[Physics FAQ] - [Copyright]

Updated by DK 2015.
Original by Don Koks 1998.

Does a clock's acceleration affect its timing rate?

It's often said that special relativity is based on two postulates: that all inertial frames are of equal validity, and that light travels at the same speed in all inertial frames.  But in real world scenarios, objects almost never travel at constant velocity, and so we might never find an inertial frame in which such an object is at rest.  To allow us to make predictions about how accelerating objects behave, we need to introduce a third postulate.

This is often called the "clock postulate", but it applies to much more than just clocks, and in fact it underpins much of advanced relativity, both special and general, as well as the notion of covariance (that is, writing the equations of physics in a observer-independent way).

The clock postulate can be stated in the following way.  First, we take the rate that our frame's clocks count out their time, and compare that to the rate that a moving clock counts out its time.  Before the clock postulate was ever thought of, all that was known was that when the moving clock has a constant velocity and speed $v$ (measured relative to the speed of light $c$), this ratio of rates is the "gamma factor" \begin{equation} \gamma = {1\over \sqrt{1-v^2}}\,. \end{equation} The clock postulate generalises this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity.  That is, this ratio depends only on $v$, and does not depend on any derivatives of $v$, such as acceleration.  So this says that an accelerating clock will count out its time in such a way that at any one moment, its timing has slowed by a factor ($\gamma$) that depends only on its current speed; its acceleration has no effect at all.

In other words, the accelerated clock's rate is identical to the clock rate in a "momentarily comoving inertial frame" (MCIF), which we can imagine is holding an inertial clock that for a brief moment slows to a stop alongside the accelerated clock, so that their relative velocity is momentarily zero.  At that moment they are ticking at the same rate.  A moment later, the accelerated clock has a new MCIF, again one that is moving momentarily to match its speed, and there is a new inertial clock that briefly slows to a stop alongside the accelerated clock.  And again, the rates of accelerated clock and the new inertial one will momentarily be the same.

So the clock postulate says that the rate of an accelerated clock doesn't depend on its acceleration.  But note: the clock postulate does not say that the rate of timing of a moving clock is unaffected by its acceleration.  The timing rate will certainly be affected if the acceleration changes the clock's speed of movement, because its speed determines how fast it counts out its time (i.e. by the factor $\gamma$).  (The clock rate won't be affected by circular motion at constant speed.)  If that all seems confusing, think of an everyday analogy.  If you ride your bicycle on an icy morning, you get very cold due to the wind chill factor.  The faster you go, the colder your hands get.  This wind chill is a function of your speed, but not your acceleration.  Nevertheless, it is affected by your acceleration when your acceleration changes your speed.  But regardless of whether you have a low or a high acceleration, the only thing that matters as far as your cold hands are concerned is what your current speed is.  And for circular motion, two cyclists who follow different-diameter circles at the same speed will feel the same wind chill, even though they have different accelerations.  So the wind chill factor does not depend on your acceleration, but it certainly can be affected by your acceleration.

The clock postulate also implies that the amount of shortening of a moving rod is independent of its acceleration.  And also, that the relativistic mass of a moving object also doesn't depend on its acceleration.

The clock postulate is not meant to be obvious, and it can't be proved.  It's not some kind of trivial result obtained by writing special relativity using non-cartesian coordinates.  Rather, it's a statement about the physical world.  But we don't know if it's true; it's just a postulate.  For instance, we can't magically verify it by noting that the Lorentz transform is a function only of speed, because the Lorentz transform is something that's built before the clock postulate enters the picture.  Also, we cannot simply maintain that an acceleration can be treated as a sequence of constant velocities that each exist only for an infinitesimal time interval, for the simple reason that an accelerating body (away from gravity) feels a force, while a constant-velocity body does not.  Although the clock postulate does speak in terms of constant velocities and infinitesimal time intervals, there's no a priori reason why that should be meaningful or correct.  It's just a postulate!  This is just like the fact that even though a 1000-sided polygon looks pretty much like a circle, a small piece of a circle can't always be treated as an infinitesimal straight line: after all, no matter how small the circular arc is, it will always have the same radius of curvature, whereas a straight line has an infinite radius of curvature.  It also won't do to simply define a clock to be a device whose timing is unaffected by its acceleration, because it's not clear what such a device has got to do with the real world: that is, how well it approximates the thing we wear on our wrist.

Although the clock postulate is just that, a postulate, it has been verified experimentally up to extraordinarily high accelerations, as much as $10^{18}$ g in fact (see the FAQ What is the experimental basis of Special Relativity?).  Of course, the postulate also speaks of more than acceleration, it speaks of all derivatives of $v$ with respect to time.  But still it can be shown to be a reasonable thing to assume, because it leads to something that has been tested in other ways, as we'll see in the next section.

