The h Bar

12:22 AM

@ACuriousMind There is an English word "orca", so maybe the question is about the biophysics of marine mammals ... — alephzero 31 mins ago

Without the moon, would there be more killer whales?

Emilio Pisanty

12:33 AM

@dmckee can we perma-star that?

@dmckee are y'all explaining the Feynman ratchet to @pZombie?

Richard Feynman Messenger Lectures at Cornell The Character of Physical Law Part 5 The Distinction of Past and Future

►

if you haven't seen it

particularly from minute ~22

ACuriousMind

@EmilioPisanty I'm sorry that I gave you the impression that I'm attacking your answer behind your back. I think that we do not differ on any formal aspects, but that we differ on what the phrase "valid state" means. Neither your nor my answer are comprehensive treatments of the nature of "rigged states". You're focusing on the aspects of it where it looks just fine as a state, I'm focusing on a position where it produces more trouble than it worth to say that its a state.

Emilio Pisanty

@ACuriousMind I'm sorry, but I don't see how you thought that text was anything other than what you describe in your first sentence here.

Apology accepted, of course.

But as I said I'm not particularly in a debate on the merits at this stage.

ACuriousMind

12:51 AM

@EmilioPisanty In reading the question I'm supposedly answering again, I agree that the first part of my answer was neither necessary nor kind. I've removed it.

Emilio Pisanty

@ACuriousMind appreciated.

ACuriousMind

I might post an answer to the question you answered later, but I feel it would at best devolve into unproductive prescriptivism about what "a state" is supposed to mean. As long as we agree that it is not physically realizable, it's probably a distinction without a difference.

Emilio Pisanty

@ACuriousMind that's probably where the disconnect is.

in that connection, it's worth noting that the position "$\delta(x)$ isn't a state, only things in $L_2(\mathbb R)$ are states", while a consistent position to take, is ultimately pretty empty.

ACuriousMind

You're gonna say that the ones that don't fall off towards infinity are not realizable either, aren't you?

Emilio Pisanty

I mean, the boxcar $\psi(x) = \theta(x)\theta(1-x)$ (for $\theta(x)$ the Heaviside function) is in $L_2(\mathbb R)$, and for huge stretches of QM it is every bit as bad as the Dirac delta

in terms of the Gel'fand triplet, if you want to get fussy, take "state" to mean only the inner-core test-function space. Otherwise, just expand it to the outer layer of functionals.

It makes no sense to me to keep it in the middle layer, and I'm having real trouble identifying any arguments that can be put forward to prop up that position.

Though, as you said, since neither is physically realizable, it's a purely-semantic distinction without a difference where it's pretty meaningless to argue.

ACuriousMind

1:01 AM

@EmilioPisanty My main reason for identifying "state" with "vector in a Hilbert space" is that if you do QM based on the abstract algebra of observables, then a state is a positive linear functional on the algebra, i.e. a prescription of how to take expectation values of operators. I have a hard time identifying why it should matter that, in the particular $L^2$-representation, the states represented by Schwartz functions should be special

Emilio Pisanty

@ACuriousMind that's a fair perspective to take (though nothing you've mentioned so far hinted at it).

Still, how is (the density operator corresponding to) the boxcar function any more acceptable to the Dirac delta?

ACuriousMind

@EmilioPisanty What unacceptable aspect of it are you thinking of?

Emilio Pisanty

@ACuriousMind it has no expectation value for $\hat p^2$?

you can juuuuuuust about stretch things to argue that it has an expectation value of $\hat p$, but if that's what you're reaching for...

ACuriousMind

@EmilioPisanty It has one for $x^2$, and since its Fourier transform is again a box, you can also compute the value for $p^2$ for it. You just have to stop insisting that the only densely defined operator $\partial_x$ is a good thing to apply to it.

Emilio Pisanty

> its Fourier transform is again a box

what?

ACuriousMind

1:09 AM

errrrr

Emilio Pisanty

the Fourier transform is a sinc function

that still makes $L_2$ (as it must)

but none of its moments are well-defined

@ACuriousMind so yes, some flavour of this.

ACuriousMind

I think I may have misinterpreted the GNS construction all these years.

I thought it was saying that if I represent the algebra of observables as an algebra of operators on a Hilbert space, then every vector in that space defines a state on the algebra. But maybe it just says that all the states map to vectors, not vice versa.

