Visual walkthrough — Electromagnetic waves — derivation from Maxwell's equations

1.8.33 · D2 · Physics › Electromagnetism › Electromagnetic waves — derivation from Maxwell's equations

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Step 0 — Teen arrows jo hum hamesha use karenge

Kisi bhi physics se pehle, teen actors se milo. Yeh arrows (vectors) hain — ek arrow ki length hoti hai (kitna strong) aur ek direction hoti hai (kis taraf point karta hai).

• $\vec{E}$ — electric field arrow. Isse blue draw karo. Agar wahan ek tiny positive charge hota, toh woh iss arrow ke saath saath push feel karta. Lamba arrow = zyada push.
• $\vec{B}$ — magnetic field arrow. Isse orange draw karo. Yeh woh field hai jis mein compass needle line up hoti hai.
• $\vec{k}$ — travel arrow (wavevector). Green. Yeh us taraf point karta hai jis taraf wave ja rahi hai. Iski length, $k$ (bina arrow ke) likhte hain, baad mein encode karega "ripple kitna squished hai". Yeh yaad rakho: bold-arrow $\vec{k}$ = direction-aur-length; plain $k$ = sirf uski length.

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Step 1 — Chaar laws, pictures ke roop mein

KYA. Hum empty space mein khade hain: koi charges nahi ($\rho =0$), koi currents nahi ($\vec{J}=0$). Maxwell ke chaar rules yahan bhi survive karte hain:

$\underset{(1)}{\underbrace{\nabla \cdot \vec{E}=0}}\underset{(2)}{\underbrace{\nabla \cdot \vec{B}=0}}\underset{(3)}{\underbrace{\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}}}\underset{(4)}{\underbrace{\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}}}$

Term by term, simple words mein:

• $\nabla \cdot \vec{E}$ — $\vec{E}$ ki divergence ($\nabla$ aur $\vec{E}$ ka dot product). Padho as: "kya blue arrows is point se bahar phail rahe hain, jaise ek tap se paani?" Equation (1) kehti hai $=0$: koi tap nahi, koi drain nahi. Vacuum mein field lines kabhi shuru ya khatam nahi hoti.
• $\nabla \cdot \vec{B}=0$ — orange ke liye bhi yahi baat. (Aur: koi akele magnetic poles kabhi nahi.)
• $\nabla \times \vec{E}$ — $\vec{E}$ ka curl ($\nabla$ aur $\vec{E}$ ka cross product): "kya blue arrows is point ke around swirl kar rahe hain, jaise paani ek plughole ke around?" Equation (3) kehti hai ki woh swirl $-\partial \vec{B}/\partial t$ ke barabar hai, yaani time mein change ho raha magnetic field electric arrows ko swirl karata hai.
• $\partial \vec{B}/\partial t$ — "orange arrow per second kitni tezi se change ho raha hai." Minus sign nature ka direction bookkeeping hai (Lenz).
• Equation (4) mirror hai: changing electric field magnetic arrows ko swirl karata hai. Aage ${\mu }_0{\epsilon }_0$ ek chhota conversion factor hai dono ke beech — isse yaad rakho, yeh light ki speed ban jaata hai.

YEH CHAAR KYUN. (1),(2) sources ko forbid karte hain; (3),(4) coupling hain — har changing field dusre ka ek swirl paida karta hai. Yahi mutual feeding poora engine hai.

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Step 2 — Trick: swirl ko dobara curl karo

KYA. Equation (3) lo aur dono sides par ek aur curl operator $\nabla \times$ lagao: $\nabla \times (\nabla \times \vec{E})=\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)$

KYUN. Right side dekho: ab isme $\nabla \times \vec{B}$ hai — aur equation (4) exactly $\nabla \times \vec{B}$ ke liye ek ready-made formula hai. Toh ek doosra curl humein (4) ko (3) mein substitute karne aur $\vec{B}$ ko poori tarah bahar nikalne deta hai. Do equations ek saath tangle hoti hain, woh $\vec{E}$ ke baare mein ek equation ban jaati hain. Yeh Faraday→Ampère–Maxwell handshake hai.

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Step 3 — Left side ko ek identity se untangle karo

KYA. Ek fixed algebraic fact hai, kisi bhi arrow field $\vec{F}$ ke liye sach: $\nabla \times (\nabla \times \vec{F})=\nabla (\nabla \cdot \vec{F})-{\nabla }^2\vec{F}$

Term by term:

• $\nabla (\nabla \cdot \vec{F})$ — "spreading-out" number lo, phir uski slope nikalo. Source-driven piece.
• ${\nabla }^2\vec{F}$ — Laplacian, defined as ${\nabla }^2\equiv \nabla \cdot \nabla =\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$. Ek vector field ke liye yeh componentwise kaam karta hai: ${\nabla }^2\vec{F}=({\nabla }^2{F}_x,{\nabla }^2{F}_y,{\nabla }^2{F}_z)$. Padho as "is point par $\vec{F}$ apne neighbours ke average se kitna alag hai?" Yeh space-mein-curvature term hai — woh wala jo har wave equation mein aata hai.

