In Wikipedia article about time dilation, it says:

"Hafele and Keating, in 1971, flew caesium atomic clocks east and west around the Earth in commercial airliners...the moving clocks were expected to age more slowly because of the speed of their travel."

Is it really true? My belief was that only acceleration/deceleration takes effect when comparing two clocks in the same frame after certain kinds of motions. According to my belief, I thought the time difference of two clocks in the aforementioned test is the same regardless of the length of the flight, if we ignore the difference of gravitational potential caused by different heights.

  • $\begingroup$ Are you asking if time dilation exists? According to definition, time dilation is the slowing of time due to moving at some speed $v_1$ relative to a rest frame, and that will also be relative to some other frame-speed $v_2$ if you are considering a third view. $\endgroup$ – Chris Gerig Nov 26 '11 at 6:48

This type of confusion is usually easiest to clear up by a geometrical analogy. You draw the time coordinate as a space coordinate, and you draw the paths of the moving clocks in the time-space continuum relative to an inertial frame, where the origin will be the center of the Earth. The moving clocks going around the Earth make a space-time spiral in two different ways--- one clock spirals with the rotation of the Earth, the other spirals against the rotation of the Earth. The two spirals in time have a different twistiness, the number of windings per unit time is different, and two spirals of the same height with different twistiness have different lengths in ordinary geometry--- the twistier one is longer.

Relativity is just like geometry, except for the minus sign in the pythagorean theorem (see here: Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman ). The minus sign ensures that the twistier spiral is shorter in relativity than the less twisty spiral, so that the clock moving in the direction of the Earth's motion will ticks less time than the clock moving against the rotation of the Earth. The clock moving against the rotation of the Earth in a modern jet is nearly stationary relative to the unrotating center of the Earth coordinates, since the jet can fly at close the rotation speed, so that the jet going against the rotation of the Earth will be ticking out true center-of-the-Earth non-rotating time, while the jet going along with the Earth will get a reduction in clock rate according to the velocity time dilation factor $\sqrt{1-v^2}$ (in units where velocity is measured in dimensionless fractions of the speed of light), the factor $\sqrt{1-v^2}$ is to be contrasted with the arclength factor $\sqrt{1+y'^2}$ in geometry, which gives the extra arc length per unit x displacement gotten by going along a curve y(x).

The geometry of relativity, aside from its counterintuitive flips of sign and qualitative shortening/lengthening effects, is exactly the same as the completely intuitive (although equally mathematically complicated) arclength effects in rotationally invariant Euclidean geometry.

  • $\begingroup$ Wow. I never knew how they did that. Well put. $\endgroup$ – John Dec 14 '11 at 21:34

One can calculate the exact time differences as follows from the general relativity. That should make it clear where the time differences come from. We start with the weak field metric: $$ d s^2=-\left(1+{2\phi\over c^2}\right)c^2 d t^2 +\left(1-{2\phi\over c^2}\right)(d x^2 +d y^2 +d z^2) $$


$$ \tau_{AB} =\int_A^B d\tau =\int_A^B\sqrt{-{d s^2\over c^2}} = $$ $$ =\int_A^B\sqrt{\left(1+{2\phi\over c^2}\right)d t^2 -{1\over c^2}\left(1-{2\phi\over c^2}\right)(d x^2 +d y^2 +d z^2)}= $$

$$ =\int_A^B d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)\left( \left(d x\over d t\right)^2 + \left(d y\over d t\right)^2 + \left(d z\over d t\right)^2\right)}= $$

$$ =\int_A^B d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)|{\bf V}|^2} $$ where $$ |{\bf V}|^2= \left(d x\over d t\right)^2 + \left(d y\over d t\right)^2 + \left(d z\over d t\right)^2 $$ is the nonrelativistic velocity. Then we expand the square root into power series and only keep terms with low powers of $c$: $$ \tau_{AB} =\int_A^B d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)|{\bf V}|^2} =\int_A^B d t\left(1+{\phi\over c^2}-{1\over 2c^2}|{\bf V}|^2\right) $$ so $$ \tau_{AB} =\int_A^B d t\left(1-{1\over c^2}\left({1\over2}|{\bf V}|^2-\phi\right)\right) $$

Now let $V_g=V_g(t)$ be the speed of the plane relative to the (rotating) Earth (positive for the eastbound flights, negative for the westbound ones), $V_\oplus={2\pi R_\oplus\over24}\,{\rm1\over h}$ the surface speed of the Earth, then the proper time for the clocks on the surface is:

$$ \tau_\oplus =\int_A^B d t\left(1-{1\over c^2}\left({1\over2}V_\oplus^2-\phi_\oplus\right) \right) $$ and for the clocks in the plane

$$ \tau =\int_A^B d t\left(1-{1\over c^2}\left({1\over2}(V_g+V_\oplus)^2-\phi\right) \right) $$ then the difference between the proper times is: $$ \tau-\tau_\oplus=\Delta\tau ={1\over c^2}\int_A^B d t\left(-{1\over2}(V_g+V_\oplus)^2+\phi +{1\over2}V_\oplus^2-\phi_\oplus\right) = $$ $$ ={1\over c^2}\int_A^B d t\left( \phi-\phi_\oplus-{1\over2}V_g(V_g+2V_\oplus) \right) $$ but $\phi-\phi_\oplus=g h$, where $h=h(t)$ is the altitude of the plane, so the final formula is: $$ \Delta\tau ={1\over c^2}\int_A^B d t\left( gh-{1\over2}V_g(V_g+2V_\oplus) \right) $$ Let's evaluate it for typical altitudes and speeds of commercial aircrafts: $$ R_\oplus=6 378.1{\rm\,km}=6.3781\cdot10^6{\rm\,m} $$ $$ V_\oplus={2\pi R_\oplus\over24}\,{\rm1\over h} ={2\pi R_\oplus\over24\cdot3600}\,{\rm1\over s} ={2\pi\,6.3781\cdot10^6\over24\cdot3600}{\rm m\over s}=463.83\rm\,{m\over s} $$

