The h Bar

NeuroFuzzy

00:03

Dangit, no popup for starring. Probably already saw it and forgot.

ACuriousMind

@NeuroFuzzy No popup for starring, what do you mean?

The only popup I know is the "Be nice." one that you get as soon as you log in for the first time after it's been introduced

NeuroFuzzy

Oh.

I see.

I'm a bit sleepy from doing general relativity.

from the 1,000s of lines of index manipulations approach

ACuriousMind

That would lull anyone into drowsiness ;P

AngusTheMan

evening everyone, long time no see :) Anyone have experience cleaning a cobalt surface for stm?

ACuriousMind

The answer is most certainly no in my case, but what does "stm" stand for?

@0celo7 To (mis)quote one of the great thinkers of the enlightenment: Habe Mut, dich deines eigenen Verstandes zu bedienen!

AngusTheMan

00:14

god you theorist :p scanning tunnelling microscopy :D

ACuriousMind

@AngusTheMan Haha, okay. I actually used one of these (well, actually an atomic force microscope) once for a lab course.

AngusTheMan

@ACuriousMind haha nice :) How have you been, how is the pHd going?

ACuriousMind

@AngusTheMan Not a PhD student yet! I've been well, though a bit lazy last semester. Currently actually trying to pin down my master's thesis topic, but my profs suck at answering their emails :P

AngusTheMan

@ACuriousMind wow sorry, I just assumed :p hmm i know that feeling! I haven't been productive recently at all. :/ I am applying to do another masters in either physics or theoretical physics and hopefully switch fields ..

ACuriousMind

00:31

What are you currently studying? Was it chemistry, or am I confusing you with someone else? Hope it works out for you! (I have no idea how difficult or easy switching is)

AngusTheMan

yes I'm a (rather crappy) chemist. cheers :)

What is it you want to go into? Academia?

ACuriousMind

@AngusTheMan Yes, but I'm well-aware most of us physicists end up somewhere else

AngusTheMan

@ACuriousMind hmm fair enough :) well best of luck with it all!

user116211

03:07

@AlfredCentauri: o/

0celo7

03:52

@ACuriousMind Ich habe keinen Mut, das wießt du.

DanielSank

@ACuriousMind 2nd quantization 4 lyfe.

Slereah

04:17

On the bus to the airport

Homeward bound

@DanielSank what about third quantization

And of course fourth quantization

"There's no such thing as fourth quantization, but if there were, it would be the same as the third-quantized one, due to the conformal symmetry."

0celo7

04:32

@ChrisWhite where do all these damn $\pi$s go in the Fourier transforms

user54412

@0celo7 In the exponential. Then they never appear as scaling factors out front.

0celo7

04:51

@ChrisWhite but in QM you don't get them in the exponential ;/

Slereah

40€ for reindeer steak no thanks :V

0celo7

@Slereah isn't it like 7 AM where you are

maybe 6

Slereah

7

At the airport going home tho

Spuvenir shop has some reindeer meat

But a bit rich for my blood

DanielSank

@ChrisWhite That's not true at all!

@0celo7 He's wrong. There are three common ways of setting up the Fourier transform.

0celo7

orly

DanielSank

05:06

One of them puts $1/\sqrt{2\pi}$ in front of the transform and its inverse.

Here are your options:

0celo7

@DanielSank I know

that's the one I'll likely use...

DanielSank

@0celo7 Ok. That one's pretty uncommon, but go for for it.

OPTION 1

$\tilde{f}(\omega) = \int dt \, f(t) \exp(-i \omega t)$

0celo7

@DanielSank that's the one I see most often

DanielSank

$f(t) = \int (d\omega/2\pi) \tilde{f}(\omega) \exp(i \omega t)$.

This is the one favored by physicists.

0celo7

@DanielSank well isn't that the one you get if you want to do QM?

DanielSank

05:08

@0celo7 You can do QM with any of them.

There's no one right choice.

And actually, this choice is sort of the opposite of what the Schrodinger equation suggests you should use.

0celo7

@DanielSank suppose I want $\langle p|\psi\rangle$ and $\langle x|\psi\rangle$ to be Fourier transforms

DanielSank

You see, Schrodinger was a jerk and used the opposite sign from the rest of the planet.

