The h Bar

Wait, how's that information? That seems like a stream of jargon which surprisingly makes sense.

Wait, information does not necessary has anything to do with energy?

I mean, information is closely linked to entropy

@BalarkaSen Don't troll people with actual questions, please :P

@YuzurihaInori I am sure actual physicists can give you a better definition or physical examples which fits in with the definition. I just pulled it out of my ass.

@ACuriousMind Would you prefer I gave an infinity-definition?

@Secret Energy functional doesn't have anything to do with energy necessarily. Just replace it with any functional.

@BalarkaSen Well, I kind of guessed that.

@Secret He just strung some random jargon together.

2:05 PM

No, it wasn't a random jargon.

Ah ok, I am more used to hearing energy functional to refer to that specific functional that we often mimise.

@YuzurihaInori For that question to be really answerable you need to be more specific what sort of information you're talking about. To get general and probably unhelpful answers, see the answers to physics.stackexchange.com/q/2685/50583

@Secret I think it's what people popularly call as the action

I just know the mathematical side of the story so I don't know if I am guessing it right

Yeah, we physicists call that action and it is path dependent (thus a functional)

fahd

hi.. which is the app for simulating granular materials?

2:07 PM

action is just one of the many kinds of functional used in physics, but from the principle of least action, out comes equations of motion

@Secret Right, so the "moduli space" should be an appropriate space of paths.

@ACuriousMind Thanks for this. But all the answers I saw, mone actually allows us to define something like "the speed of information". How do we define that?

@BalarkaSen True.

@Secret The Euler Lagrange equations if I am not wrong?

@YuzurihaInori I have no desire to define something like the "speed of information" because I think it is a meaningless concept in general.

And the "gauge group" should be (and I am guessing on this) whatever the group of coordinate transformations that preserve the Lagrangian.

@YuzurihaInori yup for classical mechanics. In GR and quantum things get a bit weirder and I have nto fully grasp it

2:09 PM

@BalarkaSen Not coordinate transformations.

@BalarkaSen just write generalised coordinates and all's good

Gauge transformations usually act on the fibers of a principal bundle "orthogonal" to spacetime and its coordinate transformations

And not every transformation that preserves the Lagrangian is gauge

@ACuriousMind Hm, I know that's what it is classical gauge theory, but I have seen, eg, the group of reparametrizations of a path to be called the group of gauge symmetries of the space of paths equipped with the arclength functional

I mostly had the Riemannian story with geodesics in mind.

@BalarkaSen Ah, well, it depends on what you mean by "coordinate transformations". But yes, reparametrization of arc-length is also a gauge symmetry

Gotcha

2:14 PM

I, however need to revise how minimising the action in a space of paths will somehow give you something related to entropy, cause I don't really know...

I read a paper a while back that proves the existence the presence of two retarded fields that satisfy four field equations from just one continuity equation.

@BalarkaSen There are several physical notions of what a "gauge transformation" is/should be, some narrow and some general. It's terribly confusing, really

arXiv:0812.4785v1

Does the paper have merit?

@Secret I interpreted "information in physics" to just be a general question about what the mathematical information a "physical system" should carry. I wasn't thinking about entropy particularly

From what I knew, information is tied to entropy, so for that description above, it will mean action is tied to entropy somehow, but either I have not seen that relation, or I don't recall it

2:15 PM

how about a link

@ACuriousMind Hah.

No wonder.

@Secret Actually, entropy is kind of loosely defined as hidden information. So there's something going in under the covers

Q: Entropy and the principle of least action

19

Is there any link between the law of maximum entropy and the principle of least action. Is it possible to derive one from the other ?

ok, turns out they have a link with each other

though not as straightforward as the link between thermodynamic entropy and energy

Niiice

The link between entropy and information needs to be carefully considered - you need to be very clear about what sort of information you're talking about. For instance, purely thermal noise has large entropy but you might claim it has very little information (just its temperature)

Conversely, systems with low entropy tend to be highly organized and contain "lots" of information from a naive human viewpoint.

2:20 PM

Tbh, I don't understand much about information other than shannon entropy quantifies uncertainty and that information is related to the number of parameters needed to specify a state

I still don't understand what the link between Shannon entropy and thermodynamic entropy is.

