The h Bar

Tfw took 30 minutes to prove a linear algebra theorem

breh

1:07 AM

@ACuriousMind What's the intuition for a degenerate critical point?

1:31 AM

@NeuroFuzzy Remember how I was telling you how to define the tangent space as the vector space of derivations of germs?

Turns out that only works in the smooth category, in the $C^r$ category you get an infinite-dimensional vector space

I will work on the proof when I have time...seems immensely interesting!

user218912

2:13 AM

@0celo7 I need to learn more first. :(

@NeuroFuzzy Ah, the reason for using germs is as follows: else it is very difficult to prove that $T_pM=T_pU$ for any open $U$ containing $p$

With curves it is pretty easy

@3750 I'm in a good mood and will help you with math tonight

user218912

I don't have any questions right now though. :'(

user218912

basically I'm gonna read through the orbifolds book and look up everything i don't know.

user218912

it seems to be an efficient way of learning.

user218912

because I find it boring to read through all the prereqs first.

user218912

2:28 AM

@0celo7 I will have loads of questions in the next few weeks.

user218912

@0celo7 can I save this as a credit to use later?

no

@ACuriousMind I might have an MO-level question on my hands.

I want to know why it's not possible to define the tangent functor on the $C^r$ category algebraically.

well, not the tangent functor proper

does MO allow for multi-part questions?

@NeuroFuzzy Good lord the construction of the tangent space for a non-smooth manifold is horrible.

2:52 AM

@0celo7 you there?

Yes

@0celo7 Using "reasons" and not just blind definitions, tisk tisk ;)

@3750 Didn't you say you were beginning a math bachelors after the summer? Are you sure you're ready for (even looking at) a book on orbifolds?

@bolbteppa Ah, you can be a voice of reason!

3:10 AM

@0celo7 Buy the new york times on friday

keep only the front page for me

I need it for a project

you can take the price off the 20 bucks you owe me

user218912

@bolbteppa lol I can do it.

how do I get a new york times?

ask bob

@3750 Are you sure you know what you're doing? :p

@bolbteppa can you prove $|ab|=|a||b|$?

No

user218912

3:15 AM

@bolbteppa yeah

user218912

@bolbteppa really?

@0celo7 Bob hasn't answered my last text about weather stations, and it's been weeks

@0celo7 Can't you find it there? It shouldn't be too hard

I don't know where one gets a newspaper

I'm a working man

I don't have time to run around looking for shit for you

user218912

but you have time to be on this chat???

I can chat at work

Measurements take time

user218912

3:18 AM

you're really lucky to have a real job in a lab in first year

user218912

how did they hire you?

I asked

Let $a = \frac{\partial }{\partial x^j}$ and $b = \vec{e}_i$ then in the space of isometries we have $|ab| = | \frac{\partial }{\partial x^j} \vec{e}_i| = dx^j(\nabla_{\partial}) dx^i (\vec{e}) = | \frac{\partial }{\partial x^j} | | \vec{e}_i | = |a| |b|$
voila

qed

huh?

You need to set this up on non-Hausdorff non-second-countable non-manifolds for this to work

3:24 AM

stop trolling

When you go on definition after definition out of thin air, is what I wrote really that hard to believe?

@bolbteppa I'm a non-Hausdorff manifold

@0celo7 Plox halp

I need that newspaper cover

@bolbteppa Yes.

You clearly do not understand modern mathematics.

Mixing "voila" with "qed" is trolling :P

3:39 AM

@ACuriousMind Suppose I have a finite-dim vector space $V$ and a map $V\to W$ where I do not know the dimension of $W$.

Suppose this map is invertible.

Does this imply $W$ is isomorphic to $V$ and thus finite-dimensional?

Well I guess if it has an inverse it's a bijection :P

user116211

Check this:

Q: If someone catches a Pokemon that is on my property, is that theft?

