Visual walkthrough — Mass-energy equivalence E² = (pc)² + (mc²)²

2.3.32 · D2 · Physics › Modern Physics › Mass-energy equivalence E² = (pc)² + (mc²)²

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Step 1 — Ek chalti hui object do alag numbers carry karti hai

KYA. Ek akeli object ko imagine karo — ek chhoti si ball — jo steady speed se page par drift kar rahi hai. Do sawaal hum pooch sakte hain:

• "Ise rokna kitna mushkil hai?" → woh number momentum kehlata hai, likha jaata hai $p$.
• "Yeh kitna kuch kar sakti hai — iske andar kitni heat, light, kaam bandi hai?" → woh number energy kehlati hai, likhi jaati hai $E$.

YEH DO KYUN. Poori physics mein, jab koi cheez kisi object par push nahi karti, yeh do numbers hamesha ke liye fixed rehti hain (yeh conserved hoti hain). Toh ek free particle ki poori kahani $(E,p)$ ki pair mein rehti hai. Hum jaanna chahte hain ki yeh dono kaise related hain.

PICTURE. Ball right taraf ja rahi hai. Ek mota arrow $p$ dikhata hai (bada arrow = rokna mushkil). Ek glowing halo $E$ dikhata hai (zyada chamakdar = zyada energy).

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Step 2 — Newton ke andaaze, aur kahan woh toot jaate hain

KYA. Einstein se pehle, rules the: $p_{\text{Newton}}=\underset{\text{mass}}{\underbrace{m}}\underset{\text{speed}}{\underbrace{v}},K{E}_{\text{Newton}}=\frac{1}{2}m{v}^2$ Yahan $m$ mass hai (kitna "stuff" hai) aur $v$ speed hai. Symbol $KE$ ka matlab hai kinetic energy — energy jo sirf chalte rehne ki wajah se hai.

$KE$ ka $E$ se relation. Step 1 se yaad karo ki $E$ total energy hai: exist karne ki energy plus motion ki energy. Newton ki duniya mein "exist karne" waala hissa ignore kiya jaata tha (shuny samjha jaata tha), toh uski total energy effectively sirf $E_{\text{Newton}}=K{E}_{\text{Newton}}=\frac{1}{2}m{v}^2$ thi. Einstein ka correction (Step 3) exactly wahi missing "exist karne" waala hissa restore karna hai — rest energy $m{c}^2$ — taaki $E$ rest energy plus kinetic energy ban jaaye. Yeh split dhyaan mein rakho: yeh Step 7 mein wapas aayega.

YEH KYUN TOOT JAATE HAIN. Yeh kehte hain: tez chalo, zyada $p$ aur zyada $E$ milega, koi ceiling nahi. Lekin nature ki ek speed limit hai — light ki speed, jise $c$ kehte hain (lagbhag $300,000$ km har second). Newton ke formulas woh limit kabhi notice nahi karte. $c$ ke paas yeh galat answers dete hain. Hume aisi formulas chahiye jo $c$ ke baare mein jaanti hon.

PICTURE. Do speedometers: left waala (Newton) seedhi line mein bina kisi wall ke chadhta hai; right waala (reality) $v=c$ par ek lal wall hai jise needle kabhi cross nahi kar sakti.

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Step 3 — Stretch factor $\gamma$: relativity ka "near-lightspeed" dial

KYA. Special relativity Newton ke formulas ki jagah leta hai: $E=\underset{\text{stretch}}{\underbrace{\gamma }}\underset{\text{rest energy}}{\underbrace{m{c}^2}},p=\underset{\text{stretch}}{\underbrace{\gamma }}\underset{\text{Newton’s }p}{\underbrace{mv}}$ Naya character hai $\gamma$ (Greek letter gamma), Lorentz Factor: $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

$\gamma$ ko symbol by symbol padhna. Square root ke andar $1-\frac{v^2}{c^2}$ baitta hai. Ratio $\frac{v}{c}$ matlab hai "light speed ke fraction ke roop mein tumhari speed." Rest par, $\frac{v}{c}=0$, toh $\gamma =\frac{1}{\sqrt{1}}=1$ — koi stretch nahi. Jaise $v$, $c$ ki taraf chadhta hai, $\frac{v^2}{c^2}\to 1$, root $\to 0$, aur $\gamma \to \infty$ — infinite stretch. Woh blow-up hi speed limit ko visible banata hai.

