Blackbody radiation from statistical mechanics — Planck distribution: Question bank

2.4.19 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Blackbody radiation from statistical mechanics — Planck dist

Poori derivations ke liye jinhe ye questions chhed rahe hain, parent note khula rakho, saath mein Density of states, Partition function, Bose-Einstein statistics, Equipartition theorem, Rayleigh-Jeans law, Wien's law aur Stefan-Boltzmann law bhi.

Definition Is page par use kiye gaye Symbols (yahaan define kiye gaye hain taaki tujhe parent mein na dhoondna pade)

$u (ν, T) d ν$ = spectral energy density: electromagnetic energy per unit volume jo un modes mein hoti hai jinki frequency $[ν, ν + d ν]$ mein hai. Yeh frequency mein ek density hai — energy paane ke liye tujhe $d ν$ se multiply karna hoga.
$U$ = total energy density: sab kuch frequency par sum kiya gaya, $U = \int_{0}^{\infty} u (ν, T) d ν$ .
$β \equiv \frac{1}{k _{B} T}$ = inverse temperature: statistical mechanics mein natural "thandak" variable; bada $β$ matlab thandak. Boltzmann weights hote hain $e^{- β ϵ}$ .
$x \equiv β h ν = \frac{h ν}{k _{B} T}$ = dimensionless energy ratio: ek photon lump $h ν$ kitna bada hai thermal budget $k_{B} T$ ke comparison mein. Chhota $x$ = sasta mode, bada $x$ = frozen mode.
$μ$ = chemical potential: system mein ek particle add karne ki energy cost. Photons ke liye $μ = 0$ (WQ7 mein explain kiya gaya hai).

Neeche teen figures hain jo visual backbone hain — answer karne se pehle unhe ek nazar dekh lo.

Figure — Blackbody radiation from statistical mechanics — Planck distribution

True or false — justify

Har answer mein reason dena zaroori hai, sirf verdict nahi. (Curves ki shape ke liye figure s01 dekho.)

TF1. "Ultraviolet catastrophe isliye hoti hai kyunki bahut zyada low-frequency modes hote hain."

False. Mode count

g (ν) = 8 π ν^{2} / c^{3}

high

ν

par unboundedly badhta hai; yeh high-frequency modes ka pile-up hai jo har ek

k_{B} T

hold karta hai, jis wajah se classical integral diverge hoti hai.

TF2. "Planck ka quantization cavity mein kitne modes exist karte hain yeh change karta hai."

False. Quantization sirf average energy per mode

⟨ ϵ ⟩

ko change karta hai. Density of states

g (ν) = 8 π ν^{2} / c^{3}

purely geometric hai aur classical aur quantum dono treatments mein identical hai.

TF3. "Bahut low frequency par Planck curve aur Rayleigh–Jeans curve agree karte hain."

True.

h ν ≪ k_{B} T

(yaani

x \to 0

) ke liye hum

e^{x} - 1 \approx x

expand karte hain, jisse

⟨ ϵ ⟩ \to k_{B} T

milta hai, jo bilkul classical (Rayleigh–Jeans) result hai — dekho kaise curves figure s01 ke left mein ek saath chalte hain.

TF4. "Kyunki high

ν

par zyada modes hote hain, ek hot body sabse zyada highest frequencies par radiate karti hai."

False.

ν^{2}

mode growth ko exponential factor

e^{- h ν / k_{B} T}

⟨ ϵ ⟩

mein hai woh beat karta hai, isliye spectrum ek finite peak

ν_{m a x} \approx 2.82 k_{B} T / h

par turn over karta hai aur baad mein girta hai.

TF5. "Bose–Einstein occupation number

\overset{n}{ˉ} = 1/ (e^{h ν / k_{B} T} - 1)

1 se zyada ho sakta hai."

True. Photons bosons hain jinmein koi exclusion nahi;

h ν ≪ k_{B} T

(saste, low-frequency modes) ke liye

\overset{n}{ˉ}

bahut bada ho jaata hai, matlab kaafi photons ek hi mode mein pile up ho jaate hain.

