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The Lorentz transformation, the factor $\gamma = 1/\sqrt{1 - v^2/c^2}$, transforming events between frames, and relativistic velocity addition.
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By the end of this lesson you will be able to compute Lorentz factors, transform events between frames, and add velocities relativistically.
From the last lesson you have the two postulates and the relativity of simultaneity. From mechanics you have the Galilean rule for changing frames, $x' = x - vt$, which works for everyday speeds. This lesson finds the rule that replaces it, the one that keeps light's speed the same in every frame, and the factor $\gamma$ that appears throughout relativity.
| Term | What it means |
|---|---|
| Speed parameter | $\beta = v/c$, a frame's speed as a fraction of light's. |
| Lorentz factor | $\gamma = 1/\sqrt{1 - \beta^2}$, always at least 1. |
| Lorentz transformation | $x' = \gamma(x - vt)$, $t' = \gamma(t - vx/c^2)$: the rule for changing frames. |
| Boost | A change to a frame moving at constant velocity. |
| Relativistic velocity addition | $u' = (u - v)/(1 - uv/c^2)$. |
| Newtonian limit | Speeds much less than $c$, where $\gamma \approx 1$ and Galileo's rules hold. |
| Light-second | The distance light travels in one second, $3.00 \times 10^8$ m. |
The Galilean rule $x' = x - vt$, $t' = t$ would make light travel at $c - v$ in the moving frame. The only linear rule that keeps light at $c$ in both frames, and treats the two frames symmetrically, is the Lorentz transformation:
$$x' = \gamma(x - vt), \qquad t' = \gamma\left(t - \frac{vx}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}.$$
The Lorentz factor $\gamma$ is $1$ for $v = 0$, $1.005$ at $0.1c$, $1.25$ at $0.6c$, and grows without limit as $v \to c$. The term $vx/c^2$ in the time formula is the relativity of simultaneity: events at the same $t$ but different $x$ get different $t'$. Velocities combine as
$$u' = \frac{u - v}{1 - uv/c^2},$$
which never exceeds $c$ and returns $c$ for light.
Another way: picture
Picture $\gamma$ as a gauge of how relativistic a situation is. For a jet airliner it reads $1.0000000000004$, and Galileo's rules are perfect. For a spacecraft at half light speed it reads $1.15$, a 15 percent effect. For particles in accelerators it reads thousands, and everyday intuitions about time, length and speed fail completely.
Another way: steps
Suppose a light flash leaves the origin at $t = 0$, when the origins of S and S' coincide. In S it spreads as $x = ct$; in S', by the second postulate, as $x' = ct'$. Try a transformation of the form $x' = \gamma(x - vt)$, a Galilean rule stretched by some factor, and by symmetry $x = \gamma(x' + vt')$.
Putting $x = ct$ and $x' = ct'$ into both equations and multiplying them gives $c^2tt' = \gamma^2(c - v)(c + v)tt'$, so $\gamma^2 = 1/(1 - v^2/c^2)$. Solving the two equations for $t'$ then gives $t' = \gamma(t - vx/c^2)$. The whole structure of special relativity follows from asking that one flash look spherical to everyone.
For small $\beta$, $\gamma \approx 1 + \tfrac{1}{2}\beta^2$. A car at $30$ m/s has $\beta = 10^{-7}$ and $\gamma - 1 = 5 \times 10^{-15}$. The International Space Station, at $7.7$ km/s, has $\gamma - 1 = 3.3 \times 10^{-10}$, enough for its astronauts to age a hundredth of a second less than people on the ground over six months. Satellites in higher orbits move more slowly and feel the effect less, while gravity, which the general theory treats, pushes their clocks the other way.
At $0.5c$, $\gamma = 1.155$; at $0.9c$, $2.29$; at $0.99c$, $7.09$; at $0.999c$, $22.4$. Near light speed, each extra nine in $\beta$ multiplies $\gamma$ by about $3.2$. Electrons at the Large Hadron Collider's predecessor, LEP, reached $\gamma$ near $200{,}000$.
