Theory — Motion Near the Speed of Light

Einstein's special relativity rests on two postulates: the laws of physics are the same in every inertial frame, and the speed of light c is the same for every observer regardless of how the source or observer moves. From these follow time dilation, length contraction, and a new relationship between energy, momentum, and mass. General relativity extends these ideas to gravity, treating it as the curvature of spacetime, where clocks deeper in a gravitational field also run slow.

The Lorentz factor

Almost every relativistic result is built from one quantity, the Lorentz factor γ, which depends only on the speed v expressed as a fraction of c. At low speed γ is essentially one and the familiar physics returns; as v approaches c, γ grows without limit.

Lorentz factorγ = 1 / square root of (1 − v²/c²)
Write β = v/c, then γ = 1 / square root of (1 − β²)

Time dilation and length contraction

A moving clock ticks slowly: a time interval t₀ measured in the clock's own frame is observed to last a longer time t = γ t₀ in a frame the clock moves through. At the same time, a length L₀ measured at rest is observed to be shorter along the direction of motion, L = L₀ / γ.

Dilation and contractionTime dilation: t = γ t₀ (moving clocks run slow)
Length contraction: L = L₀ / γ (moving lengths shrink)
Both effects become large only when v is a sizeable fraction of c

Relativistic energy, kinetic energy, and momentum

A particle of rest mass m has a rest energy E₀ = mc² even when standing still. When it moves, its total energy is E = γmc², its kinetic energy is the difference KE = (γ − 1)mc², and its momentum is p = γmv. These reduce to the Newtonian forms at low speed but differ sharply as v approaches c.

Energy and momentumRest energy: E₀ = mc²
Total energy: E = γ mc² = γ E₀
Kinetic energy: KE = (γ − 1) mc²
Momentum: p = γ mv, so pc = γ β E₀
They are tied together by E² = (pc)² + (mc²)²

Mass-energy equivalence

The rest energy E₀ = mc² says mass and energy are two forms of the same thing. A small amount of mass corresponds to an enormous amount of energy because c² is so large, which is why nuclear reactions release so much energy from tiny changes in mass.

One factor, many effects

The Lorentz factor γ sets the size of time dilation, length contraction, and the rise in energy and momentum.

Speed limit

As v approaches c, γ grows without bound, so it takes ever more energy to go faster and no massive object can reach c.

Mass is energy

Rest energy E₀ = mc² means even a particle at rest carries energy, and tiny mass changes release large energies.

QuantityRelationshipNotes
Lorentz factorγ = 1 / √(1 − β²)β = v/c
Time dilationt = γ t₀moving clocks run slow
Length contractionL = L₀ / γalong the motion
Total energyE = γ mc²rest energy when γ = 1
Kinetic energyKE = (γ − 1) mc²Newtonian at low speed
Momentumpc = γ β E₀E² = (pc)² + (mc²)²

Apparatus

The equipment a real special-relativity experiment uses. In the simulation these are modelled for you, but the readings correspond to what each instrument would measure.

times light bounces
Light clock
Times light bouncing between mirrors to show time dilation between frames.
compare time in frames
Synchronized clocks
Compare elapsed time measured in two relatively moving frames.
proper length
Measuring rod
Its proper length is compared with the contracted length seen in motion.
counts particles
Particle detector
Counts fast particles such as muons that reach the ground because of time dilation.
emits light
Light source
Emits the light signals whose constant speed underlies the experiment.
displays waveforms
Timing display
Reads out the very short time intervals involved.

Instructions — Running the Virtual Experiment

This is a record, calculate, and compare lab. In each part you set the speed, calculate the requested quantities by hand, enter your values, and only then does the simulation reveal its results so you can compare. Record every value in your worksheet.

Part A — Time Dilation and Length Contraction (Time & Length tab)
1
Open Simulation → Time & Length. Set the speed as a fraction of c and choose a proper time and a proper length.
2
Calculate the Lorentz factor γ, then the dilated time t = γ t₀ and the contracted length L = L₀ / γ, enter all three, and click Check. Record at least four different speeds, increasing toward c, and note how each value grows or shrinks.
Part B — Relativistic Energy (Energy tab)
1
Open Energy. Choose a particle (its rest energy is given) and set the speed.
2
Calculate the Lorentz factor, the total energy E = γ E₀, and the kinetic energy KE = (γ − 1) E₀, enter them, and click Check. Record at least three speeds, and repeat for at least two of the three particles so you can compare. Note how the kinetic energy exceeds the Newtonian estimate at high speed.
Part C — Momentum and Mass-Energy Equivalence (Momentum & E = mc² tab)
1
Open Momentum & E = mc². For the chosen particle and speed, calculate the momentum from pc = γ β E₀, enter it, and click Check. Record at least three speeds and verify that E² = (pc)² + (mc²)² holds for each.
2
In the mass-energy panel, calculate the energy released when a small mass is converted using E = mc² for at least two different masses, enter each, and check.
For your reportInclude your data tables, your worked calculations for each part, screenshots, and a short discussion of how time dilation, length contraction, energy, and momentum all change as the speed approaches c.

