The h Bar

Xasel

11:00

Well I think he envisioned only qualitative aspect of it and erst of Quantitative work was brought by Gauss,Ampere,Maxwell. I have strong intuition that we can derive so. @Slereah:maybe that is the answers to my question?. Is the :"definition the trajectories of a vector field have the vector field as tangents" arises from some theorem stating such or we have assumed it as axiom ?

Slereah

Well you can't prove a definition

so yes

John Rennie

@Xasel The field lines show the direction a test charge will accelerate due to the field. This is necessarily the tangent to the field.

Xasel

Nope what I meant is sometimes I have seen things are defined using some theorem as their backend and sometimes they defined by taking thems as axiom..maybe that axiom @Slereah.

@JohnRennie: Thanks for the insight but If I take that into account then I think I have some fundamental confusion :')

John Rennie

@Xasel Why?

Xasel

Lets see the first diagram...what should be the direction of a test charge exactly in that empty space between those two charges?

Balarka Sen

11:07

@JohnRennie Hehe, not really, but I know why that's relevant to the novel. Didn't remember that.

John Rennie

@Xasel There is a point where a test charge will not move at all because the forces on it sum to zero. If that's what you are asking.

ACuriousMind

@Kaumudi.H Hobbes is not imaginary, at least not obviously so. In the panels where he is not drawn as a stuffed animal, he is as real as everything else, and interacts with the world in ways imaginary friends can't. I think that the coexistence of these two realities (Calvin's and everyone else's), neither of which is definitely shown to be false or imaginary, is a major theme of C&H.

John Rennie

114

Q: Did Calvin ever realise that Hobbes was not real?

In the comic strip Calvin and Hobbes, was there any moment where Calvin realised that Hobbes wasn't real?

126

Q: Does Hobbes ever do anything that Calvin himself could not do?

So this question got me thinking that, except for pure Calvin fantasy (i.e. Spaceman Spiff, where nothing is real), is there ever a comic where Calvin claims Hobbes did something and Calvin could not have possibly performed himself? This one comes close, but Calvin could have tied himself up N...

comics calvin-and-hobbes

Balarka Sen

@ACuriousMind Why are you analyzing Calvin & Hobbes?

That's like the next stage of analytical philosophy.

ACuriousMind

@BalarkaSen Why not? It's a brillant work :P

Balarka Sen

11:13

I don't disagree

Xasel

@JohnRennie Thanx,I still have one more question bugging me: Does this implies that electric field is always differentiable or say we can never construct a system of charges whose resultant vector field is not differentiable (No existence of tangent)?

John Rennie

@Xasel The field is everywhere differentiable except at point/line/2D charges - but then they don't actually exist.

Bernardo Meurer

Heya

John Rennie

Afternoon

Bernardo Meurer

What's up John

0celouvsky

11:17

@ACuriousMind I think it's supposed to be smooth but I can't tell

John Rennie

@BernardoMeurer Drinking coffee and listening to some blues ...

... while pretending to do some work :-)

Xasel

@JohnRennie I learnt that there exist some curves which are continuous everywhere but nowhere differentiable so I was imagining if there was system of charges whose resultant electric field lines yields such curves ?

ACuriousMind

@0celouvsky How do you think I could tell when I'm not reading the book you're reading? My response would indeed be that it doesn't matter, and I think that conceptually, you should stop thinking of the different cohomologies as actually different.

John Rennie

@Xasel no

Bernardo Meurer

@ACuriousMind I'm very good at pretending to work

It is in fact my job

which I pretend to do

ACuriousMind

11:20

The cohomology group you have in the end when you write $H^q(X,\mathbb{R})$ doesn't actually know what chain complex it came from.

@BernardoMeurer Why are you telling me that?

Xasel

@JohnRennie: Can we mathematically(rigorously) prove so that such configuration of charges is not attainable then?

Bernardo Meurer

@ACuriousMind Because I pinged the wrong person

ACuriousMind

@BalarkaSen As another point, the names of C&H themselves already indicate the strip should be read with a philosophical bent ;)

0celouvsky

@ACuriousMind more on this later

Bernardo Meurer

@JaimeGallego Added more stuff, some singer-songwriter I've been listening to lately

0celouvsky

11:21

I know that simplical and smooth simplical are isomorphic when the coefficients are R, but I don't know that over Z

And the proof is absolutely awful if I remember correctly

Balarka Sen

@ACuriousMind Indeed!