Isse $\vec{E}$ par apply karo, phir equation (1), $\nabla \cdot \vec{E}=0$ use karo: $\nabla \times (\nabla \times \vec{E})=\nabla (\underset{=0}{\underbrace{\nabla \cdot \vec{E}}})-{\nabla }^2\vec{E}=-{\nabla }^2\vec{E}$

KYUN. "No charges" law (1) pehle term ko erase kar deta hai. Yahi woh exact moment hai jab vacuum assumption kaam aati hai — hum sirf clean curvature term ke saath reh jaate hain.

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Step 4 — Ampère–Maxwell ko feed karo (right side)

KYA. Step 2 ki right side par kaam karo. Pehle space-curl aur time-derivative ka order swap karo (allowed, fields smooth hain), phir equation (4) plug in karo: $\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)=-\frac{\partial }{\partial t}(\nabla \times \vec{B})=-\frac{\partial }{\partial t}\left({\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$

Term by term:

• $\frac{\partial }{\partial t}(\nabla \times \vec{B})$ — $\vec{B}$ ka swirl, time mein change hote dekha.
• (4) ke baad: ${\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$ — $\vec{E}$ ke do time-derivatives (field ki "acceleration"), ${\mu }_0{\epsilon }_0$ se scale kiya gaya.

KYUN. Yahan $\vec{B}$ hamesha ke liye gayab hota hai. Sab kuch ab $\vec{E}$ mein likha hai.

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Step 5 — Dono sides barabar karo: wave paida hoti hai

KYA. Left side (Step 3) $=$ right side (Step 4). Dono sides ke minus signs cancel ho jaate hain: $\boxed{{\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}}$

Poori cheez (3) ki jagah (4) se shuru karo, $\nabla \cdot \vec{B}=0$ use karo, aur $\vec{B}$ ke liye identical shape milti hai: ${\nabla }^2\vec{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{B}}{\partial t^2}$

KYUN yeh ek wave hai. Master Wave Equation se compare karo: ${\nabla }^{2}f=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}$

• ${\nabla }^{2}f$ — space mein shape kitna curved hai.
• $\frac{1}{v^2}{\partial }_t^{2}f$ — shape time mein kitni tezi se accelerate karta hai. Koi bhi quantity jiska space-curvature uske time-acceleration ke barabar ho (ek constant tak) ek travelling wave hoti hai. Hamaara match karta hai — toh $\vec{E}$ aur $\vec{B}$ travel karte hain.

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Step 6 — Numbers plug in karo: yeh light ki speed hai

KYA. Do lab constants daalo: $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}},{\mu }_0=1.2566\times 10^{-6}\frac{\text{T⋅m}}{\text{A}},{\epsilon }_0=8.854\times 10^{-12}\frac{\text{F}}{\text{m}}$ $c=\frac{1}{\sqrt{(1.2566\times 10^{-6})(8.854\times 10^{-12})}}\approx 3.00\times 10^8\text{m/s}$

KYUN yeh stunning hai. ${\mu }_0$ magnets aur wires se measure kiya gaya tha; ${\epsilon }_0$ static charges se measure kiya gaya tha. Kisi bhi experiment mein kabhi light involve nahi hui. Phir bhi multiply karo, invert karo, square-root lo — bahar nikalta hai measured speed of light. Usi ek coincidence ne Maxwell ko bataya: light ek electromagnetic wave hai.

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Step 7 — Geometry: E, B, k ek right-handed triad banate hain

Setup convention. Ab hum apne axes choose karte hain taaki wave $+x$ ki taraf travel kare. Matlab wavevector usi taraf point karta hai: $\vec{k}=k\hat{x}$, aur ab plain symbol $k$ sirf uski length hai. Is choice ke saath ek plane wave hai: $\vec{E}={\vec{E}}_0{e}^{i(kx-\omega t)},\vec{B}={\vec{B}}_0{e}^{i(kx-\omega t)}$

• ${\vec{E}}_0$, ${\vec{B}}_0$ — constant vector amplitudes (inke paas ek size aur ek direction dono hain). Inki magnitudes (lengths) $E_0=|{\vec{E}}_0|$ aur $B_0=|{\vec{B}}_0|$ likhi jaati hain — plain, bina arrow ke. Yeh dekho: arrow = vector, plain = uski length.
• $k$ — wavenumber ($\vec{k}$ ki length): radians of wiggle per metre. Bada $k$ = tightly squished wave.
• $\omega$ — angular frequency: radians of wiggle per second.