But this actually mirrors the information-theoretic Shannon entropy: What one needs to realize is that when people equate entropy and information, they do not think about useful information - they think about "how well could I predict the rest of this info given part of it"

@BalarkaSen No one does I guess

@BalarkaSen It's literally the same formula :P

How's life

2:21 PM

In one case your states are bytes in a string, in the other physical states, but it doesn't really matter

Howdy

@ACuriousMind but there's nothing special about that formula

It's not even the important part of entropy

Sigh why did those get sent our of order

It's just to make entropy an additive property

I don't buy the formula argument

2:22 PM

You could just write the entropy as $\Omega$

It's just that it would be multiplicative then

Q: Entropy : subjective lack of knowledge that leads to objective conclusions

I stumbled upon this answer yesterday which seemed to make the link between the two types of entropy explicit:

1

There is something I really don't get about entropy. Let's consider a newtonian system (not quantum mechanics here). We can compute the entropy of a system via the formula $S=-\sum_l Pl Log(P_l)$ where $P_l$ is the probability to find the system studies in the configuration $l$. If we work at ...

thermodynamics statistical-mechanics information

it seems the thermodyanmic entropy is the maximum possible shannon entropy of the system

@Slereah Sure, but we like additive quantities :P

To be entirely honest I understand 0 about entropy. I don't even understand why the thermodynamic entropy is defined to be - at least classically - $\delta q/T$

Which we already discussed before

You could add the number of hair on your bums

that wouldn't be related to entropy

All of physics is just a big formula pushing without understanding anything man

2

It's all bollocks

2:25 PM

Also, $\Omega$ only works for systems in equilibrium. Shannon's formula also works for out-of-equilibrium states and nicely reduces to the $\ln(\Omega)$ in equilibrium

Entropy is related to the relation between microstates and macrostates, @BalarkaSen

Basically it's what it boils down to

How many microstates have a temperature of $T$

Versus $T'$

and then you can see that there are more accessible microstates with a homogeneous temperature than microstates where the temperature is concentrated in a region

I see

That's the basic idea

That's why you define the temperature as $$T = \frac{\partial S}{\partial E}$$

"how many microstates per energy level"?

This makes some sense

2:32 PM

that's why entropy doesn't really make sense for small systems

I mean, you can calculate it I guess

But the probabilities aren't gonna be overwhelmingly in favor of a specific case

uh, slereah, it think it shoud be $\frac{\partial E}{\partial S}$, what you wrote there is thermodynamic beta without the boltzmann constant

w/e

The 1-forms $dE$ and $dS$ are scalar multiples of each other. Just define $1/T$ as the real temperature :3

well tbh, $\beta = \frac{1}{k_BT}$ is actually more general. e.g. it can explain negative temperature systems easily

$\frac{1}{T} = \frac{\partial S}{\partial E}$ because $S$ is dimensionless, while $E$ is in units of energy, hence $1$ over a quantity in units of energy

2:36 PM

@bolbteppa I am putting you on ignore

I can't take it I'm sorry

Then you wont get an insanely physical explanation of entropy in terms of density matrices and Heisenberg uncertainty :\

Lol that's tempting. Ping it with me if you write one!

I think one subtle but important bit that's usually missing from the entropy formula is like

I'm going to duck out of the chat for a while

$$S = \ln(\Omega(E))$$

It's important to remember that it's the number of microstates of a specific macrostate

2:38 PM

"microstates are energy dependent (c.f. degeneracy)"

$$S = \ln (\Omega(E, P, \mu))$$

even

Or whatever thermo variables you use

Hmm... I wonder what happens microscopically when heat flow to cold in a heat engine is that the occupancy of states with high energy level decreases in the hot end and instead populating the empty energy levels in the cold end, and that stops changing when the two (energy levels along with degeneracy) balance with each other (and thus we get the same temperature)

2:50 PM

@BalarkaSen lol shut up and calculate™ physicstoday.scitation.org/doi/abs/10.1063/1.1768652

A good advice

A lot of people worry too much about interpretation and don't actually learn QM proper

Droleulb

@vzn what's with the trademark?

