Chapter 05 · Modern Physics · Full book
Special Relativity
Core intuition: Chapter Roadmap
This is the volume's capstone. Every chapter before it assumed, without
saying so, that time and length are absolute — the same for everyone.
This chapter shows that they are not, and rebuilds mechanics around the one
thing that is absolute: the speed of light.
- § Two Postulates Why Maxwell's equations forced Einstein to choose between absolute time and a universal light speed — and which he kept.
- § Simultaneity The single idea everything else follows from: two events that are simultaneous for one observer need not be for another.
- § Time Dilation & Length Contraction Moving clocks run slow; moving rulers shrink. The light-clock derivation, and the muon that proves it.
- § The Lorentz Transformation The algebraic engine that contains dilation, contraction, and simultaneity as special cases.
- § Velocity Addition Why \(0.5c\) plus \(0.5c\) is not \(c\), and why nothing made of matter ever catches light.
- § Momentum & Energy The repair of \(p=mv\) and \(K=\tfrac12 mv^2\), ending at the most famous equation in physics, \(E_0 = mc^2\).
- § The Invariant \(E^2=(pc)^2+(mc^2)^2\): one relation that handles electrons, protons, and massless photons alike.
- § Doppler & Spacetime Relativistic Doppler shift, and the Minkowski picture in which all of the above is simple geometry.
Perplexing Questions
Perplexing questions
Hold these five questions in mind. Each one looks like a paradox; each is
answered, fully and without hand-waving, before the chapter ends.
- The Muon That Should Not Arrive: A muon is created about \(15\,\text{km}\) up in the atmosphere when a cosmic ray strikes an air molecule. It lives, on average, only \(2.2\) microseconds before decaying. Even at nearly the speed of light, that is barely enough time to fall \(660\,\text{m}\). Yet detectors at sea level catch these muons in large numbers. How do they cross the missing \(14\,\text{km}\)?
- Two Slow Clocks, No Contradiction: You watch my clock and find it running slow. I watch your clock and find yours running slow. We cannot both be right — or can we? If we can, where does the contradiction hide, and what finally breaks the symmetry?
- The Energy in a Stationary Brick: A brick sitting on a table has zero kinetic energy and zero potential energy relative to the table. Yet relativity says it contains an enormous store of energy — enough, if released, to level a city. Where is that energy, and why does ordinary physics never notice it?
- The Probe That Refuses to Reach c: A rocket flies past you at \(0.5c\) and fires a probe straight ahead at \(0.5c\) relative to the rocket. Common sense says the probe moves at \(c\) in your frame. It does not. What speed do you actually measure, and what rule replaces simple addition?
- The Wall at the Speed of Light: Push a particle harder and it goes faster — but never past \(c\), no matter how long you push or how much energy you pour in. What physically stops it? Where does all that energy go if not into speed?
The Two Postulates and the Failure of Galilean Relativity
In Volume 1 you used a rule so natural it never needed stating: if a train moves at \(v\) and a passenger walks forward at \(u'\) relative to the train, the ground sees the passenger move at \(u = u' + v\). Velocities simply add. This is the heart of Galilean relativity: the laws of mechanics look the same in every frame moving at constant velocity, and you switch frames by adding the relative velocity.
For mechanics this works flawlessly. The trouble begins with light.
Core intuition: The Crisis Maxwell Created
Maxwell's equations (Chapter 8) predict electromagnetic waves travelling at
\[
c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3.00\times10^{8}\
\text{m s}^{-1}.
\]
This is a definite number, fixed by two constants of the vacuum. But a speed is
a speed relative to something. Relative to what? If light obeyed
Galilean addition, then a beam moving at \(c\) in one frame would move at \(c-v\) in
a frame chasing it at \(v\) — and Maxwell's equations, rewritten in that frame,
would predict a different wave speed. Either the equations are wrong, or
the velocity-addition rule is.
The experiment that decided it. In 1887 Michelson and Morley tried to measure Earth's motion through the supposed “ether” — the medium light was assumed to wave in — by comparing the speed of light along and across Earth's orbital motion. They expected a seasonal difference of about \(30\,\text{km s}^{-1}\). They found nothing: the speed of light was the same in every direction, in every season, to the precision of their instrument. Light does not ride on an ether. Its speed does not depend on how the source or the observer moves.
Einstein, in 1905, took this seriously and asked what the world must be like if it is simply true. He kept Maxwell's equations and discarded Galilean addition, building everything on two statements.
Birth of the formula: Where Galilean Addition Breaks
The Galilean rule for light. Suppose a flash of light moves at \(c\) in
frame \(S\). Frame \(S'\) moves at \(v\) in the same direction. Galilean addition
says the speed in \(S'\) is
\[
c' = c - v.
\]
The conflict. Maxwell's equations are not optional rules of thumb;
they are the law in every inertial frame (Postulate 1, below). Written
in \(S'\) they still predict wave speed \(c\), not \(c-v\). The only escape is to
deny that velocities add the Galilean way at speeds near \(c\). That is
the price, and Einstein paid it. The replacement rule (§)
reduces to \(u'+v\) when speeds are small, so all of Volume 1 survives intact —
the old rule was an excellent approximation, never the exact truth.
