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F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Special Relativity

Mass–Energy Equivalence: E = mc², Total Energy and Kinetic Energy

F2.3 F2.1About 75 minutes
Learning goals
  • I can explain mass–energy equivalence and calculate rest energy with E₀ = mc².
  • I can calculate total energy E = γmc² and relativistic kinetic energy Ek = (γ − 1)mc².
  • I can convert between joules and electron volts and use MeV for particles.
  • I can use E² = (pc)² + (mc²)² to connect energy, momentum and mass.
  • I can show that (γ − 1)mc² becomes ½mv² at low speeds.

Words to know

rest energy (E₀)
The energy an object has because of its mass, E₀ = mc², even when at rest.
total energy (E)
E = γmc², the sum of rest energy and kinetic energy.
relativistic kinetic energy
Ek = E − mc² = (γ − 1)mc².
electron volt (eV)
The energy gained by an electron accelerated through a potential difference of 1 V: 1 eV = 1.602 × 10⁻¹⁹ J. 1 MeV = 10⁶ eV; 1 GeV = 10⁹ eV.
mass–energy equivalence
Mass and energy are two forms of the same conserved quantity; a change in energy ΔE of a system goes with a change in mass Δm = ΔE/c².
energy–momentum relation
E² = (pc)² + (mc²)², valid for every particle, including massless ones.

The most famous equation, all of it

Remember from Grade 11: in Work, Energy and Power you used E = mc² to find the energy equivalent of a mass and the energy released when a little mass disappears in fusion. In Grade 12 we can see where the equation comes from and what it fully says.

In late 1905, Einstein published a short follow-up paper, "Does the inertia of a body depend upon its energy content?". He showed that if a body gives off energy E as light, its mass decreases by E/c². Mass is a form of energy. Combined with relativistic momentum (lesson 10), this gives a complete picture of the energy of a moving particle.

Three kinds of energy

For a particle of mass m moving at speed v:

  • Rest energy: E₀ = mc². Every object has it, even at rest.
  • Total energy: E = γmc². This is what is conserved in collisions and decays (along with momentum).
  • Kinetic energy: the extra energy because of motion: Ek = E − mc² = (γ − 1)mc².

Low-speed check. For v ≪ c, γ ≈ 1 + v²/(2c²) (lesson 5). Then Ek ≈ (v²/(2c²))mc² = ½mv². The Newtonian formula is the low-speed limit. At high speeds, Ek rises without limit as v → c: reaching c would take infinite energy, which is another way to see the cosmic speed limit.

v/cEk / (mc²)0.00.20.40.60.81.01234Newton: ½mv²relativistic: (γ − 1)mc²v = cEk = mc² at 0.866c
Kinetic energy (in units of mc²) against speed. Newton's ½mv² (dashed) would allow any speed for enough energy. The relativistic Ek = (γ − 1)mc² agrees at low speed but becomes infinite at v = c. At 0.866c the kinetic energy equals the rest energy.

Electron volts

Particle energies are tiny in joules, so physicists use the electron volt: 1 eV = 1.602 × 10⁻¹⁹ J, the energy an electron (charge 1.602 × 10⁻¹⁹ C) gains when accelerated through 1 V. When a particle of charge q is accelerated through a potential difference V, it gains kinetic energy Ek = qV, so an electron accelerated through 50 kV gains 50 keV.

Masses are often given as rest energies in MeV, with 'MeV/c²' as the mass unit:

  • Electron: m = 9.11 × 10⁻³¹ kg, mc² = 8.20 × 10⁻¹⁴ J ≈ 0.511 MeV
  • Proton: m = 1.673 × 10⁻²⁷ kg, mc² ≈ 938 MeV
  • Neutron: mc² ≈ 940 MeV

Because γ = E/(mc²), working in MeV makes many problems quick: γ = 1 + Ek/(mc²).

Worked example

Example 1: Rest energy

Find the rest energy of an electron in joules and in MeV.

  1. E₀ = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 9.11 × 10⁻³¹ × 9.00 × 10¹⁶ = 8.20 × 10⁻¹⁴ J.
  2. In eV: 8.20 × 10⁻¹⁴ ÷ 1.602 × 10⁻¹⁹ = 5.12 × 10⁵ eV = 0.512 MeV.

8.20 × 10⁻¹⁴ J ≈ 0.51 MeV. (The accepted value, 0.511 MeV, uses c = 2.998 × 10⁸ m/s.)

