knzhou

general recent conversations

Discussion on question by Liliane: No…

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knzhou

Apr 16 16:32

Do you think there might be a chance that you’re overestimating how unique or fully-formed your ideas are, or overestimating how closely other ideas resemble it?

Discussion on question by Idiot: What…

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knzhou

Apr 7 16:09

You’re asking for advice, but when everybody gives you the obvious advice of just improving your own knowledge, you dismiss it as impossible. What are you looking to hear? Obviously, if your plan is to never learn basic undergrad material, you are doomed and should quit immediately to avoid wasting more of your life. Happy now?

knzhou

Apr 7 16:09

There’s no shame in building your understanding through dedicated study. After my undergrad, there were a lot of things I felt I didn’t understand or retain. I filled it all in over the next few years by working through a lot of books on the side.

Discussion on question by uhoh: How t…

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knzhou

Aug 14, 2024 21:44

That's the impression I got from your question -- you said that you were mainly tasked with English editing, didn't even see the manuscript as it was written, and didn't mention doing anything else. I can't imagine ever taking part in a collaboration where I can't even see the manuscript.

knzhou

Aug 14, 2024 21:44

Unrelated to the question, but how can you get coauthorship just for English editing?

Discussion on question by user366312:…

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knzhou

Oct 16, 2023 16:05

If you have no evidence to believe such an outlandish story, then maybe you shouldn’t worry about it. Old people tend to accumulate lots of half-remembered scary stories about the world.

knzhou

Oct 16, 2023 16:05

Are you sure your stories are actually true? I just searched "free Tibet" in a Chinese search engine and it worked fine. It even returned a lot of relevant results!

Discussion on answer by Ján Lalinský:…

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knzhou

Sep 9, 2023 01:25

Every magnetic field was created by charges somewhere, so you could always say that those charges are interacting with the particle of interest directly. If you follow your train of thought to its conclusion, you will end up rejecting the whole notion of a field and just thinking in terms of particle action at a distance, putting physics back about 200 years. It's an untenable position.

knzhou

Sep 9, 2023 01:25

What? You really don’t believe that accelerating single electrons radiate? Have you heard of the one electron cyclotron?

Discussion on question by foolishmuse…

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knzhou

Aug 26, 2023 01:36

Abramowicz and Bajtlik are considering much subtler situations where the spacetime itself is significantly curved. Their heuristic is good for such cases, but it doesn't apply to the usual twin paradox or your version with the Earth, which are both much simpler.

knzhou

Aug 26, 2023 01:36

I don't think that's the right resolution either. In your modified version there's an easy frame to pick, which is the Earth frame because both people start and end at rest with respect to it. But you can get the same result by working in any inertial frame, it would just be less convenient. And in all cases, you can't get the right result working in a noninertial frame, unless you're extremely careful with the turnaround.

knzhou

Aug 26, 2023 01:36

The reason the regular twin paradox is a paradox is because somebody could ask "why can't you work in Bob's reference frame and get the opposite answer?" and the answer to that question is that Bob's reference frame is accelerated. In your modified version, both Alice and Bob accelerate so there's no point in working in either's frame. There's no paradox: you just work in the Earth's frame and in that case the answer is straightforward.

knzhou

Aug 26, 2023 01:36

"They both travel the same distance in spacetime" -- no, they have the same beginning and ending points, just like in the regular twin paradox, but they take different paths through spacetime, just like in the regular twin paradox. To figure out which is younger, you just add up their aging rates along their paths accounting for time dilation.

Discussion on question by Apekshik Pa…

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knzhou

Aug 21, 2023 00:26

@JánLalinský I haven't checked, but it seems likely. With a deformation of the plank, the net normal force is effectively allowed to point in any direction, and that freedom would seem to make it easier to keep the plank up. (For example, you could have it pointing purely upward, which seems better than having part of it point leftward.)

knzhou

Aug 21, 2023 00:26

@JánLalinský Of course I agree -- it's just that usually when such problems are posed, the problem writer does not specify either (they don't mention deformation, and they draw the corner as perfectly sharp). Certainly if more detail is added then the problem is well posed.