Core intuition: The Two Postulates
Postulate 1 (Principle of Relativity). The laws of physics are
identical in all inertial (non-accelerating) frames. No experiment performed
inside a sealed, smoothly-moving laboratory can reveal its velocity. There is
no preferred frame, no state of “absolute rest.”
Postulate 2 (Constancy of c). The speed of light in vacuum has
the same value \(c\) in every inertial frame, independent of the motion of the
source or the observer.
Everything in this chapter — dilation, contraction, \(E_0=mc^2\) — is a
logical consequence of these two sentences. No new force, no new field: just
the insistence that light has one speed for everyone.
The second postulate is the radical one. It sounds almost impossible: if you chase a light beam at \(0.99c\), you still measure it receding from you at the full \(c\), not at \(0.01c\). For that to hold, the things you use to measure speed — your clocks and your rulers — cannot be the same as a stationary observer's. Constancy of \(c\) forces time and space themselves to bend. That is the thread we now pull.
Misconception alert: Relativity Means Everything Is Relative
The misconception:
“Special relativity says all motion is relative and nothing is absolute, so any
observer's view is as good as any other's and there are no facts.”
Why it is wrong:
The theory is named for what stays the same, not what changes. The speed
of light is absolute. The laws of physics are absolute. Causes precede effects
in every frame. The spacetime interval between two events
(§) is absolute, agreed on by all observers. What is
relative is a short list: simultaneity, time intervals, lengths, and
velocities. Relativity replaces one set of absolutes (universal time and space)
with another (the invariant interval and the universal speed limit). It is a
theory of invariants, not of “anything goes.”
✓ Checkpoint · The Postulates
- Galilean velocity addition predicts that a light beam chased at speed \(v\) recedes at \(c-v\). Which postulate does this violate, and what is the actual measured speed?It violates Postulate 2 (constancy of \(c\)). The chasing observer still measures the beam receding at the full \(c\), not \(c-v\).
- A laboratory floats in deep space at constant velocity. Name one experiment that could reveal its speed relative to “absolute space.”None exists. Postulate 1 (the principle of relativity) states that no internal experiment can detect uniform motion; there is no absolute-rest frame.
- True or false: relativity says “everything is relative.”False. The speed of light, the laws of physics, causal order, and the spacetime interval are all absolute. Only simultaneity, time intervals, lengths, and velocities are frame-dependent.
Worked Examples
Example 1 · A Galilean Estimate of the “Ether Wind”
Earth orbits the Sun at about \(v = 29.8\,\mathrm{km\,s^{-1}}\). If light
obeyed Galilean velocity addition, by what fraction would the one-way speed of a
light beam measured along Earth's motion differ from the value measured against
it? Compare with the Michelson–Morley result.
Solution.
Given: \(v = 29.8\times10^{3}\,\mathrm{m\,s^{-1}}\),
\(c = 3.00\times10^{8}\,\mathrm{m\,s^{-1}}\).
Find: the fractional anisotropy \(\Delta c/c\) predicted by Galilean addition.
Setup: Galilean addition would give \(c+v\) downstream and \(c-v\) upstream, a one-way difference of \(2v\); as a fraction of \(c\) the shift each way is \(v/c\).
Solve: \[ \frac{\Delta c}{c} = \frac{v}{c} = \frac{29.8\times10^{3}}{3.00\times10^{8}} \approx 9.9\times10^{-5}. \] Answer: \(\boxed{\Delta c/c \approx 1\times10^{-4}}\) — about one part in ten thousand, easily within Michelson and Morley's reach.
Check: An anisotropy of \(10^{-4}\) should have been glaring; the experiment found nothing to a small fraction of it. Postulate 2 — \(c\) is the same in every direction and every frame — is not a convenience but a measured fact, and it is Galilean addition, not Maxwell, that had to yield. ✓
Find: the fractional anisotropy \(\Delta c/c\) predicted by Galilean addition.
Setup: Galilean addition would give \(c+v\) downstream and \(c-v\) upstream, a one-way difference of \(2v\); as a fraction of \(c\) the shift each way is \(v/c\).
Solve: \[ \frac{\Delta c}{c} = \frac{v}{c} = \frac{29.8\times10^{3}}{3.00\times10^{8}} \approx 9.9\times10^{-5}. \] Answer: \(\boxed{\Delta c/c \approx 1\times10^{-4}}\) — about one part in ten thousand, easily within Michelson and Morley's reach.
Check: An anisotropy of \(10^{-4}\) should have been glaring; the experiment found nothing to a small fraction of it. Postulate 2 — \(c\) is the same in every direction and every frame — is not a convenience but a measured fact, and it is Galilean addition, not Maxwell, that had to yield. ✓
The Relativity of Simultaneity
◈ Simultaneity, re-sliced
live benchSend the flash from the middle of a moving train: in the platform frame the rear wall runs into its own pulse first — same events, different order.
If two events happen “at the same time,” surely that is a plain fact everyone agrees on. It is not. Once the speed of light is the same for everyone, simultaneity quietly loses its absolute meaning — and this single loss is the root of every relativistic effect that follows.
