Sorry @Mr.Feynman, saw the guidelines now. I didn't mean to bother. will be careful next time.
@Amit I did search on google, but most websites are oriented towards 12th grade JEE physics in India, and I couldn't find something relevent. So I thought maybe asking on SE would give help of some kind.
Clear this for me sir; are requests like these generally considered a bad practice here? I asked something like this previously also and people do not respond to such queries.
@ShikharChamoli You're free to make requests. Other people are free to help or to ignore such requests. Pinging people you're not in a conversation with is usually considered bad form.
im not sure what you mean. im having this question because theres a part of sakurai that says "the inverse operator $\frac{1}{E_n^{(0)} - H_0}$ is ill defined because it might act on $\vert n^{(0)} \rangle$" and i was assuming this meant what i guessed above but im not sure
i see -- yes that is also why i am uncomfortable with this quote in general because i dont know how to interpret the operators being in this form. so i guess my guess is wrong.
and i have one more question. sakurai also says that even for $\lambda \neq 0$ (weight factor for perturbation) we can always add to $\vert n \rangle$ (solution to perturbed problem) $c_n \vert n^{(0)} \rangle$ (solution to unperturbed problem), and i do not see why this is the case when $\lambda \neq 0$
isnt the point that $\vert n ^{(0)} \rangle$ is not a solution to the perturbed problem?
okay so if im understanding this section correctly, we are trying to find the form of $\vert n \rangle$ and we come to an expression, but this expression does not satisfy that as $\lambda \rightarrow 0$, $\vert n \rangle \rightarrow \vert n ^{(0)} \rangle$, so then he says "but we can add $c_n \vert n ^{(0)} \rangle$" and then that gives us the form of the proper eigenstate $\vert n \rangle$
I've seen the idea that the "increased voting" they saw during their tests (the stated reason for the redesign is that it increased voting in A/B testing by a lot) is just confused people clicking the button more than once to figure out whether they have already voted or not.
by the way i had a question about the resistance to this AI policy stack put out. to my understanding, the ban is against removing ai generated content on the grounds of it being ai generated. if the claim is that this ai generated content is bad or misleading, cant all the ai generated content still be removed on the grounds of being just wrong or bad? and if the content is perfect, then who cares who generated it?
so it seems it doesnt really make a difference whether its allowed or not if it can still be taken down regardless?
@Relativisticcucumber Historically we have held the firm position that moderators are not judges of correctness - we do not delete content for being "bad", we only delete plagiarized and harmful content. "Flags should not be used for technical inaccuracies, or an altogether wrong answer" is a default reason to decline flags. I tried to express this in the second paragraph of our meta post.
i guess my question then is if a user posts a question and another user answers with, theoretically, the most beautiful answer ever, but its plagiarized (lets say completely ai generated), then who does this hurt?
Hello, how is the volume of phase space exactly related to the energy of the system? Does the volume of phase space being constant mean that energy will also be constant?
@Relativisticcucumber The whole reputation system is based on the assumption that posts are original work.
And sites on academic topics like physics.SE generally have cultures based on academic integrity: Passing off someone else's work as your own is simply wrong, not because it "hurts" someone specifically but because it more generally undermines the mutual trust between scientists that science is built on.
i guess i think if someone can use ai to such a degree to generate this content where it racks up reputation reliably then they are effectively as useful a resource than someone who knows the stuff from their own brain. ofc this argument is not valid for ppl using ai to produce nonsense which i saw is an issue but yeah i guess i thought content could be removed or downvoted for being wrong
but i see the situation now so thanks for explaining
@Slereah I agree that it doesn't always work! And users with >20k rep can vote to delete posts as they see fit. My position is not so much that no one should delete posts for being wrong, but that that's not what moderators should do. Quality judgement by votes is up to the users.
@Relativisticcucumber Another response to the line of thought that "if the AI is useful why should we delete it?" is that if the users wanted a computer-generated response they could just ask the machine directly without going through an SE site as an intermediary. We're not depriving anyone of anything by taking a stance against computer-generated content: If that's what you want, you can just get it directly from the source
So I've managed to confuse myself again. So if I say the universe is isotropic and homogeneous usually I find the killing vectors which are spatially maximally symmetric and discover the flrw metric
My question is does it make sense to say the deviation vector should isotropic and homogeneous?