John Rennie

@Xasel I don't know. You would need to show that Maxwell's equations cannot have solutions that aren't differentiable (except at charges). I don't know how you would set about that.

Slereah

You can have weakly differentiable EM fields

John Rennie

10

Q: Electromagnetic field and continuous and differentiable vector fields

We have notions of derivative for a continuous and differentiable vector fields. The operations like curl,divergence etc. have well defined precise notions for these fields. We know electrostatic and magneto static fields aren't actually well behaved. They blow up at the sources, have discontinu...

electromagnetism mathematical-physics vector-fields calculus

Slereah

mine own avatar fellow

the esteemed Leonard J. Crabs

Lawyer at law

I'm not quite sure where the original photo of Leonard J. Crabs comes from

it is a mystery lost to time

Xasel

11:30

@Slereah: I am bubbling with excitement regarding weakly differentiable EM fields.Can you elaborate on it or give pointers to as such?

ACuriousMind

@Xasel Think about what sorts of discontinuities the charge density $\rho$ is allowed to have: We physically model charges either by continuous densities or by discontinuous point/line/surface charges. The discontinuities of the resulting electric field sit at the same points, so if you put in a crazy discontinuous charge distribution, you get out a crazy discontinuous electric field. What is it you want to prove?

Slereah

Weakly differentiable EM fields are when you have EM fields defined with distributions

Then you can allow discontinuities in the field

He is the best internet lawyer money can buy

Xasel

@ACuriousMind Well John said: "@Xasel The field lines show the direction a test charge will accelerate due to the field. This is necessarily the tangent to the field."

And so I followed up with the question:
@John Thanx,I still have one more question bugging me: Does this implies that electric field is always differentiable or say we can never construct a system of charges whose resultant vector field is not differentiable (No existence of tangent)?
@John I learnt that there exist some curves which are continuous everywhere but nowhere differentiable so I was imagining if there was system of charges whose resultant electric field lines yields such curves

Slereah

angelfire.com/pro/leonardjcrabs

I do miss the old internet days

John Rennie

@Xasel The direction of field lines isn't defined at discontinuities in the charge density.

ACuriousMind

11:34

In the end, the "discontinuity" issue arises because we decided to think of a density as an actual function, when densities really should be physically thought of as distributions/measures - they tell you how much stuff is inside a certain volume of space, so a density assigns a number to every volume of non-zero measure - the amount of charge inside it. That this sort of thing can often be expressed by a nice density function is a happy accident, in my view

John Rennie

But then discontinuities in the charge density are not physical because charge density is always continuous.

Slereah

@JohnRennie that's a fairly bold statement!

John Rennie

@Slereah That's true isn't it?

ACuriousMind

@Xasel Yeah, so...if your field is not differentiable, the concept of field line just doesn't make sense at that point

Slereah

@JohnRennie That is debatable

Xasel

11:36

Well I want to prove it mathematically rather than by hand-waiving

That no such system of charges is physically feasible that yields such a resultant electric field

One more thing coming to my ming: In the context of "@Xasel The field lines show the direction a test charge will accelerate due to the field. This is necessarily the tangent to the field."

Lets say such system exist and produces such resultant electric field

then what would be it's physicaly interpretation

0celouvsky

@ACuriousMind hello?

@BalarkaSen did you see my proof

Xasel

How things will go about with test charges trapped in such syatem?

ACuriousMind

@0celouvsky Yes, hello?

Balarka Sen

proof of what, @0celo

ACuriousMind

@Xasel I don't understand the question - $F=qE$ as always

But you're thinking about a completely unphysical system - I guarantee that no matter whether you can in principle find a solution to Maxwell's equation of this form, it will be utterly unfeasible to actually arrange it in practice

0celouvsky

11:44

@BalarkaSen search abelian and it'll probably be the first thing

Balarka Sen

not gonna

0celouvsky

Why not

Balarka Sen

Oh, did you mean proof of cover being abelian?

