Each observer only sees motion in others. They see themselves at rest.
Move the mouse vertically to change observers in the animation above.
(note: special relativity isn't included yet in this animation.)

Question: Imagine two cars. One is moving at 15 m/s relative to the Earth and the other at 20 m/s relative to the Earth. How fast are they moving relative to each other?

answer

$$v = 20 \, \mathrm{\tfrac{m}{s}}-15 \, \mathrm{\tfrac{m}{s}}$$ $$v = 5 \, \mathrm{\tfrac{m}{s}}$$

Question: Imagine a boat traveling up a river. The current in the river is 5 m/s downstream. The boat is moving upstream at 15 m/s relative to the water in the river. How fast is the boat moving relative to the land?

answer

$$v = 15 \, \mathrm{\tfrac{m}{s}}-5 \, \mathrm{\tfrac{m}{s}}$$ $$v = 10 \, \mathrm{\tfrac{m}{s}}$$

Question: A train moving at 20 m/s makes a sound with its steam whistle. Sound moves through air at around 343 m/s relative to the air. How fast would each observer measure the speed of sound?

Observer 1: A person at the train station.
Observer 2: A person at the back of the train.

answer

Observer 1:343 m/s.

The person on the ground measures the speed of sound as normal because they are at rest relative to the air.
Observer 2:343 m/s + 20 m/s = 363 m/s

The person on the train is moving relative to the air so they will measure the speed of sound as either higher or lower.

A Constant Speed of Light

Waves are a disruption that propagates through a medium. The speed of a wave is constant and determined by the medium. Sound's medium is atoms. Light's medium is electric and magnetic fields.

Light is different from other waves. It moves faster than any known object, and it is able to propagate through the vacuum of space. How can a wave exist in seemingly empty space? Is space filled with a medium? What is the velocity of that medium relative to Earth?

In 1887, scientists Albert A. Michelson and Edward W. Morley developed an experiment to learn about the relative velocity between the Earth and light's medium, electromagnetic fields.

Michelson and Morley split a beam of light into two perpendicular paths. These separate beams reflected off mirrors to recombine again at the splitter. By looking at the interference pattern of the combined beams, they could tell with a high accuracy if either path took less time.

They tested at different times of day and year to measure the effect of Earth's relative velocity on the speed of light. They always measured a constant speed of light in every direction. Earth's motion had no effect.

$$c = 3 \times 10^8 \, \mathrm{\tfrac{m}{s}} $$

All observers measure the speed of light as the same value even when moving at a relative velocity.

It seems like a constant speed of light for all observers leads to a paradox. How can two observers moving relative to each other both see the speed of something as the same?

Lorentz factor

These equations predict how observers in two non-accelerating reference frames can disagree on space and time if they have a relative velocity.

$$ \Delta t = \gamma \Delta t_0 \quad \quad \quad \quad L = \frac{L_0}{\gamma}$$

Before we can solve these equations, we need to calculate the Lorentz factor, written as Ɣ (gamma). The Lorentz factor is used in special relativity equations to scale how different reference frames measure space and time.

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

$ \gamma $ = Lorentz factor, gamma [no units]
$v$ = relative velocity [m/s]
$c$ = speed of light, 3 × 10⁸ [m/s]

The Lorentz factor is always greater than 1, but it grows towards infinity as the relative velocity approaches the speed of light.

Example: Calculate the Lorentz factor for two reference frames at rest relative to each other.

hint

v = 0

solution

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{0^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1}}$$ $$\gamma = 1$$

A gamma of 1 means that there aren't any relativistic effects.

Example: Calculate the Lorentz factor for two reference frames moving at half the speed of light relative to each other.

hint

v = 0.5c

solution

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{(0.5c)^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{(0.5)^2 {\color{red}c^2}}{{\color{red}c^2}} } }$$ $$\gamma = \frac{1}{\sqrt{1-0.5^2}}$$ $$\gamma = \frac{1}{\sqrt{1-0.25}}$$ $$\gamma = \frac{1}{\sqrt{0.75}}$$ $$\gamma = \frac{1}{0.866}$$ $$\gamma = 1.155$$

Relativistic effects like time dilation and length contraction will differ by 1.155 between observers at a relative velocity of 0.5c.

Example: Calculate the Lorentz factor for two reference frames moving at 0.99c relative to each other.

calculator

When the relative velocity is expressed in terms of the speed of light you can cancel the both c² inside the square root. $$\frac{(0.99c)^2}{c^2}$$ $$\frac{0.99^2c^2}{c^2}$$ $$0.99^2$$ So, I normally will cancel the c² before I solve for Ɣ as one step on a calculator. $$(1-0.99^2)^{-0.5}$$

solution

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{(0.99)^2 c^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-0.9801}}$$ $$\gamma = \frac{1}{\sqrt{0.0199}}$$ $$\gamma = \frac{1}{0.1411}$$ $$\gamma = 7.089$$

Example: Calculate the Lorentz factor for two reference frames moving at the speed of light relative to each other.

hint

v = c

solution

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{c^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-1} }$$ $$\gamma = \frac{1}{\sqrt{0} }$$ $$\gamma = \frac{1}{0}$$ $$\gamma = \text{undefined} $$

As the relative velocity approaches the speed of light, extreme effects like: time dilation, mass increase, and length contraction hinder acceleration. This means that the speed of light is also the maximum relative speed for any object.

