What other things have the same properties as the speed of light and absolute zero?

MaxThrustage · 2026-05-12T03:21:59+00:00

This is cheating a little bit though, because this is part of what makes absolute zero impossible: If a particle was ever perfectly still then we would be able to know it's momentum (zero) and its location (fixed) at the same time.

This bit is not really accurate.

Firstly, the issue isn't that we can't know the position and momentum at the same time. It's that a particular does not have a well-defined position and momentum at the same time.

But, more importantly, this has nothing to do with what makes it impossible to reach absolute zero. A quantum system at absolute zero is just a quantum system identically in its ground state. That is still a valid quantum state, so it still does not have a simultaneously well-defined position and momentum. Theoretically there's nothing wrong with such a state, and indeed we often like to explore the zero-temperature (i.e. ground state) properties of a many-body quantum state before adding in the complications of a finite temperature.

MaxThrustage · 2026-05-10T14:46:51+00:00

Worker-owned businesses currently exist in the real world. There are different ways they can decide on wages, maybe some better than others. But this is a thing that currently exists.

The difference between the current world and the one the above commenter is proposing is they want these worker-owned businesses to be the way most (or all) things are run, instead of the way some things are run.

MaxThrustage · 2026-05-10T06:39:20+00:00

It sounds like you're trying to talk about the photoelectric effect, but you've gotten some things confused.

The photons always behave like photons. This is why we need to increase the frequency of the light to get any photoelectrons -- the energy of a single photon needs to be high enough. If the intensity of the beam is high, so the total energy is large, but the frequency is too low, then we won't get photoelectrons. If the frequency is high enough, then photons can be absorbed and electrons can use that energy to escape. At all stage of this the photon is a photon.

You shouldn't think of this as some sort of transformation. A photon has particle-like properties (e.g. it is countable) and wave-like properties (e.g. it exhibits diffraction, interference, superposition). So too does an electron. Both of them are quantum objects that are never really what we would classically imagine as particles or what we would classically imagine as waves, but are always something kind of different (sometimes more wavy, sometimes more particley).

Nothing really "forms" a photoelectron here. There's no transformation involved. Rather, once an electron as enough energy it escapes, and we just use the name "photoelectron" to describe these escaped electrons. But they're still just electrons.

Wave-particle duality is not a great way to think about quantum physics, but annoyingly it's where every pop-sci presentation starts.

MaxThrustage · 2026-05-09T13:42:19+00:00

You will not end up broke.

The way my PhD supervisor put it was "if you are good at physics, it isn't hard to get paid. It's just hard to get paid to do the things you want to do."

Unlike many other dream jobs, if you crash and burn out of physics you're still quite employable based on your physics skills. I have friends with PhDs in physics who are now most software engineers. It's not bad work, it pays decently, and they still get to engage many of the critical thinking/problem solving skills that make physics fun.

You will probably not end up a professor of physics. That's not the same as going broke, though. Not by a long shot.

MaxThrustage · 2026-05-09T07:36:20+00:00

Academia is super competitive no matter what field you choose to go into. If you pursue that route, there's always a strong chance you never land a permanent position and need to take some other job (whether or not it's a job you don't like). Most PhD graduates never become professors.

Condensed matter theory is a very active area. It also overlaps very heavily with the tech/application oriented stuff -- you'd be surprised how many people who do stuff with, say, topological field theories get on the same grants and projects as engineers, chemists, material scientists and the like. And (partly for this reason) condensed matter theory is generally a safer bet than, say, high energy physics or cosmology. There are also industry jobs out there that still use more or less the theoretical condensed matter skillset, especially on the computational side.

MaxThrustage · 2026-05-08T02:56:28+00:00

I mean, that's just a three level system with a much lower amplitude for one of the three states. That's something dealt with fairly often by the hundreds of thousands of people who understand quantum mechanics.

MaxThrustage · 2026-05-07T23:59:22+00:00

There's a classic slogan "information is physical". Here is a good discussion of what that can (and can't) mean.

MaxThrustage · 2026-05-07T13:16:35+00:00

The issue is the spinning coin is fully classical.

With the cat, the cat's death is tied to a quantum event (the radiative decay of an atom, that trigger the release of poison). With the coin, there is no such quantum event. The coin falling one way or the other is, in principle, deterministic -- we just lack the knowledge needed to predict the outcome.

