What happens to a clock which is moved at some speed for some time. According to special relativity it runs slower on the move. Lets get a bit of an intuitive understanding of this.

Moving a light clock from here to there

I described the light clock earlier and showed how it runs slower when it is moving. Now, what if we have two identical light clocks, synchronized at some position. Then we move one by a distance $d$ and then stop again. Does it now

• run slower,
• lag
• and how much
• and can we actually measure this?

Consider the light speeding at $c$ along the light clock of length $l$ on the left of the diagram between $A$ and $B$.

A second, identical light clock is moved along the distance $d$. For simplicity, assume initially that we move the second clock with a speed $v$ such that it ticks exactly $n/2$ times during the move. A tick is one complete cycle of the clock where the light beam runs from $A$ to $B$ and back to $A$.

What is the speed component $v_l$ of the moving clock along $l$, the vertical axis? We have the speed $c$ along the diagonal $a$ and we have defined to move the clock with $v$ along $b$. Consider the time $T$ to be the time it takes along $a$, then $a/T = c$, $b/T = v$ and $l/T = v_l$. We also have $a^2 = b^2 + l^2$. Divide by $T$ to get $c^2 = v^2 + v_l^2$ or $v_l = \sqrt{c^2 - v^2}$.

How often does the moving clock tick until it reaches the end of the move? Since it moves with $v$ along $d$, the time is $T_d = d/v$. During this time, the total vertical movement of the light beam is $T_d v_l = v_l d/v$. Divide by $2l$ to get the count of the moving clock as $N_d = v_l d/2lv$.

Similarly the count for the stationary clock is $N_0 = cT_d/2l = cd/2lv$. The lag factor $L$ as the factor of how the moving clock ticks slower is therefore $$L=N_d/N_0 = v_l/c = \sqrt{c^2 - v^2}/c = \sqrt{1 - (v/c)^2}\,.$$ Should we not have found the Lorentz factor, not its inverse? No, it is just right, since the Lorentz factor relates the time between two ticks of the moving clock and the stationary. Since we count the ticks, we get its inverse.

Synchronized Clocks

What does this mean for clock synchronization: If I bring a well synchronized clock from here to there, I cannot avoid having the lag factor $L=\sqrt{1-(v/c)^2}$. But look, if the speed $v$ with which we move the clock is (nearly) zero, $L$ is (nearly) $1$, so there is no lag for very slow movement.

But how bad does it get? Can we measure the lag? Suppose we move a clock with an airplane and, for simplicity, assume it can do $\unit{1000}{km/h}$. We then have $$v/c = 1000\cdot1000/3600/299792458 \approx 9.27\cdot10^{-7}\,,$$ for a lag factor of $$L \approx 1-\sqrt{1-(9.27\cdot10^{-7})^2} = 4.3\cdot10^{-13}\,.$$

Now suppose we fly the clock $\unit{1000}{km}$ away, meaning the time of movement is one hour or $\unit{3600}{s}$. After the move, the moved clock lags by $\unit{3600}{s}\cdot 4.3\cdot10^{-13} \approx 1.5\cdot10^{-9}$ or $\unit{1.5}{ns}$.

Once we stop moving the clock, it runs as fast as before, but it is no longer synchronized: it runs with a lag which depends on the velocity of the move and the time it took — or the distance. Here you can experiment with different values.

The Twin Paradox describes how one twin, traveling with non-trivial speed for some time and returning back home, is then younger than his twin. This is explained in detail by Special Relativity physics. See the Wikipedia Twin Paradox page, for example.

On this page you find wording like "...clock is running slow...". How's that? The explanation I can grok most easily makes use of the Light Clock.

While in the stationary clock, light is bouncing at full speed $c$ between $A$ and $B$, in the moving clock, as viewed from the stationary frame, the light has to travel the path $A\to B' \to A'$. To travel the distance $d$ between $A$ and $B'$ it needs a time $t$ such that $d/t = c$. The distance $d$ can be partitioned into a forward move of the clock by $a$ and the clock size $b$ such that $d^2 = a^2 + b^2$. Divide by $t^2$ to get $$d^2/{t^2} = c^2 = (a/t)^2 + (b/t)^2\,.$$ The term $a/t$ is the speed of how fast the clock is moving, call it $v$. And $b/t$ is the speed left for the light clock to actually tick. Call it $c'$ to get $$ c' = \sqrt{c^2 - v^2}\,. $$

This $c' < c$ is how the light clock runs slower. And not only the clock. The clock is a viable measure of time, so time actually runs slower. How? Well! One way I intuit this is to imagine that my body, as a rough biological clock, ticks by means of metabolism, which comes down to chemical reactions, which again are based on the exchange of electrons in atoms and molecules, which again requires the exchange of virtual photons which $\dots$ are light.

A simple, striking way to say it is: when moving, the light clock gets busy with moving so it has less resources to tick.

The Car

Consider a car with wheels having a circumference of $\unit{1}{m}$, so if the wheel turns around once per second, the car has a speed of one meter per second ($\unit1{m/s}$) or $\unit{3.6}{km/h}$.

The car has a somewhat peculiar stop watch on the dashboard. I thas a single hand. The face is divided into 60 like sized parts suggesting that one round of the hand is 60 seconds or one minute. Yet it is mechnically connected to one of the wheels in the strange way such that the hand goes around once exactly for $60$ turns of the wheel.

If the car runs at $\unit1{m/s}$ it will make $60$ turns in one minute and the hand of the stop watch in the car will turn around once suggesting one minute has passed, all just fine.

Yet, obviously, if the car goes slower, the stop watch will show a "time" which is not the same time we would check on our wrist watch. In special and general relativity, the stop watch time is called the local time in the car.

In particular when computing the speed of the car based on the local time, we will find that the stop watch turns once, $1$ minute, for $60$ turns of wheel and the speed of the car is $\unit{60}{m}/ \unit{60}{s}$ = $\unit1{m/s}$, independent of how fast the car is going according to measuring the speed with our wrist watch.

Is this General Relativity?

Consider the car heading towards a curtain while being set up get slower and slower such that it comes to a halt exactly at the curtain.

If the car would be an "observer" and the curtain would be the event horizon of a black whole, you'll find descriptions a bit like: ^1⇗

As the car approaches the curtain, according to its local time, its speed is still the full $\unit{1}{m/s}$, unchanged and in the limit as it approaches the curtain. Further, an observer in the car would not find this weird. Assume their metabolism is also strongly coupled to and synchronized with the wheel's turning: if she normally breathes 15 time per minutes, now its 15 times per 60 wheel revolutions. If she normally has a heart beat of 70 per minutes, it is now 70 per 60 wheel turns. And similarly down to the chemical reactions that make a living organism. For this observer the car will continue to move through the curtain, move on as if nothing specifically happened, and will eventually crash into a singularity of all cars that already crashed into it before.

Sounds weird? I think so.

Though John Rennie says on Physics Stackexchange quite clearly: "in the coordinate system of an external observer the infalling object never crosses the event horizon", that is, simply put, for all practical purposes the car does not run through the curtain, ever. Which sounds relieving. :-)

But it leaves a conundrum about all those black holes recently detected, crashing into each other, and the singularity inside the event horizon:

• The speed of everything, not the least light, in the limit goes to zero as measured by an external observer as it approaches the event horizon, being exactly zero at the event horizon.
• Stuff passes through the event horizon to later crash into the singularity.

According to my calender and clock, when did, in the last 16 billion years (which is a lot less then infinite billion years), stuff pass through any event horizon at zero speed to join any singularity behind? Do we have singularities behind event horizons anywhere, right now, in this universe or do we rather have to wait an infinite time first?