# Tick tick tick tick boom

1. Nov 18, 2007

### Skeptick

Synchronizing clock

The purpose of synchronizing clocks is so the time on one clock is as close a possible to the time on another clock. This can be achieved by any means possible to make the error as small as possible.

In Fig 0 the green bit is a laser that surprisingly emits photons. The red bit is a detector that detects the photons. Each emitted photon pulse is encoded with the exact time on it ie minute, second, milli sec, down to femto second say, using frequency modulation.

The clocks are synchronised by emitting the encoded photon from the green bit at A. When it strikes the red bit at B the coding is deciphered and that is the time that is entered on the clock at B. If the distance between the green bit and the red bit is L then the time difference will be L/C. Obviously by making L as small as possible the time diff is minimized.

In fig 1 I have the same arrangement as in fig 0 but it is now in a moving frame of reference (MFR) at 0.5C. The MFR is moving in the direction of the red arrow that has “direction” written on it :p.

At time zero the laser and the detector are located at positions A and B. At some time in the future they are located at positions C and D. A photon is emitted from the laser at time zero. It will follow the blue path as shown and strike the detector sometime in the future. As the encoded photon has traveled a fair distance the inaccuracy of the synchronization will be large and can be easily calculated. I will leave that to you and the first right answer gets an e_freddo.

In fig 2 again at time zero the laser and the detector are located at positions A and B. At some time in the future they are located at positions C and D. A photon is emitted from the laser at time zero. It will follow the blue path as shown and strike the detector sometime in the future. As the encoded photon has traveled only a short distance the inaccuracy of the synchronization will be small. If the distance between the green bit and the red bit is L then the time difference will be L/C. Obviously by making L as small as possible the time diff is minimized.

In a MFR why cant near perfect synchronization be achieved ?

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2. Nov 18, 2007

### JesseM

The way you've drawn it, in fig. 2 the detector would have to move faster than light! If at time T0 the emitter is at A and the detector at B, and the laser is shot out from A at this moment, and at time T1 the emitter is at D and the detector is at C, and the laser arrives at C at this moment, the distance from B to C is the distance the detector moved between T0 and T1, while the length of the blue line is the distance the photons in the laser moved between T0 and T1, and you've drawn it so the blue line is shorter than the distance from B to C. Of course you could shrink the distance from B to C, but then all you're showing is that by making the distance between the two clocks very small you can also make the disagreement between frames about their synchronization very small too, which is obvious just from the L in the formula vL/c^2, which tells you how out-of-sync two clocks will be in a frame where they're moving at v if they're synchronized in their rest frame and the distance between them in their rest frame is L.

3. Nov 18, 2007

### Staff: Mentor

With the goal being simply that the clocks read the same numbers, yes, any number of clocks, moving or stationary, can be syncronized in a given inertial frame. Take, for example, the clocks aboard the GPS satellites. These clocks are moving but are synchronized with clocks in the earth-cenered inertial frame. In order to do that they have to "undo" the principal GR and SR effects. But having done so, they give the same numbers as clocks in earth's rest frame.

What does it mean that they have to correct for those effects? It essentially mean that those clocks no longer measure "time" in a physical sense. The clock does not tick off 1 s in the amount of time that it takes for a 1-s-duration physical process to occur aboard the satellite.

If your purpose is simply to get the clocks to read the same numbers, yes, that can be done. When physicists speak of synchronizing clocks that is generally not the purpose. Generally physicists synchronize clocks in order to accurately describe physical phenomenon, not just to get matching numbers.

4. Nov 18, 2007

### Skeptick

Fig 1
In Fig 1 I have a clock. The green bit is a light globe that surprisingly emits photons. The red bit is a detector that detects the photons.

In a stationary frame of reference the time taken for as photon to move from A to B will be t = square root ((H x H) + (L x L)) ( hypotenuse formula)

Fig 2
The clock is now placed in a moving frame of reference MFR and the system is accelerated up to .88C. The MFR is moving in the direction of the arrow shown. At time zero the clock will be located at point A and B at some time in the future the clock will be located at point C and D.

Fig 3
The time now taken for a photon to move from the green bit to the red bit is simply H and at all times H < = square root ((H x H) + (L x L)) ( hypotenuse formula).

Also an external observer to the frame of reference would see what is depicted in fig 3.

Please prove me wrong and demonstrate that time cannot past faster in a MFR?

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5. Nov 18, 2007

### Skeptick

the picture is not to scale, I drew it in paint, not CAD. Im sure you can choose an appropriate scale and lengths and heights to make this work.

6. Nov 18, 2007

### Skeptick

So what you are saying is the type of synchronisation used is depent on the application. So if a physicist is measuring an unknown phenonmenon he just chooses some arbitratry synchronisation, as it would be impossible for him to know the appropriate synchronisation to use. He could choose clock one to tick at the frequency of the his microwave oven and clock 2 should tick at a rate set by the heart beat of the physicist cat? Sorry I'm putting my money on matching numbers :)

If GR and SR are wrong using your system it would be impossible to prove it, as your clocks are synchronised based on the theory of GR and SR and hence this erroneous theory is only reinforced!

7. Nov 18, 2007

### JesseM

Presumably you mean the time will be this distance divided by c?
For your diagram to make sense, we must assume that the red end moves a distance equal to the horizontal separation between the red and the green (which would be $$L * \sqrt{1 - v^2/c^2}$$ due to Lorentz contraction) in the same amount of time the light beam moves a vertical distance of H, meaning:

$$\frac{L * \sqrt{1 - v^2/c^2}}{v} = H/c$$

For v = 0.88c, this means that (L*0.4749737)/0.88c = H/c, which means H = 0.5397428L. This must be the ratio between H and L or the light beam will not coincide with the red detector at the moment the light beam has travelled upwards a distance of H.