0celouvsky

@ACuriousMind I asked if simplical and smooth simplical agree for Z coefficients

Balarka Sen

I saw that. It was fine; it's what I was saying to you before

0celouvsky

11:45

@BalarkaSen yep

Ok

If there's a sheafy way to see it I'm all ears @ACuriousMind

ACuriousMind

@0celouvsky Probably? I don't think I've ever seen smooth simplicial but people wouldn't use it if it didn't agree at least in useful cases like manifolds

In the end, the "best" way to see that two cohomology theories coincide is that they fulfill Eilenberg-Steenrod, imo :P

Of course, you have to prove that E-S defines a unique cohomology theory first, which you certainly will not be willing to take as a black box...

John Doe

Hi all

Xasel

Well I want to prove or disprove(I am not sure it's prossible or not that's why I am asking you guys) it mathematically rather than by hand-waiving
That no such system of charges is physically feasible that yields such a resultant electric field which is continuous everywhere but diffeerentiable nowhere

Balarka Sen

Two cohomology theories on the category of CW-complexes, but yeah

Xasel

@ACuriousMind And in the context of @JohnRennie 's statement this could be physically interpreted as: If no derivative i.e tangent exist than a test charge can't move anywhere in that electric field .It's trapped forever.

ACuriousMind

11:51

@Xasel You can't prove mathematically that a certain system of charges is not feasible. My point was not about the mathematical impossibility of such a solution to Maxwell's equation (I'm fully prepared to believe there is such a solution if you make $\rho$ horrible enough), but that you couldn't ever create such $\rho$s in a lab in practice. That you can't do that in a lab is something you see by common sense, not by a mathematical proof.

Slereah

@0celouvsky : Looking into non-hausdorff stuff again

It is grim

“Non-Hausdorff spaces, often regarded as a technical nuisance, sometimes produce a global disaster.”

0celouvsky

@ACuriousMind you've never seen it? What do you think de Rham's theorem is o.O

Slereah

“Kelley’s book set the cat among the pigeons in 1955 by daring to omit the Hausdorff condition from many of its definitions.”

ACuriousMind

@Xasel Why wouldn't the charge move anywhere. You have $m\ddot{x} = F = qE$, no derivative/tangent of the electric field required

Slereah

everyone is very pessimistic about them

ACuriousMind

11:53

@0celouvsky Oh, you think I remember the by-hand proof of deRham's theorem? You vastly overestimate my desire to remember proofs :P

Xasel

@ACuriousMind: @JohnRennie said that "The field lines show the direction a test charge will accelerate due to the field. This is necessarily the tangent to the field."If no such tangent exist then this implies there is nowhere our test charge can move to ?

ACuriousMind

Well, @JohnRennie was wrong :P

John Rennie

@Xasel: it's not really the tangent to the field. It is the value of the electric vector at that point.

ACuriousMind

I don't know what he meant by the acceleration being "due to the tangent to the field"

John Rennie

The electric vector is either zero or points in some direction.

And the test charge accelerates in the direction of the electric vector.

It would be the tangent to the potential.

0celouvsky

11:57

@ACuriousMind de Rham gives an isomorphism between cohomology of forms and smooth simplical

It does NOT give an isomorphism to singular

John Doe

Does anyone know how one would define what the "geometric phase" is for a spin-1/2 particle under the influence of an oscillating force is?

ACuriousMind

@0celouvsky I guess I just think of "deRham's theorem" as the statement that deRham cohomology coincides with ordinary cohomology, and don't care particularly about how exactly you show that

0celouvsky

But that's too vague

What coefficients is that for?

ACuriousMind

Reals/Complex, deRham doesn't give any other coefficients

0celouvsky

So you see my confusion now?

