The h Bar - 2016-07-25 (page 2 of 3)

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5:01 PM

ok im back. @slereah who is dross :o

also does anyone know what it means for a particle to travel backwards in time?

@Obliv any particle has a world line, which is a 2D trajectory in 4D spacetime. This world line is the set of all points occupied by the particle. Particles don't travel on a world line - it's just a set of points. It's common to parameterise the world line with a proper time, and this can be done forwards or backwards as you please.

0

Q: Is there any evidence against alternate realities?

In our fantasy series, the topic of Travelling between worlds is a major theme. Is the existence of other worlds actually possible? As a writer, it often feels less like coming up with a story and more like seeing the already-existant world and already-occurred events, in order to write them down...

magic physics alternate-worlds universe

Possibly on it's way to us ...

Looking around me, alternate reality is the only explanation for much of what I see around me in everyday life.

@JohnR Could you also describe it as a 2D trajectory in 3D space?

doesn't that assume some sort of evolution of the topology of space with well-defined trajectories for matter?

@JohnRennie sorry for just barging in (thus; ignore me if you like), but why is it a 2D trajectory? shouldn't the curve be completely parametrisable by the eigentime of the particle and thus be 1D?

uhh nvm.

5:17 PM

dross
drɒs/
noun
noun: dross

1.
something regarded as worthless; rubbish.
"there are bargains if you have the patience to sift through the dross"
synonyms: rubbish, junk, debris, chaff, draff, detritus, flotsam and jetsam

this might be a gross misinterpretation of the uncertainty principle but: why is $\sigma_x \sigma_p \geq \frac{\hbar}{2}$? Shouldn't it be clear what the momentum is at any time if two measurements of position through some time interval are recorded?

You are assuming that the particle has constant velocity between measurements

is that wrong to assume?

By the way that is where the second uncertainty comes up

$\Delta E \Delta t \geq \hbar / 2$

so $E$ is not a classical definition of energy, right? It's given by $E^2 = pc^2 + ...$ or whatever?

5:27 PM

Why are we talking about physics physics

@0celo7 because physics is cool and real

also because i was thinking about the QM world while I was eating pizza. I thought, if particles can do $X,Y,Z$ then it must follow that macroscopic objects can theoretically do $X,Y,Z$ if they're comprised of the particles

Pizza?

Americans are so obese

It's unhealthy but it's cheap.. and I'm far from being obese

Does this feel like a duplicate to anyone?

1

Q: Why can the y and z-components of spin be measured simultaneously?

I have a gut feeling that this is wrong. By the uncertainty principle where $x,y,z$ are the $x,y,z$ components of spin $$ \sigma_{y}\sigma_{z}\geq \frac{\hbar}{2}\langle x \rangle $$ and it can be shown that $\langle x \rangle =0$. Therefore $$\sigma_{y}\sigma_{z}\geq 0 $$ Does this mean that I ...

quantum-mechanics quantum-spin

anyone know what would happen to ordinary matter (comprised of neutral,positive,negative charges) ,that may or may not be a conductor, if it came close to a really strong EM field?

let's say a human

5:42 PM

@Sanya oops, yes, 1D not 2D. I don't know why I said 2D - put it down to senility.

@JohnRennie :(

You're not that old are you

@Obliv your measurements of position will change the momentum, so you don't know what the moemntum was before the first measurement or after the second measurement. All you know is that the particle passed through two points in spacetime.

@EmilioPisanty Perhaps you're thinking of {this one](physics.stackexchange.com/q/165217/50583)?

@johnR how could momentum change if no net force is acting on the particle, though? The situation would be an isolated environment. Can the particle give itself energy?

You cannot perform a measurement of a particle without interacting with it.

5:47 PM

That's the observer effect, though. I'm told the uncertainty principle is strictly different from the observer effect.

@Obliv You can't say "two position measurements" and then complain that JohnRennie talks about position measurements.

You'll have to rephrase your question if you are not talking about the influence of a measurement on the state.

If you have a quantum system then to perform any measurement of it you need a second quantum system as your measuring device. The two systems have to have some non-zero interaction, so both must change.

@JohnRennie will heating elemental Ce in oxygen to 1000C kill me

Isn't cerium pyrophoric?