Measuring time in light-seconds, the transformation becomes symmetric: $x' = \gamma(x - \beta ct)$ and $ct' = \gamma(ct - \beta x)$. An event at $x = 3$ light-seconds, $ct = 5$ light-seconds, seen from a frame moving at $0.6c$, has $x' = 1.25(3 - 3) = 0$ and $ct' = 1.25(5 - 1.8) = 4$.
The inverse transformation, from S' back to S, is the same with $\beta$ replaced by $-\beta$, as the first postulate demands: S sees S' moving at $+v$, and S' sees S moving at $-v$. Neither frame is special.
An object moving at $u$ in S has speed $u' = (u - v)/(1 - uv/c^2)$ in S'. For an object moving the opposite way to S', the formula gives $u' = (u + v)/(1 + uv/c^2)$ with both speeds counted positive: two spacecraft approaching at $0.6c$ each, seen from one of them, close at $1.2c/1.36 = 0.88c$, not $1.2c$.
Put $u = c$ and the formula gives $u' = c$, whatever $v$ is: the second postulate is built in. For small speeds the denominator is nearly 1 and Galileo's subtraction returns, which is why everyday experience never hinted at the correction.
Adding velocities relativistically never produces a speed above $c$: adding $0.9c$ to $0.9c$ gives $1.8c/1.81 = 0.994c$. A rocket that fires its engine forever, gaining the same speed relative to its own momentary frame each second, approaches $c$ ever more slowly and never reaches it.
As later lessons show, the energy of a body grows as $\gamma$, so reaching $c$ would need infinite energy. Only massless things, light and gravitational waves, travel at $c$, and they cannot travel slower.
Checking an answer. $\gamma$ must be at least 1. Transforming there and back must return the original event. $(ct)^2 - x^2$ must be the same in both frames. No speed of a massive object may come out at or above $c$. And at small $\beta$ the Galilean result must return.
The transformation must be linear, or a body moving uniformly in one frame would appear to accelerate in another, contradicting the first postulate. It must reduce to Galileo's rule at small speeds, because Newtonian mechanics works there. And it must keep light at $c$. These three requirements fix it completely.
The transverse coordinates $y$ and $z$ do not change: a meter stick held across the direction of motion must look the same length to both observers, or one could tell which was moving by comparing sticks as they pass, violating the first postulate.
Hendrik Lorentz found the transformation before Einstein, in trying to explain the Michelson–Morley result with an ether that contracted moving objects. He treated $t'$ as a mathematical convenience, a local time without physical meaning.
Einstein's step in 1905 was to take $t'$ at face value: it is the time real clocks read in the moving frame. He dropped the ether altogether and derived the transformation from the two postulates alone. Hermann Minkowski then showed in 1908 that space and time together form a single four-dimensional spacetime, which lesson 4 explores.
When $\beta$ is close to 1, computing $1 - \beta^2$ directly loses precision: at $\beta = 0.999999$, $\beta^2 = 0.999998000001$, and a calculator showing eight digits would give $1 - \beta^2 = 0.000002$ with only one reliable figure. Factoring as $(1 - \beta)(1 + \beta) = 0.000001 \times 1.999999$ keeps full precision.
Physicists at accelerators work the other way round: they know $\gamma$ from the energy and want $\beta$. For large $\gamma$, $1 - \beta \approx 1/2\gamma^2$, a formula accurate to a part in $4\gamma^2$. That is how one finds that electrons with $\gamma = 1000$ travel only $150$ m/s slower than light.
There is a quantity that does add simply when frames are combined. Define the rapidity $\phi$ by $\beta = \tanh\phi$. Then $\gamma = \cosh\phi$ and $\gamma\beta = \sinh\phi$, and the velocity-addition formula becomes the identity for the hyperbolic tangent of a sum: rapidities of successive boosts along one line simply add.
