The h Bar

I have never seen such a miracle... but I have seen plenty of "blind", "deaf" and "lame" recover their senses and mobility in American mega-churches. It's on tv all the time... so it must be true! Wink, wink... :-)

0celo7

12:42 AM

@dmckee is it true that people don't connect "specialized" PCs to the internet because of viruses

1:11 AM

@0celo7 Airgapping is a way to protect systems from remote attack. But to do it seriously comes with significant costs.

@CuriousOne chomsky is not exactly so simple in the sense that the elite wants/ needs/ constricts americans to be dumb because... [x]

Most particle physics experiments make do with a couple of layers of firewalls: scary internet ===|firewall|===dmz===|firewall|===ineternal network.

agree re multiple tests, they have descended like anti intellectual plagues of locusts on the edu system

Nor are airgaps perfect without strenous discipline.

@CuriousOne humans are stupid™

1:14 AM

The Iranian ultra-centrifuges were attacked by malware designed to traverse the airgap on USB thumb-drives.

Einstein has the dibs on that, I believe... :-)

@vzn: You mean dumb Uhmericans blame cigar smoking men for keeping them dumb? Now I have seen it all...

@dmckee theres a very cool/ dramatic new documentary on that (stuxnet) gotta watch it asap theguardian.com/film/2016/jul/08/…

Really serious security systems have tightly control on anything that can move data.

And the result is that you can only work on those systems in their physical presence. It's like being back in the 19X0's (where X depends on what part of society you were on but spans 6-9).

@CuriousOne americans dont have a monopoly on stupidity that is quite clear and just maybe artificial international boundaries/ nationalities etc are part of the (highly sophisticated) distraction/ decoy system in place

It is, by the way, easy to stop the USB port from working on any motherboard... just desolder a couple of components. :-)

@czn: OK, after the Brexit the monopoly on stupid is broken...

1:17 AM

@CuriousOne Or epoxy up the port, which is apparently the recommended method for must applications.

Exactly. It's just physics.

@CuriousOne actually its a small glimmer the (unwashed) masses are restless... (JR sure had a lot of angst about it in here...) it does also show clearly an increasingly schism between elites & ~~sheeple~~ public

"But when the revolution comes, I will destroy you all. Not you, Joe!" Phoebe Buffet, Friends. :-)

@CuriousOne awww man would rather have a quote from jennifer anniston :P ... and how quintessentially american can you get? are you in europe or what?

The elites in the US are the people who made it past high school, at this point. How eerily similar that is to the Bolshevik Revolution... :-)

Phoebe is more fun.

1:22 AM

@CuriousOne there are visible elites and hidden (invisible) elites... like icebergs...

> meet the new boss! same as the old boss! —the who

1:34 AM

@vzn: The boss is always the person who was a little smarter than you or had a little more luck than you. What is really missing in the US is the opium to take the pain of the folks at the bottom away, so now they are screaming.

2:05 AM

@CuriousOne that statement seems not to grasp the advance/ extremity of wealth inequality (which, spontaneous thought, might be much more accurately referred to as wealth extremity)... anyway what "opium" is missing? ("religion is the opiate of the masses" —marx)

@CuriousOne Also the person who knows how to lead others.

This is highly nontrivial.

@dmckee Including squirting epoxy into USB connectors.

@vzn: Quite the opposite, I get all of it... the wealth inequality is what has pushed people from being slightly comfortable to painfully uncomfortable. If you keep them that way for more than a generation, you have a problem. Nothing that the top 1% couldn't fix by sharing just a little bit more, but they won't.

@dmckee Oh you already said that.

@CuriousOne but it seems you are saying (literally) if there was opium, the lower classes would not be screaming. can only interpret that figuratively, so what is the "opium"? and anyway, they have not been "screaming" noticeably for decades... are we agreeing they are now? so what has changed? there is no old/ new opium right?

@vzn: It's being slightly comfortable: fast food, game on the weekend, job security in a dead end job.

2:09 AM

@CuriousOne ok. ie "the squeeze" is finally noticeable as the dead end job security is now even deteriorated to just the so called precariat (maybe chomskys term) ie no "security" (which itself was probably a misnomer to begin with)...

You still don't need Chomsky for any of this. Marx and Lenin knew all of this. :-)

So do social democrats. :-)

Social democracy is the idea that one should not let people fall too low. Once they lose everything, they have nothing to lose, which is bad for the aristocrats. :-)

@CuriousOne coincidentally last fri bought (another) book on marx lol. agree its all quite similar however marxs bourgeoisie is technically different than the precariat which is why the latter term is increasingly used

If anybody thinks this is any different from Roman slaves revolting... please... you haven't read your history books.

@CuriousOne not much of a discernable revolt so far by historical stds, maybe very early stages... there are some other subtleties/ nuances that would have eluded marx, but it appears one of his basic predictions is finally coming into play before us, ie so called "collapse of capitalism," or at least "revolt of bourgioisie" and the outcome is uncertain/ unknown/ unpredictable at this pt.

