Intuition The Big Picture (WHY this exists)Newton said energy and momentum were separate, simple things (K E = 1 2 m v 2 KE=\frac12 mv^2 K E = 2 1 m v 2 p = m v p=mv p = m v two faces of one object — the energy-momentum 4-vector. Just like the length of a stick doesn't change when you rotate it, the "length" of the energy-momentum vector is invariant : it always equals m c 2 mc^2 m c 2 is the rest mass. The whole formula E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2 the Pythagorean theorem for this 4-vector .
E E E total relativistic energy (rest + kinetic). Units: joules.p p p p = γ m v p=\gamma m v p = γ m v m m m rest mass (invariant mass — same in every frame).c c c m c 2 mc^2 m c 2 rest energy .γ = 1 1 − v 2 / c 2 \gamma = \dfrac{1}{\sqrt{1-v^2/c^2}} γ = 1 − v 2 / c 2 1
Intuition Why it's a Pythagorean theoremThink of a right triangle: hypotenuse E E E p c pc p c m c 2 mc^2 m c 2 m c 2 mc^2 m c 2 fixed (rest energy), the leg p c pc p c E E E
Worked example Example 1 — Rest energy of an electronm = 9.11 × 10 − 31 m=9.11\times10^{-31} m = 9.11 × 1 0 − 31 E = m c 2 = 9.11 × 10 − 31 × ( 3 × 10 8 ) 2 = 8.2 × 10 − 14 J E=mc^2 = 9.11\times10^{-31}\times(3\times10^8)^2 = 8.2\times10^{-14}\text{ J} E = m c 2 = 9.11 × 1 0 − 31 × ( 3 × 1 0 8 ) 2 = 8.2 × 1 0 − 14 J ÷ 1.6 × 10 − 19 = 0.511 \div 1.6\times10^{-19} = 0.511 ÷ 1.6 × 1 0 − 19 = 0.511 Why this step? Dividing by the electron charge converts joules to electron-volts, the natural unit in particle physics.
Worked example Example 2 — Photon momentumA photon has energy E = 2.0 E=2.0 E = 2.0 p p p m = 0 m=0 m = 0 p = E / c p = E/c p = E / c p = 2.0 × 1.6 × 10 − 19 3 × 10 8 = 1.07 × 10 − 27 kg⋅m/s p = \frac{2.0\times1.6\times10^{-19}}{3\times10^8} = 1.07\times10^{-27}\ \text{kg·m/s} p = 3 × 1 0 8 2.0 × 1.6 × 1 0 − 19 = 1.07 × 1 0 − 27 kg⋅m/s Why use E = p c E=pc E = p c Photon is massless, so the full formula collapses to this — no γ \gamma γ v v v c c c
Worked example Example 3 — Total energy from momentum and massA proton (m c 2 = 938 mc^2 = 938 m c 2 = 938 p c = 500 pc = 500 p c = 500 E E E K E KE K E E = ( 500 ) 2 + ( 938 ) 2 = 250000 + 880000 ≈ 1063 MeV E=\sqrt{(500)^2+(938)^2}=\sqrt{250000+880000}\approx1063\text{ MeV} E = ( 500 ) 2 + ( 938 ) 2 = 250000 + 880000 ≈ 1063 MeV K E = E − m c 2 = 1063 − 938 = 125 KE = E - mc^2 = 1063-938 = 125 K E = E − m c 2 = 1063 − 938 = 125 Why subtract m c 2 mc^2 m c 2 Total energy includes rest energy; kinetic is whatever's left over after removing the rest part.
Common mistake "Mass increases with speed, so usem r e l = γ m m_{rel}=\gamma m m r e l = γ m Why it feels right: Old textbooks defined "relativistic mass" γ m \gamma m γ m E = m r e l c 2 E=m_{rel}c^2 E = m r e l c 2 The fix: In E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2 m m m invariant rest mass — it never changes. If you plug γ m \gamma m γ m γ \gamma γ
E = m c 2 E=mc^2 E = m c 2 Why it feels right: It's the most famous equation, so it must be universal.
The fix: E = m c 2 E=mc^2 E = m c 2 rest energy (when p = 0 p=0 p = 0 E = ( p c ) 2 + ( m c 2 ) 2 E=\sqrt{(pc)^2+(mc^2)^2} E = ( p c ) 2 + ( m c 2 ) 2 E = γ m c 2 E=\gamma mc^2 E = γ m c 2
Common mistake "Photons have no momentum becausep = m v p=mv p = m v m = 0 m=0 m = 0 Why it feels right: Newtonian p = m v p=mv p = m v The fix: Newtonian p = m v p=mv p = m v p = E / c p=E/c p = E / c does push (radiation pressure, solar sails).
