@Relativisticcucumber Consider from the isolated atoms coming together, i.e. RTL on the plot, the hybridisation starts from the first place where they join. Even after they split even further on the left, those are hybirdised, not the isolated ones. But how they hybridise is not easy to see at all; In this plot, they overlap, but actually when you plot the bandstructure, it will not show up as overlapping.
@Slereah that butterfly is definitely the one that caused the hurricane
@SillyGoose it cannot be less complicated than that because this integral is halfway to Dirac delta distribution, and you know how complicated that thing is.
@imbAF you should already have learnt how to add angular momenta. This should not be a question that you should be asking.
@SillyGoose you are just attempting this completely wrong. Using $\tau=t-t^\prime$, it was $$\begin{align}&\lim_{t\to\infty}\int_0^te^{\mathrm i\omega(t-t^\prime)}\mathrm dt^\prime\\=&\lim_{\epsilon\to0^+}\lim_{t\to\infty}\int_0^te^{\mathrm i(\omega+\mathrm i\epsilon)\tau}\mathrm d\tau\\=&\lim_{\epsilon\to0^+}\lim_{t\to\infty}\frac{\mathrm ie^{0}-\mathrm ie^{\mathrm i\omega t-\epsilon t}}{\omega+\mathrm i\epsilon}\end {align}$$ thus getting what they gave.
You are supposed to know this by yourself, even without the text saying "inserting a convergence factor and taking the limit $t\to\infty$ consistently with the Wigner-Weisskopf approximation"
you didn't even get the limits of integration correct
Of the questions I could have chosen, this one seems to be the least possible to misinterpret, so I will choose it.
I try my best to make questions as fair as possible because from my own experience physics (and any questions really) can be incredibly easy to misinterpret at lower level physics
@Obliv it is soooooooo commonly the case that, by having a LOT of words on the page, the question gets muddied up so much that a simple maths question that is impossible to get wrong, turns into a confusing and tricky question
@qwerty my sister is taking basically the same class and her professor is kind of a nut imo. He takes points off for people labeling forces incorrectly (calling force of gravity $F_g$ instead of $F_{\text{earth on X}}$ and he doesn't call friction a force, nor normal forces, it's very strange.
@Obliv push against walls. Normal reaction force is still available. And blow air out for longer than you breathe them back in. That would give you a small propulsion.
I might, but what criteria would be suitable for a question like that? Another one is: "An anti-lock braking system in a car (ABS) keeps the wheels from locking up during hard braking. Why is this important?" lol
Those kinds of questions are more like creativity questions. I'd just give anybody whose answers are nice, good grades
uugh, my poor student is giving meow horrible answers. Looks at graph that doesn't spike to infinity and still gave 1/x^2. Graph is fully above x axis yet gave answer that is below x axis
I gotta catch up on a lot of stuffs because i missed a lot of days :[
btw I forgot if you recommended ballentine (I never got a chance to get through it) but a lot of the stuff that I covered in the prelim chapter we're just now using (operators, commutators etc)
I hope meow survives. good night
I'll probably pop in from time to time frantically asking questions before deadlines as usual
I'll offer another motivation for Andy's result, one that doesn't require quite so much theory. The real part is trivial from $G+\bar{G}=2\pi\delta (k)$, but what about the imaginary part? First note that $$\Im\int_0^N\exp (ikx) dx=\Im\frac{\exp (ikN)-1}{ik}=\frac{2}{k}\sin^2 (kN/2).$$For large $...
oh yeah we recently did the spread products $\Delta_A\Delta_B$ he didn't mention it was an uncertainty principle he just did a strictly math derivation which I thought was funny
@qwerty yes. even the facts about animals that we know are filtered
@123 hi
so whether or not physics describes some objective truths is dependent on faith. i think we have to make some arguments to justify the faith
so let's take this project
the first questions are : 1. How do we argue that things other than us exist? 2. how do we argue that our observations are in some sort of bijection with the things that exist
i see a lion responding and reacting to the positions and velocities of its prey. does that mean that the lion also perceives the world in terms of positions and velocities
or does the lion perceive the world in terms of some other structure that is in bijection with the positions and velocities that we perceive in the world
@RyderRude KnK didn't give any example. They just said objects motion can be defined various ways. rather than translation and rotation by chasles' theorem
I have read this statement more than a year ago and searched a lot as well as think a lot. But didn't find any answer other than translation and rotation.
