# I Does a hot clock run slower?

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1. Aug 5, 2016

### greypilgrim

Hi.

Temperature is movement on a microscopic scale, and movement leads to time dilation. So what happens if we heat up a clock? Let's for example assume a pendulum with negligible thermal expansion, such that all other thermal effects on the period can be neglected.

Will it run slower? What would be the relevant velocity?

2. Aug 5, 2016

### Mister T

For time dilation to occur the clock's center of mass would have to move relative to you. When we increase the temperature of an object we increase the random thermal motion of its molecules. Because the motion is random, there is no motion of the center of mass.

3. Aug 5, 2016

### greypilgrim

Why is that? I've never heard this statement before.

Or let's assume a different clock that measures how long it takes for half the particles of a radioactive gas to decay. Shouldn't the mean lifetime of each individual particle increase with speed (as does for muon decay), so that it takes longer for half the particles to decay at higher temperatures?

4. Aug 5, 2016

### Staff: Mentor

Not in the general sense you're trying to use the concept of "movement" here. Instead of trying to reason intuitively using vague ordinary language, you should look at the precise math. The precise math says that, for example, if we have two pendulum clocks side by side, with the only difference between them being that one is at a higher temperature, the two will keep the same time. (There are some other technicalities here, but we don't need to go into them for this discussion.)

Once again, instead of trying to look for vague ordinary language statements, you should look at the precise math. The precise math says more or less what Mister T said if we're talking about macroscopic objects, with some technical qualifications (which I've gone into in a post in response to him). The usual precise math of relativity doesn't talk about the internal structure of objects at all, so the effects of things like temperature aren't normally modeled. But see below.

This is a different question than the one you asked in your OP. Now the question is, suppose we have two tanks full of muons sitting side by side, with the only difference between them being that the muons in the second tank are at a higher temperature. Will the decay rates seen by a laboratory observer be the same for both tanks, or not?

Here the answer (at least as best I can evaluate it with a quick calculation) is that the decay rate of the muons in the higher temperature tank will indeed be smaller (fewer muons decayed at a given point in time after we start the experiment). However, that doesn't mean that "clocks are affected by temperature" in any general sense; it just means that using a tank of muons as a "clock" if you don't control its temperature is a bad idea. But if, for example, we put the pair of pendulum clocks mentioned above in the same lab, both of them would still keep the same time--and it would be easy to use those pendulum clocks to measure the difference in decay rates between the muons in the two tanks.

5. Aug 5, 2016

### Staff: Mentor

This is too simplistic. For example, consider a storage ring containing muons--experiments using these have been done many times. The muons in the storage ring are moving in a circle in the ring at relativistic speeds (relative to the lab), and their observed mean lifetime (relative to the lab) increases accordingly. But the center of mass of the storage ring never moves relative to the lab. The CoM of each individual muon does--but so does the CoM of each individual particle in an object at a finite temperature, at a speed that increases with the temperature.

6. Aug 5, 2016

### pervect

Staff Emeritus
As others have posted, it all depends. To give some examples, if you are using atoms as clocks, heating the atoms cause them to move which causes time dilation effects. This is an important enough effect that some of the more accurate atomic clocks, cesium fountain clocks for instance, cool the atoms to improve their accuracy. See for instance https://en.wikipedia.org/wiki/NIST-F1.

But if you have, say, a pendulum clock, heating the pendulum would add energy to the pendulum, increasing it's weight and mass, but that wouldn't affect the time the clock keeps, there would be no time dilation.

7. Aug 5, 2016

### greypilgrim

Why does this thought experiment heavily depend on the type of clocks considered whereas the standard moving clock example does not? Time dilation is often derived looking at light clocks, which nobody uses in reality.

8. Aug 5, 2016

### Staff: Mentor

What "standard moving clock" example do you mean? And does the scenario in that example differ in some significant respect from the scenarios you have asked about in this thread? (Hint: if the answer to the first question is what I think it is, the answer to the second question is "yes".)

9. Aug 5, 2016

### Staff: Mentor

It's instructive to look at your original question for a light clock. For light, the analogue of "temperature" is frequency--heuristically, because temperature is "energy per particle" and frequency for light is "energy per photon". So consider two light clocks, identical except for the frequency of the light beams in them. Will they keep the same time? Yes. The frequency of the light beam inside the clock has nothing to do with the time it keeps.

So the result of your thought experiment still depends on the type of clock, even if we consider a light clock.

10. Aug 6, 2016

### pr3dator

There is no value of temperature in equation of time dilation.
The temperature may appear in this problem because the temperature can be tracked back to caotic motion.
But caotic motion means that there is no well defined direction of velocity.

11. Aug 6, 2016

### Ibix

The direction of motion isn't relevant. It's whether there is a change to the speed of the timing element - be that a pendulum, an atom, or a light pulse bouncing between mirrors. Only in the atomic clock does temperature change make the whole timing element speed up.

I think you have to idealise an atomic clock, and specialise your notion of heating, to define one that keeps (dilated) time while being heated. But it's at least possible in principle.

12. Aug 6, 2016

### Staff: Mentor

Part of the problem we're having in this thread is that the term "dilated time" is being used in two different senses.

The usual sense of "dilated time" refers to two clocks that are following different paths through spacetime; their different rates are due to the different paths.