ACuriousMind

11:59

@JohnDoe Look up "Berry phase"

@0celouvsky Not really, not

0celouvsky

This guy is writing de Rham for Z

Xasel

Now the picture is becoming clear but Is till feel a lil' haze :')

@ACuriousMind: What is the reason behind your statement : "You can't prove mathematically that a certain system of charges is not feasible. My point was not about the mathematical impossibility of such a solution to Maxwell's equation "

0celouvsky

Specifically, the first chern form of a line bundle lies in one of these spaces

ACuriousMind

@0celouvsky What? No. He's taking the map $H^q(X,\mathbb{Z})\to H^q(X,\mathbb{R})$ (which is a map that exists abstractly in cohomology), and then calling its image $\tilde{H}^q(X,\mathbb{Z})$. So using $H^q(X,\mathbb{R})\cong H^q_\text{dR}(X,\mathbb{R})$ you also get a version of that image inside the deRham version, without having to take "deRham with integer coefficients"

0celouvsky

What does abstractly in cohomology mean?

Slereah

12:03

"Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point."

what

"Among other things, Gauld references that there are two paracompact and two nonparacompact 1-manifolds, and ℵ0 paracompact and 2ℵ1 non-paracompact 2-manifolds. "

a fair bit of non-paracompact manifolds

Xasel

Can somebody enlightne me on the Curious Mind's point: You can't prove mathematically that a certain system of charges is not feasible.".What is the reason behind it or why it is so?

Slereah

though apparently the set of non-Hausdorff one-manifolds is like

Crazy huge

John Doe

In an online lecture it states that the stationary states of a time independent Hamiltonian is orthornormal $\langle \psi_n| \psi_{n'} \rangle = \delta_{n,n'}$ where each $\psi$ has it's own phase factor. Is this phase factor referring to the time dependent exponential $e^{-\frac{i E_n t}{ \hbar}}$?

Slereah

yes

John Doe

12:18

It doesn't maybe mean that each stationary states are unique up to some phase factor. Hence $\langle \psi_n e^{\phi_n} | \psi_{n'} e^{\phi_{n'}} \rangle$ still gives $\delta_{n,n'}$?

Yashas

0

A: Is the CMB rest frame special? Where does it come from?

The photons that forms the cmb come from what is called the last scattering surface. This surface is a different one depending on the position in spacetime of the observer. So each observer will have a different reference frame in which the cmb has no dipol term.

0celouvsky

@ACuriousMind I still don't know what you mean by "abstractly in cohomology." Bredon states that $H^\bullet (M;G)\cong H^\bullet_\text{smooth}(M;G)$ for any Abelian group $G$.

lol

John Doe

12:34

@Slereah Does that make sense at all?

user113223

hwllo

i have a question

Solid angle is a ratio so how can we measure it??

0celouvsky

what?

user113223

i know that its unit is steradian

Slereah

@user113223 Measure the two components of the ratio

"The relation Y, the definition 1 as well as the proof of theorem 1 can be found in [12]"

Anyone wants to guess what is reference 12?

It's not HE or Steenrod

0celouvsky

Hatcher?

Slereah

12:43

"Unpublished"

0celouvsky

lol

Slereah

That fucker

the worst part about unpublished papers is that you can't even know what they are once published

They don't even give you a working title

0celouvsky

what are you reading?

ACuriousMind

@0celouvsky Universal coefficients, once again: You have an isomorphism $H^q(X,\mathbb{R})\to \mathrm{Hom}(H_q(X,\mathbb{Z}),\mathbb{R})$, an inclusion $\mathrm{Hom}(H_q(X,\mathbb{Z}),\mathbb{Z})\to\mathrm{Hom}(H_q(X,\mathbb{Z}), \mathbb{R})$ and a surjection $H^q(X,\mathbb{Z})\to \mathrm{Hom}(H_q(X,\mathbb{Z}),\mathbb{Z})$. Put them together (inverting the first map since it's an isomorphism) to get a map $H^q(X,\mathbb{Z})\to H^q(X,\mathbb{R})$.

(That took me longer to figure out than I am willing to admit :P)

@user113223 Normal angle is also a ratio (between radius and arc length substended), why do you think we can't measure such ratios?

Slereah

Well, CAN WE?