A rocket that gains the same small speed in its own frame each second gains the same rapidity each second, so its rapidity grows steadily without limit while its speed creeps toward $c$ ever more slowly. At everyday speeds rapidity and $\beta$ are almost equal, which is why Galileo's addition works so well. Particle physicists use the closely related pseudorapidity to describe the directions of particles flying out of collisions, because distributions in it keep their shape when the collision frame is boosted.
The two-mile linear accelerator at SLAC National Accelerator Laboratory in Menlo Park, California, accelerated electrons to $50$ GeV, a Lorentz factor of about $100{,}000$. At that point they traveled only a few centimeters per second slower than light. Each further push added energy, hardly any speed.
Today the same tunnel feeds the Linac Coherent Light Source, an X-ray laser, with electrons at $\gamma$ of about $30{,}000$. Their high $\gamma$ is what makes the X-rays: the light they emit in wiggling magnets is compressed by relativistic effects into wavelengths short enough to image single molecules.
Cosmic rays strike the upper atmosphere and make muons, unstable particles that live on average $2.2$ μs at rest. Many travel at $0.995c$ or faster, with $\gamma$ of $10$ or more. Without relativity, a muon at nearly $c$ would travel only about $660$ m in its lifetime and almost none would reach the ground from $15$ km up.
Yet detectors at sea level count about one muon per square centimeter per minute. The resolution, the subject of the next lesson, is that the muon's clock runs slow by the factor $\gamma$ in the Earth's frame. The measurement, first made on Mount Washington in New Hampshire in 1940, was one of the earliest direct tests of the factor.
In everyday life, speeds add: walking forward at $1$ m/s in a train moving at $30$ m/s gives $31$ m/s relative to the ground. It is natural to expect the same at any speed, so that two probes approaching at $0.6c$ each close at $1.2c$. The relativistic formula gives $0.88c$. Speeds combine so that the result never exceeds $c$.
A related error is to think $\gamma$ can be less than 1, confusing it with its reciprocal $\sqrt{1 - \beta^2}$. The Lorentz factor is always at least 1: moving clocks run slow and moving rods are short by that factor, never the other way.
Find $\gamma$ for a spacecraft at $0.8c$.
$\gamma = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{0.6} = 1.667$
Definition.
Find $\gamma$ at $0.99c$.
$\gamma = \dfrac{1}{\sqrt{0.01 \times 1.99}} = 7.09$
Factor $1 - \beta^2$.
Find $\gamma - 1$ for a jet at $250$ m/s.
$\gamma - 1 \approx \tfrac{1}{2}\left(\dfrac{250}{3.00 \times 10^8}\right)^2 = 3.5 \times 10^{-13}$
Small-speed approximation.
Find the speed for $\gamma = 2$.
$\beta = \sqrt{1 - \tfrac{1}{4}} = 0.866$
Invert the definition.
Find the speed for $\gamma = 10$.
$\beta = \sqrt{1 - 0.01} = 0.995$
Close to light speed.
Frame S' moves at $0.6c$. An event has $x = 4$ ls, $ct = 10$ ls in S. Find $\gamma$.
$\gamma = 1.25, \quad \gamma\beta = 0.75$
At $0.6c$.
Find the position $x'$.
$x' = 1.25 \times 4 - 0.75 \times 10 = -2.5\ \text{ls}$
Behind the moving origin.
Find the time coordinate $ct'$.
$ct' = 1.25 \times 10 - 0.75 \times 4 = 9.5\ \text{ls}$
Earlier than in S.
Transform back to check.
$x = 1.25(-2.5) + 0.75(9.5) = 4.0, \quad ct = 1.25(9.5) + 0.75(-2.5) = 10.0$
Change the sign of $\beta$.
Compare $(ct)^2 - x^2$.
$100 - 16 = 84; \quad 90.25 - 6.25 = 84$
The same in both frames.
Compare with Galileo.
$x' = 4 - 6 = -2\ \text{ls}, \quad t' = t$
Off by 25 percent, and wrong about time.
A ship moves at $0.8c$ relative to Earth and fires a probe forward at $0.5c$ relative to itself. Write the addition rule for Earth's view.