This proof is insane.

user116211

9

Does something that is virtually located on your property like in Pokemon GO incur any kind of property right for the property owner?

property

@ACuriousMind I need serious help collecting my thoughts on this issue.

@ACuriousMind Help :(

user116211

4:30 AM

41

A: Rate limit suggested edits

As of a few seconds ago, this limit is now active network-wide: at-most 20 pending suggested edits per editor on beta sites at-most 5 pending suggested edits per editor on graduated sites These limits are based on my analysis of the speed at which edits are generally approved and the number o...

SHOG has spoken :P

user116211

I love Shog9 ;))

5:56 AM

@0celo7 hyperphysics.phy-astr.gsu.edu/hbase/astro/expuni.html

@JohnRennie I read that England is going to impliment south asian style maths instruction. Is that true?

I don't know the details, but I believe funding is going to be supplied for the technique to be tried at a few schools. Then the results will be compared to other schools.

Interesting.

Whether it will work I don't know. My impression is that the UK children lack the motivation of South Asian children. I think if the kids are prepared to work hard any technique will be a success, and if they're not prepared to work hard no technique will be a success.

But perhaps I am just getting old and grumpy :-)

If the results are good, it will affect how physics is taught too.

But yes, there is no substitute for hard work :-)

6:09 AM

In the UK Physics is an optional course so the children learning it are the ones who have chosen to learn it. By contrast maths is taught to everyone. I suspect it's the kids who don't want to learn maths that are the main problem.

Indeed.

What was your overall impression of the recent AMA? @JohnRennie

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ most of the discussion was too advanced for me to follow, so I can't really comment. It seemed lively, but I get the impression it was a small number of fairly expert people chatting.

Are you interested in becoming active on chem.SE?

6:25 AM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ No. I occasionally visit the Chem SE for a quick look around, but I find physics much more interesting.

user116211

@JohnRennie: Have you played Pokemon Go yet?

@MAFIA36790 The problem with PG is that it's based on the idea that it's interesting/fun to collect Pokemons.

user116211

@JohnRennie There is no trade system which I don't like it.

I do play games, but they tend to be the more mindless first person shooter type, and I only play them as a form of stress relief. Games where to the point is to achieve an arbitrary target aren't that appealing to me.

user116211

But it has got a massive response in the first week.

6:30 AM

I find myself wondering why anyone wants to collect Pokemon.

user116211

Many times servers crashed.

user116211

They need to take care of that.

I think once you start wondering what's the point? that is the death knell for a game.

Fads come and go for that^ reason :-)

user116211

It's a major step in augmented reality.

user116211

6:31 AM

@JohnRennie That maybe a point.

user116211

But with this, you can explore new realms.

user116211

But since I've Windows phone, I can't play.

@MAFIA36790 I can do that by reading books (which I do a lot) and the experience is richer than any computer game.

user116211

And frankly, I've no desire to play.

user116211

It's not that grapes are sour.

user116211

6:33 AM

I just don't have any interest.

user116211

My most favourite game is Assassins Creed.

user116211

Meanwhile @JohnRennie , I got a very old game Zork, a text-based game that ran on Commodore.

@MAFIA36790 Zork?

user116211

@JohnRennie yep.

user116211

Zork II

user116211

6:35 AM

I needed an emulator to run this on my laptop.

There was a group of us at Cambridge who wrote that type of adventure games. Mine has mercifully disappeared into the mists of time.

user116211

@JohnRennie Do you have some?

See:

ifwiki.org/index.php/Acheton

The guys who wrote Acheton wrote a game engine, and lots of us wrote games to run on it.

user116211

> Acheton is one of about fifteen classic adventure games written in the 1980s on the Phoenix IBM mainframe computer at Cambridge University.

user116211

I really cherish this period ;)

6:38 AM

The games originally ran on an IBM 370, but they were later ported to the BBC micro. Acheton is now available for Windows.

What programming language did you use?

