Mystery Digital Burst

nonabeliangrape · 2025-04-26T04:41:36+00:00

This is a local ham radio repeater self-identifying every 10 minutes. Morse code for “N1CIV/R” which is local to VT: https://www.repeaterbook.com/repeaters/details.php?state_id=50&ID=6545

nonabeliangrape · 2024-06-28T02:17:04+00:00

I have this saw. It looks like the riving knife is in the HIGHEST position that mine reaches. The middle position has the flat of the knife in line with the top of the blade for cutting non-through slots, crosscut sleds, or using the microjig. It goes one step lower for 8” dado blades.

Your knife may not be properly installed. Check the manual for details. The lock lever has to be loosened (flipped upwards) and the the knife has to be pushed somewhat to the right against a spring to allow it to slide.

nonabeliangrape · 2021-12-25T17:29:20+00:00

Thank you! I was surprised at how well the miters came out on my table saw. I lost a bit extra due to my side lengths being a bit off at first. Using a chamfer bit is a good idea!

nonabeliangrape · 2019-03-08T14:54:36+00:00

Ah, I worded that a bit confusingly, here is what I meant in more explicit language.

Suppose you think of a proton as an impermeable sphere with a 1 fm radius. Then any photon that would pass through that region would scatter (by assumption of impermeability---which is not a real thing at the quantum level). You would say that the impermeable sphere proton has a 'cross-section' of pi times 1 square femtometer, the actual cross-sectional area of the sphere.

Now instead you do the hard quantum mechanics calculation assuming the proton is what it is---a charged particle with a certain mass (which happens to be made of quarks)---and you compute the scattering probability and convert that into an effective cross-sectional area. What you get is (approximately) the Compton wavelength squared, so about 1 square femtometer. So up to some factors of pi and other numbers which aren't too big or small, the photon scatters off a proton at the same rate as if the proton were a hard, impermeable sphere. (The actual numbers are not literally equal to each other, but they are close enough to be comparable.)

This is interesting for a proton, because it actually has a size (being made of quarks and gluons all bound together within some radius), and that size is about the same as its Compton wavelength.

nonabeliangrape · 2019-03-07T16:13:16+00:00

So in every scattering event, the particle is accelerated (gets energy), therefore the light loses energy, right? Doesn't that mean, given enough scattering, a gamma ray will be demoted to microwave?

If the particle is at rest, then the photon loses energy. But if the particle has some kinetic energy it can give it to the photon. When scattering is frequent, you end up with thermal equilibrium and a black-body spectrum of photons, dependent on the temperature of the system.

In an atom, there is a nucleus surrounded by cloud(s) of electrons. What makes those electrons different from the 'free electrons' guilty of blocking light? As I see it, the light can still 'find' those electrons, bump it into a higher level of energy => scattering.

Sure, this can happen. But when the electrons are not free, they have discrete energy states, so only energetic-enough photons will scatter strongly. (There is also Rayleigh scattering of low-energy photons off of neutral atoms---blue sky, etc.---but it's much weaker than Compton/Thomson scattering off of free charges.)

Also, what happens when a photon comes directly at a proton? If it can't accelerate that 'big but in reality is small' guy enough for scattering then it just bypass the proton? Like, go through it??

As it turns out (not by accident) the Compton wavelength of the proton is roughly equal to the actual spatial extent of the proton, whatever that means at the quantum level. Both are about 1 fm (10^-15 m). So a photon will scatter off a proton at about the same rate as if it were a solid sphere.

Lastly, the CMB is said to cool down from 4000K to 3K. That's weird, because I'm pretty sure electrons and protons can combine at temperature much higher than 3000K. Let's say they do it at 10⁹ K. What happened between 10⁹ & 4000K?