Core intuition: Einstein's Train
A long railway carriage moves to the right at speed \(v\). A lamp at its exact
midpoint flashes once. The pulse travels both ways at speed \(c\).
In the carriage frame. The front and back walls are equidistant from
the lamp, the pulse travels at \(c\) both ways, so it reaches both walls
simultaneously. No surprise.
In the ground frame. The pulse still travels at \(c\) both ways
(Postulate 2). But during the flight the rear wall rushes toward the
emission point while the front wall flees away from it. The leftward
pulse has less distance to cover; the rightward pulse has more. So on the
ground the pulse strikes the rear wall first.
Two observers, the same two events, opposite verdicts on which came first —
and both are correct. Simultaneity is not a property of the events; it is
a property of the frame.
This is not an illusion or a signal-delay artefact. Each observer, after correcting for the finite travel time of light to their eyes, still concludes that the events were genuinely simultaneous (carriage) or genuinely ordered (ground). The disagreement is real and symmetric.
We can make it quantitative, but the clean derivation needs the Lorentz transformation, which we build in §. The result, derived there, is worth previewing now because it has a memorable name.
Core intuition: Leading Clocks Lag
Take a row of clocks, all synchronised in their own rest frame \(S'\), separated
by proper distance \(L_0\) along the direction of motion. Viewed from a frame \(S\)
in which they move at speed \(v\), they are not synchronised: the
leading clock (the one farther ahead in the direction of motion) reads an
earlier time, lagging behind the trailing clock by
\[
\Delta t_{\text{desync}} = \frac{L_0 v}{c^2}.
\]
The phrase to remember is “leading clocks lag.” The size of the
offset grows with separation: distant moving clocks are most badly out of step.
This term — the \(vx/c^2\) piece of the Lorentz transformation — is the part
students forget, and it is the part that resolves nearly every apparent paradox
in the subject.
Misconception alert: Simultaneous Once Means Simultaneous Always
The misconception:
“If I find two events happen at the same instant, then they happen at the same
instant for everybody — being simultaneous is a fact about the events.”
Why it is wrong:
Simultaneity is frame-dependent whenever the two events are separated
along the direction of relative motion. Events at the same place, or
separated only perpendicular to the motion, remain simultaneous for all
observers. But for events spread out along the motion, “at the same time” is a
claim that depends entirely on who is asking. This is the trap behind the train,
the pole-and-barn paradox, and most of Part C of the bank: never assume a single
universal “now.”
Worked Examples
Example 2 · How Far Out of Step Are the Clocks?
A train of proper length \(L_0 = 300\,\mathrm{m}\) carries a clock at its front and
one at its rear, synchronised in the train frame. The train moves at
\(v = 0.60c\). By how much do the two clocks appear out of synchronisation to a
ground observer, and which one is behind?
Solution.
Given: \(L_0 = 300\,\mathrm{m}\), \(v = 0.60c = 1.8\times10^{8}\,\mathrm{m\,s^{-1}}\).
Find: the desynchronisation \(\Delta t_{\text{desync}}\) seen from the ground.
Setup: “Leading clocks lag” by \(\Delta t_{\text{desync}} = L_0 v/c^2\).
Solve: \[ \Delta t_{\text{desync}} = \frac{L_0 v}{c^2} = \frac{(300)(1.8\times10^{8})}{(3.0\times10^{8})^2} = 6.0\times10^{-7}\,\mathrm{s} = 0.60\,\mu\mathrm{s}. \] Answer: \(\boxed{0.60\,\mu\mathrm{s}}\); the front clock (leading, in the direction of motion) reads the earlier time.
Check: The offset uses only proper length and speed, never a time interval, so it is a pure simultaneity effect — present even while the clocks tick at the same dilated rate. Set \(v\to0\) and it vanishes, as it must. ✓
Find: the desynchronisation \(\Delta t_{\text{desync}}\) seen from the ground.
Setup: “Leading clocks lag” by \(\Delta t_{\text{desync}} = L_0 v/c^2\).
Solve: \[ \Delta t_{\text{desync}} = \frac{L_0 v}{c^2} = \frac{(300)(1.8\times10^{8})}{(3.0\times10^{8})^2} = 6.0\times10^{-7}\,\mathrm{s} = 0.60\,\mu\mathrm{s}. \] Answer: \(\boxed{0.60\,\mu\mathrm{s}}\); the front clock (leading, in the direction of motion) reads the earlier time.
Check: The offset uses only proper length and speed, never a time interval, so it is a pure simultaneity effect — present even while the clocks tick at the same dilated rate. Set \(v\to0\) and it vanishes, as it must. ✓
Example 3 · Can a Faster Frame Reverse the Order?
In frame \(S\), event 1 (at \(x_1\)) and event 2 (at \(x_2 = x_1 + 600\,\mathrm{m}\),
farther along \(+x\)) are separated by \(\Delta t = +1.0\,\mu\mathrm{s}\) (event 2
later). (a) In a frame \(S'\) moving at \(v = 0.80c\) along \(+x\), what is
\(\Delta t'\)? (b) At what frame speed would the two events be simultaneous?
Solution.
Given: \(\Delta x = +600\,\mathrm{m}\), \(\Delta t = +1.0\times10^{-6}\,\mathrm{s}\).