knzhou

Aug 21, 2023 00:26

Anyway, the underlying reason I'm complaining so much about this is because I've noticed that these questions are pretty common in the Indian physics curriculum. They are intentionally avoided in the American, European, and East Asian curricula precisely because of the ambiguity I'm pointing out, but they seem to confuse Indian students all the time -- this is far from the first time I've seen this. From my perspective it's a flaw in the curriculum that needs to be remedied.

knzhou

Aug 21, 2023 00:26

@JánLalinský Is that so? The edge of my wooden desk is sharp to only within a few millimeters, so you probably would be right in that case. But the edge of a machined block of metal is sharp to much less than a millimeter. If you put something softer on it, like a long plank of wood, then the plank can certainly deform by more than that. The corner will leave a visible indentation in it. So I think you get different answers for metal on wood vs. wood on metal, for instance.

knzhou

Aug 21, 2023 00:26

In other words, this problem is not like other problems in mechanics, where we make idealizations like neglecting air drag, because here there are multiple possible idealizations, and you get totally different answers depending on which idealization you make. You think one idealization is obvious, I see at least 4 reasonable choices which could each hold in realistic situations. For this reason it's a badly written question.

knzhou

Aug 21, 2023 00:26

I'm saying that in real life, the answer is ambiguous because it depends on material properties. You're insisting that there's an unambiguous answer, but that's because you're making a specific implicit assumption about the material properties, which is just as likely to hold as not. Certainly you would agree that if we actually set up this system in reality, the answer would depend on the relative hardness of the two materials, and also on precisely how sharp the corner was, and how smooth the rod was?

knzhou

Aug 21, 2023 00:26

@JánLalinský You’re just saying what I’m saying. Without any deformation the problem is ambiguous. You propose to deform the corner and not the plank — that corresponds to making the nontrivial physical assumption that the plank is made of much harder material than the corner. If the hardness was reversed then you would get a different answer. The question doesn’t specify which material is harder, so it’s ambiguous.

knzhou

Aug 21, 2023 00:26

@JánLalinský If neither deform than the problem is undefined because there’s no single normal direction — in exactly the same sense that the ratio $0/0$ is undefined.

knzhou

Aug 21, 2023 00:26

@JánLalinský That’s wrong though. Look at the corner: there are three different normal directions at that point (two from the corner itself and one from the plank). If the corner is much softer than the plank, then it will deform so that the plank’s normal is the right one. But if instead the plank is much softer than the corner, then the corner will dig into it, producing two separate, independent normal forces (one horizontal and one vertical). And in general it’s between the two situations and you need to use full contact mechanics. The problem as written is ill-posed.

knzhou

Aug 21, 2023 00:26

The question is just completely wrongly posed because there’s no well defined normal direction at the corner — and the existing answers seem to be blissfully unaware of this. The actual answer depends on how the mass and corner deform, which depends on their material properties. Where did you find this question?

Discussion on question by Charles Hud…

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knzhou

Oct 2, 2022 21:02

Though if you insist on using path integral language only, you can use the fact that the field path integral is (in many cases) equivalent to one over particle worldlines, which shows the two are equally "real" in that language.

knzhou

Oct 2, 2022 21:02

This is one of the disadvantages of the path integral formulation: it's a great way to get (and define) the answers to certain technical questions, but often gives little intuition of how the dynamics actually work.

Discussion on question by Árpád Szend…

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knzhou

May 25, 2022 01:41

Really, that commonly shown table is an oversimplification... all four of those numbers are actually different things. You could definitely also compute some similarly oversimplified random number for dark energy, but the choice is pretty arbitrary.

Discussion on answer by Andrew: Why d…

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knzhou

Feb 15, 2022 00:51

Alright, I agree that a good model depends on the detector and the field! For fields where we sometimes work in the classical limit (electromagnetic, gravity, Higgs), a common detector might couple to the vev. But for fields that come in the particle limit (such as the discrete particles primarily produced at the LHC) we primarily have detectors that absorb or scatter individual particles -- which is extremely strong backreaction in terms of fields. No wonder the textbooks don't consider this case so much, I'll keep thinking about it...

