Mass-energy equivalence E² = (pc)² + (mc²)²: Question bank

2.3.32 · D5 · Physics › Modern Physics › Mass-energy equivalence E² = (pc)² + (mc²)²

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

True or false — justify

The invariant

E^{2} - (p c)^{2}

ki value har reference frame mein same hoti hai.

True. Ye

m^{2} c^{4}

ke barabar hai, aur rest mass

m

definition se frame-independent hai — alag-alag observers

E

aur

p

ko alag-alag dekhte hain, lekin combination

E^{2} - (p c)^{2}

hamesha same number par aata hai, jaise kisi stick ki length rotation se nahi badalti.

Ek photon ka mass zero hai, isliye uski energy bhi zero hai.

False.

m = 0

hone par triangle ka vertical leg gayab ho jaata hai aur formula

E = p c

ban jaata hai, toh uski energy poori tarah momentum se aati hai — bachta hua horizontal leg hi energy hai.

Agar do particles ki total energy

E

same ho, toh unka momentum bhi same hoga.

False. Same

E

(hypotenuse) alag-alag

p c

(horizontal leg) ke saath ho sakta hai, bas vertical leg

m c^{2}

alag honi chahiye — ek bhaari slow particle aur ek halka fast particle ek hi energy value share kar sakte hain.

E = γ m c^{2}

aur

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

do alag formulas hain jo kabhi-kabhi disagree karte hain.

False. Ye same relation ke do tarike hain;

p = γ m v

ko square-root form mein substitute karke simplify karne par exactly

γ m c^{2}

milta hai. Dono mein se koi "zyada correct" nahi hai.

Relativistic particle ki kinetic energy

\frac{1}{2} m v^{2}

hoti hai.

False. Ye low-speed Newtonian approximation hai. Sahi value hai

K E = E - m c^{2} = (γ - 1) m c^{2}

, jo

v \to c

jaate-jaate infinity ki taraf badhti hai, jabki

\frac{1}{2} m v^{2}

finite rehti hai aur galat hoti hai.

Massless particle ke liye triangle ka rest-energy leg zero length ka hota hai.

True.

m c^{2} = 0

matlab vertical leg gayab, toh hypotenuse horizontal leg par aa jaata hai:

E = p c

. Geometrically "triangle" ek akeli horizontal line mein degenerate ho jaata hai.

Particle ki speed double karne se uska momentum double ho jaata hai.

False.

p = γ m v

hai, aur

γ

khud

v

ke saath badhta hai, toh

c

ke paas ek chhoti si speed increase bhi momentum mein badi increase karti hai — momentum ki koi ceiling nahi hai chahe speed

c

par cap ho.

Rest energy

m c^{2}

badal sakti hai agar particle ko kaafi push karo.

False. Push karne se kinetic energy add hoti hai (hypotenuse aur

p c

leg lambi hoti hai) lekin

m c^{2}

leg ek elementary particle ke liye fixed rehti hai — yahi fixedness "invariant mass" ka matlab hai.

Spot the error

"Ek photon ka momentum zero hota hai kyunki

p = m v

aur uska mass

0

hai."

Error Newtonian

p = m v

use karna hai, jo kisi bhi relativistic cheez ke liye fail karta hai. Massless particle ke liye sahi relation

p = E / c

hai, jo nonzero momentum deta hai — isliye light solar sails par radiation pressure daalti hai.

"Jaise

v \to c

, energy

E \to m c^{2}

kyunki yahi famous formula hai."

Ulta hai. Jaise

v \to c

γ \to \infty

toh

E = γ m c^{2} \to \infty

. Value

m c^{2}

minimum (rest) energy hai

v = 0

par, high speed ki limit nahi.

"Fast particle ke liye formula use karne ke liye,

m

ko relativistic mass

γ m

se replace karo."

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

mein

m

pehle se rest mass hai;

γ m

daalne se

γ

double count ho jaata hai (kyunki

p

aur

E

mein already

γ

hai) aur identity toot jaati hai. Modern convention: "mass" matlab hamesha rest mass.

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

, toh main bas

p c

aur

m c^{2}

add karke

E = p c + m c^{2}

pa sakta hoon."

Squares addition par distribute nahi hote —

a^{2} + b^{2} \neq = a + b

. Pythagoras legs ke squares add karta hai, aur seedha sum

E

ko overestimate karta hai, sirf un degenerate cases ko chhodkar jahan ek leg zero ho.

"Formula sirf particles ke liye kaam karta hai, gas ke box jaisi systems ke liye nahi."