Emilio Pisanty

@ACuriousMind I don't know the GNS construction well enough to say.

But you don't need fancy shenanigans with boxcars - all you need is the position moments of $\psi(x) = 1/\sqrt{1+x^2}$.

ACuriousMind

I mean, on some level I guess I always knew that $L^2$ has elements with ill-defined momenta

But I never made the connection that this should mean it's not a state in the algebraic sense

Emilio Pisanty

@ACuriousMind $L^2$ has a ton of ugly-ass states, for sure

ACuriousMind

1:21 AM

And that, in turn, would mean that "the Hilbert space" is really just a representational crutch since dealing with functions has different machinery available than dealing with abstract algebras

David Z

2

I can't help it

ACuriousMind

lol

Really though, this is a "SR/GR is all about the metric" level realization for me. Nothing in these weird thought experiments ever made sense until I grasped that.

NormalsNotFar

1:59 AM

Hi people. I have my first conference presentation in a few day (25 + 5 mins, Hep-th ). I was wondering if, in your opinion, it's ok that I use a script for what I'm going to say? I will try to remember most of it, but I have a bad memory for these sorts of things, and public speaking makes me really nervous and forgetful.

dmckee

@NormalsNotFar You should deliver practice talks several times. For two reasons.

First it will help you internalize the talk, so that you don't need to carry written notes.

But perhaps more importantly, at least a couple of those practice runs should be given to audiences composed of your peers and mentors as they can provide (hopefully gently stated) feedback on what your are doing that isn't working.

ACuriousMind

@NormalsNotFar Public speaking makes most people nervous, even those who do it well. Do practice the talk - with or without a script - but reading off a written script will come across as rather wooden.

dmckee

Among particle physicists there is a strong culture of cracking the whip over students about this (and therefore of seniors coming to the practice talks even when they would rather be somewhere else).

ACuriousMind

If you can "read a script" while not appearing to do that, great. But if you can't, you should prevent that situation by only going off bullet points

dmckee

By the time you've been raked (every so kindly) over the coals three times and finally got a green light on the fourth run of the practice talk to your group, presenting to the "public" is not actually anticlimactic, but a lot less traumatic than you expect.

And your talk will be better for the experience, too.

ACuriousMind

2:07 AM

The first few presentations I had to give also required a formulated-out version of the talk - a script, if you will - to be handed in. I found that writing that was great in preparing me for the talk, but actually having it in front of me wasn't necessary in the end

dmckee

^ That was my experience as well. Between having to actually craft the words for approval and then practice them I was so prepared that I was able to handle interruptions and technological foul-ups smoothly.

The writting made me put my thoughts into better order and the practice talks gave me the confidence that I could articulate them on my feet.

ACuriousMind

So do write the script, but consider not actually bringing it on stage

NormalsNotFar

Thank you for your sharing your thoughts! I have practiced it a lot, and given it to some of my peers, hopefully my supervisor soon too.

I think I could do it from just the bullet points, but there is one issue. I feel that everything I have included in the script is important and every sentence has its purpose. So I don't want to forget anything, or say it in the wrong order.

dmckee

The slides are your visual cues, try to practice the words with the matching images in front of you.

ACuriousMind

@NormalsNotFar A talk is different from a paper. Not every sentence has to be perfectly formulated, not everything has to be perfectly argued. When you're presenting, engaging the audience is equally important as being factually accurate. By formulating your actual sentences on the fly, it will feel much more like you actually talking to the audience.

And yes, doing that can be scary. I still get stage fright five minutes before I'm supposed to talk every time I talk in front of an audience. It's normal. It's gonna be fine.

bolbteppa

2:19 AM

@vzn, you know who doesn't ignore solitons:

2

Q: M branes/D branes are solitons?

I'm really confused. In M theory/String theory, the fundamental objects are M/D branes. However, branes by defintion are just solitons. Solitons are just waves that maintain there shape. So if a brane is a soliton wave, then what is it a wave of? Would it be the excitation of a field?