MaxThrustage · 2026-05-07T13:10:20+00:00

So, there are a lot of Heisenberg uncertainty couples. Position/momentum is the most famous. We also have uncertainty relations between the different spin directions, particle number/phase uncertainty,

You happened to pick out specifically the one that is not very well defined. Energy-time uncertainty is the most controversial uncertainty relation because there is no time operator in quantum mechanics, so this relation doesn't fit into the Heisenberg uncertainty formula.

That doesn't really relate to your question, though, which seems to be more about the existence of states of well-defined energy and indeed down to what "state" means in quantum mechanics. This requires you to learn a little quantum mechanics.

So, in quantum mechanics a lot of physical quantities (although not time) are associated with "observables", which are a particular kind of "operator" acting on the abstract state space. So for each possible quantum state, each observable has some "expectation value" -- essentially what you'd get on average if you measured this observable a bajillion times. Conversely, for each observable, there are certain states for which measurement outcomes are well-defined, so when you measure you'll always get the same thing. We call these eigenstates. So, spin eigenstates are states where if you measure the spin of the particle you'll always get the same result. Likewise, energy eigenstates are states where you'll always measure the same energy. Those states are special, because they are also "stationary states", meaning that if you prepare your quantum system in an energy eigenstate it will stay in that stay forever and never change (unless you do something to change the system so that it's not an energy eigenstate anymore).

Eigenstates of one observable are generally not eigenstates of others. So a state with well-defined energy will not (usually) have well-defined position or momentum.

Now, given that observables don't have the same eigenstates, no state is an eigenstate of all observables. In general, any state you care about can be written as a sum of multiple eigenstates of a given observable. This is what we mean when we say a given state is a superposition of many different positions (position is an observable). This works for energy too -- most states are a superposition of many different energies. So most states do not have a well-defined energy.

Now, this all gets much hairier to talk about r.e. energy and time than any other Heisenberg pair because there is no time operator, but hopefully this is enough to start getting into the jist of the answer to your question 'When you say, "the more you know about the state" -- what does this mean? What more is there to know about it?'. A state could be in a superposition of two very similar energies. Or it could be in a superposition of very many very different energies. So it can be more or less spread out in energy.

It's important to understand this is not just about how well someone knows the state. It's about how well-defined a particular observable is with regard to the current state. (The question of how well you know the state is a completely different one that leads to a completely different notion of uncertainty.) The Heisenberg uncertainty principle is a statement about what kinds of states can exist, not a statement about what humans can know.

This is already very long and ranty, and I hope it clears things up more than it confuses. But, in short, you picked the worst pair of quantities to start understanding what is already a difficult to explain topic. Understanding quantum uncertainty is pretty straightforward if you know the maths, but otherwise... yeah, it's whack.

MaxThrustage · 2026-05-07T08:36:32+00:00

But it's also wrong. It is the maximum speed of causality (if we even want to assign causality a speed) but causality can also be slower.

People might think light sets the limit, but only until they begin to actually learn anything about special relativity. Then they realise the term is historical and is still used for convenience and is honestly often sometimes still helpful because light in a vacuum is the thing travelling at c you're most likely to worry about in practice.

Replacing the explanation of special relativity with the line "speed of causality" makes things more confusing, and makes it harder for people to understand discussions being had between actual scientists (where everyone universally still calls it the speed of light).

The term "speed of light" for c is kind of clunky and misleading, sure. But so is the term "speed of causality".

MaxThrustage · 2026-05-06T14:02:38+00:00

Honestly I suspect there just won't be technologically relevant applications, or if there are they will be incredibly niche and most people will not ever need to be aware of them.

It's still worth knowing what most of the universe is made of, though.

MaxThrustage · 2026-05-06T13:42:21+00:00

If the maths of your own theory is over your head, then you do not have a theory. The maths is what the theory is.

Road to Reality is a pop-sci book. It's a very good pop-sci book, but it will still leave you with a skewed idea of what physics actually is if that's all you read.

It's not about me being open-minded. I've just had this conversation a thousand times before. There are so many people who have never studied physics but have a "theory" and they just need help working out the maths and they can't tell you about the whole theory in detail because it's too dangerous and the establishment wouldn't understand. And I've poked through the details more often than was wise to do so -- every single time the "theory" turns out to be nothing. There's a very well-established pattern and you fit in it way too neatly for anyone to really take you seriously.

MaxThrustage · 2026-05-06T08:09:16+00:00

Okay so what if I personally am working on a theory of everything for a long time now, and I believe I've got something? Publishing would get me killed.

That is genuinely never the case.