The game engine was written in OS 360 assembler. That was a mighty task, and was done by a group of maths PhDs in their time off from doing scary maths :-)

The games themselves were written in a higher level language that was interpreted by the game engine.

user116211

@JohnRennie oh man ;D

I've written OS 360 assembler. It's not that complicated. Mainly it's just a pain.

I've also written in Vax assembler, which is much more fun because the VMS libraries were better. I once wrote a basic CAD program in Vax assembler.

was your official first programming language BASIC? @JohnRennie

6:44 AM

Yes, on a Commodore Pet. That was in 1978 and the weren't any other languages available at the time.

I have to drop out for a few minutes and do some real (i.e. paid :-) work ...

How long will Pokemon GO's popularity last? Will it experience a steep drop-off in usage similar to some other mobile games?

Later

user116211

@JohnRennie see ya.

user116211

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Related Quora post:

user116211

in Pallet Town, 23 mins ago, by MAFIA36790

Thanks for sharing @MAFIA36790

On an unrelated note, I like this quotation I just found:

In the midst of life, it happens that death comes

and takes a measure of man. That visit

is forgotten, and life goes on. But in silence

the suit is being sewn.

Tomas Tranströmer, Black Postcards

DanielSank

6:54 AM

@0celo7 what?

user116211

Hey @DanielSank

user116211

I'm seeing you very few times here lately.

He's getting married soon, I think?

user116211

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ yeh, he told so; in July.

\o @Jim

DanielSank

7:38 AM

@MAFIA36790 I've been busy lately.

Slereah

8:36 AM

Colombeau and Gsponer write the Fock space as $$\bigoplus L^2_S(\Bbb R^{3n}, \Bbb C)$$

Is that correct for QFT?

I thought it was over $\mathcal S(\Bbb R^{3n})$ in QFT

Well the $L^2$ of that

8:52 AM

Could someone answer a brief question?

@NoahP Fire away ...

Considering the equation for H^2

Where only $a$, $\dot{a}$ and $p$ are time dependent

With everything else constant

If i take the time derivative of H^2, the last term, $\frac{A}{3}$ should cancel?

@JohnRennie

Yes, you'll end up with $\frac{d}{dt}(H^2} = \frac{8\pi G}{3}\frac{d\rho}{dt}$

Okay

Secret

@Slereah @DanielSank
Suppose I have a hamitonian $H \sim \sigma_n$ and $\sigma_n=\sigma \cdot \vec{n}$ where \sigma is the Pauli vector
Then in index notation
\begin{align}
H \sim \sum_{ijkl}n_i\sigma_{ij}\delta_{kl}
\end{align}

Now I have some density matrix of a two qubit system
\begin{align}
\rho=\sum_{ijkl}c_{ijkl} \lvert ij \rangle \langle kl\rvert
\end{align}

Using $-i[H,\rho]$
\begin{align}
& -i[H,\rho]=-i\sum_{ijkl}n_i\sigma_ij\delta_{kl}c_{ijkl}\lvert ij \rangle \langle kl\rvert-\delta_{ij}n_k\sigma_kl c_{ijkl}\lvert ij \rangle \langle kl\rvert

9:00 AM

You can then rearrange that and substitute in $\dot{p}$

to give the second equation in that graphic

Why is the cosmological constant term still there!?

A has the units of [time]^-2

No, because $\frac{d}{dt}(\frac{\dot{a}}{a})$ turns into quite a complicated expression. Remember that both $a$ and $\dot{a}$ are functions of time.

So you are differentiating the square of a fraction.

I know, ive differentiated that and followed several derivations

Is A a function of time?

Secret

NB I timed out, thus unable to fix the non subscripted j, it should be subscripted

Where A is $\Lambda$ technically

The cosmological constant is, as the name suggests, a constant so it isn't dependent on time.

9:04 AM

How come it pops back into the equation then?

It should be removed when taking the time derivative surely?

