Is Time Defined by Relative Motion in SR and GR?

[Islam Hassan]

We only measure time via the relative movement of a standard, agreed reference, a clock for example.

If we replaced all instances of "time" in the formulation of SR and GR by "relative motion [ie, of an agreed reference]", would SR/GR remain correct/consistent? Or does SR/GR rely on a deeper, more intrinsic notion of time.IH

 

[Ibix]

We only measure time via the relative movement of a standard, agreed reference, a clock for example.

I don't think this is generally true - atomic clocks depend on electron energy levels, not motion.

If we replaced all instances of "time" in the formulation of SR and GR by "relative motion [ie, of an agreed reference]", would SR/GR remain correct/consistent?

I don't think that makes any sense.

 

[PeroK]

We only measure time via the relative movement of a standard, agreed reference, a clock for example.

You can measure time by radioactive decay. Or, by the evolution of natural processes more generally. Not just by movement.

 

[Dale]

Or does SR/GR rely on a deeper, more intrinsic notion of time.

There are two notions of time in SR/GR. One is “coordinate time”, it is just the time coordinate, usually denoted $t$, in whatever arbitrary coordinate system you have chosen.

The deeper more intrinsic notion is “proper time”. That is the physical time measured by a clock, usually denoted $\tau$. As mentioned above, not all clocks are based on motion, they can be based on any dynamic physical process.

One thing that is often missed by people philosophizing about time is the distinction between a measurement and a measurand. The measurand is the thing being measured. The measurement is the use of some device or experiment to determine the value of the measurand. So in the statement “time is the thing measured by a clock” we are identifying time as the measurand of a clock. The use of a clock to detect time is a measurement and a clock is the measurement device. So time is the thing that a clock measures, time is not a clock.

Mass is the thing measured by a scale, mass is not a scale. Length is the thing measured by a ruler, length is not a ruler.

For some reason people seem to understand the difference between the measurand and the measurement for most things. But they suddenly get weird philosophical concepts for time.

 

[vanhees71]

The reason is that time is not an observable in the strict sense, although this becomes manifest only at the level of quantum theory. This is no issue as long as you forget about the gravitational interactions. Then everything is described by a relativistic quantum field theory in Minkowski spacetime. Time is not an observable in any quantum theory and so also in relativistic QFT. Now space and time are pretty much related in relativistic physics, because of the structure of Minkowski spacetime and its symmetry group, the proper orthochronous Poincare group, and that's why on the foundational level, where you construct the quantum theory from the representation theory of this Poincare group, you start with both space and time coordinates as parameters and use the description as a quantum-field theory, where the primary "objects" are fields and local observables are distributions of energy, momentum, angular momentum, various charge and current density and so on, i.e., observables which are defined to have some value at a given spacetime point. The spacetime point is given by the corresponding coordinates due to an (for simplicity inertial) reference frame, which are the arguments of the quantum fields, and the local observable operators built from them.

 

[Mister T]

Or does SR/GR rely on a deeper, more intrinsic notion of time.

Deeper or more intrinsic than what? Than the notion that time is the thing we measure with a clock? No, I don't think so.

 

[FactChecker]

Deeper or more intrinsic than what? Than the notion that time is the thing we measure with a clock? No, I don't think so.

Although "clock" can have a very broad definition.

 

[Islam Hassan]

You can measure time by radioactive decay. Or, by the evolution of natural processes more generally. Not just by movement.

Yes that is understood, and both those processes at the microscopic level involve movement, of sub-atomic particles or anything else.

I did not mean movement in a macroscopic sense only.

 

[Islam Hassan]

Deeper or more intrinsic than what? Than the notion that time is the thing we measure with a clock? No, I don't think so.

So time is simply relative movement. Unlike gravitational mass vs inertial mass, there is no stand-alone, inertial time measure.

 

[Dale]

Yes that is understood, and both those processes at the microscopic level involve movement, of sub-atomic particles or anything else.

I did not mean movement in a macroscopic sense only.

In principle, a radioactive element can still decay even when it is stationary. And in practice, in many atomic clocks any atomic motion degrades the time keeping.

No. Ubiquitously equating time with motion is incorrect, both in principle and in practice.

 

[Mister T]

So time is simply relative movement.

No. I don't know how you got that idea from what I wrote.

Unlike gravitational mass vs inertial mass, there is no stand-alone, inertial time measure.

This sentence makes no sense to me.

 

[vanhees71]

In principle, a radioactive element can still decay even when it is in a stationary state. And in practice, in many atomic clocks any atomic motion degrades the time keeping.

