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ur-whale · 6 years ago
I wish a physicist could explain here how the very notion of "diameter" has any meaning for an object whose size (IIUC) belong entirely to the quantum realm.

Is the hydrogen atom two hard little balls of matter orbiting one another, as we were taught in primary school, or are they a probabilistic soup with various, vaguely localized extrema?

If the latter, how do you even define the notion of diameter?

lisper · 6 years ago
That is an excellent question!

The definition is somewhat arbitrary, but still has some real physical significance. In actual fact, a proton is a field, so it doesn't have sharp boundaries. But the amplitude of the field still dies off very rapidly with distance from the center, so you can pick some arbitrary small value and say "the point at which the amplitude becomes less than this value is the radius of the proton". What matters is not really the number that you get out of this, but the fact that the experimental results of measuring this value appeared to change in the presence of muons. This was a phenomenon that was not predicted by present theory, and if it had held up, would have been a major breakthrough. One of the biggest problems in physics right now is that there are no experiments (except possibly this one) whose results are at odds with the Standard Model. That makes it hard to improve the model!

jemorya · 6 years ago
There are some really good responses to lisper's post: what about 96% of the universe's mass, neutrino mass, what about gravity for that matter. (zing!)

The thing about those particular questions is that they only tell us that the standard model is incomplete. Gravity exists. The fact that it's not in the standard model doesn't necessarily mean that the model is broken, just that gravity needs to be added somehow.

What we need more of are instances where the standard model makes a precise numeric prediction and it's dead wrong. That puts a spotlight on every piece of the standard model that went into the prediction.

(Edited for the pun I didn't intend.)

pdonis · 6 years ago
The notion of "size" being referred to here is the charge radius of the proton:

https://en.wikipedia.org/wiki/Charge_radius

It is one of several distinct possible notions of "size" for a particle.

yyyk · 6 years ago
"One of the biggest problems in physics right now is that there are no experiments (except possibly this one) whose results are at odds with the Standard Model."

What about Neutrino masses? Or the Muon 'anomalous' magnetic moment?

willis936 · 6 years ago
Does the magnitude of the field wave drop at some fixed shape (like an exponential)?

Is the wavelength fixed? Is it in meters?

Is it the field of a proton made up of multiple frequencies/modes?

Is a proton’s influence on fields extend to all space or is there a point (within the hubble sphere) at which a proton does not influence fields at all?

moonbug · 6 years ago
except for an explanation of 96% of the universe's mass
DoctorOetker · 6 years ago
> One of the biggest problems in physics right now is that there are no experiments (except possibly this one) whose results are at odds with the Standard Model.

If we say that new physics comes from inconsistencies (either between theory and experiment, or simply within a theory), then there is plenty of new physics to be done, for example the pretense that an electron is a fundamental particle results in the inconsistency with predicted infinite mass. (I claim to have a solution, but I am sitting on it because I believe with a bunch more effort it can explain the dimensionless number that relates planck charge with electron charge, essentially explaining planck's constant and QM in quasiclassical terms...)

jpmattia · 6 years ago
> I wish a physicist could explain here how the very notion of "diameter" has any meaning for an object whose size (IIUC) belong entirely to the quantum realm.

Usually these are describing expectation values of radial positions.

For the well-understood example of the hydrogen electronic orbit, the Bohr radius is the expectation value of the radial position of the electron. The electron has a wavefunction Ψ; When you calculate the expectation value of the radius r using Ψ (= \int_0^\infinity Ψ* r Ψ dr ) you find a value of about 0.5 nm.

People will drop the subtlety of the "expectation value" and just say "the hydrogen atom has a radius of about 0.5 nm".

The issue for the proton radius that the article did not delve into: There is currently more than one way to measure the proton radius. One is scattering (pitch another particle at the proton, look at how it "bounces" off), and another is spectroscopically (look at the energy levels of the electron, deduce the proton radius from the interaction of the electron orbit and the proton.) These methods do not give the same answer.

whatshisface · 6 years ago
Usually the radius of something like a proton or an atom takes the form of a parameter in an equation that just-so-happens to be measured in a unit of length. It's not defined specifically to be an indication of the extent of the particle, it just happens to roughly be about the right size.