YEH EXACT FORM KYUN — dikhaaya gaya, assert nahi kiya gaya. Hum demand karte hain ki everyday slow speeds par ($v\ll c$) naye formulas Newton ke paas wapas aayein. Ise check karne ke liye hum $\gamma$ ko chhote $\frac{v}{c}$ ke liye expand karte hain. Binomial approximation $(1-x{)}^{-1/2}\approx 1+\frac{1}{2}x$ use karke chhote $x$ ke liye (yahan $x=\frac{v^2}{c^2}$): $\gamma ={\left(1-\frac{v^2}{c^2}\right)}^{-1/2}\approx 1+\frac{1}{2}\frac{v^2}{c^2}+\cdots$ Ab ise $E=\gamma m{c}^2$ mein feed karo: $E\approx \left(1+\frac{1}{2}\frac{v^2}{c^2}\right)m{c}^2=\underset{\text{rest energy}}{\underbrace{m{c}^2}}+\underset{\text{Newton’s }KE}{\underbrace{\frac{1}{2}m{v}^2}}+\cdots$ Toh total energy $E$ ek rest piece $m{c}^2$ plus exactly Newton ke $\frac{1}{2}m{v}^2$ mein split ho jaati hai — yahi woh "exist karne waala hissa" hai jo Step 2 mein promised kiya tha, ab visible ho gaya. Isi tarah $p=\gamma mv\to mv$ jab $\gamma \to 1$. Toh $\gamma$ woh minimal fix hai jo speed limit respect karta hai lekin low speed par Newton reproduce karta hai. ✓

PICTURE. $\gamma$ ka curve $v/c$ ke against: chhoti speeds ke liye flat aur $1$ ke paas, phir $v/c=1$ par vertical wall ki taraf rocket ki tarah upar jaata hai.

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Step 4 — $E$ aur $p$ ko combine karo: $E^2-(pc{)}^2$ banao

KYA. Hamare paas do formulas hain jinmein dono mein messy $\gamma$ hai. Idea: ek aisa combination banao jahan $\gamma$ khud cancel ho jaaye. Dono ko square karke subtract karne ki koshish karo. Pehle note karo ki momentum ko $c$ se multiply karne par $pc$ milta hai, jinke units energy jaisi hain (taaki hum unhe compare kar sakein): $E^2-(pc{)}^2=(\gamma m{c}^2{)}^2-(\gamma mvc{)}^2$

$c$ ki powers kahan hain (ek careful nazar). $c$ ko dhyan se dekho taaki kuch bhi trick jaisa na lage:

• Pehla term: $(\gamma m{c}^2{)}^2={\gamma }^2{m}^2{c}^4$ — yeh $c^4$ hai.
• Doosra term: $(\gamma mvc{)}^2={\gamma }^2{m}^2{v}^2{c}^2$ — sirf $c^2$ hai.

Yeh dono ek hi power ke $c$ se shuru nahi hote. Common factor nikalane ke liye hum sabse badi power lete hain jo dono share karte hain, jo hai ${\gamma }^2{m}^2{c}^2$. Pehle term se yeh $c^2$ peeche chhodta hai (kyunki $c^4=c^2\cdot c^2$); doosre se $v^2$ chhodta hai. Toh: $E^2-(pc{)}^2=\underset{\text{shared piece }(\text{one }{c}^2\text{ of the }{c}^4)}{\underbrace{{\gamma }^2{m}^2{c}^2}}(\underset{\text{left over}}{\underbrace{c^2-v^2}})$

YEH DO SQUARES SUBTRACT KYUN KARO. Bacha hua bracket sirf $c$ aur $v$ involve karta hai — exactly woh cheez jo $\gamma$ ke andar chhipi hai. Yahi hamara mauka hai $\gamma$ ko khud cancel karaane ka.

PICTURE. Do squares area ke roop mein bane — ek bada orange square jiska side $E$ hai aur ek teal square jiska side $pc$ hai. Hum teal square ko orange waale se bahar khiskaate hain; bacha hua L-shaped area wahi hai jo hum compute kar rahe hain.