It's just basic human psychology. It doesn't take much to see what's going in. The modern from of uprising is simply fascism, people seeking answers in a strong man, who, almost inevitably, makes life even worse.

2:23 AM

Can someone explain to me why, given a signal with some spectral density S, if I measure that signal for some finite period of time and compute the Fourier transform of the acquired signal, the square of the Fourier transform approximates S?

@DanielSank: You are lucky that you aren't seeing too many artifacts from the window function?

@CuriousOne right/ alas. am hoping for Think Different™ this time. o_O but humans are stupid™ :(

@vzn: History is rinse and repeat. Let's hope we don't rinse everybody down the sink, this time.

@CuriousOne agreed; the stakes are much higher this time. 300M in US, 7B on planet. have been thinking a possible (new) wildcard is social media which is growing/ having a larger sway (even to pt of influencing elections) & seems to actually favor the underclasses/ free media etc

@vzn: Social media amplify intellectual noise and fallacies much more than they can increase critical thinking. They don't make the average person smarter...

2:34 AM

@CuriousOne but Trump is a wildcard (not nec a "strongman") and his support is highly intertwined with social media. social media does have some significant "collective intelligence" possibilities eg stackexchange voting, facebook likes, electronic petitions etc

@vzn: If you think that Hitler was any smarter than Trump, the leaders of the Wehrmacht who had to listen to his "accounts" of Russian terrain (which he seem to have gotten from reading Karl May novels) have some interesting news for you. :-) OTOH, Hitler was a genius at appealing to the lowest common denominator of the masses and he had messaging geniuses working for him. :-)

Communication doesn't make people smarter. Only learning makes people smarter.

@CuriousOne right, goebels. its too soon to know whether trump might just be something like a flailing nixonian disaster or an actual dictator in the making etc... there were some rumors that Trump reads Hitler speeches o_O

Trump also reminds me of reagan in that the latter wasnt very politically or intellectually savvy etc, and supposedly delegated a lot of key decisions.... virtually delegating his entire authority way to the pt of iran-contra etc... also similar to how much Bush delegated to supposed "copresident" Cheney. (some are saying Pence might be like that)

user54412

@DanielSank Is the spectral density not the square of the fourier transform?

Yeah... where is Nixon when you need him? He would be such a refreshing choice, right now. :-)

@CuriousOne have never seen US election like this one, have seen a bunch o_O

2:50 AM

I don't want to see another one like this...

@CuriousOne what country are you in?

The US, but does it matter if an idiot destroys the US? Where are you going to go, economically? Mars?

@CuriousOne its kind of limping as it is... hopefully nobody screws it up much more... o_O

economically we have usually/ofen had a wave/ Next Big Thing™ but that sequence/ conveyor belt seems to have finally "run out of steam" as expr goes (steam engines once long ago being that Thing, nearly the 1st!) ... IT fueled it for last few decades but it looks like its finally plateaued out... in line with moores law etc...

3:12 AM

@vzn: The economy is fine... if you have higher education, current skills and can work in a collaborative environment. If you lack any of those... good luck. Now, we can't turn the time back and make everybody a coal miner or assembly worker. We could give every kid a shot at higher education... hey... isn't that what the democrats are running on? Or was that socialism? Did God and the Founding Fathers say that they want every kid to start working life with $100k debt? :-)

@CuriousOne your earlier pithy defn of social democracy was quite interesting, have heard the expr but not associated with social democracy. youve heard of Sanders right? he very nearly split the "left" vote. this is utterly unprecedented and will surely play out in future elections in a big way. its basically/ probably the Occupy Wall St movement at natl level etc

@ChrisWhite Certainly not.

Most things that have a spectral density are not square integrable.

...and therefore don't even have a Fourier transform.

In real life you Fourier transform a finite duration (and discretely sampled) segment of whatever physical process you have of interest.

@Chris is here! maybe he can share some more of that amazing philosophy knowledge... am sure ACM will luv it... hey this whole conversation was nearly inspired by it :)

@DanielSank: While carefully minding your window function... which in spectroscopy is called "instrument response" or something, right? Where, by the way, did you acquire that spectrometer with infinitely wide spectral response?

@CuriousOne

> Where, by the way, did you acquire that spectrometer with infinitely wide spectral response?

What the hell are you talking about?

3:20 AM

(uh oh) :\

@DanielSank: You were complaining about square integrability, which I rarely find a problem with real data. A finite number of samples squared and summed always seems to return a finite number on my computer.

user54412

::checks wikipedia::

user54412

ok so spectral density is the integral of the square magnitude of the function...

user54412

which is equal to the integral of the square magnitude of the transform of the function...

user54412

if there's more to this you'll have to provide definitions or examples

3:31 AM

@CuriousOne Did you even read my messages?

user54412

explain it like I'm a physicist ;)

@ChrisWhite No, that's not a great way to think about it.

Consider some noise signal.

That signal goes on forever (you know what I mean).

It's some stochastic random thing.

This noise has a correlation function.

user54412

ok

Calling one realization of the noise x(t), the correlation function is C(t, t') = <x(t)x(t+t')>

Often, C(t, t') doesn't depend on t.

