Atomic-Clock Tests of Relativity: HSC Physics Explained

Put three synchronised atomic clocks together. Leave one on the ground, fly one east around Earth, and fly the other west. Before reading on, predict the result. Do both flying clocks fall behind because they were moving? Does the eastward clock lose more time? Or should east and west make no difference at all?

In the 1971 Hafele-Keating experiment, the answer was much stranger than “moving clocks run slow”. The eastward clocks returned behind the reference time, while the westward clocks returned ahead. Relativity had predicted exactly that direction-dependent pattern: about \(-40 \pm 23\) ns eastward and \(+275 \pm 21\) ns westward. The measured shifts were \(-59 \pm 10\) ns and \(+273 \pm 7\) ns respectively, where \(1\text{ ns}=10^{-9}\text{ s}\). PubMed

The interesting HSC question is not just “did relativity work?” It is: how could a moving clock possibly end up ahead?

01The experiment is a competition between two effects

Hafele and Keating flew portable caesium-beam atomic clocks on commercial aircraft around the world, first eastward and then westward, and compared them with reference atomic clocks at the US Naval Observatory. Atomic clocks were useful because their extremely stable ticking made differences of only a few hundred nanoseconds measurable. NIST

There are two effects to keep separate:

That second row matters. Hafele-Keating is not a pure special-relativity experiment. If you ignore the altitude of the aircraft, you do not get the correct prediction.

For HSC Physics, the useful skill is to separate the effects, understand their signs, and then combine them.

02First problem: the ground clock is already moving

A very tempting argument goes like this:

Both planes have roughly the same speed relative to the ground, so special relativity should slow both clocks by roughly the same amount.

That would be a reasonable first model if the ground formed one inertial reference frame. But Earth rotates.

Picture an airport travelator moving east. A person standing still on the travelator is not stationary relative to the terminal floor. If another person walks east, their speed relative to the floor increases. If they walk west, their speed relative to the floor decreases.

For the kinematic part of the Hafele-Keating calculation, Earth’s surface behaves similarly when we use an approximately non-rotating, Earth-centred reference frame.

Let:

• \(V\) be the eastward speed of the ground due to Earth’s rotation,
• \(u\) be the aircraft’s speed relative to the ground,
• \(v_{\text{E}}\) be the eastward aircraft’s velocity in the chosen inertial frame, and
• \(v_{\text{W}}\) be the westward aircraft’s velocity in that frame.

For an idealised equatorial flight,

\[ v_{\text{E}}=V+u \]

while

\[ v_{\text{W}}=V-u. \]

If \(u>V\), then \(v_{\text{W}}\) is negative: the aircraft is actually moving west in the chosen frame. For time dilation, its speed is then \(|V-u|\).

The travelator analogy has an important limit. The terminal floor sounds like an absolutely stationary background. Relativity has no universal stationary background. We are choosing a convenient approximately inertial frame in which to do the calculation, not discovering an absolute frame of space.

03Turning speed into elapsed time

For a clock moving at constant speed \(v\), the Lorentz factor is

\[ \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}, \]

where \(c=3.00\times10^8\text{ m s}^{-1}\) is the speed of light.

If \(\Delta \tau\) is the proper time measured by the moving clock and \(\Delta t\) is the time interval in the inertial frame, then

\[ \Delta t=\gamma\Delta\tau. \]

So

\[ \Delta\tau=\frac{\Delta t}{\gamma}. \]

A clock with greater speed through that frame accumulates less proper time.

If you need a refresher on the basic Lorentz-factor calculation before continuing, see Time Dilation in HSC Physics: Using the Lorentz Factor.

Aircraft speeds are tiny compared with \(c\). That creates a practical problem: \(\gamma\) is so close to 1 that the difference can disappear in ordinary calculator rounding. For low speeds, the Lorentz result can be approximated by

\[ \Delta\tau\approx T\left(1-\frac{v^2}{2c^2}\right), \]

where \(T\) is the elapsed coordinate time.

Comparing an aircraft with a ground clock therefore gives the approximate kinematic shift

\[ \Delta\tau_{\text{k}}\approx -\frac{(v_{\text{air}}^2-v_{\text{ground}}^2)T}{2c^2}. \]

The sign tells you what happened:

• \(\Delta\tau_{\text{k}}<0\): the aircraft clock accumulated less time than the ground clock.
• \(\Delta\tau_{\text{k}}>0\): the aircraft clock accumulated more time than the ground clock.