@Droleulb memes, man :P

About the blogpost that is generating so much discussion, I think we should put the focus on the action points (which—in the end—don't give a &^%* about groups, tribes, the design of your coupling gear, or the shade of your skin):

Let’s shift from “don’t be an asshole” to “be welcoming".

Can I be welcoming but an asshole

3:01 PM

Let’s do something about comments

Let’s make it easier for new users to succeed. <=== means talking abour the homework policy again, damnit.

I'll skim a physical explanation of entropy, lets see it helps: One way I see of motivating entropy is to notice that multiparticle systems (with large numbers of particles) require density matrices, not normal quantum probability distributions, because even in time-independent systems if you tried to set up stationary states the distance between energy levels would be smaller than Planck's constant,

Let’s stop judging users for not knowing things. (We’re a Q&A site!)

Let’s reject the false dichotomy between quality and kindness.

so since you can only approximate the energy to within a certain energy interval you use density matrices, reflecting the fact that you don't know exactly how the rest of the system affects a subsystem, so now if you try to solve for the density matrix of your system, you are f***ed, you can't determine initial conditions for it, however, we can use physics -

noting that because probability for independent events is multiplicative, the density matrix of a system composed of independent subsystems is multiplicative, and so it's logarithm is additive: but the only additive functions of the system which are independent for subsystems are constants,

energy, momentum, angular momentum, and particle number, and functions of these, thus the log of the distribution function can be replaced by a simpler additive function expressible in terms of energy etc which gives 'statistical' properties of our multiparticle system. This motivates us to express the original distribution (not just it's logarithm) in terms of energy,

but to then integrate it over an interval since we can't know the exact energy values, and to then use the fact that we are looking at a time-independent system so the distribution is roughly always equal to it's average value, going to zero quickly away from it, so that we can roughly approximate the integral as being equal to it's average value times the interval containing the largest number of states such that the distribution is roughly maximized,

but because probability has to be at most $1$, we can set our approximation equal to $1$, but then dividing by the distribution function and taking a logarithm we get the famous formula for entropy, $\ln (\Omega) = \ln \frac{1}{\rho(\overline{E})} = - <\ln \rho(E) > = - \alpha - \beta E - \vec{\gamma} \cdot \mathbf{p} - \vec{\delta} \cdot \mathbf{M} - \lambda N$

Basically, if you tried to deny entropy as a concept as much as possible, this kind of thinking makes it an absolutely unavoidable artifact of multiparticle QM, something you can't deny

Anyway, I'll try to bang up a little essay on the homework issue and figure out how to get it onto meta in a meaningful way. ::sigh::

@bolbteppa Sorry, I adhere to the one-electron theory

3:10 PM

@vzn speaking of this stuff, what do you think of vacuum polarization in the Dirac sea changing the measured charge on an electron, as observed in the Lamb shift, that is, an electron causing an electrostatic repulsion with the Dirac sea of anti-particles shifting the charge in that region slightly, or the QFT explanation in terms of particle antiparticle pairs around the election acting as dipoles shifting the charge?

@JohnDuffield if you can read that, what is your explanation

This is one of the more bizarre quantum things tbh

I'd be amazed if there wasn't a vixra industry on this stuff

(Probably a better way to motivate integrating the actual distribution, expressed as a function of energy, is that you should only really work with it's average if you're expecting to relate it to that simpler function of additive quantities which represents statistical properties)

3:31 PM

Don't disturb the beast

@Slereah what's that?

@0celo7 It's a vague theory of... Wheeler, I think?