This paper[0] says that it is defined in terms of a "probability amplitude that an interaction between a photon of four-momentum q^μ (Q2=−q2) and a charged constituent of the proton can absorb such a momentum with the proton remaining in its ground state." So, not exactly a radius, but "radius" is an okay conceptual mnemonic.

[0] https://arxiv.org/pdf/1812.02714.pdf

doubleunplussed · 6 years ago
Other answers are either overcomplicated or imprecise: the proton's radius is defined as the RMS radius of its charge distribution. It's the same idea as describing the size of a probability distribution by its standard deviation.
jxdxbx · 6 years ago
One point that is so basic that physicists often forget to explain it: leptons and quarks don't really have a "size," they are considered point particles. A proton is a configuration of three point particles though whether it makes sense to say that this configuration has some spatial structure is beyond me.

Also, mass has nothing to do with size, or "stuff"-ness. A proton has about 100 times more mass than its quarks (from binding energy), but an atom has less mass than its protons and electrons added up (from the loss of potential energy). A top quark has about as much mass as an atom of gold, but again is just a point particle. And photons have no mass but the energy of photons bouncing around in a confined space does have mass.

thatcherc · 6 years ago
Depends on the particle! Electrons are considered to be point particles with an infinitesimal radius. A proton, however, is made of three bound quarks, which all whiz around each other within a small but finite region. By bouncing things off the proton, as is done in scattering experiments, a notion of how large that region occupied by the quarks is can be determined.
fdavison · 6 years ago
That reminded me of this video of NDT describing the size of the electron: https://www.youtube.com/watch?v=w6kjY6ppJRE&t=71
jarfil · 6 years ago
"Radius" is a distance at which something stops happening, defining a boundary between some "inside" and some "outside".

Interactions between particles, waves, fields, etc. are quantized, which means there is a set of distances at which a given interaction can happen, and a set of distances at which it can't. This creates a boundary, often spherical, which can be described as having some "radius", and thus some "diameter".

Depending on the interactions you choose, a particle might have multiple diameters, or even boundaries with different topologies, but they're usually somehow related to each other for any given particle.

wnoise · 6 years ago
> Interactions between particles, waves, fields, etc. are quantized,

Yes.

> which means there is a set of distances at which a given interaction can happen, and a set of distances at which it can't.

... that's not how quantization works. The exact ways in which quantum effects are actually discrete is much subtler than in most popularizations. In fact, sharp effects with distance are more likely to be seen in classical models than in quantum ones because the quantum models often allow for classically forbidden effects to happen with small probability, which decreases as the distances increase. You only really see sharp transitions for "bound states"; for everything else (e.g. scattering experiments) there are wider or narrower peaks of more likely to occur depending on both spatial and other parameters.

The radius really is measuring "over about how much space is this particle spread", and while the exact details of how you define that can give different numbers, they are all measuring interaction widths -- how close something has to be to feel its direct effect. I say direct effect, because obviously the indirect effects such as through the EM field can be felt at great distances.

Note that this measure of spread is distinct from the how the wavefunction of the center of a particle is spread, which in the right states can be highly delocalized, even though the particle hasn't gotten any wider. Electron orbitals, for instance, can have different radii in different states (or topology, as you note, if you pick a cutoff that splits high-density regions in two), but the electron still has the same negligible (usually modeled as zero) radius in comparison to any orbital.

DoctorOetker · 6 years ago
Consider a perfectly spherical earth, and a vertical tunnel perforating it through the center, with an elevator in the tunnel. suppose you are halfway towards the center, coming from the surface, and you pause the elevator. How strong will the gravitational field be? Can we pretend all of Earth's mass is concentrated in the center? No, let's see why.

