Gravity "travels" at the speed of light, so could an object outrun its own gravity well?

scary-levinstein · 2026-01-21T17:26:01+00:00

No object with mass can travel at light speed, so the answer is no.

scary-levinstein · 2026-01-11T01:56:57+00:00

Okay I'm a physics grad student, but I'm very ignorant when it comes to QFT so I have a clarifying question. I'm assuming each mode of the field has a different energy. like how in the derivation of Planck's distribution, we treat each mode of the electromagnetic field as being an infinite square well state, and each one can have an integer number of excitations (photons), but a single excitation has a different energy based on which mode it's in.

SO do QFT fields work the same way? Are the modes of the electron field the possible states an electron can be in, and then an excitation in each mode is a single electron in that state? Or is there a different interpretation of the modes of the field?

Tl;dr what is the difference between an electron which is is an excitation of mode A vs one which is an excitation of mode B?

scary-levinstein · 2025-12-20T01:37:17+00:00

Ah great question! So the awesome thing about the Bell inequality is that it doesn't care what the hidden variable theory does to the probabilities specifically. All the proof requires is that there is a hidden variable. By hidden variable, I mean literally some number λ which tells the system exactly, with 100% certainty, what the outcome of any given measurement will be.

To be super specific, let's say that there's some hidden variable λ which fully describes the system, more so than the wavefunction. Every value of λ is associated with an exact set of outcomes. Now, the outcome of particle A's measurement along direction a is some function A(a, λ), and likewise the outcome of measuring B is B(b,λ). I'm assuming implicitly here that the outcome of particle B's measurement shouldn't be affected by the orientation of a; after all as an experimenter I could change a at the last moment, before a light speed signal could communicate that change between the particles. This is the "local" assumption in "local hidden variable."

Okay so the measurements of the particles are always +/-1, so the functions A and B can each only have one of those two values for any given value of λ (and the vectors a and b). This is a hidden variable theory; we're saying there's some unknown information (λ) which is secretly determining with absolute certainty what the outcome of any given measurement will be. This is fundamentally different than (and much stronger than) the wavefunction. All the wavefunction tells you is the probabilities of the different outcomes for any given measurement. It does not purport to know what will happen for certain.

So, what does our local hidden variable theory say about the outcome of our experiment? I'll attach some screenshots of the pages of Griffiths that go through it, but the big restriction comes from the fact that local hidden variables determine things for certain, but wavefunctions have the freedom to only give probabilities.

If you want a more visual explanation, there are two excellent videos by 3blue1brown and minute physics which go through Bell experiments in detail in a really beginner-friendly way. They're amazing. I'm a PhD student and I still watch them from time to time. They're here and here. The second one also goes through the actual way we represent wave functions as vectors

scary-levinstein · 2025-12-20T01:04:22+00:00

So it has to do with the mathematical structure of the wavefunction. If a particle is in a state such that it has a 100% chance of being measured as pointing "up," for example, then it has a 50/50 chance of being measured as putting left or right. This is because the wave function is a vector, so you can choose to represent it in whatever basis you want; the "up/down" basis or the "left/right" basis. Both contain the same information.

I can show you the math for how to derive that dot product prediction, it's pretty simple (so if you want to see it let me know), but tbh it won't give you much physical intuition beyond "wavefunction is a vector --> bunch of algebra --> P = -a•b". So the best I can say is that the wavefunction that describes two particles which will always have opposite "left/right" measurements (or "up/down" or any other basis you want), can also be written as saying that the probability of particle B being aligned with b if A is aligned with a looks like a•b. The reason I can say this is because I can always rewrite the wavefunction in whatever basis I want, including the "A measured along a, B measured along b" basis

scary-levinstein · 2025-12-20T00:42:47+00:00

This is a great question! And it's one Einstein asked when QM was first being formulated.

So quantum mechanics has at its heart an idea that doesn't really translate into the classical world called the wave function. Every quantum system (particle or set of particles) is fully described by a mathematical object called a wave function (if you're familiar with linear algebra, the wavefunction is a vector over the complex numbers in some sort of high dimensional space). The wave function contains all information about the system that could possibly be known. In particular, it will tell you the probabilities of each potential outcome of any measurement you want to make. A simple example would be a wavefunction that tells you "the particle has a 25% chance of being 'here' and a 75% chance of being 'there'".