$u = \dfrac{u' + v}{1 + u'v/c^2}$
Inverse of the transformation.
Substitute the speeds.
$u = \dfrac{0.5 + 0.8}{1 + 0.40}c$
In units of $c$.
Evaluate the probe's speed.
$u = \dfrac{1.3}{1.4}c = 0.929c$
Less than $c$.
Compare with Galileo.
$0.5c + 0.8c = 1.3c$
Impossible.
The ship shines light forward. Find its speed from Earth.
$u = \dfrac{1 + 0.8}{1 + 0.8}c = c$
The second postulate, built in.
Fire the probe backward at $0.5c$ instead.
$u = \dfrac{-0.5 + 0.8}{1 - 0.40}c = 0.5c$
Signs matter.
Check the slow-speed limit with $30$ m/s and $20$ m/s.
$u \approx 50(1 - 7 \times 10^{-15})\ \text{m/s}$
Galileo's answer, to fourteen digits.
Square the speed ratio.
$\beta^2 = 0.0784$
$0.28^2$.
Take the root.
$\sqrt{1 - 0.0784} = \sqrt{0.9216} = 0.96$
A 7-24-25 triangle.
Invert for the factor.
Probe number $2$ in a test series flies past a space station at $0.6c$. What is its Lorentz factor $\gamma$?
Complete the worked solution: a frame S' moves at $0.6c$ along $x$ relative to S, where $\gamma = 1.25$. An event has coordinates $x = 2$ light-seconds and $ct = 9$ light-seconds in S. Find its $x'$ and $ct'$ in light-seconds, and the quantity $(ct')^2 - x'^2$.
Transform the position.
$x' = 1.25(2 - 0.6 \times 9) =$ x
$\gamma(x - \beta ct)$.
Transform the time.
$ct' = 1.25(9 - 0.6 \times 2) =$ t
$\gamma(ct - \beta x)$.
Square and subtract.
$(ct')^2 - x'^2 =$ s
The same as $(ct)^2 - x^2$ in S.
Check against the original frame.
$9^2 - 2^2$
An invariant, as lesson 4 explains.
Match each relativistic quantity to its formula.
| $1/\sqrt{1 - v^2/c^2}$ | $\gamma(x - vt)$ | $\gamma(t - vx/c^2)$ | $(u - v)/(1 - uv/c^2)$ | |
|---|---|---|---|---|
| Lorentz factor | ||||
| transformed position | ||||
| transformed time | ||||
| velocity addition |
A spacecraft moves at $\beta = 0.96$. Fill in its Lorentz factor $\gamma$, the factor $\sqrt{1 - \beta^2}$, and the product $\gamma\beta$, each to four figures.
| value | |
|---|---|
| Lorentz factor γ | |
| √(1 − β²) | |
| γβ |
A rocket moves at $0.2c$ relative to Earth. An object moves along the same line at speed $u$ relative to Earth, in units of $c$. Write the object's speed relative to the rocket, in units of $c$, as a formula in $u$.
Answer:
A muon made by a cosmic ray high in the atmosphere moves at $0.999c$. What is its Lorentz factor $\gamma$, to three figures?
Answer: Lorentz factor
An accelerator physicist at SLAC in Menlo Park, California, reports electrons in a beam with Lorentz factor $\gamma = 10$. How much slower than light do they travel, $c - v$, in m/s? Take $c = 3.00 \times 10^8$ m/s.
Answer: m/s short of light
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A rocket moves at $0.3c$ relative to Earth. An object moves along the same line at speed $u$ relative to Earth, in units of $c$. Write the object's speed relative to the rocket, in units of $c$, as a formula in $u$.
Answer:
You can use the Lorentz transformation. Explain to someone why two probes approaching each other at $0.6c$ do not close at $1.2c$.
21. Your turn: find $\gamma$ for a particle at $0.28c$., step 3
$\gamma = \dfrac{1}{0.96} = 1.042$
A 4 percent effect.