@manshu Thanks. I had a great party

How to understand all content of lecture?

@manshu Ohhhh...I was imagining something far more complicated after that description :D Then, yes, you can find the maximum height here just by energy conservation

@ACuriousMind Then do I have to consider the rotational energy in it?

Ok, I maybe better quit. I'm some afraid.

@user36790 : they are waves. You make them out of waves moving linearly, you can diffract them, in atomic orbitals "electrons exist as standing waves", and when you annihilate them you've got linear waves again. There is no magic. But there are spherical harmonics. Harmonics are to do with waves.

@hubot That you have to find out for yourself. I learn best when I take notes during the lecture, others don't retain anything if they have to focus on writing while listening.

@HariPrasad dont stay up all night albert...go to sleep

@manshu I'd say that at "maximum height", the block as well as the track are momentarily at rest, so there's no rotational or kinetic energy to consider

@HariPrasad : particles are waves.

@user36790 Yah.

@user36790 "Give me some sunshine! Give me some ray! Give me another chance so that i grow up once again!"

@JohnDuffield You are really great. We all love your answers.

sari umra hum...mar mar k jee liye

What about the other way around: Should we perhaps close the older physics.stackexchange.com/q/155093 as a duplicate of the newer physics.stackexchange.com/q/242499 ?

@HariPrasad I thought we were singing a song...

@manshu yes we are.

@Qmechanic Given the latter question has received far more attention and answers, that seems like the right thing to do, yes

@HariPrasad Then go to sleep...you are drunk albert

@manshu No "I have a symposium tomorrow on string theory - Albert Einstein"

@JohnDuffield Lol!

@user36790 : the electron is not some point-particle thing that exists at a certain coordinate. Its field is what it is. It's similar for a photon, only the photon is a field-variation propagating linearly at c rather than a standing-field thing.

@0celo7 His answers are worth a million lectures.

@HariPrasad Feynmann or maybe Issac can handle it...

@user36790 You can characterize spherical harmonics as "standing waves" on a sphere. To say that "atomic orbitals are standing waves" is a vacuous truth - by definition, the orbitals are solutions of the time-independent Schrödinger equation, and if you insist on calling wavefunctions "waves", the time-independent waves are the standing ones. Why?

@user36790 Do not bother me with things JohnDuffield says. I have him on ignore for a reason, and any further attempt to drag me into a conversation about or with him will result in me also blocking you.

@user36790 Heed ACM's warning.

@user36790 : for an analogy, think of the photon as something like a seismic wave in space. A seismic wave in the ground might propagate from A to B across a gedanken plain. But it isn't just the houses on top of the AB line that shake. That seismic wave takes "many paths".

No matter how much you dig you will never find any billiard-ball/pointlike particle moving from A to B. Because there isn't anything pointlike moving from A to B. And if you could make that seismic wave go round a Dirac's belt path, there still wouldn't be.

@0celo7 Do you have me on ignore?

@HariPrasad Not yet.

I'm listening.

@0celo7 I like your cat

@user36790 stop typing.

Erm, ok?

@ACuriousMind How has your day been

@user36790 lol

Ok.

@0celo7 maybe you forgot about ACuriousMinds's wrath on these kind of questions

@user36790 Listen to this before you die: youtube.com/watch?v=y6oXW_YiV6g

@0celo7 I lost a large Dwarf Fortress to a goblin invasion :'(

@HariPrasad He just ate Centre Fresh

@user36790 : no, the hard scientific evidence won. As usual. Like I said, we make electrons out of linear waves in pair production, then we can diffract them and refract them, in atomic orbitals electrons exist as standing waves, these waves are described by spherical harmonics, and when we annihilate them we've got linear waves again. There is no magic.

@ACuriousMind What game? PoE?

@user36790 hmm..so?

@0celo7 No, Dwarf Fortress

@ACuriousMind Walls and traps

@user36790 Do I need to die with you now?

@ACuriousMind oh :P

I'm fairly sure it's a standard term, I'm just too lazy to look it up.

It should be in Gorgoulhon's bible.

"surface of simultaneity" sounds like fancy talk for an achronal surface

@Slereah I had a wonderful moat but overlooked one square inside it through which they could bypass it.

@Slereah Pretty much.

JD always wins.

@user36790 Seriously, what about "So unless we've talked about the specific issue before, or you think I would find the question really interesting, don't ping me with your question. Just ask it." was unclear to you?

@Slereah A compact, even-dimensional, orientable manifold with positive sectional curvature is simply connected.

@user36790 you just pinged him

Well works in 2D at least?

