How to define a common time between two clocks? - clock synchronization in special relativity

Question

I have a basic understanding of SR, its postulates and SR math. I was going through Einstein's 1905 paper on SR, the part about synchronizing the clock goes as follows -

If at the point A of space there is a clock, an observer at A can determine the time values of events in the immediate proximity of A by finding the positions of the hands which are simultaneous with these events. If there is at the point B of space another clock in all respects resembling the one at A, it is possible for an observer at B to determine the time values of events in the immediate neighbourhood of B. But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an “A time” and a “B time.” We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A. Let a ray of light start at the “A time” $t_A$ from A towards B, let it at the “B time” $t_B$ be reflected at B in the direction of A, and arrive again at A at the “A time” $t_A^{'}$. In accordance with definition the two clocks synchronize if $t_B − t_A = t_A^{'} − t_B$.

I understand the part where he mentions that the A time and B time are local times of their respective regions and not a common time which needs to be defined. However I do not undertand the line -

We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A.

Here, why do we need to "define" that "time" for light to travel from A to B equals that for it to travel from B to A? In which clock is he defining this "time"? Why is the need to define it in this way? Using principle of relativity can't we just say that both the position A and B are equivalent and the time it would take for light to reach B from A as measured from any of the clock(A or B) would be same as for it to reach A from B as measured in any clock(A or B)?

In accordance with definition the two clocks synchronize if $t_B − t_A = t_A^{'} − t_B$

What is the logic behind synchronizing the clock in this way, i.e. by equating the travel time of light?

Answer 1

Here, why do we need to "define" that "time" for light to travel from A to B equals that for it to travel from B to A?

You need not. The choice that they be equal as opposed to one way, say from $A$ to $B$ ($A\to B$), being smaller (or larger) than the other is a convention (in light of isotropy and the simplifications it leads to, a good one). Here's a detailed answer. All that matters is $A$ and $B$ agree on the round trip times ($A\to B \to A$ and $B\to A\to B$). They must use a round trip since one needs two timestamps at the same clock for synchronization.

In which clock is he defining this "time"? Why is the need to define it in this way?

The clocks that exist at $A$ and $B$ and by extension any clock anywhere. The point of the synchronization is that you don't need to specify a location to specify a timestamp - time belongs to (is now a property of) the entire reference frame since all its clocks are mutually synced. So $A$ would say, "light has a round trip time from $B$ of so and so seconds as measured by me" and so would $B$, identically.

Using principle of relativity can't we just say that both the position A and B are equivalent and the time it would take for light to reach B from A as measured from any of the clock(A or B) would be same as for it to reach A from B as measured in any clock(A or B)?

This I think is the crux of your confusion. There are two aspects to this:

• relativity of motion itself doesn't prohibit anisotropic phenomenon. Say there was some intrinsic reason that communications between $A$ and $B$ were directionally skewed (eg. it got hotter by the time $B$ decided to reply). Then relativity must reproduce the physics responsible for the anisotropy in other inertial reference frames too, not override it.

• there is no such thing as one way synchronization. Both $A$ and $B$ need a 'bounce back' from the other before they can define a common time. So when you say "time it would take for light to reach B from A as measured from any of the clock (A or B)", you are talking about one way time of flight, a concept of time that doesn't exist yet, that is ill-defined until $A$ and $B$ have synchronized their clocks. $A$'s clock has no idea of the time light takes to get to $B$ since it hasn't received a reply. $B$'s clock has no idea of the same as it doesn't know when $A$ sent the signal.

To eliminate any confusion, one way time of flight is defined, once clocks have been synchronized. Until then, no law of physics can define or differentiate between the one-way times (from $A\to B$ and $B\to A$ ) so one might as well take them to be equal for simplicity.

Answer 2

Param_1729: Using principle of relativity can't we just say that both the position A and B are equivalent and the time it would take for light to reach B from A as measured from any of the clock(A or B) would be same as for it to reach A from B as measured in any clock(A or B)?

We can't say that, because this time interval are not being measured by a single clock.

To turn it more clear, think that way: let's suppose Alice and Bob wants to synchronize their clocks, but they don't know their positions, so they don't know the distance between them, and they don't have a sense of absolute time,but they know that light velocity in vacuum is absolute.

The procedure is described in three events:

1. Alice starts her clock $t_1^A=0$ and at the same time, she sends a light signal.
2. Bob receives Alice signal at $t_2^B$ and start his clock, so $t_2^B=0$. He knows that Alice started her clock when she emitted the signal, but he doesn't know $t_1^B$. What he can do is to emit a light signal in the moment he started his clock.
3. Alice receives Bob signal at $t_3^A$. She knows that $t_2^A-t_1^A=t_3^A-t_2^A$ (applying the symmetry argument that you mentioned), so she can infer $t_2^A$. She resets her clock so $t_3^A=0$ again.

This procedure is not enough to synchronize the clocks, because their clocks don't necessarily measures equal time intervals. Then

4. Alice sends another light sinal at $t_3^A=0$, and Bob receives it at $t_4^B$.

Bob don't know $t_3^B$, but he could infer it because he knows that $t_4^B-t_3^B = t_3^B- t_2^B$. Now, look that this procedure would be endless, if they do not assume that the inferred time $t_2^A$ and $t_3^B$ are equal. With this assumption, they know that they are resetting the clocks with the same frequency, defining equal time intervals.

Now, $t_2^A=t_3^B$ is exactly $t_3^A-t_1^A=t_4^B-t_2^B$, and this is the assumption mentioned in the text.

Answer 3

"We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A." This is absolutely false. There are many alternative definitions that are observationally equivalent to this one. You can, for example, define a common time by assuming light takes twice as long to get from A to B as from B to A.