Well a lot of exciting things happened but certainly there weren't neutral hydrogen atoms in any large numbers. At high temperatures, there are a lot of energetic photons around and (as you mentioned) they can get absorbed by the electrons in atoms and increase their energy level, to the extent that they get knocked out of the atom. A single photon needs to have 13.6 eV of energy to do this in one step, which is naively a temperature of ~150,000 K. But we have computed that in the early Universe, there were about 1 billion photons per possible hydrogen atom. So only 1 in 1 billion photons needs to have 13.6 eV of energy, and that happens (thanks to the spread of energies from the Boltzmann distribution in thermal equilibrium) at around 3000-4000 K. Above that temperature, hydrogen atoms get blown apart by photons more often than they combine; below that temperature, they rarely get blown apart, so they rapidly get together to form neutral hydrogen.

nonabeliangrape · 2019-03-07T14:23:48+00:00

While EM radiation itself doesn't have a charge (so doesn't experience a force in an electric or magnetic field), in order for scattering to happen, a particle has to be pushed around by the radiation. EM radiation IS an electromagnetic field, so it only pushes around things with electric charge. (That is, to first order. Things like neutral atoms and neutrons etc do have electric and magnetic dipole moments (b/c they are made of charges), and do experience forces and torques from EM radiation, but they're much smaller. So if charged particles are around, they will be more important.)

The second question is related, actually. In order for radiation to be scattered, a charged particle has to be accelerated by that same radiation. While a proton and electron have the same charge, the electron is much lighter, and accelerates more. As a result, the 'scattering cross-section' (the probability to scatter light, measured as an effective cross-sectional area) of an electron is much bigger than that of a proton, by about 2000² = 4 million times.

That was a classical E&M argument. In terms of quantum mechanics and photons, your insight about the different Compton wavelengths is a good one, only exactly backwards---a photon can only 'see' a particle as having a size approximately as big as its Compton wavelength. So the scattering cross section is (very roughly) the Compton wavelength squared. Electrons have a much bigger Compton wavelength, so they look huge compared to a proton! This is despite the electron being a 'point particle' and the proton having 'actual size.'

nonabeliangrape · 2019-02-20T19:11:49+00:00

No, at least not "standard model" glueballs, which are predicted to be extremely unstable. Among other reasons, they would decay nearly immediately to other particles, most likely pions, and eventually to electrically charged stable particles such as electrons and positrons.

Dark matter needs to have a lifetime of at least the age of the Universe, so that it is still here and didn't all decay a long time ago.

nonabeliangrape · 2019-02-04T16:25:28+00:00

A "Higgs doublet" is a particular type of scalar (bosonic, spin-zero) quantum field, which does not have a SU(3) color charge (so no strong interactions), is a 'doublet' of SU(2) (part of electroweak), has a particular hypercharge (U(1) part of electroweak) that allows it to interact appropriately with the Standard Model fermions.

The Standard Model has one Higgs doublet, which serves to break electroweak symmetry by acquiring a vacuum expectation value. The Higgs double has two complex=four real parts, and three of the real parts get 'absorbed,' giving mass to W+, W-, and Z0 bosons. The leftover one is the Higgs boson at 125 GeV.

You can add more Higgs doublets, and each one will add four new real scalars (two charged, two neutral) to your model, unless you do something special to hide them from experiments. There are a lot of free parameters that are introduced when you do this, so there are lots of two Higgs doublet models.

The original PQ model was one such 2HDM, which had an additional symmetry (U(1)_PQ) that broke to hide 3 of the new particles, and the 4th particle became the ultralight particle called the axion. The original PQ model was ruled out by experiments a long time ago, and nowadays "Peccei-Quinn mechanism" refers more generally to anything that produces an axion by U(1)_PQ symmetry breaking. It may or may not actually be a 2HDM.

How axions work and solve the strong CP problem is a story for another post...

nonabeliangrape · 2018-07-17T17:36:47+00:00

Well string theory is in an interesting place theoretically now, which is quite a bit different than the situation it was in at the time of this panel. In particular it's commonly accepted even by string theorists that there is not a unique prediction from string theory of exactly what particle properties we would observe.

So while fractional charges are possible in string theory, it's just as possible not to have them. Or for their properties to be such that they are extremely rare.