Find: (a) \(\Delta t'\) at \(v = 0.80c\); (b) the \(v\) that makes \(\Delta t' = 0\).
Setup: \(\Delta t' = \gamma\!\left(\Delta t - v\,\Delta x/c^2\right)\); simultaneity in \(S'\) requires \(\Delta t = v\,\Delta x/c^2\).
Solve (a): \(\gamma(0.80) = 1.667\) and \(v\,\Delta x/c^2 = (0.80c)(600)/c^2 = 480/(3.0\times10^{8}) = 1.6\,\mu\mathrm{s}\), so \[ \Delta t' = 1.667\,(1.0 - 1.6)\,\mu\mathrm{s} = 1.667\times(-0.6\,\mu\mathrm{s}) = -1.0\,\mu\mathrm{s}. \] The order is reversed: in \(S'\) event 2 happens first.
Solve (b): \(v = \dfrac{\Delta t\,c^2}{\Delta x} = \dfrac{(1.0\times10^{-6})(9.0\times10^{16})}{600} = 1.5\times10^{8}\,\mathrm{m\,s^{-1}} = 0.50c.\)
Answer: \(\boxed{\Delta t' = -1.0\,\mu\mathrm{s}\ \text{(reversed)};\quad \text{simultaneous at } v = 0.50c}\).
Check: Frames slower than \(0.50c\) preserve the order, \(0.50c\) makes them simultaneous, faster ones reverse it — exactly the behaviour allowed for spacelike events (here \(c\,\Delta t = 300\,\mathrm{m} \lt \Delta x = 600\,\mathrm{m}\)), whose order is frame-dependent but which can never be causally linked. ✓
Find: (a) \(\Delta t'\) at \(v = 0.80c\); (b) the \(v\) that makes \(\Delta t' = 0\).
Setup: \(\Delta t' = \gamma\!\left(\Delta t - v\,\Delta x/c^2\right)\); simultaneity in \(S'\) requires \(\Delta t = v\,\Delta x/c^2\).
Solve (a): \(\gamma(0.80) = 1.667\) and \(v\,\Delta x/c^2 = (0.80c)(600)/c^2 = 480/(3.0\times10^{8}) = 1.6\,\mu\mathrm{s}\), so \[ \Delta t' = 1.667\,(1.0 - 1.6)\,\mu\mathrm{s} = 1.667\times(-0.6\,\mu\mathrm{s}) = -1.0\,\mu\mathrm{s}. \] The order is reversed: in \(S'\) event 2 happens first.
Solve (b): \(v = \dfrac{\Delta t\,c^2}{\Delta x} = \dfrac{(1.0\times10^{-6})(9.0\times10^{16})}{600} = 1.5\times10^{8}\,\mathrm{m\,s^{-1}} = 0.50c.\)
Answer: \(\boxed{\Delta t' = -1.0\,\mu\mathrm{s}\ \text{(reversed)};\quad \text{simultaneous at } v = 0.50c}\).
Check: Frames slower than \(0.50c\) preserve the order, \(0.50c\) makes them simultaneous, faster ones reverse it — exactly the behaviour allowed for spacelike events (here \(c\,\Delta t = 300\,\mathrm{m} \lt \Delta x = 600\,\mathrm{m}\)), whose order is frame-dependent but which can never be causally linked. ✓
Time Dilation
◈ The light-clock and the γ dial
live benchRun the photon between the mirrors and slide v toward c: the zig-zag stretches, the tick lengthens, and γ climbs off the scale.
We now extract the first number. The tool is a deliberately simple clock — a light clock — whose ticking is governed by nothing but the speed of light, so Postulate 2 does all the work.
Core intuition: The Light Clock
Imagine a clock made of two parallel mirrors a distance \(d\) apart, with a single
photon bouncing straight up and down between them. One “tick” is one round
trip. In the clock's own rest frame the tick takes
\[
\Delta t_0 = \frac{2d}{c}.
\]
This is the proper time — the time read by a clock present at both
events (here, the photon leaving and returning to the bottom mirror), at rest
relative to them. Proper time is the shortest time any frame will assign to the
interval; everyone else measures more.
Birth of the formula: Time Dilation from the Light Clock
Setup. Let the clock move horizontally at speed \(v\) across a laboratory
frame \(S\), while the photon bounces vertically in the clock's frame.
In the clock's rest frame (\(S'\)). The photon goes straight up and back:
\[
\Delta t_0 = \frac{2d}{c}.
\]
In the laboratory frame (\(S\)). While the photon travels, the clock
moves sideways, so the photon traces a longer, diagonal zig-zag. Let one tick
take time \(\Delta t\) in \(S\). In that time the clock advances horizontally by
\(v\,\Delta t\), so each diagonal leg has horizontal extent \(v\,\Delta t/2\) and
vertical extent \(d\). The photon still moves at speed \(c\) (Postulate 2), so the
length of one diagonal leg is \(c\,\Delta t/2\). By Pythagoras,
\[
\left(\frac{c\,\Delta t}{2}\right)^{\!2}
= d^2 + \left(\frac{v\,\Delta t}{2}\right)^{\!2}.