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Step 5 — Dekho $\gamma$ cancel ho jaata hai, kuch velocity-free chhodte hue

KYA. $\gamma$ ki definition ko square karo: ${\gamma }^2=\frac{1}{1-\frac{v^2}{c^2}}=\frac{c^2}{c^2-v^2}$ Ab Step 4 ke result mein substitute karo: $E^2-(pc{)}^2=\frac{c^2}{c^2-v^2}\cdot m^2{c}^2\cdot (c^2-v^2)$

YEH PUNCHLINE KYUN HAI. Factor $(c^2-v^2)$ ${\gamma }^2$ ke bottom par aur Step 4 se top par leftover ke roop mein baitha hai. Yeh identical hain, toh yeh cancel ho jaate hain: $E^2-(pc{)}^2=c^2\cdot m^2{c}^2=m^2{c}^4$ Koi $v$ bachta nahi. Answer nahi care karta ki tum kitni tez chalte ho ya kaunse frame se dekh rahe ho — yeh ek invariant hai. Aur iska value hai $m^2{c}^4=(m{c}^2{)}^2$: rest energy, squared.

PICTURE. Top aur bottom par $(c^2-v^2)$ ko plum colour mein cross out hote dikhaya gaya, clean block $m^2{c}^4$ chhodte hue.

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Step 6 — Pythagorean triangle mein rearrange karo

KYA. $E^2-(pc{)}^2=(m{c}^2{)}^2$ lo aur $(pc{)}^2$ ko doosri taraf move karo: $\boxed{E^2=(pc{)}^2+(m{c}^2{)}^2}$

TRIANGLE KYUN. Yeh exactly ${\text{hypotenuse}}^2={\text{leg}}^2+{\text{leg}}^2$ ki shape hai. Toh hum ise draw kar sakte hain: $m{c}^2$ ko seedha leg banao, $pc$ ko flat leg banao, aur $E$ ko slanted hypotenuse jo unhe join kare.

• Vertical leg $m{c}^2$ = rest energy — fixed, yeh tab bhi nahi badalta jab tum tez hote ho.
• Horizontal leg $pc$ = motion ka contribution — tez hone par badhta hai.
• Hypotenuse $E$ = total energy — hamesha kisi bhi leg se lambi.

PICTURE. Right triangle jisme teeno sides labelled hain aur do legs ke beech right-angle marked hai.

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Step 7 — Edge case A: bilkul rukka hua ($p=0$)

KYA. Agar object move nahi kar raha, toh uska momentum zero hai, toh flat leg $pc$ kuch nahi reh jaata. $E=\sqrt{0+(m{c}^2{)}^2}=m{c}^2$

YEH KYUN MATTER KARTA HAI. Triangle apne vertical leg par collapse ho jaata hai — hypotenuse aur seedha ek hi line ban jaate hain. Jo bachta hai, $E=m{c}^2$, woh duniya ki sabse famous equation hai, aur ab tum dekhte ho ki yeh sirf triangle ka $p=0$ corner hai, poori kahani nahi. (Yahi woh "rest energy" piece hai jo humne Step 3 mein alag kiya tha — yahan yeh bilkul akela khada hai kyunki koi motion nahi bachi.)

PICTURE. Triangle sideways ek single vertical stick mein flatten ho gaya: $E$ bilkul $m{c}^2$ ke upar.

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Step 8 — Edge case B: massless light ($m=0$)

KYA. Ek photon (light ka particle) ki zero rest mass hoti hai: $m=0$, toh seedha leg $m{c}^2$ gaayab ho jaata hai. $E=\sqrt{(pc{)}^2+0}=pc⟹p=\frac{E}{c}$

LIGHT MEIN MOMENTUM KYUN HAI. Newton ka $p=mv$ deta $0\times v=0$ — galat. Triangle picture mein, koi vertical leg nahi hone par, hypotenuse horizontal leg ke upar flat ho jaati hai: $E$ exactly $pc$ ke barabar hoti hai. Toh light momentum $E/c$ carry karti hai chahe usका weight kuch bhi na ho. Isliye sunlight ek solar sail ko push kar sakti hai. Ek massless particle zaroor $c$ par travel karta hai — koi rest leg nahi hai jo "rest frame" define kare.

PICTURE. Triangle doosri taraf flatten hokar ek single horizontal stick ban jaata hai: $E$ bilkul $pc$ par.

