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Is a clock ticking at its highest possible tick rate as measured from any frame of reference at rest relative to any other clocks?

Consider 2 synchronised clocks A and B relatively stationary at the same location. Clock A is accelerated to its highest possible speed (approaching c) and then brought back to clock B. Clock A will indicate an earlier time than Clock B. The tick rate of clock A is less than that of clock B until both clocks are brought back together.

The definition of tick rate here is the quantity of events recorded by any clock during an experiment. At the end of the experiment above when both clocks are brought back together clock A has recorded less events than clock B.

Qmechanic
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jamesfairclear
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  • What is "highest possible tick rate" or "highest possible speed" supposed to mean? A massive object can't ever reach $c$, and any speeds below $c$ are just relative - e.g. the object is at rest in its own co-moving reference frame. – ACuriousMind Jan 08 '22 at 20:50
  • The tick rate of a clock varies when it is accelerated. The lowest possible tick rate of a clock coincides with it being accelerated to its highest possible speed (approaching c). The highest possible tick rate coincides with a clock being at rest relative to any other clocks. The tick rate of any other clock must be greater or equal to the tick rate of such a clock.The definition of tick rate here is the quantity of events recorded by any clock during an experiment. At the end of the experiment above when both clocks are brought back together clock A has recorded less events than clock B. – jamesfairclear Jan 09 '22 at 14:40

2 Answers2

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You appear to believe that clocks slow down when they move, which is a misconception. Accurate clocks always measure one second per second. When Clock A is accelerated away from Clock B it shows a lower reading on its return not because its tick rate has reduced but because it has measured a shorter time interval.

The distinction is subtle but crucial if you want to understand SR. Let's suppose Clock A shows 10 seconds to have passed while it was away and Clock B shows 12 second to have passed. That is because Clock A's journey has lasted 10 seconds and Clock A has faithfully measured it as such- it is not because Clock A has been slow.

The Lorentz transforms allow you to compare the duration of the interval between two events in one frame with the duration of the interval between the same two events in other frames, and generally the durations will be different. Let me repeat that, the durations will be different. That is why the clocks show different readings- because they are accurately measuring different durations, not because one of them has started running slow.

That said, I can address the underlying point of your question as follows. In SR, the duration in an inertial reference frame between two events that occur in the same place is less than the duration in any other inertial frame in which the events are spatially separated (that is the basis of time dilation). The same duration (ie between to co-located events in an inertial frame) is more than the duration along a path between the events that involves accelerations (that is the basis of the twin paradox).

Marco Ocram
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Let me draw a diagram to clarify exactly what we are talking about. We have two (or more) clocks that start at the same point in spacetime $S$ and end at the same point in spacetime $T$. So if we draw a diagram to show how the clocks moved through spacetime it might look something like this:

Clocks

And we want to compare how many ticks each of the clocks recorded so we can find out which clock recorded the most ticks. Then your tick rate is just the number of ticks the clock recorded in moving from $S$ to $T$.

The algebra involved in this can get a little complicated, but the basic principle is surprisingly simple. To explain it let's start with a slightly different situation. Suppose $A$ and $B$ are two people walking on a field and the diagram shows their position on the field as $(x,y)$ coordinates. Now change the question to:

Which person recorded the fewest steps walking from $S$ to $T$?

The answer is obviously $B$ because the number of steps is proportional to the distance and $B$ walked in a straight line while $A$ walked in a curve. We know the shortest distance between two points is a straight line, so we know $B$ walked the shortest distance so we know $B$ recorded the fewest steps, and this is always true no matter what curve $A$ walked in.

The same applies to objects moving in spacetime with one key difference. In relativity we calculate the length of a line in a different way from Euclidean geometry and as a result of this our basic principle changes to:

The longest distance between two points in spacetime is a straight line

This may seem a bit odd, but the fact the longest distance is the straight line is behind all the weird effects in relativity like time dilation and length contraction.

Anyhow, in our spacetime diagram $B$ moves in a straight line while $A$ moves in a curve so $B$ measures the longest distance. It turns out that the length of the line is just the time recorded by a clock that moves along that line. So we can immediately say that $B$ measured more ticks on their clock than $A$ did. And just as in my example of the people walking in a field this is always true no matter what weird curve $A$ moves in. So our conclusion is:

The observer who moves in a straight line through spacetime measures the most ticks on their clock

It only remains to define what moving in a straight line means in relativity and actually it means the same as in Newtonian mechanics. In Newtonian mechanics an object moves in a straight line through space if its acceleration is zero, and in relativity an object moves in a straight line in spacetime if its four-acceleration ¹ is zero.