\]
Solve for \(\Delta t\). Multiply by \(4\) and collect terms:
\[
c^2 \Delta t^2 - v^2 \Delta t^2 = 4d^2
\quad\Longrightarrow\quad
\Delta t^2 (c^2 - v^2) = 4d^2.
\]
\[
\Delta t = \frac{2d}{\sqrt{c^2 - v^2}}
= \frac{2d}{c}\cdot\frac{1}{\sqrt{1 - v^2/c^2}}.
\]
Recognising \(2d/c = \Delta t_0\) and defining the Lorentz factor
\[
\boxed{\;\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}
= \frac{1}{\sqrt{1-\beta^2}}, \qquad \beta \equiv v/c\;}
\]
we obtain the time-dilation law:
\[
\boxed{\;\Delta t = \gamma\, \Delta t_0\;}
\]
Since \(\gamma \ge 1\) always, the moving clock's tick is longer in the lab
frame: moving clocks run slow.
Reading the Lorentz factor. Everything in relativity is controlled by \(\gamma\). It is \(1\) at rest and climbs without bound as \(v\to c\):
| \(\beta = v/c\) | \(0\) | \(0.10\) | \(0.60\) | \(0.80\) | \(0.90\) | \(0.99\) |
| \(\gamma\) | \(1\) | \(1.005\) | \(1.25\) | \(1.667\) | \(2.294\) | \(7.089\) |
At everyday speeds \(\gamma\) differs from \(1\) by parts in a trillion — which is why Volume 1 never needed it. Only past a few tenths of \(c\) does it bite.
Core intuition: Proper Time Is the Invariant Clock
The phrase “moving clocks run slow” is true but loaded, because which
clock is moving depends on the frame. The frame-independent statement is:
the proper time \(\Delta t_0\) between two events is the time read by a
single clock that is present at both of them. Any other frame, in which those
two events happen at different places, must use two separated clocks and reads a
larger elapsed time \(\gamma\,\Delta t_0\). Proper time is not “the right
answer” and other times “wrong” — but it is the one quantity all observers
can compute and agree on.
The muon, settled
A muon's mean lifetime in its own rest frame is \(\tau_0 = 2.2\,\mu\text{s}\). Travelling at \(v = 0.99c\) (\(\gamma = 7.09\)), its lifetime in the laboratory frame is \[ \tau_{\text{lab}} = \gamma\tau_0 = 7.09 \times 2.2\,\mu\text{s} \approx 15.6\,\mu\text{s}, \] during which it covers \[ d = v\,\tau_{\text{lab}} = 0.99 \times (3.00\times10^8) \times (15.6\times10^{-6}) \approx 4.6\,\text{km} \] — seven times farther than the naïve \(0.65\,\text{km}\) a non-relativistic calculation gives, enough to reach the ground from the altitudes where many are made. The first perplexing question is answered; we will answer it a second time, from the muon's own point of view, in the next section.
Misconception alert: Time Dilation Is Just a Measurement Illusion
The misconception:
“The moving clock only appears to run slow because the light from it
takes time to reach us. Correct for the light delay and the effect vanishes.”
Why it is wrong:
The light-travel-time delay is a separate, ordinary effect that every careful
observer already subtracts. Time dilation is what remains after that
correction. The muon experiment proves it physically: a muon beam at \(0.99c\)
decays measurably more slowly than a muon at rest — the surviving fraction
after a fixed distance matches \(\gamma\) exactly. Particle accelerators rely on
this every day; unstable particles live longer, by precisely the dilation
factor, when fast. The slowing is real, not apparent.
◇ Figure 1
Fig 1. The light clock. In its rest frame the photon bounces straight up and
down over a distance \(d\). In the lab frame the clock has moved, so the photon
traces a longer diagonal of length \(c\,\Delta t/2\) per leg. Because its speed is
still \(c\), the lab tick \(\Delta t = \gamma\,\Delta t_0\) is longer.
✓ Checkpoint · Simultaneity and Time Dilation
- A clock moving at \(0.80c\) ticks once per second in its own frame. How long is one tick in the lab?\(\Delta t = \gamma\,\Delta t_0 = 1.667\times1.0 = 1.67\,\)s. The moving clock runs slow.
- Two events occur at the same place in frame \(S'\). Which frame records the proper time between them, and how does the lab interval compare?\(S'\) records the proper time (a single clock is present at both events). The lab, where the events are at different places, records a larger interval \(\gamma\,\Delta t_0\).
- A muon (\(\tau_0 = 2.2\,\mu\)s) moves at \(0.99c\) (\(\gamma=7.09\)). State its lab lifetime and the lab distance it covers.Lab lifetime \(\gamma\tau_0 = 15.6\,\mu\)s; distance \(v\gamma\tau_0 \approx 4.6\,\)km.
Worked Examples
Example 4 · A Pion Beam's Extended Reach
A charged pion has a rest-frame mean lifetime \(\tau_0 = 26\,\mathrm{ns}\). In an
accelerator it travels at \(v = 0.98c\). Find (a) its mean lifetime in the
laboratory and (b) the mean distance it travels before decaying.
Solution.
