# How does flying over synchronize clocks look like?

I have a relatively (pun intended) simple conceptual question that has me going in circles as I begin my course work into Modern Physics. The question is straight forward enough:

You are gliding over Earth's surface at a high speed, carrying your high-precision clock. At points X and Y on the ground are similar clocks, synchronized in the ground frame of reference. As you pass over clock X, it and your clock both read 0.

1. According to you, do clocks X and Y advance slower or faster than yours?
2. When you pass over clock Y, does it read the same time, an earlier time, or a later time than yours?

For part 1 I figure, by inertial frame of reference, all other clocks would appear to be running slower. Initially I thought this would apply to part 2, but the when over X, Y does not read 0. So my real question is: Will Y appear to run increasingly faster until I am directly over it, and slower as I get further away, and therefore have a later time? If so, does this then go against my statement that the clocks will appear to run slower from outside an inertial frame?

• Caution: the quoted question is asking about the state of the clocks at a particular time in your frame of reference, not what they would look like as you passed over and light from them entered your eye. For an intuitive idea of how those differ, you can try velocity raptor and toggle between "seen" and "measured" mode. – Craig Gidney Jan 25 '16 at 21:31

The speed at that both X and Y run only corresponds to your relative velocity, not to your distance. So from your frame of reference both X and Y run slower than your clock, at the same, constant speed.

What Y and your clock show when you pass it is easiest answered from the earths frame of reference: You are moving at high velocity, so your clock runs slower, therefore Y will show a later time.

As you pointed out correctly, what looks like a paradoxon first is resolved, once you notice that X and Y are only synchronized in the earths frame of reference. So from your point of view at the moment you pass X, clock Y will not show 0.

• Ok, if I understand correctly, my issue was with changing reference frames. From the ground, clock Y was at some time > 0 when the glider and X are at 0. And, bc of reference, the glider's clock appears slower, thus it will always show a time less than Y's.? – NotSoSN Jan 25 '16 at 3:07
• in the ground frame of reference, both clocks are synchronized, that means at the time you pass X, Y shows 0 as well. but "at the time" only refers to time in ground frame here. – Anedar Jan 25 '16 at 3:13
• Thank you for your patience and assistance. So, from the ground frame, the glider's clocks is slower, X and Y are synchronized, and therefore X and Y would both show a later time? Is it also then true that from the glider's frame, both X and Y would run slower, and thus always be behind? – NotSoSN Jan 25 '16 at 3:22
• from the ground frame: yes, after you passed X (at this point all clocks show 0) your clock will be behind the others. in your frame: yes, after you passed X (at this point your clock and X show 0) X will always be behind. however once you pass Y, as we learned from the ground frame, your clock will be behind Y (but run faster and will therefore catch up). The important point is: relativity always gives the same description for events that appear at the same position at the same time (when you pass X/Y), but not for events that appear on a different time or place (e.g. Y when you pass X) – Anedar Jan 25 '16 at 3:29

You have not specified which clock you pass first.

When you pass over the X-clock, both it and your own clock read zero. The Y-clock (according to you) reads some time earlier than 0 if you've already passed over it, or some time later than 0 if you haven't passed over it yet.

According to you, both clocks run (equally) slow, and the Y-clock is either always ahead of the X-clock or always behind the X-clock, depending on which one you pass over first.

It is always best to start by drawing the spacetime diagram. Here is the picture on the assumption that you pass the X-clock first. The thick black line is you, the thick blue line is the X-clock, the thick orange line is the Y-clock, the thin black lines are your own surfaces of simultaneity, and the thin colored lines are surfaces of simultaneity according to the earthbound clocks. Your clock and the X-clock read zero at $A$. The $Y$-clock reads zero at $B$. If you stare at the picture long enough, you'll see exactly what's happening.