You are viewing a free preview of this lesson.
Subscribe to unlock all 8 lessons in this course and every other course on LearningBro.
The Michelson-Morley null result (lesson 6) and the failure of the wave theory to fit electromagnetism into a moving-frame description together set up the most radical reshaping of physical theory since Newton. Albert Einstein, working as a patent examiner in Bern, published On the Electrodynamics of Moving Bodies in 1905 — one of four papers of his annus mirabilis and the foundation of what is now called the special theory of relativity. The paper started from two simple postulates and derived, with surprisingly little mathematics, that time intervals, length intervals, simultaneity, momentum, mass and energy are all frame-dependent. The Lorentz-FitzGerald contraction reappeared as a kinematic consequence. The aether vanished. And out of the new framework came the most famous equation in physics, E = mc² — the mass-energy equivalence whose experimental verification ranges from particle accelerators to the energy release of nuclear reactors. This lesson covers Einstein's two postulates, time dilation Δt = γΔt₀, length contraction L = L₀/γ, the Lorentz factor γ = 1/√(1 − v²/c²), the modern treatment of relativistic mass and momentum, mass-energy equivalence, and the principal experimental tests — muon lifetime in cosmic rays, accelerator energies, and GPS time correction.
Spec mapping: This lesson covers AQA 7408 section 3.12.3 — Einstein's two postulates of special relativity, time dilation, length contraction, the Lorentz factor, relativistic momentum and energy, mass-energy equivalence (E = mc² and E_total = γm₀c²), and experimental tests including muon lifetime, accelerator data, and GPS satellite corrections. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.12.3 (Michelson-Morley, lesson 6): the null result is the experimental anchor for postulate 2 (c constant in all inertial frames). Special relativity is the framework that makes the null result automatic.
- Section 3.8 (nuclear physics): the binding energy of a nucleus is ΔE = Δm × c² — a direct application of mass-energy equivalence. Nuclear fission and fusion energy releases are computed from mass defects.
- Section 3.12.2 (electron diffraction, lesson 5): at high accelerating voltages, the non-relativistic formula λ = h/√(2m_e eV) under-estimates λ. The relativistic correction uses γ and matters above ~50 kV.
Einstein's 1905 paper is structured around two postulates from which all of special relativity follows:
Postulate 1 (relativity principle): The laws of physics take the same form in all inertial frames of reference. There is no privileged inertial frame.
Postulate 2 (constancy of c): The speed of light in vacuum has the same value c in all inertial frames, independent of the motion of the source or the observer.
Postulate 1 is a generalisation of Galilean relativity (which says the laws of mechanics are the same in all inertial frames) to all physical laws including electromagnetism. Postulate 2 is the radical step. Combined with postulate 1, it forces a complete reworking of how time and space measurements relate between frames.
The reason for the radicalism is straightforward. If observer A (in a lab) measures a light pulse and gets c = 3 × 10⁸ m s⁻¹, and observer B (moving at 0.5c relative to A) measures the same light pulse, classical Galilean addition of velocities would give B's measurement as either 0.5c or 1.5c depending on direction. But postulate 2 says B measures c — same as A. Either (i) there is something special about A's frame (rejected by postulate 1) or (ii) the way time and space measurements transform between A and B is not Galilean. Einstein chose option (ii) and worked out the consequences.
Consider an observer carrying a "light clock" — two parallel mirrors separated by a distance L, with a light pulse bouncing between them. In the observer's rest frame, a single tick (one round trip) takes time
Δt₀ = 2L/c
This is the proper time between two events at the same place in the observer's frame. Δt₀ is the smallest time interval any observer measures between these two events.
Now consider a second observer in a frame moving at speed v relative to the first. To the second observer, the light pulse traces out a zigzag path: while bouncing from one mirror to the other, the clock has moved sideways. The total path length is longer, but the speed is still c (postulate 2), so the second observer measures a longer time between ticks. The geometry gives:
Δt = Δt₀ / √(1 − v²/c²) = γ Δt₀
where the Lorentz factor γ is defined by
γ = 1 / √(1 − v²/c²)
For v ≪ c, γ ≈ 1 and there is no measurable time dilation. For v = 0.5c, γ = 1.155. For v = 0.9c, γ = 2.29. For v = 0.99c, γ = 7.09. As v → c, γ → ∞ and time dilation becomes arbitrarily large.
graph LR
A["v = 0.1c<br/>γ = 1.005"] --> B["v = 0.5c<br/>γ = 1.155"]
B --> C["v = 0.9c<br/>γ = 2.29"]
C --> D["v = 0.99c<br/>γ = 7.09"]
D --> E["v → c<br/>γ → ∞"]
style A fill:#27ae60,color:#fff
style B fill:#1d4ed8,color:#fff
style C fill:#7c3aed,color:#fff
style D fill:#ef4444,color:#fff
style E fill:#1f2937,color:#fff
The pedagogical interpretation: "moving clocks run slow." A clock at rest in frame A is observed from frame B (moving at v relative to A) to tick more slowly than B's local clocks, by the factor γ. The reciprocal nature of this — frame A also sees B's clocks run slow — is the source of the apparent paradoxes (twin paradox) that resolve only when relative acceleration breaks the symmetry.
