> PS- hate the article title, pure vague clickbait
I agree. From the article:
> Yet physicists have found nothing amiss, no conclusive sign that antimatter particles — which are just the oppositely charged twins of familiar particles — obey different rules.
So no surprises. Also, the "near absolute zero" is true, but it's just to make the experiment easier, it's not important from the theoretical point of view. Anyway, it's an interesting experiment in spite of the press article title.
> Will any anti-matter/matter anhilate or does it have to be the complementary type? Like can a proton and positron annihilate?
There are a lot of conservation rules, for example
* proton + positron --> gamma rays
is impossible because the electric charge is conserved
* proton(+1) + positron (+1) --> gamma rays (0)
so +2 charge before and 0 after the annihilation, so it's imposible.
There are other number that are conserved, like the number of quarks (a quark is +1 and an anti-quark is -1) so
* proton(+3) + positron (0) --> gamma rays (0)
And the number of leptons, particles like electron are +1, and particles like positron are -1
* proton(0) + positron (-1) --> gamma rays (0)
And there are other examples of numbers that must be conserved. (And there are
numbers that are almost conserved and they can change but usually the decay/annihilation is "slower".)
It's interesting how this work with changes that are common, for example
* electron(-1) + positron(+1) --> two gamma rays(0) [charge]
* electron(0) + positron(0) --> two gamma rays(0) [quarks]
* electron(+1) + positron(-1) --> two gamma rays(0) [leptons]
Note that as said before there are more numbers that are conserved and more complicated rules. If one of these 3 rules are broken the transformation is impossible (AFAWK). If these 3 rules are ok, then you must check all the other rules.
> > PS- hate the article title, pure vague clickbait
> I agree. From the article:
> > Yet physicists have found nothing amiss, no conclusive sign that antimatter particles — which are just the oppositely charged twins of familiar particles — obey different rules.
> So no surprises. Also, the "near absolute zero" is true, but it's just to make the experiment easier, it's not important from the theoretical point of view. Anyway, it's an interesting experiment in spite of the press article title.
I'm sorry but did you actually read the article? The first quote is from the very first paragraph talking about physicists in general. And the cooling to near absolute zero was not "just to make the experiment easier." And it definitely is important.
The surprise, in case you haven't read the article, is that one expects spectral lines of atoms in an ultra dense fluid (i.e. supercooled helium) to be smudged due to the interactions with surrounding atoms. But, when performing laser spectroscopy on hybrid atoms (regular nucleus but with antiprotons taking the place of electrons) the spectral lines narrowed instead of widened. See, for example, this paragraph:
> Where the spectral lines of most atoms would have gone completely haywire in the increasingly dense fluid, widening perhaps a million times, the Frankenstein atoms did the opposite. As the researchers lowered the helium bath to icier temperatures, the spectral smudge narrowed. And below about 2.2 kelvins, where helium becomes a frictionless “superfluid,” they saw a line nearly as sharp as the tightest they had seen in helium gas. Despite presumably taking a battering from the dense surroundings, the hybrid matter-antimatter atoms were acting in improbable unison.
This was a big surprise. That's why the next two lines are:
> Unsure what to make of the experiment, Sótér and Hori sat on the result while they mulled over what could have gone wrong.
> “We continued to argue for many years,” Hori said. “It was not so easy for me to understand why this was the case.”
So, no, the article title is not clickbait and, yes, this is "important from the theoretical point of view".
<handwaving>The sharper espectral lines doesn't look too surprising, because in the superfluid liquid the Helium atoms "want" to be in the state creates the Bose condensate, so they don't "want" to colide too much with the disolved impurities. So the impurities are not disturbed too much and you get a nice spectrum.</handwaving>
> For more than two decades, encapsulation in superfluid helium nanodroplets has served as a reliable technique for probing the structure and dynamics of molecules and clusters at a low temperature of ≈0.37 K. Due to weak interactions between molecules and the host liquid helium, good spectral resolution can usually be achieved, making helium droplets an ideal matrix for spectroscopy in a wide spectral range from infrared to ultraviolet.
Side question. I understand the positive and negative charges, but when we say "+1" or "-1" what are the units? I never understood that part. an electron is -1 and a proton is +1, but what makes them equal? Why / how do all these things of varying sizes and properties have the same magnitude for charges?
Basically, quarks seem to have charges that are 1/3 e. However, they can only be stable in groups of three, and those groups must have an integer value. So the only possibilities are -1, 0, and 1.