@DanielSank: You have signals that go on forever? I really don't know what you mean.

3:34 AM

So we get lazy and write C(t) = <x(0)x(t)>.

user54412

angle brackets mean...?

@CuriousOne They go on a lot longer than I'm willing to measure on my oscilloscope. For example, the cosmic microwave background will continue to wiggle long after I stop measuring it.

user54412

@DanielSank As Jim would say, not with that attitude

@ChrisWhite Great question! They mean "average over realizations of the noise". Of course, this is in imaginary theory land so far where we imagine an ensemble of realizations of the noise.

@DanielSank: That means you don't really care what's outside of your window function... there is that word, again, that I got started with!

3:36 AM

@CuriousOne I see what you're trying to say but I think that's a red herring.

I'm going to have to just wait for you to actually explain something before I can tell for sure.

Now, @ChrisWhite, the spectral density is the Fourier transform of the correlation function.

@DanielSank: It's a red herring that window functions can screw up measurements and that one has to be careful with instrument responses? OK...

This makes a lot of sense: if x(t) is a sinewave at frequency omega, then the correlation function is also sinusoidal, and its fourier transform is a delta function at omega, as you would expect.

@CuriousOne You still haven't explained anything.

As usual, you're making snarky comments trying to come off smart but you are avoiding at all costs actually saying anything quantitative or explanatory.

Ok @ChrisWhite, now we know what spectral density is.

@DanielSank: I merely suggested that you were lucky for not having seen artifacts in your measurements that I keep seeing every time that I turn on my spectrum analyzer. Or maybe you don't know that they exist and you take what you see on your instruments for granted. I don't know. I will stand down now. :-)

this is starting to remind me of the Nyquist thm... can some of it be expressed in that form?

user54412

@DanielSank yes

3:41 AM

@vzn: Nyquist is not your friend. :-)

@CuriousOne Yes, what you said was rather confusing, as I never said anything about "not having seen artifacts in [my] measurements...".

So I have no $*@#*( idea what you're on about.

@vzn: Daniel seems to be worried about infinity, when the real problem is in the sampling and the window function, which both screw up the data if one isn't careful.

@CuriousOne Listen to me for five seconds:

I want to know why, if I acquire a noise signal for some time (with some sampling rate such that my Nyquist frequency is above the highest frequency component for whatever I'm measuring, i.e. forget about aliasing for now) is the Fourier transform of the acquired signal similar to the actual spectral density of the process.

not following exactly but was just thinking the idea of taking a fourier transform of noise seems like a problematic way to start o_O ... @Chris what are you working on anyway? did you mention it earlier in chat?

@vzn It's not.

For a given bin in your discrete Fourier transform (DFT), the mean value of the square of that bin is equal to the spectral density at that bin's frequency.

Note that I said "mean value" because every time you measure noise, even if it has the same underlying spectral density, the actual DFT values differ.

In other words, the DFT values are themselves random numbers.

You can see this if you look at a spectrum analyzer (as I'm sure @CuriousOne will attest to).

If you measure the noise of, say, a resistor, you see that the spectrum analyzer trace is not a flat line as the Johnson noise formula would suggest.

It has wiggles, for the reason just explained.

@ChrisWhite Where were we?

3:48 AM

ok yeah this sounds like the statistics of fourier sampling or something

@vzn Not sure what you mean by "Fourier sampling".

@DanielSank fourier transforms of sampling etc... the theory is sometimes defined on continuous instead of discrete fns etc... can be understood either way... the discrete approximates the continuous in the limit of many samples... lots of subtleties in how they interrelate

@vzn Yes, I know a lot about that.

What I don't know is why the Fourier transform of an acquired signal is a good estimator for the underlying spectral density.

user54412

@DanielSank We have $S(\omega) = \int_{-\infty}^\infty \mathrm{e}^{-2\pi\mathrm{i}\omega t} \langle x(0) x(t) \rangle \mathrm{d}t$ and $X(\omega) = \sum_{n=0}^{N-1} \mathrm{e}^{-2\pi\mathrm{i}\omega n/N} x_n$, and you want to know why the latter estimates the former?

@DanielSank that sounds like a (known) math thm or statistics problem. did you ask it in chat? cant figure out context/ bkg of this conversation joined in middle

3:52 AM

@ChrisWhite Yep.

Why is $|X(\omega)| \propto S(\omega)$

Here $\propto$ means "proportional to and an estimator of".

:P

user54412

if \vee is $\vee$ and \cup is $\cup$, I'm pretty sure \leftfish should have been $\propto$

2

@DanielSank think it is (good estimator) subj to nyquist thm and there is a lot of theory on that, the derivation of the thm "explains why"

@vzn The Nyquist frequency is about aliasing so I strongly doubt it has anything to do with what we're talking about.

I think @CuriousOne was more on track about the window function, except that since he never actually said anything about it we have no idea what he meant.

@DanielSank uh, your "acquired signal" is (not) a good estimator of anything depending on # of samples and frequency of the samples...?

