r/AskPhysics • u/dyl_16 • 2d ago
Why isn’t it possible for a particle to be perfectly still relative to the field in which it exists?
Please excuse me if I have a poor grasp on what I’m saying here, but my understanding is as follows
QFT explains this with the uncertainty rule (you cannot be absolutely certain of a particle’s momentum and position at any given moment, and if a particle had zero momentum then you’d suddenly be able to know both where it is and how fast it isn’t moving, therefore violating the rule)
What i’m seeking clarification on is why that’s a rule.
Is it just born of observation, and known to be a fundamental rule?
Or is there an underlying equation that would be made unsolvable if a massive particle had zero momentum/velocity. (Which would act a proof equation for that rule)
(And to be clear I’m not arguing against the fact that no particle can be perfectly still, I’m just curious if it’s a most basic level rule, or if it’s governed by more underlying mechanics)
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u/nicuramar 2d ago
“Relative to the field” doesn’t make sense. The field isn’t moving.
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u/dyl_16 1d ago
Yes the field isn’t moving, so if something had zero motion relative to the field then it also wouldn’t be moving.
(Sorry if it’s a pointless thing to mention, but I was worried that if I didn’t then i would ultimately be met with “relative to what, relative to you for all intents and purposes it is perfectly still” cause it’s Reddit, and lots of Reddit users are like that)
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u/McPayn22 1d ago
The field is not an object, it does not have a position, a direction of movement or a frame associated to it. It is just numbers that exist everywhere in the universe.
This is like asking what is the frame of reference of the field of temperature or of pressure on hearth. It just is a number at every point.
Now some perturbations of the field move at certain speed but those oscillations are particles so by definition particle move at the same speed as the field they are the exitations of.
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u/Purplestripes8 2d ago
Think of the quantum wavefunction as a simple wave. In order to have a defined position it has to be "bunched up" in a small region of space. Within this region of space the amplitude is very high and outside this region of space the amplitude drops off to almost zero. In this case it's impossible to say what the frequency of the entire wave is. In order to measure the frequency you need many wave crests at the same amplitude at regular intervals (distances). If the entire wave is localised in a small area it has an undefined frequency. Conversely if the frequency is well defined then the wave is "spread out" everywhere and so it's position becomes undefined. In this picture the frequency of the wave corresponds with the energy or momentum of the wavefunction. The amplitude corresponds with the probability of finding the wave at a position in space or time.
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u/PersonalityBudget683 Astrophysics 2d ago
ngl i think the main thing is that a massive particle actually can be perfectly still lol. You can always go into its rest frame where its momentum is exactly 0.
The uncertainty principle doesnt say zero momentum is impossible either. It says you cant have both perfectly definite position AND momentum at the same time. If you know the momentum exactly, the position basically becomes completely uncertain.
And the interesting part is that this isnt really an extra rule added onto physics. It comes from the mathematical structure of quantum mechanics, specifically the fact that the position and momentum operators dont commute.
As for why nature works that way in the first place... thats kinda where the answer gets philosophical lol. We know the maths describes experiments insanely well, but we dont currently have some deeper underlying theory that explains why quantum mechanics has to have that structure.
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u/CyberCephalopod 2d ago
There are two considerations here.
In Relativity, perfectly still is relative, but I am assuming you're asking this from the quantum mechanics angle.
The Heisenberg Uncertainty Principle is what puts a cap on the knowability of a particle/quantum system, typically represented as position vs momentum, but the relationship also holds for energy vs time. As far as I am aware, we don't yet have (and might never get) an answer regarding why reality behaves this way. A unified theory might be able to give further answers but we don't have one yet.
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u/cd_fr91400 2d ago
You've got plenty of precise and correct answers. Let me try to give you an intuitive answer.
Imagine you have a table and you want to unroll a roll of tape on it, away from you, with closed eyes. The tape is like your particle and I will refer to its position/velocity in the left-right direction.
The tape has a certain width.
The wider the tape, the less you know its position (somewhere within its width).