Given: \(\tau_0 = 26\,\mathrm{ns}\), \(v = 0.98c\).
Find: lab lifetime \(\tau_{\text{lab}}\) and mean decay length \(d\).
Setup: \(\tau_{\text{lab}} = \gamma\tau_0\) (time dilation) and \(d = v\,\tau_{\text{lab}}\).
Solve: \(\gamma(0.98) = 1/\sqrt{1-0.9604} = 5.03\), so \[ \tau_{\text{lab}} = 5.03\times26\,\mathrm{ns} = 131\,\mathrm{ns}, \qquad d = (0.98)(3.0\times10^{8})(131\times10^{-9}) \approx 38\,\mathrm{m}. \] Answer: \(\boxed{\tau_{\text{lab}} \approx 131\,\mathrm{ns},\ d \approx 38\,\mathrm{m}}\).
Check: Without dilation the beam would reach only \(v\tau_0 = 0.98c\times26\,\mathrm{ns} \approx 7.6\,\mathrm{m}\); dilation stretches this by exactly \(\gamma \approx 5\), which is why beamlines are metres, not centimetres, long. ✓
Find: lab lifetime \(\tau_{\text{lab}}\) and mean decay length \(d\).
Setup: \(\tau_{\text{lab}} = \gamma\tau_0\) (time dilation) and \(d = v\,\tau_{\text{lab}}\).
Solve: \(\gamma(0.98) = 1/\sqrt{1-0.9604} = 5.03\), so \[ \tau_{\text{lab}} = 5.03\times26\,\mathrm{ns} = 131\,\mathrm{ns}, \qquad d = (0.98)(3.0\times10^{8})(131\times10^{-9}) \approx 38\,\mathrm{m}. \] Answer: \(\boxed{\tau_{\text{lab}} \approx 131\,\mathrm{ns},\ d \approx 38\,\mathrm{m}}\).
Check: Without dilation the beam would reach only \(v\tau_0 = 0.98c\times26\,\mathrm{ns} \approx 7.6\,\mathrm{m}\); dilation stretches this by exactly \(\gamma \approx 5\), which is why beamlines are metres, not centimetres, long. ✓
Example 5 · How Fast to Halve the Tick Rate?
At what speed does a clock run at exactly half its rest rate, i.e. \(\gamma = 2\)?
Solution.
Given: target \(\gamma = 2\).
Find: the required speed \(v = \beta c\).
Setup: Invert \(\gamma = 1/\sqrt{1-\beta^2}\) to get \(\beta = \sqrt{1 - 1/\gamma^2}\).
Solve: \[ \beta = \sqrt{1 - \tfrac{1}{4}} = \sqrt{\tfrac{3}{4}} = \frac{\sqrt3}{2} \approx 0.866. \] Answer: \(\boxed{v \approx 0.866c}\).
Check: A “factor of two” slowdown already demands \(87\%\) of light speed — a vivid reminder of how flat the \(\gamma\) curve is at low speed and how steeply it climbs near \(c\). Doubling again to \(\gamma = 4\) needs \(\beta = 0.968\), barely faster still. ✓
Find: the required speed \(v = \beta c\).
Setup: Invert \(\gamma = 1/\sqrt{1-\beta^2}\) to get \(\beta = \sqrt{1 - 1/\gamma^2}\).
Solve: \[ \beta = \sqrt{1 - \tfrac{1}{4}} = \sqrt{\tfrac{3}{4}} = \frac{\sqrt3}{2} \approx 0.866. \] Answer: \(\boxed{v \approx 0.866c}\).
Check: A “factor of two” slowdown already demands \(87\%\) of light speed — a vivid reminder of how flat the \(\gamma\) curve is at low speed and how steeply it climbs near \(c\). Doubling again to \(\gamma = 4\) needs \(\beta = 0.968\), barely faster still. ✓
Solved examples
Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.
Example 1 · The Lorentz Factor at Four Speeds
Compute \(\gamma\) for \(\beta = 0.60,\ 0.80,\ 0.90,\ 0.99\), and state in one phrase
what \(\gamma\) physically controls.
Solution.
Setup: \(\gamma = 1/\sqrt{1-\beta^2}\).
