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Year 12 Science

Special Relativity Introduction

Explore Einstein's revolutionary theory that redefined our understanding of space, time, and energy -- where nothing can travel faster than light.

Einstein's Two Postulates

In 1905, Albert Einstein published his theory of special relativity, built on two deceptively simple postulates that led to extraordinary consequences for our understanding of the universe.

1

The Principle of Relativity

The laws of physics are the same in all inertial reference frames (frames moving at constant velocity relative to each other). No experiment can distinguish between uniform motion and rest.

2

The Speed of Light is Constant

The speed of light in a vacuum (c = 3.0 × 108 m/s) is the same for all observers, regardless of the motion of the light source or observer.

Historical context: The Michelson-Morley experiment (1887) failed to detect the "luminiferous aether" -- the medium through which light was thought to travel. This result was a key motivation for Einstein's second postulate.

Time Dilation and Length Contraction

The constancy of the speed of light leads to two remarkable consequences: time dilation (moving clocks run slow) and length contraction (moving objects are shortened in the direction of motion).

Time Dilation

t = t0 / √(1 - v²/c²)

t0 = proper time (measured by observer at rest relative to the event)

t = dilated time (measured by moving observer)

The Lorentz factor: γ = 1/√(1 - v²/c²)

Length Contraction

L = L0 × √(1 - v²/c²)

L0 = proper length (measured at rest relative to the object)

L = contracted length (measured by moving observer)

Contraction occurs only in the direction of motion.

Lorentz Factor at Different Speeds

Speed (v)
γ
Effect
0.1c
1.005
Negligible
0.5c
1.155
Noticeable
0.87c
2.0
Time halved
0.99c
7.09
Dramatic

Mass-Energy Equivalence: E = mc²

Perhaps the most famous equation in physics, E = mc² tells us that mass and energy are equivalent and interconvertible. A small amount of mass contains an enormous amount of energy because the speed of light squared (c²) is a very large number.

The Complete Energy-Momentum Relation

E² = (pc)² + (m0c²)²

For an object at rest (p = 0), this reduces to E = m0c², the rest energy.

Etotal = γm0

Total relativistic energy includes both rest energy and kinetic energy.

KE = (γ - 1)m0

Relativistic kinetic energy -- at low speeds this approximates to ½mv².

Real-world application: Nuclear power stations and the Sun both convert mass to energy. In the Sun, about 4 million tonnes of mass are converted to energy every second through nuclear fusion, yet this is a tiny fraction of its total mass.

Key Vocabulary

Inertial Reference Frame

A frame of reference that is not accelerating -- one in which Newton's first law holds true. All inertial frames are equally valid.

Proper Time (t0)

The time interval between two events as measured by an observer for whom both events occur at the same spatial location.

Lorentz Factor (γ)

The factor γ = 1/√(1 - v²/c²) that quantifies time dilation, length contraction, and relativistic mass increase.

Rest Mass (m0)

The invariant mass of an object measured in its own rest frame. This is the "true" mass that all observers agree upon.

Worked Examples

1

A spacecraft travels at 0.8c. A clock on board measures 10 years for the journey. How long does the journey take according to an observer on Earth?

Step 1: Identify values: v = 0.8c, t0 = 10 years (proper time, measured on the spacecraft).

Step 2: Calculate γ = 1/√(1 - 0.8²) = 1/√(1 - 0.64) = 1/√0.36 = 1/0.6 = 5/3.

Step 3: t = γ × t0 = (5/3) × 10 = 16.67 years.

Answer: The Earth observer measures 16.67 years for the journey -- the moving clock runs slower.

2

A spaceship has a proper length of 100 m. If it travels at 0.6c, what length does an external observer measure?

Step 1: L0 = 100 m, v = 0.6c.

Step 2: L = L0 × √(1 - v²/c²) = 100 × √(1 - 0.36) = 100 × √0.64.

Step 3: L = 100 × 0.8 = 80 m.

Answer: The external observer measures the spaceship to be 80 m long.

3

Calculate the rest energy of a proton (mass = 1.67 × 10-27 kg).

Step 1: Use E = m0c², where m0 = 1.67 × 10-27 kg, c = 3.0 × 108 m/s.

Step 2: E = 1.67 × 10-27 × (3.0 × 108)².

Step 3: E = 1.67 × 10-27 × 9.0 × 1016 = 1.503 × 10-10 J.

Answer: The rest energy of a proton is approximately 1.50 × 10-10 J (or about 938 MeV).

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

Which of the following is NOT one of Einstein's postulates of special relativity?

Question 2

A muon has a half-life of 1.56 μs in its rest frame. If it travels at 0.98c, what is its observed half-life on Earth?

Question 3

According to special relativity, as an object approaches the speed of light, its length in the direction of motion:

Question 4

If 1 kg of matter could be completely converted to energy, approximately how much energy would be released?

Question 5

The Lorentz factor γ for an object travelling at 0.6c is:

Key Concepts Summary

Year 12: Electromagnetic Induction Year 12: Quantum Physics