NormalsNotFar

Ok you have both convinced me. I'm going to go practice it some more without the script. Thanks very much for your help guys. Enjoy your Sunday!

dmckee

Absolutely true that stage fright is normal. Queasy stomach. Shaky knees. Cold sweats. Tunnel vision. I've had them all.

vzn

@bolbteppa lol did you read the answer? :P

Mo_

3:01 AM

If a potential field goes to infinity when $r\to \infty$ is there any possible way for a (classical) particle to get unbound and escape to infinity?!

pZombie

According to those who understand QM, does a ideal gas in a box have a finite number of states or are the states infinite?

ACuriousMind

@pZombie Even the states of a single particle in a box are infinite since energy is unbounded from above.

(If you want to be pedantic, even a two-level system has infinite states because $a\lvert 1\rangle + b\lvert 2\rangle$ is a valid state for every $a,b\in\mathbb{C}$ :P You're probably asking about the dimension of the space of states rather than the "number" of states )

Mo_

3:16 AM

@ACuriousMind ^ Any thoughts on my question? I'm solving a problem that has explicitly asked to find the temperature at which a particle can escape a logarithmic well!

ACuriousMind

@Mo_ My intuitive answer would be no, but if I was certain or had reasoned through it in any detail, I'd have responded without any prodding.

pZombie

@ACuriousMind Maybe i should rephrase then. An ideal gas inside a box with a defined total energy, does it have infinite states still?

or did you mean that a single particle can take up any energy level out of chance?

ACuriousMind

I'm not quite sure what you're getting at, and I'm afraid I'm not up to figuring it out at 4am

pZombie

3:36 AM

i guess you get what you paid for

dmckee

4:58 AM

@pZombie Computing the number of accessible states (eigen-)states (to be used in expressing the entropy) in such a situation is a standard problem in statistical mechanics (see "micro-canonical ensemble"). But notice the adjective there: "accessible" means we are looking at a restricted sub-space.

ZeroTheHero

6:25 AM

can I ask a quick questions to the mods lurking around?

I am reading this post: physics.meta.stackexchange.com/questions/4932/…

and it references a “flag” review queue.

Has that feature been disabled?

John Rennie

6:38 AM

@ZeroTheHero This page?

ZeroTheHero

yes. I don’t see any “flag” revies

at least not on my page

It’s not as if th eworld will come to an end... just curious. I did not know there was such a review queue, and I’ve certainly never accessed it.

Anonymous

7:39 AM

121

Q: Let's get rid of the 10K flag queue

The 10K tools are pretty cool... You get a birds-eye view of activity on the site, a "dashboard" view of what's happening. Some of the individual tools haven't scaled particularly well with Stack Overflow's growth, but the concept behind them is still sound: we trust you to enough to be a bee wat...

feature-request status-completed review moderator-tools flag-queue

Shing

hey guys, I need some help here. I am so confused when I calculate this:

consider two highly localized identical particles:

(i.e. right after measuring their positions)

and assume they are far enough

I calculate their probability density, and get so confused

given $\Phi(x_1,x_2)=\frac{1}{\sqrt{2}}\bigg(\psi_1(x_1)\psi_2(x_2)\pm\psi_1(x_2)\psi_‌2(x_1)\bigg)$

where $\psi_1$ is for particle 1

and $\psi_2$ is for particle 2

suppose particle 1 is highly localized at $x_1$

and particle 2 is highly localized at $x_2$

then $p(x_1,x_2) = \frac{1}{2} \big(|\psi_1(x_1)\psi_2(x_2)|^2+|\psi_2(x_1)\psi_1(x_2)|^2+2\Re[\psi_1(x_1)\psi_‌2(x_2)\psi_2(x_1)\psi_1(x_2)]\big)$

however, since they are localized and distant away

so $\psi_1(x_2)$ and $\psi_2(x_1)$ are approaching to zero.

so then we have $p(x_1,x_2) \approx \frac{1}{2} |\psi_1(x_1) \psi_2(x_2)|^2$

but that's probably wrong....

there shouldn't be a 1/2

however, I have no idea which part of steps gone wrong

would anyone be kind enough to give me a helping hand here?

John Rennie

Your expression looks fine to me. Remember that $\Phi$ contains two particles so it normalises to 2.

Shing

7:58 AM

@JohnRennie thanks! Let's try to calculate it again

I mean "let me", ha sorry, english is not my 1st language

PM 2Ring

9:55 AM

@JohnRennie Nice answer. And I guess it also applies to this question. Is it reasonable (& helpful) to say that when 2 quantum particles collide they effectively measure each other's position?

