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The spacetime interval and its sign, proper time, relativistic energy $\gamma mc^2$ and momentum $\gamma mv$, rest energy, and the invariant $E^2 - (pc)^2 = (mc^2)^2$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to classify intervals, find proper times, and compute relativistic energies and momenta.
From the last lessons you know that observers disagree about times and lengths, by the factor $\gamma$. From mechanics you know kinetic energy and momentum. This lesson finds what observers do agree on: a combination of time and distance, and a combination of energy and momentum, each the same in every frame. The second contains the most famous equation in physics.
| Term | What it means |
|---|---|
| Spacetime interval | $s^2 = (c\Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$, the same in every inertial frame. |
| Timelike | $s^2 > 0$: a clock or signal slower than light could be present at both events. |
| Spacelike | $s^2 < 0$: no signal could connect the events; their order depends on the frame. |
| Lightlike | $s^2 = 0$: only light could connect the events. |
| Rest energy | $mc^2$, the energy of a body at rest. |
| Total energy | $E = \gamma mc^2$, rest energy plus kinetic energy. |
| Relativistic momentum | $p = \gamma mv$. |
Different frames give an event pair different $\Delta t$ and $\Delta x$, but the combination
$$s^2 = (c\Delta t)^2 - \Delta x^2$$
is the same in all of them, just as a length in ordinary space is the same whatever way the axes are turned. If $s^2 > 0$, the interval is timelike: a clock moving slower than light could attend both events, it reads the proper time $\Delta\tau = s/c$, and every frame agrees which event came first. If $s^2 < 0$, it is spacelike, and no signal could connect them.
Energy and momentum form the same kind of pair. With $E = \gamma mc^2$ and $p = \gamma mv$,
$$E^2 - (pc)^2 = (mc^2)^2$$
in every frame. A body at rest still has energy $mc^2$, its rest energy; its kinetic energy is $K = (\gamma - 1)mc^2$.
Another way: picture
Picture a map where north and east change if you turn the map, but the straight-line distance between two towns does not. Spacetime works like that, except that time enters with the opposite sign. Changing frames tilts the time and space axes, and the interval is the distance that survives the tilt.
Another way: steps
Apply the Lorentz transformation: $c\Delta t' = \gamma(c\Delta t - \beta\Delta x)$ and $\Delta x' = \gamma(\Delta x - \beta c\Delta t)$. Square and subtract: the cross terms cancel, and $(c\Delta t')^2 - \Delta x'^2 = \gamma^2(1 - \beta^2)[(c\Delta t)^2 - \Delta x^2] = (c\Delta t)^2 - \Delta x^2$. The factor $\gamma^2(1 - \beta^2)$ is exactly 1.
The minus sign is what makes spacetime different from ordinary space. In space, $x^2 + y^2$ is unchanged by rotations. In spacetime, $(ct)^2 - x^2$ is unchanged by boosts, which are hyperbolic rotations. Hermann Minkowski, who introduced this picture in 1908, said that henceforth space by itself and time by itself were doomed to fade into shadows.
A positive $s^2$ means more time separates the events than light needs to cross the distance between them. Some slower-than-light traveler could be present at both, and in that traveler's frame the events happen at the same place, separated only by the proper time $s/c$. All observers agree on their order, so one can cause the other.
A negative $s^2$ means the events are too far apart for any signal. Some frame sees them simultaneous, others see either one first. Since nothing can travel between them, the ambiguity does no harm. A zero interval connects events on a light pulse's path.
For a clock moving at steady speed, the interval between two ticks is the proper time between them times $c$. For a traveler who changes speed, add the interval along each straight piece. The path with the most proper time between two events is the one that stays in a single inertial frame; any detour at high speed accumulates less.
That is the geometry behind the twin paradox: the stay-at-home twin's straight world line has the greatest proper time, and the traveler's bent one has less. In spacetime the straight path is the longest, the reverse of ordinary geometry, because of the minus sign in the interval.
Newton's momentum $mv$ is not conserved in collisions viewed from different frames once speeds approach $c$. The quantity that is conserved is $p = \gamma mv$, which grows without limit as $v \to c$. Its partner is the total energy $E = \gamma mc^2$, which for small speeds is $mc^2 + \tfrac{1}{2}mv^2 + \ldots$: rest energy plus Newton's kinetic energy.
The two combine into the invariant $E^2 - (pc)^2 = (mc^2)^2$, the energy counterpart of the interval. It gives any one of $E$, $p$ or $m$ from the other two, and it shows that a massless particle, like a photon, has $E = pc$ and must travel at $c$.
Setting $v = 0$ gives $E = mc^2$. A body at rest has energy simply by having mass. One gram corresponds to $9 \times 10^{13}$ J, about the energy of the Hiroshima bomb. Ordinary chemistry converts about one part in a billion of the mass involved; nuclear reactions about one part in a thousand; matter meeting antimatter converts all of it.