Recall Feynman: Explain to a 12-year-oldImagine energy is like a ladder leaning against a wall. The wall's height is the "rest energy" — the energy a thing has just for existing , even sitting still. How far the ladder's foot is pulled out is the "motion energy" (momentum). The length of the ladder is the total energy. If you pull the foot out more (more speed), the ladder gets longer (more energy), but the wall's height never changes — that's the unchanging mass. Light is like a ladder with no wall at all: pure motion energy, weightless, but it still leans and pushes!
Mnemonic Remember the structure"Energy is the Hypotenuse" — draw a right triangle with E E E m c 2 mc^2 m c 2 p c pc p c E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2
Worked example Predict before computingQ: As a particle's speed → c \to c → c E E E Forecast: γ → ∞ \gamma\to\infty γ → ∞ E → ∞ E\to\infty E → ∞ c c c Verify: E = γ m c 2 E=\gamma mc^2 E = γ m c 2 γ = 1 / 1 − v 2 / c 2 → ∞ \gamma=1/\sqrt{1-v^2/c^2}\to\infty γ = 1/ 1 − v 2 / c 2 → ∞ v → c v\to c v → c m c 2 = 0 mc^2=0 m c 2 = 0 at c c c
What is the full energy-momentum relation? E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2 In E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2 m m m The invariant rest mass (same in all frames)
What does the formula reduce to for a particle at rest? E = m c 2 E=mc^2 E = m c 2 (rest energy)
What does it reduce to for a massless particle (photon)? Why is E 2 − ( p c ) 2 E^2-(pc)^2 E 2 − ( p c ) 2 It equals
m 2 c 4 m^2c^4 m 2 c 4 , independent of velocity/frame
How is the formula like Pythagoras? E E E is hypotenuse; legs are
p c pc p c (motion) and
m c 2 mc^2 m c 2 (rest)
Express momentum of a photon in terms of its energy. Why can't a massive particle reach speed c c c γ → ∞ \gamma\to\infty γ → ∞ as
v → c v\to c v → c , requiring infinite energy
How do you get kinetic energy from total energy? K E = E − m c 2 = ( γ − 1 ) m c 2 KE = E - mc^2 = (\gamma-1)mc^2 K E = E − m c 2 = ( γ − 1 ) m c 2 Relativistic momentum definition? p = γ m v p=\gamma m v p = γ m v with
γ = 1 / 1 − v 2 / c 2 \gamma=1/\sqrt{1-v^2/c^2} γ = 1/ 1 − v 2 / c 2
Special Relativity — origin of γ \gamma γ Lorentz Factor — the γ \gamma γ E E E p p p Photon Momentum — the m = 0 m=0 m = 0 Relativistic Kinetic Energy — ( γ − 1 ) m c 2 (\gamma-1)mc^2 ( γ − 1 ) m c 2 Four-Momentum — the deeper structure p μ = ( E / c , p ⃗ ) p^\mu=(E/c,\vec p) p μ = ( E / c , p ) Nuclear Binding Energy — mass defect Δ m c 2 \Delta m c^2 Δ m c 2
Right triangle: E hypotenuse
Intuition Hinglish mein samjhoDekho yaar, ye formula E 2 = ( p c ) 2 + ( m c 2 ) 2 E^2=(pc)^2+(mc^2)^2 E 2 = ( p c ) 2 + ( m c 2 ) 2 Pythagoras theorem hi hai, bas energy-momentum ke liye. Socho ek right-angle triangle banao: hypotenuse hai total energy E E E m c 2 mc^2 m c 2 p c pc p c
Iski khaas baat ye hai ki E 2 − ( p c ) 2 = m 2 c 4 E^2-(pc)^2 = m^2c^4 E 2 − ( p c ) 2 = m 2 c 4 constant rehta hai — chahe aap kisi bhi frame se dekho, chahe particle kitni bhi tez chal raha ho. Isliye ise invariant mass kehte hain. Jab particle rest mein ho (p = 0 p=0 p = 0 E = m c 2 E=mc^2 E = m c 2 m = 0 m=0 m = 0 E = p c E=pc E = p c
Exam ke liye 80/20 yaad rakho: do limits — rest pe E = m c 2 E=mc^2 E = m c 2 E = p c E=pc E = p c E = m c 2 E=mc^2 E = m c 2 rest energy hai, moving particle ka total energy nahi. Moving particle ke liye poora formula ya E = γ m c 2 E=\gamma mc^2 E = γ m c 2 p = E / c p=E/c p = E / c p = m v p=mv p = m v m = 0 m=0 m = 0