Solve: \[ \gamma(0.60) = \tfrac{1}{\sqrt{1-0.36}} = \tfrac{1}{0.8} = 1.25, \qquad \gamma(0.80) = \tfrac{1}{\sqrt{1-0.64}} = \tfrac{1}{0.6} = 1.667, \] \[ \gamma(0.90) = \tfrac{1}{\sqrt{1-0.81}} = 2.294, \qquad \gamma(0.99) = \tfrac{1}{\sqrt{1-0.9801}} = 7.089. \] Answer: \(\boxed{1.25,\ 1.667,\ 2.294,\ 7.089}\). The factor \(\gamma\) sets how much a moving clock is slowed, a moving length is shortened, and a moving particle's momentum and energy are boosted — all by the same \(\gamma\). Check: At \(\beta=0\), \(\gamma=1\) (Newtonian rest); the values climb gently at first (\(1.25\) at \(0.6c\)) then explode (\(7.09\) at \(0.99c\)) — the shape that keeps everyday life Newtonian yet makes near-\(c\) physics extreme. ✓
Solve: \[ \gamma(0.60) = \tfrac{1}{\sqrt{1-0.36}} = \tfrac{1}{0.8} = 1.25, \qquad \gamma(0.80) = \tfrac{1}{\sqrt{1-0.64}} = \tfrac{1}{0.6} = 1.667, \] \[ \gamma(0.90) = \tfrac{1}{\sqrt{1-0.81}} = 2.294, \qquad \gamma(0.99) = \tfrac{1}{\sqrt{1-0.9801}} = 7.089. \] Answer: \(\boxed{1.25,\ 1.667,\ 2.294,\ 7.089}\). The factor \(\gamma\) sets how much a moving clock is slowed, a moving length is shortened, and a moving particle's momentum and energy are boosted — all by the same \(\gamma\). Check: At \(\beta=0\), \(\gamma=1\) (Newtonian rest); the values climb gently at first (\(1.25\) at \(0.6c\)) then explode (\(7.09\) at \(0.99c\)) — the shape that keeps everyday life Newtonian yet makes near-\(c\) physics extreme. ✓
Example 2 · The Muon's Two Stories
A muon with rest-frame lifetime \(\tau_0 = 2.2\,\mu\mathrm{s}\) moves at
\(v = 0.99c\). Find (a) its lifetime in the laboratory, (b) the distance it travels
in the lab, and (c) the thickness of atmosphere it sees in its own frame.
Solution.
Setup: \(\gamma(0.99) = 7.089\).
(a) Lab lifetime \(= \gamma\tau_0 = 7.089 \times 2.2\,\mu\mathrm{s} = 15.6\,\mu\mathrm{s}\).
(b) Lab distance \(= v\,\gamma\tau_0 = 0.99c \times 15.6\,\mu\mathrm{s} = 4.63\,\mathrm{km}\).
(c) In the muon frame the column is contracted to \(L = (4.63\,\mathrm{km})/\gamma = 0.65\,\mathrm{km}\), crossed in just one lifetime, \(2.2\,\mu\mathrm{s}\).
Answer: \(\boxed{15.6\,\mu\mathrm{s},\ 4.63\,\mathrm{km},\ 0.65\,\mathrm{km}}\).
Lab says “dilated time”; muon says “contracted distance” — both agree it arrives. Check: The frames must agree the muon arrives, and they do: \(0.65\,\mathrm{km}\) at \(0.99c\) takes \(0.65\,\mathrm{km}/0.99c = 2.2\,\mu\mathrm{s}\), exactly one rest-frame lifetime, matching the lab's \(\gamma\tau_0\) account. ✓
(a) Lab lifetime \(= \gamma\tau_0 = 7.089 \times 2.2\,\mu\mathrm{s} = 15.6\,\mu\mathrm{s}\).
(b) Lab distance \(= v\,\gamma\tau_0 = 0.99c \times 15.6\,\mu\mathrm{s} = 4.63\,\mathrm{km}\).
(c) In the muon frame the column is contracted to \(L = (4.63\,\mathrm{km})/\gamma = 0.65\,\mathrm{km}\), crossed in just one lifetime, \(2.2\,\mu\mathrm{s}\).
Answer: \(\boxed{15.6\,\mu\mathrm{s},\ 4.63\,\mathrm{km},\ 0.65\,\mathrm{km}}\).
Lab says “dilated time”; muon says “contracted distance” — both agree it arrives. Check: The frames must agree the muon arrives, and they do: \(0.65\,\mathrm{km}\) at \(0.99c\) takes \(0.65\,\mathrm{km}/0.99c = 2.2\,\mu\mathrm{s}\), exactly one rest-frame lifetime, matching the lab's \(\gamma\tau_0\) account. ✓
Example 3 · A Contracted Spaceship
A spaceship of proper length \(100\,\mathrm{m}\) flies past Earth at \(0.80c\). What
length do Earth observers measure?
Solution.
Setup: \(L = L_0/\gamma\), with \(\gamma(0.80) = 1.667\).
Solve: \(L = 100/1.667 = 60\,\mathrm{m}\).
Answer: \(\boxed{L = 60\,\mathrm{m}}\), contracted along the direction of motion only; the ship's width and height are unchanged. Check: A \(40\%\) loss matches \(\gamma=1.667\) (\(1-1/1.667=0.40\)); and since the crew ride with the ship, they still measure the full \(100\,\mathrm{m}\) — contraction is never felt in the rest frame. ✓
Solve: \(L = 100/1.667 = 60\,\mathrm{m}\).
Answer: \(\boxed{L = 60\,\mathrm{m}}\), contracted along the direction of motion only; the ship's width and height are unchanged. Check: A \(40\%\) loss matches \(\gamma=1.667\) (\(1-1/1.667=0.40\)); and since the crew ride with the ship, they still measure the full \(100\,\mathrm{m}\) — contraction is never felt in the rest frame. ✓
Example 4 · Lightning on a Moving Train
Lightning strikes the front and rear of a train simultaneously in the
ground frame. In the train's frame, which strike happens first?
Solution.
