Why is impossible to synchronize the clocks , along a ring globally on a uniform rotating disk?

Question

Thanks to Eric David Kramer answer I edited my question.

Let Lisa and Milhouse be two observer in rest relative to each other on a uniform rotating frame with angular velocity $\omega$. Let us suppose that they want to synchronize their clocks according to the principles of Einstein synchronization, i.e., by exchanging light signals. Lisa, at point $A$, noting that her clock registers $t_{A}$, fires a laser beam at Milhouse, her "next door neighbor"' in the counterclockwise direction, who is stationed at point $B$. At $t_{B}$, he receives and reflects the beam back to her; she receives the signal at $t_{A}^{\prime} .$ Lisa sends Milhouse a slip of paper upon which is written the value of $\left(t_{A}^{\prime}+t_{A}\right) / 2$, with instructions that his clock should have had that reading at $t_{B}$. Milhouse adjusts his clock accordingly. This procedure is followed from observer to observer around the ring, and we imagine the limit of an infinite number of observers with infinitesimal separation.

Suppose we have a inertial observer $I$ at the center of the disk using cylindrical coordinates $(t,r,\theta,z)$

Let us denote the event Liza sends light to Milhouse by $e_1$, the event reception of light by Milhouse by $e_2$ and the event reception of light by Lisa $e_3$.We denote their respective coordinates by $x_1,x_2,x_3$

Suppose that in the disk Milhouse and Liza are in the same radius $R$ and Milhouse is shifted from through an angle $\Delta \theta=\theta_0$. If $x_1=(0,R,0,0)$ we can show that $$t_2=\frac{R}{c} \frac{\theta_0}{1+v / c} \quad \quad t_{3}=2 \frac{R}{c} \frac{\theta_0}{1-v^{2} / c^{2}}$$

where $v=\omega R$

Due to the time dilation of Lisa's clock with respect to the Lab system, Lisa's clock will read $\tau_{3}=t_{3} / \gamma$ at event $e_3$ where $\gamma \equiv 1 / \sqrt{1-v^{2} / c^{2}}$, or $\tau_{3}=2(R / c) \gamma \omega_0$. According to the prescription for Einstein synchronization, Milhouse at $\theta_0$, will be given instructions to adjust her clock so that it would have read clock setting at event $e_2=\left(\tau_{3}+\tau_{1}\right) / 2=\tau_{3} / 2=(R / c) \gamma \theta_0$

Now if we take $\theta_0=2\pi$ we would have

$$\tau_{3} / 2=2\pi(R / c) \gamma \tag 1$$

According to Milhouse event $e_2$ happens at proper time $\tau_2=2\pi(R / c) \gamma $ and according to Liza $\tau_2= \frac{R}{\gamma c} \frac{2\pi}{1+v / c}$

So we have two synchronized clocks that gives the same event different times. Because of this, people often says that is impossible to synchronize the clocks along the ring globally.

I am confused about this because since Liza and Milhouse are in the same place, isn't this not a coordinate issue because we are attributing the same event to different values namely $\theta_0=0$ and $\theta_0=2\pi$ ?

Answer 1

The issue is in how you're thinking about taking $\theta_0 = 2 \pi$. You cannot actually do this as you've described-- I haven't worked them out myself, but I expect the formulas you list for $t_2$ and $t_3$ assume $\theta_0$ is very small. For a general $\theta_0$, some trig functions involved in computing the distance the light signal ends up traveling (according to $I$) must show up. Indeed, if Liza and Millhouse are separated by $\Delta \theta = \theta_0 = 2\pi$, then the events $x_1, \, x_2, \, x_3$ are all the same, so that $t_1 = t_2 = t_3 = 0$.

I expect what you really want to do is consider $n$ observers $O_k$ spaced evenly around the ring, and have them undergo the following procedure to synchronize their clocks: at event $x_1 = (0,R,0,0)$, $O_1$ sends a signal to $O_2$, who receives it at $x_2$ and immediately reflects it back while simultaneously sending a new signal to $O_3$, who receives it at $x_3$ and reflects the new signal back to $O_2$ while simultaneously sending yet another signal to $O_4$, and so on up to $O_n$, who will receive a signal from $O_{n-1}$ at event $x_n$ and immediately reflect it while sending a signal to $O_1$, who receives (and reflects it back) at event $x_{n+1}$, completing the loop.

We can imagine this as a single light ray bending around the circumference of the loop, which is partially reflected by a beam splitter at the positions $\theta_k = (k-1)\Delta \theta$ (where $\Delta \theta = 2 \pi/n$) of each of our observers. In this way, we have $n+1$ events $x_k$ at which the signals are emitted/reflected. Let's denote by $\bar{x}_k$ (for $1 \leq k \leq n$) the event at which $O_k$ receives back the reflection of the signal they sent to $O_{k+1}$ at event $x_k$.