This is a bit misleading. So let's see, how radioactive decay is described within QT. Let's take for simplicity the $\beta$ decay of an atom, which in fact on the most fundamental basis according to our present best knowledge, which is the Standard Model of elementary-particle physics, is the process that a down-quark within a neutron in the nucleus decays to an up-quark, an electron, and an anti-electron-neutrino.

Since thus $\beta$ decay is a process where particles are annihilated (the down-quark) and new particles are created (the up-quark, electron, and anti-electron neutrino).

Now this process is described in perturbation theory, with the weak interaction of the quarks and leptons treated as a small correction to the overwhelmingly much stronger strong nuclear force. So first one considers the nucleus, as if there were only the strong interaction. It's a very complicated bound state of quarks and gluons, i.e., it's a very complicated eigenstate of the QCD Hamiltonian. This implies that the nucleus in this 0th-order approximation, i.e., neglecting the weak interaction completely, were stable, because it's in an eigenstate of the Hamiltonian (of only the strong interaction).

Now, if you consider the full thing, i.e., if you add the weak interaction into the treatment, it's clear that the QCD-energy eigenstate describing the nucleus without taking into account the weak interaction is no longer an eigenstate of the full Hamiltonian, including weak interactions. This means that due to the presence of the weak interaction there's indeed some probability that the $\beta$-decay occurs, because the weak interaction couples the down quark to the up-quark, electron and anti-electron neutrino. That the corresponding decay can only happen if the down-quark is bound in a neutron and not in a proton is, because the neutron is a bit heavier than the proton, and thus the $\beta$ decay is forbidden, if the down-quark is in a proton, because of energy conservation, but for the down-quark within the neutron there's nothing (in fact no conservation law) which forbids this decay due to the weak interaction, and whatever is an allowed process in a theory, this theory predicts that this process will indeed happen spontaneously and randomly, i.e., we cannot predict, when a given nucleus will $\beta$-decay, but quantum-field theory predicts the probability rate for such an decay to occur. In this way you can derive the mean lifetime (or equivalently the half-life) of the nucleus, applying the Wigner-Weisskopf approximation, which leads to the well-known exponential-decay law.

 

[Dale]

Let's take for simplicity the $\beta$ decay of an atom, ... It's a very complicated bound state of quarks and gluons

Let's take for simplicity an actually simple system, like a muon. A stationary muon still decays.

 

[PeroK]

Yes that is understood, and both those processes at the microscopic level involve movement, of sub-atomic particles or anything else.

Radioactive decay, in terms of its frequency, does not involve movement. It's a rate of change from one state to another. You're not directly measuring the movement of something.

Also, if you want to say that "all physical processes involve motion of some description", then it doesn't say anything about time, per se, to say that all measurements of time involve movement. That's just a corollary of your main hypothesis. Then all measurements of distance involve motion. Or, all measurements of energy involve motion. Or, all measurements of electric charge involve motion.

 

[PeterDonis]

A stationary muon still decays.

By "stationary" do you mean "not moving"? As in, it's in a state with zero expectation value of momentum?

If so, I think you are simply using that term in a different way from @vanhees71. He is using the term to mean "in an eigenstate of the full Hamiltonian". A muon that decays cannot be in such an eigenstate, because if it were it could not decay; an actual eigenstate of the full Hamiltonian can never change.

 

[Dale]

By "stationary" do you mean "not moving"? As in, it's in a state with zero expectation value of momentum?

Yes.

He is using the term to mean "in an eigenstate of the full Hamiltonian". A muon that decays cannot be in such an eigenstate, because if it were it could not decay;

Would such a thing even be a muon? It may be that is what he means by "stationary", but that is not what I had intended.

 

[vanhees71]

Let's take for simplicity an actually simple system, like a muon. A stationary muon still decays.

That's of course the very same argument. The difference is that you can really calculate this in electroweak perturbation theory. The muon were stable, if you neglect the weak interaction. A single free muon (or a muon at presence of only the electromagnetic but not the weak interaction) would be stable and in an eigenstate of the Hamiltonian, not including the weak interaction. But the weak interaction couples the muon to the electron, the electron-anti-neutrino, and the muon-neutrino. Since there's no conservation law forbidding the corresponding decay, $\mu^{-} \rightarrow e^- + \bar{\nu}_e + \nu_{\mu}$, this decay really happens, and you can calculate the mean lifetime (or the decay constant, which is its inverse) from perturbation theory. Again, the muon is not stationary, because a "free muon" is not an energy eigenstate of the full Hamiltonian, including the weak interaction.