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Step 9 — Limiting behaviour: har speed par triangle

KYA. Triangle ko morph hote dekho jaise speed $0$ se $c$ jaati hai:

• $v=0$: lamba paatla triangle (sab vertical) → $E=m{c}^2$.
• moderate $v$: balanced triangle, dono legs present → $E=\sqrt{(pc{)}^2+(m{c}^2{)}^2}$.
• $v\to c$: flat leg $pc\to \infty$ (kyunki $\gamma \to \infty$), toh hypotenuse $E\to \infty$.

KUCH MASSIVE $c$ TAK KYUN NAHI PAHUNCH SAKTA. Jaise flat leg infinity ki taraf bhaagti hai, $E$ ko bhi jaana padta hai — tumhe ek massive object ko poori taraf $c$ tak push karne ke liye infinite energy chahiye hogi. Sirf massless case (bilkul koi vertical leg nahi) pehle se hi light-speed line par exist karta hai.

PICTURE. Teen triangles side by side jo ek hi fixed vertical leg $m{c}^2$ share karte hain, flat leg zyada se zyada lambi hoti jaati hai.

Recall Pehle predict karo phir verify karo

Q: Speed ek chhoti value se double ho jaati hai — kya $E$ bhi roughly double ho jaati hai? ::: Nahi — low speed par extra energy $\frac{1}{2}m{v}^2$ hai, jo chaar guna ho jaati hai; $c$ ke paas yeh bahut tezi se explode karti hai. Triangle ki flat leg nonlinearly badhti hai kyunki $\gamma$ bhi aisa hi karta hai.

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Ek picture mein summary

Is page ki sab kuch ek single frame mein: do starting numbers $(E,p)$ → $\gamma$ dial → cancellation → master triangle → uske teen limits.

Recall Feynman retelling of the whole walkthrough

Ek chalti hui cheez do locked-in numbers carry karti hai: ise rokna kitna mushkil hai ($p$) aur yeh kitna kuch kar sakti hai ($E$). Newton ke simple andaaze cosmic speed limit $c$ ko ignore karte hain, toh relativity unhe ek "near-lightspeed stretch dial" $\gamma$ se multiply karti hai jo slow hone par $1$ hoti hai aur light speed approach karne par blow up karti hai. Us dial ko slow speeds ke liye expand karo aur tum literally dekhte ho Einstein ki total energy "rest energy $m{c}^2$ plus Newton ki $\frac{1}{2}m{v}^2$" mein split ho jaati hai — purani physics nayi ke andar chhipi thi. Ab yeh trick hai: energy ko square karo, momentum-times-$c$ ko square karo, aur subtract karo. Stretch dial ek term ke top par aur doosre ke bottom par aata hai aur khud ko completely cancel kar leta hai, ek aisa number chhodta hai jisme koi speed nahi — har kisi ke liye same — aur woh number sirf rest energy squared hai, $(m{c}^2{)}^2$. Momentum term ko across slide karo aur tumhe milta hai $E^2=(pc{)}^2+(m{c}^2{)}^2$, jo exactly Pythagoras hai: rest energy seedha khadi hai (kabhi nahi badlti), momentum sideways stretch karta hai (speed ke saath badhta hai), aur total energy woh diagonal hai jo unhe join karti hai. Chalna band karo aur sideways leg gaayab ho jaata hai — tumhe milta hai $E=m{c}^2$. Saari mass hata do aur seedha leg gaayab ho jaata hai — tumhe milti hai light: $E=pc$, weightless lekin phir bhi push kar rahi hai. Mass ke saath light speed tak pahunchne ki koshish karo aur sideways leg infinity ki taraf bhaag jaata hai, $E$ ko apne saath kheeench ke — isliye tum kabhi nahi pahunch sakte.

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Connections

• Parent · Hinglish version
• Special Relativity — jahan se speed limit $c$ aur $\gamma$ aate hain
• Lorentz Factor — Steps 3 & 5 ka $\gamma$ dial
• Photon Momentum — Step 8 ka $m=0$ triangle
• Relativistic Kinetic Energy — bacha hua $E-m{c}^2=(\gamma -1)m{c}^2$
• Four-Momentum — woh deeper object jiska "length" invariant hai
• Nuclear Binding Energy — nuclei par apply kiya gaya invariant $m{c}^2$

Concept Map