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The two postulates of special relativity, events and their coordinates, timing distant events by light, and the relativity of simultaneity.
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By the end of this lesson you will be able to state the postulates of relativity, time events with light signals, and explain why simultaneity depends on the frame.
From mechanics you know reference frames, relative velocity, and that Newton's laws hold in any frame moving at constant velocity. From the electromagnetism courses you know that Maxwell's equations predict light traveling at $c = 1/\sqrt{\mu_0\varepsilon_0}$. This lesson asks the question that puzzled physicists around 1900: relative to what does light move at $c$?
| Term | What it means |
|---|---|
| Event | Something that happens at one place and one instant, labeled $(t, x, y, z)$. |
| Inertial frame | A frame of reference in which a free object moves at constant velocity. |
| Postulate | A starting assumption from which a theory's results are derived. |
| Simultaneous | Happening at the same time coordinate in a given frame. |
| Spacetime diagram | A graph of $ct$ against $x$ on which events are points and motions are lines. |
| Light-travel correction | Subtracting $d/c$ from a seeing time to find when a distant event happened. |
| Galilean transformation | $x' = x - vt$, $t' = t$: the pre-relativity rule for changing frames. |
Einstein built special relativity in 1905 on two postulates:
An event is a place and a time. To time a distant event, an observer must correct for the light's travel: an event seen at time $t$ from a distance $d$ happened at $t - d/c$. Done carefully, this reveals that simultaneity is relative: two events at the same time in one frame happen at different times in a frame moving along the line joining them. The later lessons of this unit find by how much, and what it does to clocks and rulers.
Another way: picture
Picture a train with a lamp at its exact middle. The lamp flashes, and light runs to both ends. For a passenger, it reaches the front and the back at the same moment: the distances are equal and light moves at $c$. For someone on the platform, the back of the train rushes toward the light and the front runs away from it, so the light reaches the back first. Both are right, and that is relativity of simultaneity.
Another way: steps
Before 1905, physicists assumed light moved at $c$ relative to a medium, the ether, so an observer moving through it should measure light's speed as $c + v$ or $c - v$. In 1887 Albert Michelson and Edward Morley, working in Cleveland at the Case School of Applied Science, compared the speed of light along and across the Earth's orbital motion with an interferometer that could detect a difference of a few kilometers per second. They found none.
Later experiments with light from moving stars, from particles moving at nearly $c$, and from sources on rotating turntables all agree: light's speed does not depend on the motion of its source or its observer. Einstein took this as a postulate and asked what the rest of physics must look like for it to be true.
An event is a single happening: a flash, a collision, a clock's tick. Each frame labels it with coordinates $(t, x, y, z)$. It is convenient to measure time in meters by multiplying by $c$: one nanosecond is $0.300$ m of light travel, roughly a foot, and one microsecond $300$ m.
On a spacetime diagram, with $ct$ up and $x$ across, an event is a point, an object sitting still is a vertical line, and a light pulse is a line at $45°$. A moving object is a line tilted less than $45°$ from vertical. Nothing that carries information can have a world line flatter than a light pulse's, a fact that will turn out to protect cause and effect.
You see an event when its light reaches you, not when it happens. Light from the Sun left it about $8.3$ minutes ago; a radio command to a rover on Mars takes between $3$ and $22$ minutes to arrive. To assign a time to a distant event, an observer subtracts the travel time: $t_{\text{event}} = t_{\text{seen}} - d/c$.
Clocks across a frame are synchronized the same way. Send a light pulse from clock A to clock B and back; B should read A's sending time plus half the round-trip time when the pulse arrives. Einstein's point was that this procedure, done in different frames, gives different answers for which distant events are simultaneous.
Lightning strikes both ends of a moving train at the same instant in the ground frame. In that frame, the train's midpoint moves toward the front flash and away from the back flash, so it meets the front flash first. A passenger at the midpoint sees the front flash first, and since in the train's frame both flashes traveled equal distances at $c$, the passenger concludes the front strike happened first.
Neither observer is wrong. Simultaneity depends on the frame. The effect is small for everyday speeds and distances: for a $600$ m train at highway speed, the disagreement is about $10^{-14}$ s. It becomes large only when speeds approach $c$ or distances are astronomical.
Before relativity, changing frames used the Galilean transformation: $x' = x - vt$ and $t' = t$. Velocities simply add, and time is the same for everyone. For a ball thrown forward at $20$ m/s from a truck at $30$ m/s, the ground measures $50$ m/s, correct to far better than anyone can measure.
Applied to light, the same rule predicts $c + v$ from a moving source, contradicting experiment. Something in the Galilean rule must give, and the second postulate says it is the assumption $t' = t$. The next lesson finds the replacement, the Lorentz transformation, which reduces to Galileo's rule when $v$ is much less than $c$.