@123 without knowing the context, what I can say is that any motion of point particles boils down to translation. for rigid bodies, it is translation + rotation. for non rigid bodies, it can be translation + rotation + deformation
@123 no. it is fine. 1. it only talks about rigid body motion 2. It makes it clear that ANY motion of a rigid body can be described as a combination of rotation and translation. So there is nothing else
@RyderRude Pls read the 2nd pic 3d line. "Note that the theorem does not say that this is the only way to represent the a general displacement --- merely that it is one possible way of doing so."
@123 the point of the text is that any displacement of a rigid body can be thought of as a translation followed by a rotation. but u may choose to think about it in other ways. e.g. if i give u a displacement of a rigid body, u may imagine that I first translated it and then rotated it, or u may imagine that I simultaneously translated and rotated it in a complicated motion
so there are other ways to get the same displacement (as displacement is only dependent on initial and final configurations of the body)
It means there is only one way to describe the general motion of a body translation and rotation. In sense of translation and rotation there is no other way.
We know projectile motion we decompose into x-y axes. Using x-y axes decomposition we can only speed change only due to y-axis and direction information is not there.
But if we resolve our problem into ICOR the direction information is there.
What i perceive from above information. The direction of change of motion is only limited to how we look at the problem e.g x-y axes or ICOR, r-$\theta$. Am i correct?
@naturallyInconsistent Hello , yes this is what i wanted to know. y-component of velocity changes. How can i relate this information with direction change.
Ookay. Let me try other way. I can see the motion as a whole continuously direction changes. ICOR method decompose the problem into two parts one related to magnitude change and other related to direction change. So ICOR method specifically informs us about direction change.
@123 Speaking of ICOR, the statement that you are confused about, amongst other possibilities, is definitely talking about ICOR being one of those many other possibilities. You definitely had enough information to have sorted out your own confusion.
@naturallyInconsistent Aaaaaahhh... I see... you are correct. Why i can not able to think this answer at my own. My bad... The answer was in my pocket. And you showed it to me.
Thanks NI
atan2 is angle, which is related to polar form.
Using this information i can say. If any one or both components of speed change in x-y axes it shows direction change.
@123 you can change both x y velocities in the same ratio and keep the direction the same too. You cannot just say something that is so mathematically broken and expect to have the correct answer.
@naturallyInconsistent Aaah... Yes Thanks to point out this. Is there any general of saying about direction change from x-y axes. Or it depend on situation?
I think if one of the axis change and other kept constant than definitely direction change happened. May be i can say both x-y velocities does not have the same ratio than also it shows change in direction.
We know inertial mass directly related to quantity of matter as well as type of matter. I meant to say if two objects have equal no of protons, neutron and electrons , we said they have equal resistance against change in state and indirectly masses are same.
I know how inertial mass (resistance of change of state) is related to mass. But why moment of inertia/rotational mass is considered as mass?
I meant to say resistance against change in state we termed this quantity as mass because it is directly related to quantity of matter. And we know quantity of matter is mass. But rotational mass why we termed as mass just because resistance against rotation.
@SillyGoose Sorry for not replying anymore. I've been dealing with network issues ever since
That being said, @SillyGoose the idea behind it is making $\omega$ complex $\omega\to\omega+i\varepsilon$, with a small imaginary part to that ensures convergence (exponential suppression; the sign of the imaginary part is chosen to have the integrand suppressed at $+\infty$) and then performing the calculation. Eventually you take the limit $\varepsilon\to0^+$
hm, so is the idea the following. Let $\mathcal{F}$ be some space of nice functions. we should think of distributions as functionals $\mathcal{d} \in \mathcal{F}$. Hence, the distribution $\mathcal{d}$ is defined in terms of its action on a basis of $\mathcal{F}$.
@naturallyInconsistent hm yeah i was originally thinking where they meet is hybridization "beginning" but then i was thinking that if that were true, the band gap wouldnt emerge on the left side
@naturallyInconsistent but im confused what you mean when you say that when i plot the bandstructure, there wont actually be overlap? so its not that overlap means hybridization i guess? and why is there a band gap on the left side?
