Relatively speaking, how does one clock know it's moving faster than the other?

Factual Questions
1

Acording to relativity, an object in motion will experience time slower than one at rest. The atomic clocks need to take this into effect when they are moved.

But one thing I’m confused about is how does the moving clock know it’s moving? Other than during acceleration, doesn’t each clock see itself at rest?

For example, imagine two synchronized clocks in space: Clock A and Clock B. They travel in opposite directions for the exact same time and then stop. Their clocks should still be synchronized at this point. Now, Clock A accelerates towards Clock B for a short time and then coasts for the rest of the way. At the point at which A and B meet again, which clock is slow and which clock is fast?

It would seem that A would be slow and B would be fast, but if that’s the case how did A’s clock know to slow down? Relatively speaking, how does one clock know it is moving “faster” than the other. Wouldn’t each clock think it was at rest and see the other clock moving towards it?

2

The point is, there isn’t any absolute sense of ‘fast’ and ‘slow’ - such things can only be expressed as a relative measure against something else. So if you have one clock that appears to have run fast compared to all the others, that’s exactly the same as saying all the others ran slow compared to the one.

3

But one clock accelerated and the other didn’t. That’s a not inconsiderable difference. One felt a force and the other didn’t. If you stuck an accelerometer on both, only one would record something. So they’re not identical and, relative to the frame with no acceleration, the one that did would know that it had “speeded up” by comparison.

4

Relativity is even weirder than you’re realizing, actually. When neither clock is accelerating, but they are in inertial motion relative to each other, each clock does indeed see itself at rest and the other clock as in motion. And from each clock’s point of few, the other clock is running more slowly … There isn’t a privileged reference frame from which one could state that one clock is actually running more slowly than the other.

Yeah, just chew on that for a while …

5

Is it just the act of acceleration that creates the time dialation? Is there accumulated time dialation while A is coasting?

The the positions of the clocks were like this:



1.                          AB
2. A                                                           B
3.                                                            AB

Where 1 is where they are at the beginning, 2 is where they moved apart equally, and 3 is where A moved towards B.

At step 3, what would the clocks read? Would A be slower than B? In my original post I said A would coast until it met B, but it would still be in motion. But what if A decelerated so that it stopped next to B. Now I can look at the clocks side by side in the same frame of reference. Relatively speaking, the only difference between A and B is that A went through a short time of acceleration and decleration. Let’s say A had 1 second of acceleration, 1 month of coasting, then 1 second of deceleration to come to rest next to B. Did the time dialation happen during the acceleration, coasting, or deceleration phase?

6

1.) Yup.

2.) When one is “coasting” (meaning that it is moving with constant velocity relative to the other), then both clocks are in inertial reference frames that have motion relative to each other. How could you say one is going “faster”? From the point of view of each, the other is moving with a constant velocity. There’s nothing to distinguish one from the other. But when one accelerates while the other does not there’s a clear difference.

7

There is no meaningful answer to this question. You can only meaningfully compare the clocks while they’re in the same reference frame. At time 1, the clocks agree, and at time 3, they disagree, so whatever put them out of synch had to happen at some time in between, but you can’t say exactly when that was, since you can’t compare whether they agree or not at any time in between.

8

One further conceptual nitpick: each clock “experiences” time–in terms of the rate of decay of radioactive isotopes, or the rate of equivilent thermodynamic systems, or the time required for a set distance traversed by a photon, or however you care to measure the passage of time–exactly the same, regardless of whether it is going faster or slower, under acceleration or moving “free” of obvious influence of forces, et cetera. The astronaut in the rocket won’t see his clock moving any slower as he approaches speed c than his twin back on Earth. However, the reference frames of each advance on different rates relative to the influence of net forces (resulting in acceleration, whether that causes a change in momentum or not), which causes the two brothers, when they do come back to a common inertial reference frame, to have to seriously recalibrate their watches.

As Chronos notes, there’s no meaningful comparison between the two systems until they rejoin at the same inertial state, and then when they do, you can’t tell “when” the so-called “time dilation” happened, only the difference between the two resultant states; for all you know, the twin in the rocket may have just been flying circles in a powered orbit around an imaginary star, or thrusting away at the edge of the event horizon of a black hole, or any other path which resulted in net acceleration and hence, in a different inertial frame from the twin on Earth. To put it another way, “time dilation” isn’t an event that occurs; it’s an artifact of taking a path influenced by an external force, whether it is the thrust of propellant against a rocket nozzle or the pull of gravity of a massive object.

To illustrate this using an analogy, consider two cars driven by a pair of twins, Dougal and Fingal. Dougal, a daring driver in a BMW M6, takes the high road that twists between Scottish peaks. Fingal, a more stately driver in his Mercedes E550, takes the straight and flat low road. Both drive from Edinburgh to Aberdeen, leaving and arriving at the same time. Dougal, however, has much more mileage on his M6 and has burned more fuel than Fingal due to having taken a longer and more mountainous route, even though each covers the same amount of road per mile. Until they meet back up, Dougal has no clue that he’s taken the “longer route” or driven any faster than Fin; his experience of the passage of road per mile has been exactly the same. We can’t say what his rate was relative to Fin’s speed at any given point on the road, because there is no metric for comparison between the two roads at any point; we can only say that Dougal’s average speed is greater than Fingal’s average speed. Now, this analogy clearly switches out the rate of speed for the rate of time in order to illustrate where the difference comes from and why each driver’s experience (for the passage of the rate in question) is the same, even though the system they’re in is different.