If simultaneity is relative, can the order of events flip, so that an effect comes before its cause? Only for events so far apart in space that no signal traveling at or below $c$ could connect them. Such events cannot influence each other, so their order does not matter physically.
For events that a light signal could connect, every observer agrees on which came first. The test is simple: if the time between the events is more than the light travel time between their places, $\Delta t > \Delta x/c$, one could have caused the other, and their order is fixed. Lesson 4 turns this test into the invariant interval.
Checking an answer. Times must be positive and later than emission. A result must reduce to the everyday one when $v$ is small. And statements about simultaneity must name the frame they hold in; a claim that two events are simultaneous without a frame is incomplete.
Using $c \pm v$ for closing speeds does not contradict the second postulate. In the ground frame, light moves at $c$ and the train at $v$; the rate at which the gap between them shrinks is $c + v$ or $c - v$, a statement about two motions measured in one frame. What the postulate forbids is a frame in which light itself moves at other than $c$.
Assigning times to distant events by subtracting travel time assumes light's speed is the same in both directions, which is part of how clocks are synchronized. With that convention, every inertial frame builds a consistent grid of synchronized clocks, and events get unambiguous coordinates within it.
Light takes $1.28$ s from the Moon, $8.3$ minutes from the Sun, and about $4$ hours from Neptune. Every command to a spacecraft has to allow for these delays: NASA's Deep Space Network, with antennas at Goldstone in California's Mojave Desert, near Madrid and near Canberra, times its signals to fractions of a nanosecond.
The Voyager 1 probe, launched in 1977, is so far away that its signals take nearly a day to arrive. Engineers who send it a command must wait almost two days for the reply. The delays are not a technical inconvenience but a limit of nature: no signal can go faster.
The Apollo 11, 14 and 15 astronauts left arrays of corner reflectors on the Moon. Since 1969, observatories including McDonald Observatory in Texas and Apache Point Observatory in New Mexico have fired laser pulses at them and timed the echoes. A round trip takes about $2.5$ s, and timing it to a few picoseconds measures the distance to millimeters.
The measurements show the Moon receding by about $3.8$ cm a year, as tides transfer the Earth's spin to the Moon's orbit. They also test relativity: the Moon falls toward the Sun at the same rate as the Earth, as Einstein's equivalence principle requires, to better than a part in ten trillion.
Each GPS satellite broadcasts the time from its atomic clock. A receiver compares the arrival times of signals from four or more satellites; since each nanosecond of delay means $0.3$ m of distance, the differences locate the receiver in three dimensions and fix its own clock error at the same time.
The system, run by the U.S. Space Force from Colorado, only works because it treats light-travel times with care, synchronizing satellite clocks by the procedure Einstein described. It also corrects for relativistic effects on the clocks themselves, which would otherwise throw positions off by about $10$ km a day, as lesson 3 explains.
Everyday experience suggests that "at the same time" means the same thing for everyone. It does not. Two events far apart along the direction of relative motion that are simultaneous in one frame happen at different times in the other, and each frame's judgment is equally valid.
A related error is to think that seeing two events at once means they happened at once. What you see depends on where you stand; only after correcting for light travel time can you say when events happened in your frame.
A telescope records a flare on the Sun at $12{:}00{:}00$ noon. The Sun is $1.50 \times 10^{11}$ m away. Find the light travel time.
$\Delta t = \dfrac{1.50 \times 10^{11}}{3.00 \times 10^8} = 500\ \text{s}$
Distance over speed.
Convert to minutes and seconds.
$500\ \text{s} = 8\ \text{min}\ 20\ \text{s}$
Sixty seconds a minute.
Find when the flare happened.
$11{:}51{:}40\ \text{a.m.}$
Subtract the travel time.
Its particles arrive at $400$ km/s. Find their travel time.
$\dfrac{1.50 \times 10^{11}}{4.00 \times 10^5} = 3.75 \times 10^5\ \text{s} \approx 4.3\ \text{days}$
Much slower than light.
State the warning time.
$\text{about 4 days after the flash is seen}$
Why space weather forecasters watch for flares.
Clocks A and B are $3.00$ km apart. A sends a light pulse at $t_A = 0$. Find its travel time.
$\dfrac{3000}{3.00 \times 10^8} = 10\ \mu\text{s}$
Distance over $c$.
Find what B should read when the pulse arrives.
$t_B = 10\ \mu\text{s}$
If the clocks are synchronized.
B reads $13\ \mu$s on arrival. Find its offset.
$13 - 10 = 3\ \mu\text{s ahead}$
B's clock runs ahead of A's.
B reflects the pulse; A receives it at $20\ \mu$s. Check the method.
$\text{halfway point of the round trip} = 10\ \mu\text{s}$
Assumes equal speed both ways.
Correct B's clock.
