@Danu Kinda, but she was the one who brought it up

@Danu this sounds like it belongs on Chemistry SE lol

@ACuriousMind I'm not 100% clear on the definition via equivalence classes. We define an equivalence class of curves by every curve having the same value and velocity at parameter $t=0$, right? Then the components of the vector are the components of velocity, right? We then define a linear structure?

@ACuriousMind How'd that happen? Was she like... you into me brah?

@StanShunpike 10/10, starred

@Danu Sounds like she's binned you as "friend". which is why she is comfortable. If that's not what you want, you need to let her know.

@Danu Essentially, just with a lot more stuttering and blushing :D

@Jiminion Meh, I'm fine with that. I'm just really curious

I'm mostly just happy to have found a non-weird/lame girl here in Munich to talk to. Many of my friends in Amsterdam are girls

Chemistry is the most important thing right now...give the reaction rate time ;-)

@Danu The female plants don't count ;P

@ACuriousMind Lolwut

@infinitesimal We've been hanging out a lot, like 3 times in 10 days

@Danu Cannabis -> Mary Jane

See, even the American gets it^^

@0celo7 Cannabis $\geq$ Mary Jane while Mary Jane $\geq$ Cannabis, too

@ACuriousMind Mary Jane is American slang.

@Sofia that's seems like a bad deal. You do all the work and they make money off you. Like, I feel like if they provide your article free to others, then they should be entitled to keep the revenue from ads generated. But if they sell your article, nowadays so much is free and copyright protection so poor, if they have people willing to buy, you should get a cut. Its like Spotify for articles effectively but instead you don't get any money.

Also, I think you can only smoke the female plants

The males don't bloom

No

it's just a lot better

You can smoke any plant :P

It'll just suck

Ahhh, you know what I meant :P

Yeah

I once grew some plants, turned out I had 1 male :(

@Sofia I mean, artists on Spotify benefit from attention. So that's really no different then a publication of an article. The only difference is they don't seem to compensate you, but Spotify does.

My advice now would be to back-off and let her make the next few moves on whether she wants to hang-out or not...

@infinitesimal She does, this weekend lol

fine

let her call the shots

We're going to the annual whisky festival :D

:O

That's literally calling the shots :D

(Although you shouldn't drink Whiskey as a shot)

alcohol, makes things complicated

fast

;-)

@ACuriousMind Good that you realize ;)

@Danu Noice update. Have you looked at it?

@0celo7 ?

@Danu The list.

@0celo7 You put something on there?

Hirsch. Not me.

What list

Oh god. This looks horrible.

@StanShunpike A list of books I keep

@0celo7 Oh, yeah I added that

I am the only one that can edit it, that's why I was surprised

@Danu That sounds interesting. Is it public?

No

Sawry

@StanShunpike Do you have any burning questions? I'm gonna catch up on this week's TV, and have time.

@StanShunpike I'm hesitant to share it with many people, especially those I don't know too well :\

Oh sweet! Let me collect my thoughts and I'll post some either tonight or tomorrow.

I'll ping you here once I post them

Anyone familiar with Lene Hau?

@Danu I know how you feel. Compiling information takes time and you don't want to just share it with just anyone. Then it feels like they get it for nothing.

@StanShunpike Mostly other reasons, actually :P

Really? Lol well, that would be my reason.

Also, if you want some references on some topic I could probably list you a few books, unless it's something quite specific

I'll keep that in mind. Chat is a great place for that.

SE has many avenues for info

@Danu When I answer his questions I spam references.

I love it.

I learn so much. It's very efficient.

Chat is really nice

I am WAY more active in chat than on the actual site nowadays

Chat is great place to get help with "homework-like" questions that get you burned at the stake on the main site.

@Danu Lazy bum

Don't you have a site to moderate? ;)

@ACuriousMind I love your string theory lecturer's hand drawn diagrams :D

ME TOO!

That was like the first thing I noticed.

@ACuriousMind I'll kick your lazy bum!

