Special Relativity

Speed and velocity have no meaning without something to reference. If I said we are moving at 230,000 m/s, your first question should be relative to what. A plane? The Earth? The Sun?

We are moving at 230,000 m/s relative to our galaxy, the Milky Way.

Velocity must be relative to a reference frame. A reference frame is a point of view from which events are observed.

A person on a train is at rest from the reference frame of an observer on that train. An observer at the train station will see the same person as moving.

Each observer only sees motion in others. They see themselves at rest.
Move the mouse vertically to change observers in the simulation below.
(note: This simulation doesn't include special relativity yet.)

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?
solution $$v = 20 \, \mathrm{\tfrac{m}{s}}-15 \, \mathrm{\tfrac{m}{s}}$$ $$v = 5 \, \mathrm{\tfrac{m}{s}}$$
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?
solution $$v = 15 \, \mathrm{\tfrac{m}{s}}-5 \, \mathrm{\tfrac{m}{s}}$$ $$v = 10 \, \mathrm{\tfrac{m}{s}}$$

Speed of Light

In 1676 Ole Rømer discovered that light has a speed by timing the eclipses of the Jupiter moon Io. He found the speed to be around 220,000,000 m/s. This isn't too far from the value we use today.

We know the speed of light is 3 x 10⁸ m/s, but what is that speed relative to?

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

In 1887 the Michelson–Morley experiment proved that the speed of light is constant even for observers moving at different speeds.

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 light as the same?

How can they both see the speed of light as the same? Shouldn't one of the observers see the speed of light as 3 x 10⁸ m/s + 20 m/s?

Question: How would the train experiment work if the observers were measuring the speed of sound?
Sound has a constant speed of around 342 m/s, but only for an observer at rest relative to the air.

  • An observer on a 20 m/s train yells out a window on the train and measures the speed as 343 m/s - 20 m/s. They measure sound as slower.

    (or faster depending on what direction the sound is moving)
  • An observer at the train station measure the speed of the sound as 343 m/s.

Special Relativity

The Michelson–Morley experiment and a few other experiments indicated that there was a problem with our understanding of relative velocity.

In 1905 Albert Einstein published his answer to the problem of relative velocity. Einstein suggested special relativity as a modification of the laws of physics to take into account a constant speed of light. Special relativity suggests that space and time change for different observers in a way that keeps the speed of light constant.

Special relativity is based on two postulates:
1. The laws of physics are the same in all reference frames.
2. Light moves at the same speed for all observers.

Special relativity predicted that observers moving at a relative velocity to each other don't agree on the:

  • order of events (simultaneity)
  • passage of time (time dilation)
  • length of objects (length contraction)
  • mass of objects (mass–energy equivalence)
  • Because of the high speeds needed, special relativity is difficult to observe. So far all tests support the theory to a high degree of precision.

    Here is the relative velocity simulation with modifications for special relativity. Move the mouse vertically to change observers.

    For simplicity, the doppler effect and a few other quirks of relativity aren't part of this simulation. Try this relativity game for a more accurate experience.

    Question: Are the predictions of special relativity 'real' or an 'illusion' that only affects how an observer sees things?
    Effects like time dilation and length contraction have real consequences that all observers can agree on. Time dilation can produce real differences in the ages of objects. (see twin paradox)
    Question: Does special relativity mean time travel might be possible one day?


    You can already go forwards in time by just waiting.

    If you can get up to a high speed relative to the Earth you will age slow as the Earth ages fast. You could use this to see the far future, but right now we don't have a safe way to get a person up to relativistic speeds.

    It is probably impossible to go backwards in time, but the equations don't completely rule it out.

    Lorentz factor

    We will use the Lorentz factor in equations that convert between two non-accelerating reference frames. These equations predict how two observers at a relative velocity disagree on space and time. $$ \Delta t = \gamma \Delta t_0 \quad \quad \quad \quad L = \frac{L_0}{\gamma}$$

    We will explore each of the above equations, but first we will learn how to calculate the Lorentz factor.

