Special relativity

What this unit covers

Special Relativity is about 6% of the test's approximately 70 questions, so roughly four questions (ETS, GRE Subject Test Content and Structure). ETS lists introductory concepts, time dilation, length contraction, simultaneity, energy and momentum, four-vectors and Lorentz transformation, and velocity addition (ETS, GRE Physics Test fact sheet). These are among the most reliably winnable questions on the test: the toolkit is small, the numbers are chosen to be friendly, and three or four sanity checks catch almost every wrong choice.

The postulates and the factor gamma

1. The laws of physics are the same in every inertial frame.
2. The speed of light in vacuum, $c = 299\,792\,458$ m/s exactly (NIST, CODATA 2022), is the same in every inertial frame.

Everything else follows. Write $\beta = v/c$ and

$$\gamma = \frac{1}{\sqrt{1-\beta^2}} \ge 1.$$

Memorise the friendly values, because the test uses them constantly. They come from Pythagorean triples: if $\beta = a/c'$ for a triple $(a, b, c')$, then $\gamma = c'/b$.

$\beta$	$3/5 = 0.6$	$4/5 = 0.8$	$\sqrt3/2 \approx 0.866$	$12/13 \approx 0.923$	$0.99$
$\gamma$	$5/4$	$5/3$	$2$	$13/5$	$\approx 7.1$

For small speeds, $\gamma \approx 1 + \tfrac12\beta^2$.

Time dilation and length contraction

Proper time $\Delta\tau$ is the time between two events measured by a clock present at both events (at rest in the frame where they happen at the same place). Any other inertial observer measures a longer time:

$$\Delta t = \gamma\,\Delta\tau.$$

Proper length $L_0$ is the length of an object in its own rest frame. An observer for whom the object moves along its length measures

$$L = \frac{L_0}{\gamma}.$$

Lengths perpendicular to the motion are unchanged.

The skill the test checks is deciding who measures the proper quantity. Ask: which clock is at both events? In whose frame is the ruler at rest?

Worked example (the muon problem). An unstable particle with proper mean lifetime $\tau = 2\ \mu$s in this example is made high in the atmosphere and moves down at $\beta = 0.995$, for which $\gamma \approx 10$. Without relativity it would travel about $c\tau \approx 600$ m in one lifetime. In the ground frame its lifetime is dilated to $\gamma\tau \approx 20\ \mu$s, so it travels about 6 km. In the particle's frame its lifetime is just $2\ \mu$s, but the atmosphere is contracted by $\gamma$, so 6 km of air shrinks to 600 m. Both frames agree on what matters: the number of particles that survive.

Simultaneity and the Lorentz transformation

For a frame $S'$ moving at speed $v$ along $+x$ relative to $S$:

$$x' = \gamma(x - vt), \qquad t' = \gamma\left(t - \frac{vx}{c^2}\right), \qquad y' = y, \quad z' = z.$$

The inverse transformation flips the sign of $v$. The term $vx/c^2$ is the relativity of simultaneity: two events simultaneous in $S$ ($\Delta t = 0$) but separated by $\Delta x$ are separated in $S'$ by

$$\Delta t' = -\gamma\frac{v\,\Delta x}{c^2}.$$

So the event farther ahead (in the direction of $S'$'s motion) happens earlier in $S'$.

The spacetime interval is invariant:

$$s^2 = (c\,\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2.$$

If $s^2 > 0$ (timelike), some frame sees both events at one place, and $\Delta\tau = s/c$ is the proper time between them; their order is the same in every frame. If $s^2 < 0$ (spacelike), some frame sees them as simultaneous, and their order can be reversed; they cannot be causally connected. If $s^2 = 0$, they lie on one light ray.

Worked example. In frame $S$, event A is at $x = 0$, $t = 0$ and event B at $x = 900$ m, $t = 5\ \mu$s. Then $c\,\Delta t = 1500$ m and $\Delta x = 900$ m, so $s^2 = 1500^2 - 900^2 = 1200^2$ m$^2$: timelike, with proper time $1200\text{ m}/c = 4\ \mu$s. The frame in which A and B occur at the same place moves at $\beta = 900/1500 = 0.6$, and indeed $\gamma\cdot 4\ \mu\text{s} = 1.25 \times 4 = 5\ \mu$s.

Velocity addition

If an object moves at $u'$ along $x$ in $S'$, its velocity in $S$ is

$$u = \frac{u' + v}{1 + u'v/c^2}.$$

Checks: if $u' = c$ then $u = c$; for small speeds $u \approx u' + v$. Two rockets approaching each other, each at $0.6c$ relative to the ground, close at $1.2c$ in the ground frame (a closing speed is not the velocity of anything), but each sees the other approaching at

$$\frac{0.6 + 0.6}{1 + 0.36}c = \frac{1.2}{1.36}c \approx 0.88c.$$

Energy and momentum

For a particle of mass $m$ moving at $v$:

$$E = \gamma mc^2, \qquad p = \gamma m v, \qquad K = (\gamma - 1)mc^2, \qquad E^2 = (pc)^2 + (mc^2)^2.$$

Two consequences are worth more than any other formula here:

• $\gamma = E/mc^2$ and $\beta = pc/E$. Given an energy in MeV, you get the speed without ever using SI.
• Massless particles have $E = pc$ and always move at $c$.