It never went anywhere, but the basic idea was that, since the positron was the time-reversed electron

Maybe there was only one electron in the universe

Who kept going forward and backward in time

Emitting a photon in the process

I think it was part of the whole Feynman-Wheeler trip of doing all of physics as a time symmetric exercize

3:46 PM

@Slereah I need to invent an invariant notation for higher order tensor

fuck indices

@0celo7 Penrose

I'm working with nonsymmetric things and it's bad

$$\int_\Omega A^{ij}_{\mu\nu}\theta_\alpha^2 \nabla_i u^\mu \nabla_ju^\nu$$

show me how do write this in Penrose

this is already a typo but whatever

work your magic

Droleulb

is there any material that can't take a static shock?

your soul

Droleulb

trust me i take like 30 a day

3:52 PM

Something like that, I think

loooooooooooooooooooooooooool

@Semiclassical

a bit poor but that should be the correct shapes

$\int \theta^2 \mathrm{tr} (\nabla u A \nabla u)$

I've been writing $A(\nabla u,\nabla u)$ in notes

also works

3:55 PM

it's fine when it's two $u$'s, but gets confusing when there's a $v$

the "correct" order should be $A(\nabla u,\nabla v)=A^{ij}_{\mu\nu} \nabla_j u^\nu \nabla_i v^\mu$

complete pain

4:33 PM

@bolbteppa yes a key zen question, thx for asking, there is an interpretation/ explanation in terms of spacetime fabric + fluid dynamics + pilot wave hydrodynamics + solitons + madelung fluid etc... lets chat about it sometime when you feel like discussing it without mocking it :)

@vzn thanks, would have been shocking if there wasn't some pilot Madelung fluid perspective on this haha

How can RQM be wrong if something this bizarre actually explains experimental results, this is another level crazy if you think about it

4:54 PM

@bolbteppa so do you want to hear more or is this a physics joke

@vzn what is your explanation for it

^ is the qft picture

is more rqm

5:24 PM

Greetings

@PrittBalagopal Hi, there.

Hello @dmckee

I'm learning about Schrödinger's equation and I have a doubt

Yeah ...

I know the equation is $-\frac{{\hbar}^2}{2m}\nabla^2 \psi+U\psi = E\psi$

Can we rewrite that as $\frac{{\hbar}^2}{2m}+K\psi = 0$?

Sorry if I made any mathjax mistakes, I'm on my phone, so I can't see it rendered.

Where K is kinetic energy.

Anonymous

5:29 PM

Hmmm ... the first term is the kinetic term. In classical mechanics $K = p^2/2m$, right? Then in wave mechanics $\hat{p}_x = i\hbar \frac{\partial}{\partial x}$, so the first term is the kinetic energy

Anonymous

@PrittBalagopal What is K?

@Blue New-fangled way of writing $T$.

That is the kinetic energy of a particle.

Oh I see.

BTW @PrittBalagopal, if you don't see $K = p^2/2m$ right off you should check that this reduced to the expression you are used to.

Anonymous

@dmckee I meant there seems to be something missing in the first term of this expression he wrote: $\frac{{\hbar}^2}{2m}+K\psi = 0$

Anonymous

5:31 PM

The Laplacian just vanished

@PrittBalagopal This only answer of mine offers a way to understand why the Schrödinger equation looks the way it does.

Oh yeah sorry about that

@Blue I couldn't see the rendered form, sorry.

Anonymous

@PrittBalagopal No problem

Thanks for the link dmkee, I will look at that now.

Ahh now it totally makes sense

You use the momentum operator to define the kinetic energy operator.

@PrittBalagopal Be aware that I've written $T$ for kinetic energy and $V$ for potential energy in that post. Those used to be nearly universal conventions, but in recent years that has been changing.

5:36 PM

@bolbteppa ofc have my own ideas but try starting with this, just googled for something "close/ nearby" -- Hydrodynamics of the Physical Vacuum: II. Vorticity dynamics / Sbitnev arxiv.org/abs/1603.03069

@PrittBalagopal Bingo!

Woo, I got it!

But still

We can equate it to the classical kinetic energy formula right?

Basically what I wanted to do was to solve the hydrogen atom Schrödinger's equation in one dimensions. (Trying to avoid the x y here)

Anonymous

@PrittBalagopal There are actually some subtle things going on when you write $T=(\frac{\hbar}{i}\frac{\partial}{\partial x})^2/2m=-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$

We know kinetic energy as a function of distance from the nucleus

Anonymous

Squaring the first derivative operator giving the second derivative operator...

5:40 PM

Indeed.

Anonymous

@PrittBalagopal Can you justify it? :)

You can certainly write $\hat{K}\Psi = \left[E - V(x)\right]\Psi$ as that simple re-arranges the TISE, but you seem to be asking if you can then pass to the classical situation $K = E - V$. Is that right?

