So Hafele-Keating made sense of the results by taking the center of the Earth as the reference frame. The westward flight "undid" part of the Earth's rotation and those clocks sped up relative to our reference clock on the surface.

Now, let's repeat the experiment in space. One spaceship goes forwards in Earth's orbit and the other goes backwards. To simplify the experiment, both spaceships have exactly the same amount of fuel (to travel say 500,000 km) before stopping the clocks and returning home using the secondary fuel tank.

One might think that the logical approach is to use the center of the Sun as the reference frame and expect the backwards-in-orbit travelling clocks to be faster than the Earth's.

But why not take the center of the galaxy as our reference frame? That would give different predictions.

Or the center of the Local Group? or the center of the Local Supercluster?

It ties in with the Twin Paradox. How can we know that the space-faring twin isn't just "undoing" some larger motion and will actually age faster than the stay-at-home twin?

EDIT: the trips are deliberately short to minimize differences in gravitational time dilation between the three clocks (Earth + two spaceships). In fact, the more distant the proposed center, the more insignificant the gravitational aspect.


1 Answer 1


You are oversimplifying the calculation done to calculate the time dilation.

If we choose some coordinate system, any coordinate system, then we can represent the trajectory of the clock on a spacetime diagram. If I ignore all but one of the spatial dimensions (on the grounds I can't draw four dimensional diagrams) the spacetime diagram might look something like:

Spacetime diagram

The blue line is the trajectory of the moving object in our coordinate system. So the blue line starts at the position and time the clock starts moving and ends at the position and time the clock stops moving.

Relativity (special and general) tells us that the elapsed time on the clock is equal to the length of the trajectory i.e. the length of the blue line. This is called the proper length or proper time. However the length of the curve has to be calculated using the metric tensor. If we split the curve up into infinitesimal elements $ds$:

Curve length

Then we calculate the length of the element $ds$ using:

$$ ds = g_{\mu\nu}dx^\mu dx^\nu $$

where $g_{\mu\nu}$ is the metric tensor written in an appropriate form for the coordinate system we have chosen. The length of the curve is then obtained by integrating up all the infinitesimal elements:

$$ s = \int ds = \int g_{\mu\nu}dx^\mu dx^\nu $$

Time dilation occurs when the path length $s$ is not equal to the change in our coordinate time $t$ i.e. when the elapsed time on the moving clock is different from the elapsed time shown on a stationary clock. Even given the rather complicated way the path length is calculated it should be obvious that in general the path length will not be equal to the change in the coordinate time and therefore that some time dilation will occur.

When you write down all the maths it inevitably ends up looking complicated and confusing, but for the Hafele-Keating experiment the calculation is actually pretty straightforward and easy to do, if a bit fiddly. You don't need worry about where in the universe you are, you just need to be clear what coordinate system you are using and how to write the metric in this coordinate system. In fact the calculation doesn't need the full form of the metric as you can use the weak field limit and simplify the maths quite a bit.

Response to comment:

As Benito suggests in a comment that we treat on the motion and ignore the curvature. To do this we'll consider only one space dimension again, and write the velocity of the clock relative to us as $v(t)$. Our metric is now the Minkowski metric (I'll write this in the proper time form and with the factors of $c$):

$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$

The distance $dx$ is related to the time $dt$ by:

$$ dx = v(t)dt $$

because that's just the definition of velocity. If we substitute for $dx$ in the metric we get:

$$ c^2d\tau^2 = c^2dt^2 - v^2(t)dt^2 $$

and rearranging gives:

$$ d\tau = c\sqrt{1 - \frac{v^2(t)}{c^2}} dt $$

and therefore:

$$ \Delta\tau = c\int_{t_1}^{t_2}\sqrt{1 - \frac{v^2(t)}{c^2}} dt $$

where $\Delta\tau$ is the elapsed time for the clock and will in general be different from $t_2 - t_1$.

Note that all we need for this is the relative velocity of the clock as a function of time $v(t)$. We don't care whether our coordinate frame is moving or not - only the velocity difference between us and the clock matters.

  • $\begingroup$ @BenitoCiaro: I've updated my answer to respond to your comment. $\endgroup$ Commented Dec 4, 2015 at 9:28
  • $\begingroup$ @BenitoCiaro: correct, the direction of the velocity relative to the observer on Earth does not matter. I don't understand what point you are making. If we draw a vector $\vec{r}$ between the Earth observer and the clock the velocity is just $\vec{v}=d\vec{r}/dt$. This vector does not depend on where we place the origin of our coordinate system. $\endgroup$ Commented Dec 4, 2015 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.