In particle physics, masses are quoted as rest energies: the electron's is $0.511$ MeV, the proton's $938.3$ MeV, the Higgs boson's $125$ GeV. Unit 6 uses the same idea to explain the energy released by nuclear fission and fusion.
Checking an answer. Invariants must come out the same in every frame. Total energy must be at least the rest energy. $pc$ must be less than $E$ for a massive body. And at small speeds, $K$ must reduce to $\tfrac{1}{2}mv^2$ and $p$ to $mv$.
Invariants let a problem be solved in whatever frame is simplest and the answer carried to any other. That works because the Lorentz transformation is linear and preserves the interval, just as rotations preserve lengths. It is the same reason a surveyor can measure a distance using any orientation of the map.
The energy–momentum invariant follows from the same algebra applied to $(E/c, p)$, which transforms between frames exactly like $(ct, x)$. The pair is called the four-momentum, and its conservation in collisions replaces the separate conservation of energy and momentum in Newton's mechanics.
Relativistic kinetic energy, $(\gamma - 1)mc^2$, grows without limit as $v \to c$. That is why no massive body can reach light speed: it would need infinite energy. At $0.1c$ the Newtonian formula is off by less than a percent; at $0.5c$ by $19$ percent; at $0.9c$ it gives less than a third of the true value.
Accelerator physicists work almost entirely with energies. A proton with kinetic energy of $1$ GeV has $\gamma = 2.07$ and $\beta = 0.875$; one with $1$ TeV has $\gamma = 1067$ and differs from light speed by a part in two million. Pushing it harder raises its energy and momentum, while its speed barely changes.
In Newton's physics, mass and energy are separately conserved. In relativity only their combination is. When a hot object cools, it loses a tiny amount of mass, $\Delta E/c^2$, too small to weigh. When an electron and a positron annihilate, their mass vanishes entirely into the energy of two photons.
Most of the mass of ordinary matter is not the mass of its elementary particles but the energy binding quarks inside protons and neutrons. The three quarks of a proton have rest energies adding to about $10$ MeV; the proton's is $938$ MeV. Nearly all of your mass is, in this sense, energy.
Set $m = 0$ in the energy–momentum invariant and it becomes $E = pc$. A massless particle carries energy and momentum in a fixed ratio, and since $\beta = pc/E$ for any particle, it must travel at exactly $c$, in every frame. It has no rest frame, and no clock could ride along with it: along a light pulse's path the interval, and so the proper time, is zero.
The photon is the massless particle of light, and the next unit is about it. Its momentum $p = E/c$ is why light exerts pressure, as the solar sails of the last course showed. Neutrinos were long thought massless too; experiments since 1998, including those at Fermilab and in the Sudbury mine in Canada, showed they have tiny masses, less than a millionth of an electron's, which is why they travel at very nearly, but not exactly, the speed of light.
Fermilab's Tevatron, near Chicago, accelerated protons to $980$ GeV from 1983 to 2011, a Lorentz factor over a thousand. Its successor machines at Fermilab now make intense beams of $120$ GeV protons, $\gamma \approx 128$, to produce neutrinos aimed at detectors in South Dakota, $1300$ km away.
At Brookhaven National Laboratory on Long Island, the Relativistic Heavy Ion Collider smashes gold nuclei at $100$ GeV per nucleon, $\gamma \approx 107$. In the lab frame each nucleus is flattened by length contraction into a thin disk, and the collisions briefly recreate the quark-gluon plasma that filled the universe a microsecond after the Big Bang.
A PET scanner detects the two $0.511$ MeV photons produced when a positron from a medical tracer meets an electron in the body and both annihilate. Their entire rest energy, $2 \times 0.511$ MeV, becomes light, and the photons fly apart in opposite directions to conserve momentum.
Rings of detectors around the patient record pairs of photons arriving within a few nanoseconds of each other, and a computer traces each pair back to a line through the body. Hospitals across the country use PET to find cancers and study the brain; each image is built from millions of acts of $E = mc^2$.
It is common to read $E = mc^2$ as a formula for nuclear bombs, as if mass turned into energy only in special reactions. In fact every body at rest has energy $mc^2$, and every change in energy, even warming a cup of coffee, changes its mass slightly. Nuclear reactions stand out only because the fraction converted is large enough to notice.
A related error is to use $E = mc^2$ for a moving body. The total energy is $\gamma mc^2$; $mc^2$ is only the rest energy, and kinetic energy is the difference.
In Earth's frame two events are $8$ ly apart and $10$ years apart. Find $s^2$.
$s^2 = 10^2 - 8^2 = 36\ \text{years}^2$
Timelike.
Find the proper time.
$\Delta\tau = 6\ \text{years}$
What a clock at both events reads.
Find that clock's speed.
$\beta = \dfrac{8}{10} = 0.8$
It covers $8$ ly in $10$ years.
Transform the events to a frame at $0.6c$.