@vzn I already said to assume the sampling rate is high enough to capture the highest frequency components of the measured signal.

As for number of samples (for fixed sampling rate), certainly more samples improve the estimate.

user54412

3:59 AM

well, there's two parts to Shannon sampling -- being above the Nyquist rate (the hypothesis of the theorem), and converging to the function in a very certain and often misunderstood sense

@ChrisWhite For a fixed sampling rate, if you sample more you're frequency resolution increases.

i.e. you get more bins in your DFT.

Interestingly, as you do this, the variance in each bin does not improve.

In other words, sampling more doesn't improve your spectral density estimate unless you do something like average neighboring bins together.

a quick google. suspect there are many refs on this. here is some basic theory/ examples. if not containing the answer, think it might come close. it might help to refer to terms/ pgs in here dcs.warwick.ac.uk/~feng/teaching/PowerSpectrum.pdf try sec 10 "spectral estimation by averaging modified periodograms" etc

hey guys does anyone here understand the condensed matter analogy of the higgs mechanism? I.e. how symmetry is broken in a superconductor etc.?

2

I've tried googling it but sometimes there are gaps in the explanations (or my understanding even after reading the explanations) so if there's someone here who didn't mind explaining it I'd be super grateful!

@vzn Actually I think the earlier sections cover what I wanted to know.

@DanielSank :) (the theory can be formidable but maybe not worse than QM...)

user54412

4:08 AM

on a related note, it's highly unfortunate that the wikipedia article on the shannon sampling theorem at no point gives any statement of the theorem

@DanielSank: Do you mean that oversampling correlated signals (because of finite bandwidth) doesn't increase statistics? Yes, I agree, that won't do anything at all, it only helps with avoiding sampling errors.

If you want to achieve better errors on the noise estimate, you need longer data sets.

Q: Does the range of possible trajectories of a particle change with its speed?

sec6.0 p8 seems interesting wrt the raised question/ topic

> The current discussion leads one to believe that the periodogram is an excellent estimator of the true power spectral density S(f) as N becomes large. This conclusion is false and shortly it will be verified that the periodogram is in fact a poor estimate of S(f).

Physics Meta

4:38 AM

0

Not sure if this question makes sense, but does the possible range of paths of particle get smaller or larger as the speed of a particle increases? or does speed not influence the range of possible trajectories?

discussion

Q: Does any particle ever reach any singularity inside the black hole?

@SirCumference You always ask the interesting questions after I've gone to bed :-)

This has been asked before:

25

I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and "weird things happen". Now to the question at hand, please follow my slow reasoning... As far a...

black-holes observers event-horizon singularities

Though the answers aren't that helpful. That was too early in my SE career for me to risk an answer, but note that my comment got 8 upvotes.

My take on this is that the black hole does indeed evaporate before you fall into it.

Well, more precisely the black hole never existed in the first place, in the sense that there never was a true horizon present - only an apparent horizon.

user54412

@DanielSank You know, I really do want to cast this all in terms of Shannon, but to my great frustration I can't find a good statement of the sampling theorem on the internet to reference. Insanity!

user54412

Basically, given a band-limited signal, there's the reconstruction of the signal from regularly spaced samples, and this converges in the L2 sense. It's hard enough to find this precise statement. And at the same time there's a formula for the reconstruction of the transform of the signal, and a statement about how this converges to the true transform, and I cannot find anyone stating this.

user54412

It's rather important, and I remember the convergence of the reconstructed transform not being the same as that of the function.

user54412

4:54 AM

This is all reminiscent of the 19th century fathers of analysis wasting decades talking past one another because some of them were talking about pointwise convergence, others about L2 convergence, others about uniform convergence, and so on, but they all just said "f approaches g" and didn't explain what they meant.

@JohnRennie: Do you have a mental image of what would happen to the falling observer? Is there a giant UV flash from "below" or does he just somehow "find himself dissolved" into an extremely redshifted em field by a spacetime that is stretching into a pancake?

No, and that's a tough question. The freely falling observer does not see any Hawking radiation i.e. in the freely falling frame there is no Hawking radiation. So an observer falling in cannot be evaporated by radiation that in his frame doesn't exist.

@JohnRennie: So the free falling observer gets enormously compressed by tidal forces first and then, as part of the evaporation, the process reverses (but not homogeneously, I suppose?) and parts are "dribbling out", but without being mixed with the matter/radiation from other volume elements that fell in at other times?

Hello all

Hi there.