Teen facts nikalte hain:

1. Transverse, (1) se: $\nabla \cdot \vec{E}=0$. Hamare plane wave ke liye $\nabla \cdot \vec{E}=ik{E}_{0x}$, toh $ik{E}_{0x}=0\Rightarrow E_{0x}=0$. Field ka travel direction $\hat{x}$ ke along koi piece nahi hai. (Polarization dekho baaki sideways freedom ka kya matlab hai.)
2. $\vec{E}\perp \vec{B}$, (3) se: surviving $\vec{E}$ ko $\hat{y}$ ke along choose karo, toh $\vec{E}=E_0\hat{y}{e}^{i(kx-\omega t)}$. Phir Faraday's law ko $\nabla \times \vec{E}$ chahiye. $\nabla =ik\hat{x}$ ke saath cross product compute karo (sirf $x$-derivatives survive karti hain): $\nabla \times \vec{E}=(ik\hat{x})\times (E_0\hat{y})e^{i(kx-\omega t)}=ik{E}_0(\hat{x}\times \hat{y})e^{i(kx-\omega t)}=ik{E}_0\hat{z}{e}^{i(kx-\omega t)}$ kyunki $\hat{x}\times \hat{y}=\hat{z}$. Faraday kehta hai yeh $-{\partial }_t\vec{B}=i\omega \vec{B}$ ke barabar hai, toh $\vec{B}$ $\hat{z}$ ki taraf point karna zaroori hai. Isliye $\vec{E}\parallel \hat{y}$, $\vec{B}\parallel \hat{z}$, $\vec{k}\parallel \hat{x}$: teen mutually perpendicular arrows jisme $\hat{E}\times \hat{B}=\hat{k}$ — ek right-handed set.
3. Amplitudes, (3) se: usi equation mein lengths match karo, $k{E}_0=\omega B_0\Rightarrow \frac{E_0}{B_0}=\frac{\omega }{k}=c$, isliye $E_0=c{B}_0$ (sirf lengths).

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Step 8 — Degenerate case: koi displacement current nahi, koi wave nahi

KYA. Maano hum Maxwell's term bhool jaate hain aur vacuum mein purana law $\nabla \times \vec{B}=0$ use karte hain. Step 4 dobara karo: $\nabla \times \left(-\frac{\partial \vec{B}}{\partial t}\right)=-\frac{\partial }{\partial t}(\underset{=0}{\underbrace{\nabla \times \vec{B}}})=0$ Left side se barabar karo (Step 3): ${\nabla }^2\vec{E}=0$

KYUN yeh matter karta hai. ${\nabla }^2\vec{E}=0$ mein koi time nahi hai — yeh ek static, frozen condition hai, wave nahi. Koi travelling ripple nahi, koi $c$ nahi, koi light nahi. Toh woh ek term jo Maxwell ne insert ki thi exactly ek dead universe aur ek lit universe ke beech ka fark hai. Yeh poori story ka sabse important edge case hai.

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Worked examples (har ek re-checked)

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Ek-picture summary

Recall Poora walkthrough chhe moves mein compress kiya

(1) Teen arrows aur ek tool $\nabla$ se milo. (2) Chaar vacuum laws likho; do sources ko forbid karte hain, do $\vec{E}$ aur $\vec{B}$ ko couple karte hain. (3) Faraday's law ko do baar curl karo taaki magnetic side ko Ampère–Maxwell ke zariye swap kiya ja sake. (4) Curl-of-curl identity plus $\nabla \cdot \vec{E}=0$ left side ko $-{\nabla }^2\vec{E}$ mein collapse kar deta hai. (5) Ampère–Maxwell right side ko $-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}$ mein badal deta hai; barabar karo aur ${\nabla }^2\vec{E}={\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}$ milta hai — wave equation, jiska constant $c=1/\sqrt{{\mu }_0{\epsilon }_0}\approx 3\times 10^8$ m/s padhta hai. (6) Ek plane wave wapas daalne se geometry fix hoti hai: $\vec{E}$, $\vec{B}$, $\vec{k}$ mutually perpendicular, right-handed, $E_0=c{B}_0$ ke saath — aur Maxwell's term drop karna (Step 8) wave ko poori tarah khatam kar deta hai.