Ye kisi bhi isolated system ke liye kaam karta hai:

E

total energy hai,

p

total momentum, aur

m

system ka invariant mass — jisme uske parts ki internal kinetic aur binding energies bhi shamil hain. Ek garam box thande box se thoda zyada weight karta hai.

"Kyunki

E

hypotenuse hai,

E

hamesha

p c

aur

m c^{2}

dono se bada hota hai."

Bilkul sahi,

E \geq p c

aur

E \geq m c^{2}

sabhi cases mein hold karte hain (hypotenuse kabhi leg se chota nahi hota). Strict inequality

E > p c

ke liye

m c^{2} > 0

chahiye, aur

E > m c^{2}

ke liye

p c > 0

; equality

E = p c

exactly tab hoti hai jab

m c^{2} = 0

(massless), aur

E = m c^{2}

exactly tab jab

p c = 0

(at rest). Toh "strictly dono se bada" un do boundaries par fail ho jaata hai.

Spacetime geometry time-like aur space-like parts ke beech minus sign use karta hai (Minkowski metric), ordinary space ki tarah nahi jo plus use karta hai. Wahi minus ise ek sachi frame-invariant banata hai jo

(m c^{2})^{2}

ke barabar hai.

E

mein se

m c^{2}

subtract karne par kinetic energy kyun milti hai, kuch aur kyun nahi?

E

rest energy aur motion energy dono ko bundle karta hai;

m c^{2}

pure "existence" part hai jo rest mein bhi present hai. Jo bhi bachta hai,

E - m c^{2}

, wo energy hai jo particle ke paas purely uski motion ki wajah se hai — kinetic energy ki yahi definition hai,

(γ - 1) m c^{2}

Edge cases

E

kya hoga jab

p = 0

aur

m = 0

dono ho?

Tab dono legs zero hain, toh

E = 0

— koi particle hai hi nahi. Ek massless cheez rest mein koi energy carry nahi karti aur ek real particle ke roop mein exist nahi kar sakti; light kabhi rest mein nahi hoti.

Aadhi speed of light par move karte particle ke liye energy–momentum triangle kaisa dikhega?

Dono legs present aur nonzero hain: fixed vertical

m c^{2}

aur ek modest horizontal

p c = γ m v c

. Kyunki

v < c

hai toh horizontal leg ultra-relativistic particle se chhotaa hai, isliye

E

sirf thoda sa

m c^{2}

se zyada hai.

Limit

p c ≫ m c^{2}

(ultra-relativistic) mein

E

kya approach karta hai?

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

, kyunki chhota fixed leg bade horizontal leg ke samne negligible ho jaata hai. Bahut fast massive particle almost photon jaisa behave karta hai — uska rest mass barely matter karta hai.

Kya

p c

is formula mein kabhi negative ho sakta hai?

Nahi —

p

kisi bhi direction mein point kar sakta hai, lekin formula magnitude

∣ p ∣

use karta hai, aur

(p c)^{2}

ek square hai, toh ye direction se befikar triangle mein non-negative length contribute karta hai.

Kya aisa koi frame hai jismein massive particle ka momentum zero ho?

Haan — uske rest frame mein, jahan

v = 0

p = 0

, aur

E = m c^{2}

hota hai. Har massive particle ka aisa frame hota hai. Massless particle ka nahi hota, kyunki wo har frame mein

c

par chalta hai aur kabhi rest mein nahi aaya ja sakta.

Opposite directions mein uda rahe do photons ke system ka invariant mass kya hoga?

Unka total momentum cancel ho jaata hai (

p_{tot} = 0

) jabki energies add ho jaati hain, toh

E^{2} - (p c_{tot})^{2} = E_{tot}^{2} > 0

— system ka nonzero invariant mass hota hai chahe har photon massless ho. Zero net momentum ke saath trap hui energy, mass ke roop mein dikhti hai.

Connections

Parent topic — derivation aur worked numbers
Lorentz Factor — kyun $γ \to \infty$ speed ko $c$ par cap karta hai
Photon Momentum — $m = 0$ , $E = p c$ edge case
Relativistic Kinetic Energy — $(γ - 1) m c^{2}$ leftover
Four-Momentum — jahan se minus-sign invariant aata hai
Nuclear Binding Energy — systems jinка invariant mass parts ke sum se alag hota hai
Special Relativity — neeche ki Minkowski geometry

Question bank — Mass-energy equivalence E² = (pc)² + (mc²)²

True or false — justify

Spot the error

Why questions

Edge cases

Connections