I would have to do the differentiation to see what happens. Note that the two equations are independent in the sense that one can't simply be derived from the other. It that were the case we wouldn't need two equations.

You need the equation for $\dot{p}$

If i were to make a question out of it, would you oblige?

I get $\frac{d}{dt}H^2 = 2\frac{\dot{a}}{a}\frac{a\ddot{a} - \dot{a}^2}{a^2}$. Where would you go forward from this?

@JohnRennie I'm doing a question on it as we speak

Oh hang on

I think i've got it

@JohnRennie just cracked it

thanks for the help though

9:25 AM

You're welcome - that was easy :-)

I was substituting $H^2$ back in but forgetting to include $A/3$

As you have to put $H^2$ back in to tidy up the derivation

Ugh stupid mistake

A: Accelerating expansion of the universe: so?

@NoahP possibly relevant:

3

The Hubble constant is defined by: $$ H = \frac{\dot{a}}{a} \tag{1} $$ where $a(t)$ is the scale factor as you describe. Acceleration means $\dot{a}$ is increasing, but then of course $a$ is also increasing so how $H$ behaves with time depends on which of the two increases faster. We can calcul...

Yep, the acceleration equation in your answer is what i was looking to get

9:49 AM

Is it okay to drop here a random physics question or is this chat just for random communication? :)

@Merlin1896 this is a good place to ask a question if you're not sure it would be suitable for the main site.

However things are quiet right now. Around 17:00 UTC is when it starts getting lively and there will be more people around to answer then.

I would answer but I have to dash off and do some real work.

Question is: How do I couple an electric field to a tight binding hamiltonian? Usually a magnetic field is introduced via the peierls substitution. What if I not only want the vector potential $A$ in my calculation but also the scalar potential (lets call it $\phi$). Is it just coupled to the particle density $n_i$ ?

@JohnRennie Thanks for the info :)

@Merlin1896 pretty much anything goes here (except insults, rudeness, etc.) Physics takes priority.

speed-of-light special-relativity spacetime waves time time-dilation time-travel

@DavidZ I guess there's not much a mod can do, but this user is worth a look

Pentcho Valev

17 1

also cc @JohnRennie, you'll enjoy this. (in a way.)

user116211

10:05 AM

Is there any way to stop a user posting things after getting continuous downvotes to almost every post?

2

Is there a good answer on SE which debunks all the theories and arguments of the Einstein critics? Maybe one could always link to that so that other users, especially the asker, can easily understand why the answer was downvoted and why the arguments where wrong.

user116211

@Merlin1896 You mean canonical post?

@MAFIA36790 No. You can have users with half of their posts heavily downvoted

i.e. post score distributions almost symmetrical about zero

@MAFIA36790 If that is the right term for what I mean, yes :)

which of course gives positive rep, +10 per upvote & -2 per downvote

so reaching into the thousands of rep

have a gander

data.stackexchange.com/physics/query/396403/…

user116211

10:12 AM

It's good to say even JD is somewhat right than this crackpot ;P

user116211

Oh, Jesus; there are SOME upvotes in his posts T__T

user116211

They must be crackpottiers.

user116211

@emilio, you are getting constant downvotes with severe magnitude in almost every posts. There should be some way to avert them from posting further.

user116211

Like someone being prevented from posting further question after posting continuous HW posts that get closed.

Danu

@EmilioPisanty ...or worse ;D

10:22 AM

Anyone here know anything about the cosmology deceleration paremeter?

Q: What can I do when getting "We are no longer accepting questions/answers from this account"?

10:35 AM

There are automatic bans that kick in if someone contributes enough low-quality content

209

Do not repost the question you were about to ask until you have READ EVERYTHING WE ARE ABOUT TO TELL YOU. While trying to ask a question, one could get: We are no longer accepting questions from this account. See the Help Center to learn more. Likewise, for answers: We are no longe...

support faq asking-questions post-ban

user116211

@DavidZ That is what I'm talking about.