That's of course only understandable easily in the QFT description, we are now used to. There was a long-standing problem and debate about the $\beta$-decay, because in the early days of QM creation and annihilation processes were not considered (with QFT being neglected as "overdoing things" for a few years although it was formulated already in the very first papers on modern QT by Jordan, Born, and Heisenberg). In addition the neutron hasn't been discovered until 1932, and initially physicists thought that an atomic nucleus consisted of bound protons and electrons (different from the electrons around the nucleus forming a neutral atom due to the binding through the electromagnetic interaction). This was however pretty early ruled out because of the continuous energy spectrum of $\beta$-electrons. There was also some confusion in this respect, because there were also discrete lines of electrons, but as we know today, these have nothing to do with $\beta$-decay but are due to the interaction of $\gamma$'s with the "atomic electrons". One important breakthrough was the discovery of "RaE" (Bi-210), which didn't show any discrete electron energies, because there's no $\gamma$ decay in addition to the $\beta$ (and $\alpha$) decay.

Famously the resolution came with (a) the discovery of the neutron in 1932 and (b) the assumption of the existence of a particle we know call neutrino (Pauli 1930), which together lead to Fermi's theory of $\beta$ decay in what we nowadays call an effective theory of weak interactions (1934).

 

[vanhees71]

Yes.

Would such a thing even be a muon? It may be that is what he means by "stationary", but that is not what I had intended.

Well, if you are very pedantic, you'd not call a muon a particle to begin with, because it's not stable, i.e., there's no asymptotic state, because it sooner or later decays. In this sense it's a resonance.

 

[Dale]

The muon were stable, if you neglect the weak interaction.

Why would we neglect the weak interaction? A muon without the weak interaction is not a muon.

Well, if you are very pedantic, you'd not call a muon a particle to begin with, because it's not stable, i.e., there's no asymptotic state, because it sooner or later decays. In this sense it's a resonance.

A muon is literally one of the fundamental particles of the standard model. @vanhees71 I am not going to argue further with you here. If you want to make your bizarre points somewhere then do it in a separate thread.

 

[PeterDonis]

Would such a thing even be a muon?

"Muon" is a label, heuristically, for an eigenstate of a partial Hamiltonian, one which ignores the interaction terms that cause muons to decay. So yes, the thing that decays is a muon until it decays.

Similar remarks would apply to any particle; there are always some interaction terms in the full Hamiltonian that can change it into something else, but when we name the particle and refer to it by that name we are ignoring those interactions.

Quantum field theory largely avoids this entire issue by using the Lagrangian instead of the Hamiltonian and labeling "particle" states based on perturbation theory and asymptotic free states--in other words, not ignoring the interaction but simply looking at a region of spacetime in which it isn't taking place.

 

[PeroK]

I can't see why this particular thread needed to digress into the subtleties of quantum theory. The thread is under Special and General Relativity, where it ought to be sufficient to know that radioactive decay occurs with a frequency that can be used for timekeeping. And, it is not the measure of a particle's motion in the classical sense.

 

[PeterDonis]

The thread is under Special and General Relativity, where it ought to be sufficient to know that radioactive decay occurs with a frequency that can be used for timekeeping. And, it is not the measure of a particle's motion in the classical sense.

This is true at the classical level, but the OP asserted that, while the above is true, radioactive decay still involves motion at a microscopic level. Refuting that claim requires quantum theory.

 

[Vanadium 50]

This thread seems to suffer from a lot of issues:

• Is it a personal theory? No, because it isn't even a theory. But it's close to one, and it is certainly true that the OP promotes it with the zeal and vigor also shown by personal theory proposers.
• The concept seems to be "since everything moves, everything - including time - is motion". Deep man, deep.
• It seems to rely on the idea that the way one creates a physics theory is to arrange the words in the right order, not to use mathematics. This is of course not true?

Is time relative motion? No. Two different concepts. Even the dictionary says so. Beating a dead horse won't change this.

 

[berkeman]

The concept seems to be "since everything moves, everything - including time - is motion". Deep man, deep.

Using my Mentor superpowers, I can share the OP's previous avatar before they changed it...

https://www.shutterstock.com/image-vector/vector-cartoon-illustration-man-running-clock-1285210432

 

[berkeman]

And since the OP's question has been well and truly answered, it's about time to close this thread.

Thanks to all who helped the OP.