$\text{set B back } 3\ \mu\text{s}$
Now both agree in this frame.
Note the frame dependence.
$\text{a moving observer would call these clocks unsynchronized}$
Synchronization is relative, as simultaneity is.
A $300$ m train moves at $0.60c$. Lightning strikes both ends at the same instant in the ground frame. Find $L/2$.
$\dfrac{L}{2} = 150\ \text{m}$
Distance from each end to the midpoint.
Find the front flash's time to the midpoint, in the ground frame.
$t_f = \dfrac{150}{1.60 \times 3.00 \times 10^8} = 0.3125\ \mu\text{s}$
Closing speed $c + v$.
Find the back flash's time to the midpoint.
$t_b = \dfrac{150}{0.40 \times 3.00 \times 10^8} = 1.25\ \mu\text{s}$
Closing speed $c - v$.
Find the gap between arrivals.
$t_b - t_f = 0.9375\ \mu\text{s}$
The front flash arrives first.
Check with the formula.
$\dfrac{L}{c}\cdot\dfrac{\beta}{1 - \beta^2} = 1.0 \times \dfrac{0.60}{0.64} = 0.9375\ \mu\text{s}$
Agrees.
Interpret in the train's frame.
$\text{front strike before back strike}$
Equal distances at $c$, but the front light arrived first.
Find the gap at highway speed, $30$ m/s.
$\approx \dfrac{L}{c}\cdot\dfrac{v}{c} = 10^{-6} \times 10^{-7} = 10^{-13}\ \text{s}$
Far too small to notice.
Note the round trip.
$d = \dfrac{ct}{2}$
Out and back.
Substitute the values.
$d = \dfrac{3.00 \times 10^8 \times 2.0 \times 10^{-4}}{2}$
SI units.
Evaluate the distance.
A spaceship passes Earth at $0.8c$ and shines a laser beam straight ahead. How fast do observers on Earth measure the beam to travel?
Complete the worked solution: two GPS satellites are $20100$ km and $21300$ km from a receiver. With $c = 300$ km/ms, find the first signal's travel time in ms, how much later the second signal takes to arrive in ms, and how far off the position would be for a clock error of $2$ ns, in m.
Divide the first distance by the speed of light.
$t_1 = \dfrac{d_1}{c} =$ t
In milliseconds.
Find the extra delay of the second signal.
$t_2 - t_1 = \dfrac{d_2 - d_1}{c} =$ d
Differences in delay are what locate the receiver.
Convert the clock error to distance.
$\Delta d = c\,\Delta t = 0.3\ \text{m/ns} \times 2\ \text{ns} =$ r
Every nanosecond of timing error is $0.3$ m.
Explain why relativity matters.
$\text{uncorrected, clock effects add about } 38\ \mu\text{s a day}$
Enough to drift kilometers a day, as later lessons show.
Match each term of relativity to what it means.
| a place and a time | a frame moving at constant velocity | the laws of physics are the same in every inertial frame | light travels at c in every inertial frame | |
|---|---|---|---|---|
| event | ||||
| inertial frame | ||||
| first postulate | ||||
| second postulate |
Event A happens at the origin at $t = 0$. Event B happens $6$ m away at $t = 72$ ns. With $c = 0.300$ m/ns, fill in B's time coordinate as a distance, $ct$, in meters; the time light needs to travel from A to B's place, in ns; and how much later B happens than that light could arrive, in ns (negative if earlier).
| value | |
|---|---|
| ct for event B (m) | |
| light travel time (ns) | |
| time to spare (ns) |
A flash goes off at $t = 8$ μs at the origin. A detector sits a distance $d$ (m) away. With $c = 300$ m/μs, write the time the detector sees the flash, in μs, as a formula in $d$.
Answer:
In the ground frame, lightning strikes both ends of a $600$ m train at the same instant, as the train moves at $0.8c$. Working in the ground frame, by how much does the flash from the front reach the train's midpoint before the flash from the back, in μs? Take $c = 300$ m/μs.
Answer: μs between arrivals
Astronomers at the Apache Point Observatory in New Mexico fire a laser pulse at a reflector left on the Moon by Apollo astronauts. The echo returns $2.51$ s later. With $c = 3.00 \times 10^5$ km/s, how far away is the Moon at that moment, in km?
Answer: km to the Moon
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A flash goes off at $t = 3$ μs at the origin. A detector sits a distance $d$ (m) away. With $c = 300$ m/μs, write the time the detector sees the flash, in μs, as a formula in $d$.
Answer:
You can reason about events and frames. Explain to someone why two lightning strikes simultaneous for a platform observer are not simultaneous for a passenger.
20. Your turn: a radar pulse returns from an aircraft $0.20$ ms after it was sent. How far away is the aircraft?, step 3
$d = 30\ \text{km}$
About $19$ miles.