The h Bar - 2016-06-26 (page 1 of 2)

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3:28 AM

@ACuriousMind Do you have a hint for problem 1.5.11 in GP? My first thought was that $X=f^{-1}(0)$ but that doesn't work because it could have a degenerate Jacobian on this set

Then I thought maybe I can multiply it by some bump function to take care of the Jacobian but still keep the zero set intact, but that failed.

turns out the Jacobian didn't change at all.

So I'm really at a loss on what to do

It's also a problem in Lee, but he doesn't give any more hints than GP

And since it's before PoU in GP, it doesn't use that black magic.

@ACuriousMind Huh, googling the problem gave me the answer.

user116211

4:35 AM

@dmckee @DavidZ; you wanna check this:

user116211

1

electromagnetism entropy photons quantum-electrodynamics

user116211

He is posting continuously nonsense answers.

user116211

0

A: Why don't photons split up into multiple lower energy versions of themselves?

Sorry, but science will only be allowed to observe or quantify whatever God imagines you to see, measure and/or quantify. The CERN facility or LHC is or maybe wasting time and money chasing imaginary constructs.

user116211

0

A: Why don't photons split up into multiple lower energy versions of themselves?

Additionally, since everything but God is an imaginary construct, a photon can be split by Him into an infinte number of photons with or without any momentum variations.

user116211

Too much blind belief in God or trolling ;P

user116211

4:39 AM

4 answers to the same question ;((

@ACuriousMind I'm having an epistemological crisis with "vacuous hypotheses"

For instance, the fact that a function can have a regular value outside of its image

And that nonintersecting submanifolds are automatically tansversal

Why are these "strict logical consequences" of the definitions?

5:06 AM

AHHHHHHHHHHHHH any vector subspaces have nonempty intersection

somewhat related but my question still stands @ACuriousMind

3 hours later…

user116211

8:19 AM

Combing your hair is a waste of time. — Count Iblis 8 hours ago

user116211

o.O

user54412

8:29 AM

I was wondering what that comment was supposed to mean.

11:05 AM

@0celo7 Which of these, exactly?

The thing with the "vacuous hypotheses" is that the definitions are only phrased such that they contain "for all" statements, but not "exists" statements for what in the vacuous case is the empty set.

It always depends on the specific application whether it is desirable to explicitly exclude the empty set from such definitions or not

11:50 AM

@ACuriousMind heyo

I'm in the middle of a renewed crisis of math vs. physics doubt.

@MAFIA36790 it took care of itself thanks to downvotes and 20k deletions

If I have a qubit state $\lvert A\rangle=\frac{1}{\sqrt{2}}(\lvert u\rangle - \lvert d\rangle)$ and I then trying to determine the following expectation value
$$\langle A\lvert \sigma_x\rvert A\rangle$$
I got $<\sigma_x>=\frac{1}{2}(\langle u\rvert - \langle d\rvert)(\lvert d\rangle - \lvert u\rangle)=\frac{1}{2}(-\rangle u | u\rangle - \rangle d | d\rangle)=-1$

But $\lvert A\rangle$ is not really pointing in the $\pm x$ direction, thus how can an expectation value of -1 be obtained (since to get an expectation value of -1, the probability to get +1 must be zero)?

NB: two of the \langle are typos and should be rangle. Specificlly, -<u|u>-<d|d>

Ok nvm, $\lvert A\rangle$ is pointing in the -x direction, problem solved

1:04 PM

@ACuriousMind both

@ACuriousMind No, GP says it's a "strict logical consequence" and so do Lee, Hirsch, ...

@0celo7 Yes, of the definitions they give.

ACM is saying that the way of formulating the definition is something you do with certain applications in mind

Case in point: G&P want to call non-intersecting things transversal.

So they make a definition from which it obviously follows that this is the case.

(this is a strict logical consequence of their definition)

1:28 PM

@Danu I don't see this!

It does not obviously follow

They say the map is transversal to the submanifold if the transversal equation holds for all points in the preimage

But if the preimage is empty, what do?