@0celo7 @ACuriousMind you gave out notes? ;)

WANT WANT WANT

lol

thphys.uni-heidelberg.de/~weigand/Skript-strings11-12/…

thphys.uni-heidelberg.de/~weigand/Skript-strings11-12/…

Heh, faster!

:: grumbles ::

Yeah, his drawings are great. (Sadly, he wasn't my lecturer, it was some dude who essentially taught by his notes :P )

Had him in two QFT lectures though, he's awesome

Never saw someone fill a blackboard so quickly though. Some students gave up because they couldn't write as fast as he could.

@ACuriousMind Is the tensor product of vector spaces symmetric or just isomorphic?

@ACuriousMind Hah, it's like Leeb!

@0celo7 The latter. The non-symmetry on the level of elements is what causes "braiding" - a category with a tensor product is, among other things, called a symmetric '(monoidal) braided category.

@Danu Does Leeb also lightning-fast reviews of the last lecture at the beginning of the next?

With the aid of a handwritten summary on an overhead projector? :D

Leeb?

@ACuriousMind Never

Also @0celo7 for my earlier claim on the space of derivations

from my diffgeo lecture notes

(Remark 3.9 is my claim)

@Danu Care to share the whole thing?

copy.com/s/3LVn3jhPKgvMwFiL/scripts/diffgeo.pdf

@ACuriousMind Repetition is clearly for the weak

@Danu So what you're saying is that repetition is clearly for the weak?

@Danu I see. Just like sleep

@JimdalftheGrey WEAKLING

@ACuriousMind I wouldn't be surprised if Leeb said this

@Danu Ty.

@Danu I'm not afraid of looking weak. That's only something the weak fear

@Danu What is $I_p$ and how does the claim follow?

@JimdalftheGrey Fear is weak!

@0celo7 I was loooking this up too

In the exercises we proved this I think

but I forget what $I_p$ is

I think it's just the space of germs at $p$, no?

@0celo7 found it, it's the kernel of the evaluation map on the space of germs at $p$, which is denoted by $C^k(M)_p$

@ACuriousMind See above comment

@ACuriousMind His treatment was so extremely algebraic, urgh

maximal ideals

Kernel? (On the tip of my tongue what the exact definition is.)

@0celo7 Stuff that gets mapped to zero

@0celo7 $f(x)=0\implies x\in \text{Ker}(f)$

@Danu Ideals <3

@ACuriousMind Riiiight

I had never seen taht word before that lecture

so you can understand my lack of amusement

@Danu @ACuriousMind Image of zero? Or preimage of zero? I mix those two up all the time.

@0celo7 Preimage

You can also define the kernel categorially, but that'd not be helpful here :D

@ACuriousMind I don't think I even want to know what that means.

@Danu I see. But ideals are fun. Go do some ring theory ;)

@ACuriousMind Still in chapter 2 of Vinberg: Linear algebra lol

@ACuriousMind Does it follow from the field axioms that $[f^{-1}]^{-1}$? Or is this simply by definition of the inverse?

You can even look at the space of prime ideals of a ring, the spectra. If you glue several of them together, you get schemes, and then there's this totally GAGA theorem that says that this is the same as complex analysis if you do it with rings over $\mathbb{C}$.

@0celo7 I don't see a statement there

@ACuriousMind I had to explain yesterday to a ninth grader why $\frac{1}{1/2}=2$. I wonder if this has an axiomatic reason.

@0celo7 By defintion of the inverse, $gg^{-1} = g^{-1}g = 1$

Nothing to do with a field, holds in every group

Even holds in semi-group, for those things that have inverses, I think

Here, I want you to read $gg^{-1} = 1$ as "$g$ is the inverse of $g^{-1}$" and $g^{-1}g = 1$ as "$g^{-1}$ is the inverse of $g$".

@ACuriousMind I'm confusing myself over something trivial again. Gimme a moment.

@ACuriousMind What is the most intuitive reason for $\frac{1}{1/2}=2$ then? I proved it using algebra, but that might have been a bit much.

If you think about division as partitioning up a number into chunks, it makes sense in the following way.

