Playground

Bar chart comparing rest mass energy for various particles. Slider picks mass in kg; shows energy in J and MeV.

Variables

SymbolNameSIDimensionRange
EEEnergyoutput
Rest energy equivalent of the mass
JM·L²·T⁻²1e-15 – 100000000000000000000
mmRest mass
Invariant mass of the object
kgM1e-30 – 1000
ccSpeed of light
Speed of light in vacuum (constant)
m/sL·T⁻¹299792458 – 299792458

Deep dive

Derivation
Starting from the relativistic momentum p = γmv and total energy E = γmc², compute E² − (pc)² = γ²m²c⁴(1 − v²/c²) = m²c⁴, giving E² = (pc)² + (mc²)². Setting p = 0 (rest frame) yields E = mc².
Experimental verification
Confirmed countless times: nuclear binding energies match the mass defect Δm·c² to <0.01%. Cockcroft–Walton (1932) split lithium-7 with protons and verified the energy released equals the mass deficit. Pair production e⁺e⁻ from photons (Anderson 1932) and PET scans rely on this daily.
Common misconceptions
  • E = mc² describes only rest energy; the full energy of a moving particle includes kinetic energy via the Lorentz factor.
  • Mass is not literally 'converted' to energy — rather, the system's invariant mass changes when energy leaves (e.g., a hot object weighs slightly more than a cold one).
  • It does not say 'matter and energy are the same thing'; mass is one specific property and energy is another, related by c².
Real-world applications
  • Nuclear power plants — uranium fission releases ~0.1% of fuel mass as energy.
  • Stellar fusion — the Sun converts ~4 million tons of mass to energy every second.
  • PET medical imaging — positron–electron annihilation produces 511 keV gamma rays.
  • Particle accelerators — LHC creates new massive particles from collision kinetic energy.

Worked examples

Energy in 1 gram of mass

Given:
m:
0.001
c:
299792458
Find: E
Solution

E = mc² = 0.001 × (3×10⁸)² ≈ 9×10¹³ J ≈ 21 kilotons of TNT

Rest energy of an electron

Given:
m:
9.1093837e-31
c:
299792458
Find: E (in MeV)
Solution

E = mc² ≈ 8.187×10⁻¹⁴ J ≈ 0.511 MeV

Scenarios

What if…
  • scenario:
    What if we could convert all of a 70 kg human's mass to energy?
    answer:
    E = 70 × c² ≈ 6.3×10¹⁸ J — about 1500 megatons of TNT, roughly 30× the largest nuclear weapon ever detonated.
  • scenario:
    What if c were only 1000 m/s?
    answer:
    Mass-energy equivalence would be 9×10¹⁰ times weaker. Stars couldn't fuse, atoms couldn't bind, and chemistry would dwarf nuclear energy — the universe would be unrecognizable.
  • scenario:
    What if a hot cup of coffee weighed itself?
    answer:
    It would be heavier by Δm = ΔE/c². Heating 250 g of water by 80 K adds ~84 kJ, increasing mass by ~9×10⁻¹³ kg — far below any scale's resolution but real.
Limiting cases
  • condition:
    m → 0
    result:
    E → 0
    explanation:
    A massless object has no rest energy (photons carry only kinetic/momentum energy via E = pc).
  • condition:
    v → c (relativistic)
    result:
    E → ∞
    explanation:
    Total energy E = γmc² diverges as a massive object approaches the speed of light — that is why c is unreachable.
  • condition:
    v ≪ c
    result:
    E ≈ mc² + ½mv²
    explanation:
    Low-velocity expansion of γ recovers Newtonian kinetic energy on top of the rest energy.

Context

Albert Einstein · 1905

Derived in Einstein's 'Annus Mirabilis' paper on special relativity, showing that mass and energy are interchangeable.

Hook

How much energy is locked inside a single paperclip?

A paperclip has a mass of about 1 gram. If all of its mass were converted to energy, how much energy would be released?

Dimensions: [E] = [m]·[c]² → M·(L·T⁻¹)² = M·L²·T⁻² ✓ (energy)
Validity: Universally valid in special and general relativity for any closed system. Rest-mass form applies to objects in their rest frame; for moving particles use the full relation E² = (pc)² + (mc²)².

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