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

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

    Gamma 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.
    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 one means that the effect doesn't change anything. At everyday speeds, effects like time dilation aren't noticeable.
    Example: Calculate the Lorentz factor for two reference frames moving at half the speed of light relative to each other.
    v = 0.5c

    solution $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-\frac{(0.5)^2 c^2}{c^2} } }$$ $$\gamma = \frac{1}{\sqrt{1-0.25}}$$ $$\gamma = \frac{1}{\sqrt{0.75}}$$ $$\gamma = \frac{1}{0.866}$$ $$\gamma = 1.155$$
    Example: Calculate the Lorentz factor for two reference frames moving at the speed of light relative to each other.
    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 becomes one. As the relative speed approaches the speed of light, the Lorentz factor increases towards infinity.

    Time Dilation

    Our time dilation equation is a simplified version of the Lorentz transformation. You can find a derivation for time dilation here.

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

    \(\Delta t\) = elapsed time for observer, as seen from a frame of reference where the events occur in different locations

    \(\Delta t_0\) = elapsed time for observer, as seen from 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 \(\Delta t > \Delta t_0\).

    Question: 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?
    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.

    What does the spaceship see?
    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 bound observer have to watch the spaceship while the burrito is heated up?
    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 $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = 2.29$$
    $$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (2.29)( 90s) $$ $$ \Delta t = 206 \, \mathrm{s}$$
    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 2.2 microseconds. This means after 0.0000022 seconds half of a population of muons will decay in 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 $$\gamma = 5.0$$
    $$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (5) (2.2 \times 10^{-6}s) $$ $$ \Delta t = 11 \times 10^{-6}s$$ $$ \Delta t = 11 \mu \, \mathrm{s}$$
    Example: Find the relative velocity of a clock that runs at one-half the rate of a clock at rest.
    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]

    The rest length is always longer. An object will look contracted from a moving reference frame.

    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?
    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?

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

    Mass–Energy Equivalence

    One of the non obvious consequences of special relativity is that mass and energy are equivalent. Mass can be thought of as a type of energy.

    Previously we thought that only mass determined, inertia and gravity. It is more accurate to include all types of energy in these calculations. For example, a hot object produces slightly more gravity than a cold one.

    Light has energy, so it produces and is affected by gravity. The path of light is bent by a gravitational field.

    $$ E = \gamma mc^2$$

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

    The mass of an object is not conserved, but energy is. A massive object can convert some, or all, of its mass into a different type of energy. This process is noticeable in high energy nuclear reactions.

    At rest the Lorentz factor is one. This gives us that famous equation for the mass-energy equivalence of an object at rest.

    $$E_0 = mc^2$$ Example: A massive object at rest still has a huge amount of energy. How much energy is in a 1 kg mass at rest?
    solution $$ E = mc^2 $$ $$ E = (1 \, \mathrm{kg})(3 \times 10^{8} \, \mathrm{\tfrac{m}{s}})^2 $$ $$ E = 9 \times 10 ^{16} \, \mathrm{J}$$ The energy from the nuclear bomb dropped on Hiroshima: $$6.3 \times 10^{13} \, \mathrm{J}$$
    γ γ e - e + $$ e^- + e^+ \to \gamma + \gamma $$

    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 here 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.
    Subatomic Particles Data Table
    name symbol charge (C) charge (e) mass (kg)
    proton \(p^+\) +1.602 × 10 -19 +1 1.6726 × 10 -27
    electron \(e^-\) −1.602 × 10 -19 −1 9.1094 × 10 -31
    positron \(e^+\) +1.602 × 10 -19 +1 9.1094 × 10 -31
    neutron \(n\) 0 0 ‎1.6749 × 10 -27
    neutrino \(v_e\) 0 0 1.78 × 10 -36
    antineutrino \(\bar{v}_e\) 0 0 1.78 × 10 -36
    muon \( \mu^-\) −1.602 × 10 -19 −1 1.8835 × 10 -28
    alpha \( \small{}^4_2 \normalsize \alpha\) +3.204 × 10 -19 +2 6.6447 × 10 -27
    Matter antimatter annihilation converts all the mass of a particle into energy. Find the energy of an electron at rest. That energy will be the same for a photon produced by electron and position 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 $$ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$ \gamma = \frac{1}{\sqrt{1-\frac{(0.999999991)^2c^2}{c^2} } }$$ $$ \gamma = \frac{1}{\sqrt{1-(0.999999991)^2 } }$$ $$ \gamma = 7453.6$$
    $$m_\mathrm{p} = 1.67 \times 10^{-27}$$
    $$ E = \gamma m c^2 $$ $$ E = (7453.6) (1.67 \times 10^{-27}) (3\times 10^8)^2 $$ $$E = 1.12 \times 10^{-6} \, \mathrm{J} $$
    $$ n \to p^+ + e^- + \bar{v}_e $$