In the limit of large $n$, we may apply your formulas to find the $t$ coordinate of all of our events:

$$ t_k = \frac{R}{c} \frac{(k-1) \Delta \theta}{1+v / c} \quad \quad \bar{t}_{k} = t_k + 2 \frac{R}{c} \frac{ \Delta \theta}{1-v^{2} / c^{2}}.$$

Now, for $1 \leq k \leq n-1$, observer $O_k$ can indicate to $O_{k+1}$ to shift their clock so as to assign event $x_{k+1}$ the synchronized time coordinate $s_{k+1}$ defined by $s_1 = 0$ and the recurrence relation

$$ s_{k+1} = s_k + \frac{\bar{t}_k-t_k}{2 \gamma},$$

which of course yields

$$s_k = \frac{R}{c} \gamma (k-1) \Delta \theta = \frac{k-1}{n} \left( \frac{R}{c} 2 \pi \gamma \right).$$

This works out all well and good so far, as each of our observers has adjusted their clocks so that they agree with their neighbors on the times of reception of their shared light signal, and if neighbors exchange light signals again later they'll find they're still synchronized. Each of our observers, save two: we have not checked whether the synchronization is consistent with the timing $O_1$ gives to event $x_{n+1}$ when $O_n$ and $O_1$ attempt to synchronize.

In the same way as above, $O_n$ instructs $O_1$ to assign to $x_{n+1}$ the synchronized time $$s_{n+1} = \frac{R}{c} 2 \pi \gamma.$$ But $O_1$ has already shifted their clock so that they assign $s_1 = 0$ to $x_1$, and they've since seen a proper time of $\Delta \tau = t_{n+1}/\gamma$ elapse, so this demands they assign to $x_{n+1}$ a time

$$\tilde s_{n+1} = s_1 + \Delta \tau = \frac{t_{n+1}}{\gamma} = \frac{R}{c} 2 \pi \gamma (1-v/c),$$ and we see that $\tilde s_{n+1} \neq s_{n+1}$ (differing by a factor of $1-v/c$), so the synchronization cannot be made globally consistent.

Answer 2

Your result (1) implies that a clock $2\pi$ radians around the moving disk reads a different time that the clock at 0 radians. But 0 radians and $2\pi$ radians correspond to the same point. Which means you have two synchronized clocks that read different times for the same event.

So the synchronization has to have a discontinuity somewhere, similar to the international date line.

Answer 3

I cannot follow your reasoning, so I can't say exactly where it breaks down, but I will make a general point as follows...

It is always possible to synchronise any pair of adjacent clocks in SR, in the sense of making them display the same time at the instant they are together. However, it is impossible for a set of clocks, each moving relative to the others, to be mutually synchronised.

The reason is very simple. The property of being synchronised means that the clocks show the same time at the same instant, but since the clocks are moving relative to each other, they do not have a common agreement of what constitutes 'the same instant'.

You must also bear in mind that making pairs of clocks show the same reading, does not necessarily reflect reality. Consider a long train passing at great speed a long platform just before noon. There are synchronised clocks all along the platform. The passengers are told each to look out of the window and synchronise their watch with the nearest platform clock when it shows noon. At noon precisely on the platform, all of the watches on the train will show noon; however, on the train the watches will seem out of synch, since noon all along the train is a different slice of spacetime to noon all along the platform. The fact that you have rigged the watches to suggest otherwise is irrelevant.

You might consider the following analogy. Suppose you have a line of observers along a flat road spaced a couple of metres apart, each holding an altimeter. Each observer in turn sets his altimeter to show the same reading as the next, so that they are all synchronised and showing the same height. Now consider the same procedure with each observer being on a successive balcony on a tower block. They will now set their altimeters to show the same height, apparently, but actually their settings are now out of synch with reality.

Your procedure for synchronising clocks in circular motion is rather like a synchronising altimeters on a Ferris wheel by each person setting his altimeter to match the displayed reading on the altimeter of the next person round the wheel.

Answer 4

I wouldn't say it is "impossible to synchronize clocks globally". What was demonstrated in the OP is that the specific synchronization method used for non-rotating observers doesn't work for rotating observers. With a slight modification, it is possible to make all clocks on the circle synchronized, in the sense that they all will show the same time in the non-rotating reference frame after synchronization and will remain synchronized (see Einstein synchronisation).

All that is needed to make Liza's and Milhouse clocks to show the same time is for Milhouse to set the time for event $e_2$ not $\frac{\tau_3}{2}$, but rather $\frac{\tau_3}{2(1-v/c)}$. In this case, both Lisa's and Milhouse's clocks will be showing the same time in the non-rotating reference frame (which can be verified from equations for $t_2$ and $t_3$ in the OP). The procedure, repeated for all observers on the circle would make all clocks go synchronously.

Note: The described synchronization procedure is consistent, since it satisfies the following properties (see wiki on Einstein synchronisation):

(a) clocks once synchronised remain synchronised,

(b1) the synchronisation is reflexive, that is any clock is synchronised with itself (automatically satisfied),

(b2) the synchronisation is symmetric, that is if clock A is synchronised with clock B then clock B is synchronised with clock A,

(b3) the synchronisation is transitive, that is if clock A is synchronised with clock B and clock B is synchronised with clock C then clock A is synchronised with clock C.