Since $u^\alpha \nabla_\alpha \eta^\beta = \eta^\alpha \nabla_\alpha u^\beta$
where $u$ is the 4 velocity of the geodesic and $\eta$ is the deviation
The lie derivative is $ 0 = L_V \eta = v^\alpha \nabla_\alpha \eta^\beta - \eta^\alpha \nabla_\alpha v^\beta$ where $v$ is the spatial displacement: Can I claim $v^i \propto u^i$? where $i$ are the $3$ components?
@Relativisticcucumber Write down the the angular momentum in terms of radius and tangential momentum. Show that, to conserve angular momentum, if the radius is small, then the tangential momentum is huge. Show then, that KE of the particle will become so large that it becomes classically forbidden.
@ACuriousMind I just answered a question on the main site, whereby the lack of hats contributed to a misunderstanding.
@RyderRude There is a whole bunch of authors who have decided that the two-state, spin-half, system is the correct intro to quantum theory. And they are somewhat correct. It is quite nice to see somewhat of a consensus forming.
There is little more frustrating than trying to find a paper that you read like a month ago.
I've even got a citation, though it doesn't make any sense. What the hell is PRSL A? They were apparently publishing back in 1954, though I'm guessing they shut down a while back.
My trick is to take a step back and get the probability amplitude distribution in the classical setting, so that I'm really cutting through a lot of crud and getting to the quantum world faster.
@Relativisticcucumber What does it mean for "the Jacobian matrix" to "account for" a coordinate transformation? A Jacobian matrix is just the matrix of derivatives of a differentiable map.
@Relativisticcucumber A coordinate transformation is just any diffemorphism $f: \mathbb{R}^n\to\mathbb{R}^n, x\mapsto f(x)$. Any (tangent) vector $v^\mu$ transforms under this transformation as $v^\mu \mapsto \frac{\partial f^\mu}{\partial x^\nu}v^\nu$. (Physicists for some reason love using the deeply confusing notation $x'$ for $f$ and then doing weird stuff with the prime)
@ACuriousMind but if i want to do something basic like go from spherical to cartesian coordinates i dont see how the first statement here gives me a means of doing that but it would seem that the second would?
Even though we write the individual coordinates in index notations also with a single upper index, i.e. $x^\mu$ or $x'^\mu$ or $x^{\mu'}$ or whatever, the coordinates are not (tangent) vectors, and the above formula with the Jacobian matrix does not apply to them
@Relativisticcucumber What do you mean by "a means of doing that"? Spherical to Cartesian is just the map $(r,\theta,\phi) \mapsto (r\sin(\theta)\cos(\phi), r\sin(\theta)\sin(\phi), r\cos(\theta))$.
ook and when you say "A Jacobian matrix is just the matrix of derivatives of a differentiable map." then this matrix is the derivative of the maps in the diffeomorphism you mentioned above?
Vectors can be considered as the derivatives of curves on the manifold, hence you can show that the transformation of coordinates act on that derivative through their own derivatives
Becuz then the nearby co ordinates can b identified with the tangent vectors
There is some similar idea about Riemann Normal co-ordinates, which identifies the nearby co ordinates according to the parameter of geodesic evolution in the direction of a tengent vector from that point.
I read it very long ago, maybe the name cud b wrong
for Lie groups, is there a nice expression for transforming a vector in the tangent space of the identity to a vector in the tangent space at an arbitrary point?
hm well i'd like to be able to start with the data of the lie derivatives at the identity which i understand to just be a set of generators of the lie group, and then use a mapping (which i was told that perhaps conjugation by arbitrary group elements would work) to map them into all the lie derivatives at an arbitrary point in the lie group
I think this may be related to the evolution formula $\frac{dP}{da}= [P, Q]$. It translates the generator $P$ by a parameter $a$ according to the evolution generated by $Q$
@ACuriousMind the usual form of the derivative of the exponential mapping is a bit intractable to use in this case so it would be nice to start with derivatives at the identity (which are easy to compute) and then have some nice mapping to take them to derivatives at arbitrary points. but bigger picture i need to differentiate something like $\exp(-i\Sigma_i \theta_i \Lambda_i)$ w.r.t. a single $\theta_i$. Here, $\Lambda_i$ are generators (the generalized gell-mann matrices)
and the directional derivative at the identity is a nice form because it will be $\frac{\partial}{\partial\theta_i} \exp(-i\theta_i\Lambda_i) \lvert_{\theta_i = 0}$, i.e. no "translation" term therefore no need to consider commutators and so on
Could I get feedback on this question? It's marked as needing clarity, and a couple of rounds of edits haven't fixed that. I'm not really sure where's it's unclear, honestly. https://physics.stackexchange.com/questions/754123/does-bremsstrahlung-disprove-the-classical-conceputalization-of-light-as-a-movin
@SillyGoose The canonical way to identify $T_g G$ with $T_e G\cong \mathfrak{g}$ is via $(L_g)_\ast : T_e G \to T_g G$ for $L_g : G\to G, h\mapsto gh$ is left translation by $g$.