As to why the above is true, it just seems to be a rule. We don't have any good reasoning for why that rule exists.
mesons are composed of two quarks and can have neutral or non-neutral charge. baryons are composed of three quarks and can have neutral or non-neutral charge. both mesons and baryons are classified as hadrons
charge, baryon number, isospin and strangeness are all related by the Gell-Mann-Nishijima formula[1].
the overarching principle underlying this is that in quantum chromodynamics, the theory of the strong interaction, quarks are representations of an underlying SU(3) gauge group containing color charge (a quantum number) and electric charge quantized as -1/3e and +2/3e. QCD is a gauge theory (like all other theories of nature) and the gauge boson, the gluon, also pops out of demanding the theory be invariant under local SU(3) gauge transformations.
that said, no one has any clue why all of our physical theories are gauge theories :-)
Would a theory of the history of the universe be a non-gauge theory since there was fast initial inflation, a relative slowdown, and now accelerating expansion again? Nothing seems conserved here. It seems symmetries were broken as the forces split apart, anti-particles were annihilated, etc
It should be statically typechecked of course, but especially in the quantum physics code we couldn't encode all the invariants correctly and runtime exceptions can still occur in rare occasions. A fix has been proposed and will hopefully be merged in time for the next universe release window.
I’ve always thought “annihilate” was a poorly chosen word to describe what happens here. It makes it sound like stuff is disappearing, and if you consider “stuff” to be only those particles that have rest mass, then fine, the aggregate value of the rest mass property of the system becomes zero, but most of the “substance” of everyday objects isn’t due to rest mass anyway. Also, the notion that rest mass “turns into energy” is similarly silly, as energy is just a scalar value that is conserved over time due to the time translational invariance of the laws of physics. It would be like saying I converted an onion into the number 7; it doesn’t make any sense. The energy of the system is the same after the collision as it was before the collision.
A better way to describe matter/anti-matter annihilation is that different types of particles interact in all sorts of different ways, and a collection of particles can turn into other types (and number) of particles during these interactions as long as the complete system obeys certain conservation laws. Matter/anti-matter collision is no more special in this sense than any of the other types of interactions in the Standard Model.
A star that is undergoing a "pair-instability" supernova would express rather dramatic opinions on exactly what is meant by "annihilation" as it is destroyed by positrons.
Annihilating positrons don't really consume an appreciable amount of the star's ordinary matter. Their presence mostly serves to allow a big star to compress into a much smaller space than otherwise, leading to a more powerful thermonuclear detonation.
One way to think about the gamma -> lepton-pair is that the parent (the gamma photon) must have at least the same momentum-energy as it's lepton-pair children. That is, the positron and electron each have less momentum than the gamma, and since each half of the pair can go in a different random direction, a gamma whose entire momentum may be outwards might turn into a positron that, for example, heads inwards and an electron that heads outwards, each carrying about half the gamma's original outward momentum. The outward momentum flux is what does the work of keeping the star from imploding gravitationally, and the balance is usually fairly fine; when a substantial amount of that momentum flux is turned in the wrong direction, the implosion is inevitable. There is nothing particularly special about the positrons (indeed, muons and antimuons, and other cascades of particles likely are produced too) other than that they and their partner electrons represent a redirection of mostly-outward momentum.
Another way of thinking about it is that a hot gas of gammas becomes a cooler gas of gammas+leptons, and "hot gas rises" is what keeps the star inflated. Cool down the middle, and the star contracts, because there is less hot gas rising. Initially, the sudden central cooling may re-heat the star's middle (compressive heating as outer layers sink inwards), which produces further central cooling (pair-production), which produces yet more heating, and so forth, with the swings from heating to cooling becoming more extreme over time. In some stars undergoing this process, the heating wins, and a thermonuclear detonation completely obliterates the star, scattering its remains to infinity. In other stars other factors help "lift" the outer envelope (e.g. the star's rotation, or the presence of many atomic nuclei heavier than helium) and cooling wins, with a super-dense phase (likely a black hole) holding on to some of the detonating material.
More technically, the lepton-pair has more degrees of freedom than the pair's parent gamma, and as one adds more degrees of freedom to a system, one increases the system's heat capacity. This is why one calls the process of pair production "cooling": effective temperature goes down when specific heat capacity goes up. For a nice rabbit hole, visit negative heat capacity in stars: https://en.wikipedia.org/wiki/Heat_capacity#Negative_heat_ca... with a general attitude of "momentum can condense, or become frozen into structures with large numbers of internal degrees of freedom (DoFs); black holes have enormous numbers of such DoFs inside the horizon, with number of DoFs increasing with black hole mass, which in turn is why more massive black holes are colder than small black holes".
How is the article title clickbait? They expected the spectral lines of hybrid atoms to widen the same as regular atoms when immersed in a dense fluid. Instead the spectral lines narrowed. This was so surprising they thought they made a mistake and spent years checking over their work only to conclude that, yes, the finding was in fact real.
For particles to annihilate everything has to be conserved. A proton and a positron cannot annihilate because they both have positive electric charge, but you also need a bunch of other properties to cancel out as well.
No, a positron is a lepton but a neutron is not. Since lepton number must be conserved, these two will not annihilate with each other. Similarly, one has a positive charge and the other neutral, so electric charge would not be conserved either.