Have you ever measured an angle

Seen one in the sky

Ancient japanese math had that weird thing where they didn't use angles at all

Everything was done with the length of sides

0celouvsky

12:55

as it should be

@JaimeGallego your previous picture was better

the current one looks a little too "m'lady" for me

Slereah

You can't escape the neckbeard inside, @0celouvsky

Bottles Beware: Uploaded by GooSmurf

►

Jaime Gallego

@0celouvsky lol

Slereah

speaking of neckbeards

I should probably clean up the flat

a bit messy lately

John Doe

@ACuriousMind When writing the time independent Hamiltonian $$H(t)\psi_n(t) = E_n(t)\psi_n(t).$$ As I udnerstand, we can consider that $\psi_n(t)$ is arbitrary up to a phase factor. Does that imply that $\psi_n(t)e^{i \phi}$ is considered equivalent to $\psi_n(t)$ in the contect of being a stationary state?

ACuriousMind

@JohnDoe Yes

Slereah

13:09

That is true of all wavefunctions

John Doe

Okay thanks

Slereah

that is why the real Hilbert space is the projective Hilbert space!

Katana VS Diet Coke and Mentos

►

Hm, reading the bio of Geroch

"has promoted the use of category theory in mathematics and physics."

I'm not sure I can approve of him now

0celouvsky

That has to be satire, right?

Slereah

'fraid not

Balarka Sen

@Slereah lol

Slereah

13:19

that guy has tons of such videos and swords

0celouvsky

@BalarkaSen Why is that funny?

Slereah

because category theory is a joke

that's why

Balarka Sen

extensive use of category theory in math and physics sounds like the slogan of nlab

which is inherently funny

John Rennie

@Xasel I think the point is that we all believe that the electric field cannot be discontinuous unless the charge distribution is discontinuous, and a discontinuous charge distribution is unphysical.

Slereah

You can have a discontinuous electric field without charge

John Rennie

13:22

@Xasel To prove that would presumably require starting with Maxwell's equations and deriving the result from there. However none of us want to do that because we all eblieve it's true and don't find it interesting enough to warrant the effort required.

But feel free to have a go if you want to prove it.

Slereah

Alright, I will take the totally opposite stance

0celouvsky

> We shall represent $H^1(X,\mathcal O^\bullet)$ by means of Cech cohomology

Slereah

Well, not opposite

0celouvsky

No thanks

Slereah

But I believe that ALL EM FIELDS ARE HOLDER-2 CONTINUOUS

that is what you would get by considering the measurement of an EM field on a path integral

0celouvsky

13:23

@JohnRennie Your starred post does not really make sense as stated

A solution is always differentiable, how else do you propose to plug it into the equation

Slereah

Distributions

0celouvsky

Now, if you mean a distributional solution...I don't see why you should think that's automatically smooth.

Sanya

@Xasel I think that the only condition you can take from Maxwell's equations is that there needs to be a green's function for the boundary conditions and that the convolution integral of that green's function and the charge density needs to be ... well behaved. So proving mathematically that a charge distribution is not possible would mean proving that this integral is not well behaved ... if I'm not making a mistake here

Slereah

though to be fair

The electric field is gonna be smoother than the charge distribution

So even if the charge distribution is discontinuous

The EM field will be $C^0$

0celouvsky

Proof?

Slereah

13:27

$\partial F = j$

0celouvsky

what is that $\partial$?

Slereah

$\partial_\mu F^{\mu \nu}$

0celouvsky

So you mean $\mathrm{div}$

Slereah

Well if you want to be crude about it, yes

You don't get discontinuous EM fields until you get dirac charge distributions, I think

Apparently Geroch is currently 75

Do you think he's likely to answer if I shoot him an email

Last known email was [email protected]

Let's shoot him an email

And sent

Let's hope he kept a paper he wrote 45 years ago around

Maxwell

13:45

Hello @yashas

Yashas

@Maxwell Are you Ramanujan?

Slereah

obviously not, hence the different name

user123733

If |w|=2 , then set of points $z=w -(1/w)$ is equal to ?

One of my friend help me like this

$|z| = |w - 1/w| \leq |w| + 1/|w| = 2 + 0.5 = 2.5$

$|z| \leq 2.5$

after that I am unable to proceed . can anybody help me

Maxwell

@yashas, ya

user123733

in my book it is written as an ellipse

with eccentricity 4/5

Slereah

13:57

Ahah

Geroch answered!

