@ACuriousMind thank you. out of curiosity which books actually motivate the idea? jd jackson and griffiths dont spend time on it (atleast when they introduce it)
@ACuriousMind In a topological sense, it's kind of obvious.
The integral of flux over a sphere centered at a point charge is equal to the charge (times 4pi or whatever), but the divergence of the field is zero everywhere that it is defined.
By Stokes's theorem there must be infinite divergence at that one undefined point.
@ACuriousMind They're confused because their heads are up their butts.
The define tensors as a bunch of numbers that transform in a certain way. They should start with the algebraic notion of tensors because then everything is obvious.
I only understood tensors after reading Munkres's book Analysis on Manifolds in which antisymmetric tensors are defined as linear maps with antisymmetry in their arguments. It's obvious from there that you can represent tensors as matrices (or higher order hypercubes of numbers).
But also the tensor product is just... dead simple.
@DanielSank the problem is that you shouldn't use Stokes theorem specifically because of that point. In fact, defining distributions to have Stokes theorem hold in such situations is one of the typical handwavings :P
@DanielSank I was responding to someone complaining about exactly that argument ;) I agree this is the correct motivation for accepting that something like the $\delta$ should exist, but it is true this is not mathematically rigorous - mathematically you must first define what a distribution is before you can try to take its divergence
@DanielSank 100% agree. "Tensors are multilinear maps" is simple and makes sense, "Tensors are a bunch of numbers that transform like this" is still a very puzzling line of thought to me.
and before someone else mentions that: Yes, I am aware that the "transforms like this" definition can be made rigorous and is equivalent to the usual one. That still doesn't make it make any more sense to me :)
tl;dr: People mistaking GPT for "intelligent" are making the same category error as people who think that passing an exam demonstrates mastery of knowledge of a subject
Generally I am deeply unimpressed with ChatGPT's responses, both to physics and non-physics questions
I would not trust anything it says because for every single physics question I've asked it where the answer is not ubiquitous on the internet the response has been wrong.
I've deleted plenty of GPT-generated answers by now and they're usually wrong in extremely silly ways, e.g. one claimed the friction force of an object on a table pointed downwards. Again, I would not trust any GPT responses regarding physics if you don't actually know it is correct (and if you already know, why would you ask it in the first place?)
GPT-4 can do some impressive stuff. But it still doesn't know what it's talking about. It can correctly explain the solution to a problem but be totally incapable of applying that solution. That's because it's only operating on language structure.
Also what do kind of course do u have to do to be well read like TTong? I'm guessing he took route in academia which enabled him to be basically well versed in everything?
Scott Aaronson recently gave GPT-4 a Quantum Computing exam. It scored quite well. But some of its responses were just inconsistent nonsense. See scottaaronson.blog/?p=7209
@ACuriousMind I see. Let's see if we can make a rigorous argument. We know that, given a real space with a single point removed, there is only one form that is closed but not exact. Translating to 3D, we can say there's only one vector field with zero divergence everywhere (obviously not including the missing point), yet still has a nonzero surface integral. We even know what field this is -- it's the electric field of a point charge.
We can check easily that the surface integral of this field is the same for any sphere centered on our missing point.
So I would simply argue that I literally extend my notion of charge density to include point-like charges in this specific case. There's nothing un-rigorous about that as far as I can tell and I think I just invented distributions without using any fancy words.
I didn't use Stokes's theorem or Gauss's theorem here... instead I invented a notion of point-like charge so that those theorem's still work.
That could be called hand-waving but I think it's actually how math is supposed to work.
@ACuriousMind I'm glad someone else on this green Earth agrees with me on this.
@ACuriousMind ChatGPT is moderately good at answering questions, but it's fantastic for what it was meant for: language.
Imagine you're a non-native speaker of the language you use in your profession. ChatGPT can help you write email and other correspondence without spending so much time asking people for help or looking up colloquial ways to phrase things.
It's kind of amazing that a tool designed to spit out correct language actually often spits out correct information!
@DanielSank Yes, the language ability is impressive - and arguably the problem, because we're not used to almost perfect language command paired with rather sub-par understanding of the content
BTW @ACuriousMind I think the root problem with the tensor thing is that physicists don't understand the difference between maps and variables. This fundamental problem leads to other confusions besides the tensor one, I believe.