Work in MeV, MeV/$c$ and MeV/$c^2$. The electron's rest energy is $0.511$ MeV and the proton's $938.3$ MeV (NIST, CODATA 2022).

Worked example. A proton has kinetic energy 938 MeV, equal to its rest energy. Then $E = 2mc^2$, so $\gamma = 2$ and $\beta = \sqrt3/2 \approx 0.87$. Its momentum: $pc = \sqrt{E^2 - (mc^2)^2} = \sqrt{4 - 1}\,mc^2 = \sqrt3 \times 938 \approx 1620$ MeV, so $p \approx 1.6$ GeV/$c$.

Four-vectors. $(E/c, \vec p)$ is a four-vector, like $(ct, \vec x)$. Its "length" is invariant: $E^2 - (pc)^2 = (mc^2)^2$ in every frame. For a system of particles, the same combination of the totals gives the invariant mass $M$:

$$(Mc^2)^2 = \Big(\sum E\Big)^2 - \Big(\sum \vec p\,c\Big)^2.$$

Energy and momentum are conserved; mass is not additive. Two photons, each of energy $E$, moving in opposite directions have invariant mass $2E/c^2$, though each photon is massless.

Decays and thresholds

Two-body decay at rest. A particle of mass $M$ at rest decays into two particles. Momentum conservation makes their momenta equal and opposite, so energy conservation fixes everything. For decay into two massless particles, each gets $E = Mc^2/2$. In general,

$$E_1 = \frac{(M^2 + m_1^2 - m_2^2)c^2}{2M}.$$

Worked example. A particle of rest energy 140 MeV at rest decays into two photons. Each photon carries 70 MeV, and they leave back to back.

Threshold energy. The minimum energy to create particles is set by requiring the products to be produced at rest in the centre-of-momentum frame, so the invariant mass of the initial state must equal the sum of final masses. For a beam particle of mass $m_b$ striking a target of mass $m_t$ at rest:

$$E_{\text{beam}} = \frac{(M_f^2 - m_b^2 - m_t^2)c^2}{2m_t}, \qquad M_f = \sum m_{\text{final}}.$$

Worked example. Make an antiproton in $p + p \to p + p + p + \bar p$ with a proton beam on a fixed proton target. $M_f = 4m_p$, so $E_{\text{beam}} = (16 - 1 - 1)m_pc^2/2 = 7m_pc^2$, and the kinetic energy at threshold is $6m_pc^2 \approx 5.6$ GeV. Much of the beam energy goes into the motion of the centre of mass, which is why colliding beams are more efficient.

A forbidden process. A single photon cannot turn into an electron–positron pair in empty space: in the pair's centre-of-momentum frame the total momentum is zero, but a photon has momentum $E/c \neq 0$ in every frame. A nearby nucleus is needed to absorb momentum.

Relativistic Doppler effect

A source and observer approaching each other at relative speed $\beta c$ see the frequency raised:

$$f_{\text{obs}} = f_{\text{src}}\sqrt{\frac{1+\beta}{1-\beta}}$$

(swap the signs for recession). At $\beta = 0.6$ the factor is $\sqrt{1.6/0.4} = 2$. Unlike sound, only the relative velocity enters. A source seen moving perpendicular to the line of sight (in the observer's frame) still shows a redshift by the factor $1/\gamma$ (the transverse Doppler effect), a pure time-dilation effect. For $\beta \ll 1$, $\Delta f/f \approx \pm\beta$.

The paradoxes, resolved

• Twin paradox. The travelling twin changes inertial frames at turnaround, so the situation is not symmetric. The twin whose worldline is not straight in spacetime ages less.
• Barn and pole. A 20 m pole moving at $\gamma = 2$ fits (10 m long) in a 10 m barn in the barn frame, with both doors briefly closed at once. In the pole's frame the barn is only 5 m, and the doors do not close simultaneously. Simultaneity at separated places is frame-dependent, so both accounts are consistent.
• Faster than light? Spots of light, scissor intersections and phase velocities can exceed $c$; no signal or energy can.

Fast checks

• $\gamma \ge 1$ always; any answer with a moving clock running fast, or a moving rod longer than its rest length, is wrong.
• No massive particle reaches $c$; any velocity sum above $c$ is wrong.
• Take $\beta \to 0$: the answer must reduce to Newtonian physics ($K \to \tfrac12mv^2$, $p \to mv$).
• Units: $E$ in MeV, $p$ in MeV/$c$, $m$ in MeV/$c^2$. If $c$ appears to the wrong power, the dimensions will tell you.
• If the problem gives $E$ or $K$ in units of $mc^2$, find $\gamma$ first; everything else follows.

Check yourself

• A clock moves at $0.8c$. How long does it take, in your frame, to tick off 3 s of its own time? (Answer: $5$ s.)
• A 1 m rod moves along its length at $0.6c$. How long is it in your frame? (Answer: $0.8$ m.)
• Two events are separated by 4 light-seconds in space and 5 s in time. What is the proper time between them? (Answer: 3 s.)
• An electron has total energy 1.022 MeV. What are $\gamma$, $\beta$ and $pc$? (Answer: $2$, $\sqrt3/2$, $0.885$ MeV.)
• What is the invariant mass of two photons of energy $E$ moving in the same direction? (Answer: zero.)
• Write the threshold condition for creating particles in a fixed-target collision.
• Why can a lone photon not create an electron–positron pair in vacuum?