# Clocks are not synchronized in moving system (book of David J. Griffiths) [closed]

In the electrodynamics book of David J. Griffiths, in section 12.1.2, there is a paragraph about the synchronization of the clocks on a moving train. I don't understand why " Clocks that are properly synchronized in one system will not be synchronized when observed from another system" Anyone has some ideas?

• This is a very vague question. The book already explains the answer to the question. So what specifically and in detail do you not understand about that explanation?
– Dale
Commented Jul 5 at 3:53
• It depends a bit on what you mean by "why", but if you are looking for some deeper principle by which the statement about it being impossible to synchronize clocks in different reference frames, then note that this paragraph isn't deriving that statement. It is just taking that as a given, observed fact about reality. If it makes you uncomfortable, good! It is one of the most bizarre facts about our universe. You can derive it from the postulates of special relativity, with some work. Commented Jul 5 at 5:13
• The sentence immediately following your highlight is the answer to your question. Commented Jul 5 at 11:31
• Thanks for all of your answers, it's indeed just a problem of "relativity of simultaneity". Just take the movement of clock hands as a event. It is not simultaneous when I observe from a moving frame. That's why the clocks are not synchronized. (I didn't realize this before. I thought it is not synchronized becasue the light reflected from clock need some time to reach the eyes of the observer ðŸ˜‚) Commented Jul 6 at 0:08

## 3 Answers

It might help if you create a simple thought experiment and forget about clocks altogether, because it is actually time that is out of synch, and clocks are just a tool for keeping track of time. Try this...

Suppose you are walking past me heading East at a metre per second at the exact moment I flash a light. After a second in my frame, the light has travelled one light-second away from me in all directions. Now, because you are walking at a meter per second, light that is exactly a light second away from me going East is a light second less a meter away from you. Likewise in the opposite direction, the light is a light second plus a metre away from you. If you think about that carefully, and remember that the speed of light is the same in every frame, you should be able to work out that the two instants when the light is a light-second away from me in both directions are simultaneous in my frame, but not in yours. In your frame when the light is a light second away from me to the East, fractionally less than a second has passed, while to the west fractionally more than a second has passed because the light has travelled further from you.

I don't know if you have been able to follow what I have said, but what I have described is the essence of the relativity of simultaneity, and until you really absorb it, relativity will remain confusing.

• Thanks for all of your answers, it's indeed just a problem of "relativity of simultaneity". Just take the movement of clock hands as a event. It is not simultaneous when I observe from a moving frame. That's why the clocks are not synchronized. (I didn't realize this before. I thought it is not synchronized becasue the light reflected from clock need some time to reach the eyes of the observer ðŸ˜‚) Commented Jul 6 at 0:09

In general: if two clocks are synchronized in one inertial frame, they are unsynchronized in every other inertial frame. This follows entirely from special relativity.

The most key aspect to understand is relativity of simultaneity, which essentially describes the effect that moving has on your perception of whether or not events are "simultaneous".

This graphic captures the effect well, in terms of a spacetime diagram. When you're not moving (in a particular inertial frame), A and B and C are simultaneous. But when you are, you observe space and time differently (as per the Lorentz transformations), and you see C and then B and then A, none simultaneous, or the reverse if you're moving backwards.

Another way to think about it is with a sort of relativistic Doppler effect (not to be confused with the actual relativistic Doppler effect, which affects the frequency/amplitude of photons). If two events happen simultaneously in one frame, then the photons reaching an observer who's moving in that frame will arrive at different times as that observer sees it.

The same applies for the synchronization of clocks. If each "tick" is a pulse of light, then an observer moving in a particular frame will see the ticks of two clocks that are synchronized in that frame as pulsing at different rates and at different times. Thus, if you're moving relative to the synchronized clocks (i.e. in any inertial reference frame other than the one in which the clocks were synchronized), you will not observe them as being synchronized.

• Nitpick: Your first paragraph, in its current form, assumes only one spatial dimension. Commented Jul 9 at 4:03

I also find this explanation confusing, unless you already understand it, which you don't: which is why you are reading the book. Nevertheless, the relativity of simultaneity is so far outside out Galilean intuition that it is hard to explain at first.

I really think it is best to start with the lighting striking both ends of the trains to get started. Here you can define the actual event time by when the flash reaches the origin of the coordinate (which is the "observer").

I also don't like the displayed text because it introduces the "moving" observer 1st. Of course, there is no moving observer: everyone is at rest in their own coordinates, but when we make a mental picture, the train station ($$S$$) is "at rest", for semantic purposes only. The moving train is then $$S'$$.

So the frames should be introduced in order (for clarity), and the traditional names are Alice, Bob, and maybe Charlie.

Then Alice, who is the "she", is at the station, and Bob is on the train (which is length $$2L$$).

Finally, the origins of $$S$$ and $$S'$$ have to be the same--that is, at $$(t=0=t', x=0=x')$$, Alice and Bob are at the same event.

Now if at $$t=0$$, two lighting bolts hit the train at the same time, those two events for Alice are:

$$E_C = (0, -L/\gamma)$$ $$E_E = (0, +L/\gamma)$$

here $$E$$ and $$C$$ subscripts stand for caboose and engine, respectively. Calling them $$A$$ and $$B$$ is no good, as it implies an ordering that doesn't exist in reality--just because the alphabet.

Alice measures the events to be simultaneous because she sees the flash at the same event ($$c=1$$):

$$E_{AC} = E_{AE} = (L/\gamma, 0)$$

Of course by at this time, in $$S$$, Bob has already seen the engine strike, but not the caboose strike.

If we Lorentz transform the strikes to Bob's frame, those events have coordinates:

$$E'_{C/E} = \Big(\pm v L, -L(v\pm 1)\Big)$$

so the engine strike occurred 1st, and the caboose strike followed. (Note, that if I didn't include the Lorentz contraction in Alice's frame, the lighting would have missed the train).

It's good exercise to compare that with the strikes being simultaneous for Bob, so the events are:

$$E'_C = (0, -L)$$ $$E'_E = (0, +L)$$

and then do the inverse LT to Alice's frame.

Once that makes sense, consider Bob's synchronized clocks at the ends of the train (or a lattice of many of them along train), and Alice's synchronized clocks along the platform, and Griffith's text will make sense.

• I agree with you. It seems that "Two clocks are not synchronized" and "Moving clocks run slow" are not related. The paragraphs in the picture are trying to explain why both the observers (on the ground and on the train) are right, but it seems has nothing to do with the synchronization of clocks Commented Jul 8 at 1:07