I have a question about the reason Feynman gives for why a perfect clock - a clock that remains in sync while in motion with a stationary clock, cannot exist. It is clear why the "light clock" (described in Section 15-4: Transformation of Time) while in motion must appear to run slower to the stationary observer. Feynman gives the following argument for why any clock (irrespective of how it works) must run slower.

Feynman's argument runs as follows:

Not only does this particular kind of clock run more slowly, but if the theory of relativity is correct, any other clock, operating on any principle whatsoever, would also appear to run slower. [...] Why is this so?

To answer the above question, suppose we had two other clocks made exactly alike [...]. Then we adjust these clocks so they both run in precise synchronism with our first clocks. [...] One of these clocks is taken into the space ship, along with the first kind. Perhaps this clock will not run slower, but will continue to keep the same time as its stationary counterpart, and thus disagree with the other moving clock. Ah no, if that should happen, the man in the ship could use this mismatch between his two clocks to determine the speed of his ship, which we have been supposing is impossible. We need not know anything about the machinery of the new clock that might cause the effect—we simply know that whatever the reason, it will appear to run slow, just like the first one.

My question:

Let me call the two light clocks $L_1$ (stationary), $L_2$ (moving), and the two supposedly perfect clocks of unknown mechanism as $P_1$ (stationary), $P_2$ (moving). The observer in the space ship will never see a mismatch between his two local clocks ($L_2$ and $P_2$). It is the the stationary observer who sees $L_2$ slowing down with respect to his local clocks ($L_1$ and $P_1$).

Nothing in the argument seems to prevent $P_2$ appearing normal to the stationary observer (i.e., the stationary observer can see $P_2$ in sync with $L_1$ and $P_1$). To the stationary observer $L_2$ and $P_2$ appear to disagree, but the observer in the space ship sees them in sync. Hence there is no violation of the principle of relativity.

  • $\begingroup$ " It is clear why the "light clock" (described in Section 15-4: Transformation of Time) while in motion must appear to run slower to the stationary observer." This is a huge misconception. The clock does not appear to run slower to the stationary observer. It doesn't appear to him neither late, nor early, nor on time. Because the stationary observer simply cannot see this light. In order to see light, the photon must hit your eye. Feynman's diagram there is just a speculation that has nothing to do with reality. $\endgroup$ Commented Oct 16, 2015 at 11:54

3 Answers 3


The typical undergraduate explanations of time dilation using light clocks are a reasonable introduction, but they give you an oversimplified and ultimately misleading idea of what time dilation is and how it happens. The key fact you need to understand is that the elapsed time shown by a clock is equal to the length of the clock's worldline, and that length is an invariant that has the same value for all observers.

Suppose some clock moves a distance $dx$ in a time $dt$ (for simplicity we'll consider only one spatial dimension) then the length of the path traced out by the clock is calculated using the metric:

$$ d\tau^2 = dt^2 - \frac{dx^2}{c^2} \tag{1} $$

where $d\tau$ is the length of the path and is called the proper time. Equation (1) is called the Minkowski metric and it is the fundamental equation that defines special relativity. If the coordinates $t$ and $x$ are the rest frame of the clock then $dx = 0$ because in the rest frame of the clock its position isn't changing. In this case equation (1) simplifies to:

$$ d\tau^2 = dt^2 \tag{2} $$

and we have the result that the length of the clock's worldline, its proper time $\tau$, is just equal to the time shown on the clock, which is where we started.

But now suppose I am watching the clock moving and my coordinates are $t'$ and $x'$. In my coordinates the clock is moving so $dx' \ne 0$ and I calculate the proper time to be:

$$ d\tau^2 = dt'^2 - \frac{dx'^2}{c^2} \tag{3} $$

But the path length $d\tau$ is an invariant that has the same value for all observers in all frames no matter how they are moving. That means I can equate equations (2) and (3) to get:

$$ dt^2 = dt'^2 - \frac{dx'^2}{c^2} \tag{4} $$

Remember that $dt'$ is my time coordinate and $dt$ is the clock's time coordinate, so equation (4) immediately tells us that my time and the clock's time must be different. We can be more precise by noting that in my coordinates the clock's velocity is given by:

$$ v(t') = \frac{dx'}{dt'} $$

and therefore that:

$$ dx' = v(t')dt' $$

where the velocity that I measure, $v(t')$, can be any function of time - it doesn't have to be a constant velocity. Substituting this in equation (4) we get the relationship between the clock's time and my time:

$$ dt^2 = dt'^2 \left(1 - \frac{v(t')^2}{c^2}\right) \tag{5} $$

This is the key result because it shows that the clock's time $dt$ must be less than my time $dt'$ for any velocity $v \ne 0$. This is what time dilation means. Note that we haven't said anything about what type of clock it is, and we haven't had to construct any complicated diagrams involving light rays. This result follows immediately from the starting point that the spacetime geometry of general relativity is described by the Minkowski metric.

  • $\begingroup$ This answer (minus some text) is very closely related to this old one that you used to answer a question of mine. So if you enjoyed reading this, then the linked answer may also be useful to you. $\endgroup$
    – Danu
    Commented Oct 16, 2015 at 11:20
  • 1
    $\begingroup$ @Danu: these answers are excerpts from my definitive explanation of time dilation that I'll one day post as a canonical Q/A. I kep trying out bits of the article to see how they go down :-) $\endgroup$ Commented Oct 16, 2015 at 12:08
  • $\begingroup$ I see: My proposal of canonical Q/A sets is still alive... somewhere ;) $\endgroup$
    – Danu
    Commented Oct 16, 2015 at 12:13

The key fact is that the observer on the spaceship observes his light clock $L_2$ tick in sync with his perfect clock $P_2$, which means that the each tick of $L_2$ happens simultaneously with one of $P_2$ at the same spatial location. Relativity forces us to give up the notion of temporal simultaneity, which is what allows all of the wonky management of space and time to make sense in the end, but this only holds for spatially separated events. If one observer sees two events happen at the same time and in the same place, then all observers must agree with her.

In this particular case, the 'stationary' observer must therefore observe $P_2$ to tick simultaneously with $L_2$, and therefore to tick at the same rate. When you say

Nothing in the argument seems to prevent $P_2$ appearing normal to the stationary observer

what prevents that from happening is the pegging of $P_2$'s ticks to those of $L_2$ by the fact that they're observed as simultaneous by the moving observer.

  • 1
    $\begingroup$ Your answer, which is same as the one given by @Crimson, resolves the issue I had with Feynman's argument. $\endgroup$
    – Ravi
    Commented Oct 19, 2015 at 4:46

You are right that if L2 and P2 appear to disagree for the stationary observer, but seem in sync for the observer in the space ship, the proof would break down.

However, it is difficult to imagine such a situation. Imagine that both clocks emit a short light pulse every second. By comparing the number of flashes from each clock, one can then check whether they run in sync. Since the light pulses from the two clocks have to travel the same path to the stationary observer, it will take them the same amount of time to arrive. The stationary observer will therefore always come to the same conclusion as a local observer about the synchronization of the clocks.


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