I think that Ehrenfest's theorem gives you $\langle K \rangle = E - V$, but that does not guarantee any given measurement.

Unfortunately no, you got me there. I never really looked into that. @Blue

Yes @dmckee that's what I'm asking

Sorry for delayed response, it's nerve wracking to render all that latex mentally.

And I don't have a PC :(

Anonymous

You can render latex on your mobile browser too

I can?

Please enlighten me

Anonymous

5:46 PM

math.ucla.edu/~robjohn/math/mathjax.html

Anonymous

Scroll down

The energy-eigenfunction of the 3D system are not kinetic energy eigenfunctions (unless the particle is free $V(x) = V < E$), so you won't generally measure the same kinetic energy at any given attempt.

Yes, I understand that.

But we can find kinetic energy alone from the Bohr's model formula right? We know $mvr = n\hbar$

So we can get v, which we substitute in $0.5mv^2$

Is this approach incorrect?

@Blue it's still not working :(

Anonymous

@PrittBalagopal Weird

Anonymous

Works on mine

Anonymous

5:53 PM

I use the standard stock android browser

Anonymous

Not chrome

@Blue Why have you deleted your SE account?

Anonymous

@Abcd Making it look professional by deleting embarrassing questions history XD

@Blue When will you activate it then?

Anonymous

@Abcd Enable what?

5:56 PM

@Blue your account

Anonymous

My account is active on Stack Overflow, Math Overflow and Quantum Computing SE. Just this chat profile is not linked to those accounts.

Anonymous

I hardly use the Physics SE main site. Don't think I will be needing it anytime soon

@Blue Whenever you will enable it, the questions will be back

Anonymous

@Abcd Nope. Deletion is permanent

Anonymous

The questions still exist. But they are anonymized

5:58 PM

If you want to get rid of your past (questions), you may create a new "Blue" named account

@Blue No, I have had the experience

All my questions came back non anonymised

(my previous (now deleted) account)

Anonymous

@Abcd I have some experience too O:) : physics.stackexchange.com/users/194046/blue?tab=profile

though they had got anonymised for the time I had deleted my account

@Blue wow

you lost 2k reputation

@PrittBalagopal Bohr's model is deeply flawed and you should probably move away from using it as a reference, but in this instance I believe it gives you the correct expectation value (i.e. agrees with Ehrenfest's theorem).

Anonymous

@Abcd I think what you're talking about is account hiding.

Anonymous

That's different

Anonymous

6:01 PM

@Abcd Yup :P

Ahh okay @dmckee I will remember that. I'll try to avoid using the Bohr model.

@PrittBalagopal Are you a chemical engineer?

Anonymous

@Abcd Well, a student :P

@Blue Student of chemical engineering?

I am entirely made of chemicals

6:06 PM

who isn't?

Those energy being in Star Trek

Anonymous

@Abcd Yes

Anonymous

@Slereah You're made of dark energy

@Blue wow, I wanted to ask him what they teach in Chemical Engineering

Anonymous

That makes you so mean

Anonymous

6:07 PM

;)

Anonymous

@Abcd You definitely should! Always good to talk to relevant people to get an idea of the field you're about to dive into

I am having some problems with azure

It is refusing install

Anonymous

(I find chemical engineering boring though)

I can not do it as root either

My stack is python based

what do I do?

@Blue do you have any idea about it

6:09 PM

I am attempting to integrate and modify some software with my stack

I guess I have to solve this myself

ok

Anonymous

@Abcd Well, one thing; my friends who went into chemical engineering hoping that they'd get to learn lots of chemistry, were disappointed. It's more of applied physics, thermodynamics and stuff

Anonymous

Also, you should check if you're interested in its sub-branches like petroleum engineering, renewable energy engineering, etc

Anonymous

Oh and also pharmaceutical, textile, and all those

Anonymous

@Abcd Check out a sample syllabus if you wish

@Blue you are suggesting such boring branches

I dont even know the names of basic fibres lol

like what is polyster

Keep these mind

6:24 PM

Psychopharmacology

Psychopharmacology (from Greek ψῡχή, psȳkhē, "breath, life, soul"; φάρμακον, pharmakon, "drug"; and -λογία, -logia) is the scientific study of the effects drugs have on mood, sensation, thinking, and behavior. It is distinguished from neuropsychopharmacology, which emphasizes the correlation between drug-induced changes in the functioning of cells in the nervous system and changes in consciousness and behavior. The field of psychopharmacology studies a wide range of substances with various types of psychoactive properties, focusing primarily on the chemical interactions with the brain. The term...

