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1:00 PM
@Kenshin I don't know (not being a poli-sci person) of reliable metrics for ideology between countries through time.
@Secret A small-c conservative thinker would point out (a) 1940ish is an unusually broad middle, so regression to a mean's naturally going to take hold; (b) looking carefully at the note on the "5th party system" you'll see that we're probably measuring the "wrong" axis for the past 20-30 years in that ^^ chart--who knows how mixed/centrist it might seem along a more-informative axis?
 
Can someone please help me with a quantum mechanics question
 
@Semiclassical too bad I have class right now. I think this epsilon trick might work.
 
@Kenshin Just go ahead and ask.
 
Ok so I worked through the SE in an infinite potential well between 0 and a
 
okay
 
1:09 PM
the solution was psi = A sin (kx)
the problem is, doesn't this imply that the wavefunction persists even beyond the infinite potential at 0 and a?
Shouldn't we expect a solution that looks like a gaussian distribution (bell curve) that contains the wavefunction within the potential well?
 
@Kenshin That's the solution for 0<x<a, no?
 
nope
 
You're telling me that psi satisfies the TISE outside [0,a]?
(It's been twenty years, but I don't think so...)
 
I dunnno
I mean I'd hope it doesn't, but it appears to me that it does
I've probably made an error somewhere tho
 
Speaking of
1
Q: Has the definition of "statistical mechanics" changed from its original meaning?

Steven HattonI guess this never registered with me when I read the Feynman Lectures on Physics in the past. But I have wondered, from time to time, what distinguishes statistical mechanics from, say, kinetic theory. Like most people, I assumed the distinction was that statistical mechanics is a probabilisti...

Due to often/daily mistagging of posts as by newbies, it may be a good idea to make a synonym of the tag. Keep in mind that there is no point in having a refined tag system if it is applied wrongly. Thoughts?
2
 
1:14 PM
kinetic-theory and statistical mechanics are seaparate in my opinion
 
@Kenshin Just try it. Take the second derivative, multiply by whatever, add V (infinite!), and see if you get a multiple of psi. If not, that psi is not the solution in that region of the potential.
 
@nitsua60 well V isn't infinite
I have V infinite only at the points 0 and a
V is like a dirac-delta distribution at those points in a sense
 
@Kenshin Oh! You've got two delta-potentials?
 
yea
 
I thought you were talking about infinite square well.... Sorry.
 
1:16 PM
np, I probably should have been doing a square well
so do you think in the delta-potential well, the state just tunnels straight through?
 
@Kenshin Well, think about the region outside of [0,a]. If there's behavior out there, what would you expect?
 
yeah id expect to equally likely find the particle inside or outside the box
because i haven't explicitiy specified whether the particle is inside or outside in my set up
I think that makes sense, thanks bro
 
@Kenshin That's not what I'm saying. (And be careful with "equally" there.)
 
yeah
I mean, it wouldn't be a normalisable solution
 
@Kenshin Yeah, but we won't let that stop us =)
 
1:19 PM
but the fact I've only specified two potentials in the SE, the particle will form a wavefunction that isn't necessarily in the box
and I think that is why the wavefunction appears to just pass straight through the potentials
 
@Kenshin Right. So you've got to think about your free-space solution, then about transmission/reflection, &c.
 
yea
too hard for now
I think I'll just change the problem to square potential
but at least I understand the one I was working on a bit more
 
@Kenshin Mind if I ask what you're working on? ("Change the problem" raises an eyebrow.) Some sort of presentation or something?
@Kenshin glad to hear it =)
 
just working through a textbooks that's all
for independent learning
it actually was a square potential in the textbook i just didn't read it properly ;)
but then I still wanted to understand how my interpretation would work
 
@0celo7 lol, it’ll still be there when you finish
 
1:23 PM
@Kenshin Okay, that makes sense. That's typically tackled pretty early, then at some point a single delta potential, and by then you should be able to tackle a collection of deltas (like having two at x=0 and x=a) without too much trouble.
 
One way to look at it is that a delta-potential is what people use to describe a “hard-core” interaction between, say, an electron and the nucleus
So the two only interact if there’s a possibility of the two coming into “direct contact” so to say
 
@Semiclassical Hey, let's keep it family-friendly. (he he he)
 
As such, the interaction does nothing if the wavefunction happens to vanish at the delta-potential since the probability density of finding the particle there is zero
Lol
 
@Semiclassical I thought we decided it was insane to expect that
 
1:29 PM
0
Q: Would it be possible to travel from A to B on spacetime with a gravity generator?

TschallackaSay you have two points A and B on the plane of space, represented in 2D by the black line. Let's also say you have a gravity generator that can generate infinite amounts of gravity without destroying itself and can be turned on and off at will. When you place that in the middle it will caus...

@Slereah you happen to know some wormhole solutions that behave like this?
 