This equation is a low-speed approximation to the Lorentz-factor result. It is useful here because the effect is tiny. It is not a replacement for the exact time-dilation equation at relativistic speeds.

04Second problem: the aircraft is higher

Now imagine two identical clocks that are not moving relative to each other, but one is several kilometres above the other.

Which should run faster?

General relativity predicts that the higher clock does. Near Earth’s surface, a useful approximation for the accumulated gravitational time difference is

\[ \Delta\tau_{\text{g}}\approx\frac{ghT}{c^2}, \]

where:

• \(g\) is gravitational field strength in \(\text{m s}^{-2}\),
• \(h\) is the height difference in metres,
• \(T\) is the elapsed time in seconds, and
• \(c\) is the speed of light.

For an aircraft above the ground, this contribution is positive.

So the simplified total shift is

\[ \Delta\tau_{\text{total}} = \Delta\tau_{\text{k}} + \Delta\tau_{\text{g}}. \]

That is the key idea. The velocity effect and gravitational effect can reinforce one another or compete.

05Why east and west give different answers

For the actual Hafele-Keating flights, the predicted contributions were approximately:

The exact gravitational contributions were different because the two real journeys did not have identical flight times, altitudes, or routes. The important pattern is the sign. HyperPhysics

For the eastward flight, the plane moved faster than the already eastward-moving ground in the chosen non-rotating frame. Its kinematic loss was large enough to beat the positive altitude effect. The clock therefore finished behind.

For the westward flight, the plane’s motion reduced its speed compared with the rotating ground. It therefore accumulated more proper time from the kinematic effect than the ground clock did. The altitude effect also pushed its reading ahead. The two effects reinforced each other.

So “moving clocks run slow” is not wrong. The mistake is asking whether the aircraft is moving without also asking: moving at what speed, compared with which other worldline, in what reference frame?

06The misconception to avoid: “westward clocks run faster”

That statement is too broad.

In the historical experiment, the westward clocks gained time relative to the ground reference. But the word “westward” itself does not make clocks run faster.

Suppose you kept increasing a westbound aircraft’s speed. Initially, travelling west reduces its speed in the non-rotating Earth-centred frame. At one particular ground-relative speed, the aircraft could even be momentarily stationary in that frame while Earth’s surface moved east beneath it.

Make it much faster still, however, and the aircraft would be racing west through the inertial frame. Its speed magnitude could become greater than the ground clock’s speed. Its kinematic effect would then become negative again.

So the decision rule is not:

• east = slower clock,
• west = faster clock.

The better rule is:

1. choose one suitable reference frame,
2. calculate both clocks’ speeds in that frame,
3. compare the squares of those speeds,
4. add any gravitational contribution, and
5. only then decide which clock accumulates more time.

07What makes the experiment strong evidence?

A useful experiment does more than produce a strange result after the fact. The theory must make a prediction that could have been wrong.

Hafele-Keating had several useful features.

First, relativity predicted the signs before the clocks were compared: an eastward loss and a much larger westward gain. The asymmetry was not added later to rescue the model. PubMed

Second, the experiment produced quantitative predictions. “The clocks might differ a bit” would be weak. Predicting changes on the scale of tens or hundreds of nanoseconds gives the measurement something specific to test.

Third, the calculation had to include the actual physical conditions. Speed, direction, altitude, duration, and Earth’s rotation all mattered. A model that ignored a relevant effect could easily give the wrong sign.

Finally, agreement does not mean one experiment has “proved relativity forever”. Experiments support models by surviving tests that could have exposed a disagreement. Later clock experiments have tested relativistic time effects with much greater precision. NIST

That is the evidence skill worth carrying into an HSC response: state the prediction, state the observation, compare them with uncertainty, and explain why the agreement is significant.

08Questions and solutions

Question 1

An atomic clock on a spacecraft moves at \(0.80c\) relative to an inertial observer. The spacecraft clock records \(3.0\) h between two events that occur at the same place on the spacecraft.

Calculate the time interval measured by the inertial observer, and state how much less time the spacecraft clock records.

Solution 1

The inertial observer measures \(5.0\) h, so the spacecraft clock records \(2.0\) h less.

