stuck at 8991 rep - seems it's difficult to get over 9000

@KyleOman I like your vector calc/multivar calc answer.

+1

:)

and there it is :P

9001 :P

lol why did you link the "know your meme" page

I don't think we have anyone that out of touch here

you never know!

anyway, going home, thanks for humouring me :)

I don't think we've covered the dot product in physics...am I allowed to use it on the homework?

Damn, I have to talk to the TA...

@ACuriousMind I just sent the TAs an email if I can use the dot product for this assignment. My reason: I don't like coordinates. They probably think I'm insane :/

@0celo7 Using the dot product is not sufficent for declaring you insane, I think

I guarantee I'm the only person in the class who likes coordinate free descriptions.

@0celo7 That's completely normal in physics :P

(being in a tiny minority who likes coordinate-free descriptions)

i LOVE them

but you read Weinberg G&C

there's coordinates on like every page

True, but I also read pure math

@ACuriousMind When does your semester start?

@0celo7 mid-October

Did you fulfill your summer goals of better understanding string theory and TQFT?

@0celo7 Not yet.

Oh, you're working on it?

@ACuriousMind Geometry question: this has been bugging me for a while.

How is $i$ defined along each geodesic?

And how is it defined along these "broken geodesics?"

@0celo7 I don't quite understand what you are asking

$i$ is not defined "along geodesics", it is the isometry that is defined on the entire manifold

I don't know how to justify "Thus, the above information...through p."

@ACuriousMind I know this.

I don't understand how the action of $i$ on the geodesics is determined.

@0celo7 Ah. Well, take any geodesic $\gamma$. $i^*$ tells you what the tangent vector at $p$ maps to at $i(p)$, so you know which geodesic through $i(p)$ it maps to (tangent vectors determine geodesics uniquely). Call that geodesic $\delta$. Now, $i(\gamma(\tau)) = \delta(\tau)$.

OK, what about "broken geodesics"?

@0celo7 C'mon now, that's not hard to see.

@ACuriousMind What precisely does "piecewise geodesic" mean?

@0celo7 It's a curve $[0,1] \to M$. You may partition $[0,1]$ into finitely many closed intervals $I_i = [b_i,e_i]$ (with $b_0 = 0$, $e_N = 1$) such that $I_i \to M$ is a geodesic.

(What did you think it means?)

Not that, I honestly have no clue what that means.

But you see that it matches the meaning? It's a curve, and when you chop it into pieces, each piece is geodesic

Ok...but why are we chopping it into pieces?

Did you never encounter "piecewise smooth function" or "piecewise continuous curve" or somesuch in other contexts?

I did.

@0celo7 Because you can't generally find a geodesic connecting every two points, but a broken geodesic.always works

@ACuriousMind Proof?

@0celo7 Obvious. :P

I feel like I know this, but the why escapes me.

Seriously, it's not hard to come up with

What do you know about two points and connecting them with a geodesic?

I'm thinking so hard I just swallowed my gum.

I have no clue what to do about this.

@ACuriousMind I feel like I should know something.

Every point has a neighbourhood in which every other point can be connected to it with a geodesic.

Yeah, a convex normal nbd.

Oh.

Ok, so I still don't get how $i$ is defined on such a broken geodesic.

@0celo7 Uh, it's defined on every piece?

Dude, you're saying words.

And each word I understand.

Yeah, so, I don't know what the problem is

You can also do the proof without "broken geodesics":

Cover $M$ with convex normal neighbourhoods

Then iteratively propagate the definition of $i$ to the entire manifold

Knowing it is defined along geodesics, you can define $i$ on the nbd of $p$

Ok, that makes more sense.

Then you pick points in the intersection of the nbd of $p$ with the others, and extend the definition to those

etc., etc.

But let's go back to these broken geodesics.

Let's take the simple case $\mathbb{R}^2$.

Would a broken geodesic be e.g. a "corner"?

Bad example, because all points can be connected by geodesics directly

@0celo7 Yes, a "broken geodesic" is just some straight lines glued together.

@ACuriousMind ohhhh

Ok

Can we just delete all of this?

I see you now have the picture in your mind :D

Yes, this is all quite elementary.

I don't know why you were so confused.

(Thank you.)

np :)

I need to find a really tough geometry problem to ask Freire and stump him.

Maybe ask him about a random proof in HIM?

> I strongly encourage you to break it into components, as it's very helpful in understanding the forces and what exactly they're doing to the beam they're attached to. Breaking forces into components will be a very handy skill later this semester so it's best to practice it now. It's also easier for me to grade and give you partial credit if you break it into components.