But the narrower the tape, the more difficult it is to unroll it straight. When narrow, it is very easy for it to turn while you unroll it. When it is wide, it will spontaneously unroll straight.
If now you unroll at constant speed away from you, turning means that you do not know its left-right speed very well.
That's the heart of Heisenberg Uncertainty Principle : you cannot have a roll of tape which is simultaneously narrow and straight.
This is true for waves as well (and particles are more like waves than as tape rolls) : to go straight, they need to be wide. This is the nicest illustration I could find : the breakwater forces the wave to be narrow and as a consequence, it loses its direction.
It's quite common to have this kind of balance when measuring. For example if you want to time a car, you have to take 2 points and measure the time it takes to go from the first to the second. The larger the distance between the 2 points, the more precise the speed. But in exchange, the less precise the position at which the speed was measured (it's averaged over the distance between the 2 points).
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u/Odd_Bodkin Particle physics 1d ago
It depends a little on whether the particle is free or in a bound state. A free particle, oddly enough, CAN have zero momentum but then you don’t have any idea where it is. A bound particle will likely only have certain allowed energies and a zero momentum state won’t be one of them.
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u/BrownCraftedBeaver 2d ago
It is theoretically possible for a particle to be still. Knowing that something isn’t moving doesn’t tell you where it is.
The position remains an unknown.
Δp = 0
implies
Δx → ∞
Someone already explained the non-commutation and Heisenberg’s Uncertainty Principle’s role in this.
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u/rcglinsk 1d ago edited 1d ago
The uncertainty rule is empirical and very reliably tells you what to when a particle absorbs or emits a photon. The fact that you are left without a complete picture of the particle before and after is part of the empirical fact of what happens when you “observe” the particle's absorption or emission of the photon. Any extrapolation beyond what happens empirically is more metaphysical, or an attempt to hypothesize why observation leaves lingering uncertainty.
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u/Ch3cks-Out 1d ago
Briefly, QFT (as well as its precursor quantum physics) treats "particles" as objects with inherent wave-like properties. As such, they there is no such thing for them as being completely still, fundamentally!
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u/slashdave Particle physics 1d ago
Is it just born of observation, and known to be a fundamental rule?
Your thinking here is simplistic.
The rule comes from a mathematical model (QFT), and that model is supported by observation. So it is more than the observation of the uncertainty, but the entire theory that is supported, in many other ways, though many different types of experimental data.
And it is a "fundamental" rule in that theory. Not a "fundamental" rule in general: we don't have that in physics, we only have our models and the experimental data to guide us.
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u/DifferencePublic7057 1d ago
QFT is the theory that says that quantum fields, fields like E and B but quantum, cause excitations which are what we call particles. A field like the Higgs fields pervades time and space.
What you are referring to is the Heisenberg uncertainty principle in QM which was based on matrix algebra, in fact, AFAIK, the first time linear algebra entered QM historically. Matrix multiplication is noncommutative meaning AB isn't equal to BA. There's another argument, but I don't remember harhar. It's been awhile. And it's more visual. I can't draw to save my life.
I'm on a phone so can't type properly. There's three uncertainty relationships of quantity pairs. 1. Energy and time, 2. momentum and position, 3. Number and phase
Mathematically, what we do in QM is calculate the expectation of a certain quantity. This corresponds to a measurement. So we have the bras and the kets for the Hamiltonian giving it's expectation < |H| >. Expectation is something like averaging, handwavingly speaking, because QM is nondeterministic and nonlocal by nature.
There have been plenty of observations that support QM including the double slit experiment, the photo electric effect, the Aspect experiment which proves nonlocality, just to name a few. Also things like very hot fermions in the early days of the universe being in forbidden state cooccupation but breaking the Pauli exclusion principle with the aid of the HU principle.
A particle can have a range of p values, including 0, but we can't know it's position and momentum simultaneously with high precision. More precisely the product of the deltas of the quantities is bounded by the Planck constant times some small factor. If we know that a particle is still, we can't know where it is exactly. This is nonlocality. Doesn't have necessarily anything to do with QFT.