The spacetime metric, or interval

This something is the idea of a spacetime metric.  Typically when learning special relativity, at some stage we note that the "interval" between two events, $\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$ (where I write $t$ and $T$ as shorthand for $ct$ and $cT$), is independent of the inertial frame in which we make our measurements.  So although relativity has taught us to discard our ideas of the absoluteness of space and time, this "something" that is observer independent is a new absolute thing called spacetime.  And since the interval is observer independent, we can also say immediately that the time $\Delta T$ elapsed on an inertial clock connecting two events is just $\Delta T^2 = \Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$, since in its own inertial frame, the clock doesn't move, and so increments in its own spatial coordinates are zero.

But, what is the time as shown on a noninertial (i.e. accelerated) clock connecting two widely separated events?  There's no reason why it should be $\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$ (and it certainly isn't), because that expression looks only at the end points and doesn't take into account how the clock moved.  In fact, what we do to get the right answer is this: we assume the clock postulate holds.  So, when an accelerating clock moves from one event to another that is infinitesimally close, we can say that the infinitesimal time $\text{d}T$ elapsed is given by \begin{equation} \text{d}T^2 = \text{d}t^2 - \text{d}x^2 - \text{d}y^2 - \text{d}z^2, \end{equation} since this is the time that elapses on a clock in the MCIF.  And we can now integrate this $\text{d}T$ along the accelerated clock's world line to get the actual elapsed time that it shows.

This sort of idea is exactly analogous to the idea of calculating the length of a curve by dividing it into a large number of short segments, each of which is "almost straight", and then adding up the lengths of each of those using Pythagoras: $\text{d}\ell^2 = \text{d}x^2 + \text{d}y^2 + \text{d}z^2$.  That's an exercise in calculus.  Sure, each of those short segments is definitely not straight—it still has some curvature that doesn't go away as we break it into smaller and smaller parts.  But the total error that we introduce by taking the segments as being straight gets ever smaller as we make the number of segments larger and larger; again, that's standard calculus in action.

The idea of calculating the time elapsed on a clock that accelerates from one event to another is just the same idea: we divide its path into small segments that the clock postulate lets us work with.  And just as the curvatures of those small segments in the previous paragraph never become zero (and yet we can ignore them), so too the accelerations of the clock along its path never become zero either—and yet the clock postulate allows us to ignore them too.

This is why we can give some structure to spacetime, because it's possible to talk about the "length" of a curved world line as being given by the time shown on a clock moving along that world line, even though the clock itself is accelerating (since the world line is curved—we are still only talking about flat space here).  If the "length" between two infinitesimally separated spacetime points was something that depended on how a clock connecting them moved, then it wouldn't be an intrinsically geometric thing.  The clock postulate geometrises relativity by saying that we don't need to consider how the clock moves; the time (squared) shown by any clock connecting two closely separated events is just $\text{d}t^2 - \text{d}x^2 - \text{d}y^2 - \text{d}z^2$, and it doesn't matter how the clock moves.

Ultimately, that's why the interval is written using infinitesimals.  It has nothing to do with curved spacetime; we use infinitesimals even if the spacetime is flat.  We use them because in this form the interval embodies the clock postulate.  But now the grand thing is that this idea of geometrical spacetime structure allows us to make the transition to general relativity, and the great success of that theory lends plausibility to the parts that make it up—and one of these is the clock postulate.

The clock postulate is one of the ideas we use to build the theory of gravity known as general relativity.  People sometimes think (and books sometimes say) that a proper description of accelerating clocks demands a theory of gravity: general relativity.  (The historical names "special relativity" and "general relativity" might possibly have changed over the years, so that where once general relativity might have referred to all noninertial motion, nowadays it's generally accepted to refer only to gravity, and both inertial and noninertial motion are now labelled as special relativity.)  But no, analysing acceleration does not need to refer to gravity.  General relativity is built on a foundation that includes the clock postulate.  Once we've built general relativity, we cannot then turn around and start using it to "explain" the physics of accelerating clocks!  That would be circular reasoning.  General relativity (gravity) is built upon the ideas of special relativity, but the physics of accelerating clocks is completely handled by special relativity alone.

But what about the Equivalence Principle?

Sometimes people say "But if a clock's rate isn't affected by its acceleration, doesn't that mean the Equivalence Principle comes out wrong?  If the Equivalence Principle says that a gravitational field is akin to acceleration, shouldn't that imply that a clock isn't affected by a gravitational field, even though the textbooks say it is?"