Consider a perfectly spherical hollow shell of mass, and an observer inside the shell, will she be attracted to the center of mass of the shell? No, let's see why.

We know that gravitational field falls of with distance squared ( 1 / r^2 ). Consider a random observer position, and a random direction. Now consider a cone tipped at the observer with the random direction as its axis, and also consider a second cone with the same cone angle but the opposite direction as its axis. So the observer position is where the 2 cone tips meet. Then consider the distance between the observer towards each patch of the spherical shell. The area (and thus mass) of this patch will scale with the distance squared, so if one patch is say 3 times further away than the other, it will be 9 times weaker due to 1 / r^2 gravitational fall off, but also 9 times heavier due to geometric scaling of the patch, so gravity due to the patches will cancel, and since the direction of the axis was arbitrary, the gravity of each part of the shell will be balanced by the gravity of a corresponding opposite part of the shell.

So when you are in an elevator halfway down the radius of the hypothetical spherical Earth, all the layers of Earth above you will cancel gravitationally, and the gravitational field will only be due to the mass of the Earth that is contained in the sphere up to your altitute with respect to the center of the Earth.

Similarily, when the electron is far enough from the (assumed spherical) proton, we can somewhat pretend the proton is a point particle. but when the electron is inside the proton, then the effective charge of the proton will be lower because of all the charge of the proton that is farther from the proton's center than the electron is invisible to the electron (since the electric field also falls of like 1 / r^2 ).

I hope that answers the question?

chasd00 · 6 years ago
forgive my ignorance, but don't you have to assume uniform density in both cases? It seems to me the person in the earth elevator halfway down would feel more gravity toward the center of the earth because there's actually more mass due to the higher density. In a similar way, the proton would have higher mass toward the center because of the higher wave amplitude.
blhack · 6 years ago
This is true for every physical object though. You cannot tell me with perfect certainty the width of a ball of tungsten, because the edges aren’t actually rigidly defined. In fact: it doesn’t really have an “edge” the way you might think of it geometrically. Just a boundary across which forces start interacting with each other.
pbhjpbhj · 6 years ago
Well if we're going to be pedantic why not do it properly - surely there is no force boundary, instead there is a proximity at which the forces become significant. Significance being determined according to the situation being analysed.
calhoun137 · 6 years ago
This is a surprisingly hard question to answer! An orthodox point of view says "diameter" is just a word for a physical quantity which can be measured by experiment and plugged into a formula to predict the result of other experiments. In my personal opinion, theoretical physics is not currently capable of providing a satisfactory answer, but if we knew more about the relationship between quantum field theory and general relativity it would probably be possible to say more.

There are plenty of ways to wave your hands about it, and I have been doing this for a long time and could give 100 different explanations. At the end of the day what do I see in my head when I think about the "diameter" of the proton? Basically a probabilistic soup which is confined to a small area.

The "wave function" of a quantum particle can tell you where it is but is not the same as the particle itself. Standard QM assumes particles are infinitesimal points, in QFT this leads to infinities which can be resolved if you assume particles have a "smeared out" radius. This "smeared out" radius is what the proton diameter should be related to when the more complete quantum and gravity theory is discovered.

WhitneyLand · 6 years ago
It may be simpler to start our not thinking of particles as matter, but instead thinking of them as little energy fields. Like a magnetic field, a particle has properties and interacts with some things and not others. This is why it's a "field theory".

Say you had ten different light bulbs. Could it make sense to talk about their differing light output as "diameters"? It wouldn't be defined in terms of a physically measured length of the bulbs, rather a diameter could be defined to be about properties of a light bulb.

Remember all these words are in some ways arbitrarily chosen by people. If you replaced the word diameter everywhere with "fat content" would all the math still work?

The math would still work fine. The problem would be it's even less understandable than diameter, and it makes it less efficient as an abstraction that you can grok and reason about.

dkersten · 6 years ago
> It may be simpler to start our not thinking of particles as matter, but instead thinking of them as little energy fields.