@MAFIA36790 There are automatic bans, it's just hard to know exactly when they kick in.

That's intentional and it should be kept that way, because otherwise it would be much easier to game.

No matter what system you make up; there are always going to be "gamers."

10:57 AM

@DavidZ Are question / answer bans invisible to mods too?

@EmilioPisanty why do you ask?

Is a question/answer ban the same as a suspension?

No, a suspension is a manual, temporary penalty during which a user cannot ask questions, nor post answers. All other privileges, including commenting and voting, are also revoked by temporarily setting the reputation to 1. Such suspension is publicly visible to other users.

A post ban is enforced automatically, and only prevents posting questions or answers. It is invisible to others.

That last bit

We can find out if a user is Q/A banned, but we have no control over the ban (because it's automatic)

@DavidZ Sure.

But it might be worthwhile changing 'It is invisible to others' to e.g. 'It is only visible to moderators'.

Let me bring that up privately to the team, because I'm honestly not sure if the fact that it's visible to mods is supposed to be public knowledge.

I can't imagine why it would have to be hidden, but better safe than sorry I guess

I'll keep you posted

11:00 AM

@DavidZ cool.

hah

It's locked to edits anyway

Still might be worthwhile seeing if that language can be sharpened a bit without making it less clear to new users.

Why have I never heard of a chat suspension from just one room?

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ doesn't exist AFAIK. Chat suspensions apply to the whole system.

Should I bring it up as a question on meta?

@EmilioPisanty FWIW, it's probably best to assume that any statement about something being invisible to others may come with an implicit "except moderators". When it becomes necessary for moderation, I think we can get access to almost any information SE has about a user. Though of course we cannot see exactly which posts they voted on.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ you could, but I'd be surprised if it didn't exist already. Have a look on Meta Stack Exchange.

Certainly you're not the first to think of single-room chat bans.

A: Should chat room owners be given power to ban rude, aggressive users from chat-room?

11:16 AM

13

There is a "Flag for moderator attention" option separate from the spam/offensive flags. You can use that one to get a moderator in there and deal with the situation. Moderators can see deleted messages, so the deletion won't stop them from investigating the situation. There is no tool that bans...

I seem to recall mods/owners being able to "boot" someone from a room?

Yeah, there's a kick-mute function which removes the person from the room and stops them from chatting in that room for... 15 minutes I think?

I believe it's room-specific

Either 15 minutes or 1 minute, I forget which :-P

(but I was able to get right back in)

That's meant for cases where someone is posting rapid-fire inappropriate messages and they won't respond to people telling them to stop

We don't really use it here. It's not a problem we tend to have.

That's close enough to what I meant.

Thanks for the info :-)

no problem

Jim

12:02 PM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ yes?

Just welcoming you back @Jim

:-)

12:19 PM

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ the kickmute times escalate. If I kickmute you now you're out for 1 minute. Íf after you return I do it again you're out for five minutes and a third time it's half an hour.

DanielSank

@Secret What's the question?

Good to know :-) @JohnRennie

@NoahP I got the impression you had this all worked out ...

May I again ask my question from earlier today? How do I couple an electric field to a tight binding hamiltonian? Usually a magnetic field is introduced via the peierls substitution. What if I not only want the vector potential $A$ in my calculation but also the scalar potential (lets call it $\phi$). Is it just coupled to the particle density $n_i$ ?

12:36 PM

@EmilioPisanty his reply to the speed of light question isn't actually wrong, but it's a comment not a reply, hence my vote to delete.

skullpatrol

Wow! So much negative rep?

From time to time we get people with, shall we say, insular views on modern physics. Most of them eventually get bored and go away. I have to say I think we should vote to delete more often than we do. Leaving rubbish answers in place reflects badly on the whole site.

skullpatrol

Why isn't there a least bound on the amount of negative rep?