It seems absurd that this implies transversality

But even in the critical points case (of which transversality is a generalization) it's absurd

How can we talk about the rank of the derivative when the derivative does not exist

@Danu @ACuriousMind I just read the wiki article on vacuous truths and I still don't get it

Anyone know the answer to this?

0

Q: Why don't globular clusters flatten with a galactic disc?

Globular clusters lie in the galactic halo, outside of the disc. However, galaxies are more or less a collection of material and objects — why is it, then, that most stars form a plane due to the angular momentum, but some patches of stars do not? In fact, globular clusters often contain some of...

star gravity galaxy galactic-dynamics globular-clusters

@0celo7 Read a book on basic set theory?

Yeah I knew you'd just be a dick

This is why I asked ACM

"Every point with property X also has property Y." If there is no point with property X, then this is always true.

It's really just that simple

Maybe a more "pure" formulation: Any element of the empty set has property X (where X is an arbitrary property), because there are none.

If you are serious that this is beyond you, then I can honestly not say much except to just take a step back, read a few pages of either basic set theory or logic in some book, or just stop working for now---it may be obvious in an hour or so.

1:52 PM

@Danu I've always thought vacuous truths were some informal stupidity, like saying "literally 120 degrees"

It's been beyond me for 18 years.

But the Wiki article is just confusing

> For example, a child might tell his or her parent "I ate every vegetable on my plate"

Impossible because you have to eat something to have eaten

> "You're my favorite nephew. In fact, you're my only nephew."

This, otoh, makes perfect sense.

And they seem like two completely different statements.

@0celo7 Wrong, because the logical statement the child made was:

Every vegetable that had the property of being on my plate, now has the property of being in my stomach.

Since the first set is empty, the statement is true.

It's always easy to reduce vacuous statements to properties of elements on the empty set.

@Danu So the claim is that every element of the empty set has every property?

@0celo7 Sure

@TrollTerminator the same number of good games the Raiders have had in the past 15 years

In the sense that, if you fix a property $P$, the following is true: $\forall x\in \varnothing, Px$

2:03 PM

Anyone want to weigh in on this discussion about the size of the electron? chat.stackexchange.com/rooms/41685/…

@GerbenVenken There's no size of an electron

Yes I know

I think you're pretty much correct there @GerbenVenken

That's what I've been trying to explain to the other person

At least in QM

2:04 PM

@Danu What's the proof?

Use the properties of implication

the "truth table"

When is $A\implies B$ false?

In some sense, the concept of vacuous truth is exactly this: $\neg A\implies (A\implies B)$.

Oh snap, what is the negation symbol?

\neg

Right

Troll Terminator

~

@Danu I don't know what you're asking for here

2:11 PM

@0celo7 The statement I wrote earlier is equivalent to $x\in \varnothing \implies Px$.

What is the only way this can fail?

@Danu ...no clue

@0celo7 So you need to understand the basic properties of logical symbols. This is what I meant when I said earlier that you have to backtrack to basic set theory/logic.

but $x\in \emptyset$ is not possible

that's what I don't understand

Is anyone here an astro-nerd?

@Danu I understand the symbols

2:13 PM

It's not possible, so the first statement is always false.

@0celo7 Then what I'm saying should be obvious.

$x\in\varnothing$ is never fulfilled

@Danu obviously not

@Danu exactly

@0celo7 Then you don't understand.

so the statement is itself incorrect

I'm serious man, this is just a basic fallacy in your logic.

how can something which does not exist have a property

2:15 PM

If I am in the US right now, then you are in Asia right now (random example). This is true.

35

Q: Why the galaxies form 2D planes (or spiral-like) instead of 3D balls (or spherical-like)?

Question: As we know, (1) the macroscopic spatial dimension of our universe is 3 dimension, and (2) gravity attracts massive objects together and the gravitational force is isotropic without directional preferences. Why do we have the spiral 2D plane-like Galaxy(galaxies), instead of spherical...

gravity astrophysics angular-momentum galaxies

@Danu How if the premise is false?