TF6. "Temperature double karne se total radiated energy density double ho jaati hai."

False. Stefan–Boltzmann deta hai

U \propto T^{4}

, isliye

T

double karne par

U

2^{4} = 16

se multiply hoti hai, na ki 2 se.

TF7. "

u (ν)

ka peak aur

u (λ)

ka peak ek hi frequency/wavelength pair ko point karte hain."

False. Dimensionless ratios

x_{ν} = h ν / k_{B} T

(frequency form) aur

x_{λ} = h c / λ k_{B} T

(wavelength form) likhne par, Jacobian

∣ d ν / d λ ∣ = c / λ^{2}

maximum ko shift kar deta hai:

x_{ν} \approx 2.82

lekin

x_{λ} \approx 4.97

. Physics same hai, peak location alag hai.

TF8. "Ek perfect blackbody isliye black hota hai kyunki woh kabhi koi light emit nahi karta."

False. "Black" ka matlab hai ki woh sab incident radiation absorb karta hai; thermal equilibrium mein woh poora Planck spectrum bhi emit karta hai. Ek hot blackbody khub chamakta hai.

TF9. "Zero-point energy

\frac{1}{2} h ν

thermal radiation spectrum ki shape ke liye mayne rakhti hai."

False. Equilibrium photon picture mein

⟨ ϵ ⟩ = h ν \overset{n}{ˉ}

use hota hai bina kisi zero-point term ke; constant

\frac{1}{2} h ν

per mode

T

-independent hai aur observable radiated spectrum mein drop out ho jaata hai.

Spot the error

Neeche har statement mein ek flaw chupi hai. Use naam do aur correct karo. (Figure s02 SE1 ke liye $n$ -space octant dikhata hai.)

SE1. "Hum modes ko

n

-space mein radius

n

ki ek full sphere ke roop mein count karte hain, volume

\frac{4}{3} π n^{3}

Octant error. Kyunki har

n_{i} = 1, 2, 3 \dots

positive hai, sphere ka sirf

1/8

hissa physical hai:

N = \frac{1}{8} \cdot \frac{4}{3} π n^{3} = \frac{π n ^{3}}{6}

SE2. "Har wavevector

k

ek mode contribute karta hai, isliye

g (ν) = 4 π ν^{2} / c^{3}

Polarization factor missing hai. Har

k

do independent transverse polarizations carry karta hai, isliye 2 se multiply karo:

g (ν) = 8 π ν^{2} / c^{3}

SE3. "Equipartition se har oscillator

\frac{1}{2} k_{B} T

carry karta hai, isliye

u = \frac{4 π ν ^{2}}{c ^{3}} k_{B} T

Har 1-D oscillator ke do quadratic degrees of freedom hote hain (kinetic + potential), jo dete hain

2 \times \frac{1}{2} k_{B} T = k_{B} T

, na ki

\frac{1}{2} k_{B} T

. Aur equipartition khud invalid hai jab levels discrete hoti hain: theorem derive hota hai ek continuous energy par integrate karke jo quadratically appear hoti hai; jab allowed energies

0, h ν, 2 h ν, \dots

par stuck hoti hain toh woh integral ek sum se replace ho jaata hai, aur jab gap

h ν ≫ k_{B} T

hota hai toh mode apne pehle excited level tak bhi nahi pahunch sakta, isliye woh

k_{B} T

se bahut kam carry karta hai.

SE4. "Partition function hai

Z = \sum_{n} e^{- nh ν β} = \frac{1}{1 - e ^{- x}}

(

β = 1/ k_{B} T

aur

x = β h ν

ke saath), isliye

⟨ ϵ ⟩ = - \frac{\partial ln Z}{\partial x}

Energy average hai

⟨ ϵ ⟩ = - \partial ln Z / \partial β

β

ke respect mein differentiate karke, na ki

x = β h ν

ke. Variable galat choose karne se

h ν

ka factor drop ho jaata hai (kyunki

\partial x / \partial β = h ν

SE5. "Kyunki

⟨ ϵ ⟩ = h ν / (e^{h ν / k_{B} T} - 1)

, high

ν

par har mode lagbhag

h ν

hold karta hai."

