Relativistic momentum p = γmv — Hinglish notes

2.3.31 · Physics › Modern Physics

YEH kya kehta hai

KAISE derive karte hain (first principles se)

Step 1 — $d t$ aur $d τ$ ko relate karo. Moving clock slow chalta hai (time dilation): $d t = γ d τ \Rightarrow \frac{d t}{d τ} = γ$ Yeh step kyun? Particle ki apni clock $d τ$ tick karti hai; lab dekhta hai $d t = γ d τ$ guzar jaata hai. Yeh ek hi relativistic ingredient hai jo humein chahiye.

Step 2 — Lab time ki jagah proper time se momentum define karo. $p \equiv m \frac{d x}{d τ}$ Yeh step kyun? Kyunki $d x$ (ek 4-vector ka space part) ko invariant $d τ$ se divide karne par yeh ek proper 4-vector ki tarah transform karta hai — isliye ek frame mein conservation ka matlab hai sab frames mein conservation.

Step 3 — Jis velocity ko hum actually measure karte hain uspe convert karo. Chain rule use karo: $p = m \frac{d x}{d τ} = m \frac{d x}{d t} \cdot \frac{d t}{d τ} = m v γ$ $p = γ m v$ Yeh step kyun? $d x / d t = v$ ordinary lab velocity hai; $d t / d τ = γ$ Step 1 se. Extra $γ$ hi poora relativistic correction hai.

Energy connection (ek extra line of insight)

Worked examples

Common mistakes (Steel-man + fix)

Flashcards

Relativistic momentum formula kya hai?

p = γ m v = \frac{m v}{1 - v ^{2} / c ^{2}}

Lorentz factor

γ

define karo.

γ = 1/ 1 - v^{2} / c^{2}

, hamesha

\geq 1

Momentum ko relativistically kyun redefine karna padta hai?

Taaki Lorentz transformation ke baad har inertial frame mein momentum conserved rahe.

Clean 4-momentum ke liye hum kis invariant time se differentiate karte hain?

Proper time

d τ

(particle ke saath chalne wali clock).

Lab time aur proper time ka relation?

d t = γ d τ

v \to c

par

p

kya banta hai?

γ \to \infty

, toh

p \to \infty

—

c

tak nahi pahuncha ja sakta.

Energy–momentum invariant?

E^{2} = (p c)^{2} + (m c^{2})^{2}

Photon (

m = 0

) ka momentum?

p = E / c

Modern convention mein

γ

kahan hota hai?

Velocity/kinematics par, mass par nahi (mass invariant rest mass hai).

p = γ m v

ki Newtonian limit?

γ \to 1

toh

p \to m v

Recall Feynman: 12-year-old ko explain karo

Socho tum ek toy car ko push kar rahe ho. Normally, double push karo toh double speed milti hai. Lekin yeh ek magic car hai: jitni zyada fast already chal rahi hai, dhakka dene mein utni hi bhari lagti hai, aur light speed ke paas toh almost infinitely sluggish lagti hai — toh tum kabhi bhi light speed tak nahi pahunch sakte. Yeh "magic stubbornness" ek number hai jise $γ$ kehte hain. Slow car: $γ = 1$ , ekdum ordinary. Super-fast car: $γ$ bahut bada. Momentum = $γ \times m \times v$ , toh stubbornness tumhari normal push-amount ko multiply kar deti hai.

Connections

Lorentz factor and time dilation — $d t = γ d τ$ step ka source
Mass-energy equivalence E = γmc² — is momentum ka energy partner
Energy-momentum four-vector — kyun dono saath transform hote hain
Why nothing exceeds the speed of light — $γ \to \infty$ ka direct consequence
Newtonian momentum p = mv — low-speed limit
Photon momentum and radiation pressure — $m = 0$ case