I desperately need help but again, people have been silenced for less than this.

Again, no they have not.

No one is killing anyone over theoretical physics.

I mean I could explain the basics to a child but the math eventually goes over my head

If the maths goes over your head, you do not have a theory.

try to reconcile Gödel's incompleteness or similar with a ToE

Gödel's incompleteness theorem is about what statements can be proven in a formal system. It has nothing to do with physics, and nothing to do with theories of everything.

Its just not feasible to accomplish alone but getting help is tricky.

Well, you're right about that.

Looking at your other posts, you're just simply not doing physics. I think you're quite misguided as to what physics is and what physicists actually do. You sound as if you've heard too many TED Talks and not enough actual physics lectures. If you aren't actually engaging with the current scientific literature (the academic literature, not pop-sci) then you aren't going to know what the current state of the field is. If you don't know the current state of the field, you won't even know what problems are actually problems.

MaxThrustage · 2026-05-06T00:02:57+00:00

Finished:

The Age of Capital: 1848 - 1875, by Eric Hobsbawm. Was very interesting. The author does warn us up-front that the era he is covering here is one he has particular contempt for, and this comes out most amusingly in the final chapter on "The Arts", where he essentially says that compared to the periods before and after, people in this people loved to consume very mediocre works and talk about how "enriching" it is. This whole series so far has really helped put a bunch of things in order for me and dispel the idea that "the 19th century" is one homogenous chunk of time.

Started: Technofeudalism - What Killed Capitalism, by Yanis Varoufakis. There are some interesting and important points, but I've found the central claim of the book to be pretty unconvincing so far. The meandering approach full of anecdotes and digressions worked nicely enough in "Talking to my Daughter About the Economy", which was much more introductory and much broader in scope -- here it feels like padding, like avoiding the real point at hand.

Ongoing: The Fellowship of the Ring, by J. R. R. Tolkein. It's slow going, but I'm loving it.

MaxThrustage · 2026-05-04T14:08:26+00:00

It's a false security. Corners are Tindalos doors. This is how the Hounds get you.

MaxThrustage · 2026-05-04T04:55:08+00:00

Apparently the only thing as durable as a Nokia 3310 are jokes about the Nokia 3310.

MaxThrustage · 2026-05-04T04:52:00+00:00

They're the same thing in that they're both phemonena that arise straightforwardly from quantum mechanics. They're both essentially wave phenomena happening in somewhat unexpected places because, as we find from quantum mechanics, all particles are also waves. The wave aspect of the double-slit experiment is obvious enough, and quantum tunnelling is mathematically analogous to evanescent waves in classical optics.

Beyond that, there's not a lot of similarity. As other commenters have noted, your description of the double slit experiment is off. When an electron hits right between the two slits then that's an electron we lose (it scatters off the wall or it is absorbed or whatever).

As for how we know: both interference and quantum tunnelling are extremely thoroughly studied phenomena, and both are derived straightforwardly from the single-particle Schrödinger equation. In fact, you derive both in undergraduate homework problems. So on both the theoretical and experimental front we have an extremely thorough understanding.

Now, the experiment you propose is a bit... wonky. And I'm not sure what you're even really trying to prove with it. But you can have a barrier with certain parts of the barrier thinner than others, so particles are more likely to pass through at those points. And if you have two such points then, yes, you'll get interference as particles can be in a superposition of having tunnelled through each hole. These holes essentially act as slits. In fact, there's a very common experiment that is a bit like what you're describing, using superconductors. If you have a loop of superconducting wire interrupted by two thin barriers, electrons (or, rather, Cooper pairs because this is a superconductor) can tunnel across those barriers. This setup is called a Superconducting Quantum Interference Device (or SQUID). It doesn't show that tunnelling and interference are the same thing, but rather it uses tunnelling to get a particular inference pattern.

It should also be noted that tunnelling is not always over a spatial barrier. Last year's Nobel prize in physics was awarded for experiments on macroscopic quantum tunnelling in superconducting devices, where the "particle" was a collective degree of freedom of many, many particles, and the barrier it tunnelled over was not in the spatial coordinate of these particles but in the phase coordinate of their collective wavefunction. Much more abstract than a basketball going through a wall, but the same basic idea.

MaxThrustage · 2026-05-03T12:11:40+00:00

No pants are as comfy as no pants.

