Mass-energy equivalence E² = (pc)² + (mc²)²: Question bank

Figure — Mass-energy equivalence E² = (pc)² + (mc²)²

True or false — justify

The invariant

E^{2} - (p c)^{2}

has the same value in every reference frame.

True. It equals

m^{2} c^{4}

, and rest mass

m

is frame-independent by definition — different observers disagree on

E

and

p

separately, but the combination

E^{2} - (p c)^{2}

always lands on the same number, like the length of a stick being unchanged by rotation.

A photon has zero mass, therefore it has zero energy.

False. With

m = 0

the vertical leg of the triangle vanishes and the formula collapses to

E = p c

, so its energy comes entirely from momentum — the surviving horizontal leg is the energy.

If two particles have the same total energy

E

, they must have the same momentum.

False. Same

E

(hypotenuse) can pair with different

p c

(horizontal leg) provided the vertical leg

m c^{2}

differs — a heavy slow particle and a light fast particle can share one energy value.

E = γ m c^{2}

and

E = (p c)^{2} + (m c^{2})^{2}

are two different formulas that sometimes disagree.

False. They are the same relation written two ways; substituting

p = γ m v

into the square-root form and simplifying gives

γ m c^{2}

exactly. Neither is "more correct."

Kinetic energy of a relativistic particle is

\frac{1}{2} m v^{2}

False. That is the low-speed Newtonian approximation. The correct value is

K E = E - m c^{2} = (γ - 1) m c^{2}

, which blows up toward infinity as

v \to c

, while

\frac{1}{2} m v^{2}

stays finite and wrong.

For a massless particle the rest-energy leg of the triangle has zero length.

True.

m c^{2} = 0

means the vertical leg vanishes, so the hypotenuse collapses onto the horizontal leg:

E = p c

. Geometrically the "triangle" degenerates into a single horizontal line.

Doubling a particle's speed doubles its momentum.

False.

p = γ m v

, and

γ

itself grows with

v

, so near

c

a tiny speed increase causes a large momentum increase — momentum has no ceiling even though speed is capped at

c

The rest energy

m c^{2}

can change if you push the particle hard enough.

False. Pushing adds kinetic energy (lengthens the hypotenuse and the

p c

leg) but the

m c^{2}

leg is fixed for an elementary particle — that fixedness is exactly what "invariant mass" means.

Spot the error

"A photon has no momentum because

p = m v

and its mass is

0

The error is using the Newtonian

p = m v

, which fails for anything relativistic. The correct relation for a massless particle is

p = E / c

, giving nonzero momentum — this is why light exerts radiation pressure on solar sails.

"As

v \to c

, energy

E \to m c^{2}

because that's the famous formula."

Backwards. As

v \to c

γ \to \infty

E = γ m c^{2} \to \infty

. The value

m c^{2}

is the minimum (rest) energy at

v = 0

, not the limit at high speed.

"To use the formula for a fast particle, replace

m

with the relativistic mass

γ m

The

m

E^{2} = (p c)^{2} + (m c^{2})^{2}

is already the rest mass; plugging in

γ m

double-counts

γ

(since

p

and

E

already contain it) and breaks the identity. Modern convention: "mass" means rest mass, always.

E^{2} = (p c)^{2} + (m c^{2})^{2}

, so I can just add

p c

and

m c^{2}

to get

E = p c + m c^{2}

Squares don't distribute over addition —

a^{2} + b^{2} \neq = a + b

. Pythagoras adds the squares of the legs, and the straight sum overestimates

E

except in the degenerate cases where one leg is zero.

"The formula only works for particles, not for systems like a box of gas."

It works for any isolated system:

E

is total energy,

p

total momentum, and

m

the system's invariant mass — which includes the internal kinetic and binding energies of its parts. A hot box weighs slightly more than a cold one.

"Since

E

is the hypotenuse,

E

is always larger than both

p c

and

m c^{2}

Precisely,

E \geq p c

and

E \geq m c^{2}

hold for all cases (a hypotenuse is never shorter than a leg). Strict inequality

E > p c

requires

m c^{2} > 0

, and

E > m c^{2}

requires

p c > 0

; equality

E = p c

occurs exactly when

m c^{2} = 0

(massless), and

E = m c^{2}

exactly when

p c = 0

(at rest). So "strictly larger than both" fails at those two boundaries.

