Mass–Energy Equivalence

Equivalent forms

E^2 = (pc)^2 + (mc^2)^2

Two symbols, one constant — the equation that explains why stars shine and bombs detonate.

Symbol	Name	SI unit	Dimension	Range
$E$	Energyoutput Rest energy equivalent of the mass	J	M·L²·T⁻²	0 – 1000000000000000000
$m$	Rest mass Invariant mass of the object	kg	M	0 – 1
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Natural
$E$	J	erg	MeV
$m$	kg	g	MeV/c²
$c$	m/s	cm/s	1

Where it holds

Valid for any object at rest in the chosen reference frame. For moving objects, use the full relativistic energy-momentum relation

E^{2} = (pc)^{2} + (mc^{2})^{2}

.

Dimensional analysis

E \to [M\cdot L^{2}\cdot T^{-2}]

m\cdot c^{2} \to [M]\cdot [L\cdot T^{-1}]^{2} = [M\cdot L^{2}\cdot T^{-2}]

Both sides are

[M\cdot L^{2}\cdot T^{-2}]

= Joules.

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Discovery

Albert Einstein · 1905

Einstein derived this relation in his special relativity paper 'Does the inertia of a body depend upon its energy-content?'

Try this

How much energy is locked inside a paperclip?

A 1 g paperclip is fully converted to energy. How many joules are released?

Research status: stable

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 $\sim 4.3$ million tonnes of mass to energy per second via pp-chain fusion)
Nuclear weapons (both fission and fusion devices)

Common misconceptions

$E = mc^{2}$ 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 The concept of 'relativistic mass' that increases with speed is outdated and discouraged in modern physics.
$E = mc^{2}$ does not by itself explain how to release nuclear energy — that requires understanding nuclear binding energies and reaction mechanisms.

Experimental verification

Confirmed by precision mass spectrometry of nuclear reactions (Cockcroft–Walton 1932), annihilation of electron-positron pairs producing gamma rays with energy

2 \times 0.511\,\mathrm{MeV}

, and modern measurements of atomic mass differences in nuclear fission and fusion reactions agreeing with released energy to parts per billion.

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(\gamma - 1) \approx Lv^{2}/(2c^{2})

for small v.

Comparing with

\tfrac{1}{2}\Delta mv^{2}

shows the body's inertial mass decreased

by \Delta m = L/c^{2}

, hence

E = mc^{2}

.

Limiting cases

m \to 0

⟶

E \to 0

A massless object at rest carries no energy; massless particles like photons are never at rest.

m \to \infty

⟶

E \to \infty

Infinite mass would require infinite energy — there is no upper bound in principle.

m = 1\,\mathrm{kg}

⟶

E \approx 9.0 \times 10^{16} J

Even 1 kg of matter contains roughly 21.5 megatons of TNT equivalent energy.

What if…

What if the speed of light were

10\times

smaller?

$E = mc^{2}$ would give $100\times$ less energy per unit mass, making nuclear energy far less potent and stars unable to shine as brightly.

What if we could convert 1 kg of matter entirely to energy?

We would release $\sim 9 \times 10^{16} J$ , equivalent to about 21.5 megatons of TNT — roughly the yield of the largest thermonuclear weapons.

What if the object is moving?

$E = mc^{2}$ gives only the rest energy. The total energy becomes $E = \gamma mc^{2}$ , where $\gamma = 1/\sqrt{1 - v^{2}/c^{2}}$ , or equivalently $E^{2} = (pc)^{2} + (mc^{2})^{2}$ .

1

Energy in a paperclip

Given ·

m:: 0.001
c:: 299792458

Find · E

Steps

Step 1: Identify values — $m = 1\,\mathrm{g} = 0.001\,\mathrm{kg}$ , $c = 299792458\,\mathrm{m/s}$ .
Step 2: Compute $c^{2} = (299792458)^{2} = 8.988 \times 10^{16} m^{2}/s^{2}$ .
Step 3: $E = mc^{2} = 0.001 \times 8.988 \times 10^{16} = 8.988 \times 10^{13} J \approx 21.5$ kilotons of TNT.

Result ·

E = mc^{2} = 0.001 \times (299792458)^{2} = 8.988 \times 10^{13} J

2

Energy from electron-positron annihilation

Given ·

m:: 1.82187674e-30
c:: 299792458

Find · E

Steps

Step 1: Total rest mass $= 2 \times m_e = 2 \times 9.1093837015 \times 10^{-31} = 1.822 \times 10^{-30} kg$ .
Step 2: $c^{2} = 8.988 \times 10^{16} m^{2}/s^{2}$ .
Step 3: $E = 1.822 \times 10^{-30} \times 8.988 \times 10^{16} = 1.637 \times 10^{-13} J$ .
Step 4: Convert to MeV: $1.637 \times 10^{-13} / 1.602176634 \times 10^{-13} = 1.022\,\mathrm{MeV}$ (two 511 keV photons).

Result ·

E = mc^{2} = 2 \times 9.1093837015 \times 10^{-31} \times (299792458)^{2} = 1.637 \times 10^{-13} J = 1.022\,\mathrm{MeV}

Nuclear Binding Energy→Q-Value of Nuclear Reaction→Relativistic Energy–Momentum Relation→