@0celo7 $\frac{1}{1/2}$ is the symbol for the inverse of $\frac{1}{2}$, i.e that which multiplied by $\frac{1}{2}$ gives $1$. And that's $2$.

@0celo7 Take two half-cakes and put them together:D

The quantity $p/q$ means how much of $p$ each person gets when it is split evenly between $q$ people.

@MarkMitchison How does that generalize to $\frac{1}{p/q}=q/p$? @ACuriousMind That's algebraic.

Oh, it is? :D

Damn, I'm a mathematician

If you give $1$ to $1/2$ a person, then a whole person gets twice that, i.e. 2.

@ACuriousMind I'm Cartan compared to 9th grade algebra kids.

I'm not sure how nicely that will generalise to the concept of inverting rational numbers in general.

But it should be possible

I think for any rational the following way is pretty simple

Consider an entire cake, and say you want to divide it into pieces of size $r = p/q$. How many pieces do you get?

The first sentence describes $1/r=1/(p/q)$, and the second $q/p$

@Danu "pieces of slice $r$"

Sorry, what?

$r$ pieces?

size, sorry, fixed the typo

no, $1/r$ pieces of size $r$

@Danu I'm really dumb. What do you mean by "size $r$"?

e.g. size 1/2 (obviously implicitly multiplied by the entire size of the cake), just a fraction

at least for all positive fractions, this works

@Danu What about fractions greater than 1?

You can make 1/2 piece of size 2

I didn't understand the comments to a [question(physics.stackexchange.com/questions/167054/…) whether the electronic charge should be a constant of the nature. It seems that it isn't because the quarks have partial charge, just they are unstable. So, the electron charge is not really a constant of the nature. Is someone here who can explain?

@Sofia The "electric charge" appears in QFT as a prefactor of a term in the Lagrangian, and if you do QFT, you find that in the course of renormalization, this charge becomes dependent on the energy the process you are looking at has. What we usually call "the elementary charge" is the low-energy value of that dependence.

Essentially, in quantum field theory, there really aren't "constants of nature", there are "flows of nature" that give the constants at different energy scales

@ACuriousMind but how does all this mathematics arrange with the natural fact that we are aware that in the nature, the quarks have less than the electron charge? I mean Lagrangian O.K., but the experiment beats the Lagrangian. Isn't that so? Why don't we adapt the Lagrangian? Something here doesn't go well. The Lagrangian doesn't predict the quark?

@Sofia This has nothing to do with the quark

The quark charges run in the same way as the electron charge

all electromagnetic charges run like this, in fact, no matter whether you look at the electron, the quark, or composite particles.

And the running couplings are well-tested by collider experiments, by the way, they match experiments perfectly

By the way, what is the status of predicting bound states in QCD from the Lagrangian? Is it possible to theoretically demonstrate the existence of the proton starting from the Standard Model? Perhaps I am being silly, but I heard at one time this was an unsolved problem.

I was just reminded by Sofia's comment.

@ACuriousMind you say, "What we usually call "the elementary charge" is the low-energy value of that dependence." I just guess that one day we may find that there exists a subquark? With an even smaller charge?

@Sofia I think you two are talking about two different things.

The running couplings is not related to quarks

The point is just that the "charge of an electron" is energy dependent

Although I guess at high energies you wouldn't really call it the charge of an electron

@MarkMitchison Well, you get stuff like pion condensation at the chiral symmetry breaking scale. I'm not sure what the status of the baryons is.

But the mesons are in principle understood, I think

"condensation" meaning what exactly?

Or perhaps that's too complicated to explain here...

@MarkMitchison The $\bar\psi\psi$ terms acquire a vev, and the fluctuation around that is the pion field

@MarkMitchison can you tell me what are running couplings in short? I won't understand much from article.

So it's like Cooper pairing in a superconductor?

(A bit)

In a way, I think you can say that

OK, that's pretty cool

@Sofia Well, I expect @ACuriousMind can probably explain better. I would just say that the charge quantifies the strength of the coupling between elementary particles, in the sense of how likely scattering processes are. It turns out that this coupling changes when the particles that you are scattering have more energy.