Boring^^^

Anonymous

@Abcd It is boring, unless you're interested in entrepreneurship later on. :P Chemical engineers normally take more business and economics courses than other engineers

Anonymous

At least, here

Anonymous

So it's useful perhaps if you wish to transition into industry

yaaaaaaaaaaaaaaaaaaaaaassssssssssss!!!!! Solved it like a boss

I am king of the cloud!!!

hehehehe

Anonymous

6:26 PM

Otherwise if you're interested in Chemistry wholesale then better take it up as a stream

@Blue what do you mean by "wholesale"

@Blue ya, chemistry is the most interesting subject

Anonymous

@Abcd Chemistry without the "engineering" part

Anonymous

Like there are organic chemists, physical chemists and so on

@Blue how many years does it take to become one

Anonymous

I think Pilani offers a dual degree along with chemistry. So maybe if you're interested in computer science along with chemistry then you could take that

Anonymous

6:30 PM

@Abcd Well, a PhD is mostly necessary if you wish to become a "chemist"

@Blue BITS Pilani was so low in this year's ranking :(

Anonymous

On average 10 years maybe

DId you see the college ranking of 2018

Keep these mind

@Abcd I'd say that, in general, the fun is in becoming rather than being.

Anonymous

@Abcd Rankings are useless

Anonymous

6:31 PM

Don't trust them

Anonymous

Last year VIT was ranked above so many good colleges

Anonymous

Hilarious

@Keepthesemind Yeah, but it shouldn't take like 5-6 years. I want to complete it in 4 years

Anonymous

@Abcd In IISERs you can choose a major in your 3rd year, be it chemistry or physics or any other science subject. If you're unsure, that's a good option too

Anonymous

They're doing really good research nowadays

Anonymous

6:33 PM

I know some fourth yearites from IISER Kolkata. They all have terrific research profiles

Anonymous

@Abcd In 4 years you can't become a "chemist". That's for sure.

Anonymous

Getting an undergrad degree doesn't make you a chemist

@BalarkaSen listen to the new JRE with Abby Martin

Not Nice

Mithrandir24601

@Keepthesemind Another way of putting it is that if you don't like the journey, you're not overly likely to like the destination either

Anonymous

6:51 PM

@Mithrandir24601 Except that there isn't any legitimate destination ;) graduation -> higher degree -> post grad period -> waiting for tenure -> publish or perish -> death :P If one doesn't like their subject during graduation they'll hardly like it later on

Anonymous

Well, we'll die anyway. Nothing matters :P

Mithrandir24601

@Blue Replace 'destination' with [whatever goal someone is aiming towards] :P

7:32 PM

@Blue @Abcd BITS Pilani, Goa Campus has one of the finest Theoretical Physics faculties in the country and offers MSc in Physics in 4 years!

Anonymous

@SwapnilDas Yes, that is true. By theoretical physics you're referring to high energy physics I guess though.

Yup. CM is also good

Anonymous

I wonder who does string theory stuff in India

Anonymous

Other than the math institutes like IMSc and CMI

TIFR

HRI, ICTS too are homes to stalwarts.

Anonymous

7:37 PM

They have a string theory group in TIFR?

Anonymous

Never heard of them

Of course!

Anonymous

Maybe a couple of individuals

theory.tifr.res.in/members.php

ICTS is a branch of TIFR anyway.

Anonymous

@SwapnilDas I see only Mandal Gautam

7:38 PM

Shiraz Minwalla?

They don't have detailed page actually. TIFR DTP is the most sought after in the country.

@Blue Do you know something about Non-Linear Dynamics?