Or am I thinking of cerium powder?

oh I see. @acuriousmind Okay then the uncertainty principle is the principle stating that there is a fundamental limit on the precision of the measurements of both momentum and position simultaneously due to the reasoning described above, then?

5:51 PM

@Obliv I prefer to not talk about "precision" because it implies there is a "true" value that the measurements approximate

@JohnRennie I didn't want to be offensive or anything :D I was just puzzled by it; thank you

@0celo7 angstromsciences.com/sites/angstromsciences.com/files/files/…

also, does anyone know whether singularities must be neutrally charged? I imagine if it were strictly positive or negative charges being amassed to such a small region, the repulsive force of EM would disallow any further compression of the mass

@Obliv that is a fascinating question without a simple answer.

The objects in the uncertainty principle are standard deviations. If you perform many measurements on identically prepared states, those tell you how much the results will scatter around the average, but the average is not a "true" value in the sense that the measurements far from it would be "imprecise".

5:53 PM

If you take a Reissner-Nordstrom black hole it has a mass and a charge. However it doesn't contain any matter or charge. Work that one out :-)

@acuriousmind but we're assuming that the quantum systems interact with each other, changing the state of each system. If there was no interaction between them, they would have had different states?

@Obliv I'm sorry, I don't understand the question.

@0celo7 tl;dr - yes

Wow that's the first time you replied with that response instead of "....what?"

@Obliv What, exactly, do you mean by "singularity"?

5:55 PM

:D

an infinitely dense object with finite mass

Point charges in classical electromagetism, for instance, are singularities of the fields/charge density

@Obliv and your mathematical model to decribe it is ... ?

@Obliv Those things don't exist.

Aren't they observed to exist because of observations of intense gravitational lensing and some prediction from GR?

@Obliv well black holes don't contain any matter

5:57 PM

@JohnRennie Shit, I have to heat elemental cerium to 1000C :D

they are vacuum solutions - the stress-energy tensor is zero everywhere.

@0celo7 do it in a vacuum!

@JohnRennie I need it to oxidize :/

1000C will speed it up greatly

30 minutes @ 1000C = years of oxidation at room temp

@0celo7 if you're just trying to oxidise it you don't need 1000C. Start at a lower temp and ramp up gradually.

@johnR Is there a difference between a black-hole and a singularity? It seems like the latter is a point in space and the former is just a really dense object?

For us physicists the term black hole means a specific solution to the Einstein equations i.e. a spacetime geometry.

This geometry contains a singularity.

6:01 PM

@JohnRennie When will it kill me?

hmm okay.. So then @acuriousmind "they don't exist" is not really fair to say, as they may exist if we can prove that the spacetime geometry we live in is the one that john described right?

@0celo7 I have to say I'm a bit nervous about this. You run the risk of the experiment running away, and you'll only find out when it's too late. If you can't find a procedure for doing it start with a small piece of Cerium.

@Obliv A black hole is $M-J^-(\mathscr J^+)$. Not trolling, this is the definition.

@Obliv Note that John hasn't said anything about "infinitely dense".

Thus far you've talked about black holes and singularities

I don't think it will explode unless it's in a sealed container, but the combustion may get, erm, vigorous :-)

6:03 PM

@JohnRennie I don't want to ruin the furnace :/

@0celo7 It is trolling to throw symbols at people when you know they don't know what they mean.

I need a decent sized sample of mixed phase cerium oxide

@ACuriousMind Like when you explain algebra to me?

Can you tell me what that mathematical description says @0celo7 what is $M$ and $J$ and that weird squiggly thing

$M$ is spacetime

@0celo7 I use words that you can look up. One can't look up $\mathscr{J}$

6:04 PM

$J^-$ is the causal past

$\mathscr J^+$ is future null infinity

@0celo7 Just buy some:

sigmaaldrich.com/catalog/product/aldrich/…

$-$ is setminus

@JohnRennie that's pure CeO2

I need CeO2-x

so a blackhole is the spacetime setminus the casual past multiplied by the future null infinity

No, $J^-$ is a function

@ACuriousMind It feels like there's an even closer duplicate somewhere.