Why questions

Why does the combination

c^{2} - v^{2}

vanish from the final formula, leaving no velocity?

Write each square explicitly:

E^{2} = (γ m c^{2})^{2} = γ^{2} m^{2} c^{4}

and

(p c)^{2} = (γ m v c)^{2} = γ^{2} m^{2} v^{2} c^{2}

. Subtracting,

E^{2} - (p c)^{2} = γ^{2} m^{2} c^{2} (c^{2} - v^{2})

. Now

γ^{2} = \frac{c ^{2}}{c ^{2} - v ^{2}}

, so the factor

(c^{2} - v^{2})

cancels the one hiding inside

γ^{2}

, leaving

m^{2} c^{4}

with no

v

at all — that cancellation is why the result is frame-independent.

Why is

E^{2} = (p c)^{2} + (m c^{2})^{2}

preferred over

E = γ m c^{2}

for photons?

For a photon

v = c

γ = 1/ 1 - 1

is undefined (divide by zero), while

m = 0

makes

γ m c^{2}

an indeterminate

\infty \times 0

. The Pythagorean form sidesteps

γ

entirely and cleanly gives

E = p c

Why can massless particles travel at exactly

c

but massive ones cannot?

Reaching

c

needs

γ \to \infty

, i.e. infinite energy, for any

m > 0

. A massless particle has no rest-energy leg to lift, so it isn't governed by

γ m c^{2}

at all — it must move at

c

to carry any energy via

E = p c

Why do we call

E^{2} - (p c)^{2}

"the length of the 4-momentum" rather than

E^{2} + (p c)^{2}

Spacetime geometry uses a minus sign between time-like and space-like parts (the Minkowski metric), unlike ordinary space which uses plus. That minus is what makes the result a true frame-invariant equal to

(m c^{2})^{2}

Why does subtracting

m c^{2}

from

E

give kinetic energy and not something else?

E

bundles rest energy and motion energy together;

m c^{2}

is the pure "existence" part present even at rest. Whatever remains,

E - m c^{2}

, is the energy the particle owes purely to its motion — that is the definition of kinetic energy,

(γ - 1) m c^{2}

Edge cases

What is

E

when both

p = 0

and

m = 0

Then both legs are zero, so

E = 0

— there is no particle at all. A massless thing at rest carries no energy and cannot exist as a real particle; light is never at rest.

What does the energy–momentum triangle look like for a particle moving at half the speed of light?

Both legs are present and nonzero: the fixed vertical

m c^{2}

and a modest horizontal

p c = γ m v c

. Since

v < c

the horizontal leg is shorter than for an ultra-relativistic particle, so

E

only modestly exceeds

m c^{2}

In the limit

p c ≫ m c^{2}

(ultra-relativistic), what does

E

approach?

E = (p c)^{2} + (m c^{2})^{2} \to p c

, because the tiny fixed leg becomes negligible beside the huge horizontal leg. A very fast massive particle behaves almost like a photon — its rest mass barely matters.

Can

p c

ever be negative in this formula?

No —

p

can point in any direction, but the formula uses the magnitude

∣ p ∣

, and

(p c)^{2}

is a square, so it contributes a non-negative length to the triangle regardless of direction.

Is there a frame in which a massive particle's momentum is zero?

Yes — its rest frame, where

v = 0

p = 0

, and

E = m c^{2}

. Every massive particle has such a frame. A massless particle does not, since it moves at

c

in every frame and can never be brought to rest.

What happens to the invariant mass of a system of two photons flying apart in opposite directions?

Their total momenta cancel (

p_{tot} = 0

) while energies add, so

E^{2} - (p c_{tot})^{2} = E_{tot}^{2} > 0

— the system has nonzero invariant mass even though each photon is massless. Energy trapped with zero net momentum shows up as mass.

Connections

Parent topic — the derivation and worked numbers
Lorentz Factor — why $γ \to \infty$ caps speed at $c$
Photon Momentum — the $m = 0$ , $E = p c$ edge case
Relativistic Kinetic Energy — the $(γ - 1) m c^{2}$ leftover
Four-Momentum — where the minus-sign invariant comes from
Nuclear Binding Energy — systems whose invariant mass differs from the sum of parts
Special Relativity — the Minkowski geometry underneath