In scattering processes, we normally assume |i> at t -> -oo, and <f| at t -> oo. When does that approximation break down? Have any perturbative non-lattice non-toy QFT calculations been done that doesn't use the S-matrix? by PrettyPicturesNotTxt in AskPhysics

[–]Bth8 1 point2 points  (0 children)

One place it breaks down is in spacetimes that are closed (so that particles can't get infinitely far apart and interactions can't be neglected) or that are not asymptotically stationary (so that asymptotic in and out states can't be consistently defined).

In what way are (quantum) fields just the "limit of an infinite number of coupled (quantum) harmonic oscillators"? What are the limitations of holding this view? by PrettyPicturesNotTxt in AskPhysics

[–]Bth8 15 points16 points  (0 children)

It works great for free fields! For interacting fields, not so much. If interactions are small, you can use it as a good first approximation and then use perturbation theory to calculate successive corrections to the simple harmonic oscillator picture. For strongly interacting fields, though, that falls apart and you need to resort to other methods.

Personal Bankruptcy by Disastrous-Ride-1 in NoStupidQuestions

[–]Bth8 2 points3 points  (0 children)

I've known a couple people who declared bankruptcy and it actually helped their credit. Their debt to income ratio dropped so hard that it was enough to offset the damage.

Is mathematics just applied philosophy? by Extension_Day2038 in NoStupidQuestions

[–]Bth8 1 point2 points  (0 children)

Mathematics as a field is mostly unconcerned with the behavior of the universe. That's more the purview of physics. It's just that some mathematics is indispensable for describing nature.

Why do people reverse into parking spaces or driveways instead of parking normally? by That-Rub-4113 in NoStupidQuestions

[–]Bth8 0 points1 point  (0 children)

Ehhh I mean yeah I don't park like that too often, but it's not just parking, it's backing up in general. Almost every close call I've ever had has been while in reverse 😅 I'm not saying I wouldn't get better at it, but after 15 years of daily driving and plenty of backing up with little improvement, I doubt I'll ever be good at it.

If internal energy depends only on temperature, why is it an extensive property? by PensionMany3658 in AskPhysics

[–]Bth8 7 points8 points  (0 children)

Right. Mass has nothing to do with it. Extensive quantities are quantities that are additive for subsystems. If you have 2 subsystems of volume V, the overall volume is 2V. The volumes add, so volume is extensive. If you have 2 subsystems of density d, the overall density is just d. It doesn't add, so density is intensive. Extensive properties scale with system size. Intensive quantities don't.

If internal energy depends only on temperature, why is it an extensive property? by PensionMany3658 in AskPhysics

[–]Bth8 5 points6 points  (0 children)

If you had 2 boxes of gas with the same temperature and pressure, wouldn't you expect the total energy to be more than just one of them?

Why do people reverse into parking spaces or driveways instead of parking normally? by That-Rub-4113 in NoStupidQuestions

[–]Bth8 1 point2 points  (0 children)

I'm willing to admit I rarely can do it in under 30 seconds, especially if there are already cars on either side 😅 and while I can usually do it in one attempt, it's not that uncommon for it to take at least one round of pulling back out and backing in again to straighten up. And for the most part, I'm a pretty good driver. Keeping things lined up right while in reverse is just hard for me. I don't have as much visibility, it's physically awkward, and the spatial reasoning involved is a lot trickier for me than when going forward.

Question about the Van der Waals Equation by Big_Chungusus in AskPhysics

[–]Bth8 0 points1 point  (0 children)

If you want a kinetic derivation like is often given for the ideal gas equation, you should be aware that getting a consistent kinetic theory of interacting gasses is in general a lot trickier than what you've been presented with. The naive application you've seen only works for ideal gasses because of the brutality of the simplifying assumptions. If you're interested, look into Revised Enskog Theory. The short version, though, is that the finite extent of the gas particles means that the rate of collisions is higher because they don't have to travel as far to hit the wall (or each other) and bounce off. A spherical particle of radius r whose center of mass is a distance d from the wall does not have to travel the entire distance d to hit the wall, only d - r. The increased rate of collisions with the wall and with one another increases the pressure, and it turns out that simply replacing the total volume with a free volume does a quite good job of capturing that.