0celo7

@DanielSank huh?

DanielSank

@0celo7 Yeah... you can do that with any FT pair.

0celo7

@DanielSank ok but $\langle p|x\rangle$ has a square root, doesn't it?

or am I insane

DanielSank

05:10

hahah yeah I mis spoke, the transforms I wrote have a sign opposite from what physicists like.

0celo7

dood

DanielSank

@0celo7 It simply does not matter as long as you're consistent.

@0celo7 dood what?

0celo7

@DanielSank nothing

Shankar and Cahill define their stuff with $(2\pi)^{-1/2}$ out front.

DanielSank

@0celo7 k. That's a choice.

I'm used to physicists putting the $2\pi$ underneath the $d\omega$.

0celo7

I've seen that too

DanielSank

05:12

That is far and away the most common choice I've seen.

0celo7

I think $f(\omega)=\int_\mathbb{R}f(t)\mathrm{e}^{-\mathrm{i}\omega t}\,\mathrm{d}t$ is the most common

DanielSank

There's a variant of OPTION 1 in which the sign of the $i$'s is switched. This is the choice I think is most common in physics.

0celo7

I don't know what sign flip you're talking about

DanielSank

I just said it: flip the sign of the $i$'s. What's not to understand?

0celo7

all of it

DanielSank

05:14

How do you not understand $i \rightarrow -i$?

0celo7

hmm

what's the difference

DanielSank

@0celo7 Well, for one thing the equation for impedances of stuff such as $Z = i \omega L$ for an inductor depend on the choice of this sign.

If you change the sign of the $i$ in the transforms all the impdance signs flip.

0celo7

eww

DanielSank

Second, things like the time dependence of various operators, i.e. $a(t) = a(0) \exp(-i \omega t)$ have an implicit sign choice in there, and it's opposite what is used for impedance. This is a major headache if you work with quantum electronics like me :-)

0celo7

@DanielSank BUT for PDE it's best to have $f(t)=\int\mathrm{e}^{+}\cdots$

so when you take derivatives you don't get minuses flying around

DanielSank

05:18

@0celo7 Well, you get minuses either way.

If you pick $\exp(i\omega t)$ as your basis set then the annoying thing is that the hermitian operator you have is $-i (d/dt)$.

No matter how you slice it you get a minus somewhere.

0celo7

QM is just as bad as GR

DanielSank

I should say, the Hermitian operator with positive eigenvalues is that thing.

Anyway, that's normalization OPTION 1 with two possible sign choices.

There are two other normalization choices, each with both sign choices.

0celo7

ugh...

DanielSank

Yes.

Fortunately, there are really only two choices in practical use.

Oh, heh, actually there are more.

You can use $\exp(i 2 \pi \nu t)$ instead of $\exp(i \omega t)$ and that removes the $2\pi$ normalization factors entirely.

The annoying thing there is that derivatives bring down $2 \pi \nu$ instead of $\omega$.

You should just learn to always be paranoid and insist on knowing what convention is being used.

I always write my convention explicitly in anything I write.

gtg

0celo7

@DanielSank well my whole thing is showing that the stuff we use in solid state (Born approximation, pair distribution functions) is just Fourier transform shit

and there's no $2\pi$ in those exponentials

@Slereah Can you come up with a spacetime in which $\partial I^+(p)$ is not differentiable?

bolbteppa

05:54

@JohnDuffield I just have to ask again - you continually deny the idea that a rotation can be decomposed into a product of reflections, despite visual evidence sureshemre.files.wordpress.com/2014/07/… and cannot seem to comprehend the idea this is linked to the idea of spin, does this also mean you deny the Cartan-Dieudonne theorem en.wikipedia.org/wiki/Cartan%E2%80%93Dieudonn%C3%A9_theorem or are you maybe just not understanding any of this?

0celo7

oh come on

Slereah

07:53

@0celo7 Not sure?

Usually GR assumes that the metric is at least $C^2$

So that you don't end up computing $\delta^2(x) (\partial_x \delta(x))^2$

user507974

hi ladies

Slereah

Hey

Not sure if the light cone can be non-differentiable when the metric is

user507974

@Slereah sometimes I look at what others are talking about here and I really remember the only basic topic I dont know at this point is GR

Slereah

Well time to learn then!