The sense of "dilated time" being used in the above quote refers to two clocks that are following the same path through spacetime, but differ in some property (in this case temperature) that affects their "rate". But in the usual sense of "time dilation" (the one described above), it's impossible for two clocks following the same path through spacetime to be time dilated relative to each other. Proper time is the same for both of them, since they're following the same path through spacetime. The only difference between them is the "scale" of time they use, so to speak.

To put it another way, in the case of the two atomic clocks in this example, we could make them tick at the same rate by making sure they are both at the same temperature. But that's only possible because they are following the same path through spacetime; if they were moving relative to each other, following different paths through spacetime, we could not do this.

13. Aug 6, 2016

### vanhees71

Let's take an atomic clock, which is used to define the unit of time in the SI. It's defined by the giving the frequency of a certain finestructure transition in Cesium a certain value. From Wikipedia

[CITE]Under the International System of Units (via the International Committee for Weights and Measures, or CIPM), since 1967 the second has been defined as the duration of 9192631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.[1] In 1997 CIPM added that the periods would be defined for a caesium atom at rest, and approaching the theoretical temperature of absolute zero (0 K), and in 1999, it included corrections from ambient radiation.[1] Absolute zero implies no movement, and therefore zero external radiation effects (i.e., zero local electric and magnetic fields).[/CITE]

There are many effects of "temperature". The first that comes to mind is the Doppler shift of the frequency used to define the time unit, which leads to a broadening of the corresponding spectral line, which means an uncertainty in the frequency. A very detailed review about accurate time measurements and the definition of the time unit can be found in

https://arxiv.org/pdf/1401.2378

14. Aug 6, 2016

### Ibix

Actually, I think it's "clock" I'm cheating with. A clock in relativity is usually just a bare pendulum powered by who-cares and with no counter mechanism (e.g. a pair of mirrors with a light pulse bouncing back and forwards). But here I'm adding a counter mechanism and using it to define a frame where the "pendulum" of the atomic clock is in motion after heating, and calling that a clock too (and glossing over a lot of the practical details @vanhees71 refers to).

15. Aug 6, 2016

### Staff: Mentor

Not really. It's perfectly possible to have "clocks" with different scalings. It's just that two clocks having different scalings is a different thing from two clocks being time dilated relative to each other.

16. Aug 6, 2016

### greypilgrim

Assume I take a clock (whatever its working principle) and move it back and forth repeatedly. For simplicity, the magnitude of the velocity shall be constant and close to c and the turnarounds infinitely fast. Similar to the twin paradoxon, the oscillating clock will run slower than a stationary clock, because it doesn't remain in an inertial frame.

Now we decrease the amplitude of the oscillation, maintaining the magnitude of the velocity (i.e. increasing the frequency). I wonder what happens in the limit where the amplitude goes to zero, i.e. the clock doesn't move at all. But I'm not quite sure how to compute this.

17. Aug 6, 2016

### PAllen

Why not just posit circular motion? Then, for a given proposed speed, kept the same as radius shrinks, you are taking the limit of constant time dilation, thus trivial.

18. Aug 6, 2016

### jack476

Yes, because expansion of the metal in the gears and softening of the springs will mean the mechanisms work more slowly. It will, however, have nothing to do with relativity because there is no net motion of the clock relative to the reference frame of the observer.

19. Aug 6, 2016

Staff Emeritus
Well, thermal expansion would alter the pendulum's period. Which highlights the problem that this thread is struggling with: there is a difference between effects that make particular clocks keep bad time, and effects that change the rate at which all (good) clocks keep time.

20. Aug 10, 2016

### M L N Rao

In pendulum clocks the time of each oscillation depends on the length of a pendulum. Shorter the length, shorter the time and longer the length, longer the time. This principle was taken into account and they have manufactured cheap clocks with wooden stick suspension and for more expensive and accurate clocks using different metals which compensate the temp. variations using a very long pendulum, and enclosed in a glass box to isolate the air influences. clocks have evolved from sand glass to electronic and atomic varieties. Each has its own accuracy and quality.

21. Aug 11, 2016

### Staff: Mentor

That limit is not well-defined because it leads to infinite accelerations.

As a simplified answer: time dilation is relevant if your whole clock moves, it is not relevant if parts of your clock mechanism move.
The muons are a clock - averaged over a large sample they "measure" time via their decay rate, heating them makes them move faster which leads to measurable time dilation (at least in principle). The atoms in a pendulum are not a clock - the clock is the motion of the whole pendulum mass, which is independent of temperature.

Doppler shift/broadening is more important than time dilation, as it is a first order effect (v/c) while time dilation is second order (v2/c2).

22. Aug 11, 2016

### Staff: Mentor

Not only that, but the time dilation of the atoms in the atomic clock, relative to the lab, is irrelevant to the clock's function, whereas the Doppler shift broadening is not. The clock does not measure how long it takes for the atoms to emit a particular spectral line; it only measures the frequency of that spectral line.

23. Aug 11, 2016

### Ibix

I was thinking that either the time dilation effect is subsumed into the (relativistic) Doppler effect, or it is an additional source of change to the rest frame emission frequency on top of the (naive) Doppler effect.

24. Aug 11, 2016

### Staff: Mentor

It is. But as mfb pointed out, it's a second-order correction to the first-order Doppler effect.

25. Aug 11, 2016

### Ibix

Agreed. So you're just pointing out that first order (i.e. the non-relativistic limit) will break the clock before the second order effects become significant, right?