@ACuriousMind

"Practice," seriously?

I just completed this course in high school and got the highest score on the standardized test for it...

Thats a joke

Wtf

It's so much easier with the dot product.

@obe down for GR chat

When did Ron Maimon's account get suspended?

Fun question

@0celo7 You want a cookie?

@DanielSank thanks for linking that in chat!

@ChrisWhite I think you're mostly on the money there.

I'm not 100% sure reversibility is enough to get fluctuation-dissipation. I have to think about that.

@ChrisWhite I think it's only "mysterious" because, as you say, we often forget one side of the coin.

@NeuroFuzzy Asking if @0celo7 wants a cookie?

@ChrisWhite I think there is a bit more going on.

Reversibility itself doesn't seem like enough to get the full relation of the FD theorem.

@ChrisWhite Ha! You really helped me understand that. Remember when I was asking about the lossless transmission line ladder?

@ChrisWhite Yeah. I was going around preaching that to my friends.

None of them seemed quite as impressed as I was...

~Sigh~

@ChrisWhite Well, you don't have temperature, etc. just from reversibility.

@ACuriousMind I see your point, it is true "rest" and "zero acceleration" make a difference. I am quite interested in that topic honestly, but it sounds strange to me. To my knowledge, Newton's I law is to define* and induce* inertial frame. and if one can define inertial frame in different ways, and different ways do give you different results or predictions, then it doesn't make sense. Physics or nature should be independent of how you define things.

Hey

Hi Slereah

@DanielSank Depends on what kind of cookie...but I'm not sure why you asked that.

@Shing Well, even with "rest", there is not a single moment in the moving solution for Norton's dome that violates the law "Things upon which no force act, are at rests/uniform motion" since in the interval $[T,-\infty]$, no force acts on the particle, and it is not accelerating, and in the interval $[\infty,T)$, it is accelerating, and the force on it is non-zero because it is no longer at the top of the dome.

The problem here is not in the laws, it is in the shape of the dome - if you can construct a situation such that the conditions for the uniqueness theorems for the solutions of differential equations are not satisfied (in this case Lipschitz continuity), then you get "classical non-determinism".

washingtonpost.com/local/education/…

@DanielSank No, I'd like to not have to resort to heathen methods like "coordinates" and "components."

Doesn't "do you want a cookie for that" mean as much as "that's irrelevant"?

Maybe.

But telling me I need to use components to "practice" them is dumb because I've already taken this course.

And I could have skipped it if I weren't in this major.

Coordinates are best

Out, heretic :D

I also prefer to do QFT with REAL FUNCTIONS

None of that Hilbert Space vector!

Yes!

GR SHOULD BE DONE WITH ADM BECAUSE MIXING TIME AND SPACE IS BAD

Down with distributions!

Spacetime Jim Crow!

Although really, the greatest way to start a war on a physics channel

Just start talking about QM interpretation

MANY WORLDS FOR LYFE

ONLY BOHM MAKES SENSE

u wanna fite

get yo shitty pilot waves outta here

Many worlds is the only logical choice.

Also, look who supports it: Sean Carroll, Michio Kaku

You can't argue against them.

Can't I

No, you don't even have a PhD.

That's a low blow!

low blow

It's correct though.

Brb lecture

low blow

& you need to be lectured to :P

I may not have a PhD but at least

I HAVE A DEGREE

That's irrelevant

You're a lot older than me

go to lecture

I am. He's not here yet.

I am

find a chick to sit next to :P

You little whippersnapper :V

"Little whippersnapper"

You're old enough to have 2 PhDs.

Tell me about it

Hi @JohnDuffield

:D

but why?

this is a chatroom

@vzn Some of the things some physicists take for granted are wrong. For example see this answer about the speed of light.

Then that evil is just as much a part of the world as goodness.

Well yes, but that's no reason to unleash it

You don't go eating poison mushrooms just because they're natural

We all should just chat nicely to one another, as the room was intended to be :-)

Well said Rigor.

Thank you Sir.

@Rigor : where was that?

@Rigor : I thought it was perfectly fine myself.

Yo @Danu

find any cheap textbook sites yet?

Hi

No

:(

@Slereah and @JohnDuffield "If you ain't got nothing nice to say, then don't say nothing" You are both getting a short timeout from me, and I suggest you simply ignore each other in the future (use the "ignore" function of chat, perhaps?).

That was uncalled for.

You just made me feel like an instigator.

@Rigor No, I think that it is time that this nonsense stops.

my last message

@Danu The ignore function is rather useless if you see the conversation of others with the user you blocked.