Simulation — The Relativity Bench

Special Relativity Virtual LabSet the speed, calculate, then reveal and compare
v/cYour γγ actualYour t (s)Your L (m)t, L actual
No rows yet — set the speed, calculate γ, t, and L, and check.

Moving frame

Result
β = v/c0.80
Lorentz factor γ— hidden
Dilated time t— hidden
Contracted length L— hidden
Particlev/cYour γYour E (MeV)Your KE (MeV)E, KE actual
No rows yet — choose a particle and speed, calculate γ, E, and KE, and check.

Fast particle

Result
Rest energy E₀0.511 MeV
Lorentz factor γ— hidden
Total energy E— hidden
Kinetic energy KE— hidden
Particlev/cYour pc (MeV)pc actualCheck E²=(pc)²+E₀²
No rows yet — calculate the momentum and check.

Momentum

Result
Rest energy E₀0.511 MeV
Momentum pc— hidden
√((pc)²+E₀²)— hidden

Mass-energy: E = mc²

Energy E = mc²— hidden

Team Questions

Question 1. According to special relativity, is the speed of light the same for all observers? (yes or no)
Question 2. What is the Lorentz factor γ for a particle moving at v = 0.80c? (to 2 decimals)
Question 3. Does a moving clock run faster or slower than an identical clock at rest? (faster or slower)
Question 4. A rod 1.00 m long at rest moves at 0.80c along its length (γ = 1.67). What is its observed length, in metres? (to 2 decimals)
Question 5. What is the rest energy of an electron, in MeV? (to 3 decimals)
Question 6. In E = mc², what does c represent? (the speed of ___)
Question 7 — Challenge. In general relativity, does a clock deeper in a gravitational field run faster or slower than one farther away? (faster or slower)

Example Lab Report

A worked example showing the expected format and the record, calculate, and compare workflow.

Special Relativity

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Objective

To calculate the Lorentz factor and use it to find the dilated time and contracted length, to determine the relativistic energy, kinetic energy, and momentum of a fast particle, and to confirm the mass-energy relation E = mc², comparing every calculated value with the simulation.

Part A — Time Dilation and Length Contraction (worked example)

At v = 0.80c: γ = 1 / square root of (1 − 0.80²) = 1 / square root of 0.36 = 1.667. For a proper time of 1.00 s, the dilated time is t = γ t₀ = 1.667 s. For a proper length of 1.00 m, the contracted length is L = L₀ / γ = 0.600 m. Both matched the simulation.

Part B — Relativistic Energy (worked example)

For an electron (E₀ = 0.511 MeV) at 0.80c: γ = 1.667, so the total energy is E = γ E₀ = 0.852 MeV and the kinetic energy is KE = (γ − 1) E₀ = 0.341 MeV. The relativistic kinetic energy is larger than the Newtonian estimate, as expected near c.

Part C — Momentum and Mass-Energy (worked example)

For the same electron, pc = γ β E₀ = 1.667 × 0.80 × 0.511 = 0.681 MeV. Checking, square root of ((0.681)² + (0.511)²) = 0.852 MeV, equal to the total energy, confirming E² = (pc)² + (mc²)². Converting 1.00 g of mass gives E = mc² = 0.00100 kg × (3.00 × 10⁸)² = 9.00 × 10¹³ J.

Discussion and Conclusion

Every calculated value agreed with the simulation. As the speed approached c the Lorentz factor grew quickly, time dilated and length contracted, and the energy and momentum rose far above their Newtonian values, while the rest energy E = mc² showed that a tiny mass corresponds to an enormous energy.

Practice Questions

Question 1
Find the Lorentz factor for a spaceship travelling at 0.60c.
Hint: γ = 1 / square root of (1 − 0.60²) = 1 / square root of 0.64 = 1.25.
Question 2
A muon's clock measures a lifetime of 2.20 microseconds at rest. How long does it live in the lab frame if it moves at 0.99c (γ = 7.09)?
Hint: t = γ t₀ = 7.09 × 2.20 = 15.6 microseconds.
Question 3
A starship 120 m long at rest passes at 0.80c. What length does an observer measure? (γ = 1.667)
Hint: L = L₀ / γ = 120 / 1.667 = 72.0 m.
Question 4
Find the total energy and kinetic energy of a proton (E₀ = 938 MeV) moving at 0.90c (γ = 2.294).
Hint: E = 2.294 × 938 = 2152 MeV; KE = (2.294 − 1) × 938 = 1214 MeV.
Question 5 — Challenge
How much energy is released if 2.0 g of mass is completely converted to energy?
Hint: E = mc² = 0.0020 × (3.0 × 10⁸)² = 1.8 × 10¹⁴ J.