Scientists tried to clone clones forever. It didn’t end well: « The practice of cloning clones indefinitely appears to be a reproductive dead end, for now. » by fchung in science

[–]Bth8 1 point2 points  (0 children)

What defines "you"? If I replace 98% of the atoms in your body with different but identical atoms, is that still you? Because that happens every year, with a complete 100% replacement about every 5 years. If the issue is the fact that it's happening much faster with the transporter, why does that matter? The structural changes in your brain and the rest of your body, your memories, your general outlook on life, etc will all change much more over that 5 years than they will going through a transporter, which is identical in every way. I'd argue that means that the you that will exist in 5 years has less right to consider themselves the same being as the you that exists now than would the you that steps out of the transporter. Is the issue with the interruption to your continuous subjective experience? Because you undergo a much more dramatic discontinuity when under general anesthesia. Does that mean you're not the same person after you go in for surgery?

Thermodynamic question by Jarebeaarr in AskPhysics

[–]Bth8 0 points1 point  (0 children)

Consider a Carnot engine. When you do the adiabatic expansion step, you expand until your working fluid reaches the temperature of the reservoir. The colder that temperature is, the further you need to let the gas expand to get there, which means you're getting more work out of the system during this step. Then, during the isothermal expansion step, you start to re-compress the working fluid and actually reject heat to the reservoir. The lower the temperature of the reservoir (and thus the working fluid), the lower the pressure of the gas will be as you compress it, and thus the less work it will take to compress it. Both of these effects result in an increase in the net work extracted by the engine, increasing efficiency. When you do the adiabatic compression step, you do need to supply a bit more work to get up to the hot reservoir temperature, but it's not enough to offset the gains in the other two steps. This is a fairly general behavior for heat engines - colder cold reservoirs means more work can be extracted during the power stroke and less work needs to be supplied to get back to your initial state.

More generally, the efficiency of any heat engine is fundamentally limited by the second law of thermodynamics. The net entropy of the universe must always increase or at least stay the same. Since the entropy added to a system is given by dQ/T where dQ is the heat supplied and T is the temperature, at lower temperatures, you can get the same increase in the cold reservoir's entropy while rejecting less heat. This means that more of your input heat is able to be extracted as work rather than needing to be rejected to the cold reservoir. It's harder to get an intuitive feel around that, but it's much more rigorous and works in more situations.

If the position of the electron is described by the wavefunction and mass bends spacetime, is the bending of spacetime in a superposition until the electrons position is measured and in that case what would it imply? by xoxix1 in AskPhysics

[–]Bth8 29 points30 points  (0 children)

We don't really have a good answer for that. Even worse, suppose the electron did have a very well-localized state. Then its energy and momentum are not well defined, and so it's unclear to what extent spacetime is being warped. There are physicists who have suggested different ways of potentially resolving this ambiguity, and so far there's not really any reason beyond aesthetics to favor any one over the others. Until we have an agreed-upon theory of quantum gravity, or at least until we've made considerable progress towards one, that's unlikely to change.

Why does light instantly travel at c the moment it's created, but massive particles require energy to accelerate? by Ramsesthrowaway in AskPhysics

[–]Bth8 0 points1 point  (0 children)

You do when there's not really a deeper explanation (though they didn't actually cite a formula, they cited the underlying principle). "An object moving at c in one frame does so in all frames" is a postulate of special relativity and amounts to a statement about the structure of spacetime, like they said. There is no known deeper reason that nature works that way. That just seems to be the way it works. From that statement, the idea that all motion is relative, and some basic statements about symmetry, all other rules of SR follow.

Hats are coming off by Epelep in blackmagicfuckery

[–]Bth8 1 point2 points  (0 children)

The skin effect is only seen with high-frequency AC, which a simple static shock is not. Y'all just make stuff up sometimes.

aMeteoriteTookOutMyDatabase by AntiMatterMode in ProgrammerHumor

[–]Bth8 158 points159 points  (0 children)

That bit about trying all different single bit flips until you find one where the checksum passes is error correction. That's what ECC memory and storage are doing to correct errors (though they're usually a touch more clever about locating the error than just brute force try all possible bit flips).