John Rennie

@PM2Ring I think the second question is a bit different because it's best explained using quantum field theory. QFT would have been overkill for the first question :-)

PM 2Ring

Oh, ok. Is that because in the 1st question we can simplify things by pretending the atom is at a fixed location, so we're basically just firing a probe electron at an electron in a potential well, but in the 2nd question we don't have a simple potential well, we have 2 sets of particles and we need to use creation & annihilation operators to say what the field is doing?

John Rennie

10:11 AM

Let me have a go at answering the second question ...

PM 2Ring

Thanks!

John Rennie

10:34 AM

@PM2Ring done, although I'm not sure it's worded as well as it could be. When I have more time I'll revisit it.

PM 2Ring

@JohnRennie It looks ok to me, although I really don't have the QM knowledge to judge it properly. ;)

Physics Meta

12:51 PM

0

Q: I am not able to understand what should I do on this question to avoid it from being downvoted?

Please help me understanding about How I should improve my questions like this- Which torque is making this disc to rotate and how does it operate? from being downvoted? 1-I think I have given the specific difficulty How does the torque operate and which torque is it? 2-Shown my work. 3-Have...

discussion questions specific-question down-votes

Anonymous

1:28 PM

In case someone is interested in some Bell stuff: Why is $P(1,2)_{\text{same}} = \frac{1}{4}$ and not $\frac{1}{2}$ in Preskill's Bell experiment?

Anonymous

I'm sure I'm missing something trivial... :/

Anonymous

1:43 PM

Hmmm, maybe $P(1,2)_{\text{same}}$ doesn't mean what I think it means. This is confusing.

Physics Meta

2:12 PM

-3

Q: Please see if this question is suitable to ask on the main Physics.SE

How does torque operate? I am not asking you to solve this. Just give me the situation idea-what will be going on with disc? Please see if now this question is okay to be asked about? I have edited the previous question (suggested by Kyle Kanos). And what other things should I modify to make it ...

discussion questions specific-question homework-and-exercises down-votes

Jake Rose

2:41 PM

@JohnRennie do only quadratic dispersion relationships have pulse broadening?

actually, scratch that

do linear dispersion relationships have pulse broadening?

John Rennie

3:39 PM

@JakeRose I have no idea. Sorry.

I think any dispersion causes pulse broadening.

pZombie

3:55 PM

Starting with a box containing an idle gas at some very low entropy state. The gas is all gathered at one corner of the box. As i understand it, the gas now has some so called accessible states with a higher entropy value. I assume each of those accessible states have a probability assigned to them.

If the gas takes one of those accessible states with a higher entropy value, is the former state with a lower entropy value where the gas was sitting at the corner now one of the accessible states (with a low probability) OR are only states with a higher entropy value accessible from this point?

If the former state is accessible still then everyone writing that entropy always increases is simply wrong. That would be like saying that a coin never lands 20 times on heads in a row

Also, once the gas in the box reaches a so called maximum entropy state, from there on, are there only maximum entropy states available? Do lower entropy microstates become unavailable completely by some magic law once the gas is at its maximum entropy state?

If lower entropy microstates are available at a maximum entropy state, then what is the ratio of available lower entropy states to maximum entropy states once the gas reaches its maximum entropy state?

if, when the gas is at a maximum entropy state, the total of accessible lower entropy states are higher than the total of accessible max entropy states, then the 2nd law would simply be false. Because now we have a system where it is more likely for entropy to decrease rather than increase

actually, we have a system where the only way is for entropy to decrease, since you cannot increase the entropy from a max entropy state by definition. At best, it would remain the same indefinitely

again, the 2nd law would have to be stated more precise as in entropy always increases or remains static

Emilio Pisanty

4:34 PM

@JakeRose linear dispersion is a delay

anything else will (in general) cause pulse broadening

Jake Rose

Delay?

Emilio Pisanty

@JakeRose temporal delay

or what exactly do you mean by the question?