High

ν

par,

e^{h ν / k_{B} T} ≫ 1

isliye

⟨ ϵ ⟩ \approx h ν e^{- h ν / k_{B} T} \to 0

. High-frequency modes almost kuch bhi hold nahi karte,

h ν

nahi.

SE6. "Stefan–Boltzmann integral mein,

x = h ν / k_{B} T

substitute karne par ek

T

-dependent integral bachta hai."

Us substitution ka poora point yahi hai ki integral

\int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x = π^{4} /15

ko ek pure number banana hai jisme koi

T

na ho; saara

T

dependence front mein

T^{4}

ke roop mein nikal aata hai.

SE7. "Wien's law

⟨ ϵ ⟩

maximum set karne se aata hai."

Tum poori spectral density

u (ν) \propto ν^{3} / (e^{x} - 1)

(density of states × energy) maximize karte ho, sirf

⟨ ϵ ⟩

ko nahi. Yeh deta hai

3 (1 - e^{- x}) = x

, yaani

x \approx 2.821

Why questions

WQ1. Classical result kyun diverge karta hai lekin quantum result finite kyun rehta hai?

Classical mein har mode ko

k_{B} T

milta hai, isliye

\int ν^{2} k_{B} T d ν

diverge karta hai. Quantization

⟨ ϵ ⟩

ko large

ν

par

e^{- h ν / k_{B} T}

ki tarah decay karwata hai, aur exponential decay

ν^{2}

growth ko overwhelm kar deti hai, isliye integral converge ho jaata hai.

WQ2. Energy

h ν

ke lumps mein quantized kyun hai (frequency ke proportional) — yeh specific fix kyun hai, bas koi bhi discreteness kyun nahi?

Lump size

h ν

ka frequency ke saath scale karna exactly wahi cheez hai jo high-frequency modes ko expensive banati hai: ek fixed thermal budget

k_{B} T

saste low-

ν

lumps excite kar sakta hai lekin costly high-

ν

wale nahi, jinhe woh freeze out kar deta hai. Frequency-independent lumps UV modes ko selectively suppress nahi karte.

WQ3. Hum har standing-wave mode ko ek independent harmonic oscillator kyun treat kar sakte hain?

Linear cavity mein EM field normal modes mein decompose ho jaata hai jo directly energy exchange nahi karte; har mode ka amplitude time mein sinusoidally oscillate karta hai bilkul ek 1-D harmonic oscillator ki tarah, isliye uske energy levels ek quantum oscillator ke hote hain.

WQ4. Ek hotter star zyada blue kyun dikhta hai?

Wien displacement deta hai

ν_{m a x} \propto T

, isliye

T

badhane par spectral peak higher frequency (shorter wavelength) ki taraf shift ho jaata hai, perceived color red se blue-white ki taraf shift ho jaata hai.

WQ5. Density of states

ν^{2}

par kyun depend karta hai,

ν

ν^{3}

par nahi?

Modes 3-D

k

-space mein ek shell fill karte hain;

\propto ν

radius ki shell mein modes ki sankhya uski surface area

\propto ν^{2}

ke saath scale karti hai. 2-D mein yeh

\propto ν

hogi, 1-D mein constant — exponent spatial dimension minus one track karta hai.

WQ6. Partition function mein factor geometric series kyun hai, aur ratio 1 se kam kyun hona chahiye?

Allowed energies

nh ν

equally spaced hain, isliye

Z = \sum_{n} (e^{- x})^{n}

ratio

e^{- x}

wali geometric series hai. Kyunki

x = h ν / k_{B} T > 0

hai isliye

e^{- x} < 1

, aur sum converge hota hai

1/ (1 - e^{- x})

par; ratio

\geq 1

matlab unphysical infinite

Z

. (In Boltzmann weights ka geometric decay figure s03 mein draw kiya gaya hai.)