6:06 PM

okay. got it.

@0celo7 OK, but unless you can control the combustion precisely i suspect you'll get a rather random product i.e. you won't be able to make the same stoichiometry twice.

a blackhole is the spacetime setminus the causal past of the future null infinity.

@JohnRennie Perhaps.

Someone coming all surprised that they'd managed to simultaneously "diagonalize" all three $L_k$ because the $l=0$ state is a common eigenstate

I need a random product...have to check a suspicious blip on some spectrometer readings from some other stuff

6:07 PM

@0celo7 you could buy pure CeO2 and ask on chemistry stackexchange about a good way to get the oxigen out? :D

@Sanya I have pure Ce and pure CeO2

@EmilioPisanty hmmmm, yeah, there might be a question like that around

I want to randomly oxidize the Ce and get CeO2-x

@ACuriousMind but for the life of me I couldn't find it

well

OK, I wouldn't put my life on the line for that

but still

I like how I have an answer on the duplicate

But I don't understand the question

Either question

@JohnRennie Can I be going senile at 18?

6:09 PM

yeah, I got that, but from my time in the chemistry lab I know that reactions in the solid phase suck ... maybe there's a good alternative?

I could look up the pattern I'm trying to create

But there's no fun in that, and I've got nothing else to do this week

@0celo7 I think senility is pretty much defined as what happens when you get old. However there are all sorts of delightful diseases that can eat your brain at age 18.

@ACuriousMind Are you deliberately ignoring that algebra question of mine...

Are you secretly trying to dissuade me from reading further

@Obliv For an explanation of that definition, see Wald.

@Obliv: didn't you have an exam in three hours, three hours ago?

2

Or are you posting from the exam room using your phone? :-)

@JohnRennie ACM is ignoring me, do you know anything about exact sequences

6:13 PM

No. Next question?

I proved a very interesting topology theorem yesterday

Though I do know they are less likely to kill you than molten cerium in pure oxygen.

@0celo7 Did you test whether your idea for the maps worked?

:)

@ACuriousMind Yes.

@0celo7 Good. So what do you need me for?

6:14 PM

Because when you apply the two maps you get $0\otimes z=0$.

So the exactness is preserved

@ACuriousMind To confirm!

@johnR I have it in an hour

@JohnRennie Remember the time difference

@JohnRennie So anyway, I proved a very interesting topology theorem yesterday

@0celo7 ...how is "in three hours" subject to time difference?

2 hours ago, by Obliv

instead of doing practice problems for the exam i have in ~3ish hours i'm just gonna think about the theory and hope i'm competent enough to apply it. So I'm gonna see if my notions of gauss' law, electric potentials, and electric fields are correct. You just stand there and look pretty, ok? @johnR

You can always choose charts on a manifold so that any finite overlaps are diffeomorphic to $\Bbb R^n$

(assuming the overlap is nonempty)

6:16 PM

Hey, that was at 15:53. That's not two hours ago!

I think I'm the only one super excited about that theorem

@JohnRennie It uses a very cool Riemannian geometry theorem

And I think it uses the Axiom of Choice if the manifold is noncompact.

But I'm not sure about that.

even though that's not my time, isn't it only ~18:00 for you right now @johnR

@ACuriousMind Oh right, time is like a gauge theory

Moving time zones is a gauge transformation

@Obliv it is currently 19:18 in the UK

As indeed shown by the timestamps on the chat messages

it's in UTC and it's 18:18 in UTC right now

6:19 PM

@0celo7 that's...a good analogy!

@0celo7 I'm still chasing methods for preparing CeO2 ...

Most people seem to do it by precipitation.

@ACuriousMind (please do not think I actually thought time zones change time differences)

@JohnRennie Does Alfa have mixed phase?

Not even sure if that's what chemists call it

@0celo7 Alfa?

Alfa Aesar

Biggest scientific chemical company?

Maybe besides Sigma-Aldrich

There is something magical about proving black box theorems

user218912

I wonder if I can ever beat my 17 ricekrispies in a day record.