@DanielSank I mean its not just the warehouses that have the toxic work culture. While I have heard its different from team to team. In mine its a bit extreme imo
@ACuriousMind Maybe my simplified example was too simple. My actual expression looks more like $\int dx' \partial_x[ \delta(x-x') f(x,x')]g(x')$. It arises from a Poisson bracket term in my equation of motion. The Poisson bracket has the derivative of the $\delta$-function. Does it still look strange? If so, in what way?
There is a fundamental difference between form and meaning. Form is the physical structure of something, while meaning is the interpretation or concept that is attached to that form. For example, the form of a chair is its physical structure – four legs, a seat, and a back. The meaning of a chair is that it is something you can sit on.
This distinction is important when considering whether or not an AI system can be trained to learn semantic meaning. AI systems are capable of learning and understanding the form of data, but they are not able to attach meaning to tha…
@B.Brekke My problem with the expression is that on a technical level it is not exactly clear what the $\delta(x-x')f(x,x')$ is even before we come to the derivative. Like, isn't that just $f(x,x)$?
quantum bee theory: all bee larvae can become queen but in the end only one does, and this is what happens when the wavefunction collapses, so the underlying structure must be bees
My confusion about quantum theory is twofold:
I lack an adequate understanding of how the mathematics of quantum theory is supposed to correspond to phenomena in the physical world
I still have an incomplete picture in my mind of how cause and effect relationships occur at the quantum level of...
It's forced upon us the moment we have two electrons or a continuous medium.
In the two electron case, we can use the antisymmetrized wave function but that is insane and it's way more sensible to recognize that "electron" means "excitation of the electron field".
I think that GPT does have a pretty good understanding of language structure. And I think it has some degree of conceptual understanding, but it's pretty tenuous, and mostly a side-effect of how the concepts are associated with the language structures in its training data and its prompts.
@B.Brekke Maybe this is the right way (leaving out the $g$ again): Just use the Leibniz rule to get $\partial_x\delta(x-x')f(x,x') + \delta(x-x')f_1(x,x')$, where by $f_1$ I mean the derivative of $f(-,-)$ with respect to its first slot. Then we can integrate the second term without issues to get $f_1(x,x)$, and switch the integration and differentiation in the first term to get $\partial_x f(x,x)$, so I claim that $\int \partial_x(\delta(x-x')f(x,x')$ is $f_1(x,x) + \partial_x f(x,x)$.
@Mr.Feynman Which is stupid, and if it weren't for the horrifying fact that a state in my home country is literally burning books right now I'd make a joke about what I think should happen to books calling "relativistic quantum field theory" by the name "QFT".
To me that's like always saying "non elephant animals" when you want to talk about dogs.
It's just wacko.
Physics is already a minefield of weirdly inconsistent terminology and notation. We should do future students the courtesy of using words in ways that make sense.
@DanielSank Mainly historical inertia: relativistic QFT was developed to "fix" relativistic QM, the realization that this reorganization into fields is a general principle applicable also to non-rel. many-body physics is a later development as I understand it.
so at the point at which people understood that relativistic QFT is a more annoying version of non-rel. many body physics, the term "QFT" for the relativistic QFT was already well-entrenched
But, in principle, one could have also written down the non-relativistic Schrodinger equation for, say, all the electrons and lattice ions in a solid and described the same physics. This situation is quite different from relativistic quantum theory, wherein the basic principles demand the existence of antiparticles, to which there is no simple non-relativistic analogue. Obviously, this has extra implications that go beyond the fact that we need to deal with sys- tems having variable number of particles — which, by itself, can be handled comparatively easily.
I sure agree about that, I'm just saying that in the case of QFT it's not that dramatic. At least, in my case that was the very first thing we said in out relativistic QFT course and one of the first things I've read on P&S
@Mr.Feynman ...and yet it's almost impossible to find a readable book that teaches you quantum field theory without the reader having to learn special relativity at the same time.
This is a disaster of pedagogy in my opinion.
It's kinda like when you want to write a web page and you have to learn HTML, CSS, javascript, and a backend server language just to do one thing.