It’ll still be there for you to obsess over when class finishes:P
It’ll still be there for you to obsess over when class finishes :P
 
@Secret Yeah, I was just looking at that. In fact, at all three that just came across the ticker. You-all get quite the variety of questions around here.
 
Our not perfect, but still managed to do the job homework policy helps retain weird and interesting questions
meanwhile MSE is flooded with homework vampire questions
 
great question
@Semiclassical thanks that helps
 
Based on what Slereah taught so far, we cannot simply change the topology of spacetime, thus I will imagine there will be issue at that fold
 
1:33 PM
@Secret Do you-all tend to handle homework-help in here, or is that forwned upon?
 
If I have a hardcore interaction, like you mention, will it always be positive roots
for coefficient A
 
@nitsua60 Well, I am lazy, the other guys here does msot of the question management, thus I am a bad person to ask
 
(I ask because I was around last night when someone posted a question from their HW, got it closed, came in here, and just got crickets. Rather, got everyone else continuing their conversation while they were all "any help? anyone?")
(And I noticed that this room isn't listed as a resource in the meta on "where can I ask for HW help," so I figured I'd ask before I chimed in and helped that kid.)
 
well as far I recalled, we do sometimes help answering people homework questions, but onyl if they show enough work. That said, there are other JEE and whatever exam comrades that helps. In addition, there'sa problem solving room for that purpose
 
@Secret Well, your activity graph says you're here a bunch, so I figured you'd have seen.
@Secret Okay, cool. Thanks.
 
1:37 PM
above all, Johnrennie is usually very nice and answer most people queries which is how he goes stardom

 Problem Solving Strategies

General chat for high school physics. For MathJax see [here](m...
but for main, we usually will close homework like questions in order to keep the main relevant for researchers
 
@nitsua60 there is a site called physics.qandaexchange.com that they also use
 
@Secret Sure--that makes sense. I just wasn't sure if I'd have been "polluting" this room if I'd helped them out in here. In Math's main chat, for instance, they're always willing to walk a student through where they're having trouble with something.
 
yeah we are usually free to help them as long they show enough effort and really stuck in chat, though sometimes when conversations flow fast they get swept away by the chat flow
 
if there's a few conversations competing I do like to go to a separate room
and if I'm not interested in answering a question I usually just don't engage with it
(probably could deal with it more constructively but tbh being seen as someone who is willing to answer questions gets draining)
 
I am usually strict when it comes to question. The person has to show enough effort for me to notice it (and myself free enough of course)
and I usually don't like to answer newtonian questions because those contraptions are just tedious
 
1:46 PM
yeah, there's a level of due diligence I usually look for
 
Interestingly though, sometimes there are homework questions that are somehow interesting and it inspires me a completely new idea to test at some higher level
 
oh, I think I managed to find an error in Griffiths E&M (or at least something that's misleading as written)
 
Hi. Anyone having an idea on this: Let's say we have a trace over four vectors, lets call them k,l,m,n, so we have Tr[klmn]. If the vectors have extra indices on another space, let's say a and b, so that we have in fact Tr[k_a l_b m_a n_b], can I somehow simplify this quantity more? Thanks.
 
use mathjax please
also, I have no idea what the trace over four vectors means.
you can't multiply vectors together.
(not in without saying more about the product at any rate)
 
1:55 PM
For example, that ball hitting spring question 2 days ago will make you think about the case when the ball hits the spring too hard such that even if it fully contract cannto store all the excess kinetic energy. To find out where it will go turns out to be not so trivial as the original question since the spring will be pushing as the ball bounce away from it
 
I knew what a trace of a diagonal matrix mean, but not a vector
 
@vzn : reddit shadowbanned me. I have a special place in physics hell reserved for them.
 
loooooooooooool
 
Ok, sorry, on my pc mathjax doesn't show up on screen for chat.
The product comes from the calculation of Feynman amplitude, so in fact we have 4 spinors contracted. That's what I meant. Srednicki, chapter 46 has an example.
Let's say we have a trace over four vectors, lets call them k,l,m,n, so we have $Tr[klmn]$. If the vectors have extra indices on another space, let's say a and b, so that we have in fact $Tr[k_a l_b m_a n_b]$, can I somehow simplify this quantity more? Thanks.
Does it show up ok now?
 
for mathjax in chat use the link in the room desc
I have vague memories of those kinds of calculations, but it's been literal years so I'll pass
 
1:58 PM
ok, spinors are waaaayyy over my legue, I ma pass this question
 
:) Come onnnnn!!!! :) It's not that hard. OK, thanks anyway guys.
 
Can someone help me with this thermodynamics question?
 