Anonymous

@SwapnilDas Ah theory.tifr.res.in/~minwalla

Anonymous

Nice

I really donno what exactly is that:/

Anonymous

@SwapnilDas From the math side a little bit

(NLD)

What is that about?

Anonymous

7:40 PM

You should read Hirsch Smale

Anonymous

It's a good introduction for dynamics

Ok. But what does the subject study?

I mean what kind of mathematical/physical systems?

@Blue Is it true that those who pursue string theory should have a UG/PG degree in Maths?

Anonymous

@SwapnilDas Well. Tons! For example in basic control theory we assume the initial conditions to be relaxed and the output to be proportional to the input. And that extends to all electrical circuits where given a certain range of inputs you can make predictions about the whole system. But real circuits tend to be non-linear

Anonymous

Also maybe you heard of the Navier Stokes equation

Anonymous

Finding an exact solution to that is an open problem

7:47 PM

Millenium Prize Problem right?

Anonymous

It is not only useful in fluid dynamics but also other areas like predicting trajectories of objects in flight motion

yes.

Anonymous

@SwapnilDas Yup!

:)

Anonymous

Control theory is a very recent subject though

7:49 PM

Hey, can you answer that string theory thing I asked just now?

skull

get well soon @SirCumference

Anonymous

@SwapnilDas Well it's not a necessity. But surely string theory is (one of) the most math heavy branch of physics out there. A math graduate is mostly better suited. You should really talk to ACM about this :P

Anonymous

I don't claim to have any knowledge about string theory

ACM ?

Anonymous

@SwapnilDas ACuriousMind

7:54 PM

Oh yeah.

He made a transition to Software Industry right?

( It made me sad :( )

Anonymous

Yeah, he sold his soul

Yeah.

What are your plans?

I mean what would you do after Btech?

Anonymous

@SwapnilDas From what I know pursuing string theory academically is a very difficult and demanding career path.

Yup, I'll never try :P

@Blue So you'll be doing Engg or Phd in Physics in future?

Anonymous

@SwapnilDas reddit.com/r/Physics/comments/271apx/… (I don't know how much of that is true, but an interesting view nevertheless)

Anonymous

7:59 PM

@SwapnilDas Haven't thought very far yet. Maybe mathematics or computer science rather than pure physics

oh great. Good luck!

Anonymous

quantum computing, information, engineering etc. look interesting too

Anonymous

I seem to like interdisciplinary subjects more

Oh. nice

Physics Meta

8:38 PM

How can we "make it easier for new users to succeed"?

4

Emilio Pisanty

8:58 PM

@PhysicsMeta @dmckee I'm supremely un-fond of this

> 1. Change the policy in a way that reduces the barrier.

please consider changing it to

1. Change the policy in a way that reduces the barrier, without opening the floodgates of venue-consuming low-level problem-solving

@EmilioPisanty Well, I'm going to assuming (for argument's sake) that there is a way to smooth the barrier more obvious for users willing to work at it without opening the floodgates for those who aren't.

I'm not sure I really believe it.

@EmilioPisanty Done.

Emilio Pisanty

@dmckee ah, good. My phrasing was maybe a bit hyperbolic =P, but it does work.

my concern is that there's plenty of folks out there who read the homework policy as mean-spirited evil people who are out to stop people from learning, and they're often completely unwilling to consider the fact that a stream of low-level homework does erode any venue.

so the phrasing on a post like this should give as few footholds to "just allow all homework always, you evil people"

Well, personally, I think the answer has to lie with scaffolding and guiding question askers—I just don't see another way—but coming from a mod the question has to invite constructive input from people who feel differently or we'll never make progress toward a broad consensus.

Not that we've had much luck butting our heads against that wall in the past. But my headache has cleared up enough to have another go, and the blogpost gives some context for it.

I really hope that the team will consider us as an early target for any experimental question-asking wizard they put together.

2

But I suspect we'd have to provide rather a lot of thoughtful input for such a thing.

Emilio Pisanty

@dmckee yeah, I think I'm still nursing some forehead bruises from the last go-round, I'll probs sit this one out

9:20 PM

9:46 PM

@skull Thanks a lot :)