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I was about to watch Interstellar for the third time when I decided to do some research on its phenomena. Time dilation is the one that struck me most. I have read and read about it and understand -- as far as that is possible for me -- how it works.

But I don't get how a mechanical, or quartz clock could tick slower as they are composed of gears and other parts that are always moving at a constant speed, right. The example with the two-mirror 'light clock' is enlightening but how does time dilation work on a 'real' clock?

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    $\begingroup$ Of course it would tick at the same rate in the frame where it itself it as rest. An observer in relative motion would find it ticks slower, but gearing of the clock itself is unaffected by the relative motion of this observer. $\endgroup$ Apr 5, 2018 at 12:50
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    $\begingroup$ I recommend getting the book, today, for $10: amazon.com/dp/0393351378 (The Science of Interstellar), by Kip Thorne who is a brilliant explainer of space-time and its curvature. $\endgroup$
    – JEB
    Apr 5, 2018 at 13:55
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    $\begingroup$ I think this question is pretty much asking if length(amplitude) contraction will occur in oscillation along the direction of movement. $\endgroup$
    – Shing
    Apr 5, 2018 at 13:59

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I've rarely seen explicit demonstrations along these lines. It would certainly be very difficult; for most clocks one would at the very least have to use relativistic quantum mechanics.

In general we don't want to re-prove time dilation for every kind of clock. Instead the reasoning runs the other way: if a light clock and another clock are ticking side by side, they'd better also be ticking side by side in another frame, so the other clock must also experience time dilation. Since clocks can rely on classical mechanics, or quantum mechanical effects, or anything else, that means that all our theories should be compatible with special relativity. So we build the theory to be relativistic from the very start. Once you've done that it's not really worth bothering to double-check relativity works in a specific case, as we already know it's going to work for all cases.

Still, this is a neat question, so let's do the check anyway! For simplicity I'll consider a 'magnetic' clock. The idea is that a vertical magnetic field $B_z$ makes a particle move in circles, and the clock ticks every time one circle is completed. In the clock's frame, supposing the particle moves slowly enough to neglect relativity, we have $$q v B = \frac{dp}{dt} = ma, \quad a = \omega v$$ since we're dealing with circular motion, and combining gives $$\omega = \frac{qB}{m}.$$ Now consider a frame where the clock is moving along the $z$ direction with time dilation factor $\gamma$. Then the momentum is now $\mathbf{p} = \gamma m \mathbf{v}$, so the only modified equation is $$\frac{dp}{dt} = \gamma ma$$ and we instead find $$\omega = \frac{qB}{\gamma m}$$ which is exactly as we'd expect by time dilation. Physically, the reason the clock ticks slower is that the particle is harder to turn around, by virtue of its relativistic motion in the $z$ direction. With a bit of handwaving, this explanation works for the quartz clock too. The gears in the clock have nothing to do with it -- what matters are the resonant oscillations in the piece of quartz, which control everything else. So it's plausible these oscillations get slower as each particle participating in them gets harder to accelerate.

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  • $\begingroup$ Nice. Except now you have to convince me the magnetic field in the moving frame is the same is in the stationary frame, and you sorta rectus pluckus'd $\gamma$ in the momentum equation. $\endgroup$
    – JEB
    Apr 5, 2018 at 14:01
  • $\begingroup$ @JEB That's the rub with these intuitive explanations, right? I'm trying to derive a simple consequence of relativity in a fully relativistic theory, from more complicated consequences (the field transformations). The logic can't ever be perfect, because this just isn't the direction it runs -- we really just make everything relativistic from the start and forget about it. The logical structure is not $A \to B \to C \to D$ but rather $A \to \text{everything}$. $\endgroup$
    – knzhou
    Apr 5, 2018 at 14:05
  • $\begingroup$ @JEB I toyed with an explicit calculation with a solenoid. The issue is that the solenoid length contracts, so the only way to keep the magnetic field constant... is to apply time dilation to the current! Which is what I'm trying to show. I could manage to do it without time dilation using a more complicated setup, but then we're really hitting diminishing returns. $\endgroup$
    – knzhou
    Apr 5, 2018 at 14:08
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    $\begingroup$ This has the usual problem with transverse versus longitudinal inertial ($\gamma m$ versus $\gamma^3m$ and intermediate values for angles between), but also the usual resolution of that difficulty (length contraction along the direction of relative-motion fixes up the output). I worked through the hairy details for light clocks in another answer. $\endgroup$ Apr 5, 2018 at 14:21
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    $\begingroup$ Sure, there is no worry about it as long as you keep the motion transverse, which you have rigged it to do. But when I was learning that always seems a very arbitrary restriction, so being able to say with confidence that it still works for longitudinal clocks is a nice additions. $\endgroup$ Apr 5, 2018 at 14:27
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It makes sense to talk about time dilation only if you're comparing your clock to another clock that is or has been in different reference frames. The clock is not ticking slower because it is in motion. Otherwise it will be possible to tell if you're moving or not. And the whole idea of relativity is that The laws of physics are the same in all inertial frames of reference. Meaning that regardless if you're at rest or moving with constant velocity in a straight line, you cannot perform any experiment that will determine your state of motion (at rest or moving). So if you're moving with constant velocity in a straight line you will experience everything that a person at rest is experiencing. So the clock it will tick at a "normal rate" in its proper frame.

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    $\begingroup$ Except that in the movie, the dilation occurs near a black hole, so replace the SR specifics in your answer with GR specifics. $\endgroup$
    – JEB
    Apr 5, 2018 at 13:51
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Say you're driving a car at a constant velocity (constant speed, constant direction) and on the passenger seat next to you is an accurate clock. A stationary roadside observer would measure your time, as recorded by your clock, as slowing down. However, you wouldn't notice your clock doing anything different. Weirdly, you would also measure the roadside observer's clock as slowing down. Hence, the description that moving clocks run slow, one of the fundamental results of special relativity. The effect is barely noticeable at everyday speeds, but if your car was very, very fast (approaching the speed of light) the slowdown becomes significant. The twin paradox (which isn't a paradox at all) explores this notion of moving clocks running slow in greater detail.

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It is a mistake to assume that time dilation is an effect which interferes with the working of clocks. The time interval between two events is frame and path dependent, and depends on the geometry of spacetime. For example, in flat spacetime, the interval between two events that occur in the same place in one inertial frame is less than the interval between the same two events in any other inertial frame in which the events occur in different places. For example, the interval might be four seconds in the first frame and ten seconds in the other. Note that it is the actual duration of the interval which changes- an accurate clock in the first frame will tick off the four seconds between the events, while accurate clocks in the other frame will suggest that ten seconds have passed. In both frames the clock run at the same rate, ticking off a second every second.

Consider your watch, which ticks off a second every second. In the frame of a passing muon, the interval between each tick of your watch might be fifty seconds, say, while in another frame it might be nine seconds, or three years, or any other greater value depending on the relative speed of the other frame. Clearly it is impossible for your watch to be affected mechanically, in or any other way, to account for the fact that the duration of its ticks can take an arbitrary value depending on the frame of reference.

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