PM 2Ring

4:49 PM

@pZombie 20 heads in a row is just a 1 in a million chance. Try doing a million heads in a row. And even that's nothing compared to a cubic millimetre of gas.

pZombie

@PM2Ring decrease the gas volume further then. I doubt that any mathematician would ever say that 1 million heads in a row is impossible. Just unlikely yet bound to happen given enough coin flips

John Rennie

@pZombie the entropy has statistical variations about its average value, with the timescale of the fluctuations being of the order of the time between collisions between the gas molecules.

The second law is generally taken to mean the average value (averaged on timescales that are long compared with gas molecule collision times) never decreases.

pZombie

even if it is about averages, never decreases would be wrong still

John Rennie

In principle you are correct that never decreases is really exceedingly unlikely to decrease, but given that the expected time required to observe a significant decrease in entropy is longer than the age of the universe most of us are happy to go with never.

I should note that while there is a certain amusement factor in being pedantic it will not win you friends.

pZombie

@JohnRennie what about when the gas is at its maximum entropy state? Are there more states accessible with the same maximum entropy value than there are states accessible with a lower entropy value?

i am not trying to be pedantic, just precise

but even if there were not, at a maximum entropy state, since you agreed that lower entropy states are available still, the entropy can only either remain the same or decrease

John Rennie

5:04 PM

When we talk about a state we don't really mean a state in the sense of a position in configuration space (a microstate). We mean a set of indistinguishable microstates. That is, if we swapped some gas molecules around the gas overall would still look the same.

pZombie

which means that the 2nd law of TM simply does not apply to a gas at a max entropy state

John Rennie

Then if $W$ is the number of microstates the entropy is the famous $k \ln W$.

So the maximum entropy state is the one that has the highest number of indistinguishable microstates.

ACuriousMind

@pZombie What do you mean by a "state of maximum entropy"? Since temperature is unbounded from above, so is entropy.

pZombie

@JohnRennie And that is what i am asking. Are the number of accessible microstates with a max entropy value higher than the number of accessible microstates with a lower entropy value once the gas reaches max entropy?

John Rennie

If a gas is at the maximum entropy state then by definition the entropy can't increase, but that doesn't cause us any problems with the second law because all the second law tells us is that the entropy won't decrease.

pZombie

5:08 PM

@JohnRennie Then you contradicted yourself. Formerly you said that there is a chance, even if unlikely, for a gas to take on a lower entropy state

@JohnRennie If that was the case, and since the gas is at a max entropy state, the only way would be towards lower or remaining equal

Emilio Pisanty

@ACuriousMind well, if you've got a finite system, then entropy does max out, and if you keep increasing the temperature then eventually it will be infinite and then negative

(though to be clear, a gas is not a finite system.)

John Rennie

@pZombie You may assume that when I say entropy never decreases I mean you'd have to wait many times the age of the universe to see any measurable decrease. If you are unwilling to accept my terminology we should probably both find something more productive to do.

ACuriousMind

@EmilioPisanty sure

pZombie

No, i might not have to wait several times the age of the universe. What if we just happen to live in a universe where the unlikely happens more often than in your average universe? Don't all possible universes exist according to QM? Isn't the same argument often brought up by physicists to explain why some constants seem to be tuned so finely?

Emilio Pisanty

> What if we just happen to live in a universe where the unlikely happens more often than in your average universe?

that's not how probability works.

you're just asking "what if we just happen to live in a universe where probabilities are utterly meaningless, because every time you get close to quantifying how unlikely an event is, a magical force comes along and makes it more likely"

pZombie

5:16 PM

but setting that aside, my final question question was quite precise and wasn't answered. When a gas inside a box is at a max entropy state, then are there more accessible microstates with a lower entropy value or are there more microstates accessible with the same (max entropy) value?

ACuriousMind

@EmilioPisanty "Million-to-one chances...crop up nine times out of ten." - Terry Pratchett

Emilio Pisanty

@pZombie sorry, I stopped reading when you based your question on the existence of a thing that doesn't exist.

there are no "max entropy states" for a gas in a box.

pZombie

@EmilioPisanty Can a gas inside the box be at a thermal equilibrium state?

@EmilioPisanty if yes, then thermal equilibrium for a gas inside a box is not the max entropy state for the gas inside the box system, correct?

ACuriousMind

What about "there is no maxmum entropy state because there are always states with higher temperature and higher entropy" is unclear?