5:00 AM

I am a student of meteorology

i feel more at home in Physics SE than ES SE

i can tell you that is because most of questions or discussion points are very mathematical

Time to switch classes?

no

just bring in more Math into Meteorology

I thought you guys are highly model driven, already?

we are but a lot of times in practice we end up only running code

i am interested in the Math behind the code

:-)

5:03 AM

i mean the Math gets glossed over

like as in read it up in some paper for God's sake

user54412

So sometimes I type ph <enter> too fast and before my browser autocompletes physics.SE, it decides to google search "ph." Today it gave me this interesting picture:

user54412

get the code goin' !!

i had actually a valid reason for coming here today

i am interested in analogies in Physics

like for instance the same equations which are used in Electrostatics are used in Potential Vorticity

user54412

I always knew the east coast had acidic rain (some even say this removes sulfites from the soil and affects the wine in the area), but I didn't realize it was that acidic

the Laplacian , method of images, dielelectrics in the atmosphere

hopefully later this month i can bring myself to ask a few questions based on this paper - climate-dynamics.org/wp-content/uploads/2015/08/gpv.pdf

brings in Maxwell's equations into Meteorology

gauge invariance

5:08 AM

@ChrisWhite As I knew, rain water is drinkable

@peterh - acid rain is what is causing the Taj Mahal to turn yellow

when that sulfite or sulfate in acid rain combines with the calcium carbonate in the building it turns yellow

The good thing about reading more physics oriented journal articles (except computational chemistry) is there are a lot of equations that provide me enough empty space to rest my eyes that is otherwise drowned in the wall of text

@Secret - one equation = 1000 words ?

still better than seeing a never ending wall of text. Because equations are somewhat "visual" in a sense

plus I process things slightly faster when presented vertically

when you look at a computer software representation of a partial differential equation so much of the visualization is lost

you need to go back to the original reference to understand what it really means

5:21 AM

I think you will be disappointed to learn that the surface of Earth doesn't have sufficient symmetries to make the analogy. You aren't playing on an empty, homogeneous background but with non-trivial topography. I don't think symmetries will do much for you.

@CuriousOne - how about the region between the upper atmosphere and stratosphere? would that have sufficient symmetry?

onlinelibrary.wiley.com/doi/10.1002/qj.49712051710/abstract

user54412

@peterh yes, and? Soda and citrus juice have pH below 4, and sometimes below 3

@gansub Yes, it will be gypsum

@peterh though the yellow colour will be due to other contaminants. Gypsum is colourless.

@JohnRennie Monocrystallic gypsum with crystallic water is yellow opaque.

@JohnRennie Probably there is a layer CaCO3 - CaSO4 - CaSO3 + n H20 mix there

5:31 AM

@peterh proof?

@JohnRennie I think with spectroscopy it would be possible even from light years away.:-)

Gypsum is colourless

@JohnRennie I suspect it is for the pure powder, and not where the crystall is built up with water molecules between the ions. Here is a visual argument :-)

@JohnRennie Do you know, which nucleus has the highest electric dipole moment?

@gansub: Is the system sufficiently isolated from below and above? Maybe.

@JohnRennie I can't find any data about that on the net.

5:41 AM

@CuriousOne - yes it is because it is above the planetary boundary layer the layer where friction impacts the wind

the stratosphere is very stable as well

@gansub: Interesting. I wasn't aware of that.

@peterh those show samples of gypsum contaminated with trace amounts of transition metals. The colour comes from the contaminant not the gypsum.

user54412

gypsum is quite white

@JohnRennie Ok. Then why is becoming the Taj Mahal yellow from the acid rain?

@JohnRennie And why this gypsum desert isn't?

user54412

5:46 AM

^ ?

user54412

White Sands, New Mexico, is the largest gypsum desert in the world, and the point is it is as white as newly fallen snow.

user54412

oh I see -- why doesn't the desert turn yellow?

@peterh because there are other atmospheric contaminants, which is what I said about 300 comments ago.

user54412

I'm willing to bet the rainfall in India is both substantially more acidic/contaminated and substantially more existent than the rainfall in the southwestern US.

Trust me, I did a (physical) chemistry degree then worked for 12 years as a physical chemist.

A yellow tinge comes from absorption at the uv end of the spectrum and pretty much any degraded organic matter will go yellow.

So going yellow is the default colour change mode for almost everything.

5:51 AM

Yellow is also the complementary color for violet thus absoprtion of violet colors necessary means the object will reflect yellow

@ChrisWhite - rainfall is substantially more existent ?

what does that mean ?

@gansub it rains more

so more rainfall means more acidity ?

@JohnRennie I trust you. :-)

I think the point is that if the contaminant level in the rainwater is constant then more rainwater means more contaminant.

5:55 AM

@JohnRennie that is a possibility. The Taj Mahal is surrounded by oil refineries

and coal plants

@gansub The Taj Mahal is surrounded by oil refineries - look no farther :-)

so these aerosols move depending on the wind direction

there could be non local sources as well

@JohnRennie Ok-ok. Sorry for the sudden switch but the google seems surprising silent. Do the nuclei have an electric dipole moment?

Well, maybe I make a question from that.

Q: Why are pear-shaped nuclei possible?

In general nuclei don't have an electric dipole moment. To have an EDM requires parity violation which mostly doesn't happen.

However the weak force violates parity, and if the weak force is important in the nucleus it can result in an asymmetric nucleus with an EDM.

27

In a recent question, Ben Crowell raised an observation which really puzzled me. I obtained a partial answer by looking in the literature, but I would like to know if it's on the right track, and a fuller explanation for it. It is a well-known fact in atomic and molecular physics that electronic...

nuclear-physics specific-reference parity

[a question coming out of the blue] I wonder what happens when we fuse two pear shaped nuclei together...?

can we get more interesting shapes?