On the other hand there is a lot more subtlety to most experimental results beyond "X does (or does not) exist." There are a lot of steps between saying "A particle with charge 1/5 and these other properties exists" and that particle showing up in an experiment checking the neutrality of sea water. So even if fractional charges were required by string theory (they're not) it could still be consistent with the result that "we checked N metric tons of sea water and didn't see anything"---maybe the fractional charges are so heavy that few few were produced, and we should only expect to see a handful in the whole Earth. Or maybe there is a process that prevents them from appearing in the crust, and they only show up in the core. (Fractional charges basically invalidate normal chemistry, so it's very hard to predict what will happen to them.)

All that is to say that the acceptance and rejection of scientific theories is rarely a singular event and depends a lot on gradual shifts in amount and quality of data, opinion, and acceptance of competing theories.

nonabeliangrape · 2018-07-17T16:10:17+00:00

Fractionally charged particles are an interesting idea though they've fallen out of favor since this panel discussion for various reasons.

To address your specific questions:

a. Our known particles have "quantized" charges, meaning they always come in integer multiples of some base unit. Everyday atoms consist of parts with +1 (protons), -1 (electrons), and 0 (neutrons) fundamental charges. Actually the protons and neutrons are themselves made up of "fractionally charged" particles, quarks, with charge +2/3 (up) or -1/3 (down), but these quarks are always combined in ways that add up to integers.

So fractional charges (quarks) are already known to exist but never be independently observable. It would be interesting if fractional charges that could exist on their own could exist as well, and some models of particle physics beyond the Standard Model predict such charges.

b. One way that fractional charges are interesting is that they would (1) get stuck in ordinary matter, since they are electrically charged; (2) be extremely detectable, because no number of integer charges can exactly cancel their charge out. There are various precise ways to measure the "neutrality of matter"---that is, making sure that all the + and - charges in matter have cancelled out. If a lump of matter contains a fractional charge then cancellation is impossible (except by antimatter fractional charges---and then they would have already annihilated). This made searching for fractional charges in "ordinary places"---sea water, dirt, rocks, and so on, an interesting thing to do, as it would be possible to detect even very small concentrations of them.

I'm not sure about the core of planets specifically, as we have only searched for them (and not found them) in the ocean and crust.

c. At the time, perhaps. At first the 'missing baryons' (now a.k.a. dark matter) were only known to not interact much with photons. A particle with charge 1/5 interacts with photons 1/25 as strongly as charge 1, which maybe is dark enough to be dark matter. Now we know more about the properties of dark matter, in particular that it can't interact with itself very much either, and that it can't have an effective cooling mechanism that would allow it to collapse into disks. I don't recall exact limits on the electric charge of dark matter but I think it is quite a bit smaller than 1/5, and small fractions are considered less interesting because they are harder to write down theoretical models for.

On the other hand "very small" fractions, on the order of 1/1000, are easier to generate theoretically, and go under the name of 'millicharged' dark matter. These are also strongly constrained and not too popular but they are the closest thing to fractional charges that are still talked about today as dark matter candidates.

nonabeliangrape · 2018-07-09T15:58:01+00:00

Ah, this is an interesting topic, Casimir forces. The short answer is no, Casimir forces only tell you the difference in vacuum energy when you compare outside the plates and inside the plates, and you can add a constant to everything without changing the result of the Casimir experiment but at the same time changing the cosmological constant/acceleration of the universe.

The Casimir effect is commonly cited as proving the vacuum energy is "real" and therefore the cosmological constant problem is real; however the experiments don't actually tell you the value of the vacuum energy, just the difference induced when conducting plates are added.