At a low relative velocity, the effects of special relativity aren't noticeable because the Lorentz factor is one. As the relative speed approaches the speed of light, the Lorentz factor increases towards infinity.

Time Dilation

Time dilation is amazing because it disagrees with our natural intuition. Be careful when solving these problems. Keep a clear idea of the observer and their velocity relative to the events.

derivation of time dilation

To understand where the time dilation equation comes from we will explore a hypothetical situation. Imagine light clocks that work by firing light at a mirror and measuring the time for the light to return. Each light clock's path is marked in blue.

On the left, a light clock is not moving relative to the observer.

$ \Delta t_0 $ is the time for one tick of the light clock at rest.
On the right, a light clock is moving at v relative to the observer.

$ \Delta t $ is the time for one tick of the light clock at velocity v.

First we apply $ v= \frac{\Delta x}{\Delta t}$ to each side of the triangle highlighted in red.

$$ c = \frac{2D}{\Delta t } \quad \quad c = \frac{2L}{\Delta t_0 } \quad \quad v = \frac{2W}{\Delta t}$$ $$D = \tfrac{1}{2} c \Delta t \quad L = \tfrac{1}{2} c \Delta t_0 \quad W = \tfrac{1}{2} v \Delta t $$

Next, we apply the Pythagorean theorem.

$$ D^2 = L^2 + W^2 $$ $$ (\tfrac{1}{2}c\Delta t)^2 = (\tfrac{1}{2}c \Delta t_0)^2 + (\tfrac{1}{2}v\Delta t)^2 $$ $$ \tfrac{1}{4} c^2 \Delta t^2 = \tfrac{1}{4}c^2 \Delta t_0^2 + \tfrac{1}{4}v^2 \Delta t^2 $$ $$ \Delta t^2 = \Delta t_0^2 + \tfrac{v^2}{c^2} \Delta t^2 $$ $$ \Delta t^2 - \tfrac{v^2}{c^2} \Delta t^2 = \Delta t_0^2$$ $$ \Delta t^2 (1 - \tfrac{v^2}{c^2} ) = \Delta t_0^2 $$ $$ \Delta t \sqrt{1 - \tfrac{v^2}{c^2}} = \Delta t_0 $$ $$ \Delta t = \frac{\Delta t_0}{ \sqrt{1 - \tfrac{v^2}{c^2}}} $$ $$ \Delta t = \gamma \Delta t_0 $$

$$ \Delta t = \gamma \Delta t_0$$

$\Delta t$ = elapsed time for an observer in a frame of reference where the events occur in different locations

$\Delta t_0$ = elapsed time for an observer in a frame of reference where the events occur in the same location, proper time

$ \gamma $ = Lorentz factor, gamma [no units]

Gamma must be greater than one. This means that $\Delta t> \Delta t_0$.

$\Delta t$ represents the time passed for an observer watching events that are moving relative to the observer.

This is the point of view where there is a relative velocity between events and observer.

For example, if you watch an airplane fly past, events on the plane occur in different locations because the airplane is moving.
$\Delta t_0$ represents the time passed for an observer while watching events that are at rest relative to the observer.

This is the point of view where there is no relative velocity between events and observer.

For example, if you are flying in an airplane, the events on the plane occur in the same location, because there is no relative motion.

Thought Experiment: You are on Earth, and a spaceship zooms past near the speed of light. When the ship is overhead, will you see the ship in slow motion or fast forward?

results

You will see them in slow motion. Observers always see moving objects in slow motion.

Although, this gets more complicated when an object is moving towards or away from you because of the Doppler effect. The doppler effect makes it so that when something is moving towards you they will seem to you to be in fast forward, and when they move away from you they seem to be in slow motion.

What does the spaceship see?
Are you in fast forward, slow motion, or normal?

results

From the ship's point of view you are in slow motion! This is similar to how two people at a distance will see each other as smaller.

Example: A spaceship is moving at 0.9c relative to Earth. A person on the spaceship cooks a burrito for 90 seconds. How long does an Earth based observer have to watch the spaceship while the burrito heats up?

disclaimer

These problems about spaceships are all science fiction.

Current spaceships can't travel near the speed of light. As of 2025, the fastest manmade object is the Parker Solar Probe at 191 000 m/s (430 000 mph) relative to Earth. This is still much less than 300 000 000, so you wouldn't be able to sense the relativistic effects. $$\frac{191000}{300000000} = 0.0006366$$ $$ \gamma = 1.0000002 $$

hint

What is the event? (the 90s on the ship)
Who sees the event at rest? (the ship observer)
Who sees the event as moving? (the Earth observer)

solution

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = 2.29$$
$$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (2.29)( 90s) $$ $$ \Delta t = 206 \, \mathrm{s}$$

Does this make sense? Should it take more than 90s for the Earth observer?

Yes. If you watch something in slow motion it will last longer.

