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vzn
12:29 AM
claims intuition in physics vzn1.wordpress.com/category/physics
 
1:03 AM
Hmm, ok, so that means we have an n electron system $\bigotimes_{i=1}^{N_e}\Bbb{R}^3$ and we get the density matrix of this system. The one electron RDM is then tracing out all except one electron

$$\gamma (\vec{x},\vec{x}')=\int \Psi^*(\vec{x}',\vex{x}_2,\dots ,\vec{x}_{N_e}) \Psi^*(\vec{x}',\vex{x}_2,\dots ,\vec{x}_{N_e})$$
typo:
$$\gamma (\vec{x},\vec{x}')=N_e\int \Psi^*(\vec{x}',\vec{x}_2,\dots ,\vec{x}_{N_e}) \Psi(\vec{x},\vec{x}_2,\dots ,\vec{x}_{N_e})d\vec{x}_2d\vec{x}_3\cdots d\vec{x}_{N_e}$$
So it gives us the phase information and probability of any one electron regardless on where all other $N_e - 1$ electrons are
Meanwhile, in the DFT book I am reading, we talked about the 2 electron RDM
$$\gamma (\vec{x}_1,\vec{x}_2;\vec{x}_1',\vec{x}_2')=N_e(N_e-1)\int \Psi^*(\vec{x}_1',\vec{x}_2',\dots ,\vec{x}_{N_e}) \Psi(\vec{x},\vec{x}_2,\dots ,\vec{x}_{N_e})d\vec{x}_3\cdots d\vec{x}_{N_e}$$
 
 
2 hours later…
2:44 AM
@BernardoMeurer Quavo is too damn good
It's actually a really good video
 
I'm becoming weirdly attracted to Richard Stallman
 
3:08 AM
@BernardoMeurer That is weird. Unless you're just impressed by the beard. It's a good beard.
 
@dmckee It is indeed a majestic plumage
I think I might be turning into a free software freak
I've been planning on replacing my BIOS chip with one that supports a FOSS alternative
I'm thinking of ditching my TPM and it's binary blob firmware
I'm toying with not using proprietary driver anymore
I've ditched proprietary formats
Screw it really, I'm purging all proprietary software from this machine
To hell with it, my hardware
Fuck you Nvidia
 
 
2 hours later…
6:07 AM
So
$\mathcal{L}$ or $\mathscr{L}$?
 
Everybody is welcome to join the NCAA bracket challenge Password is quora.
For added banter:

 The Clubhouse

General discussion for sports.stackexchange.com
 
6:29 AM
Brackets lock in 8 hours.
 
 
1 hour later…
7:33 AM
@skillpatrol What is that?
A game or something?
 
7:51 AM
@2017 yup, college basketball
 
8:14 AM
Oh! I don't like playing online games that much :-P
 
8:42 AM
These are 64 real college basketball teams playing for the national US championship.
 
how many fake college basketball teams?
 
Dunno.
Those would be between the teams where the game result is "fixed."
But the NCAA has a strict ant-gambling policy.
Not like the NBA.
 
@skillpatrol If that's a real life game then why did you say :"Everybody is welcome to join the NCAA bracket challenge Password is quora." :O
 
@2017 The games haven't started yet.
The idea is to guess what will happen
64 teams---> 1 champion
 
@skillpatrol So the challenge is to guess who has highest probability of winning ?
 
8:52 AM
yup
 
That's even more boring for a non-US citizen :P
 
true
 
9:33 AM
@Slereah for?
 
Lagrangian
 
Cal.
 
9:49 AM
 
Wildberger is no Penrose.
imho
 
"As a topological space, Fx is homeomorphic to GL(k, R) although it lacks a group structure, since there is no "preferred frame"."
How does this follow
$F_x$ is the space of all frames at $x$
 
10:11 AM
Jeff Lee is a good manifolds book
 
what makes a math book "good" for you
 
@Slereah because you go from one frame to another via an invertible transformation
 
@skillpatrol That question reminds of the fact that I haven't used any maths book for the last 4 years. I find most maths books to be too cluttered and repetitive. I hope the situation changes at the university level.
 