About

El metrico

user507974

@Slereah a metric is basically a differentiable manifold right

Slereah

08:08

No

user507974

hmm

Slereah

You can have a manifold with no metric

Some manifolds don't even admit a metric

user507974

what was the relationship between a metric and the jacobian

Slereah

Coordinate change is done via a jacobian?

And the metric is covariant wrt that

user507974

i always forget covariant and contravariant

Slereah

08:13

Covariant = varies with the basis

Contravariant = varies opposite to the basis

user507974

so both are kind of one to one mappings?

Slereah

Vectors are contravariant, dual vectors are covariant

Well, the components are, anyway

The vector itself, $V^a \partial x_a$, is invariant

So if you vary the basis, $\partial x_a \rightarrow \frac{\partial x_a'}{\partial x_a} \partial x_a$

Then the vector components have to vary the opposite way

$V^a \rightarrow \frac{\partial x_a}{\partial x'_a} V^a$

That way the vector itself remains unchanged

user507974

hmm

that example was invariant you said

Slereah

You might recognize the basis vector $\partial x_a$ as $e_a$, as it is usually noted in most other fields in physics

user507974

so what would be an example of covariant

Slereah

08:19

V is contravariant, $\partial x_a$ is contravariant

The whole vector is invariant

user507974

oh ok

Slereah

$V^a \partial x_a \rightarrow V^a \partial x_a \frac{\partial x_a}{\partial x'_a}\frac{\partial x_a}{\partial x'_a} = V^a \partial x_a$

user507974

@Slereah what does the differential symbol correspond to here specifically, just the usual idea of a tiny sliver of space

Slereah

$\partial x_a$ is a basis

For instance, in Cartesian coordinates

$\partial x_0 = (1,0,0,0)$

$\partial x_1 = (0,1,0,0)$

$\partial x_2 = (0,0,1,0)$

$\partial x_3 = (0,0,0,1)$

That is the basis of the tangent plane

user507974

@Slereah basically are those coordinates describing a space that doesnt vary as you traverse it or do they correspond to the actual physical vectors that would span a cartesian space?

Slereah

08:24

Well, spacetime isn't a vector space

Unfortunately, you can't have a vector space structure in general on a manifold

So you have to define your vector in the tangent plane

That is, at every point of the manifold, you define a copy of $R^4$, where you do your vector business

This is similar to the notion of a derivative being the tangent of a curve

user507974

@Slereah so is $\partial x_a$ just describing space locally?

Slereah

Every curve in the spacetime has a "tangent" in this plane

Basically, yes

That is why we say manifolds are locally flat

user507974

great, I had a feeling that the partial was indicating something like thta

Slereah

There's a theorem that says that, in a spacetime, there is a small neighbourhood where what happens in the spacetime is approximately what happens in the tangent plane

With an arbitrarily small error

Welp

user507974

basically the equivalent of the theory behind limits but for a physical system

Slereah

08:27

Computer's about charged

And 1 hour 'til I go on the plane

Let's skedaddle

user507974

well thanks for the primer

once I finish up with my finals I just hope I take some time and go through Dirac's book

Sometimes I really am sad I took a group theory class last year instead of tensor algebra =\

@Slereah have a safe flight

Physics Meta

09:11

0

Q: Why this post is against the guidelines and especially not about physics itself?

As the title stays: It got alot of downvotes and put on hold while recieving just jerkish or rude comments that don't take my post serious. Why is it that way? I was serious with the post, tryed to pretty miuch express what is my motivation about asking, and what aspects I'm refering to. while i...

discussion

Physics Meta

09:45

1

Q: Why is my question not appearing in the unanswered questions list?

I recently asked this question on Physics SE: Is there an underlying physical reason why the Coriolis force is similar to the magnetic component of the Lorentz force? Although I didn't accept any of the answers, I can't find it in the unanswered questions list. Is there a possible reason why th...

bug unanswered-questions

DanielSank

10:04

@0celo7 Physicists rarely, if ever, put the $2\pi$ factor in the exponent.