6:22 PM

There you go:

Cerium(IV) oxide

Cerium(IV) oxide, also known as ceric oxide, ceria, cerium oxide or cerium dioxide, is an oxide of the rare earth metal cerium. It is a pale yellow-white powder with the chemical formula CeO2. It is an important commercial product and an intermediate in the purification of the element from the ores. The distinctive property of this materials is its reversible conversion to a nonstoichiometric oxide. == Production == Cerium occurs naturally as a mixture with other rare earth elements in its principal ores bastnaesite and monazite. After extraction of the metal ions into aqueous base, Ce is separated...

Just heat pure Cerium oxide.

Right, I want a mixed powder

@JohnRennie Proof?

Wait

And it releases oxygen and goes non-stoichiometric.

I have CeO2

@JohnRennie oooo

No, I need en.wikipedia.org/wiki/Cerium(III)_oxide

I think

> Cerium(III) oxide is produced by the reduction of cerium(IV) oxide with hydrogen at approximately 1,400 °C (2,550 °F) to make air stable cerium(III) oxide.

Uhhhh

Have fun with that :-)

It was nice knowing you.

2

We have hydrogen...

And a 1700C furnace

:3

6:26 PM

20 mins ago, by 0celo7

I need CeO2-x

Yeah?

To get CeO_{2-x} start with CeO2 and heat it.

Link

The Wikipedia article I posted above - did you even read it!!

I did

Where does it say it

6:27 PM

> At high temperatures it releases oxygen to give a non-stoichiometric, anion deficient form that retains the fluorite lattice.

Oh :P

I skipped that line

:: slaps forehead ::

Ok, and that will not explode, right

Right.

Good

6:28 PM

And it should be a lot more reproducible

High temperature?

> has been shown to predict the equilibrium non stoichiometry {\displaystyle x} x over a wide range of oxygen partial pressures (103 - 10−4 Pa) and temperatures (1000-1900 °C).

whoa

where's that

oh

> Bulfin, B.; Lowe, A. J.; Keogh, K. A.; Murphy, B. E.; Lübben, O.; Krasnikov, S. A.; Shvets, I. V. (2013). "Analytical Model of CeO2 Oxidation and Reduction". The Journal of Physical Chemistry C 117 (46): 24129–24137. doi:10.1021/jp406578z

@JohnRennie I'm illiterate btw

user218912

6:30 PM

lol I realized the chat logs for the h bar are indexed by google search. 0.o

@0celo7 If we could only stop you getting distracted by mathematics you have the complete lack of any concern for your own safety that is essential for a glittering career in synthetic chemistry.

I do have a concern for my safety

I asked you if I would die

you said yes

so I turned off the furnace

What!

You were actually doing it?

QED!!!

...

joke

it was a joke

I need a drink

6:32 PM

lololol

@JohnRennie I am about to put some Ce in LN2 and try to smash it with a hammer.

Will that kill me?

No.

Great

But the Ce fragments are pyrophoric if they are small enough

Ok, maybe Cu then

6:33 PM

So do it away from anything inflammable

that will not kill me for sure

Copper is definitely fine

user218912

Lol once a grad student in the lab I used to work at touched the furnace when he was preparing a substrate while it was on and his nitrile gloves melted to his hands.

user218912

blue hands!!!

But prepare for disappointment.

6:35 PM

1

Q: What is the expression for a Lorentz invariant wave function?

I was recently thinking about Lorentz invariance description of a particle. In special relativity is based on the invariance of $$ \text{constant} = c^2t^2 - x^2 - y^2 - z^2$$ I was wondering if there was a similar expression in relativistic quantum mechanics for the invariance of wave functi...

quantum-mechanics special-relativity

what

@JohnRennie What do you mean

Will it not shatter?

Brittle means the metal will break when you bend it, but I doubt it means the metal will shatter into a million pieces when you hit it.

Worth a try though

I have time

What if I break it into small pieces

And then keep freezing them and breaking them

I need to make a powder for reasons

I'm unconvinced that would be a good way of making a powder. You'd get a very poorly controlled particle size and contamination from the hammer.

True

I don't think I need a controlled particle size though

But I probably won't get it small enough for what I'm trying to do, anyway

@3750 17 rice crispies, or 17 bowls (boxes?) of rice crispies? The former isn't any great achievement :-)

6:40 PM

@0celo7 We tried that with cobalt in an inorganic beginner's course. It didn't work. ;-)

user218912

@JohnRennie ricekrispy bars.