Reasoning: Let the train move forward (toward the front). In the ground
frame the two strikes share \(\Delta t = 0\) but are separated by
\(\Delta x = L\) (the train's length) along the motion. By the Lorentz
transformation, \(\Delta t' = \gamma(\Delta t - v\,\Delta x/c^2)
= -\gamma vL/c^2 \neq 0\). With \(\Delta x \gt 0\) measured front-minus-rear,
\(\Delta t' \lt 0\): in the train frame the front strike registers
earlier.
Answer: \(\boxed{\text{The front strike happens first in the train frame.}}\) Simultaneity is frame-dependent because the events are separated along the motion. Check: The offset \(\gamma vL/c^2\) grows with the train's length and speed and vanishes as \(v\to0\), recovering the Newtonian “simultaneous for all”; strikes separated across the motion would stay simultaneous. ✓
Answer: \(\boxed{\text{The front strike happens first in the train frame.}}\) Simultaneity is frame-dependent because the events are separated along the motion. Check: The offset \(\gamma vL/c^2\) grows with the train's length and speed and vanishes as \(v\to0\), recovering the Newtonian “simultaneous for all”; strikes separated across the motion would stay simultaneous. ✓
Example 5 · Rest Energy of a Proton
Take the proton mass as \(m_p = 1.67\times10^{-27}\,\mathrm{kg}\). Find its rest
energy in joules and in MeV (\(1\,\mathrm{MeV} = 1.602\times10^{-13}\,\mathrm{J}\)).
Solution.
Setup: \(E_0 = m_p c^2\).
Solve: \[ E_0 = (1.67\times10^{-27})(3.0\times10^8)^2 = 1.50\times10^{-10}\,\mathrm{J} = \frac{1.50\times10^{-10}}{1.602\times10^{-13}}\,\mathrm{MeV} = 938\,\mathrm{MeV}. \] Answer: \(\boxed{E_0 \approx 1.50\times10^{-10}\,\mathrm{J} \approx 938\,\mathrm{MeV}}\). This is the natural energy scale of nuclear physics. Check: Dividing by the electron rest energy, \(938/0.511 \approx 1836\) — the known proton-to-electron mass ratio, an independent confirmation of the arithmetic. ✓
Solve: \[ E_0 = (1.67\times10^{-27})(3.0\times10^8)^2 = 1.50\times10^{-10}\,\mathrm{J} = \frac{1.50\times10^{-10}}{1.602\times10^{-13}}\,\mathrm{MeV} = 938\,\mathrm{MeV}. \] Answer: \(\boxed{E_0 \approx 1.50\times10^{-10}\,\mathrm{J} \approx 938\,\mathrm{MeV}}\). This is the natural energy scale of nuclear physics. Check: Dividing by the electron rest energy, \(938/0.511 \approx 1836\) — the known proton-to-electron mass ratio, an independent confirmation of the arithmetic. ✓
Problem bank
Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.
✎ Free practice questions
- Lorentz factor:
Compute \(\gamma\) for \(\beta = 0.60\) and for \(\beta = 0.80\).\(\gamma(0.60) = 1/\sqrt{1-0.36} = 1.25\); \(\gamma(0.80) = 1/\sqrt{1-0.64} = 1.667\). - Time dilation:
A particle with rest-frame lifetime \(100\,\mathrm{ns}\) moves at \(0.80c\). Find its lifetime in the laboratory.\(\tau = \gamma\tau_0 = 1.667 \times 100 = 167\,\mathrm{ns}\). - Length contraction:
A rod of proper length \(2.0\,\mathrm{m}\) moves lengthwise at \(0.60c\). Find its measured length.\(L = L_0/\gamma = 2.0/1.25 = 1.6\,\mathrm{m}\). - Event transform:
In \(S\) an event is at \(x = 6.0\times10^8\,\mathrm{m}\), \(t = 3.0\,\mathrm{s}\). Frame \(S'\) moves at \(0.80c\). Find \((x', t')\).\(\gamma = 1.667\), \(v = 2.4\times10^8\,\mathrm{m\,s^{-1}}\). \(x' = 1.667(6.0\times10^8 - (2.4\times10^8)(3.0)) = -2.0\times10^8\,\mathrm{m}\); \(t' = 1.667(3.0 - (2.4\times10^8)(6.0\times10^8)/9\times10^{16}) = 1.667(3.0 - 1.6) = 2.33\,\mathrm{s}\). - Electron in a field:
An electron at \(0.99c\) moves perpendicular to \(B = 0.10\,\mathrm{T}\). Find its radius of curvature (\(\gamma = 7.09\)).\(r = \gamma m_e v/(qB) = (7.09)(9.11\times10^{-31})(0.99\times3\times10^8)/(1.6\times10^{-19}\times0.10) = 0.12\,\mathrm{m}\).
Chapter test
A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.
Pattern
Questions
Clock
§ 4 – 12 · The rest of the chapter
The chapter continues.
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Still ahead in this chapter
- Length Contraction
- The Lorentz Transformation
- Relativistic Velocity Addition
- Relativistic Momentum
- Relativistic Energy and E₀ = mc²
- The Energy–Momentum Invariant
- The Relativistic Doppler Effect
- Spacetime: The Minkowski Picture
- Common Pitfalls and Exam Strategy
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