The spacecraft clock measures the proper time,

\[ \Delta\tau=3.0\text{ h}. \]

The Lorentz factor is

\[ \begin{aligned} \gamma &=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ &=\frac{1}{\sqrt{1-(0.80)^2}}\\ &=\frac{1}{0.60}\\ &=1.67. \end{aligned} \]

Therefore,

\[ \begin{aligned} \Delta t &=\gamma\Delta\tau\\ &=(1.67)(3.0\text{ h})\\ &=5.0\text{ h}. \end{aligned} \]

The difference is

\[ 5.0-3.0=2.0\text{ h}. \]

The tempting mistake is to multiply or divide by \(\gamma\) without first deciding which interval is proper time. The clock travelling with the spacecraft measures the proper time because the two events occur at the same place in its own frame.

Question 2

An idealised pair of aircraft fly for \(12.0\) h near the equator at an altitude of \(9.00\text{ km}\). Each has a ground-relative speed of \(300\text{ m s}^{-1}\), one eastward and one westward.

Take the ground’s rotational speed to be \(465\text{ m s}^{-1}\), \(g=9.8\text{ m s}^{-2}\), and \(c=3.00\times10^8\text{ m s}^{-1}\).

For each aircraft, calculate:

• its velocity in the non-rotating Earth-centred frame,
• its kinematic clock shift relative to the ground,
• its gravitational clock shift, and
• its total clock shift.

Use

\[ \Delta\tau_{\text{k}} \approx-\frac{(v_{\text{air}}^2-v_{\text{ground}}^2)T}{2c^2} \]

and

\[ \Delta\tau_{\text{g}}\approx\frac{ghT}{c^2}. \]

Solution 2

The eastward clock finishes about \(46.2\) ns behind the ground clock, while the westward clock finishes about \(87.7\) ns ahead.

First convert the duration:

\[ T=(12.0)(3600)=4.32\times10^4\text{ s}. \]

For the eastward aircraft,

\[ v_{\text{E}}=465+300=765\text{ m s}^{-1}. \]

Its kinematic shift is

\[ \begin{aligned} \Delta\tau_{\text{k,E}} &=-\frac{(765^2-465^2)(4.32\times10^4)} {2(3.00\times10^8)^2}\\ &=-8.856\times10^{-8}\text{ s}\\ &=-88.6\text{ ns}. \end{aligned} \]

For the westward aircraft,

\[ v_{\text{W}}=465-300=165\text{ m s}^{-1}. \]

Its kinematic shift is

\[ \begin{aligned} \Delta\tau_{\text{k,W}} &=-\frac{(165^2-465^2)(4.32\times10^4)} {2(3.00\times10^8)^2}\\ &=4.536\times10^{-8}\text{ s}\\ &=45.4\text{ ns}. \end{aligned} \]

Both aircraft have the same altitude and duration, so they have the same gravitational contribution:

\[ \begin{aligned} \Delta\tau_{\text{g}} &=\frac{(9.8)(9.00\times10^3)(4.32\times10^4)} {(3.00\times10^8)^2}\\ &=4.2336\times10^{-8}\text{ s}\\ &=42.3\text{ ns}. \end{aligned} \]

Hence the eastward total is

\[ \Delta\tau_{\text{E}} =-88.6+42.3 =-46.2\text{ ns}, \]

while the westward total is

\[ \Delta\tau_{\text{W}} =45.4+42.3 =87.7\text{ ns}. \]

The trap is treating \(300\text{ m s}^{-1}\) as the relevant speed for both aircraft. That is only their speed relative to the rotating ground. The relativistic comparison requires both clock speeds to be expressed consistently in the chosen inertial frame.

Question 3

A hypothetical aircraft flies westward near the equator at \(1200\text{ m s}^{-1}\) relative to the ground for \(6.00\) h at an altitude of \(10.0\text{ km}\).

The ground moves eastward at \(465\text{ m s}^{-1}\) in a non-rotating Earth-centred frame.

Using the low-speed kinematic and gravitational approximations from this article, determine whether the aircraft clock finishes ahead of or behind the ground clock.

Explain why the answer does not contradict the westward Hafele-Keating result.

Solution 3

The westward aircraft clock finishes about \(15.4\) ns behind the ground clock.

This is deliberately different from the historical westward result. “Westward” does not guarantee a clock gain.

Take eastward as positive. The aircraft velocity in the Earth-centred frame is

\[ v_{\text{air}}=465-1200=-735\text{ m s}^{-1}. \]

Its velocity is westward, but its speed is

\[ |v_{\text{air}}|=735\text{ m s}^{-1}. \]

That is greater than the ground clock’s speed of \(465\text{ m s}^{-1}\).