@Loong crap

@3750 ah, they have long since ceased to form a major part of my diet.

user218912

My room used to be full of ricekrispy wrappers.

user218912

like 50-70 of them

6:42 PM

ew

Back in the mid 80s there was a vending machine in the Cambridge computer science lab that would issue unlimited supplies of chocolate/rice crispy bars if you jiggled it just right. It took several years before I could bring myself to eat another chocolate/rice crispy bar.

2

user218912

@0celo7 how

How?

Trash laying around is gross

user218912

well it's just plastic.

user218912

and it's clean otherwise

6:44 PM

There's residue

Not the complex analytic kind, either.

user218912

well you're the one who had ants on his desk so you shouldn't be talking.

What

When did I have ants in my room?

user218912

long time ago I remember you saying that.

We had an infestation in the house because of wetness

Different situation

user218912

still, ricekrispies < ants

6:47 PM

@Slereah yes?

user218912

the only difference is that mine was self inflicted.

@yuggib is algebraic topology better than geometry

@yuggib in the wavefunctional representation

Is the vacuum of a free field just $z$

With $z z^* = 1$

And how does one express the ladder operators in it

@ACuriousMind Suppose I have two vector bundles $E_1\to M$ and $E_2\to M$, and $f\in\mathrm{Diff}(M)$. If I also have an isomorphism $L$ on each fiber viewed as a vector space in its own right, can I stitch together a vector bundle isomorphism $E_1\stackrel{\sim}{\to}E_2$ out of $f$ and $L$?

@0celo7 No

6:53 PM

If the vacuum is just a phase it would certainly work out alright for $\hat \pi |0\rangle = 0$, but then I'm not sure I can find $\langle 0 | \hat \varphi | 0 \rangle = 0$, as is the case

How could there be non-isomorphic bundles of the same rank if that were true?

@ACuriousMind Of course...what else is needed?

However, the "converse" is true - a bundle map that is a fiberwise isomorphism is an isomorphism.

What actually determines if two vector bundles are isomorphic

@ACuriousMind Hmm

@0celo7 zee said $GM$ is a length and hence an inverse mass. Why does that make 0 sense to me? $GM$ in units is $\frac{N\cdot m^2 \cdot kg}{kg^2} = \frac{m^3}{s^2}$ I don't get how it's a length let alone an inverse mass.

6:54 PM

Is that not the definition of vector bundle isomorphism?

@Obliv DON'T YOU HAVE AN EXAM

30 minutes

@0celo7 I'm not sure what you mean - you simply can't build a global map out of data on the fibers

@ACuriousMind Doesn't a vector bundle isomorphism have to descend to a diff of the bases?

How does a random homo that's an iso on each fiber do that

@0celo7 You might take it as a definition, but I always have the general definition of isomorphism as a morphism that has a two-sided inverse in mind.

@Obliv what has he set =1

6:57 PM

$c$ and $\hbar$

so $m/s=1$

@0celo7 Uh...bundle maps are by definition compatible with the projections.

that doesn't make sense

@ACuriousMind what does that mean

@Obliv then you're not reading correctly

@0celo7 That they descend to the identity on the base, always.

6:58 PM

@ACuriousMind uhhh

so $E_1$ and $E_2$ are assumed to have the same base?

You can relax that and demand they descend to a diffeo, though.

He's relating the temperature of a blackhole to the inverse of $GM$

where is the motivation for that?

@0celo7 ...how could they be isomorphic if they don't have the same base?

@ACuriousMind no, I mean in a general vector bundle homo

And you wrote $E_1\to M$ and $E_2\to M$!

6:59 PM

had long ago revealed to us that temperature,
a highly mysterious concept at one time, is merely the average energy of the microscopic
constituents of macroscopic matter. Hence temperature has the dimensions of energy, that
is, of a mass in units with $c = 1$.
It follows immediately that $T_h ∼ \frac{1}
{GM}$.

