Time Dilation and different clock types

[Mister T]

It has nothing to do with appearance and locallity. A moving clock runs slower even locally.

I would prefer "observed" to "appears" to run slow from a different reference frame.

 

[Mister T]

How about: both clocks measure proper time along their worldlines. This is also the coordinate time in a frame in which the clocks are at rest, which is why we can say they tick at a rate of one second per second in that frame. In that sense, neither is running slow. However, both are running slow if we compare the amount of proper time between ticks with the amount of coordinate time between ticks in any frame in which the clock is not at rest.

An excellent way to explain it to even a beginner who is first struggling with mastery of the concept. But often we find ourselves trying to explain it to nontechnical types who will likely never engage in that struggle. For them it's harder to find a way to say it that's accurate and doesn't induce misconceptions.

 

[SlowThinker]

We just need to adjust the clock if we want to measure the Earth's rotation - because it is irregular indeed.

Have there been any adjustments recently? I don't recall, and neither do I pretend to understand, all the reasons behind it. But there is a strong argument among many metrologists that these adjustments should not be made.

See the graph and table in https://en.wikipedia.org/wiki/Leap_second
The rotation is slowing down, leading to the leap seconds being more and more common (in the long run).
Also the rotation is irregular, this can be seen as irregularities in the graph.
Last adjustment has been made on June 30, this year.

 

[PeterDonis]

The Earth's rotation will look simple however the laws of physics won't

Using which time standard?

Using a time standard based on fundamental physics, like an atomic clock, the Earth's rotation looks complicated (it's certainly not a straight sinusoidal function of time), but the laws of physics look simple. Using a time standard based on the Earth's rotation, the Earth's rotation looks simple (since now it is the time standard), but the laws of physics don't.

 

[facenian]

Using a time standard based on fundamental physics, like an atomic clock, the Earth's rotation looks complicated (it's certainly not a straight sinusoidal function of time), but the laws of physics look simple. Using a time standard based on the Earth's rotation, the Earth's rotation looks simple (since now it is the time standard), but the laws of physics don't.

Yes, that's what I meant, I thought you meant otherwise.

 

[A.T.]

I would prefer "observed" to "appears" to run slow from a different reference frame.

Both are potentially misleading. I would leave such terms out completely, and just say that this is how the clock runs in a certain frame,

 

[A.T.]

A.T., you know better than to ever talk about a "moving" anything without saying what the motion is relative to.

Relative to any chosen reference frame.

 

[A.T.]

Oh. I just assumed "locally" MEANS at rest relative to the clock, or "in the frame in which the clock is stationary".

"Locally" usualy means "within a smlal spatial extend". And that is completely wrong here, because position is irrelevant to kinetic time dialtion.

 

[phinds]

"Locally" usualy means "within a smlal spatial extend". And that is completely wrong here, because position is irrelevant to kinetic time dialtion.

But my point was that if you are sitting next to a clock, or holding it in your lap, you see the clock tick at one second per second totally regardless of your velocity relative to anything else and how any other observer sees the clock. How is that not "locally"?

 

[A.T.]

But my point was that if you are sitting next to a clock, or holding it in your lap

Here again you are conflating distance to the clock (next to it), with relative velocity (holding it). Being local to a clock says nothing about its speed relative to you, and only that speed is relevant for time dialtion, not the locality.

 

[phinds]

Here again you are conflating distance to the clock (next to it), with relative velocity (holding it). Being local to a clock says nothing about its speed relative to you, and only that speed is relevant for time dialtion, not the locality.

OK, then I've been using "locally" wrong and I guess I need to always specify "in a frame in which the clock is at rest". Thanks for that correction.

 

[A.T.]

OK, then I've been using "locally" wrong and I guess I need to always specify "in a frame in which the clock is at rest". Thanks for that correction.

The other problem is the use of terms like "appears" and "see", which again is not what SR is about. It's about what actally happens according to different frames.

 

[phinds]

The other problem is the use of terms like "appears" and "see", which again is not what SR is about. It's about what actally happens according to different frames.