Hmmm . . . really? I guess one of the perks of getting older is that
you get credit for all sorts of things you didn't do! I did sit down
and tried to understand non-Hausdorff space-times at one point; and
I wouldn't be surprised if I wrote some small set of notes, largely
for my own consumption. So, I think I know the basic facts about
this subject, but I don't have anything in writing (at least,
anything I can find!).

Sorry I couldn't be of more help. If you have some questions --
specific or general, precise or vague -- in this area, I'd be happy

:D

Glad to see Geroch is still around

0celouvsky

14:14

What did you email him about?

Slereah

Some paper on non-Hausdorff spacetime claims that he had an unpublished paper on the topic

0celouvsky

he seems nice

But non-Hausdorff spacetimes seem quite horrible

Slereah

Well it's not too bad

0celouvsky

Can you have a Lorentzian metric? I know you can't have a Riemannian one.

Slereah

The structures aren't on the entire manifold itself

You pick $H$-manifolds, which are maximal Hausdorff submanifolds

that's where the real action happens

For instance for the branching line, the $H$-manifolds are two copies of $\Bbb R$

so for the most part you can just deal with everything in a Hausdorff way

paracetamol

14:24

@JohnR Sorry about that, my brother cut me off ._.

Slereah

"Let $M$ be a $Y$-spacetime whose $Y$-boundaries are all three-dimensional hypersurfaces (not necessarily smooth). Then the following two requirements are not compatible :
1) There are no timelike bifurcate curves with bounded acceleration in $M$
2) M is strongly causal"

neat

Bernardo Meurer

@Slereah What exactly do you work with?

Balarka Sen

@0celouvsky You can't have Lorentzian metric on most manifolds in general, let alone non-Hausdorff chaps

But of course you know that.

Bernardo Meurer

@BalarkaSen Hello

Slereah

You can't even have a partition of unity on non-Hausdorff spacetimes

Balarka Sen

14:26

Hi @Bernardo.

paracetamol

Open question: Is there any difference between the terms "Spin quantum number" and "Magnetic spin quantum number"? A cursory Google search seems to suggest they're the same, however, my copy of University Physics implies otherwise .-.

Bernardo Meurer

I'm a non-Hausdorff manifold

paracetamol

Slereah

The paper does say that a $Y$-spacetime has a Lorentz metric but I assume he refers to the $H$-manifold cover rather than the manifold itself

paracetamol

Oh wait, @Acurious, say weren't you into QM?

ACuriousMind

14:29

@paracetamol Just ask your questions, if somebody wants to answer them, they will. I don't like commiting to answering questions before I've heard them.

Slereah

Although...

One thing I wonder is

Take the branching real line

How many Hausdorff submanifolds that cannot be extended does it contain?

Is it two or three?

paracetamol

@ACuriousMind Just scroll up a bit...my question's already there ;)

Slereah

I can think of three

paracetamol

4 mins ago, by paracetamol

Open question: Is there any difference between the terms "Spin quantum number" and "Magnetic spin quantum number"? A cursory Google search seems to suggest they're the same, however, my copy of University Physics implies otherwise .-.

3 mins ago, by paracetamol

Slereah

$R^- \cup R^+_1$, $R^- \cup R^+_2$ and $R^+_1 \cup R^+_2$

For the third case, if this was a Lorentz manifold, do you run into the risk of having your structures meshing up together poorly

Like the time orientation being discontinuous

ACuriousMind

14:33

@paracetamol In that case the answer is that you shouldn't get so hung up on terminology. Your book decides to call the total intrinsic spin "spin quantum number" and a specific value of intrinsic spin the "magnetic spin quantum number". Someone else might be sloppy and just call the latter the "spin quantum number", silently assuming everything has total spin 1/2 to begin with. Neither is "right" or "wrong", these are simply differing conventions.

user123733

0

Q: complex number with determinant

Let $z_1$ and $z_2$ be two distinct complex numbers and let $\,z=(1-t)z_1 +tz_2\,$ for some real number $t$ with $0<t<1$. Then we have to prove $$\begin{vmatrix} z-z_1 & \overline{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1} \end{vmatrix}\;=\;0$$ I thought about it, but don't ge...

complex-analysis

paracetamol

@ACuriousMind So what conventions do you folks at Heidelberg use? I guess I'll adopt that ^_^

dmckee

Cases where the two possibly meanings of a casual "spin quantum number" are both sensible and you need to ask for clarification are rare.