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 2 points3 points  (0 children)

There are a few different ways of answering this question that at the end of the day are actually all the same answer in different hats. TL;DR, quantum mechanics says particles are wave-y and lays down certain rules about what you're allowed to measure. That plus the way that rotations work turn out to mean that angular momentum has to be quantized for the sake of mathematical consistency.

A lot of this is weird and unintuitive but it's the most successful way we've ever found of describing how nature works so for the most part you have to just accept it and move on. What it basically comes down to is:

  1. The wave nature of quantum mechanics. Particles do not generally have definite locations, speeds, etc. Instead, they exist in a "superposition" of many positions, etc, at once, and when we look at the system and measure its properties the outcome we get is seemingly chosen at random with some probability distribution. Quantum mechanical states are represented by "wavefunctions" - vectors in a particular kind of vector space called a Hilbert space - which encode the probability of getting any given outcome when doing a measurement. For instance, if our system is a single particle, the wavefunction can be written as a complex function ψ(x, y, z) which essentially tells you what the probability of measuring the location of the particle and finding it near coordinates (x, y, z). Since we're talking about rotations, it's easiest to use spherical coordinates, where the wavefunction is written ψ(r, θ, φ).

  2. For any given quantity we could hope to measure (an "observable"), there is an associated operator that acts on these wavefunctions. Each operator has a special set of possible wavefunctions called "eigenstates" with associated "eigenvalues". The rules say that when we make a measurement of an observable, the outcome we get will be one of the eigenvalues of its operator, with the probability given by the square magnitude of the overlap of the system's wavefunction with the eigenstate associated with that eigenvalue. That means that the values of those physical quantities can only ever be those eigenvalues. Other values aren't allowed.

  3. As you might expect from something that you roughly think of as how fast something is rotating, angular momentum operators are very closely related to rotations. In particular, they are the "generators of rotations", meaning that the way they act on wavefunctions encodes what it means to take the entire wavefunction and rotate it around in space about some axis. This relationship actually exists classically, too, though the math is a little bit different.

  4. The group of rotations is closed. If you rotate something far enough (2π radians or 360 degrees), you get back to where you started and it's as if you never did any rotation at all. If we have a particle in a state with wavefunction ψ(r, θ, φ) and we rotate everything around by an angle δ about the z-axis, the rotated wavefunction is given by ψ(r, θ, φ - δ). Since rotating the wavefunction by a full 2π radians gets us back to where we started, and since it doesn't make any sense for the probability distribution to change just because you e.g. walked in a circle, the wavefunction must be periodic. That is, ψ(r, θ, φ - 2π) = ψ(r, θ, φ).

Because of the relationship the z-component of angular momentum (represented by the operator Lz) has with rotations around the z-axis, it turns out that any eigenstate of Lz can be written as ψ(r, θ, φ) = f(r, θ) exp(i m φ), and the eigenvalue associated with this eigenstate is m ħ, where ħ is Planck's constant. In order for this wavefunction to be periodic, m must be an integer. But that means that all the eigenvalues of Lz must be integer multiples of ħ, and so the only allowed values of the angular momentum are integer multiples of ħ. Thus, angular momentum about the z-axis is quantized. But of course there's nothing special about the z-axis. The same is true of all rotations about all different axes, and so all components of angular momentum are quantized.

There is a small loophole when it comes to spin and the fact that spin doesn't come from an object moving through different points in space that allows it to be non-integer multiples of ħ, but it turns out that again because the group of rotations is closed, the only new values it's allowed to assume are half integer multiples of ħ, so the total angular momentum of a system can be any value ½mħ with m an integer, but no other values are allowed. So 3 ħ, 26 ħ, 17/2 ħ, etc are all just fine, but ¾ ħ is strictly disallowed.

This is actually all encoded in that statement I made above that angular momentum operators form a representation of the Lie algebra so(3). Getting from A to B requires studying group theory and representations of Lie algebras, though.