Example: A spaceship is moving at 0.9c relative to Earth. A person on the ship can see a live Earth broadcast of an episode of Star Trek, but the episode lasts 100.9 minutes. How long is the episode when viewed at rest on Earth?

solution

$$\gamma = 2.29$$
$$ \Delta t = \gamma \Delta t_0 $$ $$ 100.9 \, \mathrm{ min} = (2.29) \Delta t_0 $$ $$ \Delta t_0 = 44 \, \mathrm{ min} $$

Example: What fraction of the speed of light is $ \small 2\times 10 ^{8} \tfrac{m}{s} $?

solution

$$ 2\times 10 ^{8} \, \mathrm{\tfrac{m}{s}} \left( \frac{c}{3 \times 10^8 \tfrac{m}{s}} \right) = \tfrac{2}{3} c = 0.\overline{6}c$$

Example: A spaceship is moving at $ \small 2\times 10 ^{8} \tfrac{m}{s} $ relative to the Earth. A person on the spaceship watches a 74 minute episode of the show Black Mirror at normal speed. How long would an Earth observer spying on the spaceship have to wait for the episode to be over?

solution

$$ 2\times 10 ^{8} \, \mathrm{\tfrac{m}{s}} \left( \frac{c}{3 \times 10^8 \, \mathrm{\tfrac{m}{s}}} \right) = \tfrac{2}{3} c$$ $$\gamma = 1.34$$
$$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (1.34) (74 \, \mathrm{min}) $$ $$ \Delta t = 99.16 \, \mathrm{min} $$

Muons are elementary particles similar to electrons, but more massive. Muons are produced in Earth's upper atmosphere by cosmic rays, but they have a half-life of 1.5 microseconds at rest. This means after 0.0000015 seconds half of a population of muons will decay into other particles.

Example: As muons enter the Earth's atmosphere, they typically have a speed of 0.98c. How long would the half life of these high speed muons look to an observer on Earth?

solution

$$\gamma = 5.0$$
$$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (5) (1.5 \,\mu\mathrm{s}) $$ $$ \Delta t = 7.5 \,\mu\mathrm{s}$$

Example: Find the relative velocity of a clock that runs at one-half the rate of a clock at rest.

strategy

Use the time dilation equation to solve for the Lorentz factor.
Use the Lorentz factor to find the velocity.

solution

$$ \Delta t = \gamma \Delta t_0 $$ $$ 1 = \gamma \tfrac{1}{2} $$ $$ \gamma = 2 $$
$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$2 = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\tfrac{1}{2} = \sqrt{1-\frac{v^2}{c^2} }$$ $$\tfrac{1}{4} = 1-\frac{v^2}{c^2}$$ $$\tfrac{3}{4} = \frac{v^2}{c^2}$$ $$\sqrt{\tfrac{3}{4}} = \frac{v}{c}$$ $$0.866c=v$$

Length Contraction

Length contraction only occurs in the direction of the relative velocity. Objects don't look smaller; they look shorter.

An observer sees lengths contracted along the relative velocity.

$$ L = \frac{L_0}{\gamma}$$

$ L$ = contracted length, as seen from a relative velocity
$ L_0$ = rest length, as seen from its own rest frame
$ \gamma $ = Lorentz factor, gamma [no units]

An object will be contracted as seen by a moving reference frame, so the rest length is always longer. L₀ > L.

Example: When you are holding a meter stick it looks like it is 1 meter long. How long would the meter stick look if it was moving lengthwise at 0.5c relative to you?

solution

$$\gamma = 1.15$$
$$L = \frac{L_0}{\gamma}$$ $$ L = \frac{1 \, \mathrm{m}}{1.15} $$ $$ L = 0.87 \, \mathrm{m} $$

Example: I see myself as 1.84 m tall. If I ran past you at 0.4c how tall would I look to you?

solution

I would still look 1.84 m tall because the length contraction is in the direction you are traveling. I would look thinner though!

Example: You see a spaceship pass over head and you quickly record some information about the ship. How long would the ship be if it landed on Earth?

width (side to side) = 40 m
windows = 87
length (front to back) = 134 m
relative velocity = 0.99c
color = fuchsia
passengers = at least 2 dogs

solution

$$\gamma = 7.09$$
$$L = \frac{L_0}{\gamma}$$ $$ 134 \, \mathrm{m} = \frac{L_0}{7.09} $$ $$ L_0 = 950 \, \mathrm{m}$$

Mass and Energy

One of the implications of special relativity is that mass is a type of energy. Mass is not the same as energy. Mass is a type of energy, just like how kinetic and gravitational potential are types of energy.

We originally thought that mass produced gravity and had inertia. We now know that all types of energy have these properties, although it does take a large amount of energy to measure them.