I dunno about simpler, but it certainly helps explain many of the weird effects we’ve observed in the universe. I recently went down the Quantum Field Theory rabbit hole (thanks to PBS Space Time on YouTube) and it’s an absolutely fascinating topic (even if the maths is beyond me) and QFT especially both makes sense to me and explains a lot of the “problems” with particle physics as I learned it in school.

fsh · 6 years ago
As far as atomic physics is concerned (except for hyperfine structure), the proton is simply a blob of charge which can be to first order described by a position-dependent charge density \rho(r). The RMS charge radius r_p is defined by r_p^2 = (\int r² \rho(r) d³r) / (\int \rho(r) d³r).
davrosthedalek · 6 years ago
Almost, but not quite. The rms radius is defined via the slope of the proton electric form factor G_E. Via a Fourier transform you can connect the form factor to the distribution of charge in some space-like coordinate. That works out to be (very close) to the rest frame space for heavier atoms, because relativistic effects are small. For the proton, this is not true -- the Fourier transform gives a spacial distribution in the so called Brick-wall or Breit-frame, not in the rest frame of the proton. (There are additional complications about waveforms not being the same before and after).
Someone · 6 years ago
https://en.wikipedia.org/wiki/Proton_radius_puzzle#Problem (not yet updated with this result, btw) gives 3 methods.
analog31 · 6 years ago
In addition to the many good answers here, I'd add that whatever parameter represents "diameter" for a proton, probably also represents the classical diameter of something macroscopic like a billiard ball if carried to that extreme.
BurningFrog · 6 years ago
They're a soup.

I suppose you have to pick a probability criteria for the diameter.

robotresearcher · 6 years ago
From the article:

> an electron [...] spends part of its time inside the proton (which is a constellation of elementary particles called quarks and gluons, with a lot of empty space).

mncharity · 6 years ago
Not a physicist, but I spent some time on this, exploring the possibility of a "teach scale down to nuclei, then nuclei up to matter" learning progression for primary school.

Nuclei are quasi-classical little balls. It's electrons that are wacky quantum weirdness.

Which makes the current pedagogical emphasis on electrons, rather nuclei, as a foundation for understanding atoms... perhaps miss some opportunities for greater clarity. And that's even before getting to state standards that require teaching both "atoms are conserved in chemical reactions" and "atoms are electrically neutral, so if charged, they are no longer atoms, but ions". Sigh.

Nucleons overlap (not orbit), but nuclei aren't homogeneous, especially light ones. Light nuclei are well described as clusters of alpha particles, heavier ones as liquid drops. Ground states are generally more or less spherical. Though Neon looks like an old-style wine bottle, with a tetrahedron of alphas in the bowl, and a fifth making a neck. Non-ground states for lights are mostly rearranged alphas, for heavies, variously distorted balls. Fission is droplets stretching and necking apart.

A proton is a mix of odd shapes, but directly measurable properties are spherical. It's just a ball.

Nuclear mass and charge density are, to a good first approximation, colocated. You don't have mass concentrated one place and charge somewhere else. No surprises.

Electrons are a peak with exponential falloff, making choosing a radius quite arbitrary. What's the diameter of a cartoon volcano, that blends seamlessly into plains? Wacky, wacky electrons.

Nuclei have a flat-ish density plateau, then a steep slope, and only then a tiny exponential footlands (Woods–Saxon potential). So it's just a ball with a fuzzy edge, rather than a wacky exponential cloud thing.

Note the narrow range of sizes being discussed. No one is suggesting 10 femtometers, or 0.1 femtometers, or even 1.0 fm or 0.8 fm. "GEOID-Next committee tables discussion and heads to a bar, unable to agree on the shape of the Earth!!! News at 11!!!" But for most everyone, primary school up, that's "a femtometer ball".

A superball can be bounced, and crushed, and burned. You need polymer physics to understand just why it behaves as it does. But you can describe that behavior to a Kindergartener. They don't need polymer physics. Nuclei are just little balls.