Yes, but I was talking about what actually happens in the frame in which the clock is at rest. I do admit I have a tendency to dismiss kinetic time dilation as an "optical illusion" when, as you point out, it IS what actually happens in an observer's frame of reference. My calling it that is to emphasis that it is not what is happening in the rest frame of the clock, but I recognize the validity of your objecting to the way I phrase it.

EDIT: some of this is because my emphasis is on combating the frequent and serious error of believing that the clock is ticking slow in the frame in which it is at rest. You see people here asking what happens when our biological processes slow down due to time dilation and clearly they believe that the slowing is what happens in the rest frame of the clock.

 

[Dale]

Yes, it is a difficult pedagogical challenge.

 

[AlejandroDes]

Hello PF.
A moving clock is seen by the platform observer to run slow. This applies to all clock constructions. A light clock runs slow because the light path lengthens and the clock takes longer to tick over.

Other types of clocks don't have a light path to lengthen. By what mechanism do they run slow?

If the answer is, there is no mechanism, time just slows and the clocks record that, then why does the light clock not only slow but also shows the mechanism by which it slows i.e. longer path?
Regards.

Imagine you have 2 pendulums (they are exactly alike, same period) that you use to measure time, then you take the pendulum #2 and move it at certain speed relative to the pendulum #1. If you record the total amount of distance that the ball on pendulum #2 traveled over a period we willl see that it is larger than the one on pendulum #1 because in the same period the ball travel it's normal path plus the distance traveled by the relative speed. If you define an "absolute speed" as the total distance recorded over time you'll see that the pendulum #2 has an absolute speed larger than #1. That all makes sense.

The thing about litght is that no matter how we measure it, it's speed will always remain the same! So if instead of pendulums we use light we have a problem beacuse the absolute speed cannot be different between the 2 light clocks. So if we know that the speed is same for both light clocks and we know that the total distance recorded is larger for the moving clock then there is no other explanation than relative to the standing clock, the time slowed down for the moving clock

So there isn't a mecanism by which the clocks (or time) slows down. It's only what needs to happen in order for matter to comply we our "universal speed limit".

 

[MidiMagic]

Here is a thought experiment to show that it is relativity and the properties of spacetime, not something affecting the mechanism of the clock.

Case 1 is the case the original post posited: One clock is put into a vehicle moving relative to you at a constant velocity. You and an identical clock are left behind. Then you observe the clocks from your frame of reference.

Case 2 is a case that should be equivalent in relativity. You get into a vehicle moving at a constant velocity, taking one clock, and leaving an identical clock behind. Then you observe the clocks from your frame of reference.

Relativity says that both observations should produce the same result. You should observe the clock in the other frame of reference running slower than the clock you have with you runs.

 

[Raymond Potvin]

OK, then the old Twin's paradox is solved: reunited in the middle of the two observers, the two clocks should show the same time, which would contradict the light clock predictions. Here is some thinking I made about that light clock mind experiment, and it seems to contradict relativity, so you tell me where I am wrong.

A light clock runs slow because the light path lengthens and the clock takes longer to tick over.

Hi guys, hi DAC,

Elapsed time is not measured with light paths, but with frequencies, thus with light waves if the clock is a light clock. The light clock mind experiment shows a longer path for the light ray, but it doesn't show how this longer distance would be accounted for by the clock. If we replace the two mirrors by a light source and an observer, there would be no way for the observer to measure that distance, because even if, to travel in the direction of the future position of the observer, the ray was emitted at an angle to the direction of motion, thus producing doppler effect at the source, this effect would be nullified at the observer because he would be meeting the same ray at the same angle and at the same speed but in the opposite direction. Moreover, he would have no way to measure the real direction of the ray either because aberration at the observer would indicate that the source was not in motion. Here is a drawing I made about that:

Fig. 1 and 2 show how aberration and doppler effect occur for two bodies moving in different reference frames. Fig. 3 shows how aberration and doppler effect could actually be occurring at the source but would later be nullified at the observer, thus would always be unobservable, for bodies moving in the same reference frame.