Shadow banning (also called stealth banning, ghost banning or comment ghosting) is the act of blocking a user or their content from an online community such that the user does not realize that they have been banned. By making a problem user's contributions invisible or less prominent to other members of the service, the hope is that in the absence of reactions to their comments, the problematic user will become bored or frustrated and leave the site. == History == Michael Pryor of Fog Creek Software described stealth banning for online forums in 2006, saying how such a system was in place in the...
Shadow banning is super effective because it is very hard to circumvert it
 
The answer isn't nRTln2 as I'd have expected
 
2:00 PM
@Semiclassical That's certainly fair. One shouldn't feel compelled to help if they don't want.
 
In general, any policies and solutiom methods of the form: "Do A without realising it is A", "A without A", "A that is not A" etc. are super effective because we understood little how these "hollows" work
 
I need to revise what an isotherm is like
 
@Secret : it isn't super effective. People soon work out when they've been shadowbanned. But it is super nasty.
 
@JohnDuffield no I mean, the point is, even if they work out, they cannot unban themselves or similar
It's effectiveness is kinda similar to the desolate state when someone failed to gain justice, there is simply no way out without e.g. nuking the server for example
Me and some of my friends are very interested in these measures because of their close association to the ghosting phenomenon. We believed that once a crack is found in these perfectly hopeless measures, ghosting phenomenon will be discourage thus help facilitate confrontations and quick resolutions of conflicts
 
In this case, why won't the work done by the atmosphere on the piston be equal to the isothermal work done on the gas to compress it?
 
2:07 PM
@nitsua60 It's also annoying because people will try to guilt trip you if you register your activity but decline to help
14 mins ago, by Semiclassical
oh, I think I managed to find an error in Griffiths E&M (or at least something that's misleading as written)
^wrooooong
now I'm annoyed at myself
 
Ghosters think they can just solve the problem and not confront them by shadow blocking people mainly hinge on the difficulty to crack theseshadow banning techniques
 
@MasterYushi Talk for a moment about the heat in the problem.
 
and I hate unexplained ghosting so much I will want the world end just to get rid of them
(Long paragraph censored because it contains too much superheated stuff)
but anyway, back to the thermodynamics question
 
@nitsua60 All the heat will be used up in compression of the gas since it is an isothermal process
 
(whoops)
@MasterYushi That sounds more like adiabatic than isothermal. (Unless I'm misrememberung everything from 20 years ago.)
 
2:13 PM
@Semiclassical : what error in Griffiths??
 
It's not an error.
 
If my memory serves, isothermal process involve heat exchange in order to keep the temeprature constant
 
(sorry for all the pinging: first tripped up by a new plugin, then trying to untangle the mess I made)
 
Or, rather, it's not an error on Griffiths part
 
@nitsua60 adiabatic is when heat exchange is zero, isothermal means temperature is constant, so internal energy wont change, hence according to first law Q = W
 
2:14 PM
yup
 
I was being silly and forgot that the gyroradius of an electron scales r ~ v not r ~ 1/v i.e. a faster moving electron has a larger radius of curvature
 
@Secret this is my point. to be isotherm, there'll be heat exchange @MasterYushi needs to account for.
 
(the Lorentz force scales linearly with velocity, but the required centripetal acceleration scales as v^2)
rather annoyed at myself for that
 
@Semiclassical : OK then.
 
How do I account for that? The work done in an isothermal process is nRTln(V1/V2), my question here is that why won't this be equal to the work that the atmosphere does on the piston, where would the "extra" work go into?
 
2:23 PM
$U=c_vnRT$
(fail)
 
yup, in this case =0
 
what's the answer, is it of opposite sign to your answer?
 
nope
it is nRT/2
 
$dU=\delta W + \delta Q$
$dU=TdS-pdV$
$p=\frac{nRT}{V}$
 
and..?
 
2:29 PM
thinking...
 
@Semiclassical It works. Oh my god.
Yau was right (of course)
 
so it goes
 
P_gas = P_atm, initially, so P_atmV = nRT.
And W = FS = PA*l/2 (let l be the length of cylinder) = PV/2 = nRT/2
 
@Semiclassical According to the literature, he'd been using this thing for 13 years before he wrote the paper I'm reading. So he just left out the details I guess
 
This gets the answer but still doesn't answer my question as to why this is not equal to the isothermal work
 
2:32 PM
right
 
I'm surprised by advsor didn't know about it though, he was Yau's student at the time
 
this is a fun picture:
Ooo, but I can make it better
 
@Semiclassical what am I looking at
 
@MasterYushi wait a minute, is part of the work from the atmosphere get dissipated to heat in order to keep the chamber at constant temperature?
 