So chuck out the notion that the clocks are measuring time “at different rates.” Observers always measure time at the same rate (in their own reference frame), but the relationship between one frame and another is dependent upon its path in spacetime, and all movement occurs in four dimensions; three in space and one in time. A frame that moves “faster”–was at some point under acceleration–through space will have less movement in the time direction, and an object that moves faster in time by necessity goes “slower” (i.e. has seen less acceleration) in space, although this is only apparent relative to another frame. Another way of thinking of it is that all objects are always moving at c, but that is divided between all four dimensions. Massy objects always have some component of speed along the time direction, and objects that are at rest with respect to the observer’s frame are moving exclusively through time, while massless particles like photons move only in space and have no measurement of or movement through time from their inception to terminus.

Does that make it all as clear as eggnog?

Stranger

9

Here’s an analogy to help you figure out why this question is ill-posed: Suppose you and I are at the same point in a city. I walk a block north and stop. You walk a block east, a block north, and then a block west, and stop. Both you and I are at the same point now, but you’ve covered more distance than I have. Where did you cover the extra distance? During the turns, or during the walking? Neither answer is wholly satisfactory; the best answer is just that we took different paths and so covered different distances.

Now here’s the analogy: instead of two different paths in space, your two clocks are describing two different paths in spacetime. They both set out from the same point in spacetime, and they both end up at the same point in spacetime, but the “distances” (namely, elapsed times they measure on their clocks) are different. And just as in the case of you & I walking through the city, you can’t really say that there was a specific place where the time discrepancy “happened.”

Edit: and, of course, Stranger beat me to exactly the same analogy, only with more colour.

10

>there’s no meaningful comparison between the two systems until they rejoin at the same inertial state

Nitpick, maybe, but of course there are meaningful such comparisons. You can shoot one clock past the other one and stand where you will be equidistant from them at their closest mutual approach, and take a picture with your camera flash. Since we are talking velocities something like c to make the situation interesting, you will of course in any experiment take account of the fact that it takes the light you are photographing with some time to move around, too. So far as I have heard, nobody’s come up with a situation where Einsteinian general relativity failed to predict what you’d observe in experiments as close to this as anybody has thought of a way to do. Fast moving astronomical bodies that do repetitive things are a favorite target.

One major stumbling block is the idea that time is advancing everywhere in unison. Time is a local phenomenon, and there is no unique version of my “now” anyplace other than my “here”. Observers moving in different ways will report different versions of what events were simultaneous, if they take proper account of all the obvious light transmission delays (if they don’t, it will be even more confusing). Oddly, while there is no special significance to “stationary” and “now”, there is special significance to the speed c, and just about the only thing all observers will agree on is that something moving at c is moving at c.

For the sake of thought experiments, you can imagine a nice powerful pointer and a huge screen to play it on. You can move the spot around on the screen at faster-than-c speeds if you want - or at least from your perspective thats what it will do. Some observers will be able to see, truly see, the spot sitting in one place, then another spot appear elsewhere, while a third spot moves from the destination to the origin, where it and the first spot will disappear together. The spot your laser pointer beam hits the screen doesn’t carry information or mass around with it and can do all these things if it likes.

11

Actually, permit me another dalliance. Spraying a stream of water into the air with a garden hose nozzle is a useful model, in some ways, for the laser spot on a screen. If you paint figure 8’s in the air, lying on their sides, the point on the ground where the stream lands can move around in funny ways. It’s possible, for example, to have several simultaneous places where the stream is landing at one moment, and no such places a moment later.

12

Would the clocks agree at step 2? They start together, they apply exactly the same thrust in opposite directions, travel for an exact amount of time according to their own clock, and then decelerate. In that instance, would both clocks still be synchronized? Imagine that the clocks send out their own time radio signal and there is a receiver exactly in the middle (where they started). Would the receiver show the clocks synchronized.

In step 3 while clock A is coasting towards clock B, aren’t they moving through space-time at the same rate? That is, clock A would see clock B coming closer, and clock B would see clock A coming closer. During the coasting phase, isn’t it just as correct to say that B is traveling towards A as it is to say A is traveling towards B?

Hmmm… I just thought of another scenario that might make this clearer…

What if A sent out a radio signal every clock tick with it’s current time and B had a receiver. When the clocks are at rest far apart (step 2), I would think that B would receive A’s ticks at the same interval as B’s clock. That is, 1 tick of A’s clock would exactly match 1 tick of B’s clock. Even if the clocks were not in sync, I would think the spacing of the ticks would be identical.

Now, say the clocks know that at a pre-determined time, A will apply 1 second of thrust. B will continue to receive radio signals the whole time A is traveling towards B. Relative to his own clock, B should be able to see the delta in A’s ticks as it travels towards B.

Since B knows exactly when A applied the thrust and for how long, B should be able to exactly compute the distance to A at any time and how long it will take for A to arrive relative to his own clock. Since it can compute this so exactly, it should be able to take into account the shorter path that the radio signals will need to travel to reach it and factor that in to the tick delta. The radio signals travel at c, so they should not be affected by A’s velocity.

It would seem like that experiment could reveal the time dialation. If you looked at B’s time delta log, I would think it would look like this:

(2 ticks per second)
1 tick of A took 1 tick of B (before acceleration)
–thrust begins—
1 tick of A took 1.1 ticks of B (beginning acceleration)
1 tick of A took 1.2 ticks of B (maximum acceleration)
1 tick of A took 1 tick of B (no acceleration)
1 tick of A took 1 tick of B
…
So wouldn’t that show the spread of the dialation? A and B’s clock will not agree on the number of ticks that have passed, but I would think that B could accurately measure the change in A’s tick time relative to B’s own clock.