No, the Equivalence Principle is fine.  Again, the confusion here is the same sort of thing as above where we spoke about the wind chill factor.  Let's try to see what is happening.  Imagine we have a rocket with no fuel.  It sits on the launch pad with two occupants, a couple of astronauts who can't see outside and who believe that they are accelerating at 1 g in deep space, far from any gravity.

One of the astronauts sits at the base of the rocket, with the other at its top, and they both send a light beam to each other.  Now, general relativity tells us that light loses energy as it climbs up a gravitational field, so we know that the top astronaut will see a redshifted signal.  Likewise, the bottom astronaut will see a blueshifted signal, because the light coming down has fallen down the gravitational field and gained some energy en route.

How do the astronauts describe what is going on?  They believe they're accelerating in deep space.  The top astronaut reasons "By the time the light from the bottom astronaut reaches me, I'll have picked up some speed, so that I'll be receding from the light at a higher rate than previously as I receive it.  So it should be redshifted—and yes, so it is!"  The bottom astronaut reasons very similarly: "By the time the light from the top astronaut reaches me, I'll have picked up some speed, so that I'll be approaching the light at a higher rate than previously as I receive it.  So it should be blueshifted—and yes, so it is!"

As you can see, they both got the right answer, care of the Equivalence Principle.  But their analysis used only their speed, not their acceleration as such.  So just like our wind chill factor above, applying the Equivalence Principle to the case of the rocket doesn't depend on acceleration per se, but it does depend on the result of acceleration: changing speeds!

Is the clock postulate in fact not a postulate?

There are some who state that the clock postulate isn't a postulate at all; that it's simply a theorem that can be proved in the following way.  They use some standard reasoning about inertial frames that you'll find in most books on relativity, but then they apply that reasoning incorrectly to an accelerated frame.  Here's their reasoning.  I'll make sure to include all factors of $c$ explicitly here.

First, they appeal to a standard and useful description of a "light clock" in relativity, as follows.  Sit yourself on a train that is moving with constant velocity relative to the platform, with a torch aimed at a mirror on the ceiling a distance $L$ above you.  You send a pulse of light up from the torch at speed $c$ (light has speed $c$ in all inertial frames), and let it bounce off the mirror so that it returns to you some short time $\Delta t'$ later.  You have a receiver that emits a tick when it detects the light.  Applying "time taken = distance the light travelled divided by light speed", you write down \begin{equation} \Delta t' = 2L/c\,. \end{equation} On the platform, I watch you move past me with speed $v$, and I note that the light moved with speed $c$ along two diagonal paths in a total time of $\Delta t$, which may or may not equal your $\Delta t'$: that's what we want to work out.  As the light climbs to the top of the train, the train moves a distance $v\Delta t/2$, and similarly for the light coming back down.  Pythagoras tells me that the length of that whole path is twice the square root of $(L^2 + (v\Delta t/2)^2)$, so when I apply "time taken = distance the light travelled divided by light speed", I get \begin{equation} \Delta t = {2\over c} \sqrt{L^2 + (v\Delta t/2)^2} \,. \end{equation} I solve this for $\Delta t$, getting \begin{equation} \Delta t = {2L\over c\sqrt{1 - v^2/c^2}} \,. \end{equation} But remember that $2L/c$ is just $\Delta t'$, so finally I have \begin{equation}\label{postulate-analysis1} \Delta t = {\Delta t'\over\sqrt{1 - v^2/c^2}} > \Delta t' \,. \end{equation} In other words, the time $\Delta t$ taken for the pulse of light to do a round trip "torch-to-mirror-to-torch", as seen from the platform, is greater by a factor of $1/\sqrt{1 - v^2/c^2}$ than the time $\Delta t'$ taken on the train.  This is traditionally one way in which time dilation is derived.

All well and good.  Now here's where they mis-apply the above argument to the case of an accelerating train.  They suggest having the train accelerate with some acceleration $a$, so that the distance moved by the train as the pulse goes up is now not $v\Delta t/2$, but instead $v\Delta t/2 + a\Delta t^2/2$.  That's okay so far.  Then they replace "$v\Delta t/2$" by "$v\Delta t/2 + a\Delta t^2/2$" in the above analysis, and then reason that as $\Delta t$ goes to zero, the effect of the acceleration term $a\Delta t^2/2$ becomes negligible, and so arrive once more at equation \eqref{postulate-analysis1}, without ever needing to invoke the clock postulate, saying that this all follows from the first two postulates of special relativity, and the clock postulate is therefore not a third postulate after all, but just a theorem.

Where they went wrong was to assume, with no justification, that the speed of light on the accelerating train is also $c$.  Is it?  What do Maxwell's equations say?  Maxwell's equations are valid only for inertial frames, so they don't tell us the speed of light on the accelerating train.