Now look at the giant hairy mess of this thread. It's like, we teach "4/3 pi r^3, 4 pi r^2", but not "the volume and area of a ball are half of its box". But dialed way way up. Science education, variously distracted, failing to provide a coherent briefing on the physical world.

If anyone has ideas on how to describe this point, I'd appreciate them, as I've never come up with something nice.

Here's a tiny Neon pic. I just think Neon is cute. https://arxiv.org/abs/1406.2473 page 4. There's a nicer one somewhere...

scriptkiddy · 6 years ago
From my admittedly entry-level understanding, particles like electrons, quarks and gluons are largely considered to be points with energy fields. So, not exactly "stuff", but the stuff that "stuff" is made out of.
conistonwater · 6 years ago
> If the discrepancy was real, meaning protons really shrink in the presence of muons, this would imply unknown physical interactions between protons and muons — a fundamental discovery. Hundreds of papers speculating about the possibility have been written in the near-decade since.

This reminds me of a joke. An experimental physicist walks into a theoretical physicist's office with a really cool experimental result, shows the printed out graph to the theoretician. He thinks for a while and says, "this is perfectly in line with theory, let me explain how". The experimentalist looks at the graph, scratches his head, says "uh, this is upside down", and rotates the paper 180 degrees. The theoretician thinks a while more and says, "well this can also be explained".

WhitneyLand · 6 years ago
The clever thing about these kinds of jokes as when it happens in politics is, they manage to mock two groups of people such that each group reads the joke and tends to laugh because they infer it as poking fun at the other group.
AnthonBerg · 6 years ago
We laugh even if it is rotated by 180 degrees.
jpmattia · 6 years ago
Probably worth noting: The author writes the article as though the latest experiment "solves" the issue.

A more nuanced description (which is not exactly quantamagazine's forte) would note there is a conundrum about the proton size. The article describes a measurement that falls under "spectroscopic methods" in the wiki [1].

Why eg scattering measurements should yield a different value is not at all clear.

[1] https://en.wikipedia.org/wiki/Proton_radius_puzzle

raldi · 6 years ago
This was a great writeup. Accessible, yet I feel like I learned some genuine not-dumbed-down science.
FuckButtons · 6 years ago
Quanta has a lot of good articles like this one, definitely worth looking at every now and then.
fsh · 6 years ago
The article doesn't mention that since 2010 two more measurements on atomic hydrogen that determine the proton size with similar accuracy have been published. The first one (https://science.sciencemag.org/content/358/6359/79) agrees well with the muonic value, while the second one (https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.12...) disagrees with the muonic value and agrees with the old, larger value.
davrosthedalek · 6 years ago
Yes, very true. I think it's a shame that some people just declare victory and ignore the discrepancy in these modern experiments. At the same time, a similar split seems to happen using the scattering method, but the data are not published yet.

The results from Hessels were announced as preliminary results more than a year ago -- it's great that they are finally out peer-reviewed. But the community at large didn't stop back then when the results came out, with more experiments planned all over the world, especially but not only in the scattering sector. We made progress, but the puzzle isn't dead yet. There is actually a conference starting next week on this topic: http://ecsac.ictp.it/ecsac19/

It's a rare occasion where nuclear physics and AMO overlap. Another one is the Zemach radius, which also requires knowledge of the proton magnetic form factor.

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fastaguy88 · 6 years ago
One has to wonder why the original "muon" paper, which produced such controversial results, did not include the control of measuring with the same methods conventional ("electron") hydrogen. Doing the control would have cleared things up a lot sooner.
fsh · 6 years ago
It took Eric Hessel's group eight years of hard work to do the measurement in electronic hydrogen. The experiments are completely different.
dschuetz · 6 years ago
I don't understand how physicists cannot see the obvious: the size of a proton appears not to have a fixed/natural value. It seems to depend on the quantum state of their system, energy levels, and the coupling partner's quantum state, energy levels and mass. "Physicists finally nail..." sounds like a part of a joke: "Physicists finally nail the coffin with the research funds shut"