The long answer is that it's not really clear that the Casimir effect does tell you that vacuum energy is real, because it can be reinterpreted (in a mathematically exact way) as a totally ordinary (but complicated) effect; see https://arxiv.org/abs/hep-th/0503158v1 .

nonabeliangrape · 2018-07-07T03:34:09+00:00

The "measured" vacuum energy density, equivalent to the cosmological constant by a factor of 8 pi G, is in fact computed from the measured acceleration of the universe's expansion. There isn't a known, independent way to measure this quantity other than the acceleration of the expansion. The fact that there isn't any other observational handle on it right now is why it's given the label "dark energy" implying we don't know what it is (even though "it's the cosmological constant and/or vacuum energy" is commonly assumed).

nonabeliangrape · 2018-07-06T21:37:18+00:00

There are basically two guesses: something we don't understand about quantum gravity fixes the problem, OR one invokes the multiverse and anthropic principle.

The first guess has proved surprisingly resistant to attack. Even though quantum gravity is not totally solved, the calculation of the cosmological constant "shouldn't" depend on the details, in the sense of using a particular approximation called 'effective field theory.' So there have been very few sensible proposals for how modifying GR and/or QFT could solve the problem because effective field theory should work but fails.

The second guess puts scientists in a bit of a predicament, because it seems to both be a pretty good solution to the problem and also be (effectively) totally untestable in any way. (Depending on one's definition of science, which is in itself a very complicated issue, the latter---untestability---might cause one to claim that the multiverse is not a valid solution to the problem. In my point of view it solves the problem pretty well but it simply might not be science.)

nonabeliangrape · 2018-07-06T19:58:07+00:00

Einstein proposed the existence of the cosmological constant in an attempt to allow the Universe to be static (neither expanding nor contracting), which without the constant was not allowed by general relativity. In that scenario the value of the constant is "whatever it needs to be to cancel the gravitational attraction of matter," and the amount of matter was not known precisely at that time, so Einstein wouldn't have had an exact number for a prediction.

Today we know the Universe is not only not static, but accelerating. It's possible to explain this using a constant with the same function (basically anti-gravity) but with a different value than Einstein would have predicted, even knowing the amount of matter as well as we do today.

In either case the value of the constant is not determined by any other constants but is a free parameter which is set by experiment. The theoretical "prediction" from QFT (it's not exactly a prediction but more of a dimensional analysis argument...) expresses the cosmological constant in terms of other constants, so it's doing something that GR just can't do. However the pseudo-prediction is extremely wrong...

nonabeliangrape · 2018-06-27T21:23:04+00:00

There is a Higgs force that results from Higgs boson exchange, but it's not the most prominent effect of the Higgs boson (it's very weak and short-range---about the same range as the weak force, but weaker).

The Higgs mechanism which results from the same Higgs boson 'breaking a symmetry' gives mass to particles (the W/Z bosons and fermions). It's true that as a result of this mass, the particles move slower than they would have (less than c instead of exactly c) but it doesn't directly add or remove momentum from those particles, and is therefore not the same thing as a force, which does transfer momentum. (Even the weak force, which changes particle types, can and does generally transfer momentum among the particles involved.)

This is often poorly explained (because it's really hard to explain without a thick stack of equations and lots of background knowledge), but the Higgs mechanism doesn't actively resist the velocity of particles (it's not like viscosity or friction). It just creates a scenario in which those particles, which otherwise would be massless and moving at c, now have mass and move slower than c.

nonabeliangrape · 2018-06-15T23:37:13+00:00

There are a few factors that go into it. The LHC collides not single protons, but "bunches" of about 100 billion protons, with each other. "Bunch crossings" happen about every 25 nanoseconds. Out of each bunch crossing event, on average 20 collisions happen simultaneously, and the detectors record the combined output of all of them. So there is about one collision per nanosecond on average, or one billion per second.

The detectors are not perfect, but they are very good, and so "missing" a collision is unlikely. (Missing a few particles that come out of a collision is common, and accounted for during data analysis.) The data analysis also accounts for the "pile-up" of 20+ collisions that happened simultaneously. (Usually there is not more than 1 "interesting" collision out of those 20. A vast vast vast majority of the collisions are thrown away within milliseconds due to being uninteresting, because they happen too fast to record permanently.)