@ACuriousMind I mean in general

@0celo7 Sorry, I don't know what we're talking about now

You started questioning me about isomorphic bundles. That notion only makes sense if the bundles are over the same space

If I have a general vector bundle homo $E_1\to E_2$, does that assume $\pi E_1=\pi E_2$?

no

@ACuriousMind then what does this mean

@ACuriousMind and this

7:02 PM

That the diagrams on this Wikipedia page commute.

a mass in units with $c = 1$ <- what does this mean.. a planck mass is $M_p = \sqrt{\frac{hc}{G}}$ so if $c, \hbar = 1$ the planck mass is $\sqrt{\frac{1}{G}}$ the paper apparently relates $T_h$ to $\frac{M}{M_p}$ so $ = M\sqrt{G}$?

And you can convince yourself that you lose nothing by not allowing a general $f$ on the base in the case of bundles on the same space.

You made it articificially difficult for you to follow me by switching to the general case over different bases without properly notifying me :P

:)

Bott & Tu's definition of vector bundle morphism is too curt

I'll refresh my memory with Lee before I read on

Or maybe Jost

@ACuriousMind You mean that $f$?

Is there another f on that page?

I'm illiterate!

7:17 PM

@yuggib why hast thou forsaken me

Weeee $300 physics book for next semester

@ACuriousMind Oh, is that diagram just saying that the map is fiber preserving?

Wait...what does "fiber preserving" even mean

> Countrywide disruptions to flight services are expected in France due to a strike by unionised cabin crew, affliated with Air France...

wait, you can ionise cabin crew?

and if these ones are neutral, then why are they on strike?

XP

7:34 PM

@ACuriousMind I have a question for you pesky German

@ACuriousMind Are you there?

Did the AfD finally capture you?!

@0celo7 Basically, yes, that diagram is the definition of fiber-preserving, since it implies that the fiber at one point is mapped into the fiber at its image

@BernardMeurer I'm here

@ACuriousMind So, I was reading a discussion on watercooling SoCs, and it's pretty straightforward, Waterblock on the SoC -> (tubing) -> Radiator -> Pump -> SoC. What they were discussing was pump speed. Some people were claiming that if your liquid was cycling too fast the heat wouldn't have time to properly transfer to the water through the block, but that doesn't make any sense to me

...why would you ask me that? What's an SoC?

System on a Chip, basically a CPU

Read: It gets hot

Because this is just thermodynamics I guess

Yeah, and I have no particular expertise in thermodynamics

7:41 PM

I mean, it's not like the block-water heat exchange is happening in a frame-like fashion, it's continuous so it wouldn't be like after the water passing faster it'd have to wait for another 'frame' of water, that's stupid, there will always be water passing so the speed shouldn't matter too much right?

@ACuriousMind I'm giving AfD your address, they gon' get ya boy

@JohnRennie Halp

@EmilioPisanty Maybe the airline usually uses electric fields to move the ionised crew to their positions?

@ACuriousMind Is this good enough for a question Mr.Downvote?

@BernardMeurer I would agree with you it sounds kinda weird. And I guess you could ask that as a question, yes

@ACuriousMind Will you downvote it? :p

Probably not

:P

7:46 PM

@ACuriousMind You better not or I'll make my potatoes target skinny, long haired, glass-wearing, white, young, germans who know what a hamiltonian is exclusively

Good thing I broke my glasses, then :D

@ACuriousMind That only makes you easier to detect! Less matching to calculate MUAHAHAHAHA

You probably underestimate the amount of skinny long-haired German students here!

@ACuriousMind Do you have a class picture? :p

So you can train your potatos on better test data? You wish!

7:54 PM

Dang it

You got me this time

(Also we don't take class pictures at university)

@ACuriousMind LAME

#makeunigreatagain

Well, after the courses in the first few semesters that everyone takes together there's nothing one does as a "class" together, anyway

The solution to that is clearly to build a wall and not allow students in anymore

@BernardMeurer The solution to what? Where's the problem?

7:57 PM

@ACuriousMind The solution to not having a wall :)

And clearly, if the problem is that students don't take class pictures, the obvious solution is to get rid of students altogether!

Your approach to problem solving seems...drastic

@ACuriousMind Btw I managed to fry eggs yesterday and not murder anyone in the process

or start a fire

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