John Rennie

@ACuriousMind I think we're assuming an isolated box so no heat can get in or out.

pZombie

5:23 PM

@ACuriousMind how can there always be states with a higher temperature if i limit the total energy available to the gas?

ACuriousMind

Ah, so you're talking about the maximum entropy state subject to some constraint. The equilibrium state is precisely the one with maximum entropy under that constraint.

pZombie

if you cannot handle gas in a box. Think about a mini-universe with just a few gas particles with a total finite energy in it

@ACuriousMind usually that constraint is implied when people talk about "gas in a box"

Anonymous

@pZombie As for the entropy in an isolated system decreasing thing, you could read this thread.

pZombie

So now that it became clear what kind of box and system i am talking about and we seem to have established that the thermal equilibrium state in this case would also be the max entropy state, can someone finally answer my final question?

would at thermal equilibrium/max entropy state of the gas, the number of accessible microstates with same entropy (max entropy) be lower or higher than the number of accessible microstates with lower entropy?

ACuriousMind

Entropy is not a property of microstates, but of macrostates.

The higher the entropy of a macrostate, the higher the number of microstates it corresponds to.

pZombie

5:31 PM

But there are accessible microstates for the gas when being at max entropy still, are there not?

ACuriousMind

I don't know what you mean by "accessible microstate". There's a bunch of microstates corresponding to the "max entropy" macrostate.

pZombie

oops

ok, then let me try it your way

what is the probability of a gas at max entropy/thermal equilibrium to access a macrostate corresponding to lower entropy vs the probability of remaining in the macrostate it is in (max entropy)

Mo_

Is it possible that a journal editor does not agree to extend the revision time for a few weeks, and reject the paper?

ACuriousMind

@pZombie 0

Anonymous

@pZombie Nope.

Anonymous

5:37 PM

12

Q: How strict are paper revision deadlines?

I recently received a "revise and resubmit" decision for a submission to an Elsevier journal, and the deadline to resubmit is in about 5 weeks. Since I have to consult with my coauthors for my revision, and one of them is a professor who's not exactly fast in replying emails, I'm worried whether ...

publications journals deadlines

pZombie

In physics, the Poincaré recurrence theorem states that certain systems will, after a sufficiently long but finite time, return to a state very close to, if not exactly the same as, the initial state.

i guess the gas inside the box ignores poincare

Mo_

@Blue But I have emailed the editor Friday (morning) and he hasn't replied yet (tomorrow is the deadline)

I'm worried

Anonymous

Uh. I guess it obviously depends on the journal. But if it's a reputed journal (and they've interacted with your before) I doubt they'll reject it immediately. Also, it's the weekend! Many people don't check their mails during the weekend.

Anonymous

This shouldn't be a rare issue at all! My prof submitted our paper just 2 hours before the deadline. I suspect many people miss the deadline by a few days.

ACuriousMind

@pZombie Poincaré recurrence holds only for finite-dimensional phase spaces, but standard thermodynamics works in the limit $N\to\infty$, so it doesn't apply. Since reality doesn't have $N=\infty$, of course reality is slightly ill-described by ideal thermodyamics, but the recurrence time of a typical system is many times the age of the universe.

Boltzmann and Zermelo already debated this more than hundred years ago when discovering the H-theorem

Anonymous

5:48 PM

@pZombie IIRC there's a big story about that.

ACuriousMind

^too slow :P

Anonymous

Damn. I was doing other things. :P

pZombie

@ACuriousMind but you said the probability was exactly zero.

ACuriousMind

In ideal thermodynamics it is.

Anonymous

@pZombie Note the statement "standard thermodynamics works in the limit $N\to \infty$" carefully.

ACuriousMind

5:52 PM

Look, you can protest all you want that thermodynamic's statement are "technically" not completely true for realistic systems. But it is a matter of fact that it is a good model for these systems nevertheless, and you cannot practically exploit the difference in any way, shape or form because you can't predict when the deviation from ideal thermodynamics happens.

Anonymous

To be fair, standard thermodynamics has a lot of assumptions. There are probably advanced versions which take into account real world effects. But yeah, the standard version works pretty well for us.

pZombie

@ACuriousMind it is hard to consider your statements as facts, when nowhere did i mention i was asking my questions in respect to some kind of "ideal" thermodynamics, a theory which obviously does not fully represent what exactly is happening in the "real world". I will always have to assume from now on that your answers might be with the consideration of some other "ideal" system i did not include as part of my question.