Are you thinking of a neutron plus an antiproton? Those can annihilate with each other. They are not direct antiparticles of each other, but the neutron is really an up quark and two down quarks, while the antiproton is really two anti–up quarks and an anti–down quark. There are several different ways that they could annihilate with each other.
Worse, all hadrons (like neutrons and protons) contain a confined gluon field which you can think of as containing additional quark–antiquark pairs. There’s a chance that the gluon fields of the two particles will interact as if a quark–antiquark pair from one had annihilated with a quark–antiquark pair from the other, further complicating the results. This can cause the creation of other exotic particles as the quarks combine and recombine.
lepton number does not need to be conserved. it is an approximate symmetry of nature. if lepton number were conserved, neutrinos could not oscillate.
quarks, the particles composing protons and neutrons, have fractional charge; these would be the particles that would interact with an electron or positron. the charges wouldn't work out (charge is conserved (as far as we known...)) so there wouldn't be a fundamental electromagnetic interaction between a single quark and a e+/e- (i.e., an annihilation). But there are fundamental weak interactions between quarks and e+/e-; these processes are known as inverse beta decay and are used for pet scans.
To quote Wikipedia: “Lepton flavor is only approximately conserved, and is notably not conserved in neutrino oscillation.[6] However, total lepton number is still conserved in the Standard Model.“
The beta decay gives rise to e.g. a positron and a neutrino (or an electron and an anti-neutrino).
What is going on in neutron stars? The layman's answer is that the electrons get squeezed into the protons due to the extreme gravity, leaving only neutrons. But I suppose (?) there needs to be a anti-neutrino or similar that comes along to "complete" the reaction?
The electrostatic repulsion isn't really an issue. If you shoot an electron hard enough, it will overcome the electrostatic repulsion.
The bigger issue here is that a positron is a lepton, and a neutron has a 0 lepton count. So they can't react directly while conserving lepton count.
However, a neutron can decay into a proton, an electron, and a anti-neutrino. The neutrino is a lepton. So there is a path where the neutron and positron react to form a proton, anti-neutrino, and two gamma rays.
I don't know if that actually occurs. There are even more factors to be considered. But it illustrates how you have to go about conserving all of the things, including obscurer factors like leptonicity.
I skimmed the article looking for something about the spectroscopy of antiprotonic helium. I’d assume that the spectroscopy of such an atom is very different from normal helium but the article says nothing about how the spectroscopy is different or if the spectroscopy observed is what we expect (thinking of anti Hydrogen here). The article does point to Helium made from a pion though where the atoms spectroscopy was observed.
Very cool. I had no idea antiprotons -- well, an antiproton and an electron -- could orbit a nucleus. Does it fill up the usual quantum state of the first electron, or does it have its own set of quantum states to fill? Can they get a second antiproton to take the place of the second electron? I imagine that should be at least somewhat stable as the electrons of the enclosing container should repel the antiprotons just fine.
Classical mechanics has F = ma. Quantum mechanics has Schrodinger's Equation. Given a probability distribution for the location of your particle, and the potential (as in classical mechanics, just the integral of the force), it describes how that probability distribution evolves over time.
For something orbiting a nucleus, the only parameters that matter are electric charge and mass.
This stuff was figured out in the 1920s, and is taught in a year or two of undergraduate physics. The examples are usually an electron orbiting a proton (the hydrogen atom), but work just as well for muons, or anything else. It's only once you get inside the atom that you get into more advanced stuff.
> Does it fill up the usual quantum state of the first electron, or does it have its own set of quantum states to fill?
Not a physicist. But if I understand correctly, it fits the same equations the electron would, except you have to use the proton mass instead of the electron mass, which means it's much closer in. The remaining electron (I think) doesn't have Pauli exclusion with the anti-proton, so it looks like the electron in a hydrogen atom. (Since it's much further out, it sees the effective charge of +1 for the "nucleus" of two protons and an anti-proton.)
It took reading it a few times to figure out that you can slam antiprotons into liquid helium and end up with a shelf stable helium/antimatter hybrid atoms.
That is the part that amazes me... I had no idea that was even possible. I wonder what other atoms this would work with?
Well. Depends on the scale you consider to be shelf stable: The antiproton can thus orbit the nucleus for tens of microseconds, before finally falling to its surface and annihilating. I think that is a relatively long time in terms of particle-physics experiments, but it's not something you could stockpile.
It makes no reference to lifetimes in the article, other than to say the outer electron shields it.
The Wikipedia article mentions that it has a high energy, which aligns with the original article, which also states they figured out how to lower the energy.
No. The problem is that once the antiproton reaches the ground state, its wavefunction has significant overlap with the nucleus. That means that it's going to encounter the proton in short order.
PS- hate the article title, pure vague clickbait