The flight duration is

\[ T=6.00\times3600=2.16\times10^4\text{ s}. \]

The kinematic shift is

\[ \begin{aligned} \Delta\tau_{\text{k}} &=-\frac{(735^2-465^2)(2.16\times10^4)} {2(3.00\times10^8)^2}\\ &=-3.888\times10^{-8}\text{ s}\\ &=-38.9\text{ ns}. \end{aligned} \]

The gravitational contribution is

\[ \begin{aligned} \Delta\tau_{\text{g}} &=\frac{(9.8)(1.00\times10^4)(2.16\times10^4)} {(3.00\times10^8)^2}\\ &=2.352\times10^{-8}\text{ s}\\ &=23.5\text{ ns}. \end{aligned} \]

Therefore,

\[ \begin{aligned} \Delta\tau_{\text{total}} &=-38.9+23.5\\ &=-15.4\text{ ns}. \end{aligned} \]

The aircraft is westbound, but it is moving so quickly west that its speed magnitude in the inertial frame exceeds the ground’s speed. Its kinematic time loss therefore outweighs its gravitational gain.

The historical westward aircraft were in a different regime: travelling west reduced their speed relative to the non-rotating frame, so their kinematic contribution was positive. The direction label is not the governing principle. The actual speeds are.

Question 4

Two aircraft fly in opposite directions for \(8.00\) h with equal ground-relative speed \(u=250\text{ m s}^{-1}\) and equal constant altitude. After all corrections associated with the experiment are made, their clock shifts relative to the same ground reference are

\[ \Delta\tau_{\text{E}}=-10.64\text{ ns} \]

and

\[ \Delta\tau_{\text{W}}=+53.36\text{ ns}. \]

Assume the simplified low-speed model

\[ \Delta\tau_{\text{E}} = G-\frac{(2Vu+u^2)T}{2c^2} \]

and

\[ \Delta\tau_{\text{W}} = G+\frac{(2Vu-u^2)T}{2c^2}, \]

where \(V\) is the eastward speed of the ground in the inertial frame and

\[ G=\frac{ghT}{c^2}. \]

Using \(g=9.8\text{ m s}^{-2}\) and \(c=3.00\times10^8\text{ m s}^{-1}\), infer both \(V\) and the aircraft altitude \(h\).

Solution 4

The data imply \(V=400\text{ m s}^{-1}\) and \(h=1.00\times10^4\text{ m}=10.0\text{ km}\).

The useful move is to combine the two measurements in different ways.

First subtract the eastward result from the westward result:

\[ \begin{aligned} \Delta\tau_{\text{W}}-\Delta\tau_{\text{E}} &=53.36-(-10.64)\\ &=64.00\text{ ns}\\ &=6.400\times10^{-8}\text{ s}. \end{aligned} \]

Subtracting the two model equations makes \(G\) cancel:

\[ \Delta\tau_{\text{W}}-\Delta\tau_{\text{E}} =\frac{2VuT}{c^2}. \]

The duration is

\[ T=(8.00)(3600)=2.88\times10^4\text{ s}. \]

Therefore,

\[ \begin{aligned} V &=\frac{(\Delta\tau_{\text{W}}-\Delta\tau_{\text{E}})c^2}{2uT}\\ &=\frac{(6.400\times10^{-8})(3.00\times10^8)^2} {2(250)(2.88\times10^4)}\\ &=400\text{ m s}^{-1}. \end{aligned} \]

Now average the two measured shifts:

\[ \frac{-10.64+53.36}{2}=21.36\text{ ns}. \]

A tempting argument is that averaging east and west should remove every kinematic effect. It does not.

Adding the two model equations and dividing by two gives

\[ \frac{\Delta\tau_{\text{E}}+\Delta\tau_{\text{W}}}{2} = G-\frac{u^2T}{2c^2}. \]

The terms involving \(Vu\) cancel, but the \(u^2\) term survives.

Calculate that remaining term:

\[ \begin{aligned} \frac{u^2T}{2c^2} &=\frac{(250)^2(2.88\times10^4)} {2(3.00\times10^8)^2}\\ &=1.00\times10^{-8}\text{ s}\\ &=10.0\text{ ns}. \end{aligned} \]

Therefore,

\[ G=21.36+10.0=31.36\text{ ns}. \]

Now use

\[ G=\frac{ghT}{c^2}. \]

Rearranging,

\[ \begin{aligned} h &=\frac{Gc^2}{gT}\\ &=\frac{(31.36\times10^{-9})(3.00\times10^8)^2} {(9.8)(2.88\times10^4)}\\ &=1.00\times10^4\text{ m}\\ &=10.0\text{ km}. \end{aligned} \]

The deeper point is that the east-west difference isolates one combination of the velocity effects, while the average still contains another. Opposite directions do not simply make special relativity disappear because time dilation depends on \(v^2\), not directly on the sign of \(v\).