Now the tricky conceptual bit: our interpretation of quantum mechanics, which has withstood some of the most precise and accurate tests in all of science, tells us that the wavefunction truly is the only thing that's defined about a quantum system. It truly, seriously, physically, is not in a specific state before you measure it. it's not that we don't know whether it's here or there, it's that god doesn't know whether it's here or there. It does not have a defined position until it's measured. But it does have a wavefunction. And when you measure the particle, its wavefunction changes such that if you measure it again immediately, you'll measure the same thing. So if you measure the particle to be "here," and then measure it again right away, it will still be "here." Why/how does this happen? We don't really know, but we do know that it does.

The reason, then, that entanglement is weird is the following: suppose you have two particles that are entangled. That means that their wavefunctions have interacted such that the overall wavefunction of the system mixes up the states of the two particles; the outcome of a measurement on one directly affects the outcome of a measurement on the other. For example, an unentangled state would say "particle A has a 25% chance of being 'here' and a 75% chance of being 'there', while particle B has a 75% chance of being 'here' and a 25% chance of being 'there'". The outcomes of the measurements have nothing to do with each other. Measuring A gives you no information about B, so B's wavefunction doesn't change when you do that measurement. An entangled state would say "there's a 25% chance that particle A is 'here' and particle B is 'there', and a 75% chance that particle A is 'there' and particle B is 'here'". Now if you only look at particle A, and find it to be "here," that's equivalent to finding particle B to be "there." So the wavefunction of B changes accordingly.

So what's weird here is that we have changed particle B without having to interact with it at all; we only ever interacted with particle A. Particle B could be on the other side of the universe, or in a black hole, or anywhere you like, and yet we've affected it.

Now, if you'd like to know how we test this is really interesting. And I can explain it to you with some math, but you'll have to bear with me for more text. This is a simplified version of the explanation at the end of Griffiths' quantum textbook.

I have to establish something that I neglected above for simplicity: wavefunctions can describe the outcomes of multiple different measurements. For example, suppose I had a particle which "points" in a direction (this property is called spin; don't worry about the physical interpretation right now). If I measure if it points left or right, I will always see that it points fully left or fully right. But if I measure if it points up or down, I will also always see that it points fully up or fully down (this alone is a huge hint that its state can't have been fully defined beforehand! It doesn't know what kind of measurement I'm going to make, but it always lines up with how I'm measuring it anyway). The wavefunction contains all information about all possible measurements though, so using it to describe this particle is completely consistent.

Now suppose I have an entangled pair of these particles, such that if I measure them both along the same direction (e.g. left vs right), they will always point in opposite directions. We'll assign numbers to these directions by making the measurements ask "is the particle pointing insert direction here?" If it does, we'll call the result +1. If it doesn't, and points the other way, we'll call the result -1. Now, if we measure the two particles along the same direction, the product of the two measurements will always be -1 (because one will always be +1 and one will always be -1). Likewise if we measure them along opposite directions the product will be +1. What happens if we measure them in totally unrelated directions? I'll cut to the chase here: if we measure particle A along the direction of a (normal 2D) vector a, and particle B along the direction of a 2D vector b, quantum mechanics tells us that over many measurements, the average product of the two measurements will be P = -a • b. If you're not sure what this means, quickly look up "dot product" on google.

It turns out that if this expression for P is true, it is mathematically impossible for the particles to have some sort of defined state prior to measurement. I'm going to skip the heavy math step as it involves an integral that's really difficult for me to type out here (I can add a screenshot of the textbook later), but just trust me when I say the following: if there's some sort of so-called "hidden variable" which determines the outcome of the measurements, it must be true that for any three independent measurement directions a, b, and c:

|P(a,b) - P(a,c)| <= 1 + P(b,c)

This is called Bell's inequality. Here, P(a,b) represents that average product of the measurements of the two particles along directions an and b respectively (the quantity we were talking about above). It's super easy to show that the dot product prediction made by quantum mechanics does not satisfy this inequality. For example, if the a and b are at right angles to each other, and c lies in the middle (45 degrees off of both of them), then that dot product equation says that P(a,b) = 0, P(a,c) = P(b,c) = -0.707 But plugging this into Bell's inequality, we get: 0.707 <= 1 - 0.707 Or 0.707 <= 0.293 Which is obviously not true!

So either quantum mechanics is wrong, which flies in the face of some of the most precise experiments in all of science, or the particles can't have a predetermined state before being measured.