6:02 AM

@JohnRennie Thanks!!!

@Secret I think it doesn't really depend from which nuclei was it fused from. Nuclei have a ground state, it tends to be more symmetric, and a lot of excited states, which tend to have a short life time, but more asymmetric.

@Secret What I find surprising, that they don't have electric dipole moment. They are compound particles from a lot of charged components, having complex interactions between them. Logic dictates at least some of them should have a significant EDM, especially the larger ones.

@ChrisWhite I can tell you all about Shannon, but it's not really what I'm trying to understand here.

Shannon sampling theorem is rather simple: if you sample a continuous signal such that the sampling rate is twice as fast as the highest frequency component in the sampled signal, then the original analog signal can be reconstructed exactly from the samples.

In the simplest case, the reconstruction involves a sum of sin(x)/x functions.

It has to do with the fact that the sin(x)/x function is the Fourier transform of the top-hat function.

user54412

6:25 AM

@DanielSank But it's not just that. There's a part of the theorem that tells you exactly, mathematically how $X \to S$ is what I'm saying.

user54412

But no one seems to be aware of this.

o_O

user54412

its almost like i'm imagining it, though, with how little it shows up in my searches

A: How to derive non-degenerate Bose-Einstein statistics in the canonical ensemble?

2

You have to remember that Bose-Einstein statistics only arises from the grand canonical ensemble, i.e. for a system where the energy and the number of particles are not fixed ; only their mean energy $\langle E \rangle $ and mean number of particles $\langle N \rangle $ are fixed respectively by ...

One to look at for possible VLQ flagging (under our experiment)

user54412

6:40 AM

Can't even understand the question

user54412

One day I'll be called upon to teach stat mech, and when that day comes, I swear I will make no reference to "grand canonical ensembles" or "partition functions" or any of that vestigial nomenclature from the 19th century.

what are the modern updates to the concepts "grand canonical ensembles" and "partition functions" (especially the latter, cause we physical chemists use them a lot)?

user54412

I also will never write down a formula with an extensive quantity in it. Outside of those relatively clueless guys in ~1750 who played around with compressing gasses, who on Earth thinks in terms of "volume" as a property of the gas?

user54412

@Secret "partition function" is just overly fancy wording for the normalization factor that makes your total probability come out to 1

right, makes sense

user54412

6:46 AM

and while there might be use for "grand canonical ensemble" the concept, none of those words means anything so the phrase is worthless

user54412

my theory is that the "grand" is the same word as in "grand jury" -- that is, it's taken directly from French for "big" and thus is a false friend already (besides being still a useless word in this context)

I think if they ever come up with the "super-grand canonical ensemble", we'll know for sure. (Must be pronounced with a French accent)

7:03 AM

Honestly, I got a bit anti hype about SUSY in recent years...

hey guys

anyone mind clearing up something about gauge invariance (and actually DavidZ this relates to your explanation on the higgs mechanism)

David you said that in order to maintain invariance of the field under a phase rotation you redefine the derivative operator to D(phi(x)) = d(phi)-iqA(x)phi(x)

but I don't understand how you can just redefine the derivative operator and just 'create' terms to satisfy the invariance

Well, maybe you shouldn't think of it as redefining the derivative operator. We're not redefining $\partial$.

It's more like creating a new operator which is gauge invariant but which acts the same as the derivative to leading order in $q$.

And then you use that operator instead of the plain old derivative $\partial$ in the Lagrangian, because the Lagrangian has to be gauge invariant.

sorry what does $q$ represent?

Electric charge. (I was going off the definition you used a few messages up.)

Specifically, the charge associated with the field that the new operator acts on.

Oh I got the 'q' from your answer, but I don't actually know what it represents

okay that makes sense, but isn't the Kinetic term of the Lagrangian defined to be a simple derivative and nothing else? So the new operator you create wouldn't be the same as the Kinetic term right?

Unless... is anything with a derivative essentially a kinetic term?

7:15 AM

People aren't really precise about the usage of the word "kinetic term", but it generally refers to whichever term involves two derivatives.

oh okay so you can introduce any term with derivatives, and then you pick the one which satisfies gauge invariance

I'm not sure what you mean

I was just saying, you could probably see either $\partial_\mu\phi\partial^\mu\phi$ or $D_\mu\phi D^\mu\phi$ referred to as the kinetic term, depending on how the Lagrangian is written, who's writing about it, or so on

Ok I think I understand now

thanks a lot!!

Now I just have to find a condensed matter theorist

to understand the condensed matter analogue of the h.m. :P

You could ask a followup question on the site. (But do a bit of background research first. I'm sure Wikipedia has something.)

True

7:22 AM

Is it true that the kinetic term does not always have something to do with kinetic energy due to the way it is defined?

I think it's called the kinetic term because if you were going to define a kinetic energy (or kinetic energy density), it would make sense to do so using that term. But we're talking about systems that go beyond classical mechanics, where the familiar notions of kinetic and potential energy come from.