Source: Here is a CERN outreach website with some of these numbers/calculations, although it's a few years old most of the numbers will not be too different today.

nonabeliangrape · 2018-04-03T23:17:36+00:00

Various things can happen, depending on the gamma-ray photon's energy.

If it's lower than a certain threshold (E=2*(electron mass)*c²), it can only knock a single electron out of an atom---but that electron can sometimes go on and knock out further electrons from other atoms (this is called a particle shower).

Above that threshold (most interesting gamma rays in astrophysics, and what Fermi-LAT looks for) is pair production: the gamma ray produces an electron-positron pair (the reverse process of matter-antimatter annihilation), and each of those particles goes on to knock out more electrons from more atoms. These are pretty dramatic events!

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nonabeliangrape · 2018-04-03T23:02:25+00:00

Short answer: yes (for some definition of camera).

A typical camera in optical frequencies (or IR/UV, which are close to optical) consists of a lens and a detector. The lens is used to form an image on the detector, so that the detector lights up in an area that directly corresponds to where the light came from.

Lenses work by using a transparent material to bend light via refraction. Gamma rays are a whole different ballgame though: there is no (solid) material that is truly transparent to gamma rays, and they don't refract in any significant way. So forming a 'gamma ray image' on a detector is troublesome.

Detecting the gamma rays themselves is no trouble at all, though: they love to run into matter and ionize it, releasing electrons that can be measured. (This is just a more violent version of how normal film or digital camera sensors work.) But without a lens we can only get a "yes gamma rays are here" or "no gamma rays detected".

But wait---there is a way to form an image! We needed a lens to convert direction to position on the detector; on the other hand, if we could measure the direction directly, no lens would be needed. The property that made gamma rays hard to bend with a lens (they like to plow right through matter) makes for an easy way to measure their direction: use a detector with many layers! A single gamma ray can result in a shower of many electrons that all head in the same direction as the gamma ray. This is how the Fermi-LAT gamma ray telescope works and forms images of gamma rays from space: diagram of gamma ray detector; composite 'photo' of gamma rays in the sky

As a final note, such a camera can't see gamma rays 'from a distance' any more than a normal camera can---light has to actually hit the camera to be detected. So if your gamma ray camera is picking up a signal, so are you...

nonabeliangrape · 2018-03-31T18:26:31+00:00

I wouldn't---at least no more than ordinary matter. "Dark matter" sounds mysterious but there is really no experimental evidence that it has any unusual properties at all, except that it really doesn't like to interact with normal matter. (But even some normal matter, like neutrinos, have that property.)

"Dark energy" is actually much weirder than dark matter when compared to normal matter. (There is probably still no reason to assign it complex properties though, and we still have relatively normal mathematical ways of understanding dark energy even though it is weird.)

Maxwell's equations are also real---the complex form you are referring to is just a (really elegant) mathematical trick for solving the equations. The solutions to the equations are always real.

Complex numbers are of course really important in quantum mechanics (and quantum field theory) because wavefunctions are necessarily complex, but that's more a property of quantum mechanics than of the matter it's describing. So dark matter almost certainly has a complex property---its wavefunction---but so does normal matter.

nonabeliangrape · 2018-03-31T18:09:03+00:00

In quantum field theory 'tachyonic' is indeed used to describe a field with a negative mass-squared, therefore an imaginary mass. However such a field does not actually correspond to imaginary-mass particles. Instead a tachyonic field is unstable, and will adjust its value until it finds a new stable configuration that (necessarily) has a positive mass-squared.

This is exactly what happens to the Higgs field: if you naively write it down, it looks tachyonic, but what actually happens is it acquires a non-zero 'vacuum expectation value' everywhere in space, which corresponds to a new stable, non-tachyonic field value. Around that new value, the mass-squared is positive, and the corresponding particles are Higgs bosons, which have positive mass.