In special relativity (things get more complicated in general relativity) a stationary object is obviously not accelerating, so in your example where $B$ is stationary it is moving in a straight line (in fact it is moving straight up the time axis at one second per second). By contrast $A$ is accelerating so it cannot be moving in a straight line. So from our argument above we can be confident that $B$ always measures more time on their clock than $A$ does, no matter exactly how $A$ moves ².

This is of course the twin paradox, and what I have described above is how it is resolved. A more detailed description of this is given in What is the proper way to explain the twin paradox? though the maths in that are probably a bit too hard for people just starting to learn relativity.

¹ Describing exactly what the four-acceleration is can get a little involved, but a useful simplification is that your four-acceleration is zero if you feel no g-forces. So an observer floating weightless in space has a zero four-acceleration while an observer being shaken about inside a rocket has a non-zero four-acceleration.

² This is not true if $A$ is allowed to move faster than light, but then all sorts of strange things happen if we allow faster than light travel.

John Rennie
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  • "Which person recorded the fewest steps walking from S to T?" - I would caution that, whilst this seems to be couched in the familiar language of everyday activity, a person is likely to ask *what does it mean to be walking through time?". Walking is an act that humans perform, and I can walk forwards, backwards, across, and can stop. And when I walk forward in one direction, it doesn't imply walking to the left in another. Why would walking in space mean I am walking in time? And why would I be walking in time when I am stopped in space? I think these points would need to be addressed. – Steve Jan 09 '22 at 12:02
  • @john rennie. Hi John, thanks for your response. You conclude " always measures more time on their clock than does, no matter exactly how moves". From this basis then that one clock may record more events than another it follows that there can be a clock that records the highest possible number of events which can be considered to be at rest relative to all other clocks. – jamesfairclear Jan 09 '22 at 17:03
  • @jamesfairclear We are considering only clocks that start and end at the same point, and for all such clocks there is one clock that ticks the most i.e. the clock that moved in a straight line though spacetime between the points. However you cannot say this clock is "at rest". What you can say is that this clock didn't accelerate i.e. that its velocity was constant. That constant velocity could be zero or non-zero since all velocities are relative to the observer. – John Rennie Jan 09 '22 at 18:05
  • @john rennie "We are considering only clocks that start and end at the same point". I am considering a clock that is accelerated in n directions until it is found to record a higher quantity of events than any other clock and can therefore be considered to be at rest relative to any other clock. – jamesfairclear Jan 10 '22 at 09:32
  • @jamesfairclear you cannot usefully compare clocks unless they are at the same place. In my diagram the clocks start at the same place and end at the same place, so we can compare them to see how time was recorded by the clocks. This is a frame independent comparison because all observers in all frames will agree on the difference between the clocks. If the clocks are headed off in different directions then the problem is that there is no frame independent way to compare their rates because the difference between the recorded times will depend on the observer. – John Rennie Jan 10 '22 at 09:48
  • @john rennie from an experimental perspective the clocks are all synchronised at one location, accelerated away and then brought back to the same location. The clock that has recorded the highest quantity of events is the clock that can be considered to have been at rest relative to the other clocks at some points during the experiment. We can predict that a clock travelling at 0.99c will record a lower quantity of events than any clock travelling at < 0.99c. We can also predict that a clock recording a higher quantity of events than any other clock is at rest relative to to any other clock. – jamesfairclear Jan 10 '22 at 11:32
  • I feel as though we are going in circles. The clock that has recorded the maximum ticks is the one that has not accelerated i.e. has had a constant velocity. That velocity may or may not be zero depending on the observer. If I am speeding past the experiment at 0.999c none of the clocks will be at rest in my frame, but I will still agree that the clock which didn't accelerate recorded the most ticks. – John Rennie Jan 10 '22 at 11:42
  • @john rennie all that we know about the initial conditions is that the clocks are at rest relative to each other in the same location. All the clocks are accelerated away and then brought back to the original location. If for example the initial reference frame is in motion at 0.99c then one of the clocks in the experiment might be accelerated in the opposite direction of motion in which case it will be recording a higher quantity of events until it is brought back to the original location. – jamesfairclear Jan 10 '22 at 13:31