@JohnDuffield ::falls off of chair sofa::

I'll trust you on more dimensions

@Slereah does it

That was in response to @0celo7

what are the compact 2D manifolds?

$S^2$?

@0celo7 : the orientable ones are $S^2$ and various flavors of toruses

The only one with positive curvature is the sphere

toruses?

You mean tori?

Which is simply connected

Torises

@user36790 Aaaaand there's your spot on my ignore list. I'm fed up with people flat out ignoring what I'm saying.

@ACuriousMind Like a boss

lol

People are really out to ignore ACM lately.

What's up with that?

@0celo7 : shrug. It's the truth. If there was something I'd said that was wrong there would be a barrage of comments castigating my error. But there isn't. Because I'm not wrong. ACM knows this.

@0celo7 who is like galileo?

"ACM knows it"; the new version of "I'll tell mommy!"

have you googled it @user36790

@user36790 related to spin too

@user36790 what?

a bit

If people keep annoying @ACuriousMind he'll just stop coming around. Please stop annoying him.

@0celo7 I can't decide how much of that is nice and how much selfish ;)

@ACuriousMind My love for you is not selfish.

d'awwwwww

romantic love?

@manshu Maybe if he gets a haircut.

why only one hair?

what is so special about that one?

lol

what?

...please learn English.

didn't you just see lol word, people

@user36790 : be wary of Purcell. AFAIK this is where the myth began that length-contraction is the cause of magnetism. It isn't. See this question. NB: the bounty was an honour thing, awarded to the least-worst answer.

omg...this room is more fun than counter strike

@0celo7 The day I cut my hair is the day I grow up. That day is far away.

@ACuriousMind Your hair is twice as long as my gf's...that's pretty crazy.

Do you have a beard, too?

@0celo7 Yes

Are you short?

@0celo7 Stop it...he is not Tyrion Lannister

@0celo7 1.77m, decide for yourself

I hate links not working in this chat...

vignette2.wikia.nocookie.net/lotr/images/5/57/…

they work

ACM:

There we go.

Sadly, my beard does not grow to dwarven proportions

@ACuriousMind ahem...your comment about silly questions....what happened now huh?

@ACuriousMind Would you mind giving this a quick read? Dr. Freire said I can send his proof to Ellis (via Sam), but does it make sense?

@user36790 : good. The cause of magnetism is the "screw" nature of the electromagnetic field. See Minkowski's Space and Time:

"In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy to the force-screw in mechanics; the analogy is, however, imperfect".

@manshu what? I didn't say anything about silly questions. I said people should stop asking me random questions, i.e. those that do not relate to me specifically.

Yes, managed to use the chocolate banana in another answer

send the link

understood nothing

@manshu why not?

\o

@user36790 good morning

@HariPrasad Well...that's bcz i have no knowledge about computer languages

@0celo7 \0/

@0celo7 The beginning is a bit unclear. The first sentence is what you want to prove, right? And what is the "this" that $\Sigma$ is supposed to satisfy in the second sentence?

@ACuriousMind Yes, the first sentence is the claim, "this" is that $\Sigma$ separates $\mathcal{M}$. I'll fix that.

@user36790 ahh I live somewhere near to a black hole. So for me time runs slower than yours.

@ACuriousMind Now I want to know if that "if it is properly embedded" can be strengthened to "iff"

But I really need an algebraic topologist for that.

@user36790 we can't catch the signals...coz radiation can't escape the black hole...duh

@0celo7 Yeah, I recall that, but I don't know

@ACuriousMind You recall what?

Will Dr. Elvish Knots know anything about that?

I have to stop by his office anyway

@user36790 Black hole updates: Its not that black, but small, its cold and i can't resist its gravity!

@0celo7 I recall you wanting to know that. I think.

@ACuriousMind Oh, oh. Yeah.

Sadly the proof of the generalized theorem like that is way beyond my reach. My prof admitted to having little to no idea how it's done.

It's in a book that contains "functors" in almost every chapter title...

@ACuriousMind See, the essential ingredient is that $\Sigma$ is the boundary of $D_-$, so to get into $D_-$ you have to cross $\Sigma$ again.

In Ellis' proof I see no reason for having to cross $\Sigma$ more than once.

But that's nonsense.

@ACuriousMind Steiner is hunting spiders

Well, good for him!

@ACuriousMind Does my objection to the Ellis proof make sense or not?