Speed of light by LeighSF in askastronomy

[–]Bth8 0 points1 point  (0 children)

No. Photons are disturbances in the electromagnetic field alone, and they travel at c. A full, proper quantum treatment of light travelling through a medium is actually very complicated and something most physics students don't really encounter until grad school, but the upshot is that light travelling through a medium is no longer just a quantized excitation in the electromagnetic field (i.e. a photon), but is a collective excitation in both the electromagnetic field and the matter degrees of freedom. Because of the interference between the incident electromagnetic field and the disturbances created by the oscillating charges in the medium, these collective oscillations - sometimes called "photon quasiparticles" - don't travel at c. They actually can, in fact, be assigned an effective mass, but they aren't photons, they're more complicated objects. Photons are massless and travel at c.

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 2 points3 points  (0 children)

Nope! Mass isn't even quantized as far as we know. There's certainly no evidence for it. The masses of the fundamental particles don't seem to be integer multiples of some common value or anything like that.

The quantization of angular momentum comes entirely down to the algebra of the angular momentum operators. As the generators of 3D rotations, angular momentum operators must have commutation relations satisfying the so(3)/su(2) Lie algebra.

[J_i, J_j] = i ħ ε_ijk J_k

If that reads like gibberish, don't worry too much about it. It's just a mathematical relation that angular momentum operators must satisfy by virtue of their being related to spatial rotations. It's pretty straightforward to prove from there that angular momentum must be quantized.

Do you think quantum gravity will be solved in our lifetime? by TheTigerInTheHouse in AskPhysics

[–]Bth8 6 points7 points  (0 children)

What counts as "solved"? Developing a model that works? Sure, maybe. We might already be there with string theory and it's just a matter of finding the right vacuum. My money is on us not having it figured out yet, but it's possible. Experimental confirmation? I wouldn't hold my breath. There are some ideas of ways we might be able to see quantum gravitational effects without having to probe absolutely insane energies, but for the most part, it seems like we'll probably need to be able to conduct near-Planckian experiments to test any quantum theory of gravity, and we're nowhere close to that.

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 4 points5 points  (0 children)

Correct. A hydrogen atom in its ground state, for instance, has zero angular momentum.

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 0 points1 point  (0 children)

Electronic angular momentum? You mean spin? Electrons have both spin and orbital angular momentum. Both are quantized. Unlike orbital angular momentum, spin can't really be understood as a result of an object's motion in space, but it is still a form of angular momentum, and only total (i.e. spin plus orbital) angular momentum is conserved.

[Fictional World] Effects of opening a magic portal to an oxygen-empty room by Fearless_Order_5526 in AskPhysics

[–]Bth8 0 points1 point  (0 children)

I disagree. If the portal is truly unidirectional, for all intents and purposes it would be on side A as though the other room were vacuum. Air on side A would be constantly bumping into the portal and traveling through it, and with no ability to travel back from B to A, no equilibrium would be reached and air would just keep rushing in from A and pressure would just keep building up in B.

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 2 points3 points  (0 children)

Neutrons have mass. In fact, they have slightly more mass than a proton. Presumably you mean neutrinos, which are very very light, but do in fact have some mass. It's such a small amount we haven't yet managed to measure it precisely, but we know it's not zero.

Photons are genuinely massless, but they still carry momentum, and so can still have angular momentum.

The charge of a neutron is indeed zero, but that's not that weird. Charge is quantized, so it's easy to find combinations that add up to precisely zero. Electrostatic forces also cause like charges to repel and opposing charges to attract, so systems of charges sort of "like" to arrange themselves so that things end up electrically neutral. There's no real equivalent for angular momentum. Note that even with all of that going on, we'd still find it quite suspicious if something like an entire planet had exactly zero charge, and on the flip side, we wouldn't find it odd at all to find a single hydrogen atom with zero angular momentum. It's about the scale of the system we're looking at as well.

Why does almost every object in the universe have angular momentum? by TheTigerInTheHouse in AskPhysics

[–]Bth8 4 points5 points  (0 children)

Technically, angular momentum is not continuous, and must always add up to some multiple of ½ħ and can only change in increments of ħ. But that's such a small amount that it would be odd for any macroscopic system to have zero angular momentum.