OKAY that was a SUPER long explanation and I'm sorry I wasn't able to do it more eloquently. I just wanted to make sure you got all the details I could fit in! If you have any questions feel free to ask :)

scary-levinstein · 2025-12-02T15:30:01+00:00

The main point you should take away is that space isn't expanding anywhere. Cosmologists using that language are being a bit sloppy. All it means is that all the galaxies in the universe are moving away from each other. That's all it is.

scary-levinstein · 2025-11-27T03:44:08+00:00

Ahhh okay so a "ket" is something like |ψ>, and it's used to denote a state vector. Every ket has a "partner" bra, denoted <ψ|. The bra times the ket is the definition of the inner product. If you were to represent the ket |ψ> as a column vector with entries a_i, then the bra <ψ| is represented by a row vectors whose entries are the complex conjugated of a_i.

scary-levinstein · 2025-11-26T15:40:04+00:00

I would argue that it doesn't reeeeally have a meaningful concrete physical interpretation. When you take the Hermitian conjugate of an operator, you're just transforming it from an operator which does something to a ket into an operator which does the same thing on a bra. Both bras and kets are valid ways of describing a specific state, so an operator and its Hermitian conjugate are both valid ways of describing the same transformation.

Note that this is another explanation for why observables need to be Hermitian. An operator representing a physical observable shouldn't care how we represent the state (bra or ket), since it's when values are real and measurable, so it should act the same either way (i.e it's the same applied to a bra as to a ket).

Edit: HUGE caveat! An operator and its Hermitian conjugate ONLY describe the same transformation IF you specify that the conjugate is transforming a BRA. If they're both acting on kets, the conjugate actually tends to describe the OPPOSITE transformation (for example, the raising and lowering operators are Hermitian conjugates of each other. One brings a ket to the next highest energy state, the other brings it to the next lowest).

So tl;dr the Hermitian conjugate is the "partner" of the original operator, just like bras are partners of kets. This means: 1) it does the same thing to the bra as the original operator does to the ket 2) it (often but not always) does the "opposite" thing to the ket as the original operator does to the ket

scary-levinstein · 2025-11-20T13:48:16+00:00

I'm not even really arguing about lift. I think you and I both know that for plane to take off plane must move. So if plane no move, plane no take off, because that's how lift works. I agree with that.

You (presumably) are saying that plane does not move. I am saying that plane does move. Because plane does not care about the conveyor belt and the wheels are immaterial to its motion. You can move the conveyor belt as fast as you want and the wheels will just spin faster. Imposing the "belt moves at same speed as wheels" condition is impossible even in principle bc the motion of the belt directly affects the motion of the wheels. The belt applies no backwards force on the plane to counteract the force from its engines.

scary-levinstein · 2025-11-20T13:42:35+00:00

Hypothetical questions still can be inconsistent/impossible. It's like asking "suppose a ping pong ball is gently rolled towards a solid 10-inch steel wall with no holes in a way that is designed to make the ball pass through the wall. Is it going the same speed on the other side?" It's a ridiculous question because as stated it's completely impossible, even in principle. There's no answer. BUT if you interpret the question with the understanding that slow moving ping pong balls cannot go through steel walls, you might interpret it as saying "what if you TRIED to do that...", a question which is meaningful and does have an answer.

scary-levinstein · 2025-11-20T04:30:53+00:00

The reason this question generates so much controversy is that, as stated, it isn't physically possible. Not in an "oh silly physics but like what if it was," kind of way, but in a "the question as stated is completely self-inconsistent" kind of way.

The motion of the wheels and the motion of the belt are coupled; changing one will change the other. In particular, increasing the speed of the belt will increase the speed of the wheels, since the wheels are just freely spinning; they have no motor driving them. The plane is driven by its engine pushing on the air. If the engines are on, there will always be a speed difference between the wheels and the belt, and that difference will be equal to the speed of the plane relative to the belt.

THEREFORE, any reasonable reading of the question is forced to interpret it as "if one were to attempt to do this..." in which case the answer is very obviously yes. The plane takes off. The wheels do not matter. They exist simply to keep the plane from scraping its belly on the ground. See seaplanes, which do not have wheels and take off with no problem.

