
How Big Is a Proton? - conse_lad
https://sparkonit.com/2019/11/04/how-big-is-a-proton/
======
symplee
>> The new measurement puts the size of the proton to around 0.833
femtometers, against the previously accepted figure of 0.842 femtometers.

1 femtometer = 10^−15 metres. Also commonly called a fermi.

It's a quick read, worth the click to read how they measured it, and at least
see CERN's imagined rendition of what the "turmoil of quarks and gluons" that
make up a proton might look like.

~~~
arberavdullahu
> The new measurement puts the proton's radius to around 0.833 femtometers,
> which is 4 percent less than the previously accepted figure.

How is 4 percent less equal to 0.833?

~~~
slavik81

      0.96 * 0.842 = 0.808
      1.04 * 0.833 = 0.866
    

Well, that rules out the obvious possibilities.

~~~
jonsen
0.842 is also an experimental result, not the official figure.

~~~
slavik81
I should probably have reread the full section. It's more clear in context
that 0.96 * 0.8768 = 0.842.

------
witherk
I find this somewhat baffling. 4% just seems like such a huge disparity,
especially for such a ubiquitous particle. It seems like any experiments
before 2010, that used the size of a proton in the calculations should have
noticed something was wrong. Fermilab has spent the last couple years trying
spruce up an old experiment to measure the dipole-moment of the muon down to
0.14 ppm[1][2] Apparently that small of a discrepancy in theory and experiment
was interesting enough to throw a huge amount of time into. How is it that we
are trying to verify the 9th significant digit in the dipole moment of an
obscure particle, while something as mundane as the size of the proton was off
by 4%? Obviously, I'm missing something.

[1][https://en.wikipedia.org/wiki/Muon_g-2](https://en.wikipedia.org/wiki/Muon_g-2)

[2][https://www.youtube.com/watch?v=UckuqHDB08I](https://www.youtube.com/watch?v=UckuqHDB08I)

~~~
fsh
The proton is so small that its finite size only shows up in a few purpose-
built experiments such as the one explained in the article. Since the effect
is very small, measuring it to high precision is extremely difficult.

It is also worth noting that protons are much more complicated particles than
muons and their theoretical description is much less well developed.

------
JohnJamesRambo
>When a hydrogen atom is in its lower energy state, the electron not only
orbits around the proton, but rather stays inside it.

This is the first time I have ever learned this is possible in all my years of
schooling and working in science. Fascinating, this article changed how I
think about the universe.

~~~
Yajirobe
You should not think of the electron as occupying some fixed point in space.
It makes more sense to talk about the probability distribution of the
position.

~~~
MisterTea
We have decades of ball and stick models along with solar system like atomic
diagrams to thank for that. And it's compounded by the fact that we say
electrons orbit so people immediatly think spherical planet like objects
orbiting a sun like nucleus.

------
fsh
Another article that fails to mention two out of three accurate measurements
done on regular hydrogen since 2010. The news article about the most recent
measurement gets it right: [https://news.yorku.ca/2019/09/05/scientists-
measure-precise-...](https://news.yorku.ca/2019/09/05/scientists-measure-
precise-proton-radius-to-help-resolve-decade-old-puzzle/)

------
omegaworks
If a proton is a [probability] cloud, what exactly does the size measure?
where that cloud's probability goes to zero?

~~~
davrosthedalek
It's defined as the slope of the electric form factor at Q^2=0. Since that is
not a very helpful definition, with some hand-waving, you can say it's the
root-mean-square radius of the electric charge distribution. The proton has
also a magnetic radius, defined via the magnetic form factor. And something
called a Zemach radius, which is a combination if them both.

------
hurrdurr2
It's really incredible what our everyday mundane world is made of...
especially taking into account the quantum realm.

~~~
baggy_trough
Yes. For example it’s mind boggling that most of your mass is due to kinetic
energy of quarks zooming around each other at relativistic speeds pulled by
the strong force.

~~~
ReaLNero
This seems really interesting. Could you explain more about the strong force
and this "kinetic mass" effect? I thought quarks _had_ mass, and that's where
my mass comes from?

~~~
chaotic_clanger
the quarks have it own (rest) mass, but quite tiny compared to when you
include binding energy (9 Mev vs 900 Mev)

~~~
davrosthedalek
Rest-mass being a somewhat difficult concept if your particle only "exists" in
bound systems, i.e. is confined. But yeah, forget the Higgs, almost all the
mass you see around you is generated dynamically by non-perturbative QCD.

Edit: More explanation: If you take the quark masses as unbound rest mass,
it's not clear why the proton is bound -- it would have higher energy (=mass)
than the unbound constituents. This is different from atoms, as a hydrogen
atom has slightly less mass than the sum of a free proton and free electron
(13.6eV in the ground state).

------
PaulHoule
Physicists have "observed quarks" for a long time in the sense that if you
scatter protons against protons or other particles, you can see that the
proton scatters particles as if it was made up of three little particles.

------
bufferoverflow
Not a physics guy, so my understanding is probably wrong. Aren't all particles
infinite in size and are just blips in the fields behind our universe?

~~~
antonvs
The "wave function" that we use to model a particle's state has indefinite
extent, but when the particle actually interacts with something else, it does
so in a single, small location (from our perspective.)

When they talk about the proton radius, they're talking about the radius of
that interacting object, not its wave function. The wave function can be
thought of as describing the probability of an interaction finding the
particle in a particular location.

~~~
bufferoverflow
Isn't there a contradiction in your answer? Since the probability of
interaction far away is not zero, why is it not included in the "radius of
that interacting object".

~~~
antonvs
"Radius of the interacting object" was poorly worded. I was referring to the
radius of the interaction itself, which is much smaller than the extent of the
corresponding wave.

For example in the double slit experiment, the interaction occurs on a
detector screen in a single, small location with a radius on the order of
10^-15 meters (subatomic), whereas before it interacts with the screen, the
wave extends throughout the experimental apparatus.

------
known
A neutron star is so dense that a matchbox sized portion of one would weigh 3
billion tons.

[https://clearlyexplained.com/neutron-
stars/index.html](https://clearlyexplained.com/neutron-stars/index.html)