    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 (C) charge (e) mass (kg)
    proton \(p^+\) +1.602 × 10 -19 +1 1.6726 × 10 -27
    electron \(e^-\) −1.602 × 10 -19 −1 9.1094 × 10 -31
    positron \(e^+\) +1.602 × 10 -19 +1 9.1094 × 10 -31
    neutron \(n\) 0 0 ‎1.6749 × 10 -27
    neutrino \(v_e\) 0 0 1.78 × 10 -36
    antineutrino \(\bar{v}_e\) 0 0 1.78 × 10 -36
    muon \( \mu^-\) −1.602 × 10 -19 −1 1.8835 × 10 -28
    alpha \( \small{}^4_2 \normalsize \alpha\) +3.204 × 10 -19 +2 6.6447 × 10 -27
    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 $$E_0=mc^2$$ $$E_p = (1.672621 \times 10^{-27})(9 \times 10 ^{16}) = 1.5053589 \times 10^{-10} \, \mathrm{J}$$ $$E_e = (9.109383 \times 10^{-31})(9 \times 10 ^{16}) = 8.1984447 \times 10^{-14} \, \mathrm{J}$$ $$KE = 0.78 \, \mathrm{MeV} \left(\frac{1.602 \times 10^{-19} \, \mathrm{J}}{ \, \mathrm{eV}}\right) = 1.24 \times 10^{-13} \, \mathrm{J} $$ $$ E_p + E_e + KE$$ $$ \small 1.5053589 \times 10^{-10} \, \mathrm{J} + 8.1984447 \times 10^{-14} \, \mathrm{J} + 1.24 \times 10^{-13} \, \mathrm{J} = $$ $$ \color{blue} 1.507418 \times 10^{-10} \, \mathrm{J} $$ $$ E_n = (1.674927 \times 10^{-27})(9 \times 10 ^{16}) = \color{blue} 1.5074343 \times 10^{-10} \, \mathrm{J}$$
    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 subtracting the energy of an object at rest from the energy at a relative velocity.

    $$KE = E - E_0$$ $$KE = \gamma mc^2 - mc^2 $$ $$KE = (\gamma -1)mc^2$$ Example: How much kinetic energy would an observer measure for a 1kg object moving at half the speed of light?
    solution $$ \gamma = 1.15$$
    $$KE = (\gamma -1)mc^2$$ $$KE = (1.1547 -1)(1)(3 \times 10^8)^2$$ $$KE = (0.15)(1)(9 \times 10^{16})$$ $$KE = 1.35 \times 10^{16} \, \mathrm{J}$$
    It would take that much energy to accelerate 1kg to half the speed of light.


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

    $$ \Delta s^2 = \Delta t^2c^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$$

    \( \Delta s \) is the spacetime interval. Observers at different relative velocities disagree about the position of events in space and time, but they will agree on the value of the interval.

    The speed of light puts a hard limit on causality. This means that there are places in space and time that could never interact with each other. A light cone divides space and time into separate regions to help visualize the limits of causality.

    The value of the spacetime interval tells us about the possible causal relationship between two events.

    Derivation of Velocity Time Dilation

    To understand where the time dilation equation comes from we will explore a hypothetical situation. Imagine a light clock that works 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 at rest relative to the observer.
    On the right, a light clock is moving at v relative to the observer. every

    \( \Delta t \) is one tick of the light clock at velocity \(v\).
    \( \Delta t_0 \) is one tick of the light clock at rest.

    Δtₒ Δt L W D D L v

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

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

    Next, we apply the Pythagorean theorem to the triangle.

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

    Lorentz Transformation Solver