Anonymous

Yes, that would be a fair assumption when you ask questions to physicists. We deal with mainstream physics here, which involves a lot of idealizations and assumptions as far as modelling the real-world goes. And you will get familiar with them over them.

pZombie

I think it was pretty clear where i was going with my questioning.

Anonymous

Also, I don't know whether you have noticed, but your tone comes across as rather hostile and entitled. We're here for discussing interesting physics and not for debating. So when someone is trying to help, it's best to be patient and explain your question elaborately. We can't read your mind.

ACuriousMind

6:01 PM

All physical models are idealizations that in one way or another do not exactly reflect realistic systems, but reduce the system under consideration to its relevant properties and have proven time and again to deliver accurate predictions nevertheless. If you don't want idealizations, then you're not interested in what physics actually does.

Jake Rose

@pZombie very little (if any) of physics is perfect. You can rarely prove that nature works exactly that way. Its an approximation verified by experiment to certain limits. This is the only way we can talk about nature, so realistically it is you who is at fault for not being able to accept those assumptions

I agree with @Blue you came acros considerably rude

ACuriousMind

Time to stop this conversation here. Let's talk about something else.

PM 2Ring

@ACuriousMind Oh, ok. I just want to make a brief reply I just typed up...

@pZombie Sorry about the delay. The probabilities get outrageously small, very quickly. As John said, you'd need to wait far longer than the age of the universe to get a detectable entropy reversal, and it would be very brief. Even for a few hundred particles, the numbers are ridiculous.

Let's expand the coin tossing experiment to 320 coins. 2^10 is a little over 10^3, so 2^320 is around 10^96. The universe is just under 14 billion years old, and there's about 31 million seconds in a year. There's around 10^80 protons in the observable universe, so let's pretend we can flip each proton like a coin, 320 times per second. We'd expect a single run of 320 heads by now.

ACuriousMind

@ZeroTheHero Ah, the problem with meta - old posts about tools are rarely useful because the site has gone though several iterations of redesigns in the meantime

The idea of a permanent repository of knowledge doesn't work very well when you're describing a moving target

But trudging through meta and closing/correcting obsolete posts would be a monumental task with very little actual benefit.

Emilio Pisanty

@pZombie you're playing pretty loose with language, and it makes it extremely hard to have that conversation.

a gas in a box does have states of maximum entropy if you restrict the energy to a fixed value.

If you don't, then it doesn't.

ACuriousMind

6:11 PM

8 mins ago, by ACuriousMind

Time to stop this conversation here. Let's talk about something else.

PM 2Ring

Sorry, we're stuck in a macrostate of maximum entropy, and are finding it difficult to escape. :D

ACuriousMind

dammit, that actually made me laugh :P

PM 2Ring

My pleasure!

rob

6:27 PM

Escaping from maximum-entropy states tends to be a messy business.

Emilio Pisanty

@ACuriousMind yeah, I only saw that later.

Mo_

@ACuriousMind OK, the exact potential I intended was zero in $0<r<a$ and $V_0\ln(r/a)$ afterwards to some $R$, and $R\to\infty$. Is there any possible way (classically, say by increasing temperature) for a particle to escape to infinity?

I don't think there is anything that I'm missing
When $V\to \infty$ then an infinite amount of energy would be required

ACuriousMind

Yes

Emilio Pisanty

6:47 PM

@Mo_ that sounds like a big round no, I should think.

change $\ln(r/a)$ for any other potential such that $\lim_{r\to\infty}V(r)=\infty$ like, say, $V(r)=r^2$, and ask the same question.

Wernher von Braun explains the possibility to reach the Moon. "Man and the Moon", Dec. 28, 1955

►

is he using a slide rule as a pointer?

cc @dmckee who I think will enjoy it

PM 2Ring

@EmilioPisanty It looks like a slide rule to me.

Emilio Pisanty

@PM2Ring I mean

as you do

right?

PM 2Ring

I used a slide rule in my early high school years. In my final 2 years I had a scientific calculator. I've still got a circular slide rule hiding somewhere

Sadly, I lost the slide rule I inherited from my father.

dmckee

@PM2Ring I used a (my Dad's high-end engineering job, actually) slide rule for one year in high school as well. But not because I'm old enough to pre-date adequate electronic calculator.