If you add kinetic energy to an object it will have more gravity and more inertia. This also means that objects with no mass, like light, have inertia and gravity.

$$ E = \gamma mc^2$$

$ E$ = Energy [J]
$ \gamma $ = Lorentz factor, gamma [no units]
$ m $ = mass [kg]
$c$ = speed of light, 3 × 10⁸ [m/s]

At rest, the Lorentz factor is one. This gives us Einstein's famous equation for mass-energy equivalence at rest.

$$E = mc^2$$

Example: A massive object at rest still has a huge amount of energy. The book "Surely You're Joking, Mr. Feynman!" by Richard Feynman has a mass of 0.27 kg. How much energy does the book have at rest?

solution

$$ E = mc^2 $$ $$ E_{\text{book}} = (0.27 \, \mathrm{kg})(3 \times 10^{8} \, \mathrm{\tfrac{m}{s}})^2 $$ $$ E_{\text{book}} = 2.43 \times 10 ^{16} \, \mathrm{J}$$

Richard Feynman was part of the Manhattan Project that developed the first nuclear bomb. The energy released by the nuclear bomb dropped on Hiroshima is 400 times less than the energy stored in his book:

$$E_{\text{bomb}} =6.3 \times 10^{13} \, \mathrm{J}$$

Mass is not a conserved value. Mass can be created and destroyed when it changes into a different type of energy. For example, the mass of a particle is converted into energy when it is annihilated.

$$ e^- + e^+ \to \gamma + \gamma $$

When a fundamental particle and its antiparticle collide they annihilate to form two photons. The photons have the same energy as the original particles.
(The symbol gamma is used to represent a photon, not the Lorentz factor.)

Example: The antimatter version of an electron is a positron. Calculate the energy of a photon produced by the annihilation of an electron and positron. (Assume the electron and positron aren't moving at relativistic speeds.)

Subatomic Particles Data Table

name	symbol	charge (e)	mass (kg)
proton	$p^+$	+1	1.6726 × 10^-27
electron	$e^-$	−1	9.1094 × 10^-31
positron	$e^+$	+1	9.1094 × 10^-31
neutron	$n$	0	‎1.6749 × 10^-27
neutrino	$v_e$	0	1.78 × 10^-36
antineutrino	$\bar{v}_e$	0	1.78 × 10^-36
muon	$ \mu^-$	−1	1.8835 × 10^-28
alpha	$ \small{}^4_2 \normalsize \alpha$	+2	6.6447 × 10^-27

strategy

Matter antimatter annihilation converts all the mass of a particle into energy.

Calculate the energy for an electron and positron. You can look up their mass on the data table. The electron and positron energy will equal the energy of the 2 gamma rays produced by the annihilation.

solution

$$ m_e = 9.109 \times 10^{-31} \, \mathrm{kg} $$
$$ E = mc^2 $$ $$ E = (9.109 \times 10^{-31})(3 \times 10^{8})^2 $$ $$ E = 8.198 \times 10^{-14} \, \mathrm{J}$$ Photons produced by annihilation are typically considered gamma rays.

Example: The Large Hadron Collider can accelerate a proton to 0.999999991c. How much energy would an observer measure for a proton at that speed?

solution

$$ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$ \gamma = \frac{1}{\sqrt{1-\frac{(0.999999991)^2c^2}{c^2} } }$$ $$ \gamma = \frac{1}{\sqrt{1-(0.999999991)^2 } }$$ $$ \gamma = 7453.6$$
$$m_\mathrm{p} = 1.67 \times 10^{-27}$$
$$ E = \gamma m c^2 $$ $$ E = (7453.6) (1.67 \times 10^{-27}) (3\times 10^8)^2 $$ $$E = 1.12 \times 10^{-6} \, \mathrm{J} $$

Question: If mass and energy are the same thing, why do we still use different words for them? Are they really the same?

answer

Mass and energy are interchangeable concepts with different units. When we talk about the 'mass' of an object in modern physics, especially for particles, we're usually referring to its rest mass. This is the energy an object has when it's not moving.

Mass gets more complex for composite particles, like protons and neutrons. These composite particles have energy in the form of mass and also the internal binding energy holding groups of particles together.

For example, a proton is made of 3 particles called quarks, but the rest mass of 3 quarks is less than the mass of a proton. It turns out that 99% of a proton's mass comes from the binding energy of the strong nuclear force that holds the quarks together.

It turns out that the rest mass of fundamental particles isn't where most energy is stored. Protons, neutrons, atoms, molecules, people, planets, and stars get most of their mass from the binding energy of the strong nuclear force. Of course those things only make up about 5% of the total energy in the universe. The rest of our observable universe is dark matter(27%), and dark energy(68%). Dark matter and dark energy aren't well understood, but there is strong evidence they exist.

$$ n \to p^+ + e^- + \bar{v}_e $$

A free neutron is unstable with a half-life of about 15 minutes. It decays into a proton, an electron and an antineutrino. This process is called beta decay.

Example: The beta decay products are measured to have a total kinetic energy of 0.78 MeV. Show that conservation of energy holds for beta decay.

Subatomic Particles Data Table

name	symbol	charge (e)	mass (kg)
proton	$p^+$	+1	1.6726 × 10^-27
electron	$e^-$	−1	9.1094 × 10^-31
positron	$e^+$	+1	9.1094 × 10^-31
neutron	$n$	0	‎1.6749 × 10^-27
neutrino	$v_e$	0	1.78 × 10^-36
antineutrino	$\bar{v}_e$	0	1.78 × 10^-36
muon	$ \mu^-$	−1	1.8835 × 10^-28
alpha	$ \small{}^4_2 \normalsize \alpha$	+2	6.6447 × 10^-27

strategy

Convert the masses of each particle into energy. Conservation of energy holds if the left side of the reaction equals the right side plus the kinetic energy.