@ACuriousMind Yeah, probably true. I guess it's just a matter of self-discipline, then.

Rate the following sentence for its accuracy on a scale from 1 to 10: "In conventional quantum field theory the elementary particles are mathematical points."

@ACuriousMind 1?

Why?

@Danu Yes, that is my impression as well

In conventional classical field theory, the elementary particles are mathematical points

That'd be more accurate

@Danu Well, it's one of the first things I just read when opening BBS.

(...but then the photon isn't an elementary particle)

@ACuriousMind what

Sigh... I guess they're just trying to contrast with strings?

@Danu It's in the introduction, first sentence in the paragraph titled "The size of strings"

@ACuriousMind Already found it

...now they called Lie groups Lie algebras.

Not sure deciding to read this was the best idea :P

Maybe they're just using the (nonstandard) notation that the algebra is denoted by allcaps? :P

(which, in view of the actual usage of notation for groups/algebras in e.g. QCD circles, may as well be the truth lol)

There's "Lie algebra $SO(32)$" and a little later "the gauge group is $SO(32)$".

sigh

Why is the kernel of a homomorphism of Lie groups a closed subgroup?

@ACuriousMind Perhaps GSW? Or is that not accessible as an introduction?

@Danu It's an embedded submanifold, isn't it? (Preimage of a regular point)

@Danu That's from '87, I wanted something modern

If it's not an introduction, it's not a problem, I know the basics already.

@ACuriousMind Not all embedded submanifolds are closed.

In my notes, we first show that Lie group homom's are constant rank (thus the kernel is an emb. submnf.), and then there's a remark that it also follows from the kernel being closed (and then applying the closed Lie subgroups theorem thingy)

@Danu A subgroup is closed if and only if it is embedded Lie

@ACuriousMind Oh, it goes the other way too. My notes didn't mention that :P

@Danu BBS is an updated GSW...in a sense.

They don't distinguish algebra/group either IIRC.

deep sigh

I don't think Polchinski does either, but I haven't read much of his books.

BLT distinguishes, but it's also the driest book in the business.

Also that whole section on spinors left me wondering why I was bothering to read it.

@ACuriousMind I think you've explained this once before...why is a harmonic function on a compact Riemannian manifold constant?

@0celo7 Spinors are important for pure mathematics too---I'd like to learn a lot about them.

@Danu Lol, don't try BLT for that.

^^ACM after my questions about that chapter.

Also, who are BLT

Blumenhagen, Lüst, Tiesen.

Can't guarantee spelling of that last one.

Theisen, isn't it?

Perhaps.

Book is in the dorm, I'm not.

Oh, that one

It should be with an "h", at least :P

The first few chapters are excellent.

yes, Theisen

I kinda stopped reading it because it turned into "let's calculate partition functions for 100 pages."

I'll be taught from it next semester... by Lüst himself ;)

@ACuriousMind According to Dr. Freire, the trick is to examine the integral $$\int_\mathcal{M}\operatorname{div}(f\operatorname{grad}f)$$

doot doot

I recall you saying something different.

Hullo

@Slereah Frenchie.

murcan

What's a "murcan?"

You kidding?

Perhaps.

Speaking of that...

@ACuriousMind Should I try for German citizenship? Is it hard? What benefits does it bring?

American $\rightarrow$ Merican $\rightarrow$ Murican $\rightarrow$ Murcan

@Slereah you watch your tongue or I'll have to come deliver some freedom to your country

@ACuriousMind 10...after all, it's in a textbook!

@0celo7 The manifold has to be connected, but then it follows from "Every cohomology class contains one harmonic form" and "The zeroth cohomology vanishes".

::looks at erratum page::

Nope, it's correct. So 10.

@0celo7 I have not the slightest idea

@0celo7 Why would you?

So anyway

@Danu I was born in Germany, speak the language and have German heritage?

What's a cohomology

I saw it in my topology class and all

But I had already given up by that point

I remember a lot of arrows on things

CHAINS

2 CHAINS

Cohomology = math hip-hop

@Slereah The dual theory to a homology theory

;)

What's a homology

The dual of the cohomology

@0celo7 youtube.com/watch?v=zn7-fVtT16k

It's about counting topologically different "holes" and "surfaces" that fit into the space, heuristically

(Co)homology is a topological invariant, but tells you surprisingly much - cohomology on smooth manifolds tells you how many forms there are that are closed but not exact, and, more generally, how many different bundle structures you may construct

Mathematicians have invaded the chat again

brb homework that I didn't finish last night

@0celo7 I'll stop now ;)