$$ V_g = 870\,{\rm km\over h}=241.67\rm\,{m\over s} $$ $$ h = 12{\rm\,km}=12000\rm\,m $$ $$ t = {2\pi R_\oplus\over V_g} = {2\pi\,6.3781\cdot10^6\over 241.67}{\rm\,s} =165824.41{\rm\,s}\approx 46{\rm\,h} $$ $$ c = 3\cdot10^8\rm\,{m\over s} $$ For eastbound flights we get: $$ \Delta\tau ={t\over c^2} \left( gh-{1\over2}V_g(V_g+2V_\oplus) \right) =-4.344\cdot10^{-8}{\rm\,s}=-43.44{\rm\,ns} $$ and for westbound flights we get: $$ \Delta\tau ={t\over c^2} \left( gh-{1\over2}V_g(V_g-2V_\oplus) \right) =3.6964\cdot10^{-7}{\rm\,s}=369.63{\rm\,ns} $$ By neglecting gravity, one would get: eastbound flights: $$ \Delta\tau ={t\over c^2} \left(-{1\over2}V_g(V_g+2V_\oplus) \right) =-260.34{\rm\,ns} $$ and for westbound flights: $$ \Delta\tau ={t\over c^2} \left(-{1\over2}V_g(V_g-2V_\oplus) \right) =152.73{\rm\,ns} $$ By just taking the clocks to the altitude $12\rm\,km$ and staying there for 46 hours (without moving with respect to the inertial frame, e.g. far galaxies), one gets: $$ \Delta\tau ={ght\over c^2}=216.90{\rm\,ns} $$

  • 1
    $\begingroup$ +1 I'm sure you're right, and the math is beautiful. It's just that, if I don't take a swig of coffee, unplug my phone, and slow down to read/understand each equation, I'm not going to understand you. C'est la vie. $\endgroup$ – Mike Dunlavey Nov 29 '11 at 13:58

You may be thinking of the "twin paradox" in special relativity where one twin gets in a rocket that accelerates to a speed close to the speed of light (c), travels some distance away from earth, decelerates and then accelerates back towards earth where he decelerates once again to land on the earth. The twin that stayed on earth will be much older than the twin who traveled at close to c. So it is tempting to think that the accelerating and decelerating causes the asymmetry in aging. Actually the asymmetry is purely due to the fact that the traveling twin uses two different inertial frames compared to one frame for the twin on earth. See Twin Paradox for a full explanation. So the length of the flight does matter because that is where the time dilation occurs.

As you say, in addition to the special relativity time dilation for moving reference frames, there is also a general relativistic time dilation effect due to gravitational fields. The GR principle of equivalence between accelerating reference frames and gravitational fields implies that acceleration will also cause time dilation. So all of these effects (both special and general relativity) will need to be taken into account in this kind of experiment.


The exact text quoted from Wikipedia contains:

"...the moving clocks were expected to age more slowly because of the speed of their travel"

Your question is "Is it really true?"

The answer is "No, it is not true." That is, it is not true that the clocks were expected to age more slowly due to their speed of travel. What is true is that one clock was expected to age more slowly, and the other more quickly. That is not only what was expected, but it is also what was in fact found.

"Ohh, please... OP surely was questioning whether there would be any difference between the clocks, not whether the exact wording in that passing reference was correct."

Maybe. But when you are basically in over your head, trying to make sense of things as foreign as relativity, even small mis-statements like that can really trip you up; it sure trips me up, anyway.

OK, but on to your real question, which I suppose is really this:

Where is the asymmetry coming from that causes one plane to age differently than the other?

You can find a frame of reference in which the experiment is entirely symmetrical; both planes accelerate the same amount, travel the same distance, at the same speed. This frame of reference is not the one that was used for the calculations, but why not? Why is the particular frame of reference where the earth is spinning any more valid than any other? You could even come up with a frame of reference (one that is spinning twice as fast as the earth) in which the asymmetry is reversed. What's so magical about the frame of reference where the earth is spinning at 360 deg/24hr?

And the answer, I eventually came to understand, is that that frame of reference in which the experiment is symmetrical is not an inertial frame of reference, and the one that was used for the calculations is (or is more so, at least).

Rotation is not "relative" it has some absolute meaning (not that I yet understand exactly what that meaning is). The earth is spinning really. For a plane to travel with the earth's spin, it first needs to accelerate. To travel against the earth's spin it first needs to decelerate, in real, not relative terms. Therefore, the experiment is not symmetrical.

To at least begin to understand why rotation is "absolute" in some way, or at least to believe it, if not understand it, think about this:

Consider two balls on the ends of a string. Neglecting the gravity between them, if the balls and string are at rest, or moving uniformly, the string between the balls will be slack. However, if you were to spin the balls about the center of the string, the balls would try to fly apart from each other, and the string would be taut. There is no way to look at that scenario, from any reference frame, where the string is slack.

One other thing: I don't believe this is correct: "...only acceleration/deceleration takes effect when comparing two clocks in the same frame after certain kinds of motions. ... the time difference of two clocks in the aforementioned test is the same regardless of the length of the flight"

As I understand it, after you are done accelerating, you are in a new inertial frame, with a different time scale. The longer you remain in that new frame, the more difference your clock will accumulate, and will be apparent when you finally return to your original frame.


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