Although really we are talking about virtual particles I guess.

So in that sense the value of the elementary charge is actually a function of the state of the Universe, and so shouldn't really be considered a constant of nature.

@Sofia: It's pretty difficult because the QFT picture is so different from usual QM in this respect. You see, naively, you write a term like $e\bar\psi\psi A$, where $\psi$ is some fermion field (a quark, for example), $A$ is the EM field, and $e$ is meant to be the electromagnetic charge of the particle that's associated to $\psi$. It measures "how strong" the EM field couples to the fermion.

There are now two ways to go - one is the "old" picture of renormalisation where you get a whole lot of infinities, and to get finite results, you find that you must allow the $e$ to depend on the energy scale of the interaction, so the charge of $\psi$ becomes a function rather than a constant.

@ACuriousMind do you say that the charge is a constant of coupling?

@Sofia Yes. The charge measures how strongly the photons interact with the charged particle

@ACuriousMind hmmm! so, other interactions, other constants!

The "newer" Wilsonian picture is that the theory is defined at some energy scale, and gives a rule how to go from one energy scale to another. This views every energy scale as a different theory, and so, within every theory, $e$ is a constant, but a different one.

@Sofia No, it stays the same interaction

It's still quark-antiquark-photon, all the time

Right but quark-gluon interactions have a different "constant"

Just, if the quark-antiquark pair have 100 TeV energy, they are more likely to emit a photon than if they only have 1 TeV. (This doesn't always happen, the strong interaction becomes less likely with rising energy, for example)

@MarkMitchison Oh, yes, right - every force field has, in principle, a different constant for every matter field

All constants belonging to the same force will "run" in the same way, though

Hi!

@ACuriousMind but we have a full Noach-arch of particles. Everywhere we find the electron charge.

I mean, there are mesons, and other animals, but everywhere we find the full electron charge.

@Sofia Ah, that is related to group theory, and has nothing to do with running couplings or energy scales

You could as well choose the quark charge as fundamental, then you'll find the quark charge everywhere!

Surely you'll find three times the quark charge everywhere?

@MarkMitchison Yes, right

@ACuriousMind why? I ask, in experiments - leaving the theory, elem. particle have the electron charge. How so?

What is this * conspiracy*. Why all these animals have the electron charge?

@ACuriousMind there is something deeper in this.

@Sofia Yes, group theory. Charges are just representations of groups, and these are heavily constrained :)

It comes from the way the electroweak group breaks to give rise to the electromagnetic charge, essentially

@ACuriousMind I know that the quarks aren't able to be free, that they immediately do trouble and group together in trios.

@Sofia That's something completely different

@Sofia @ACuriousMind There is also an interesting argument due to Dirac about magnetic monopoles and charge quantisation

Confinement has nothing to do with electric charge, but with the strong force, it's completely unrelated

@MarkMitchison Ah, yes, you can get the fact that charge is quantized in some unit by geometric arguments, essentially

Right. So that if there exists a single charge with magnitude $q$, then all charges have magnitude $nq$, with integer $n$

@ACuriousMind @MarkMitchison I saw an article speaking of electron internal structure. There didn't seem to be quarks inside.

But that also requires the existence of magnetic monopoles!

It works only if there are magnetic monopoles, though

Yeah, right :D

Yup

@Sofia Current QFT does not predict any internal structure of the electron, and current experiments do not indicate any.

@MarkMitchison @ACuriousMind I speak from just a feeling. The magnetic dipole of the electron, should be based on some internal structure of charges.

@MarkMitchison The other quantization by group theory does essentially the same thing though - if you choose U(1) as the group such that the charge is quantized, then infintesimally thin Dirac strings are monopoles for all practial purposes.

Not a spherical charge rotating - God knows what structure, but a structure with charges.

@Sofia The magnetic dipole moment of the electron is fully predicted by higher-order Feynman diagrams within Quantum Electrodynamics. No substructure needed

@Sofia @ACuriousMind But don't quarks also have a magnetic moment? So your feeling also implies that quarks have sub-structure.

