Speed of light c = 1 - √(ε₀ μ₀)

1.8.34 · Physics › Electromagnetism

WHAT — claim kya hai

Hairaan kar dene wali baat ye hai: ${\epsilon }_0$ electrostatics (Coulomb's law) se aaya tha aur ${\mu }_0$ magnetism (wires ke beech forces) se aaya tha. Light ko ek bilkul alag cheez samjha jaata tha. Maxwell ne dono ko combine kiya aur ek aisi speed nikli jo light ki measured speed ke barabar thi — ye prove karta hai ki light hai hi electromagnetism.

WHY — ye combination speed kyun deta hai?

HOW — Maxwell's equations se derive karo (scratch se)

Hum $\vec{E}$ ke liye wave equation derive karte hain aur speed padhte hain. Vacuum mein kaam karo (koi charges $\rho =0$ nahi, koi currents $\vec{J}=0$ nahi). Maxwell's equations:

$\nabla \cdot \vec{E}=0,\nabla \cdot \vec{B}=0$ $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}(\text{Faraday})$ $\nabla \times \vec{B}={\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}(\text{Ampeˋre–Maxwell})$

Step 1 — Faraday's law ka curl lo. $\nabla \times (\nabla \times \vec{E})=-\frac{\partial }{\partial t}(\nabla \times \vec{B})$ Ye step kyun? Hum sirf $\vec{E}$ mein ek akela equation chahte hain. Curl lene se hum $\nabla \times \vec{B}$ ke liye doosri Maxwell equation substitute kar sakte hain aur $\vec{B}$ ko eliminate kar sakte hain.

Step 2 — vector identity use karo $\nabla \times (\nabla \times \vec{E})=\nabla (\nabla \cdot \vec{E})-{\nabla }^2\vec{E}$. Kyunki vacuum mein $\nabla \cdot \vec{E}=0$ hai, pehla term khatam ho jaata hai: $-{\nabla }^2\vec{E}=-\frac{\partial }{\partial t}(\nabla \times \vec{B})$ Ye step kyun? Ye awkward double curl ko Laplacian ${\nabla }^2$ mein convert karta hai, jo wave equation mein appear hota hai.

Step 3 — Ampère–Maxwell substitute karo $\nabla \times \vec{B}={\mu }_0{\epsilon }_0\partial \vec{E}/\partial t$: $-{\nabla }^2\vec{E}=-\frac{\partial }{\partial t}\left({\mu }_0{\epsilon }_0\frac{\partial \vec{E}}{\partial t}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}$ Ye step kyun? Ye woh move hai jo do source-free Maxwell equations ko couple karta hai aur constants ${\mu }_0{\epsilon }_0$ ko ek saath appear karaata hai.

Step 4 — clean up karo: $\boxed{{\nabla }^2\vec{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\vec{E}}{\partial t^2}}$

Ampère–Maxwell ka curl lene pe bhi, $\vec{B}$ ke liye same equation aata hai — toh $\vec{E}$ aur $\vec{B}$ dono ek hi speed $c$ pe saath chalte hain.

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Connections

• Maxwell's Equations
• Displacement Current
• Faraday's Law of Induction
• Ampère–Maxwell Law
• Electromagnetic Waves
• Refractive Index and n = c/v
• Permittivity and Permeability
• Wave Equation

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