Worked example

Example 2: An X-ray tube electron

An electron is accelerated from rest through 50 kV, as in an X-ray tube. Find its speed (a) relativistically and (b) with Newtonian physics.

  1. Ek = qV = 50 keV = 0.050 MeV.
  2. (a) γ = 1 + Ek/(mc²) = 1 + 0.050/0.511 = 1.0978. v/c = √(1 − 1/γ²) = √(1 − 0.8297) = 0.413.
  3. (b) ½mv² = Ek → v = √(2Ek/m) = √(2 × 50 000 × 1.602 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹) = 1.33 × 10⁸ m/s = 0.442c.

(a) 0.413c (1.24 × 10⁸ m/s); (b) 0.442c. Newton overestimates by about 7%, so X-ray and electron-microscope engineers use relativistic formulas.

Worked example

Example 3: A TRIUMF proton

TRIUMF's cyclotron gives protons kinetic energies up to 520 MeV. Find γ, the speed and the momentum (in MeV/c) of a 520 MeV proton (mc² = 938.3 MeV).

  1. γ = 1 + 520/938.3 = 1.554.
  2. v/c = √(1 − 1/1.554²) = √(1 − 0.4141) = 0.765.
  3. E = 520 + 938.3 = 1458.3 MeV. From E² = (pc)² + (mc²)²: pc = √(1458.3² − 938.3²) = 1116 MeV.
  4. So p = 1116 MeV/c (in SI: 1116 × 10⁶ × 1.602 × 10⁻¹⁹ J ÷ 3.00 × 10⁸ m/s = 5.96 × 10⁻¹⁹ kg·m/s).

γ = 1.55, v = 0.765c, p ≈ 1.12 × 10³ MeV/c.

Energy and momentum together

Square E = γmc² and p = γmv and subtract: E² − (pc)² = γ²m²c⁴(1 − v²/c²) = m²c⁴. So

E² = (pc)² + (mc²)²

Think of it as a right triangle: the legs are the rest energy mc² and pc, and the hypotenuse is the total energy E. Two special cases:

  • At rest (p = 0): E = mc².
  • No mass (m = 0): E = pc. A photon has no mass but carries energy and momentum and moves at c. (The sibling unit Quantum Mechanics uses E = pc for photons in the Compton effect.)

For fast particles (E ≫ mc²), E ≈ pc. Particle physicists often treat electrons with energies of many GeV as nearly massless.

rest energy mc²pctotal energy EE² = (pc)² + (mc²)²At rest: p = 0, so E = mc²Photon: m = 0, so E = pcKinetic energy: Ek = E − mc²Example: electron at 0.99cmc² = 0.511 MeV, E = 3.62 MeV,pc = 3.59 MeV, Ek = 3.11 MeV
The energy–momentum relation as a right triangle: the rest energy and pc are the legs, the total energy is the hypotenuse (not to scale). For an electron at 0.99c, 3.59² + 0.511² ≈ 3.62² (in MeV).
Worked example

Example 4: Energy for a starship

How much energy is needed to accelerate a 1000 kg probe from rest to (a) 0.10c and (b) 0.90c? Compare (a) with ½mv².

  1. mc² = 1000 × 9.00 × 10¹⁶ = 9.00 × 10¹⁹ J.
  2. (a) γ = 1/√(1 − 0.01) = 1.005038. Ek = 0.005038 × 9.00 × 10¹⁹ = 4.53 × 10¹⁷ J. Newton: ½(1000)(3.00 × 10⁷)² = 4.50 × 10¹⁷ J.
  3. (b) γ = 2.294. Ek = 1.294 × 9.00 × 10¹⁹ = 1.16 × 10²⁰ J.

(a) 4.53 × 10¹⁷ J (within 1% of ½mv²); (b) 1.16 × 10²⁰ J, about 3.2 times the Newtonian ½mv² = 3.65 × 10¹⁹ J. Reaching high fractions of c costs enormous energy.

Energy toolkit

  • γ = E/(mc²) = 1 + Ek/(mc²): a fast way to get γ from energies.
  • Work in MeV for particles, joules for everyday objects. 1 eV = 1.602 × 10⁻¹⁹ J.
  • E² = (pc)² + (mc²)² links everything. Momentum in MeV/c: p = pc ÷ c.
  • Kinetic energy is not ½mv² at high speed, and not ½γmv² either. Use (γ − 1)mc².
Just for fun

The rest energy of a single raisin (about 0.5 g) is about 4.5 × 10¹³ J, more than the energy released by burning roughly a million litres of gasoline. Unfortunately for snack-powered cars, nobody knows how to turn a raisin's whole mass into energy.