But assuming the ratio is correct, then yes, the time in the observer's frame will only be H/c. However, as in previous experiments, if there are clocks at the green emitter and the red detector, and the clock at the emitter stops at the same moment the laser is sent out, and the clock at the detector stops at the same moment the laser reaches the detector, then the external observer will still predict that the difference in the readings of the two clocks when they stop is not H/c but $$\frac{\sqrt{H^2 + L^2}}{c}$$, because of the way the two clocks are slowed-down and out-of-sync in the observer's frame. I can prove this if you like.

Now, suppose the observer--call him "Bob"--has his own similar detector/emitter/clock setup which is at rest in his frame (the only difference is that his setup is a mirror image of the one that's moving in his frame, with the green emitter to the left of the red detector rather than to the right--I'll explain why later). Of course as we've already established, since the setup is at rest in his frame, the time in his frame for the light to travel from the emitter to detector will be $$\frac{\sqrt{H^2 + L^2}}{c}$$.

But now switch perspectives, and take the point of view of a second observer, "Alice", who is traveling alongside the detector/emitter/clock system which is moving at 0.88c relative to Bob. In Alice's own rest frame, it is her system which is at rest, while Bob's is moving to the left at 0.88c. And since Bob's system is a mirror image of her own, her view of Bob's system will look just like a mirror image of fig. 3, so that in her frame the time for the laser to go from the green emitter to the red detector on Bob's setup will be exactly H/c. But of course, she will not predict anything different about the readings on the clocks on Bob's system after they stop--she'll predict the difference in readings is exactly $$\frac{\sqrt{H^2 + L^2}}{c}$$ because of the way Bob's clocks are slowed-down and out-of-sync in her frame.

So you see, the situation is perfectly symmetrical. Each observer says that they are at rest while the other one is in motion, and that the time for the light to go from emitter to detector on their system is $$\frac{\sqrt{H^2 + L^2}}{c}$$ while the time for the light to go from emitter to detector on the other person's system which is moving relative to them is H/c, but they both agree on the predictions about the readings of clocks after they stop--for both systems the difference in readings will be $$\frac{\sqrt{H^2 + L^2}}{c}$$. So there is no way you can use this type of setup to determine which of them is "really" moving, or whose clocks are "really" slowed down or which pair of clocks are "really" out-of-sync.

8. Nov 18, 2007

### Skeptick

Im sorry about this I should have been more precise. What I meant to say is that the time on clock 1 is exactly the same as on clock 2 and the rate at which time passess in clock 1 and clock 2 is also exactly the same. As the clocks were identical and thus have identical internal workings I thought this would have been implied.

9. Nov 18, 2007

### JesseM

So you agree that if the distance the light travels is very small, the distance between clocks must be very small too? I thought you were trying to argue that we could synchronize clocks a large distance apart while keeping the distance the light must travel arbitrarily small.

Either way, the point is that your method won't "synchronize" clocks in any universal sense--if the horizontal distance between the clocks in their rest frame is L, then no matter what method you use, in a frame where the clocks are moving at speed v they will appear out-of-sync by vL/c^2. By making L very small you can make this difference small too, but it never disappears entirely, and it is key to understanding why there is no experiment you can do to determine whether the clock system is "really" at rest or "really" in motion, you can calculate things from the perspective of a frame where the clock system is at rest or a frame where the clock system is in motion and you'll get exactly the same predictions about physical questions like the readings on two clocks after they stop (as I did in the Alice and Bob scenario).

10. Nov 18, 2007

### JesseM

If the scientist wants to analyze things in the context of an inertial frame in SR, he must always use the Einstein synchronization method based on light-signals, or something equivalent (one way of doing this would be to set off a light signal at the midpoint of two clocks and set them to read the same time at the moment the light reaches them). You can use other synchronization methods of course, but then you're no longer dealing with an inertial frame and you can't assume that the usual rules of SR, such as the constant speed of light, will hold in your non-inertial coordinate system.
No, it would be easy to prove SR wrong, because SR makes a specific and broad physical prediction--if you do an experiment while at rest in one inertial frame, and then you repeat the exact same experiment while at rest in a different inertial frame, then you must get identical results, there can be no experimental way to show that the laws of physics work differently in one frame than they do in another (it can be show that this will be true as long as the equations for all the fundamental laws of physics have a property known as 'Lorentz-invariance', and all the fundamental laws found so far have this property, but we could certainly imagine a law which didn't). So, for example, if you have a setup where the clock at the emitter stops at the moment light is emitted from it, and where the clock at the detector stops at the moment the light is detected, and you look at the difference in readings between these two clocks after they have both stopped, this difference should not vary depending on which frame the setup is at rest in. If it did, then relativity would be wrong. But as I've been trying to show you, in all the setups you've proposed so far we'll make the same prediction about the difference in readings between the two clocks after they stop regardless of whether they are at rest of moving relative to some observer.

Last edited: Nov 18, 2007
11. Nov 18, 2007

### Staff: Mentor

No, what I am saying is that the FRAME of synchronization can be dependent on the application. In the case of GPS the goal is to measure the position of the detector (a physical phenomenon) on or near the surface of the earth, so synchronizing the clocks in the earth-centered inertial frame is a sensible choice. Thus the goal is not simply to get matching numbers on the clocks but to accurately describe physical phenomena in a specific reference frame.

That is a pretty silly view for a physics forum. Then let's just run the numbers on our clocks backwards and call it time travel.

I challenge you to find a prediction that has been in error. It has been subjected to numerous experimental challenges, by supporters and detractors, and the experimental evidence in support is overwhelming: http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

Last edited: Nov 19, 2007
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