@Merlin1896 this is a futile thing to attempt. The Einstein critics (that's a nice polite term for them :-) are not motivated by physics. They are generally motivated by a conviction that the world is conspiring against them. In the face of this any form of reasoned argument is pointless.

user116211

@Skull patrol

user116211

12:43 PM

Who got pinged ;P

user116211

Ah!

user116211

patrol and petrol

skullpatrol

Me :-)

user116211

@JohnRennie Reminds me of a famous guy ;P

@MAFIA36790 you might say that but I couldn't possibly comment :-)

user116211

12:45 PM

@JohnRennie Although he is not that extreme ;))

@JohnRennie new question

You there?

Hi Noah

Hi

I need help

What do you know about the deceleration parameter?

user116211

12:47 PM

@0celo7 star?

@JohnRennie can you be of assistance

Isn't that the same as the acceleration parameter i.e. $\dot{a}/a$

@0celo7 I can try, though history suggests the answer is no

It's $q_0$

@JohnRennie I would agree. It seems pointless to argue with them, but I think it is not pointless to have a "canonical answer" to these questions, where new users (or readers not familiar with these kind of critics) can easily find good arguments why the physics community usually does not waste much time with these "ideas".

@NoahP as in:

Deceleration parameter

The deceleration parameter q {\displaystyle \!q} in cosmology is a dimensionless measure of the cosmic acceleration of the expansion of space in a Friedmann–Lemaître–Robertson–Walker universe. It is defined by: q = d e f − ...

12:49 PM

If we take the current density of the universe to be matter dominated, how would one show the current value of $q_0$, including the cosmological constant. And yes, that's the one.

I have no idea

how to obtain the equation:

$q_0 = \frac{\omega_M}{2}-\omega_\lambda$

user116211

@NoahP Complete the command.

$

@JohnRennie I'm trying to determine if it's possible to define the tangent functor from the $C^r$ category to the $C^{r-1}$ category algebraically.

Preliminary investigations appear to say: no.

user116211

@NoahP You can edit your earlier comment ;)

user116211

within 2 mins

12:52 PM

@MAFIA36790 Like so?

user116211

@NoahP good.

@0celo7 absolutely no idea. Let me go run some more oil on my butt and see if that helps.

4

@JohnRennie By "algbebraically", I mean define the tangent space at $p\in M$ as the derivation of the algebra of functions.

@MAFIA36790 What about the question then?

I know for a smooth manifold you have to take the derivations of smooth functions.

If you just do $C^r$ functions you get an infinite-dimensional tangent space.

I think it follows that on a $C^r$ manifold, where only $C^r$ functions make sense, you cannot define it algebraically.

But there might be some weird functional analytic way of salvaging this construction.

12:55 PM

@0celo7 I don't think there's an intuition for them - the nice thing about the non-degenerate ones is that they are local extrema and isolated, but the degenerated ones can be that, too. I think the "generic" degenerate point should be thought of as not being isolated, though.

In particular, I know that $\operatorname{dim}\operatorname{Der}C^r(M)$ is infinite, but I have not investigated $\operatorname{dim}\operatorname{Der}C^r_p(M)$, i.e. germs. But I suspect these are secretly the same vector space.

@0celo7 Is your map a linear map?

@ACuriousMind I think so.

I'm apparently studying functional analysis now

@ACuriousMind Well, I thought that the nondegenerate ones would be the ones which look locally parabolic

Or hyperbolic

Using the Morse lemma here

@0celo7 The derivative is not a derivation on $C^r$-function, since the derivative of a $C^r$-function need not be $C^r$. So you can't use the derivations on $C^r$-functions to define the tangents because the thing you want to capture - partial derivatives/derivatives along curves/something like that - isn't a derivation in the non-smooth case.

but there are all sorts of examples of degenerate critical points in Milnor

@ACuriousMind Maybe I don't want to capture that.