I don't really know
 
2:36 PM
that is W_atmo=W_gas+Q_gas
 
the red is a magnetic field going into the page which is spherically symmetric; the black lines show where the field goes from being positive to negative and vice versa. (I used the zeroth Bessel function for the field b/c why the hell not)
I put a particle in the center with some initial velocity
 
actually that can't be, because isothermal work is more, W_iso = 0.693nRT, whereas W_atm = 0.5nRT
 
and that's the trajectory you get
oh, and outside the red disk there's no field at all
the fun bit is that the net flux of the field through the disk is zero
and as a consequence the particle exits the disk on a ray emanating from the origin
i.e. the straight-line trajectory it lies on after exiting passes through the origin
which is neat.
 
Guys , I have a doubt.
If $|z_1|=12$ and $|z_2-3i-4|=5$ , then what is the minimum value of $|z_1-z_2|$ , $z_1$ and $z_2$ are complex numbers
 
What is z?
 
2:44 PM
Do I solve it diagramatically ? Or use the formulae ?
 
you should ask in the maths exchange chat
 
A geometric solution seems apt.
But, for sake of not distracting people, let's indeed move this to the math chat
 
Hey
I was going to ask a question
Can I solve an equilibrium question without using Lami's Theorem?
 
@MasterYushi Suppose the process is not isothermal, do we expect ideal gas to heat up as we compress it?
 
Is it important?
 
2:46 PM
@Semiclassical yup , fine.
 
@Secret of course, the work being done on it has to go somewhere, it goes into heating the gas up
and some into changing the volume
 
right, then (0.5-0.693)nRT make sense because its a negative number, meaning that heat is lost from the system to the surroundings to maintain constant temperature
 
better picture:
 
@Secret but the gas is being compressed, it has to absorb heat, not release it
 
underneath the plot of the trajectory there's a density plot of the magnetic field
 
2:48 PM
system gains heat from the surrounding
 
so in the yellow you get positive radius of curvature but in the blue you get negative curvature
 
We previously deduced if there is no heat loss, as the gas compresses, it will heat up and hence increase in temperature. But in an isothermal process, the gas has constant temperature, meaning that the heat escaped instead of heating the gas
 
The heat does not escape, all of it goes into changing the volume
and none of it is used to change the temperature of the gas
 
but heating should expand gases by giving them more kinetic energy, hence pressure, but here the opposite happens
 
expanding isn't the only thing it can do when heated, heating can increase pressure if volume is kept constant
in isothermal process, PV is kept constant
both can't increase at same time because this follows P inversely proportional to V
 
2:58 PM
ok I have no idea where the other part of the energy go in compressing that piston
$$nRT(\ln 2 - \frac{1}{2})$$
somehow this amount is lost in some unknown source. If this is a realisti piston, then that could be friction, but if this is a frictionless piston, I have no idea
 
Well, i think i got it
Initially the system is in equilibrium, and there is no given reason why it changes it's state, so we have to assume there is an external force on the piston which initiates the state change
That extra 0.193nRT of work must be the mechanical work of the external agent
I can't think of any other thing
 
3:18 PM
Whats the question
 
1 hour ago, by Master Yushi
user image
 
And whats the answer ?
 
nRT/2
 
@MasterYushi Option 1?
I feel it should be option 4.
 
I second that @Abcd
 
3:23 PM
@MasterYushi I think the answer in your worksheet is just wrong.
 
rob
Interesting: there's a chess.se and the markup supports a neat chess-game simulator in questions and answers (e.g.)
 
The answer isn't wrong I have confirmed, and it is from a reknowned source
Solution is also given
 
@MasterYushi Share it please! I'd like to see how they have done it.
 
It's wrong.
This solution is only valid for an irreversible process.
While your question clearly mentions "slowly" implying reversibility.
 
3:32 PM
It is work done by the atmosphere, not the gas
 
Both are equal in reversible process
Though, experts can correct me if I am wrong.
 
but they aren't in this case for some reason which I am trying to figure out
I have asked my teachers, the answer and solution both are correct
@JohnRennie can you help?
Apparently, that extra 0.193nRT of work goes into something which i can't figure out
 
@MasterYushi it's kind of a trick question
 
@JohnRennie Yeah that is expected from the entrance exams here
 
The atmospheric pressure is fixed So the work is just $_\text{atm}\Delta V$.
 
3:39 PM
But why isn't that equal to the isothermal work?
 
^^^
 
There must be some other force keeping the motion slow and reversible e.g. fiction in the piston.
 
so you mean there is an external force involved?
because initially, the system is in equilibrium so why would it change it's state until it is disturbed, probably by an external force?
 
Yes. If you let the atmosphere compress a low pressure gas container you get an explosive compression.
 
I see, thanks for your help
 
3:56 PM
Hello, guys. What is the rigorous definition of energy in physics?
This is the best one I have found so far. All the others seems to me rely on circular reasoning.
 
@Mockingbird360 Whatever you want it to be
A tool for determining the motion of a system, I guess
I don't think it has a strict definition
 
Well, I agree with you. @SirCumference
 

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