Danu

@bolbteppa Why are you looking for a confrontation again?

@ACuriousMind What do you mean by "really is"?

ACuriousMind

11:37

@Danu Like...why is a quadratic dependence of the mass of resonances on the angular momentum a "trajectory"? How does one even assign a proper angular momentum to a "resonance"? Why is this supposedly mysterious? What has it to do with string theory?

@DanielSank You're coming from a different angle here. You say "it's completely natural to consider the Fock space instead of the one-particle space", but that is only natural in a situation where the particle number is not fixed, and in all usual situation one considers at first in QM, particle number is conserved. It's unnatural to introduce an n-particle space when all you've got is 2 particles.

And as soon as you use the creation/annihilation operator formalism (which I think is what you mean by "second quantization"), you have to admit arbitrary n-particle states.

yuggib

12:18

@ACuriousMind In some sense, it is a natural generalization to consider a space where you can have all possible numbers of particles at once, even if the number of particles is conserved. Essentially, it is similar to superselection: you have a lot of possible sectors, and fixing a conserved number of particles superselects the right Hilbert space (and eventually the correct symmetrization) where the theory lives completely

ACuriousMind

@yuggib Well, but I could imagine having distinguishable particles, then there is no symmetrization happening. Just generalizing to n particles is not enough, you need to either impose the creation/annihilation operator formalism or to figure out on the level of the particles that they're indistinguishable and n particles states should be symmetrized.

(This is about Daniel's comments here, btw)

yuggib

12:33

@ACuriousMind well, you are not obliged to do the symmetrization/anti-symmetrization in the Fock space

you can define it anyways, as well as the creation and annihilation operators

(or more properly, the number and other particles preserving operators in that case)

(and anyways, I do not agree with him that QM is "useless")

Danu

12:58

@ACuriousMind Eh, I think it's nothing super deep

It is not very interesting/hard to see that, in the quantization of the bosonic string, the states all lie on straight lines in the $(J,M^2)$ plane

I think the trajectories were one of the reasons why people liked string initially---the Regge trajectories were known and string theory could reproduce them

See e.g. pages 46-48 of Lüst et al

The "trajectory" is just the straight line

The angular momentum is the maximum $J$ allowed for given $M^2$

No idea why it's mysterious

ACuriousMind

Wait...so the states of the bosonic string are supposed to correspond to resonances in the QFT?

John Duffield

@bolbteppa : re I just have to ask again - you continually deny the idea that a rotation can be decomposed into a product of reflections. I understand it all. Let me reiterate: a rotation is a rotation, and a reflection isn't. Being able to get from state A to B via a rotation or a series of reflections doesn't change this. Electron intrinsic spin is a rotation, not some flip-flip series of reflections. It's a real rotation too, as evidenced by the Einstein-de Haas effect.

ACuriousMind

I guess I should refresh my memory of what exactly a resonance is before trying to understand what those Regge trajectories signify...

bolbteppa

13:18

@JohnDuffield But since Heisenberg's Uncertainty Principle says 'there is no concept of the path of a particle', isn't the idea of a particle rotating from A to B more plausibly described by reflections than by rotations?

@JohnDuffield quantum mechanics doesn't analyze rotations, it analyzes representations of rotations since the idea of a path doesn't even exist, so yes it does matter whether you are talking about a representation of a rotation directly or whether you are considering it as a composition of reflections right?

John Duffield

13:55

@bolbteppa : no. And the HUP doesn't say that, it's just a wave thing: "It has since become clear, however, that the uncertainty principle is inherent in the properties of all wave-like systems,[8] and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects". But note that the rotating wave is the particle. Think tornado. It has intrinsic spin, that spin makes it what it is.

@bolbteppa : the scientific evidence says the path does exist. You know a solenoid is a magnet because the electrons are rotating round the coils. And that a solenoid is like a bar magnet. The electron itself is like a tiny bar magnet. There's a rotation in there, and electron diffraction tells you you're dealing with a wave. How QM analyzes it is secondary.

bolbteppa

oh my god

Do you not realize you've just denied the most basic aspect of quantum mechanics by saying a path exists, displayed you have no understanding of what Heisenberg's uncertainty principle says, and are literally using tornado's and bar magnets over math???