Question 5

Consider an idealised experiment in which two aircraft spend the same time \(T=10.0\) h at the same altitude. They travel with equal ground-relative speed \(u=220\text{ m s}^{-1}\), one eastward and one westward. At their latitude, the ground has an eastward inertial-frame speed \(V=350\text{ m s}^{-1}\).

A student makes three claims:

1. Because the aircraft have equal speeds relative to the ground, special relativity predicts no difference between their clocks.
2. Because both aircraft have the same altitude and flight time, gravitational time dilation cancels when their two clock offsets are compared directly.
3. If the Earth-centred inertial-frame calculation correctly predicts a difference, that proves this frame is the universe’s physically preferred frame.

Calculate the predicted difference between the two airborne clock offsets using the low-speed model, then assess all three claims.

Solution 5

The westward clock is predicted to finish \(61.6\) ns ahead of the eastward clock; claim 1 is false, claim 2 is correct under the stated idealisations, and claim 3 is false.

For equal altitude and flight time, let the common gravitational contribution be \(G\).

The eastward shift is

\[ \Delta\tau_{\text{E}} = G-\frac{(2Vu+u^2)T}{2c^2}. \]

The westward shift is

\[ \Delta\tau_{\text{W}} = G+\frac{(2Vu-u^2)T}{2c^2}. \]

Subtracting gives

\[ \begin{aligned} \Delta\tau_{\text{W}}-\Delta\tau_{\text{E}} &= \frac{(2Vu-u^2)T+(2Vu+u^2)T}{2c^2}\\ &=\frac{2VuT}{c^2}. \end{aligned} \]

Notice what has disappeared. The common gravitational contribution \(G\) cancels, and so do the two \(u^2\) terms.

With

\[ T=10.0\times3600=3.60\times10^4\text{ s}, \]

the difference is

\[ \begin{aligned} \Delta\tau_{\text{W}}-\Delta\tau_{\text{E}} &=\frac{2(350)(220)(3.60\times10^4)} {(3.00\times10^8)^2}\\ &=6.16\times10^{-8}\text{ s}\\ &=61.6\text{ ns}. \end{aligned} \]

So the westward clock is \(61.6\) ns ahead of the eastward clock.

Claim 1 fails because the ground-relative speeds \(+u\) and \(-u\) are not the complete velocities required by this calculation. The rotating ground itself has velocity \(V\). The two aircraft therefore have inertial-frame velocities \(V+u\) and \(V-u\), whose squared magnitudes are generally different.

Claim 2 is correct here. Because altitude and duration are stated to be identical, both clocks receive the same gravitational contribution, so that term disappears when their offsets are subtracted. In a real experiment, unequal routes, altitudes, and durations would need to be accounted for rather than assumed away.

Claim 3 is the subtle trap. Using an Earth-centred inertial frame does not make that frame absolute. It is simply a convenient coordinate system in which the kinematic calculation is clean.

The quantity eventually compared by the experiment is the elapsed proper time along each clock’s path. Different inertial observers may assign different coordinate velocities and coordinate times during the journey, but a complete relativistic calculation must agree on the actual readings when the clocks are compared. If one frame produced a genuinely different reunion reading from another, relativity would be internally inconsistent.

The failed shortcut was not “using the wrong preferred frame”. It was treating Earth’s rotating surface as though it were one global inertial frame and then using only the aircrafts’ local speeds relative to that surface.

09What this lets you understand next

Atomic-clock experiments turn time dilation from an equation into a testable prediction. The important method is to compare whole clock paths: identify the correct speeds, include any gravitational effect that matters, predict the size and sign of the difference, and then compare prediction with measurement and uncertainty.

That same evidence logic appears in a very different experiment involving short-lived particles. Muon evidence for time dilation gives you a useful comparison because the observable is not two clock displays but the number of unstable particles that survive long enough to reach Earth’s surface. The apparatus changes, but the question stays the same: does nature produce the elapsed time that relativity predicts?