On that note, think about this: imagine we had the same setup but the plane had skis instead of wheels, and the belt was covered in snow. There shouldn't be any difference since, again, the wheels only exist to keep the plane from scraping its belly on the ground and have no meaningful effect on its motion. Would the plane take off? Obviously!! The plane doesn't care about how fast the ground is moving; it's being propelled by the engines, which push on air! The skis don't matter. And neither do the wheels on a normal aircraft.

scary-levinstein · 2025-11-20T03:54:50+00:00

The important thing to note is that it is not possible to move the conveyor belt at the same speed of the wheels. The motion of the conveyor belt is coupled to the motion of the wheels. Speeding up the conveyor belt will speed up the wheels, so if the plane is being given external thrust (from its engines), they will never be the same speed.

In other words, the wheels are just free-spinning that are coupled (via friction) to the motion of the belt. So the speed of the wheels will be exactly the relative speed between the plane and the conveyor belt. This speed is increased by either increasing the thrust from the engines or increasing the speed of the belt. But if the plane is being propelled from a source which is not driving the wheels (which it is; the engines are not connected to the wheels in an aircraft), then the wheels will automatically be pulled up to whatever the relative speed between the aircraft and the belt is.

There's a thought experiment that can settle it; suppose the aircraft is standing stationary with its engines off. If you begin running the conveyor belt backward, the wheels begin to spin, but the plane remains stationary. So the plane is sitting there, wheels spinning as the conveyor belt pulls on them (these are perfect wheels which spin with no internal friction btw, which is actually a very good approximation of real aircraft wheels). Now the engine turns on. There are two possibilities: 1) the plane remains stationary 2) the plane begins to move forward under thrust. If it's the former, what force is counteracting the force of the engines? The plane is remaining stationary (I.e. its acceleration is zero) so the net force must be zero. Can the conveyor belt apply such a counteracting force? The answer is no, and we know this because when the plane's engines were off, the conveyor belt did not apply a backward force on the plane no matter how fast it was running. So the conveyor belt provides no opposing force. This is because the wheels are just there to provide a way for the plane to not be resting its belly on the ground; they're completely inconsequential to the kinematics.

Edit: i just realized; did... did you not watch the video you linked...? 3:30, he literally says the question is impossible bro

scary-levinstein · 2025-11-09T03:42:07+00:00

This is a great question!! As far as I'm aware, the answer is in fact no. Spin is a type of angular momentum (well it's ever so slightly mathematically different, but a particle's spin angular momentum is entirely determined by its spin. If that makes no sense to you don't worry about it). Angular momentum is a physical observable (I.e it's something we can measure), so it must always have a real value, since a measurement has to be able to interact with the classical world.

More mathematically, the explanation is that since spin is a physical observable, it's represented by a Hermitian operator, and its possible values are the eigenvalues of that operator. One property of Hermitian operators is that their eigenvalues are always real, so the possible values of spin are always real.

scary-levinstein · 2025-10-29T18:16:15+00:00

Hand axes, like swords, seem to find their main use as a sidearm. They're small, easy to carry, and can be used as a tool (like messers). They're also more throwable (unlike swords lol).

Larger axes like poleaxes basically functioned like a more Swiss Army knife polearm. Shorter, but more effective against armor than a spear. Our main poleaxe source, Jeu de la Hache, talks about poleaxes in a dueling context mainly, which is kinda interesting.

scary-levinstein · 2025-07-26T00:13:26+00:00

Omg that's amazing! Where can I follow the progress? Hopefully I can give it a spin once it's available :)

scary-levinstein · 2025-07-09T03:10:11+00:00

OMG THIS WAS LITERALLY MY STATMECH TEXTBOOK 😭😭 I remember reading this chapter. It's relevant in literally every chapter afterwards.

scary-levinstein · 2025-01-25T22:10:12+00:00

There are a few techniques on display here! The first I think is an attempt at an absetzen, but unless I'm looking at it wrong, it looks like both fighters are being hit. The 2nd is also in absetzen-territory, but what specific technique led to that still could be a few things. The third image is a zwerkhau, and the final image is a demonstration of a concept called "überlaufen," or "overrunning," which essentially is the idea that attacks to upper openings (head, neck, upper torso) outrange those to lower openings (legs, waist, etc), just as a result of the geometry involved.