knzhou

Feb 15, 2022 00:51

Hey, thanks a lot for this detailed answer! I hate to say "but" again, but... this is the issue I highlighted in the sentence starting "Another variant". Your causality argument works for detectors that measure $\langle \phi \rangle$, which is all that field theory textbooks ever seem to consider, but almost no real detectors do that. The crux of my question is that when you have a detector model that can absorb quanta, which more closely resembles a real detector, the causality argument breaks down. In fact, if you just plug my $H_{\text{det}}$ into your argument, you'll get a nonzero result!

knzhou

Feb 15, 2022 00:51

On the other hand I think you're on to something about the idea of "neglecting backreaction"... it's possible that my "direct" detector coupling $H_{\text{det}}$ does something to the field's dynamics, so that it can't be treated as a free field anymore. In that case the resolution would have to involve some subtle limit on how well you can couple an absorbing detector to a field without changing its mode structure?

knzhou

Feb 15, 2022 00:51

Hey, thanks for the elaboration! I think we're making progress, and I agree with your final expression. But the catch is, your final commutator doesn't actually vanish!

knzhou

Feb 15, 2022 00:51

Under your definition, $H_{\text{int}} = \epsilon_s \phi(x) \delta(t) + (\epsilon_d |e \rangle \langle g| \phi_-(y) + \text{h.c.})$ and $H_{\text{det}, 0} = E_e |e \rangle \langle e| + E_g |g \rangle \langle g|$, so you don't have a simple field commutator anymore. In particular, the second term of your $H_{\text{int}}$ doesn't commute with $H_{\text{det}, 0}$.

knzhou

Feb 15, 2022 00:51

It's kind of confusing to have two terms in $H_{\text{int}}$ where one is singular in time, so things might look simpler if you start at time $t = 0^+$. At that point the source has driven the field to $|\psi(0^+) \rangle = |0 \rangle - i \epsilon_s \phi(\mathbf{x}) |0 \rangle$. The analogue of your final result is then $it \langle \psi(0^+) | [H_{\text{det}}, H_{\text{det}, 0}] | \psi(0^+) \rangle$ which is nonzero; up to constants, it's just the norm squared of my $\mathcal{M}$. I think the paradox stands...

knzhou

Feb 15, 2022 00:51

I'm trying to understand what you're saying, but I just don't think what I'm doing is unreasonable... as far as I know, it's standard in atomic physics, and in undergraduate physics in general. Also, to describe the detector excitation, you don't calculate the expectation value of $H_{\text{det}}$, which describes the detector-field coupling. Either you directly calculate the amplitude of the excited state, $\langle e | 0, t \rangle$, or you calculate the expected value of the detector's energy, $H_{\text{det},0} = E_e |e \rangle \langle e | + E_g |g \rangle \langle g|$.

knzhou

Feb 15, 2022 00:51

What I'm doing is just a notch more complicated than, say, driving a harmonic oscillator in QM 101. In that context, the amplitude to go from the ground state to $|n \rangle$ is $\langle n | \psi(t) \rangle = \langle n | U(t) |0 \rangle$. There's no need to "evolve the bra", in fact I'm not even sure what that would mean...

knzhou

Feb 15, 2022 00:51

I trust that the "in-in" formalism works, but I had something simpler in mind... why can't you just throw out all this formalism and treat my setup using ordinary time evolution under the Schrodinger equation, as in a first course in quantum mechanics? The amplitude to be in $|e \rangle$ starts out zero, then becomes nonzero, which seems to violate causality. And there's no need to invoke the evolution of operators, or distinct "in" or "out" Hilbert spaces... I mean, the explicit, full time evolution in my setup is not much more complicated than driving a harmonic oscillator.

knzhou

Feb 15, 2022 00:51

In other words, while I trust that there's a more complicated, systematic way of calculating the answer that automatically guarantees causality-respecting results, what I'm primarily confused about is why the simple calculation doesn't.

Discussion on question by John McSimo…

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knzhou

Jul 21, 2021 16:47

Lots of gleefully vindictive comments here with nobody applying common sense. If people submitted the exact same project without even trying to make it look different, the obvious explanation is that they misunderstood "group assignment" as meaning any amount of collaboration is allowed. You should still penalize them for it (e.g. make them all turn in a new one) but the amount of moral grandstanding here is excessive in my opinion.