$c\Delta t' = 1.25(10 - 4.8) = 6.5, \quad \Delta x' = 1.25(8 - 6) = 2.5$
Lorentz transformation.
Check the interval there.
$6.5^2 - 2.5^2 = 42.25 - 6.25 = 36$
Unchanged.
An electron has kinetic energy $1.00$ MeV. Find its total energy.
$E = 1.00 + 0.511 = 1.511\ \text{MeV}$
Kinetic plus rest.
Find the Lorentz factor $\gamma$.
$\gamma = \dfrac{1.511}{0.511} = 2.957$
$E/mc^2$.
Find the speed ratio $\beta$.
$\beta = \sqrt{1 - \dfrac{1}{2.957^2}} = 0.941$
From $\gamma$.
Find the momentum $pc$.
$pc = \sqrt{1.511^2 - 0.511^2} = 1.422\ \text{MeV}$
The invariant.
Check with $\gamma\beta mc^2$.
$2.957 \times 0.941 \times 0.511 = 1.422\ \text{MeV}$
Agrees.
Compare with Newton's speed.
$v = \sqrt{\dfrac{2K}{m}} \Rightarrow \beta = \sqrt{\dfrac{2}{0.511}} = 1.98$
Faster than light: Newton fails badly here.
A pion at rest, $mc^2 = 139.6$ MeV, decays into a muon, $105.7$ MeV, and a nearly massless neutrino. Write momentum conservation.
$p_\mu = p_\nu = p$
Equal and opposite.
Write energy conservation.
$139.6 = E_\mu + pc$
The neutrino has $E = pc$.
Use the muon's invariant.
$E_\mu^2 = (pc)^2 + 105.7^2$
Energy–momentum relation.
Substitute $E_\mu = 139.6 - pc$.
$(139.6 - pc)^2 = (pc)^2 + 105.7^2$
One unknown.
Solve for the momentum.
$pc = \dfrac{139.6^2 - 105.7^2}{2 \times 139.6} = 29.8\ \text{MeV}$
The squares cancel.
Find the muon's kinetic energy.
$K_\mu = (139.6 - 29.8) - 105.7 = 4.1\ \text{MeV}$
Total minus rest energy.
Check the total energy.
$109.8 + 29.8 = 139.6\ \text{MeV}$
Conserved.
Compute the squared interval.
$s^2 = 12^2 - 13^2 = 144 - 169$
Years and light-years.
Evaluate the squared interval.
$s^2 = -25\ \text{years}^2$
Negative.
Classify the interval.
Two events are $5$ light-years apart and $9$ years apart in time, in Earth's frame. What kind of interval separates them?
Complete the worked solution: two events are $8$ light-years apart and $10$ years apart in Earth's frame. Find the squared interval in years squared, the proper time between them in years, and the speed, as a fraction of $c$, of a clock that attends both.
Square and subtract.
$s^2 = 10^2 - 8^2 =$ a
Years and light-years.
Take the root for the proper time.
$\Delta\tau = \dfrac{s}{c} =$ b
What a clock at both events reads.
Find the clock's speed.
$\beta = \dfrac{\Delta x}{c\Delta t} =$ c
It must cover the distance in the time.
Check with time dilation.
$\Delta t = \gamma\Delta\tau$
The two routes agree.
Match each relativistic quantity to its expression or meaning.
| $(c\Delta t)^2 - \Delta x^2$ | $s^2 > 0$ | $\gamma mc^2$ | $E^2 - (pc)^2 = (mc^2)^2$ | |
|---|---|---|---|---|
| spacetime interval | ||||
| timelike | ||||
| total energy | ||||
| energy–momentum invariant |
An electron, with rest energy $0.511$ MeV, moves at $0.6c$, where $\gamma = 1.25$ and $\gamma\beta = 0.75$. Fill in its total energy, its kinetic energy, and its momentum times $c$, each in MeV.
| value | |
|---|---|
| total energy (MeV) | |
| kinetic energy (MeV) | |
| pc (MeV) |
Two events are $8$ light-years apart in Earth's frame and $t$ years apart in time. Write the squared interval $s^2$ between them, in years squared, as a formula in $t$.
Answer:
A proton, with rest energy $938.3$ MeV, has kinetic energy $7000$ MeV. What is its momentum times $c$, in MeV?
Answer: MeV for pc
Brookhaven's Alternating Gradient Synchrotron accelerates protons to a total energy of $33$ GeV. With a proton's rest energy $0.938$ GeV, what is their Lorentz factor?
Answer: Lorentz factor
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two events are $5$ light-years apart in Earth's frame and $t$ years apart in time. Write the squared interval $s^2$ between them, in years squared, as a formula in $t$.
Answer:
You can use relativistic invariants. Explain to someone why all observers agree on the order of two events one of which caused the other.
21. Your turn: two events are $13$ ly apart and $12$ years apart. Classify the interval., step 3
$\text{spacelike: no signal could connect them}$
Their order depends on the frame.