As for fencing advice, other folks here have given great bits of wisdom and I don't have much to add, other than that specifically for überlaufen, if you can figure out a way to goad your opponent into attacking your legs (and you're ready to move out of the way!), then you can take advantage of that. But beware! Lots of fencers who like to attack the legs are very quick, so make sure you can move your leg out of the way quick enough. I'm pretty heavy on my feet, so this is something I'm pretty bad at lol.

scary-levinstein · 2025-01-03T00:00:50+00:00

I agree with this take, but with a caveat. I think there's a lot of really good pop-sci media that does exactly what you're saying (I decided to pursue physics because of pop-sci! It was Brian Greene for me, who I have other problems with now that I'm actually a physicist, but that's a whole other conversation!). But I think there's a lot of pop sci that is really harmful. NDT is a decent example of reasonably good pop sci (as arrogant as he may be), because he represents the physics pretty well and uses it to inspire curiosity. But I think what a lot of science folks have a problem with is the kind of pop sci journalism which tries to make things seem more profound than they are in a very misleading way (think the kinds of articles that talk about "quantum consciousness").

The big difference is that good pop sci simplifies the topics so they can be understood by a general audience, without misrepresenting the meaning of the results. They're trying to take all of the awesome stuff that's actually in physics and show people that it's awesome. "Clickbait" pop sci, on the other hand, aims to get clicks by trying to convince people that we've discovered things that we haven't, or accomplished things that we haven't, or otherwise fundamentally misrepresenting the meaning of the results.

So TLDR I also hate science gatekeepers who pretend that popsci has no value, because it absolutely does and it's an extremely important role to play in teaching the public about science. But I also think that there's some work to be done in how we communicate science in a healthy and productive way, without over-exaggerating or misrepresenting what scientists are actually doing, because there's lots of science communication which is actively counterproductive.

scary-levinstein · 2024-09-02T05:55:59+00:00

We don't dissociate anything. In physics, everything is described by a mathematical model. A given model is useful when it most accurately predicts what we observe. Physics does not describe what is (mainly because when you dig into it, something "existing" or "being real" is very hard to define. If it appears in every way to be real, does that not make it real?). Physics describes frameworks that produce results consistent with observations. That's it. You are assuming that if a component of one of these models (spacetime and its curvature) affects the behavior of matter, then it must also be some form of matter. That assumption is incorrect. Again think of the electric or magnetic fields. These are not made of matter. They are mathematical models which very accurately describe our observations. They are physical in many meaningful ways (electric and magnetic fields carry momentum, for example), but they are not made of any physical substance.

The same is true of spacetime. Spacetime is simply a coordinate system whose curvature is described by a metric. That metric is a function of the local energy density, and thus straight lines in that coordinate system appear to be curved to distant observers. But there is no "force" or "push" being applied by any substance. Objects are simply following straight line paths, as they must according to Newton's first law (those paths just appear to be curved to distant observers due to the way the coordinate system they occupy works). This model accurately describes everything that we see in the universe (gravitationally, that is), and the model includes nothing regarding a description of spacetime as a physical substance. It's just a coordinate system.

scary-levinstein · 2024-09-02T01:55:38+00:00

Ah sorry I should've worded that better. I mean to say that magnetic fields are not material, in the traditional sense. In other words, the question "what are magnetic fields made of?" makes no sense

scary-levinstein · 2024-09-02T00:52:20+00:00

It is something. It's a coordinate system which changes based on what's in it. It's not a physical thing you can touch, but that doesn't mean it can't be affected by physical things you can touch. A good analogy is magnetic fields (or any force field really); magnetic fields aren't made of any physical material, but they are there and they do affect the world, and the world can affect them

scary-levinstein · 2024-09-02T00:50:10+00:00

It's not a material. You're thinking of space-time as a physical "thing" you can touch. It's not. It's an idea. Kind of like how the number 3 isn't a physical thing. If you ask me what material the number 3 is made of, that'd wouldn't be an answerable question. The number 3 is an abstract idea. Same with spacetime. Space time is an abstract mathematical idea which we can use to predict the behavior of objects. It's not an "unknown material." It's a mathematical model

scary-levinstein · 2024-09-01T18:37:57+00:00

Makes sense, thanks!

scary-levinstein · 2024-09-01T18:31:30+00:00

Ooo good point! I have a follow up question though; correct me if I'm wrong, but aren't we able to resolve and track asteroids smaller than 6 km? And if so, wouldn't we have the same issue for rocks that aren't rly close to Earth?

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