Is there an agreed upon/conjectured theoretical limit to our ability to resolve the accuracy of time?
2 Answers
The smallest resolvable time can be related to the smallest resolvable length using the maximum possible speed, the speed of light $c$. The ability to resolve a time $\Delta t$ should be theoretically equivalent to the ability to resolve a distance $\Delta x = c\,\Delta t$, by timing how long light takes to cross the distance in question.
In quantum physics, time and energy are related to each other by the Schrödinger equation $$ i\hbar\frac{\partial}{\partial t}\psi=H\psi, \tag{1} $$ where the operator $H$ represents the total-energy observable, and $\psi$ is a time-dependent state-vector in the Hilbert space.$^{[1]}$ Equation (1) implies that in order to resolve a time $\Delta t$, we need the state $\psi$ to involve a spread $\Delta E$ of energies at least as large as $\Delta E\sim\hbar/\Delta t$. (In contrast, a state with a sharply-defined energy would be a "stationary state" that does not change in time, so it would not provide any time-resolution at all.) And if the spread of energies is $\Delta E$, then a typical energy $E$ will also be of order $E\sim \Delta E$. So, to resolve a time $\Delta t$, we need to use energies of order $E\sim \hbar/\Delta t$.
This implies that to resolve a distance $\Delta x$, say by timing how long light takes to cross it, we need to use energies of order $$ E\sim \frac{\hbar}{\Delta t} \sim \frac{\hbar c}{\Delta x}. \tag{2} $$ Having that much energy far away from where the measurement is occurring wouldn't do any good. We need that much energy where the measurement is occurring, so we need something like that much energy crammed into a distance like $\Delta x$.
According to general relativity, energy makes gravity (loosely speaking). Normally, we don't use enough energy in our measurements to create a significant amount of gravity. But the question here asks for a theoretical limit, so we shouldn't assume that gravity will be negligible. If we cram an energy $E$ into a space of width $$ \Delta x\sim \frac{GE}{c^4}, \tag{3} $$ we get a black hole of that width. ($G$ is Newtons' constant, and (3) is the Schwarzschild radius of a black hole of mass $M=E/c^2$, ignoring a factor of $2$ because this is only meant to be a rough estimate.) If we try to cram more energy into that space, we defeat our own purpose: we get a bigger black hole, not a smaller one. Since we can't see inside a black hole,$^{[2]}$ this effectively represents the smallest distance that we can resolve, and therefore (via $\Delta x=c\,\Delta t$) also the smallest time.
What is this smallest resolvable time, quantitatively? Solving equation (3) for $E$ and equating it to the expression for $E$ in equation (2) gives $$ \frac{c^4}{G}\Delta x\sim \frac{\hbar c}{\Delta x}, \tag{4} $$ which implies $$ \Delta x\sim \sqrt{\hbar G/c^3}\sim 10^{-35}\text{ meter}. \tag{5} $$ This is the Planck length. If this is the smallest resolvable length, then the smallest resolvable time is $$ \Delta t = \frac{\Delta x}{c}\sim 10^{-44}\text{ second}. \tag{6} $$ This is the Planck time. The amount of energy required to resolve that time is $$ E\sim \frac{\hbar}{\Delta t}\sim 10^9\text{ Joules}. \tag{7} $$ This is the Planck energy. By human standards, this is not an especially mind-boggling amount of energy: a 10 000 kg mass dropped from a height of 10 km would hit the earth with roughly that amount of kinetic energy (ignoring atmospheric effects). However, cramming that much energy into a space of size $10^{-35}$ meters would be quite an achievement, one that is far beyond our current capabilities. So while (6) may represent the smallest resolvable time theoretically, this is not a limit that we'll be encountering in the near future.
As emphasized in footnote [2], this doesn't mean that time or space is "discrete", with the Planck time or Planck length representing some kind of minimum step-size. On the contrary, we have strong (theoretical) indications that the very concepts of time and space are mere approximations that break down when we try to approach those tiny scales. Exactly how they break down, and what takes their place, is one of the big questions that drives quantum-gravity research, with the AdS/CFT correspondence being one of the most fascinating insights so far.
Footnotes:
$^{[1]}$ Equation (1) is most familiar from non-relativistic single-particle quantum mechanics, but it is also valid in relativistic quantum field theory, with a more complicated $H$ and with a state $\psi$ that depends on many — nominally infinitely many — abstract variables, none of which are directly associated with "particles." Those details are beside the point; the point is simply that equation (1) is a legitimate starting point for the present analysis.
$^{[2]}$ The holographic principle suggests that the usual concept of space (even curved space) breaks down in a black hole, leading to weird non-local effects. When our attempts to resolve a small distance involve enough energy to make a black hole, the idea isn't that smaller distances can't be resolved, but that smaller distances don't even make sense.
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1$\begingroup$ And if the spread of energies is $\Delta E$, then a typical energy $E$ will also be of order $E∼\Delta E$ What assumptions go into this statement? Certainly there exist distributions where the mean is not on the same order as the standard deviation. $\endgroup$ Commented Mar 17, 2019 at 5:36
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1$\begingroup$ Actually shouldnt be the energy where problems start be ard 1MeV (pair production)? $\endgroup$– lalalaCommented Mar 17, 2019 at 9:13
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$\begingroup$ @AaronStevens You're right. We could have a distribution where $E$ is concentrated near zero and (in equal proportion) near $2\times\Delta E$, but then we'd be relying on energies $E\sim 2\times\Delta E$, which is only a factor of $2$ off. We can make that factor bigger by shifting the balance of the low and high ends of the distribution, but such unnatural distributions (even if balanced) probably don't correspond to any usable physical configuration of equipment... probably. Yeah, that's a hole in my argument, and so is the fact that I'm extrapolating far beyond what's ever been done. :) $\endgroup$ Commented Mar 17, 2019 at 13:37
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$\begingroup$ @DanYand Fair enough. Thanks for the honesty :) still gave +1 though $\endgroup$ Commented Mar 17, 2019 at 13:45
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1$\begingroup$ @hisairnessag3 If we can trust these extreme extrapolations of our current understanding, then yes: trying to improve the resolution further by adding more energy would just give us a bigger black hole, not better resolution. $\endgroup$ Commented May 22, 2019 at 23:42
" When millisecond pulsars were first discovered, they rivaled the stability of atomic clocks. Today, atomic clocks have surpassed pulsars in accuracy. But if you were to compare them over a longer period of time - say, decades - pulsars can be just as good" (http://www.bbc.com/earth/story/20150805-the-most-accurate-clocks-in-space)
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$\begingroup$ I edited the question per Ben's comment but you are welcome to reanswer the question $\endgroup$ Commented Mar 16, 2019 at 17:40