There's a set of lecture notes I've seen which says the uncertainty principle is delta_x * delta_p >= h

Can I assert that this must be incorrect because it's missing the factor of 2pi?

It's common to be a bit vague about the HUP. I've seen it written with ℎ, ℏ and ℏ/2 on the right side.

Strictly speaking ℏ/2 is correct, but we so rarely approach that limit that it doesn't really matter.

No @Allure do not begin your studies by asserting that the uncertainty principle is incorrect. Instead, accept it as true, for now, and read on to see what develops from that principle.

@user20458579510081670432 I don't think Allure meant the HUP was incorrect. They just meant writing it as ΔxΔp ≥ ℎ is incorrect because it should be ℏ/2 on the right side.

Okay, perhaps I jumped to the wrong conclusion.

Sorry 😔

Thanks =)

@Slereah what do they mean vague hahaha? There is a concept for this called curvature

They just mean curving a little and curving a lot

as you can see

that's exactly what it means tho

osculating circle and stuff

although I must admit I like this explanation hahaha

very straightforward to say the least, I appreciate this kind of approach

I don't quite understand why $\mathbb{B}^3/\mathbb{Z}_2$ is not simply connected

I must also admit that I don't get why the rotations to get to the antipodal points are equivalent

I think the axis of rotation is the same but the angles are different

Oh I see, if two antipodal points $P,P'$ are identified by the same rotation, as they say, then you can't contract a curve smoothly to a point

again I don't see why $P,P'$ are the same point

@Slereah That's nearly a Cornu spiral, aka Euler spiral or clothoid en.wikipedia.org/wiki/Euler_spiral

sagecell.sagemath.org/…

> SO(3) is a schematic of the group of rotations in three dimensions. Any rotation can be specified by a vector pointing along the axis of rotation, with a length equal to the amount of rotation; using this correspondence, each cube here has been rotated by its own position vector, relative to the central cube.

hi

@MoreAnonymous this answer is from.the perspective of Hamiltonian mech physics.stackexchange.com/a/726064/156987

In Lagrangian mech, u hav some technicalities, it seems physics.stackexchange.com/questions/107921/… @MoreAnonymous

@Obliv i feel this pushes the problem to the definition of "choosing"

Illusions of choices exist in deterministic universes.. these can still be called free will in the compatibilist philosophy

in Many worlds philosophy, all choices come true

Many worlds philosophy is still deterministic.. all futures are bound to become real. so it's not "True free will"

“ We estimate that in 2023 MDPI ($681.6 million), Elsevier ($582.8 million) and Springer Nature ($546.6) generated the most revenue with APCs”. See arxiv.org/abs/2407.16551

After adjusting for inflation, we also show that annual spending almost tripled from $910.3 million in 2019 to $2.538 billion in 2023

Hello everyone, this is my first time here. I'm upset that my question has received so little attention. How can I improve?

.. and for a lighter shade of irony on a weekend: science.org/content/article/…

How long have you waited @Lagrangiann?

To confess, I erased it once and asked again, so it lasted about a day and a half. I'm sorry if I'm impatient.

your current question on superradiance was posted 2hrs ago…

“… Specifically, could you please explain in detail the full derivation of the relationship between the reflection coefficient”. This is not good. Contributors are rarely here to “explain in detail” a full derivation by someone else. Moreover, you question as stated is not conceptual, just technical.

I'm sorry... where should I ask this question?

"What are the detailed steps for deriving the reflection and transmission coefficients for both bosonic and fermionic cases as described?" This kind of derivation can be found in multiple textbooks. Maybe you can start there and, if there is an issue with the physics or the assumptions behind a step, then produce a question based on this. Right now you're just asking people to redo for you what's already found in textbook derivation...

.. without specifying how an answer can add physical contents to a technical derivation.

anyways... best of luck.

I'm gonna sound stupid as hell but I'll ask this anyway

I'm trying to follow some steps on how to visualize the SO3 group and I have some doubts about a sentence which is written in it

I consider the rotation of a right-handed cartesian coordinate system from $x,y,z \to X,Y,Z$. On the unit sphere, the initial coord. system is such that the z-axis points to the North Pole and I wanna make it slide along a geodesic(don't know how to define those) which conicides with the 0° meridian. I want to preserve the angles between the x and y axes and I want the x axis to always be tangent to the meridian

the article says this: The prescribed recipe works in all cases except when the Z-axis is in the direction of the south pole, so we exclude this case for the time being

my question is: why doesn't this work when z points to the South-pole? I can picture this very well, I don't know why it shouldn't work anymore. I don't know the definition of a geodesic honestly, but it is said that those coincide with the meridians and the equator

It doesn't seem to stop working, provided that my sketch is correct. The only problem I see is that maybe the poles are not like the other points, since one can decide to follow a different-angle meridian, but then again I have fixed that to be 0° already

accurate?

@Claudio whats the article in question

idk what you're doing but when you change hemispheres, right becomes left

so maybe that has something to do with it

assuming we orient 'up' to mean the pole and down to mean the equator

@ACuriousMind what a lineup! Idk if I could handle all that metal but hope you had fun.

I recognize maybe like 10 of all those bands lol

Are there non-flat spacetimes whose isometry group is Poincare? I am looking for non-trivial examples, not cylinder, cone, flat torus... blah blah.