The antineutrino has such a low mass it can be ignored

Be sure to include enough precision in the masses.

solution

$$E_0=mc^2$$ $$E_p = (1.672621 \times 10^{-27})(9 \times 10 ^{16}) = 1.5053589 \times 10^{-10} \, \mathrm{J}$$ $$E_e = (9.109383 \times 10^{-31})(9 \times 10 ^{16}) = 8.1984447 \times 10^{-14} \, \mathrm{J}$$ $$K = 0.78 \, \mathrm{MeV} \left(\frac{1.602 \times 10^{-19} \, \mathrm{J}}{ \, \mathrm{eV}}\right) = 1.24 \times 10^{-13} \, \mathrm{J} $$
$$ E_p + E_e + K$$ $$ \small 1.5053589 \times 10^{-10} \, \mathrm{J} + 8.1984447 \times 10^{-14} \, \mathrm{J} + 1.24 \times 10^{-13} \, \mathrm{J} = $$ $$ \color{#f02} 1.507418 \times 10^{-10} \, \mathrm{J} $$
$$ E_n = (1.674927 \times 10^{-27})(9 \times 10 ^{16}) = \color{#f02} 1.5074343 \times 10^{-10} \, \mathrm{J}$$
The energy of a neutron is very close to the total energy of its products. The small difference can be attributed to the detail in the measurements.

Kinetic energy is the energy an object has because of its velocity. We can find a relativistic equation for kinetic energy by finding the difference between the energy of an object at a relative velocity and at rest.

$$K = E - E_0$$ $$K = \gamma mc^2 - mc^2 $$ $$K = (\gamma -1)mc^2$$ Example: How much kinetic energy would an observer measure for a 1 kg object moving at half the speed of light?

relativistic solution

$$\gamma = 1.15$$
$$K = (\gamma -1)mc^2$$ $$K = (1.15-1)(1)(3 \times 10^8)^2$$ $$K = (0.15)(9 \times 10^{16})$$ $$K = 1.35 \times 10^{16} \, \mathrm{J}$$

non-relativistic solution

$$K = \tfrac{1}{2}mv^2$$ $$K = \tfrac{1}{2}(1)(1.5 \times 10^8)^2$$ $$K = 1.125 \times 10^{16} \, \mathrm{J}$$

Spacetime

Another implication of relativity is that space and time are unified to form a single four-dimensional continuum called spacetime.

$$ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$ $ \Delta s^2$ = spacetime interval [light-years², ly²]
$ \Delta t $ = time [years, yrs]
$ \Delta x $ = distance [light-years, ly]
$c$ = speed of light, 3 × 10⁸ [m/s]

$ \Delta s^2 $ is the spacetime interval. Not $ \Delta s $.

Observers at different relative velocities will disagree about ∆t and ∆x, but they will always agree on the value of ∆s².

Example: What is the spacetime interval for an event 7 years in the future and 8 light-years away?

light-years

A light-year [ly] is a unit of distance. It's the distance that light travels in one year. Light-years are a great unit for the spacetime interval because they can be used as a unit of distance or time when moving at the speed of light.

Multiplying years by the speed of light converts the units into light-year. $$ c = \frac{\Delta x}{\Delta t} $$ $$ c \Delta t = \Delta x$$ $$ (\tfrac{\mathrm{m}}{\mathrm{s}}) (\mathrm{yrs}) = \mathrm{ly} $$ So years times the speed of light can be added to light years with out the need for conversions.

solution

$$ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 $$ $$ \Delta s^2 = c^2 (7 \, \mathrm{yr})^2 - (8 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = (7 \, \mathrm{ly})^2 - (8 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = 49 \, \mathrm{ly}^2 - 64 \, \mathrm{ly}^2 $$ $$ \Delta s^2 = -15 \, \mathrm{ly}^2 $$

Remember, $ \Delta s^2 $ is the spacetime interval. Not $ \Delta s $. So we can stop here.

The spacetime interval's sign tells us about the possible causal relationship between two events. In other words we can tell if two events in spacetime could communicate.

$$ \Delta s^2 = \mathrm{interval} $$ $$ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 $$ When only a single space dimension is needed we can use a shorter equation for the spacetime interval.
$$ \Delta s^2 = 0 $$ $$ c^2 \Delta t^2 = \Delta x^2 $$ A light-like interval is zero. This means that two events could be connected by a photon of light.
$$ \Delta s^2> 0$$ $$ c^2 \Delta t^2> \Delta x^2 $$ A time-like interval is positive. This means that there could be a causal relationship between two events.
$$ \Delta s^2 < 0$$ $$ c^2 \Delta t^2 < \Delta x^2 $$ A space-like interval is negative. This means that there could not be a causal relationship between two events.

An event can cause another event to occur, but the speed of light puts a limit on causality. There are places in space and time that could never interact with each other, because nothing can move fast enough to connect them. A light cone divides space and time into separate regions to help visualize the limits of causality.