At fig. 1, the observer is considered moving while the source is at rest. While the observer at B travels to B', light travels from A to B', so for the observer at B', the light ray suffers the aberration angle α and has the apparent direction of the dotted red arrow, but it also suffers doppler effect because the observer is moving at an angle to the incoming ray.

At fig. 2, the source is considered moving while the observer is at rest. While the source at A travels to A', light travels from A to B, so for the observer at B, the light ray does not suffer aberration but has the same apparent direction as in fig. 1. For us, it makes the same angle α with A'B, and it does not suffer doppler effect because the ray was emitted at a normal to the motion of the source, but it suffers relativistic doppler effect, which gives exactly the same number as in fig. 1 where the observer's speed was producing doppler effect while moving at an angle to the ray.

At fig. 3, both source and observer are considered moving at the same speed and in the same direction, so the only difference with fig. 1 is that the source is also traveling. For the observer at B', the light ray still has the same apparent direction as in fig. 1 since it suffers the same aberration angle, but this time, its direction points to the actual position of the light source because it has traveled the same distance as the observer, there is no measurable doppler effect because the one produced at the source nullifies the one produced at the observer, and there is no measurable relativistic doppler effect either because the two sources travel at the same speed. Of course, it would give the same result if the observer was at A and the source at B.

To me, fig. 3 means that if the source was a laser beam aimed perpendicularly to its motion, this beam would never hit the observer, which seems to contradict the reference frame principle. Does it?

 

[MidiMagic]

"OK, then the old Twin's paradox is solved: reunited in the middle of the two observers, the two clocks should show the same time, which would contradict the light clock predictions."

Not so fast.

1. When something is accelerated from one frame of reference to another, clocks that were in synchronization before the acceleration will not be in sync after the acceleration takes place from the point of view of either frame of reference.

2. To bring the two clocks back together, at least two more accelerations must take place. These further unsynchronize the two clocks.

One thing to think about in the light clock and light propagation drawings: Try making the drawings again as though you are sitting in the other frame of reference.

 

[Raymond Potvin]

To bring the two clocks back together, at least two more accelerations must take place. These further unsynchronize the two clocks.

If both accelerations would be the same, the clocks should stay synchronized.

One thing to think about in the light clock and light propagation drawings: Try making the drawings again as though you are sitting in the other frame of reference.

Drawing 1 and 2 show the two viewpoints, and drawing three shows both at a time, because we can interchange observer and source.

 

[Janus]

If both accelerations would be the same, the clocks should stay synchronized.

It is not just the magnitude of the acceleration that counts, it is also the distance between the clocks and there direction with respect to each other relative to the acceleration that factors in.
If if both clocks start at the same point and one clock then undergoes a short intense acceleration, due to the fact that there is no to little separation distance between the two clocks, the acceleration causes only a small additional difference in the clock readings. However, when that clock reaches a point some distance from the other clock and then decelerates to a stop and then accelerates back to the other clock, the large separation distance between the clocks will cause the other clock to run fast from his perspective. If he then decelerates upon his return, the small distance between the two clocks again results in only a small difference.

From the accelerating clock's view, the acceleration he undergoes when the clocks are separated at the turn around has a much more profound effect on the relative clock readings than the acceleration he undergoes when the clocks are adjacent or very near each other.

 

[Raymond Potvin]

Hi Janus,

If both clocks were departing directly from one another prior to getting back together, then their first deceleration could be the same, their acceleration towards one another also, and their last deceleration also, so it should produce no difference in their timing, what would sort of solve the paradox. But my questioning was about the light clock. What do you think of that reasoning?

 

[Dale]

@Raymond Potvin please do not hijack other people's threads. This would be far more appropriate as its own thread.

In a light clock the time is not measured using the frequency of the light, the Doppler shift is 0 and the actual frequency is not relevant. The time is measured by measuring the "echo time" from a target at a known range.