Early in my sophomore year I lost my first scientific calculator (a HP 11c), which was back then a considerable investment.

PM 2Ring

I can't recall seeing any of my teachers using a slide rule as a pointer, but I guess it's not unheard of. Although not too practical, since if you gesture too excitedly the slide would probably slide out. :)

dmckee

7:01 PM

I was afraid to tell my folks, so I started using the log tables in the back of my math book (fortuitously we had just studied the properties of logarithms).

Dad caught me at that and said "You know, there is a better way".

God knows I was already known as a geek around school, but using a slide rule in the mid 1980s must have pushed that reputation to a new level.

PM 2Ring

I had a couple of friends with HP calculators, and I fell in love with RPN. A few years later, a friend had a programmable HP, which was even more fun to play with.

dmckee

Anyway, I got a new HP 11c for my junior year.

The 11c is one of the early programmable, breast-pocket sized ones.

PM 2Ring

@dmckee Definitely! :)

dmckee

They still produce the financial variant from that generation (the 12c) because it is one of very few machines approved for use on some certified public accountant exam.

Well, the produce a newer, lower power, faster "gold" version of the 12C.

Efforts to get them to produce a faster version of the 11c, 15c, or 16c included an on-line petition, but they refused.

PM 2Ring

One of my chemistry teachers had a fairly rare cylindrical slide rule, with helical scales. It didn't have many scales, just the 2 main ones for multiplication & division, and a log scale. But those scales were 60" long, so it gave you an extra digit or so of precision.

dmckee

7:12 PM

I think they have one of those on display at the Museum of Science and Industry in Chicago.

PM 2Ring

@dmckee Once upon a time, HP had a reputation for excellent scientific instruments. How the mighty have fallen... I guess their printers are still pretty good

dmckee

The reason they gave for not going back into production on the breast-pocket sized calculators was that their new, graphing ones were better in every way.

PM 2Ring

My chem teacher was from the States. IIRC, he bought the cylindrical slide rule from an ad in Scientific American. Oh, I forgot to mention that it collapsed like a portable telescope, which made it convenient to carry in a pants pocket, but a bit too bulky for a shirt pocket.

dmckee

But I know a dozen colleagues who have stuck by their c series machines to this day: the form factor is fantastic, the buttons give exactly the right amount of tactile feedback, the battery life is excellent (months of heavy use on 3 #357 coin cells), and survivability of the line is amazing.

PM 2Ring

I'd like to get a decent scientific calculator app for my Android phone, the basic calculator it ships with is a joke. It can't even do constant calculations or square roots. I've looked at lots of reviews in Play Store, but most of them are infested with ads, or have too much stuff I don't really need, like financial calculations and units conversion.

I've got a programmable Casio that's still going strong after almost 30 years. It currently needs new batteries, but they normally last more than a year with moderate use.

Mo_

7:25 PM

@PM2Ring Use Taylor series for square roots and other functions

PM 2Ring

On my phone, I tend to use the Google online calculator, and Wolfram Alpha for graphs & integrals. And if I need to program an algorithm I use an online Python interpreter.

@Mo_ Maybe, but it's a PITA if the calculator isn't programmable, and doesn't even have a proper memory. And for square roots (& other poly roots), I tend to use Newton's method.

And for basic stuff, I just use my brain. It's not as fast as it once was, but it still works. :)

ZeroTheHero

8:03 PM

@ACuriousMind No big deal. @Blue provided the appropriate historical link.

Anonymous

@ZeroTheHero Hey...about the search tips thing: I've been thinking about it and will write it up sometime soon. I'm not getting much time these days due to uni coursework. Sorry for the delay.

Anonymous

Meanwhile Chair added the relevant Math SE meta posts. Those have some very good tips (mostly from Martin Sleizak). You could check those out. :)

ZeroTheHero

@Blue yes I have been inspired by this.

Physics Meta

11:39 PM

0

Q: what do I do if I started a bounty on accident

I started a bounty and I did not mean to can I do anything to undo it. if there is not I request that there is a way to retract the bounty on the same day.

discussion feature-request bounty

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