Example: What is the spacetime interval for an event 3 years in the future and 2 light-years away? Is the interval space-like, time-like, or light-like?

solution

$$ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 $$ $$ \Delta s^2 = c^2 (3 \, \mathrm{yr})^2 - (2 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = (3 \, \mathrm{ly})^2 - (2 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = 9 \, \mathrm{ly}^2 - 4 \, \mathrm{ly}^2 $$ $$ \Delta s^2 = 5 \, \mathrm{ly}^2 $$

The interval is greater than zero, so it is time-like. One could potentially effect events 3 years in the future and 2 light-years away.

The nearest star, not counting the Sun, is Proxima Centauri. It is 4.22 light-years away from Earth.

Example: If the current date is January 1st 2049, could you make it to Proxima Centauri by the year 2052? Use the spacetime interval to prove your answer.

solution

$$\Delta t = t_f - t_i $$ $$\Delta t = 2052 \, \mathrm{yr}- 2049 \, \mathrm{yr}$$ $$\Delta t = 3 \, \mathrm{yr} $$
$$ \Delta s^2 = c^2 \Delta t^2 - \Delta x^2 $$ $$ \Delta s^2 = c^2 (3 \, \mathrm{yr})^2 - (4.22 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = (3 \, \mathrm{ly})^2 - (4.22 \, \mathrm{ly})^2 $$ $$ \Delta s^2 = 9 \, \mathrm{ly}^2 - 17.8 \, \mathrm{ly}^2 $$ $$ \Delta s^2 = -8.8 \, \mathrm{ly}^2 $$

The spacetime interval between Earth in 2049 and Proxima Centauri in 2052 is negative. The interval is space-like.

Nothing could travel between the events in that amount of time. This means that these two locations in spacetime couldn't be casually linked.

Lorentz Transformation Solver

Relative Velocity
v = c
v = × 10⁸ m/s
Lorentz Factor
γ =
MOVING RELATIVE TO SPACESHIP
time = s
energy = J
length = m
AT REST RELATIVE TO SPACESHIP
proper time = s
rest energy = J
rest length = m

Frame of reference where an observer sees the spaceship in different locations.
Frame of reference where an observer sees the spaceship in the same location.

Practice printout.pdf

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Question: Two observers move at constant velocity relative to each other. Each observer says the other observer's clock runs slow. Why is that not a contradiction?

answer

Each observer is using their own reference frame as the frame at rest.

Special relativity is symmetric for inertial observers. Each observer sees the other's moving clock as slow, and the disagreement is tied to how each frame measures time and simultaneity.

Example: A research probe passes Earth at 0.60c. Earth is the observer frame, and the probe is moving relative to Earth. What Lorentz factor does Earth use for the probe?

solution

The relative speed is already written as a fraction of c.

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$\gamma = \frac{1}{\sqrt{1-(0.60)^2}}$$ $$\gamma = 1.25$$

Earth uses gamma = 1.25 for this probe.

Example: A particle beam moves at 2.40 × 10⁸ m/s relative to a laboratory detector on Earth. What fraction of c is that, and what Lorentz factor does the laboratory observer use?

solution

First compare the speed to the speed of light.

$$\frac{v}{c} = \frac{2.40 \times 10^8\,\mathrm{m/s}}{3.00 \times 10^8\,\mathrm{m/s}}$$ $$\frac{v}{c} = 0.80$$

Now calculate gamma.

$$\gamma = \frac{1}{\sqrt{1-(0.80)^2}}$$ $$\gamma = 1.67$$

The beam speed is 0.80c, and the laboratory observer uses gamma = 1.67.

Example: A clock inside a spacecraft records an 8.0 minute engine burn. Earth observers see the spacecraft moving at 0.80c. According to Earth, how long does the burn last?

solution

The burn starts and ends at the same place in the spacecraft frame, so the spacecraft clock measures proper time.

$$\Delta t_0 = 8.0\,\mathrm{min}$$ $$\gamma = 1.67$$ $$\Delta t = \gamma \Delta t_0$$ $$\Delta t = (1.67)(8.0\,\mathrm{min})$$ $$\Delta t = 13.4\,\mathrm{min}$$

Earth observers measure the moving burn lasting about 13.4 minutes.

Example: Earth observers measure a high-speed trip lasting 30 years. The spacecraft moves at 0.95c relative to Earth. How much time passes for an astronaut riding with the spacecraft?

solution

Earth sees the spacecraft moving, so Earth measures the dilated time. The astronaut is at rest with the spacecraft clock, so the astronaut measures proper time.

$$\gamma = \frac{1}{\sqrt{1-(0.95)^2}}$$ $$\gamma = 3.20$$ $$\Delta t = \gamma \Delta t_0$$ $$\Delta t_0 = \frac{\Delta t}{\gamma}$$ $$\Delta t_0 = \frac{30\,\mathrm{yr}}{3.20}$$ $$\Delta t_0 = 9.4\,\mathrm{yr}$$

The astronaut ages about 9.4 years during the trip.

