Playground
Interactive mass-to-energy converter: drag the mass slider and watch the energy bar explode upward via E = mc².
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Energyoutput Rest energy equivalent of the mass | J | M·L²·T⁻² | 0 – 1000000000000000000 | |
| Rest mass Invariant mass of the object | kg | M | 0 – 1 | |
| Speed of light Speed of light in vacuum | m/s | L·T⁻¹ | 299792458 – 299792458 |
Deep dive
Derivation
Starting from Einstein's 1905 special relativity postulates, consider a body emitting two light pulses in opposite directions. In the rest frame the body loses energy L. Analyzing in a boosted frame using the Lorentz transformation, the kinetic energy change equals L(γ − 1) ≈ Lv²/(2c²) for small v. Comparing with ½Δmv² shows the body's inertial mass decreased by Δm = L/c², hence E = mc².
Experimental verification
Confirmed by precision mass spectrometry of nuclear reactions (Cockcroft–Walton 1932), annihilation of electron-positron pairs producing gamma rays with energy 2 × 0.511 MeV, and modern measurements of atomic mass differences in nuclear fission and fusion reactions agreeing with released energy to parts per billion.
Common misconceptions
- E = mc² does not mean mass can be 'converted' into energy as if they are different substances; mass IS a form of energy.
- The equation applies to rest mass only. The concept of 'relativistic mass' that increases with speed is outdated and discouraged in modern physics.
- E = mc² does not by itself explain how to release nuclear energy — that requires understanding nuclear binding energies and reaction mechanisms.
Real-world applications
- Nuclear fission reactors (mass defect of uranium-235 fission products powers electrical grids)
- PET medical imaging (positron-electron annihilation produces 511 keV gamma rays)
- Stellar energy production (the Sun converts ~4.3 million tonnes of mass to energy per second via pp-chain fusion)
- Nuclear weapons (both fission and fusion devices)
Worked examples
Energy in a paperclip
Given:
- m:
- 0.001
- c:
- 299792458
Find: E
Solution
E = mc² = 0.001 × (299792458)² = 8.988 × 10¹³ J
Energy from electron-positron annihilation
Given:
- m:
- 1.82187674e-30
- c:
- 299792458
Find: E
Solution
E = mc² = 2 × 9.1093837015 × 10⁻³¹ × (299792458)² = 1.637 × 10⁻¹³ J = 1.022 MeV
Scenarios
What if…
- scenario:
- What if the speed of light were 10× smaller?
- answer:
- E = mc² would give 100× less energy per unit mass, making nuclear energy far less potent and stars unable to shine as brightly.
- scenario:
- What if we could convert 1 kg of matter entirely to energy?
- answer:
- We would release ~9 × 10¹⁶ J, equivalent to about 21.5 megatons of TNT — roughly the yield of the largest thermonuclear weapons.
- scenario:
- What if the object is moving?
- answer:
- E = mc² gives only the rest energy. The total energy becomes E = γmc², where γ = 1/√(1 − v²/c²), or equivalently E² = (pc)² + (mc²)².
Limiting cases
- condition:
- m → 0
- result:
- E → 0
- explanation:
- A massless object at rest carries no energy; massless particles like photons are never at rest.
- condition:
- m → ∞
- result:
- E → ∞
- explanation:
- Infinite mass would require infinite energy — there is no upper bound in principle.
- condition:
- m = 1 kg
- result:
- E ≈ 9.0 × 10¹⁶ J
- explanation:
- Even 1 kg of matter contains roughly 21.5 megatons of TNT equivalent energy.
Context
Albert Einstein · 1905
Einstein derived this relation in his special relativity paper 'Does the inertia of a body depend upon its energy-content?'
Hook
How much energy is locked inside a paperclip?
A 1 g paperclip is fully converted to energy. How many joules are released?
Dimensions:
- lhs:
- E → [M·L²·T⁻²]
- rhs:
- m·c² → [M]·[L·T⁻¹]² = [M·L²·T⁻²]
- check:
- Both sides are [M·L²·T⁻²] = Joules. ✓
Validity: Valid for any object at rest in the chosen reference frame. For moving objects, use the full relativistic energy-momentum relation E² = (pc)² + (mc²)².