Discussion on question by LoneAcademi…

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knzhou

Apr 24, 2021 02:30

If you work on Chegg's "advanced physics", you've possibly helped people cheat on exams I've written! I recommend you don't admit to doing any such thing. In terms of the damage you've done, you're worse than an actual cheater.

Discussion on answer by AccidentalFou…

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knzhou

Jan 30, 2021 22:11

There are very few people on this site that are able to answer your questions as it is, if you keep this up in the long term they're all just going to ignore your questions.

knzhou

Jan 30, 2021 22:10

To give some unsolicited advice, @MadMax, you're way too fast to declare answers to be wrong. Reasonable people should not be expected to put up with an instant dismissal after answering an obscure question, and the only reason I bothered relying to your dismissal of my answer to your other question was because I was bored and in a good mood at the time. (I guess the same is true for AccidentalFourierTransform.)

Discussion on answer by knzhou: Does …

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knzhou

Jan 29, 2021 15:48

@MadMax Sure! And I suppose you were right to be skeptical, since you already received 3 now-deleted answers.

knzhou

Jan 29, 2021 15:48

@MadMax We're going in circles here. If you really insist on setting it up that way, then it's not impossible. But you literally asked in this question if it was "conceptually loopy" to do this, and my answer was yes, because it's extremely inconvenient. Nobody works with structure constants that can flip signs.

knzhou

Jan 29, 2021 15:48

@MadMax Of course you are completely free to set it up whatever way you want, but if you insist on doing it that way... why did you even ask the question in the first place?

knzhou

Jan 29, 2021 15:48

@MadMax I'm saying that if $A_0 = A_0^x \sigma_x + A_0^y \sigma_y + A_0^z \sigma_z$, then $A_0^T = A_0^x \sigma_x - A_0^y \sigma_y + A_0^z \sigma_z$. The point is that I'm not expanding $A_0^T$ in terms of transposed generators, because it's much more clear to always work with the same set of generators. I could have written $A_0^T = A_0^x \sigma_x^T + A_0^y \sigma_y^T + A_0^z \sigma_z^T$ but this is a useless expression; the $\sigma_i^T$ don't even satisfy the right commutation relations.

knzhou

Jan 29, 2021 15:48

@MadMax Because it's actually not the easy way. It gets really confusing, which is exactly why you asked the question in the first place! The quantities $\sigma_i^T$ don't even satisfy the $SU(2)$ commutation relations anymore.

knzhou

Jan 29, 2021 15:48

@MadMax The expression in components depends on the choice of generators, unfortunately. For example, specializing to $SU(2)$, let the generators be the Pauli matrices and suppose that the coefficients of $A_0$ are $A_0^i = (A_0^x, A_0^y, A_0^z)$, so that $A_0 = A_0^i \sigma_i$. Well, because $\sigma_x$ and $\sigma_z$ are symmetric and $\sigma_y$ is antisymmetric, the coefficients of $A_0^T$ are $(A_0^x, - A_0^y, A_0^z)$. In other words, the second coefficient gets flipped in sign and nothing else, a result which clearly depends on convention.

knzhou

Jan 29, 2021 15:48

@MadMax The fact that the result is convention dependent and annoying to write down is probably the exact reason why most resources avoid being specific about how $A_\mu^a$ transforms.

knzhou

Jan 29, 2021 15:48

@MadMax I wouldn't put it that way. I would say that $A_\mu^c = - A_\mu^T$ and that $A_\mu^c = A_\mu^{a, c} T^a$, which implies some transformation between the coefficients $A_\mu^{a, c}$ and $A_\mu^a$. It's the coefficients that transform, not the generators. The specific transformation of the coefficients can be messy to write down, and depends on how the generators are defined, which is why I suppressed it.

knzhou

Jan 29, 2021 15:48

@MadMax $A_\mu^T$ always means the transpose of the quantity $A_\mu$. I wouldn't call its coefficients $A_\mu^{a, T}$ because that makes it look like you're transposing $A_\mu^a$, which classically is just a number, not a matrix. It is just some new set of coefficients.

knzhou

Jan 29, 2021 15:48

@MadMax No, I think it's a mistake to think of "conjugating" $T^a$ or $f^{abc}$ at all; these are numbers, not operators. It's like saying that if $L_+$ time reverses to $L_-$, then $1+1 = 2$ time reverses to $1 - 1 = 2$.