@JohnRennie I already know why

I see that your example is equivalent to the set-theoretic one

But I'm equally inclined to believe neither.

I'm asking why globular clusters don't flatten out with the disk

Troll Terminator

2:20 PM

what do you think "vacuous" means when it is describing "truth"?

@0celo7 read this

@SirCumference oops, yes, sorry, I read your question in a hurry and misread it

0

Q: Why don't globular clusters flatten with a galactic disc?

Globular clusters lie in the galactic halo, outside of the disc. However, galaxies are more or less a collection of material and objects — why is it, then, that most stars form a plane due to the angular momentum, but some patches of stars do not? In fact, globular clusters often contain some of...

star gravity galaxy galactic-dynamics globular-clusters

Yeah

@0celo7 You're just falling victim to one of the most elementary misconceptions in logic. I can't really say much more; consult a book.

I know I'm stupid

No need to tell me

@TrollTerminator Now I'm convinced the empty set is not a subset of every set.

Thanks.

Troll Terminator

np

2:29 PM

@Danu If $x\in\emptyset\implies P(x)$ for any property $P$, then we must also have $x\in\emptyset\implies\neg P(x)$, which is a contradiction.

@0celo7 No, it's no contradiction because the empty set is empty.

In fact, even $\forall x\in \varnothing, P\wedge \neg P$ (directly)

Ok, so why aren't disjoint submanifolds transverse and nontransverse at the same time?

Just define both properties and it'll roll right out

@JohnRennie So, uh, do you have any idea?

2:45 PM

@Danu Because nontransversal means "not transversal" not "the transversal equation does not hold"?

@0celo7 I'm not sure what the definition are. But could be.

Has any of you guys ever seen a theory of a 2-form field coupling through its field strength to a 1-form (instead of to a 2-form, and directly)?

2 hours later…

4:31 PM

@EmilioPisanty By the way, that DOI resolver of yours is not working for me.

Try e.g. dx.doi.org/10.1098%2Frsbm.2006.0011

@dmckee You know things about astronomy, right?

@Danu Do you know how to determine the tangent bundle of the circle without using the fact that it's parallelizable?

4:49 PM

@SirCumference Uh ... not at a professional level, but I've used a telescope and I took an astrophysics class in my undergrad days.

@0celo7 Eh, no I never really thought about it.

@Danu Tangent bundle stuff is no fun without the tools Lee develops...

@Danu seems me here rsbm.royalsocietypublishing.org/content/roybiogmem/52/137

@EmilioPisanty Yes, but if you replace the dx.doi thing with doai.io, you get an error :(

Oh, lemme try

4:54 PM

@Danu My current sketch:

user image

Yeah, it looks outta whack at the moment

Now proving that $g$ is actually a diff will be stupid.

@0celo7 Actually, I do know one way. Just write down an explicit isomorphism.

The main domain is also down

It's not very hard IIRC

4:55 PM

@Danu which is?

That's kinda what I'm doing right now

By "isomorphism" do you mean "diffeomorphism"?

Isomorphism of vector bundles

@Danu I'm not working in the framework of vector bundles so for me "isomorphism" has a different meaning right now

But I think I'm doing what you're suggesting.

ok

@0celo7 I imagine it should be doable to write down an explicit iso between the normal bundle of $S^1\subset \Bbb R^2$ and the tangent bundle.

and I know an explicit iso between the former and $S^1\times \Bbb R$

@Danu ooo, can you please show me

I want to say that it's just a 90 degree rotation in each point.

5:06 PM

well actually, you just need to "rotate" up the thingies

yeah

so that'd be a linear iso on each tangent space

and that's pretty much all you need

And the tangent space is just a rotation of the normal space

That's what I said, yes

@Danu No, you said that the normal bundle is a rotation of $S^1\times\Bbb R$

@0celo7 No, I didn't.

