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An inertial frame is one that is not accelerating.
Common notation:
$S$ = unprimed frame
$S'$ = frame moving at constant velocity $v$ relative to $S$
If the motion is along the $x$-axis, the standard analysis uses one-dimensional Lorentz transformations.
Difficulty
State the two postulates of special relativity.
Solution
The two postulates are:
The laws of physics are the same in all inertial frames.
The speed of light in vacuum is the same for all inertial observers.
These are the starting point for special relativity.
For a speed of $v = 0.6c$, compute the Lorentz factor $\gamma$.
Solution
Use
Substitute $v = 0.6c$:
Two events are separated by $\Delta t = 4\ \text{ns}$ and $\Delta x = 1\ \text{m}$ in one spatial dimension. Classify the spacetime interval.
Solution
Compute the light-travel distance for the time separation:
Since
the interval is timelike.
A clock has proper time $\Delta \tau = 8\ \text{s}$ and moves at $0.6c$ relative to you. How much time do you measure?
Solution
First find $\gamma$ from $v = 0.6c$:
Then use time dilation:
A rod has proper length $L_0 = 12\ \text{m}$ and moves at $0.8c$ relative to you. What length do you measure?
Solution
For $v = 0.8c$,
Use length contraction:
In one frame, two events are simultaneous and separated by $12\ \text{m}$ along the $x$-axis. If $S'$ moves in the $+x$ direction at $0.6c$, what is $\Delta t'$?
Solution
If the events are simultaneous in the first frame, then $\Delta t = 0$ and
With $\gamma = 1.25$, $v = 0.6c$, and $\Delta x = 12\ \text{m}$:
Since $c = 3.00\times 10^8\ \text{m/s}$,
A probe moves at $0.5c$ relative to a ship, and the ship moves at $0.5c$ relative to Earth in the same direction. What speed does Earth measure for the probe?
Solution
Use relativistic velocity addition:
Substitute $u' = 0.5c$ and $v = 0.5c$:
What is the rest energy of a $1\ \text{kg}$ object?
Solution
Use
So
What do you call the time measured by a clock that travels with two timelike-separated events?
Solution
That time is called the proper time.
It is the time measured in the rest frame of the clock.
For a photon with energy $E$, what is its momentum?
Solution
For light, the relation is
So the momentum is
Difficulty
In frame $S$, an event occurs at $t = 2.0\times 10^{-8}\ \text{s}$ and $x = 3.0\ \text{m}$. If $S'$ moves at $v = 0.6c$ in the $+x$ direction, find $x'$ and $t'$.
Solution
For $v = 0.6c$, we have $\gamma = 1.25$.
Use the Lorentz transformations:
and
First compute $vt$:
So
Next,
Therefore,
A spaceship moves at $0.8c$ for $10\ \mu\text{s}$ as measured in Earth frame. How much proper time passes on the ship, and how far does it travel in Earth frame?
Solution
For $v = 0.8c$,
The proper time is
The Earth-frame distance is
So the ship experiences $6\ \mu\text{s}$, and it travels $2400\ \text{m}$ in Earth frame.
Two events are separated by $\Delta t = 20\ \text{ns}$ and $\Delta x = 9\ \text{m}$. Can a light signal connect them?
Solution
Compute the light-travel distance for the time separation:
Since
the interval is spacelike. A light signal cannot connect the events, so there is no causal connection at or below light speed.
What is the kinetic energy of a $2\ \text{kg}$ object moving at $0.6c$?
Solution
For $v = 0.6c$, the Lorentz factor is
Use
So
That is
A particle has rest mass $m$ and momentum $p = \frac{3}{4}mc$. Find its total energy in terms of $m$ and $c$.
Solution
Use the invariant energy-momentum relation:
Substitute $p = \frac{3}{4}mc$:
Factor out $m^2c^4$:
So
A particle has proper lifetime $2.2\ \mu\text{s}$ and moves at $0.8c$. It is created $500\ \text{m}$ above a detector. Does it reach the detector before decaying?
Solution
For $v = 0.8c$, we have
So the lab-frame lifetime is
The distance traveled is
Since $880\ \text{m} > 500\ \text{m}$, the particle reaches the detector before decaying.