@DanielSank Well, I think I can see what you mean as you might want a book which focuses on other fields but I don't think having to study QFT and not knowing SR is that common
I mean you can do that just as you can study general relativity without knowing electromagnetism but why would you do that?
@DanielSank There are 3 issues I can think of that makes it messier
The second complication is the following: Combining relativity with quantum theory also forbids the localization of a particle in an arbitrarily small region of space.
@Mr.Feynman Because maybe I'm in a lab with a silicon membrane cooled to 10 milliKelvin and I need to understand the quantum mechanics involved.
That's a quantum field theory problem.
And there's no special relativity in sight.
Or maybe I study superconductivity and would like to try to compute the critical temperature of a metal from it's crystallographic data and data about the electron-phonon interaction.
@MoreAnonymous Well, in some sense QFT is when you have a little QM system at each point in a continuous region, each one interacting with its neighbors.
QFT is to a violin string as QM is to a harmonic oscillator.
Or better said: QFT is to QM as a violin string is to a harmonic oscillator.
@DanielSank What I mean by annoying is stuff like: No proper position operators hence no fallback to position wavefunctions, forced spin-statistics, the Hamiltonian formulation is suspect because it splits time from space,...
@ACuriousMind You are right. This is the ambiguity that confuses me, and it pops up earlier in my derivation. After fixing this, the Leibniz rule seems more applicable.
@MoreAnonymous it's not forced upon us by relativity, it's just crazy not to use it
Technically you could work with wave functions $\Psi(x_1,x_2,...)$ for systems of identical particles with an infinite number of coordinates, and completely avoid fields, in the field case we just transfer this to the states the field operators act on, though relativity changes the very interpretation of the wave function making it more ridiculous, and there is great motivation to change to occupation numbers as your variables with systems of identical particles regardless of relativity
The minute we have identical particles, we can remove our heads from our butts by using second quantization and understanding the whole business as excitations of fields.
Funny enough, my field of research (superconducting electrical circuits), is pretty familiar with nonrelativistic QFT
We have for example little wires whose length is say half of the wavelength of light at 5 GHz.
We excite these things with 5 GHz radiation and the wires support oscillation modes of charge just like a violin string supports mechanical oscillation modes.
I think that the most important difference is that in a non relativistic framework it is more of a practical necessity, while with relativity one can't do otherwise
@DanielSank This is something I completely disagree with. In fact QFT was born in a relativistic framework first, I mean before it became popular in condensed matter as far as I know. I don't think that learning many body theory beforehand would change that much
The problems it leads to are so obvious: in QM you work with wave functions, you then go do QFT and things look completely different, wave functions turning into (single particle) operators, no explanation of how it really relates to the baby stuff you're used to, multi-particle wave functions look like some anomaly, usually all justified by harmonic oscillator analogies at best, it's always better to see the non-relativistic case first where it exists
There are still basic things in QFT I need to sync up with normal QM, it's really no joke how important this is (link to notes?)
@Mr.Feynman Your historical description is correct, but I agree with Daniel that the historical method of teaching stuff can be deeply damaging to students' understanding
my personal pet peeve is that we still often start teaching QM with the double-slit and "wavefunctions" when developing the basic ideas of QM in a finite-dimensional setting first (qubits, Stern-Gerlach, whatever) would feel to me so much clearer
@ACuriousMind Again, in general I agree too (imagine for example teaching GR the historical way)
@ACuriousMind I think that a lot of people teaching these courses don't really appreciate the difference between finite dimensional and infinite dimensional and genuinely think that just replacing sums with integrals and promoting discrete indexes to real variables is all you need to do
Even if you don't realize all the functional-analytic subtleties of infinite-dimensional Hilbert spaces, it seems obvious to me that it would be easier to talk about a 2-dim space than an $\infty$-dim space :P
just on a practical level
you can draw the Bloch sphere!
you can't draw whatever the heck is going on in the projective space of $L^2(\mathbb{R})$
I agree with you on that matter, that's what Feynman's introduction to QM does in the third volume of FLP
Most of the book focuses on finite dimensional systems
Is it ok to say that in GR the trajectories of light rays are null paths $x^{\mu}(s)x_\mu(s)=0$? Shouldn't we also require that they are (null) geodesics?