7:41 AM

the kinetic term is very similar to classical ke yeah because they both involve derivates with respect to position, I think that's why it's called the KE term

7:57 AM

Is there exists ghost fields that are not instabilities nor terms added to cancel something?

Hi there! May I go off topic here?

everything is on topic here @The_Vinz

I'm Italian, and while reading in English I found something unclear: what does "message sent via ITT" mean? I found it on a old ad of the ITT company (1968), but I don't understand what is that :/

Can you give us some more context?

https://books.google.it/books...

8:10 AM

international telephone and telegraph company :D it's explained about half a page above

"When she cabled her husband in London to confirm the date and time of her arrival, the message was sent via ITT."

(let me try to fix that formatting)

Yes, but I was wondering if telegraph was still used in 1968, or maybe it was something else, like telephone, fax or some ancient form of communication ahah

I think the use of "cabled" implies that it was a telegraph, but we can't be sure.

I don't know of any other form of communication that would be described as "cabled", at least not in modern usage.

Well, thank you at least for assuring me that it's not the "barrier of language" that keeps me from understanding :D

Thanks :)

8:16 AM

It's hard to imagine a world before e-mail and SMS messaging, but it did exist :-)

I know the feeling ... always being like "is it my bad english or do they just not make any sense"

@JohnRennie I dunno if a modern kid would be that familiar with the concept of SMS anymore :D :D

@DavidZ: would you be willing to cast your eye over a proposed new answer of mine?

@BernardMeurer More concrete example here math.stackexchange.com/questions/1442099/…

@JohnRennie Sure

physics.stackexchange.com/questions/147096/…

I've left it deleted while I seek reassurance that it isn't complete bobbles

8:23 AM

Ah. Well... I wouldn't say "complete bobbles" at all, but the way you're using operators is kind of different from how they get used in QFT and I'm not sure if it's equivalent.

OK thanks. When our resident QFT head shows his face I'll see what he thinks.

The point I was trying to get at is to show how virtual particles arise in a calculation. Maybe there is a better way to do this.

I'm thinking about it. You may be on to something, but it takes me some time to understand the meaning of what you're saying.

And actually, on second thought maybe your use of the operators is fine. The normal way is that you apply the creation and annihilation operators to the vacuum state to create the initial waves and destroy the final waves; in other words, for $AB\to CD$ you would do $\langle 0\rvert A_C A_D (\text{dragons}) A_A^\dagger A_B^\dagger\lvert 0\rangle$

But when we calculate the actual cross section or scattering probability, that gets multiplied by its complex conjugate. So you could conjugate it from the start and it would be just as effective. That would be $\langle 0\rvert A_A A_B (\text{snogard}) A_C^\dagger A_D^\dagger\lvert 0\rangle$, as you have in the answer.

Thanks. Do you think the approach I've used helps explain what virtual particles are?

What I was trying to do is provide a perspective of what exactly we mean by a virtual particle.

And in the process explain why we say virtual particles don't really exist.

That I'm still not sure about. It seems like what you're describing is not the same thing as what I know as a virtual particle. What you're talking about - the difference between $a$ and $A$ - seems more like what I would call "dressing".

It probably would be good to get another perspective on this.

Where is ACM when you need him? :-)

8:33 AM

I could try to give you my rendition of what virtual particles are, but there's no way it would translate into an accessible answer.

@DavidZ that's the problem. It's extraordinarily difficult to explain QFT in Discovery Channel terms without it becoming a caricature.

(I have not start QFT yet!) I am not really sure about this. But recently I managed to explain my nonscience friends the gist of quantum mechanic's mathematical formulation in some simple terms
"
There are 3 basic things about quantum weirdness:
1. You have states that can superimpose and interfere with each other and they live in a abstract space called helibert space
2. Your outcome is determined by a probability obtain from this and you only ever seen one outcome at a time
3. States can be correlated (quoting the two qubit entangled state example experimental observations)

(Basically, my way to deal with quantum mechanics mathematical formalism is talking about the quantum state as if it is something you can touch with your hands, and then the maths becomes easier to interpret)

@JohnRennie no kidding. If I ever figure out this particular issue, maybe I'll do a blog post.

This is because while quantum mechanics is nondeterminstic, it is random and this randomness evolves determinstically.

That means if I perform experiments in a way to indirectly illustrate this determinstic property, then quantum becomes a lot more visual

e.g. measuring 1000 replicates of a system and using that piece of tick tape as "representation" of the probability distribution

@Secret if your nonscience friends didn't throw up their hands the moment you said "superimpose" then you have a very different kind of friend group from the rest of us

8:49 AM

I thought "superimpose" is a really common thing that the gist is populated rather accurately by popsci media (save for the "many places all at once thing)?

