Electromagnetic waves — derivation from Maxwell's equations

1.8.33 · Physics › Electromagnetism

Starting point: Maxwell's equations in vacuum

YEH KYA kehte hain: divergence laws (1),(2) kehte hain ki vacuum mein field lines ka koi source nahi hota (koi isolated start/end points nahi). Curl laws (3),(4) coupling hain — ek type ka time-varying field doosre type ka field curl up kar deta hai.
WHY $μ_{0} ε_{0} \partial_{t} E$ term matter karta hai: woh displacement current term, jo Maxwell ne add kiya, missing ingredient hai. Iske bina (4) vacuum mein $\nabla \times B = 0$ ho jaata hai aur koi wave exist nahi karti. Maxwell ka correction literally wahi hai jo light ko possible banata hai.

Key vector identity (derivation ka engine)

$E$ ke liye wave equation derive karna

Step 1. Faraday's law (3) ka curl lo. $\nabla \times (\nabla \times E) = \nabla \times (- \frac{\partial B}{\partial t})$ Yeh step kyun? $E$ equation ka curl lene se right side par $\nabla \times B$ aa jaata hai — aur humare paas $\nabla \times B$ ka formula hai, yaani (4). Isi tarah hum loop close karte hain.

Step 2. Left side: curl-of-curl identity apply karo, phir (1) se $\nabla \cdot E = 0$ use karo. $\nabla (= 0 \nabla \cdot E) - \nabla^{2} E = - \nabla^{2} E$ Yeh step kyun? Equation (1) pehle term ko khatam kar deti hai — yeh woh moment hai jab "no charges" assumption kaam aati hai, ek clean Laplacian reh jaata hai.

Step 3. Right side: space-curl aur time-derivative ka order swap karo (fields smooth hain), phir (4) insert karo. $\nabla \times (- \frac{\partial B}{\partial t}) = - \frac{\partial}{\partial t} (\nabla \times B) = - \frac{\partial}{\partial t} (μ_{0} ε_{0} \frac{\partial E}{\partial t}) = - μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}}$ Yeh step kyun? Hum $B$ -wali cheezein poori tarah $E$ -wali cheezein se replace kar lete hain Ampère–Maxwell use karke — ab equation mein sirf $E$ hai.

Step 4. Dono sides equate karo; minus signs cancel ho jaate hain. $\nabla^{2} E = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}}$

Identical procedure se (curl of (4) lo, $\nabla \cdot B = 0$ use karo) tumhe milta hai: $\nabla^{2} B = μ_{0} ε_{0} \frac{\partial ^{2} B}{\partial t ^{2}}$

Figure — Electromagnetic waves — derivation from Maxwell's equations

Plane wave ki structure ( $E$ , $B$ , $k$ ki geometry)

$+ x$ mein travel karne wali plane wave try karo: $E = E_{0} e^{i (k x - ω t)}$ , $B = B_{0} e^{i (k x - ω t)}$ .

Transverse, (1) se: $\nabla \cdot E = 0 \Rightarrow ik E_{0 x} = 0 \Rightarrow E_{0 x} = 0$ . Toh $E$ ka travel direction ke along koi component nahi. $B$ ke liye bhi same. → EM waves transverse hoti hain.
$E ⊥ B$ , (3) se: $\nabla \times E = - \partial_{t} B$ . $E = E_{0} \overset{y}{^}$ ke saath, curl $\overset{z}{^}$ ke along point karta hai, forcing $B ∥ \overset{z}{^}$ . Toh $E, B, k$ ek right-handed orthogonal triad banate hain ( $\hat{E} \times \hat{B} = \hat{k}$ ).
Amplitude ratio, (3) se: $k E_{0} = ω B_{0} \Rightarrow \frac{E _{0}}{B _{0}} = \frac{ω}{k} = c$ .

Worked examples

Common mistakes (Steel-manned)

Recall Feynman: 12-saal ke bacche ko samjhao

Imagine karo do bacche ek seesaw par hain jo chup nahi reh sakte. Jab bhi ek upar jaata hai, woh doosre ko neeche dhakelta hai, aur woh pehle wale ko wapas upar push karta hai — hamesha ke liye. Electricity aur magnetism aise hi hain: ek hilta hua electric field ek magnetic field banata hai, aur ek hilta hua magnetic field ek electric field banata hai. Yeh ek doosre ko push karte rehte hain aur hilna aage empty space mein run karta hai. Woh running hilna light hai. Cool twist yeh hai: do numbers jo bilkul alag experiments mein measure hue (ek magnets ke liye, ek static charge ke liye) aapas mein multiply hoke bataate hain ki hilna exactly kitni tez run karta hai — $300, 000$ km har second.

Flashcards

Kaun se do Maxwell equations E aur B ko couple karke wave banate hain?

Faraday

\nabla \times E = - \partial_{t} B

aur Ampère–Maxwell

\nabla \times B = μ_{0} ε_{0} \partial_{t} E

Maxwell ka add kiya kaun sa term EM waves ko possible banata hai?

Displacement current

μ_{0} ε_{0} \partial_{t} E

Ampère's law mein.

Curl-of-curl identity batao.

\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F

Derivation mein

\nabla (\nabla \cdot E)

term kyun vanish ho jaata hai?

Kyunki charge-free vacuum mein

\nabla \cdot E = 0

hota hai.

Vacuum mein E ki final wave equation?

\nabla^{2} E = μ_{0} ε_{0} \partial_{t}^{2} E

Vacuum mein EM wave ki speed (formula)?

c = 1/ μ_{0} ε_{0} \approx 3 \times 1 0^{8}

m/s.

c

ki prediction historically important kyun hai?

μ_{0}, ε_{0}

magnetism/electrostatics se aate hain (optics se nahi) phir bhi speed of light dete hain — prove karta hai ki light ek EM wave hai.

EM waves transverse hain ya longitudinal, aur kaun sa equation prove karta hai?

Transverse;

\nabla \cdot E = 0

k

ke along koi field component force nahi karta.

E, B, k ke beech geometric relation?

Mutually perpendicular, right-handed:

\hat{E} \times \hat{B} = \hat{k}

E aur B ke beech amplitude relation?

E_{0} = c B_{0}

Kya E aur B equal energy carry karte hain?

Haan:

\frac{1}{2} ε_{0} E^{2} = \frac{1}{2 μ _{0}} B^{2}

Plane EM wave ki dispersion relation?

ω = c k

Connections

Maxwell's Equations — woh chaar laws jisse yeh derivation shuru hoti hai
Displacement Current — woh term jo wave ko enable karta hai
Faraday's Law of Induction — equation (3)
Ampère–Maxwell Law — equation (4)
Wave Equation — universal $\nabla^{2} f = \frac{1}{v ^{2}} \partial_{t}^{2} f$ template
Poynting Vector and EM Energy — $S = \frac{1}{μ _{0}} E \times B$ (energy kahan jaati hai)
Electromagnetic Spectrum — same equations, alag $λ$
Polarization — transverse $E$ ka consequence