Question: A student says, "Time dilation just means the moving clock looks slow because the light from it takes longer to reach us." What is missing from that explanation?

answer

Signal delay is real, but it is not the whole effect.

Time dilation is about what observers calculate after accounting for light travel time. It is not just an optical delay. A moving clock actually measures less proper time between the same pair of events than another inertial frame measures for those events.

Example: Muons created high in the atmosphere have a rest-frame half-life of 2.2 microseconds. Earth observers see the muons moving at 0.98c. What half-life do Earth observers measure?

solution

The muon is at rest with its own decay process, so the muon half-life is proper time.

$$\gamma = \frac{1}{\sqrt{1-(0.98)^2}}$$ $$\gamma = 5.03$$ $$\Delta t = \gamma \Delta t_0$$ $$\Delta t = (5.03)(2.2\,\mu\mathrm{s})$$ $$\Delta t = 11.1\,\mu\mathrm{s}$$

Earth observers measure a half-life of about 11.1 microseconds.

Example: A moving clock ticks once every 5.0 s in its own frame. A lab observer measures 9.0 s between those ticks. How fast is the clock moving relative to the lab, as a fraction of c?

solution

The moving clock is at rest with the two ticks, so 5.0 s is proper time.

$$\Delta t = \gamma \Delta t_0$$ $$9.0\,\mathrm{s} = \gamma(5.0\,\mathrm{s})$$ $$\gamma = 1.80$$

Use gamma to solve for speed.

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$1.80 = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ $$0.556 = \sqrt{1-\frac{v^2}{c^2}}$$ $$0.309 = 1-\frac{v^2}{c^2}$$ $$\frac{v^2}{c^2} = 0.691$$ $$\frac{v}{c} = 0.831$$

The clock moves at about 0.83c relative to the lab.

Example: A survey ship is 120 m long in its own rest frame. It passes Earth lengthwise at 0.75c. What length does an Earth observer measure for the moving ship?

solution

The ship's own frame measures rest length. Earth sees the ship moving, so Earth measures contracted length.

$$\gamma = \frac{1}{\sqrt{1-(0.75)^2}}$$ $$\gamma = 1.51$$ $$L = \frac{L_0}{\gamma}$$ $$L = \frac{120\,\mathrm{m}}{1.51}$$ $$L = 79.5\,\mathrm{m}$$

Earth measures the ship length as about 79.5 m.

Question: A spacecraft is moving horizontally across your screen. Does special relativity make its height contract?

answer

No. Length contraction only occurs along the direction of relative motion.

If the spacecraft moves horizontally, its horizontal length contracts. Its vertical height does not contract from this effect.

Example: Earth observers measure a passing experimental railcar as 72 m long while it moves at 0.88c relative to Earth. The railcar has 12 windows, but that does not matter here. What is the railcar's rest length according to an observer riding with it?

solution

Earth measures contracted length. The rider measures rest length.

$$\gamma = \frac{1}{\sqrt{1-(0.88)^2}}$$ $$\gamma = 2.11$$ $$L = \frac{L_0}{\gamma}$$ $$L_0 = \gamma L$$ $$L_0 = (2.11)(72\,\mathrm{m})$$ $$L_0 = 152\,\mathrm{m}$$

The railcar's rest length is about 152 m.

Example: A spacecraft moving at 0.90c relative to Earth flashes a nose light and a tail light. The lights are 40 m apart in the spacecraft frame, and the ship clock says the flashes happen 3.0 ns apart. What time interval does Earth measure between the flashes?

solution

This cannot be solved with the equations on this page.

The two flashes happen at different locations, so the simple time dilation equation does not apply. The page's time dilation equation needs events that occur in the same location in one frame.

To solve this fully, you would need a Lorentz transformation equation for events separated in space and time. That equation is not taught in this practice section, so the problem is intentionally not solvable from the given tools.

Example: A sealed 4.0 g sample sits at rest on a lab bench. The lab observer wants its rest energy. What is the sample's rest energy in joules?

solution

Convert grams to kilograms first.

$$m = 4.0\,\mathrm{g}\left(\frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}\right)$$ $$m = 0.0040\,\mathrm{kg}$$

At rest, gamma is 1, so use rest energy.

$$E = mc^2$$ $$E = (0.0040\,\mathrm{kg})(3.00 \times 10^8\,\mathrm{m/s})^2$$ $$E = 3.6 \times 10^{14}\,\mathrm{J}$$

The sample has 3.6 × 10¹⁴ J of rest energy.

Question: If a stationary object has rest energy, why does it not explode or release that energy just by sitting there?

answer

Rest energy is not automatically available energy.

The energy is part of the object's mass. It only changes into other forms of energy during processes such as nuclear reactions, particle annihilation, or other interactions where mass changes.

Example: A proton has mass 1.67 × 10^-27 kg and moves at 0.80c relative to a detector. What total relativistic energy does the detector observer assign to the proton?

solution

The detector sees the proton moving, so use total relativistic energy.

$$\gamma = 1.67$$ $$E = \gamma mc^2$$ $$E = (1.67)(1.67 \times 10^{-27}\,\mathrm{kg})(3.00 \times 10^8\,\mathrm{m/s})^2$$ $$E = 2.51 \times 10^{-10}\,\mathrm{J}$$

The detector observer assigns the proton about 2.51 × 10^-10 J of total energy.