@Danu I'm sorry I don't know what to say, it's just hard to believe this is real

Emilio Pisanty

14:12

Folks around to help handle this one?

-4

Q: i am not going to understand the polity of a magnet after break it into 2 pieces??? please help

kcndrfirfncflkndlwjhfuo3hwlejkofwuihgowerujgvjbkfdsvmndfcliwhflwcnf lwjkenf

electrostatics

John Duffield

@bolbteppa : the path exists, I understand the HUP totally, and I'm taking careful note of the hard scientific evidence, such as the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". You're dismissing that hard scientific evidence because it doesn't fit with the math.

Hari Prasad

@EmilioPisanty what the hell is that?

0

Q: is this not workdone in physics when we move from initial point and back to the intial point,,,w=f.d?

Here's a link! And a reference-style link to a panda. References don't have to be numbers. Visit http://area51.stackexchange.com/ regularly! Use angle brackets to force linking: Have you seen http://superuser.com?

homework-and-exercises electromagnetism newtonian-mechanics thermodynamics

John Duffield

@bolbteppa : also note the [Poynting vector](https://en.wikipedia.org/wiki/Poynting_vector#Static_fields) for static fields: _"while the circulating energy flow may seem nonsensical or paradoxical, it is necessary to maintain conservation of momentum"_. You'd think the electron has a static field. But it moves the way it does in a magnetic field for a very good reason. Because it's a "dynamical spinor". Not because of quantum magick.

Hari Prasad

@ACuriousMind @JohnRennie Is that a virus?

yuggib

14:38

arxiv.org/pdf/1203.1920.pdf

0celo7

@yuggib my advisor dissed Godel yesterday

bolbteppa

@JohnDuffield my guess is that your interpreting that quote from wikipedia about the Einstein-de Haas experiment to try to imply spin is some classical non-magic thing to do with tornadoes?

yuggib

@0celo7 why?

for he was a psycho that starved to death?

0celo7

@yuggib because he's a logician and his proof of some GR thing is "probably wrong"

@yuggib what

bolbteppa

@JohnDuffield the fact that you think a path exists is flat out denial of physics, the very first thing in quantum mechanics

yuggib

14:44

@0celo7 he voluntarily starved to death being afraid of poisoned food

0celo7

@yuggib huh

@JohnDuffield I like how "dynamical spinor" is less magical to you than quantum mechanics :D

yuggib

@0celo7 a correct mathematical proof is never wrong

0celo7

@yuggib wat

Well it's not a correct proof...

Also dat tautology

A correct proof is correct

yuggib

I mean that your advisor is probably questioning the physical content of Gödel's solution to Einstein equations

because as far as I know, the fact that it is a correct mathematical solution is not disputed

but I do not know any details about that, so I can very well be wrong

0celo7

No, he's questioning a proof that Godel wrote which claims there are no complete spacelike hypersurfaces

yuggib

14:48

@0celo7 ah ok, I don't know that stuff at all

0celo7

Because completeness is a global property in Riemannian geometry and the "proof" uses a local integrability criterion

yuggib

anyways if you want I can give you the complete works of godel

if you'd like to check

0celo7

I have the paper.

ny advisor says he'll read it and we'll discuss it after spring break

yuggib

give me the year's reference and/or title

(the papers in the collected works are commented)

0celo7

Ok, when I get to class.

It's 1949 and probably has "spacetime" in the title

@yuggib > An . xam~i. .e oi a

So..utions oI .

'.4'ew ', yjve oi Cosmo. .ogica. .

& inst:ein s .

Gravitation

& ie..~ . .

Huh

An Example of a New Type...

@yuggib that enough for you to find it?

yuggib

15:01

yeah

is the one on Gödel solution to Einstein eqns

with a preface by Hawking

(that says he discusses the solution on HE, in two pages)

0celo7

Yeah, and its BS in HE

No one understands the proof Hawking gives

Timaeus claims to, but he's a bit of a dick

yuggib

there is an expanded version of the paper that he gave as a lecture

Danu

@ACuriousMind Well, the string model of the strong force never really got too far

yuggib

it is in the third volume of the collected works

if you want it, let me know

0celo7

letting you know

Danu

15:06

@bolbteppa Get over it! :)

Why does it even bother you?