The companion result to time dilation is length contraction. An object of proper length L₀ in its rest frame, observed from a frame moving at speed v parallel to its long axis, appears contracted to
L = L₀ √(1 − v²/c²) = L₀ / γ
For v = 0.5c, L = L₀ × 0.866 (13.4% shorter). For v = 0.99c, L = L₀ × 0.141 (7× shorter). Note: length contraction occurs only along the direction of motion; transverse dimensions are unchanged. (Otherwise, one observer could see an object as wider than another, contradicting the relativity principle.)
The Lorentz-FitzGerald contraction of lesson 6 reappears here as a kinematic consequence of the postulates — not a mechanical effect of an aether wind, but a frame-dependent feature of how space measurements relate between observers in relative motion.
Newtonian momentum p = mv must be replaced at high speeds. The relativistic expression is
p = γ m₀ v = m₀ v / √(1 − v²/c²)
where m₀ is the rest mass (the mass measured in the particle's rest frame). At v ≪ c, γ ≈ 1 and p ≈ m₀v, recovering the Newtonian formula. As v → c, p → ∞.
The total energy is
E = γ m₀ c² = m₀ c² / √(1 − v²/c²)
At v = 0 (γ = 1), this gives the rest energy:
E₀ = m₀ c²
— the famous mass-energy equivalence. Even at zero velocity, a particle has energy m₀c² stored in its mass. For an electron, m₀c² = 511 keV. For a proton, m₀c² = 938 MeV. These are the rest energies that appear in nuclear- and particle-physics tables.
The kinetic energy is the difference between total energy and rest energy:
KE = E − E₀ = (γ − 1) m₀ c²
For v ≪ c, expanding γ in a Taylor series gives KE ≈ ½m₀v², the Newtonian formula. For v = 0.5c, KE = 0.155 m₀c² (15% of the rest energy); for v = 0.99c, KE = 6.09 m₀c². At ultra-relativistic speeds, KE > m₀c² — a particle accelerated to "twice its rest energy" is travelling at v ≈ 0.94c, illustrating that doubling the energy does not double the speed.
The relation E₀ = m₀c² is the most famous equation in physics. It says: mass and energy are interchangeable, with the conversion factor c² (a huge number: 9 × 10¹⁶ J kg⁻¹).
Experimental verifications:
In every case, the prediction E = mc² matches experiment to high precision. The 2014 atom-trap measurements of the equivalence at the 0.0004% level represent the most precise direct verification to date.
The cleanest A-Level example. Muons are unstable particles with a rest-frame lifetime of τ₀ = 2.2 µs. They are produced when high-energy cosmic-ray protons strike the upper atmosphere, typically at altitudes of 10-15 km. Their typical speed at production is v ≈ 0.995c.
Without relativity, in the muon's 2.2 µs lifetime, even travelling at c they would cover only c × τ₀ ≈ 660 m — most would decay before reaching ground level. Yet ground-level muon detectors observe a substantial flux of cosmic-ray muons.
With time dilation: in the lab (ground) frame, the muon's lifetime is τ = γτ₀, where γ = 1/√(1 − 0.995²) ≈ 10. So in our frame the muon lives for ~22 µs, travelling 0.995c × 22 µs ≈ 6.6 km — easily enough to reach the ground.
In the muon's own rest frame, time runs at the proper rate but the atmosphere is length-contracted. The 10 km of atmosphere overhead contracts to 10 km / γ = 1 km — and the muon, travelling at 0.995c for τ₀ = 2.2 µs, covers ~660 m, plenty to traverse the contracted distance. Both descriptions agree.
Atomic-clock variants of this experiment (Hafele-Keating 1971, putting caesium clocks on commercial airliners and circumnavigating the Earth) have verified time dilation directly to ~10⁻¹⁰ precision.
The Global Positioning System uses satellites at altitude ~20,200 km, moving at ~3.9 km s⁻¹. Two relativistic effects must be included:
The net effect is +38 µs/day. Uncorrected, GPS positions would drift by ~11 km/day. The system's nanosecond-level timing accuracy is therefore a continuous, working verification of both special and general relativity.
In the LHC, protons are accelerated to v ≈ 0.999999991c, giving γ ≈ 7,500. The required energy is (γ − 1) m_p c² ≈ 7 TeV per proton — exactly what the LHC provides. The fact that the protons follow their designed circular trajectory at this energy (which depends on γ through the relativistic momentum p = γm_p v in the magnetic-bending equation r = p/eB) is a continuous verification of special relativity in a working facility.
A muon produced in the upper atmosphere travels at v = 0.995c. Calculate (a) the Lorentz factor γ, (b) the muon's lifetime as measured in the lab (ground) frame, given the rest-frame lifetime is τ₀ = 2.2 µs, (c) the distance the muon travels in the lab frame during its lifetime.
Solution.
(a) γ = 1/√(1 − 0.995²) = 1/√(1 − 0.990025) = 1/√(0.009975) = 1/0.0999 = 10.0 (to 3 s.f.).
(b) τ = γτ₀ = 10.0 × 2.2 × 10⁻⁶ = 2.2 × 10⁻⁵ s = 22 µs.
(c) Distance = v × τ = (0.995 × 3.00 × 10⁸) × (2.2 × 10⁻⁵) = 2.985 × 10⁸ × 2.2 × 10⁻⁵ = 6.57 × 10³ m ≈ 6.6 km.
This is much greater than the ~660 m that a non-relativistic muon would travel and is comparable to the height at which cosmic-ray muons are produced. Therefore, a substantial fraction of upper-atmosphere muons reach ground level — exactly what experiments observe.
Subscribe to continue reading
Get full access to this lesson and all 8 lessons in this course.