@WaveInPlace Like we've been telling you with your other questions about photons and classical EM, the question doesn't really make sense because you seem to be working with a model of what a "photon" is that is not the model of mainstream QED, but there is not enough information to discern what exactly that model is, so no one knows what misconception to correct.
so let's say i have $SU(2)$ and take the pauli matrices as my infinitesimal generators. Let $g \in G$. Then will $g \cdot \sigma_x$ be an infinitesimal generator in $T_g G$?
@ACuriousMind, I thought this hewed pretty close to the mainstream. The genesis was me watching a youtube on the Thomson derivation of the Larmor formula, and wondering how that fit with the "changing electric field leads to changing magnetic field" adage.
@WaveInPlace You are simultaneously talking about photons and definite values of the electric and magnetic fields as if these existed at the same time. But in the description where you have photons, the electric field is an operator, not a definite value. In the description where you have definite EM fields, there are no photons.
This is not specific to EM: If you try to mix the classical and quantum descriptions instead of properly thinking about the limiting processes that turn one into the other, you're going to arrive at nonsense. That's not a failure of physics.
@ACuriousMind hm so when would the usual way of computing the derivative of the exponential mapping come in handy? this method of translating the lie derivative seems much easier?
@WaveInPlace Then what is the question? All you're saying is "the classical description cannot explain this" but we already know that the classical description can't explain everything light does!
The thing is also, if you like something, you don't care whether you're smart enough to do it well or not, as long as you can do it at all. It's not like you need to pass an entrance exam to open Wald's book, or see an online lecture, or browse Physics.SE, etc. :)
Oh, I see. I thought I knew triangles! But then someone posted this thing on Math.SE... long story short, the challenge was to do it without Trig. I couldn't do it. But I feel a bit better 'cause it took a week or two for someone to post a correct answer :)
The $\ast$ is a limb: $f_\ast$ is the pushforward - the $f$ is kicking something forward with its leg. $f^\ast$ is the pullback - the $f$ is pulling something backward with its arm
@Slereah Pretty sure the $\mathcal{D}$ is supposed to be a derived category, which is in particular a triangulated category - hence "geometry is triangles"
hm so is this an accurate drawing of how to compute the directional derivatives at an arbitrary point in a lie group given the data of the directional derivatives at the identity?
But ive read that differential geometry can b seen as a generalisation of algebraic geometry, in one sense. But in another sense, algebraic geometry is a generalusation of differential geometry
I also thought this is very intriguing stuff. It also relates to the Schrodinger picture and the Heisenberg picture. In the former, we transform the state vector using the group elements. But in the latter, we transform the algebra elements using the group elements.
The former application of a group is more natural to me.... as a group of transformations that act on some object like state vectors
The latter is somewhat hard to digest. But i guess you could see the "generators" as the "objects" on which the group can act on.
Another way to see this is that if you have a curve passing through the neutral element on your Lie group, the left action will move this entite curve to the point g
@Slereah hm so let's say we want to deal with $SU(2)$ and use the pauli matrices as generators. Then, let's associate "moving in a direction" with each pauli matrix. E.g. $\sigma_x$ is associated with "moving in the x direction". Then, will $g \cdot \sigma_x$ be mapped to the directional derivative at point $g$ in the $x$ direction?
ok so if coordinates are defined as specified above, i have another question :P in carroll, he says that the lorentz boost transformations are $t' = tcosh\phi - xsinh\phi$ and $-tsinh\phi + xcosh\phi$ then he says "from this we see that the point defined by $x' = 0$ is moving and has a velocity $\frac{x}{t}$." but i do not understand what this claim means given the previous discussion of coordinates