@bolbteppa Hello. I thought about ur e^(0/0) stuff. I think one mistake is in setting h=0. h is an experimentaly determined non-zero constant. U can only make it negligible compared to the action. But action wud beed 2 b infinite for h to b completely negligible
So, a perfectly classical system does not exist becuz h is always the same non zero constant. We cant set h=0
@Mr.Feynman Isn't that technically just a special case of $ds^2 = 0$ for $dx^{\mu} = x^{\mu} - 0$, and in general it should be $ds^2 = 0$, in which case one has to ask where the geodesic equation is coming from since the action is $S = - m \int ds = - 0 \cdot \int 0$...
@RyderRude No mistake, you know what book to check where this is discussed
Wait, no it's technically just $S = - m \int ds = - 0 \cdot \int ds$ (can't assume $ds = 0$ in $S$...)
@ACuriousMind I thought about a precise way to couple the probability density to the metric. It is mathematically a cool idea. Let's say u take the probability density evolution on the 2D phase space of a single particle, i.e. $\rho (x, p, t) $. We treat this thing mathematically as if it were a classical field. We define a Field lagrangian for this such that the Euler Lagrange eqn agrees with the Poisson bracket evolution of this
And then we add the coupling to the lagrangian : coupling constant times curvature times probability density
So if we make the coupling constant zero, we recover the usual Hamiltonian mechanics
Curvature is the Ricci scalar derived from the metric. We use this thing in the interaction term
@Slereah I'm fine with the idealization at this level :P
@bolbteppa I'm not sure what you mean here. I was thinking of finding the null paths by setting the metric $ds^2$ equal to zero and then combine this with the geodesic equation
I don't know if we can use the action in that case: when deriving the geodesic equation from an action principle one has to use the proper time parametrization, which gives problems for null paths as $u_\mu u^\mu=0$, so we should use another parameter
It's a small point, because you can just assume $ds^2 = 0$ directly and analyze it, but how do you derive the Geodesic equation in GR? You extremize the action $S = - m \int ds$. Where does the geodesic equation come from if $m = 0$ and $S$ is trivially zero to begin with?
Setting the covariant derivative of the tangent vector to zero basically amounts to promoting the flat-space Euler-Lagrange equation of motion to curved space, but where did the flat-space eom come from if $m = 0$
Everything in classical physics technically has to be reduced to the principle of Least action, yes books ignore this in the massless particle case because it's so obvious to think of light as satisfying $ds^2 = 0$ in flat space and then doing a coordinate transform to curved space, but that doesn't mean it's logically right, when you try to apply to the action you can see there's a problem
Where is this tangent vector coming from, I see a condition on $ds^2 = 0$ (i.e. square of $dx^{\mu}$), where is the derivative of the tangent vector being equal to zero condition coming from in flat space without an action leading to eom
But now I'm curious about the relation it has with the vierbein. The latter appears when we consider non coordinate basis in the tangent bundle such that the metric tensor is Lorentzian, what about einbein?
I recently noticed an answer which, according to the author, was written by ChatGPT. The question was very poor (about quantum entanglement, from a person who have known about entanglement from a popular magazine). I thought it can't be answered, but ChatGPT understood what kind of answer is expe...
@Mr.Feynman It takes about 30 seconds to set this up. Really you are talking about $S = - m \int ds$ subject to the constraint $\phi = p^2 - m^2 = 0$ coming from the eom. The canonical Hamiltonian associated to $S$ is $H_c = \phi = 0$. In Hamiltonian form the Lagrangian reads as $L = p dx - H_c - \frac{e}{2} \phi$, where I also added the constraint to $L$ using a Lagrange multiplier $e/2$. Note $H_c = 0$ in this so we ignore it.
If you now use the $p_{\mu}$ eom in $L = p dx - \frac{e}{2} \phi$ to eliminate $p_{\mu}$ you get the einbein action, and in this form you can then set $m = 0$ and set $e = 1$ (technically related to fixing a gauge under 1D diffeomorphisms on the worldline) and derive the flat geodesic equation. I haven't checked the signs/signature in all of the above.
My confusion about quantum theory is twofold:
I lack an adequate understanding of how the mathematics of quantum theory is supposed to correspond to phenomena in the physical world
I still have an incomplete picture in my mind of how cause and effect relationships occur at the quantum level of...