I guess it could be, but my impression has been that despite being a word that's thrown around a lot by the popular media, most people don't really understand what it means in this context

O, for that I describe them using colors as an analogy, because red+blue=magenta light

i think that approximate superposition quite well in the mathematical formalism

I think most layperson have some very basic understanding of colors as different wavelengths of light, thus that will help convey the message

I actually like the science talks done by brain greene and other scientists. They have over the years expanded the vocabulary of science terms available to the layperson with somewhat high accuracy, thus allowing even more technical terms to be described without losing too much rigor

My preferred method on illustrating complex things is to 'zoom out' from that thing until it reached a point where it becomes somewhat 'unchanging'

One example being chaos, It looks very messy despite deterministic because trajectory diverge from each other exponentially as revealed by the Lyapunov exponent. But for some types, if you zoom out or in enough, there's actually a relatively invariant property in these systems, such as how they often have some kind of fractal structure. This structure, no matter how long you run this system, will be relatively unchanged as a whole

John Duffield

10:02 AM

@vzn : interesting stuff.

Q: Is (magical) mass decrease would cause antigravitation?

10:36 AM

-1

Suppose some mass $M$ is located in rest in some point of space. It creates gravitational potential well, which attracts test bodies towards $M$: Now suppose that the mass of a body drastically decreases several times to $m << M$ (by magic). This causes potential well become shallower. But thi...

general-relativity gravity black-holes newtonian-gravity

Ignore that he said magic causes the potential to drop. I guess the actual question that need to be solved are:
1. if something looses mass and we knew that any information that carries that change in the potential cannot move faster than the speed of light, then by continuity, we should expect something like a smooth version of the kink shown there to happen in the gravitational potential. However I don't think that necessary means that kink will be antigravity because gravity in GR is not just $-grad(\phi)$

I think a relevant question that can address his question, is the comonly asked question on what happens when the sun is suddenly removed, how will the geometry of spacetime changes in e.g. the solar system

(might have to revise a bit or perhaps PSE has that question somewhere)

@JohnRennie I'm afraid I don't get you - there are no creation/annihilation operators for interacting fields. We have to stipulate that the particle come/go to the asymptotic past/future, and that in those asymptotic times, the fields can be treated as free. Then we compute the overlap of the in-state $\langle A,B\vert = \langle 0\rvert a_A a_B$ with the overlap of the out-state $\vert C,D\rangle = a_C^\dagger a_D^\dagger$ where we first evolve the past in-state into a future state.

It's in that evolution of the asymptotic past state to the asymptotic future state that the virtual particles hide - and you can't talk about $a$ and $a^\dagger$ there because during the evolution the fields are interacting

so under this formalism, is the maths that describes what happens in the middle is done by whatever that evolve the state from in to out (I think that is done by the propagator)?

Each vertex of a Feynman diagram corresponds to a particular term in the fields, like a four vertex of the field $\phi$ corresponds to a $\phi^4$

@Secret What? The evolution happens by the same means as all time evolutions in quantum mechanics.

@ACuriousMind but aren't the field operators we introduce at the beginning of QFT essentially creation/annihilation operators from 2nd quantisation in spacetime basis? :O

@Sanya If you come from the Weinberg point of view, yes they are

But you construct free fields from those operators

And then you use that notion of "field" to write down interaction terms in the fields

10:51 AM

ah ok - yeah, at that point things get different, I can see that - thank you @ACuriousMind

I'm glad to see that I at least learned a little bit during my qft lectures :D

Well, you should be aware that building the field from the particles is one way to do it, but not the only one. One can also start from a classical field theory, wonder how to quantize it, and arrive first at the field operators and then decompose the free fields into the c/a operators, and only introduce the notion of particle after all that is done

This is one of the reasons the notion of particle is so slippery in QFT - the computations come out the same, no matter if "particle state" is a fundamental or derived notion in your approach

yeah, I think our lecture tried showing both ways, but I didn't put thaaaaaat much effort into the lecture, so ...

I am curious, Is there exists a formalism that does not use particles nor classical fields as the building blocks. Similar to how in quantum mechanics of finite dimensional hilbert spaces, one start with the postulates of quantum mechanics and just work from there and only reference to classical quantities when the calculation finishes?

I mean, I heard from the discussions here that quantum fields are operator valued fields. Is there a formalism that begins with using these as the intrinsic(?) mathemtical objects so that one can work with the fields directly, isntead of building them from …

Another way to ask that is. are there abstract formulation of QFT such that the notion of classical fields and particles are derived from it?

@Secret The "abstract" formulation of QFT is simply quantum mechanics with infinitely many degrees of freedom

But we would actually not call all such theories "QFT"

The very characteristic of a QFT is that there are fields, if you want the axiomatics for the fields look at Wightman or Haag-Kastler

(and be aware that we can't actually show in any but the simplest cases that our QFTs fulfill those axioms :P)

11:06 AM

https://ncatlab.org/nlab/show/Haag-Kastler+axioms
O wow, so this is the AQFT you guys talked about 6 months ago

Mikael Kuisma

11:29 AM

@ACuriousMind Ah, that is how I pictured QFT, not the abstract QFT. I thought that QFT is going from countable degrees of freedom to uncountable ones. And this brings all the trouble mathemathical trouble such as renormalization and such. Like I said here physics.stackexchange.com/questions/273032/… in my section 5.