@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?
I 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 kinematics posts as kinetic-theory by newbies, it may be a good idea to make kinetic-theory a synonym of the statistical-mechanics tag. Keep in mind that there is no point in having a refined tag system if it is applied wrongly. Thoughts?
@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.
@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.
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
Say 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...
@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.
(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 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
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
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.
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
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.
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
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
@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
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
@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
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
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?
@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
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
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
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
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
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