Example: A 1200 kg spacecraft moves at 0.20c relative to Earth. Earth observers use the relativistic kinetic energy equation from the page. How much kinetic energy do they assign to the spacecraft?

solution

First calculate gamma for the Earth observer.

$$\gamma = \frac{1}{\sqrt{1-(0.20)^2}}$$ $$\gamma = 1.021$$

Use relativistic kinetic energy.

$$K = (\gamma - 1)mc^2$$ $$K = (1.021 - 1)(1200\,\mathrm{kg})(3.00 \times 10^8\,\mathrm{m/s})^2$$ $$K = 2.3 \times 10^{18}\,\mathrm{J}$$

Earth observers assign the spacecraft about 2.3 × 10¹⁸ J of kinetic energy.

Question: Why should a relativity solution say which observer measures the time or length?

answer

Different inertial observers can measure different times and lengths for the same physical situation.

The equations use different symbols for different observers. Proper time belongs to the observer at rest with the events. Rest length belongs to the observer at rest with the object. If the observer is not named, the problem can become ambiguous.

Example: A space station observer measures two events separated by 6.0 years and 4.0 light-years in the x direction. There is no y or z separation. What spacetime interval squared does that observer calculate?

solution

When time is in years and distance is in light-years, c is 1 light-year per year.

$$\Delta s^2 = c^2\Delta t^2 - \Delta x^2$$ $$\Delta s^2 = (1)^2(6.0\,\mathrm{yr})^2 - (4.0\,\mathrm{ly})^2$$ $$\Delta s^2 = 20\,\mathrm{ly^2}$$

The spacetime interval squared is 20 ly².

Example: An electron and positron are nearly at rest before they annihilate. Use electron mass 9.11 × 10^-31 kg. If the annihilation makes two identical photons, what energy does each photon have in joules?

solution

Each particle's rest energy becomes photon energy. Since there are two equal-mass particles and two identical photons, each photon gets one electron's rest energy.

$$E = mc^2$$ $$E = (9.11 \times 10^{-31}\,\mathrm{kg})(3.00 \times 10^8\,\mathrm{m/s})^2$$ $$E = 8.20 \times 10^{-14}\,\mathrm{J}$$

Each photon has about 8.20 × 10^-14 J.

Example: A spacecraft moves at 0.92c relative to Earth. An astronaut at rest inside the spacecraft measures a repair taking 3.0 hours, and the spacecraft's rest length is 80 m. What repair time and ship length do Earth observers measure?

solution

The astronaut measures proper time because the repair happens at rest in the spacecraft frame. The astronaut also measures the ship's rest length.

$$\gamma = \frac{1}{\sqrt{1-(0.92)^2}}$$ $$\gamma = 2.55$$

Find the Earth-observer time.

$$\Delta t = \gamma \Delta t_0$$ $$\Delta t = (2.55)(3.0\,\mathrm{h})$$ $$\Delta t = 7.7\,\mathrm{h}$$

Find the Earth-observer length.

$$L = \frac{L_0}{\gamma}$$ $$L = \frac{80\,\mathrm{m}}{2.55}$$ $$L = 31\,\mathrm{m}$$

Earth observers measure the repair taking about 7.7 h and the ship length as about 31 m.

Reading (8 minutes): Read Einstein taught us: It's all 'relative' from Science News Explores. Then answer these questions.

Why can two observers moving at a constant speed relative to each other both use the same laws of physics?

answer

Neither observer has a special claim to being absolutely at rest. The laws of physics work the same way in each inertial frame.

Why did Einstein's idea of a person falling from a roof help him think about gravity?

answer

A falling person feels weightless even while gravity is acting. This suggested a deep connection between gravity and accelerated motion.

Why is it important to say which observer measures a time interval or length in a relativity problem?

answer

Observers in relative motion can measure different times and lengths for the same events. Naming the observer tells which measurement belongs in the equation.

Special Relativity

A Constant Speed of Light

$$c = 3 \times 10^8 \, \mathrm{\tfrac{m}{s}} $$

Special Relativity

YES! and no

Lorentz factor

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

Time Dilation

$$ \Delta t = \gamma \Delta t_0$$

Length Contraction

$$ L = \frac{L_0}{\gamma}$$

Mass and Energy

$$ E = \gamma mc^2$$

Spacetime

Lorentz Transformation Solver

Practice printout.pdf

name	symbol	charge (e)	mass (kg)
proton	\(p^+\)	+1	1.6726 × 10^-27
electron	\(e^-\)	−1	9.1094 × 10^-31
positron	\(e^+\)	+1	9.1094 × 10^-31
neutron	\(n\)	0	‎1.6749 × 10^-27
neutrino	\(v_e\)	0	1.78 × 10^-36
antineutrino	\(\bar{v}_e\)	0	1.78 × 10^-36
muon	\( \mu^-\)	−1	1.8835 × 10^-28
alpha	\( \small{}^4_2 \normalsize \alpha\)	+2	6.6447 × 10^-27