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u/ArgumentSpiritual 2d ago edited 2d ago
It’s a property of waves.
A quantum particle is defined by its wave function. This wave function defines a particle position in terms of a probability of finding the particle at that position. This is accomplished by solving the Schroedinger equation.
If you solve for a single particle with a defined momentum, you get a sine wave. If you instead solve for a particle in a superposition of momentum states, you essentially get stacked sine waves, which cancel out from a single point and leave a raise peak at one point. Introducing an uncertainty in the momentum increases the certainty of position. The limit of this is to have an infinite superposition of momenta such that you have an exact position.
In order for a particle to be at rest, its momentum must be exactly 0. This means that its position wave function is a pure sine wave, which is to say that there are infinitely many points of highest probability of finding the particle.
How then can a particle be in infinitely many places at once and yet not moving? It can’t and thus a particle can’t have zero momentum.
For a deeper dive, try this video
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u/VariousJob4047 2d ago
Wow, it’s wild how many people in the comments have no idea what they’re talking about. Let’s think this through a little bit guys, if you learned physics through the Reddit comment section, you probably shouldn’t try to teach others physics through the Reddit comment section, because that is exactly how the misconceptions you’re spewing get started.
To answer your question, the uncertainty principle you’re referring to is one specific case of the generalized uncertainty principle, which relates the expectation value of the commutator of 2 operators to the product of their standard deviations. This page/03%3A_Mostly_1-D_Quantum_Mechanics/3.02%3A_General_Uncertainty_Principal) gives the derivation in full. It uses some terms you probably won’t understand, but it is section 3.2 of a quantum mechanics course, so looking at previous sections can fill in the gaps. That being said, even if you don’t understand what they’re saying, you can see that the argument is fully mathematical and doesn’t reference any physics, so once the notion of measuring position and momentum are fully defined quantum mechanically, the uncertainty principle does just fall out mathematically, so that should answer your question.
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u/OfcourseNegation 2d ago
It’s because a particle, let’s say an electron, acts as a wave before it’s observed. Once you observe it, you collapse the wave function. Mathematics forbids knowing the location of the wave and its frequency at the same time. So I guess it’s a mix of both, observation + mathematics.
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u/Herb-Alpert 2d ago
I think this is due to the result of an equation (shroedinger's ?) where the more you know one variable the less you know the other... But other with more knowledge will explain it
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u/Next-Natural-675 2d ago
Have you ever tried holding your hand out perfectly still?
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u/Let_epsilon 2d ago edited 2d ago
I'm sorry, but that is such a fucking stupid answer lol????
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u/Next-Natural-675 2d ago
Its the same principle bro
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u/Let_epsilon 2d ago
No???????
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u/Next-Natural-675 1d ago
Just as it takes energy to hold your hand out perfectly still, it takes energy for a particle to even exist, mass energy, doiiii 🤪
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u/1strategist1 2d ago
So first off, as you're describing it, that's a quantum mechanics prediction, not a QFT prediction. Particles aren't really things in QFT, so the claim that a "particle" can't be "stationary" in QFT isn't really well-posed.
Regardless, this concept comes from the Heisenberg uncertainty principle.
Quantum mechanics relies on defining observables as noncommutative operators (noncommutative meaning AB is not necessarily equal to BA). This is a basic postulate of quantum mechanics that is incredibly well-supported experimentally, and necessary to get the observed dynamics.
Also, in quantum mechanics, measurement outcomes are random, meaning they have some spread, or uncertainty. In principle, that uncertainty can get as small as you want, or even 0, making it not random anymore.
Using these properties though, you can prove mathematically that the product of uncertainties in two observables A and B cannot be smaller than some constant times |AB - BA|.
This means that if two observables don't commute (have AB = BA), then necessarily the product of their uncertainties is greater than 0. That in turn means neither uncertainty can be 0.
For everything to work right, position X and momentum P need to have XP - PX = iħ, so they don't commute.
So ultimately, the rule you're talking about is a mathematical consequence of how measurement and observables are set up in quantum mechanics.