Two lightning flashes happen simultaneously in Earth's frame and are separated by $20\ \text{m}$ along the $x$-axis. If a train moves at $0.6c$ in the $+x$ direction, what time separation does the train measure?
Solution
Use the simultaneity formula with $\Delta t = 0$:
For $v = 0.6c$, $\gamma = 1.25$. Then
Since $c = 3.00\times 10^8\ \text{m/s}$,
So the flashes are not simultaneous in the train frame.
A photon is emitted with frequency $6.0\times 10^{14}\ \text{Hz}$ and later observed at $5.4\times 10^{14}\ \text{Hz}$. By what percentage did its energy change?
Solution
For a photon,
so energy is proportional to frequency.
The frequency changed from $6.0\times 10^{14}$ to $5.4\times 10^{14}$, which is a factor of
So the energy dropped to $90\%$ of its original value, a decrease of $10\%$.
Difficulty
A problem asks for the clock-rate correction needed for a GPS satellite relative to a ground clock. Which theory is essential, and why?
Solution
General relativity is essential, because the satellite and the ground clock are in different gravitational potentials.
Special relativity handles motion in inertial frames, but it does not account for gravitational time dilation. GPS needs both motion and gravity, so the gravitational effect requires general relativity.
A muon has proper lifetime $2.2\ \mu\text{s}$ and moves at $0.6c$. It is created in the upper atmosphere $500\ \text{m}$ above the ground. Does it reach the ground before decaying?
Solution
For $v = 0.6c$, the Lorentz factor is
The lab-frame lifetime is
The distance traveled is
So the muon travels about $495\ \text{m}$, which is just short of $500\ \text{m}$. It does not quite reach the ground.
Two lightning strikes happen simultaneously in Earth's frame at opposite ends of a moving train. Explain why a passenger on the train can disagree about whether they were simultaneous.
Solution
The time coordinate transforms as
If the two strikes occur at different positions, the $vx/c^2$ term is different for each event. That means equal $t$ in Earth frame does not imply equal $t'$ in the train frame.
So simultaneity depends on the frame.
Two identical particles each have rest mass $m$ and move directly toward each other at $0.6c$. They collide and stick together. What is the rest mass of the final composite object?
Solution
For each particle moving at $0.6c$, $\gamma = 1.25$, so each has energy
The total energy before the collision is
Because the particles have equal and opposite momenta, the total momentum is zero. After they stick together, the composite object is at rest, so its rest energy is
Therefore,
A photon climbs out of a gravitational field and its frequency drops by $12\%$. By what percent does its energy drop?
Solution
For a photon,
so energy changes by the same percentage as frequency.
If the frequency drops by $12\%$, the energy also drops by $12\%$.
Difficulty
In frame $S$, two events are simultaneous and separated by $24\ \text{m}$. Another frame $S'$ measures them to be $80\ \text{ns}$ apart in time. What is the speed of $S'$ relative to $S$? Give the magnitude.
Solution
With $\Delta t = 0$,
Take magnitudes:
Substitute $|\Delta t'| = 80\ \text{ns}$ and $\Delta x = 24\ \text{m}$:
So
which simplifies to
where $\beta = v/c$.
Thus
Square both sides:
so
Therefore,
Two events occur $10\ \text{ns}$ apart and $1.8\ \text{m}$ apart. Find the invariant interval and the proper time between them, if it exists.
Solution
First compute the light-travel distance:
Since $3\ \text{m} > 1.8\ \text{m}$, the interval is timelike.
The invariant interval is
For a timelike interval,
So the proper time is $8\ \text{ns}$.
Two identical particles each have rest mass $m$ and move at $0.6c$ in opposite directions. They collide and form one object at rest. Show that the final object's rest mass is larger than $2m$.
Solution
Each particle has $\gamma = 1.25$, so each carries energy
The total energy is
Because the momenta cancel, the final object is at rest. Its rest energy is therefore
so
That is larger than $2m$ because some kinetic energy has been converted into rest mass.
Why does light bend near a massive object in general relativity, even though light has no rest mass?
Solution
General relativity says mass-energy curves spacetime, and light follows the curved geometry of spacetime.
So light is not being pulled by a Newtonian force in the usual way. Instead, its path is a geodesic in curved spacetime.
The equivalence principle is the key idea behind this: locally, gravity and acceleration have the same physical effects, so even light is affected by the geometry associated with gravity.