WQ7. Photon picture mein chemical potential

μ = 0

kyun hota hai (atoms ki gas se alag)?

Yaad karo

μ

ek particle add karne ki energy cost hai. Atoms ke liye particle number fixed hota hai, isliye

μ

adjust hota hai use enforce karne ke liye aur positive ya negative ho sakta hai. Lekin photon number conserved nahi hota — walls freely photons create aur destroy karte hain — isliye system free energy minimize karta hai number ke respect mein khud, jo

μ = 0

force karta hai. Exactly yahi wajah hai ki Bose–Einstein deta hai

\overset{n}{ˉ} = 1/ (e^{h ν / k_{B} T} - 1)

bina kisi

e^{- β μ}

factor ke.

Edge cases

EC1. Fixed

ν > 0

ke liye strict limit

T \to 0

mein

⟨ ϵ ⟩

kya hota hai?

x = h ν / k_{B} T \to \infty

, isliye

\overset{n}{ˉ} \to 0

aur

⟨ ϵ ⟩ \to 0

. Har mode apne ground state mein fall ho jaata hai; thandhi cavity kuch bhi radiate nahi karti.

EC2. Fixed

T

par

ν \to 0

hone par

⟨ ϵ ⟩

ka kya hota hai?

x \to 0

se milta hai

⟨ ϵ ⟩ \to k_{B} T

, classical equipartition value — bahut low-frequency corner hi woh jagah hai jahan quantum aur classical predictions merge hoti hain.

EC3. Kya total energy density

U

finite rehti hai jabki

g (ν)

large

ν

par diverge karta hai?

Haan. Haalaanki

g (ν) \to \infty

, product

g ⟨ ϵ ⟩

large

ν

par exponentially decay karta hai, isliye

U = \int_{0}^{\infty} g ⟨ ϵ ⟩ d ν

converge hota hai

\frac{8 π ^{5} k _{B}^{4}}{15 c ^{3} h ^{3}} T^{4}

par.

EC4. Kya exactly

ν = 0

wala koi mode hota hai?

Koi physical standing wave cavity mein zero frequency nahi rakht (uske liye

n_{x} = n_{y} = n_{z} = 0

chahiye, yaani koi wave hi nahi).

ν \to 0

ek limit hai, actual mode nahi, aur woh negligibly contribute karta hai kyunki wahaan

g (ν) \to 0

hai.

EC5.

h \to 0

ki limit mein (Planck's constant ko shrink karne ki kalpana karo), hum kaun sa spectrum recover karte hain?

h \to 0

ke saath, saare

ν

ke liye

x = h ν / k_{B} T \to 0

, isliye har jagah

⟨ ϵ ⟩ \to k_{B} T

aur hum divergent Rayleigh–Jeans law recover karte hain — classical catastrophe wapas aata hai, jo confirm karta hai ki

h

hi use tame karta hai.

EC6. Jab exactly

h ν = k_{B} T

(

x = 1

) ho toh ek mode ka occupation kya hota hai?

\overset{n}{ˉ} = 1/ (e^{1} - 1) \approx 0.58

, yaani average mein lagbhag half a photon per mode — "saste, populated" aur "mehengate, frozen" regimes ke beech ka crossover region

x \sim 1

ke paas baith ta hai.

Recall Har trap ki one-line summary

Count $g \sim ν^{2}$ classical geometry hai; energy $⟨ ϵ ⟩$ woh jagah hai jahan quantum mechanics enter karta hai aur jahan catastrophe khatam hoti hai. Factor of 2 (polarization), octant (1/8), $β$ -vs- $x$ derivative, $T^{4}$ (na ki $T$ ) scaling, aur $ν$ -peak vs $λ$ -peak Jacobian par dhyan rakho.