The feeling is simply too classical - the higher-loop effects of QFT produce anomalous magnetic moments (even electric moments!, but very small ones) without implying anything about the structure of the thing having the moment

@ACuriousMind Well, the existence of an electron magnetic moment pops straight out of the Dirac equation, right? No Feynman diagrams needed. It's only for the correct $g$ factor you need quantum theory.

@MarkMitchison Ah, right again :)

So the implication of that is just that spin-1/2 fields with electric charge must have a magnetic moment?

Or indeed fermion fields more generally.

Where "must" means in order to be consistent with Lorentz symmetry.

(I'm guessing here)

I think you're right, Lorentz symmetry forces fermions to fulfill the Dirac equation, and that forces them to have a magnetic moment

Yeah that's what I was getting at. That's interesting

When you dig down deep enough, it all becomes just symmetry so often that it's amazing

Hmm

I do agree

Perhaps we're just good at inventing symmetries, though :P

But I also question the willingness of many physicists to attribute properties to symmetries

e.g. I would say that it is a fact that charges exist.

Rather than the Lagrangian has a U(1) symmetry

The second is a consequence of the former. But perhaps I am wrong. There was a question on SE asking exactly that, whether the relationship goes both ways

There seemed to be some disagreement

Certainly the standard view is symmetry $\Rightarrow$ conserved charges

Not the other way round

@MarkMitchison In a way, it does - the existence of the charge implies the existence of a potential (by the force law for the charge), and potentials have U(1) symmetries

You have to postulate the force law though, rather than derive it, which many don't like.

That's an interesting viewpoint. I'm not sure if I like it.

ha

And it always goes both ways - symmetries imply conserved quantities, and conserved quantities imply symmetries

But does it though?

Let's see

In the Hamiltonian formalism, it does

The symmetries implied are global, not gauge though - for gauge, you need the force law with its potential

Ah, I see you looked at this recently

Lol, I protected it today

Qmechanic's answer is: Yes, if your system is nice :D

It always seems like these protection acts come too late

Whenever I see a question that has been recently protected, I can be sure that I will find a low-quality answer from a new user at the bottom of the list

@ACuriousMind OK, well thanks for the summary. I wasn't up for trawling through all that. I've never seen QMechanic post such a long answer.

@MarkMitchison Yeah, that's why I protect them in the first place. We should probably make an effort to go through all reasonably popular questions and decide whether to protect them

@MarkMitchison Ron Maimon's answer is more straightforward, and contains the essence of the idea, but Qmechanic apparently wanted to make sure to satisfy all the nitpickers

OK

Well in that case I feel my interpretation is probably justified.

The central role of symmetry in physics is because it help physicists solve problems.

But Nature is built on conservation laws.

Obviously that's just a philosophical preference. It may not even really be a meaningful distinction given the one-to-one correspondence

The main difference is that Qmechanic starts from a Lagrangian, and Lagrangians are far worse for dealing with symmetries and such than Hamiltonians and their phase space, and he must ensure that the procedure of obtaining the Hamiltonian preserves the symmetry of the Lagrangian in a suitable way, I think.

@MarkMitchison I am inclined to agree - what we observe are the conserved quantities, not the symmetries

Which is evident from the puzzled look an anyone's face when you tell them that momentum conservation is the same as translational symmetry :P

Ha

But Noether's theorem is just such a beautiful result, it's hard not to get carried away with it.

It's kind of surprising and obvious at the same time. Surprising at first, but then suddenly explains so many things when you grok it.

@Danu please don't take any of my comments too seriously about your new friend, I was just taking the chemistry analogy to the next level of abstraction :)

@MarkMitchison Well, it at least trains you to respect the Lagrangian and Hamiltonian formulations compared to Newton's laws

Because you could never see it in this clarity in Newtonian mechanics

That's true

Although I respected Lagrangian/Hamiltonian way before that

Trying to solve the bead on a spinning hoop problem :)

I wouldn't like to try that using F = ma

Hah, yes, it's certainly not the only thing that does that