Mass-Energy Equivalence

Equivalent forms

E_0 = m_0 c^2

E = \gamma m c^2

Three symbols compress the deepest unification in physics: mass and energy are one.

Symbol	Name	SI unit	Dimension	Range
$E$	Rest Energyoutput Energy content of a body at rest	J	ML^2T^-2	0 – 100000000000000000
$m$	Rest Mass Invariant rest mass of the object	kg	M	0 – 1
$c$	Speed of Light Speed of light in vacuum (constant)	m/s	LT^-1	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Imperial
$E$	J	erg	ft·lbf
$m$	kg	g	lb
$c$	m/s	cm/s	ft/s

Where it holds

Universal in special and general relativity. Applies to rest energy of any body with rest mass m.

Dimensional analysis

E \to [ML^2T^-2]

[M] * [LT^-1]^2 = [ML^2T^-2]

Both sides give energy.

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Discovery

Albert Einstein · 1905

Derived in Einstein's third 1905 Annus Mirabilis paper as a consequence of special relativity: a body losing energy L by radiation loses mass L/c^2.

Try this

How much energy is locked inside a single paperclip?

A 1-gram paperclip has rest energy E = mc^2 = 0.001 * (299792458)^2 ≈ 8.99e13 J — roughly the yield of a 21-kiloton atomic bomb.

Research status: stable

Real-world applications

Nuclear fission reactors and weapons
Stellar fusion powering the Sun
PET medical imaging (positron annihilation)
Particle accelerator energy budgets

Common misconceptions

$E = mc^2$ does not mean mass turns into energy magically — it states they are the same quantity in different units
The c^2 factor is just a unit conversion between kg and J
Photons satisfy $E = pc$ , not $E = mc^2 (m = 0$ for photons)

Experimental verification

Pair production

(\gamma \to e+ + e-)

and annihilation, nuclear binding energies, and stellar nucleosynthesis confirm

E = mc^2

to extraordinary precision (parts in 10^7 in Penning trap measurements).

Derivation

From the relativistic energy-momentum relation

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

, at rest

p = 0

gives

E = mc^2

.

Equivalently, integrating

dE = v dp

with

p = \gamma m v

yields total energy gamma m c^2; subtracting the rest term gives kinetic energy and isolates

E_0 = mc^2

.

Limiting cases

m \to 0

⟶

E \to 0

Massless particles (photons) have no rest energy; their energy comes from momentum:

E = pc

.

v > 0⟶

E = \gamma m c^2

Including kinetic motion, total energy scales with the Lorentz factor.

nuclear binding⟶

\Delta E = \Delta m c^2

A 1% mass deficit in fission releases

\sim 1

% of mc^2 — the basis of nuclear power.

What if…

What if you converted 1 kg fully to energy?

$\approx 9e16\,\mathrm{J}$ — enough to power the entire United States for about a day.

What if c were 10x smaller?

Rest energy would shrink 100x; nuclear processes would release far less energy and stars would burn dimmer.

What if mass could be negative?

Negative rest energy is unphysical for known particles; it would allow runaway energy extraction and violate causality.

1

Energy in 1 gram of mass

Given ·

m:: 0.001

Find · E

Steps

$c^2 = (299792458)^2 \approx 8.99e16\,\mathrm{m}^2/s^2$
$E = m * c^2 = 1e-3 * 8.99e16$
$E \approx 8.99e13\,\mathrm{J} \approx 21$ kilotons of TNT

Result ·

E = 0.001 * (299792458)^2 \approx 8.99e13\,\mathrm{J}

2

Solar luminosity mass loss

Given ·

E:: 3.828e+26

Find · m

Steps

Luminosity $L = 3.828e26\,\mathrm{W}$
$m_dot = L / c^2 = 3.828e26 / 8.99e16$
$m_dot \approx 4.26e_{9} kg/s$

Result · Sun loses

\approx 4.26e_{9} kg

per second to radiation.

Lorentz Factor→Relativistic Momentum→energy momentum relationRelativistic Kinetic Energy→