Then, should "regularizing" the integral be understood as actually restricting the domain of the distribution?
It is like saying "this integral can converge if we think of the integrand as a distribution and if we only consider its action on a restricted domain of "nice functions"
@Relativisticcucumber the energy of a certain Bloch state is a function of band index n and also the k point in 1BZ. This plot ignored the k dependence. When you plot the bandstructure with the k dependence, the overlapping turns into much more interesting curves within the band. Typically, wide energy bands imply a lot of hybridisation.
hm i think i have in mind what you are referring to. in the bandstructure diagrams, they are just curves, right? so is there a way to see hybridization on these diagrams?
im thinking smth like this:
this is only the 3d orbital, but in principle you can plot all orbitals right?
i mean if I am to treat this "quantized EM field" as in the QFT sense, then $a, a^\dagger$ should act on a different part of Hilbert space as $\sigma_\pm$?
@imbAF It has never been explained. Those L have 4 indices, 2 of which are LABELS and 2 of them are tensor indices. And then people tend to use the same symbols for both types of indices. Students can never hope to understand them by themselves.
@SillyGoose I don't quite know the context here, but usually this works like this: The multiplications are implied tensor products - you have some space $H_1$ on which the $a$ act, and some space $H_s$ on which the $\sigma_i$ act. The full state space is $H_1\otimes H_s$, and all occurrences of $a$ and $\sigma_i$ are then implicitly $a\otimes 1$ and $1\otimes \sigma_i$. Then the multiplications just follow the rule $(A\otimes B)(C\otimes D) = (AC)\otimes (BD)$.
@B.Brekke make sure you read down to Giorgio's answer, which the author of the top-scored answer also found the best (it's lower-scored because it was posted after the question left HNQ)
states of matter are one of these things where we first learn the simple but wrong thing (that there are three), then learn there might be more ("X is a fourth state of matter" for some values of X), then increasingly realize the whole idea of categorizing stuff into clear finite bins labeled "state of matter" might not be as clear-cut as we thought :P
He says: "This implies the possibility of a continuous transition from “liquid-like” to “gas-like” states, and the continuity implies that there is no point where it is possible to put a sharp border between gas and liquid."
Does this mean that there has to be a continuous phase transition? My intuition says yes, but I never see a continuous transition line plotted. Only a horizontal and a vertical dotted line specifying the supercritical fluid phase. I do not understand if the dotted lines are physical or not.
@B.Brekke I'm not sure which dotted lines you mean, but what is meant here is just that in this diagram you can move from gas to liquid "around" the critical point without crossing a line of phase transition
the line of sharp phase transition stops at the critical point
Yes, the first-order transition is clearly plotted by a solid line. I wonder if there is a continuous phase transition hiding in the white area. I could, for example, be defined by some discontinuity of a higher-order derivative of the free energy.
Ah, I think there might be a terminology mismatch here - you're right that "continuous phase transition" is also a technical term replacing the older "second-order phase transition", but I think the answer in that sentence is just using it to mean that the "motion" of the system along the path does not cross a sharp/first-order phase transition
the answer even links to the supercritical boundaries as candidates for where to put the line along this continuous transition
I just started my new job as an engineer. It is on some transportation of CO2 in liquid and gas form. My colleagues were surprised when I said I did not know the difference, but now I do!
Hello everyone. I have a question, about the notation in four vector algebra. I want to derive, show that for a LT the time component is either bigger than 1 or smaller than -1. So I consider $\Lambda^\nu_{\ \ \sigma}\Lambda^\mu_{\ \ \rho}g_{\nu\mu}=g_{\sigma\rho}$. And I consider the $g_{0,0}$. So in the RHS I would have $(\Lambda^\nu_{\ \ \sigma}\Lambda^\mu_{\ \ \rho}g_{\nu\mu})_{0,0}$
deriving the specific form of the Lorentz transformations is much easier in block matrix form than index notation, but sure, you can do that in index notation if you want to suffer. What problem do you have in expanding the l.h.s.? It's just a matrix multiplication.
