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.




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

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?

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.

Thought Experiment: Imagine a train moving at half the speed of sound. Whistles at the front and the back of the train blow at the same time.

results
  • A train passenger measures the sound from the front of the train as moving faster, and the sound from the back as moving slower.
  • An observer on the ground measures both sound waves moving at the same speed, but the train is moving to the right.

Both frames of reference agree that the sound waves meet towards the back of the train.

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?

Thought Experiment: Imagine a train moving at half the speed of light. Imagine lights on the front and back of the train flash simultaneously for both observers.

results
  • The train isn't moving from the passenger's frame of reference. This means they see the light meet in the middle of the train.
  • A frame of reference on the ground sees the train moving to the right. They see the light waves meet closer to the back of the train.

The frames of reference don't agree on the location and timing of events. This happens when two frames of reference have a relative velocity between them. These contradictory results indicate there is a problem with our model of physics.

Special Relativity

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

In 1905 Albert Einstein published his answer to the problem. He 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 predicts 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, but so far all tests support the theory to a high degree of precision.




    Here is a relative velocity simulation with time dilation and length contraction. Move the mouse vertically to change observers.

    The doppler effect and a few other quirks of relativity aren't part of this simulation. Try this game or this video for a more accurate experience.

    Question: Are the predictions of special relativity 'real' or an 'illusion' that only affects how an observer sees things?
    solution

    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?
    answer

    YES! and no

    You can already go forwards in time by just waiting, but you will get old.

    If you have a velocity near the speed of light 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.

    Thought Experiment: Remember the train paradox? The ground based frame of reference and the train based frame of reference disagreed on where the flashes of light met. Special relativity removes this inconsistency by not letting different frames agree on simultaneity.

    It is only possible for the light flashes to be simultaneous for one of the observers. If the flashes occur at the same time for the train observer, the ground observer will see the flashes at different times.


    Question: What relativistic effects can you observe in the the simulation.
    answer

    simultaneity and length contraction

    The trees are narrow from the train's reference frame. The train is shorter from the ground's reference frame.

    Time dilation should also be present, but there isn't a way to see it in the simulation.

    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 be 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.
    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.

    0.10.20.30.40.50.60.70.80.91.01.11234567 v/cLorentz factor

    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.

    Δtₒ Δt L W D D L

    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 \(\Delta t> \Delta t_0\).

    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.



    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 bound 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 2018, the fastest manmade object is the Juno space probe at 73,800 m/s relative to Earth.

    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 $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$$ $$\gamma = 2.29$$
    $$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (2.29)( 90s) $$ $$ \Delta t = 206 \, \mathrm{s}$$
    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 $$\gamma = 5.0$$
    $$ \Delta t = \gamma \Delta t_0 $$ $$ \Delta t = (5) (2.2 \,\mu\mathrm{s}) $$ $$ \Delta t = 11 \,\mu\mathrm{s}$$
    Example: Find the relative velocity of a clock that runs at one-half the rate of a clock at rest.
    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]

    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?
    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?


    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 implications of special relativity is that mass and energy are equivalent. Mass is energy. Energy is mass.

    $$ E = \gamma mc^2$$

    \( E\) = 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.

    $$E = mc^2$$
    feynman 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–energy equivalence means that 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.

    Time 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 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 the same thing with different units. An accurate measurement of mass includes all energies. I'm not sure why we continue to use separate terminology. Maybe it's tradition, or maybe different fields need different units.


    Rest mass is the energy of single particle at rest. Most particles are actually made of several fundamental particles that are held together by a force. When we say "mass" we typically mean the sum of rest mass and the internal energies 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.

    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.

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

    \( \Delta s^2 \) is the spacetime interval. Not \( \Delta s \). 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.

    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.

    worldline

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

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

    \( \Delta s^2 \) is the spacetime interval. Not \( \Delta s \).

    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