5:08 PM

there's two rotations here

The "I want to say [...]" was about the iso between tangent and normal bundles. The one from normal to trivial bundle is just $(x,t)\mapsto (x,tx)$.

Of course you did. You first rotate $T_xS^1$ into $N_xS^1$ and this should give the first isomorphism $TS^1\approx NS^1$.

This is a trivialization of the normal bundle of any sphere

@dmckee Cool

Then you rotate up each fiber of $NS^1$ into the $z$-direction and that gives you a fiber of $S^1\times\Bbb R$

5:09 PM

1

Q: Why don't globular clusters flatten with a galactic disc?

Globular clusters lie in the galactic halo, outside of the disc. However, galaxies are more or less a collection of material and objects — why is it, then, that most stars form a plane due to the angular momentum, but some patches of stars do not? In fact, globular clusters often contain some of...

star gravity galaxy galactic-dynamics globular-clusters

Can you answer that?

@0celo7 I don't think you need any rotation.

I just gave you a trivialization

@Danu I like to picture $S^1\times\Bbb R$ as a cylinder

Only tangent to normal is a rotation, in my mind.

@0celo7 OK, well I don't like that here because then you're embedding in 3-space, which isn't the natural setting if you're talking about the normal bundle of $S^1$ as a subset of $\Bbb R^2$.

@Danu In GP the tangent bundle is defined as embedded in $\Bbb R^4$

@Danu Can you explain what $(x,t)\mapsto (x,tx)$ does?

@SirCumference Not with any certainty. Flattening of a gravitational bound system like that requires lots of interaction between mass elements. That happens faster in a cloud of gas and dust than in a system of discrete and compact objects like stars and planets.

5:12 PM

@0celo7 It's an iso from the product to the normal bundle.

@Danu I know, but I don't see that

what is $tx$?

Maybe the early formation of starts in those entities simply put them on a longer scale for thermalization of velocities.

Is $x$ here a vector of $\Bbb R^2$ with unit length?

@0celo7 Multiplication by a scalar in the vector space $\Bbb R^n$

I'm not going to spell this out more for you---I think it's a nice exercise.

@dmckee Why not comment that? You'd help other people with the same question

5:13 PM

@Danu I have my own solution cooking

I'm trying to understand what you're doing

If $x$ is a 2-vector then $(x,tx)$ is an element of $\Bbb R^4$

so you're embedding, which you said you didn't want to do

If $x$ is not a vector, then I don't know what $tx$ is

0

A: Would a light or a heavy ball roll fastest down a slope?

Hi my name is Eugene Flanders and I am a professor at Colombia university. I think I might have what you're looking for if you need to explain roller coasters or balls on a track. Here's the equation why: f=y6H+x%L Assuming f=fastest H=heaviest L=Lightest Use this in Science homework's and pro...

user image

Ahh

So, as I said, I'm embedding $S^1$ in $\Bbb R^2$.

yes it makes sense

5:18 PM

@Mostafa That's funny.

now I'm trying to figure out if my approach is viable

I'm having no luck computing the inverse of my diff :/

I've deleted both of those posts, now, so for the benefit of users without 10K rep:

user image

LOL

...how is he a Columbia prof and a Yale student at the same time?

I'm stuuuupid

@Danu I would appreciate feedback on

user image

Sorry, I don't really feel like spending time proof-reading. Ask in the math chat, perhaps. They're way more qualified than me, anyways.

They're also discussing G&P haha

5:30 PM

I've done 22 exercises in GP for the first problem set

not even close to being done 0.0

@Danu what happens if I complain about vacuous truths on there

@0celo7 What do you want me to tell you?

@Danu that vacuous truths are a farce.

@dmckee I remember Wikipedia had a page to archive funny vandalism that get deleted :)

Here it is:
https://en.wikipedia.org/wiki/Wikipedia:-%29

It's shortcut is WP:-)

5:57 PM

FTFY

@Numrok By minimal substitution, do you mean the "I want my derivative to transform in the right way so I have to modify it" line of reasoning?