You can use the clock postulate to show that the speed of light does turn out to be $c$ locally in the accelerating train: in other words, the light's speed as measured by the train passenger actually turns out to be a function of its distance from that passenger, but as the light gets closer to the passenger, its speed as measured by the passenger gets closer to $c$.  But those who apply the above inertial calculation to an accelerating train cannot know or claim a-priori that the speed of light is $c$ in the light clock on the accelerating train (and, of course, they're not allowed to use the clock postulate to show that).  They are only assuming that the speed is $c$; and by assuming what turns out to be true locally (and they never knew anything about "local"), they fluked the right answer, that $\gamma$ has no dependence on $a$.  But because they were not allowed to assume a-priori that light travels at speed $c$ in an accelerated frame, their calculation has failed to show that the clock postulate is a theorem.

Generalising the clock postulate

On a more advanced note, the clock postulate can be generalised to say something about measurements we make in a noninertial frame.  First, it tells us that noninertial objects only age and contract by the same gamma factor as that of their MCIF.  So, any measurements we make in a noninertial frame that use clocks and rods will be identical to measurements made in our MCIF—at least, if these measurements are made over a small enough region.  This is because, in general, different regions of the noninertial frame have different MCIFs, a complicating factor that makes the construction of noninertial frames very difficult.  In fact, it can only readily be done for constant-velocity and constant-acceleration frames; and for the last, that is rather difficult.

But we now choose to extend the clock postulate to include all measurements (though perhaps it can be argued that all measurements only ever use clocks and rods anyway).  This idea leads onto "covariance", which is a way of using tensors to write the language of physics in a way that applies to all frames, noninertial as well as inertial.  How is this done?

Take, for example, the electromagnetic field.  Suppose some measurements were made in an inertial lab frame, and it was determined to have the four spacetime components $A^\alpha$ (up to a choice of gauge).  Now, we know from other considerations that $A^\alpha$ are the components of a four-vector.  That means they transform between inertial frames identically to the way $(\text{d}t, \text{d}x, \text{d}y, \text{d}z)$ transform.  In the language of tensors (where repeated up and down indices are summed over), we can say that from one inertial frame, $(t, x, y, z)$, to another, $(t', x', y', z')$, the $A^\alpha$ transform via \begin{equation}\label{tensor1} A^{\alpha'} = {\partial x^{\alpha'}\over\partial x^\beta} A^\beta \, \end{equation} (where repeated indices are always summed over).  This much can be shown without the clock postulate, because it deals only with inertial frames.  The question is: in this language, how do the $A^\alpha$ look in a noninertial frame?

Suppose we are in a noninertial frame (written with double-primed coordinates), and wish to know what values the field will take for us.  The clock postulate tells us that the field in our frame, $A^{\alpha''}$, will be identical to the field measured by our MCIF, $A^{\alpha'}$.  But that frame is inertial, so we know how to calculate the $A^{\alpha'}$: these are just given by equation \eqref{tensor1}.  Next step: how can we relate our double-primed coordinates to the way in which the $A^\alpha$ transform?  After all, the MCIF always exists, but it's just a tool; we don't want to consider it (those single-primed coordinates) explicitly.  That is, how do we relate double-primed coordinates to non-primed coordinates?

We use the clock postulate.  It says that our measurements of distance and time are unaffected by any pseudo forces we feel, so that they are identical to those of our MCIF: \begin{equation} \text{d}x^{\alpha''} = \text{d}x^{\alpha'} \,. \end{equation} This is just another way of saying that \begin{equation} {\partial x^{\alpha''}\over\partial x^\beta} = {\partial x^{\alpha'}\over\partial x^\beta} \,. \end{equation} So, we can put it all together to write \begin{align} A^{\alpha''} &= A^{\alpha'} &&\text{(by our extended clock postulate)}\\ &= {\partial x^{\alpha'}\over\partial x^\beta} A^\beta &&\text{(Lorentz transform between MCIF and lab frame)}\\ &= {\partial x^{\alpha''}\over\partial x^\beta} A^\beta &&\text{(the clock postulate again).} \end{align} Finally \begin{equation} A^{\alpha''} = {\partial x^{\alpha''}\over\partial x^\beta} A^\beta, \end{equation} which is just like \eqref{tensor1} above, except now it applies to a noninertial (double-primed) frame!  So, we've managed to write our original transformation law \eqref{tensor1}, which held only between two inertial frames, in a way that looks just the same as \eqref{tensor1}, except that it's using different coordinates, some from a noninertial frame.  And this is the idea behind covariance and tensor language, that it's possible to write the laws of physics for arbitrary frames, even noninertial ones, in one way: the way of equation \eqref{tensor1}.