Our cat loved hunting insects (spiders were boring for some reason) and demolished everything in his way trying to get to them

@0celo7 I don't actually have the Ellis proof memorized, so I can't tell

@ACuriousMind They argue that a closed curve must cross the hypersurface an odd number of times, but the curve is homotopic to a point, which crosses zero times. Thus such a closed curve cannot exist.

And your objection is?

@ACuriousMind Because the curve could cross once, I guess.

And there's no contradiction there.

I'm not convinced about the homotopy thing any more.

If that were true, then no hypersurfaces could exist in any $\mathbb{R}^n$, because I can find a boundaryless hypersurface and a closed curve that crosses it once.

Well...there's still the thing that they might mean boundaryless and closed (as a topological subset), right?

@ACuriousMind Perhaps.

@ACuriousMind What exactly does "closed" for a manifold mean?

For a manifold, that usually means "compact and boundaryless"

Define "boundaryless" please.

In the manifold sense

That is, strictly speaking, since manifolds with boundary are not manifolds, a closed manifold is just a compact manifold

The terminology in this field is horrible

But in the manifold sense it's the part that gets mapped to the boundary of half-space under the charts, right?

Yep

Ok, so I know all of that.

Now what do they actually mean...

Maybe all boundaryless (in the manifold sense) hypersurfaces are properly embedded?

from an email

Does that have a boundary?

it's an embedding of the real line $(0,\infty)$, right?

which has a boundary, namely $\{0\}$

@0celo7 OP's account vanished, I guess we'll never know what they meant...

@dmckee: Is the automatic answer ban currently enabled on Physics?

@ACuriousMind Yes.

Huh, then I guess it must be more lenient than I thought. Thanks, anyway.

@ACuriousMind Huh?

@0celo7 Huh?

@ACuriousMind It apparently takes relatively few positive scored post in a category (question of answer) to protect a user from the consequences of many negatively scored ones.

Not that I know who you're thinking about, but I've marveled at that a time or two myself.

@ACuriousMind i don't know what lost your referencing

*post

Then get on a real computer and find out :P

@dmckee Yeah, must be. It's just that I see the posts from this user in the VLQ queue almost every day, and they never improve.

@ACuriousMind dilbert.com/strip/1995-06-24

Heh

Hello guys

What is spectral flow symmetry? I can't find it on wikipedia

Here's an interesting milestone: if you want to be on Top Reviewers All Time for Close Votes, since recently, you need the full Steward badge on that queue

physics.stackexchange.com/review/close/stats

i.e. the lowest such user is at 1000 reviews

I wonder whether it's usual across the SE network for the close queue to be the queue with the most all time reviews?

Hi, in the standard configuration of the Lorentz transforms why do y' and z' not depend on any other variables and why do x' and t' only depend on x and t? My question is more generally is ' what is the justification of whittling down from potentially 16 degrees of freedom to 4'?

@GridleyQuayle What do you mean "standard configuration"? Are you talking about a boost in the x-direction?

Yeah

Well, a boost in the x-direction doesn't affect the y- and z-direction. It's not the most general Lorentz transform.

What would be the names of more general transformations?

Uh, "Lorentz transformation"?

can anyone tell me what spectral flow symmetry is?

please?

do you know what spectral flow is

@ACuriousMind Ah.

oh right, I was confused when the topic was first introduced. The lecturer introduced the x boost as a more general transformation but I must have heard it as 'the most' general for some reason. Thanks

@GridleyQuayle You can write every Lorentz transformation as the product of a spatial rotation, a time reflection (or not), and a Lorentz boost. Since spatial rotations and the possible reflection are not "relativistic effects", and you can choose your coordinate system freely, you can always declare the boost is "in the x-direction", so to see what relativity "does", it suffices for many things to think just about such boosts.

Proof?

@ACuriousMind What should I drink tonight?

@0celo7 Section 4.1 here.

That...was a joke, but ok. Did you write those?

@0celo7 Notes of my group theory course that I never finished typing up.

Too complicated for me

I've eaten a lot of random things today. Bagel, cheese, some almond bar thing, pizza, hot dog, peanuts, crab.

Oh and a liter of soda (diet ofc).

@0celo7 Brightly colored cocktails with fruit in them.

That sounds awk.

Deleting just to be sure.

Vodka, then.

Not an alcoholic.

Do you just randomly drink straight vodka?

Expensive vodka, then.

@0celo7 Nope

I think we have Smirnoff and Retel One. Nothing expensive.

I'm currently enjoying a Schwarzbier.

Verdammt, ich wollte meine Mama fragen was Feldsalat auf Englisch heißt.

Moment.

@ACuriousMind Gibt's hier nicht.

Yeah, I suspected as much