Grade 12

Advanced

9 min

Lesson 4 of 12

Relativistic Momentum and Mass-Energy Equivalence

Not Started

Understanding why objects with mass cannot reach the speed of light and the derivation of the world's most famous equation.

Imagine pedaling a bicycle with all your might, but instead of going faster, you simply feel yourself getting 'heavier' and harder to move. Why does the universe have a cosmic speed limit that no object with mass can ever cross?

Learning Objectives

1.Explain why classical momentum $p=mv$ fails at high velocities and how the Lorentz factor $\gamma$ corrects it.
2.Derive the relationship between mass and rest energy to interpret the physical meaning of $E=mc^2$ .
3.Calculate the total energy and kinetic energy of particles moving at relativistic speeds.

The Breakdown of Classical Momentum

In classical mechanics, we define momentum as

p = mv

. This suggests that if you keep applying a force, an object's velocity will increase indefinitely. However, Einstein’s Special Relativity shows that as an object approaches the speed of light

c

, its momentum grows much faster than its velocity. To account for this, we introduce the Lorentz factor

\gamma

. The relativistic momentum is defined as:

p = \gamma mv = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}}

v

approaches

c

, the denominator approaches zero, causing the momentum to approach infinity. This is why it would require an infinite force to accelerate an object with mass to the speed of light.

Quick Check

What happens to the Lorentz factor $\gamma$ as an object's velocity $v$ approaches the speed of light $c$ ?

Look at the denominator of the

\gamma

equation:

\sqrt{1 - v^2/c^2}

Answer

The Lorentz factor $\gamma$ approaches infinity.

An electron ( $m = 9.11 \times 10^{-31}$ kg) travels at $0.8c$ . Calculate its relativistic momentum.

1. Calculate $\gamma$ : $\gamma = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{0.6} \approx 1.67$ . 2. Use the momentum formula: $p = \gamma mv$ . 3. $p = (1.67)(9.11 \times 10^{-31} \text{ kg})(0.8 \times 3.0 \times 10^8 \text{ m/s})$ . 4. $p \approx 3.65 \times 10^{-22} \text{ kg}\cdot\text{m/s}$ .

Mass-Energy Equivalence: $E=mc^2$

Perhaps the most famous realization in physics is that mass is not just 'stuff'—it is a highly concentrated form of energy. Einstein proposed that even an object at rest possesses rest energy (

E_0

). This is expressed by the equation:

E_0 = mc^2

Here,

c^2

(the speed of light squared) is a massive conversion factor (

9 \times 10^{16} \text{ m}^2/\text{s}^2

), meaning a tiny amount of mass contains a staggering amount of energy. This principle explains the energy released in nuclear fission and the fusion powering the sun. Mass and energy are two sides of the same coin; mass can be converted into energy, and energy can be converted into mass.

Quick Check

If a particle is completely at rest, does it still have energy?

Think about the definition of

E_0

Answer

Yes, it has 'rest energy' equal to $mc^2$ .

Total Energy and Relativistic Kinetic Energy

When an object moves, its total energy (

E

) increases because it now possesses both rest energy and energy of motion. The total energy is given by:

E = \gamma mc^2

In classical physics, kinetic energy is

K = \frac{1}{2}mv^2

. However, at relativistic speeds, we must define relativistic kinetic energy (

K

) as the difference between the total energy and the rest energy:

K = E - E_0 = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2

At low speeds, this complex-looking formula mathematically simplifies back to the familiar

\frac{1}{2}mv^2

through a Taylor series expansion, showing that relativity encompasses classical physics.

A proton ( $m = 1.67 \times 10^{-27}$ kg) is accelerated to a speed where $\gamma = 10$ . Find its total energy and kinetic energy.

1. Find Rest Energy: $E_0 = mc^2 = (1.67 \times 10^{-27})(3 \times 10^8)^2 \approx 1.50 \times 10^{-10} \text{ J}$ . 2. Find Total Energy: $E = \gamma E_0 = 10 \times (1.50 \times 10^{-10} \text{ J}) = 1.50 \times 10^{-9} \text{ J}$ . 3. Find Kinetic Energy: $K = E - E_0 = 1.50 \times 10^{-9} - 0.15 \times 10^{-9} = 1.35 \times 10^{-9} \text{ J}$ .

How much work is required to accelerate an electron from $0.5c$ to $0.9c$ ?

1. Calculate $\gamma_1$ at $0.5c$ : $\gamma_1 = \frac{1}{\sqrt{1-0.5^2}} \approx 1.155$ . 2. Calculate $\gamma_2$ at $0.9c$ : $\gamma_2 = \frac{1}{\sqrt{1-0.9^2}} \approx 2.294$ . 3. Work done ( $W$ ) equals the change in kinetic energy: $W = \Delta K = \Delta E$ . 4. $W = (\gamma_2 - \gamma_1)mc^2$ . 5. $W = (2.294 - 1.155)(9.11 \times 10^{-31} \text{ kg})(3 \times 10^8 \text{ m/s})^2$ . 6. $W \approx 9.34 \times 10^{-14} \text{ J}$ .

Key Takeaways

1Relativistic momentum $p = \gamma mv$ approaches infinity as $v \to c$ , preventing objects with mass from reaching the speed of light.
2Mass is a form of energy; $E_0 = mc^2$ represents the energy of an object at rest.
3Total energy $E = \gamma mc^2$ includes both rest energy and kinetic energy.
4Relativistic kinetic energy is calculated as $K = (\gamma - 1)mc^2$ , not $\frac{1}{2}mv^2$ .

Quick Check

Multiple Choice

Why can't an object with mass reach the speed of light?

Multiple Choice

Which formula represents the total energy of a moving particle?

True/False

At very low speeds (much less than $c$ ), the relativistic kinetic energy formula gives approximately the same result as $\frac{1}{2}mv^2$ .

What's Next?

Review Tomorrow

In 24 hours, try to write down the formulas for relativistic momentum and kinetic energy from memory and explain why $p=mv$ is insufficient.

Practice Activity

Look up the mass of a Sun-like star and calculate how much energy would be released if just 1% of its mass were converted directly into energy using $E=mc^2$ .

Browse

Browse

Relativistic Momentum and Mass-Energy Equivalence

Learning Objectives

The Breakdown of Classical Momentum

Mass-Energy Equivalence: $E=mc^2$

Total Energy and Relativistic Kinetic Energy

Key Takeaways

Quick Check

What's Next?

Relativistic Momentum and Mass-Energy Equivalence

Learning Objectives

The Breakdown of Classical Momentum

Mass-Energy Equivalence: $E=mc^2$

Total Energy and Relativistic Kinetic Energy

Key Takeaways

Quick Check

What's Next?