Hands-on activity

How much energy is hiding in your kitchen?

You need: a kitchen scale, three household objects (for example an apple, a cup of water and a pencil), a recent electricity bill or the number 750 kWh per month (a typical Ontario home), and a calculator.

  1. Weigh each object in grams and convert to kilograms.
  2. Calculate each object's rest energy E₀ = mc² in joules.
  3. Convert to kilowatt-hours: 1 kWh = 3.6 × 10⁶ J.
  4. How many months could each object's rest energy run your home (if it could all be converted to electrical energy)?
  5. Now go the other way: a fully charged phone battery stores about 15 Wh = 5.4 × 10⁴ J. How much more mass does the charged battery have than the empty one? (Δm = ΔE/c².)

What to look for: a 150 g apple: E₀ = 0.150 × 9.00 × 10¹⁶ = 1.35 × 10¹⁶ J = 3.75 × 10⁹ kWh, enough for 750 kWh/month for 5.0 million months (over 400 000 years). The battery gains only 6 × 10⁻¹³ kg, far too little for any scale. Write a sentence explaining why nobody noticed E = mc² in chemistry or everyday life before 1905.

Your turn

1

Find the rest energy of a 2.0 g paper clip in joules.

2

An electron moves at 0.80c. Find its total energy and kinetic energy in MeV.

3

A proton has kinetic energy 938 MeV. What is γ, and what is its speed?

4

Convert 2.5 MeV to joules.

5

An electron is accelerated from rest through 1.0 MV. Find its kinetic energy, γ and speed.

6

Use E² = (pc)² + (mc²)² to find pc for an electron with total energy 1.511 MeV.

7

Show that (γ − 1)mc² ≈ ½mv² for v ≪ c.

More practice: warm-up

1

What is the rest energy of 1.0 kg of anything?

2

Write the formula for relativistic kinetic energy.

3

What is 1 eV in joules?

4

What is the rest energy of an electron in MeV?

5

What is E for a photon with momentum p?

More practice: on your own

1

An electron moves at 0.99c. Find E, Ek and pc in MeV.

2

A proton has a kinetic energy of 2.0 GeV. Find γ and its speed.

3

Hospital linear accelerators for cancer treatment give electrons about 6 MeV of kinetic energy. Find γ and v/c.

4

How much kinetic energy (in J) does a 1.0 kg mass need to reach 0.60c? Compare with ½mv².

5

At what speed is a particle's kinetic energy equal to its rest energy? Its total energy three times its rest energy?

Challenge

1

Derive E² = (pc)² + (mc²)² from E = γmc² and p = γmv.

2

A particle has momentum 500 MeV/c and total energy 1000 MeV. Find its mass (in MeV/c² and kg) and its speed. (Hint: v/c = pc/E.)

3

Show that v/c = pc/E for any particle, and explain what it implies for massless particles.

Check your understanding

Quiz

1.

What is the total energy of a particle with mass m moving at speed v?

2.

An electron (mc² = 0.511 MeV) has kinetic energy 0.511 MeV. What is its speed?

3.

Which formula gives the kinetic energy of a particle moving at 0.9c?

4.

How much energy is 1.0 MeV in joules?

5.

For a photon, which relation holds?

6.

Why can't an object with mass reach c?

For parents and teachers

Big idea (F2.3): a particle's total energy is E = γmc²: its rest energy mc² plus kinetic energy (γ − 1)mc². At low speeds the kinetic energy reduces to ½mv². Energy, momentum and mass are linked by E² = (pc)² + (mc²)².

How to help: The kitchen activity makes the size of c² vivid. Ask your teen to explain why an apple's huge rest energy does not mean we can power a house with an apple. (We have no process that converts all its mass to usable energy; fission converts only about 0.1% of the fuel's mass, and fusion less than 1%.)

Common difficulties:

  • Using ½mv² or ½γmv² for fast particles. Use (γ − 1)mc².
  • Unit confusion: keep MeV with MeV, joules with joules. Convert with 1 eV = 1.602 × 10⁻¹⁹ J.
  • Forgetting to square c (9.00 × 10¹⁶) or squaring 3.00 × 10⁸ incorrectly on the calculator.
  • Confusing total energy E with kinetic energy Ek. Ek = E − mc².