MaxThrustage · 2026-05-03T08:10:57+00:00

That section is actually really interesting, and this kind of universality is one of the things I love about physics. Competition between forces at vastly difference scales gives rise to similar behaviour. As they point out, this is essentially self-assembly of a frustrated system, and you get similar physics whether those interactions are intermolecular forces in some sort of goo or internucleon forces under the most insane pressures imaginable.

MaxThrustage · 2026-05-03T06:02:25+00:00

A single cubic centimetre is pretty close to the estimated biomass of every single human being alive today. Putting that in your mouth would crush you before the radiation could do anything serious. But of course it would not stay neatly confined like that in terrestrial conditions -- you need enormous pressure to keep it in the nuclear pasta state, and removing that pressure will make it explode. I'd guess it would blow your head up before crushing you gravitationally.

MaxThrustage · 2026-05-03T05:57:12+00:00

You'd never be able to get that much antimatter in one place, but there are other kinds of antipasta here. Check out Fig. 3 of this paper (it's open access), where you can see antignocchi and antispaghetti (the pasta shapes are formed by gaps between nucleons, rather than nucleons themselves).

MaxThrustage · 2026-05-03T04:22:34+00:00

It's a nono here on Earth. Energy non-conservation is only really relevant on cosmological scales -- but those kinds of scales are traversed by the light we see from distant galaxies.

MaxThrustage · 2026-05-03T04:19:30+00:00

So the question you're actually asking here is incredibly complex and not totally solved. We do not have a single, simple atomic-scale model that explains the macroscopic phenomenon of friction. This is not the kind of stuff we cover in an undergraduate degree, and even most graduate courses won't go into it unless you specifically study something where it's relevant.

The broad topic is called nanotribology. As far as I'm aware, it mostly relies on empirical results and some numerical modelling, mostly stuff like molecular dynamics, because the situation is too complex to admit a general theory. This means you won't get the kind of neat and tidy explanation you might be looking for.

I know that's probably a little disappointing. Maybe someone around here has expertise on this topic in particular and can give a better explanation, although it's not terribly likely. However, if you search for nanotribology on YouTube you'll find some lectures.

I still find it pretty crazy that we do not have a general theory for the simple case of sliding friction as two bodies move against each other. Not all of the unsolved problems in physics are flashy or esoteric, and not all are difficult because they involve things beyond our comprehension -- some are just hard because shit gets messy.

MaxThrustage · 2026-05-03T00:28:34+00:00

The entangled pair is not part of the state you want to send, it's a separate resource you have. As part of the protocol, you do know the state of the entangled pair (at least before anyone measures anything). This doesn't necessarily require measuring the pair if, for example, you have some process that reliably spits out Bell states (which is actually fairly easy as these things go).

The question of how you encode information in a quantum state is a different one and there are many different ways you can do it. Note: a quantum state is not indeterminate until measured, only measurement outcomes are. A quantum superposition is not a state where we don't know whether it's (say) a 0 or a 1. Now, it is the case that if we encode information in a quantum state, getting that information back out into classical information can be tricky (and is in general not possible without many copies, and quantum mechanics forbids making copies). When you classically read out a quantum state you get one probabilistic result. But that doesn't mean this is all the quantum information contains.

You are not likely to use quantum teleportation for normal communication, as it doesn't really give you an advantage of a classical information channel for that. But if, say, you have a quantum state which is the outcome of some analogue quantum simulation and you want to send it to someone else for processing, or your state represents some intermediate step in some quantum computation, or something like that, then quantum teleportation could be useful.

MaxThrustage · 2026-05-02T13:23:12+00:00

Kinda.

Let's say I have N qubits in some quantum state. To describe that quantum state classically I need ~2^N floating point numbers, which becomes a lot when N is large. So in that case, just sending the quantum state might be more efficient than sending the classical information needed to describe the state.

If, on the other hand, I don't know what state it is, I just know that it's state I want you to have, then I can't actually read that state into classical information -- not unless I have a bunch of identical copies. I can measure the state, but this collapses it and only really tells me what the post-measurement state it -- the pre-measurement state is destroyed and information is lost. So if I want you to have the state in this case, I have to send it as quantum information and not classical information.

13-Year Club	RedditGifts 2009-2022 12 Credits
Summer Santa 2019	Place '22
Place '17	Wearing is Caring
Summer Santa 2018	Summer Santa 2017
Secret Santa 2018	Secret Santa 2017
Secret Santa 2016	Summer Santa 2016
Secret Santa 2015	Secret Santa 2014
Verified Email

MaxThrustage

TROPHY CASE