6:38 PM

@Danu Life as usual, then ;P

@0celo7 What? What do you mean "no"? I said it depends on the situation whether it is desirable to explicitly exclude the vacuous case - if you don't, it is of course included. I'm not sure what the issue is.

@ACuriousMind mother duck, could you please check my circle tangent bundle proof

I think it's pretty legit

@ACuriousMind This time I'm pretty sure I will jump ship :P

@ACuriousMind See my discussion with the Dutchman

(don't bother @ACM)

What's this "of course" business

6:42 PM

The conclusion was that 0celo7 is not willing to accept that $x\in \varnothing\implies Px$ for any property $P$

I must be retarded because every fiber of my being rejects this principle

in the end, I told him to revisit a basic logic/set theory book

I think the room is telling you somethinge @0celo7

Who the fuck keeps starring his insults

They're actually not insults. I genuinely tried to help you.

At some point there is nothing more I can do---books will explain it better than me.

@0celo7 On each tangent space $T_p S^1$ in $S^1$, choose the unit tangent vector which positively orients $T_p S^1$. This is a smooth nowhere zero vector field on $S^1$, i.e., a nonzero section of $TS^1$.

6:45 PM

@0celo7 It appears to be correct, but why do you not just show existence of a nowhere zero vector field?

Having a nonzero section means trivial.

@0celo7 I'm afraid I have to agree with him - that $x\in\emptyset\implies P(x)$ is true for all properties $P$ follows just from the way implication is defined. If you don't see why this is true, then you don't know what $\implies$ actually denotes.

@BalarkaSen Sigh

I'm looking for an elementary proof.

I do not even have the concept of "fiber bundle" or "section" at my disposal.

@ACuriousMind What does $\implies$ denote?

Maybe try to prove that a bundle which admits a nonzero section is trivial. It's worth knowing.

Am I the only one who, when doing an exercise, will try to use only what I'm given in the book?

6:52 PM

@0celo7 $A\implies B$ is just a shortcut for $\neg A\vee B$.

Of course $TM$ is trivial for $M$ parallelizable, it's in Lee somewhere.

But I'm not doing problems in a book where I have that theorem available

I am not referencing a fact: I am giving a proof.

@ACuriousMind That's not the right way to explain it :P

@BalarkaSen If I wanted to use your fact I would have. But that fact is not elementary and is not in GP.

If you don't know bundles and sections and stuff, try proving that a manifold which admits a nonzero tangent vector field has trivial tangent bundle.

6:54 PM

Forget it.

I disagree that's not elementary. Whether you do not like it is a different matter entirely, and is acceptable.

It's just a fact worth knowing.

@Danu There is no "right" way to explain it. But I think $\neg A\vee B$ exhibits directly why the implication is true when $A$ is false.

I know the fucking fact.

@ACuriousMind Sure, you can just state this. But then the burden is upon you to motivate why this is "implication".

(you've just shifted the problem)

2 hours ago, by 0celo7

@Danu Do you know how to determine the tangent bundle of the circle without using the fact that it's parallelizable?

6:56 PM

@Danu No, that's the classical definition of the implication.

@BalarkaSen Does this suggestion of mine work? I know an explicit trivialization of the normal bundle, and I just write down an iso between normal and tangent bundle (each fiber is just rotated by 90 degrees so that's not too hard, right?).

I did not use the fact "parallelizable". I gave a proof.

@BalarkaSen What are you going on about?

@ACuriousMind ...which is wholly unmotivated, right now.

I'm sure you see what I'm saying.

@0celo7 I mean I gave you a prove that there exists a nonzero tangent field on S^1.

6:58 PM

@BalarkaSen That was never in question?

@Danu That is completely fine. Note that the proofs are not too different, as we both used orientation to construct a nonzero tangent field/outward normal field.

@Danu Not really. Logic is not about "motivation".

@BalarkaSen Hmm.

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