0celo7

Because JD is a nice person

brb class

Probably

Danu

@bolbteppa It's called the internet ;)

yuggib

@bolbteppa and you really should not care too much

0celo7

(removed)

Danu shows up and stuff gets deleted

Danu

@0celo7 The people fear me

bolbteppa

15:15

Anybody vaguely know the link between twistors and spinors?

0celo7

@Danu Whoa Mak

@bolbteppa I wish

All twistor stuff is either wandwavy physics or crazy sheaf cohomology

Slereah

I am home

0celo7

👍🏻

bolbteppa

15:32

"Twistors as higher-dimensional spinors - The shortest, but hardly the most transparent, way to describe a (Minkowski-space) twistor is to say that it is a reduced spinor (or half spinor) for O(2, 4)." The space of spinors for an even-dimensional space splits into two spinor spaces, so in the QFT case it's like taking the left or right spinor part of the dirac equation solution.

Wow that's insane, he goes to this 6 dimensional space, somehow uniting conformal symmetry with the Poincare group, but still does a spinorial decomposition of an 'isotropic'($||v||^2 = 0$) vector like I was talking about yesterday

John Duffield

@bolbteppa : I referred to the Einstein-de Haas effect because it "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics". It demonstrates that electron spin isn't magic.

FenderLesPaul

Magic is all around us

bolbteppa

Nobody said electron spin was magic, everybody knows it's of "the same nature" as angular momentum, I don't get what point you're going for so I'm assuming it's probably denying something about quantum mechanics

John Duffield

@bolbteppa : the path exists. Spherical harmonics map it out for an electron in an orbital. It still exists for a free electron. Which you can annihilate with a positron. Then the path exists for a 511keV gamma photon too. Only then it's linear. Think in terms of a seismic wave. It moves from A to B, but it isn't just houses on top of the AB line that shake. Because it takes many paths.

Gotta go, bye for now.

@bolbteppa : no, I'm saying it's a real rotation. The "rotating faster than light so spin isn't classical" thing is a non-sequitur. Now do excuse me, I must go.

0celo7

Lol

It would be tragic if it weren't so funny

Danu

15:42

This discussion is completely unconstructive

Please, stop it.

John Duffield

:28182299 : whatever I can write down is academic. Spherical harmonics aren't something I made up. Thomson and Tait coined the phrase. Go and read On Vortex Particles by David St John. Bye.

bolbteppa

lol

Ok

FenderLesPaul

@Danu hai

how's life?

Danu

Hi, good.

Reading a book on Riemann Surfaces

Thanks @bolbteppa.

bolbteppa

Ok done

0celo7

15:49

@bolbteppa forever??

bolbteppa

Yes, as soon as I read "the path exists" I began to resign myself from all hope

0celo7

He can still be saved!!!

Don't give up :(

I don't have a scribd subscription

bolbteppa

Have twistors actually done anything apart from Witten finding some string theory use for them?

Danu

@bolbteppa They're now finding application in QCD cross-sections. They also turn out to be relevant to the whole amplituhedron story.

See here: twistordiagrams.org.uk

bolbteppa

The motivation for them sounds amazing in a sense, like going from Newton (local) to Lagrange (global):
"Since twistors refer to an active group of spacetime transformations (the conformal group) where spacetime points get sent to other spacetime points under the action of the group, twistors are seen to be entities that refer globally to the spacetime, rather than to individual points in the spacetime. Local quantities such as vectors, tensors, or ordinary spinors refer to the symmetry group that acts at a point, e.g. the rotation or Lorentz group. Although this makes twistors more difficu…

0celo7

15:57

@Danu Zee talks about them. That's what I meant by "handwavy physics"

bolbteppa

Waving my hands really hard, the amplituhedron is basically using differential forms which is basically a way to go global

Zee has a new group theory book coming out

Danu

Another physicist's group theory book? Just what we need...

0celo7

@bolbteppa oh god

Danu

That'll be handwavy$^2$, judging by his QFT book.

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