A_{0} = L_{0}^{2}$ — cold area (magenta core).
$k^{2}$ — area growth factor.

Ab $k = 1 + α Δ T$ substitute karo aur multiply out karo:

$A = A_{0} (1 + α Δ T)^{2} = A_{0} (core 1 + do strips 2 α Δ T + tiny corner α^{2} Δ T^{2})$

$1$ — original square (magenta).
$2 α Δ T$ — do orange strips (ek per direction).
$α^{2} Δ T^{2}$ — woh violet corner jahan strips overlap hoti hain.

Step 5 — Corner ko delete kyun karte hain

KYA ISLIYE. Agar $α Δ T = 1 0^{- 3}$ , toh uska square $1 0^{- 6}$ hai — strips se hazaar guna chhota. Ise rakhna aisa hai jaise road trip measure karo aur ek atom ki width ki chinta karo. Ise drop karna "first-order approximation" hai.

PICTURE. Step 4 jaisa hi square, lekin do orange strips bold hain (rakha gaya) aur corner ek whisper tak fade ho gaya hai (drop kiya gaya), sizes labelled $1 0^{- 3}$ vs $1 0^{- 6}$ .

Step 6 — Volume: teen badhti sides

$k^{3}$ kyun? Volume = length $\times$ width $\times$ height hai — teen badhti lengths ka product. Teen factors, har ek apna $α Δ T$ lekar, isliye teen first-order slabs appear hote hain.

PICTURE. Thanda cube magenta core hai. Heating right face par ek slab, top par ek slab, aur front par ek slab add karti hai — teen orange slabs, ek per direction. (Patli edge-bars aur tiny corner cube woh higher-order leftovers hain jo hum drop karenge.)

$L_{0}$ cold side ka ek cube lo. Uska cold volume magenta core hai:

$V_{0} = L_{0}^{3}$

$V_{0}$ — cube ka cold volume (side $\times$ side $\times$ side).
$L_{0}$ — cold side length, wahi jo Step 2 se hai.

Har side $k L_{0}$ tak badhti hai, isliye:

$V = (k L_{0})^{3} = V_{0} k^{3} = V_{0} (1 + α Δ T)^{3}$

$V = (k L_{0})^{3}$ — hot volume, teenon sides $k$ se multiply.
$V_{0} k^{3}$ — cold volume times volume growth factor $k^{3}$ .

Multiply out karo:

$V = V_{0} (core 1 + teen slabs 3 α Δ T + teen edge-bars 3 α^{2} Δ T^{2} + corner cube α^{3} Δ T^{3})$

$1$ — original cube $V_{0}$ .
$3 α Δ T$ — teen face-slabs (woh parts jo actually matter karte hain).
$3 α^{2} Δ T^{2}$ aur $α^{3} Δ T^{3}$ — edge-bars ( $1 0^{- 6}$ ) aur corner cube ( $1 0^{- 9}$ ): delete kiya gaya, Step 5 jaisa hi reason.

Step 7 — Degenerate cases (kabhi uncover nahi chhode jaate)

Case $Δ T = 0$ (koi heating nahi). Toh $k = 1$ , isliye $L = L_{0}$ , $A = A_{0}$ , $V = V_{0}$ . Kuch nahi hilta. Achha hai — formulas kuch nahi karte jab tum kuch nahi karte.

Case $Δ T < 0$ (cooling). Toh $α Δ T < 0$ , isliye $k < 1$ aur sab kuch exactly usi rule se shrink hota hai. Expansion aur contraction ek hi formula hai $Δ T$ ke do signs ke saath.

Case of a hole. Hole sirf "missing metal" hai jo real metal se ghira hua hai. Har real atom $α Δ T$ fraction se outward move karta hai, isliye hole ki boundary bhi outward move karti hai — hole badhta hai, solid area jaisa hi $β = 2 α$ se. Woh band nahi hota.

Case $γ < 0$ (water, 0–4 °C). Zyaadaatar solids ke liye $α > 0$ , lekin paani $0$ aur $4^{\circ} C$ ke beech $γ < 0$ hai: woh garm hone par sikodhta hai. Hamaari algebra unchanged hai; sirf coefficient ka sign flip hota hai. Dekho Anomalous expansion of water.

PICTURE. Left panel: hole wala thanda ring. Right panel: garm ring — outer edge badha (orange) aur hole badha (dashed magenta circle bada hai), arrows har point se outward radiate karte hain.

Ek-picture summary

α L = L_{0} k^{1} β = 2 α A = A_{0} k^{2} γ = 3 α V = V_{0} k^{3}

Keep only the first-order term of each and the coefficients fall out clean: $α : β : γ = 1 : 2 : 3.$

Recall Feynman retelling — walkthrough plain words mein

Ek lopsided spring do atoms ke beech imagine karo; use heat karo aur do atoms thoda sa drift kar jaate hain (Step 1). Rod inhi springs ki ek lambi chain hai, isliye rod usi tiny fraction se badhta hai — us fraction ke "sab rakho aur thoda add karo" multiplier ko $k$ kaho (Steps 2–3). Ab ek flat plate ki do sides hain, dono $k$ se stretch hoti hain, aur area = side times side hai, isliye area $k$ se do baar multiply hota hai — $k^{2}$ — jo purane square plus do patli strips mein split hota hai (Steps 4–5). Ek solid block ki teen sides hain, isliye uska volume $k$ se teen baar multiply hota hai — $k^{3}$ — purana block plus teen patle slabs deta hai (Step 6). Extra corners aur edges itne ridiculous level par chhote hain ki hum unhe rubout kar dete hain. Strips aur slabs count karo: area ke liye do, volume ke liye teen. Yahi poora $1 : 2 : 3$ hai — yeh sirf yeh hai ki shape kitni directions mein badh sakti hai. Cooling? Same story, $k$ one se neeche jaata hai. Hole? Sab kuch outward scale hota hai jaise zoomed photo, isliye hole bhi bada hota hai (Step 7).

Connections

Thermal expansion — linear, area, volumetric (index 1.7.7) — parent result jise yeh page draw karta hai.
Interatomic potential energy curve — Step 1 ka lopsided spring.
Kinetic theory of gases — same "zyada jhatke, phail jaao" idea, gases mein kaafi zyada strong.
Anomalous expansion of water — Step 7 ka $γ < 0$ case.
Stress and strain — kya hota hai jab garm rod ko badhne se roka jaata hai.
Pendulum and simple harmonic motion — ek lamba rod clock ko slow kar deta hai.
Gas laws — jahan $Δ T$ absolute hona chahiye, yahan ke unlike.