ACuriousMind

@paracetamol It's...not a convention one needs to talk about. If you can't tell from context what someone using these terms is talking about, you have larger issues than such a minor terminology quibble.

dmckee

Usually it is clear if you are talking about the magnitude of intrinsic angular momentum or the projection in some direction from the context.

paracetamol

14:37

Roger that! Danke @ACurious and @dmck

dmckee

The former is part of the identity of a particle and can only be an open question if the identity of a particle is not known , if particles are being created and destroyed or if you are measuring this value.

Slereah

Oh wait

I don't actually think you can make a submanifold out of $R^+_1 \cup R^+_2$

At least one that's Hausdorff

Also how does this work, exactly

Geroch makes some vague notes in Chicago

Hijacek, in Bern, writes a paper that references those notes

What did he do

Did he steal Geroch's notes???

ACuriousMind

@Slereah That's...a good question

Especially if Geroch himself doesn't know where he might have those notes!

Slereah

Well I'll give him some slack on that

It was 45 years ago

11 presidents ago

12, even

I'm guessing that he shared the content of these notes with some people

but didn't have enough to make a paper out of it

paracetamol

Oh, another question @ACurious? Doesa certain "Bohr-Bury (n+l) rule" with regard to orbital energy exist? A teacher mentioned it at school when he took up the topic of Atomic Structure, but I can't seem to find any reference to it anywhere else ._.

Anonymous

14:48

@paracetamol That's a part of Aufbau's rule iirc

Anonymous

Aufbau principle

The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible. Aufbau is a German noun that means construction or "building-up". The Aufbau principle is sometimes called the building-up principle or the Aufbau rule. The details of this "building-up" tendency are described mathematically by atomic orbital functions. Electron behavior is elaborated by other principles...

ACuriousMind

I don't know what a Bohr-Bury rule is and that's vague chemistry rules, not QM :P

Slereah

All elements are composed of various proportions of fire, water, air and earth

paracetamol

@ACuriousMind And yeah... it was during a Chem. class :3

AHA! They call it the Madelung Rule too. Nice one @blue ;)

Anonymous

@paracetamol That rule is an empirical rule btw. It might not hold true always....

paracetamol

14:51

I knew un-ignoring you was a smart idea... ~~wish I could've said the same for someone else~~

Anonymous

@paracetamol You love me. You just don't know it yet ;)

Anonymous

lol

Anonymous

That someone else....

paracetamol

But...I'm straight ._.

Hullo @JohnR o/

Anonymous

lol^

John Rennie

14:53

@paracetamol Hi

paracetamol

@JohnRennie Mind if I tickle you?

I guess you do...

:'(

John Rennie

@paracetamol I'm currently eating lunch, but I will still answer questions - just more slowly

Bernardo Meurer

Anyone here know who Slavoj Zizek is?

paracetamol

@JohnRennie Great! Question 1: Mind if I tickle you? 0:)

Slavoj Žižek

Slavoj Žižek (/ˈslɑːvɔɪ ˈʒɪʒɛk/ SLAH-voy ZHIZH-ek; Slovene pronunciation: [ˈslaʋɔj ˈʒiʒɛk]; born 21 March 1949) is a Slovenian psychoanalytic philosopher, cultural critic, and Hegelian Marxist. He is a senior researcher at the Institute for Sociology and Philosophy at the University of Ljubljana, Global Distinguished Professor of German at New York University, and international director of the Birkbeck Institute for the Humanities of the University of London. His work is located at the intersection of a range of subjects, including continental philosophy, political theory, cultural studies,...

You haven't Googled? 0_0

John Rennie

@paracetamol I must be suffering a sense of humour failure because I'm finding puerile comments less amusing than usual.

Bernardo Meurer

14:58

@paracetamol No, I was asking because I just got two books from him for my birthday

2

And I thought I'd brag

paracetamol

^ :D

Bernardo Meurer

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