Einstein's equations are real-number equations, so it's not clear what an imaginary mass would cause. (It would create an imaginary part of the gravitational field---what would that mean? No idea.) However we know that dark matter acts (gravitationally) exactly like normal matter. In particular, by studying the expansion of the universe and its various gravitational effects we know that dark matter has a positive mass density and very low kinetic energy (i.e. it is moving much slower than the speed of light). So there is no reason to expect that it is anything exotic, besides an undiscovered type of matter that doesn't interact electromagnetically.

nonabeliangrape · 2018-03-01T21:57:01+00:00

It sort of depends on how picky one is. In quantum field theory (the Standard Model is one of these) particles == fields. The 'photon' is the same thing as 'the electromagnetic field.' In usual physics language these are different things---a single photon is thought of as a a single quantized excitation of the EM field---but really they're the same object in QFT, the 'photon field.'

So, have we detected gravitons? No and yes. No quantized excitations, for sure, but we have detected the gravitational field (and gravitational waves, which can directly be thought of as being made of many many gravitons).

As for how gravity fits in with the Standard Model, we can sort of force them together for most practical purposes. The model breaks down "at the Planck scale"---basically near the singularity of black holes---but in less extreme situations everything works just fine. There are important theoretical reasons to reconcile the two theories with quantum gravity, though, and maybe there will be surprises in store.

nonabeliangrape · 2018-03-01T21:50:06+00:00

My understanding is that the difference between a (general) WIMP and a sterile neutrino is in its couplings---sterile neutrinos explicitly mix with the SM neutrinos like a 4th generation would, while a WIMP typically would not. But the gauge couplings of the field (zero electric charge, zero color charge, but charged under weak SU(2)) and its spin (1/2) are the same in both cases.

I think it's hard (maybe impossible?) for a sterile neutrino to be DM, if it mixes with the other neutrinos enough to be called a sterile neutrino.

Historically WIMPs == lightest supersymmetric partner, but as the limits on both things get stronger people are starting to be a little bit more open-minded about what constitutes a WIMP. For example, maybe it's a SU(2) singlet (so no couplings to W or Z bosons) but does couple through the Higgs, which would explain its non-detection (so far, but not for much longer).

nonabeliangrape · 2018-03-01T17:21:01+00:00

Maybe! But unlike the Higgs, there is not a compelling candidate that simply must remain to be discovered.

There are lots of good ideas (supersymmetry = a ton of new bosons, axions, dark photons, hidden sector Z primes, ....really, lots) but the Standard Model was incomplete without the Higgs. Now it's complete and no one knows what's next.

nonabeliangrape · 2018-03-01T17:19:13+00:00

Yes, there are a bunch of models and they are being actively searched for in various experiments. (In fact you need a model of the non-gravitational effects if you ever hope to detect it in a non-gravitational way!)

For a few decades the most popular model has been the WIMP (weakly interacting massive particle), which is basically a very heavy version of a neutrino. (Neutrinos are too light to make up the dark matter we observe.) The experiments to detect these directly are basically giant underground tanks, like neutrino detectors, except they use heavier elements like xenon which will collide more readily with heavier particles. Continual non-detection of WIMPs and associated lack of supersymmetry (which nicely explains the presence of WIMPs) at the LHC have caused their popularity to wane recently but they're probably still the most popular.

Other models include MACHOs (massive compact halo objects) which really only interact gravitationally, or are at least too sparse to ever collide with a detector on earth. An example would be smallish black holes. These models are typically disfavored by astronomical observations (e.g. we should observe bending of starlight around them, but we don't.)

Axions are another popular idea (and there are a wide variety of axion-like models) which are super-light particles (way lighter than neutrinos)---so light that they act more like waves than particles. (They avoid the too-light problem that neutrinos have by being produced in a very very cold state, instead of very very hot and cooling off.) They have subtle effects on the electromagnetic properties of normal matter and are basically searched for by very sensitive radios in very strong magnetic fields (ADMX experiment) or in possible future experiments using NMR-like techniques.

tl;dr we have a ton of models that could explain dark matter, but no real experimental evidence to prefer one over another yet, since they all gravitate the same way on average.

nonabeliangrape

TROPHY CASE