you probably don't need a math book for high school
shit's easy
 
@2017 it will change
 
10:21 AM
@Slereah High school maths is indeed very easy. But competitive maths (at the olympiad level) isn't very easy.
 
psh
 
I hope to write a good olympiad level maths book sometime in the future (when I am able enough :P)
 
cool
 
@2017 does maximum oxidation state = number of valence electron
p-block
 
10:37 AM
Hi all
 
@skillpatrol Do you know any atomic physics by any chance?
Okay no prob will post a question here in case anyone can answer. It's quite basic. I just started learning some atomic physics.
In atomic physics book it states "An electron oscillating parallel to a magnetic field $\vec{B}$ radiates an EM wave with a linear polarization and angular frequency $\omega_0$. This $\pi$-component of the line is observed in all directions except along the magnetic field. In the special case of the transverse observation ($xy$ plane) the polarization of the $\pi$-component lies along $\hat{e}_z$." Does anyone know what is meant by $\pi$-component?
 
11:05 AM
@Koolman Right...(in most cases...i guess there are some exceptions like flourine....
 
So this is apparently a thing:
Ludwig Georg Elias Moses Bieberbach (German: [ˈbiːbɐˌbaχ]; 4 December 1886 – 1 September 1982) was a German mathematician. == Biography == Born in Goddelau, near Darmstadt, he studied at Heidelberg and under Felix Klein at Göttingen, receiving his doctorate in 1910. His dissertation was titled On the theory of automorphic functions (German: Theorie der automorphen Funktionen). He began working as a Privatdozent at Königsberg in 1910 and as Professor ordinarius at the University of Basel in 1913. He taught at the University of Frankfurt in 1915 and the University of Berlin from 1921–45. Bieberbach...
also this
In the area of modern algebra known as group theory, the Tits group 2F4(2)′, named for Jacques Tits (French: [tits]), is a finite simple group of order 211 · 33 · 52 · 13 = 17971200 ≈ 2×107. It is sometimes considered a 27th sporadic group. == History and properties == The Ree groups 2F4(22n+1) were constructed by Ree (1961), who showed that they are simple if n ≥ 1. The first member of this series 2F4(2) is not simple. It was studied by Jacques Tits (1964) who showed that it is almost simple, its derived subgroup 2F4(2)′ of index 2 being a new simple group, now called the Tits group. The group...
 
And other titterworthy mathematical objects such as the Cox ring and the Cox Zucker machine
 
Miriam Butt is Professor of Linguistics and Chair of the Department of Linguistics (Fachbereich Sprachwissenschaft) at the University of Konstanz. She is best known for her theoretical linguistic work on complex predicates and on grammatical case, and for her computational linguistic work in large-scale grammar development within the ParGram project. Butt earned her doctorate in linguistics in 1993 at Stanford University. She subsequently held research and teaching positions at the Institut für Maschinelle Sprachverarbeitung at the University of Stuttgart, University of Manchester Institute of...
Butt, Miriam, Tracy Holloway King, Maria-Eugenia Niño & Frederique Segond. 1999. A Grammar Writer’s Cookbook. Stanford, CA: CSLI Publications.
The famous Butt King
 
Staaap!
 
11:10 AM
@JohnRennie Cox ring is nowhere near as good
Cox Zucker is pretty good, though
 
@EmilioPisanty I wouldn't know, having tried neither, but I'll bow to your experience :-)
 
we're gonna get in trouble...
 
@skillpatrol I was just linking to Wikipedia
is that problematic nowadays?
 
I know pal, I'm just saying...
 
Somewhere in Googlespace is a web page with dozens of those mildly smutty names. Anyone would think mathematicians have a sense of humour.
 
11:12 AM
 
@JohnRennie It was Bieberbach that tipped me over the edge this time
 
not that he could have foreseen what would happen to his name a mere thirty years after his death
 
Why is Bieberbach funny? Am I so old and boring I have missed a meme somewhere?
A Welsh version of Justin Bieber?
 
A cross between Justin Bieber and Johann Sebastian Bach?
 
11:18 AM
In mathematics, the Tits alternative, named for Jacques Tits, is an important theorem about the structure of finitely generated linear groups. == Statement == The theorem, proven by Tits, is stated as follows. Let G {\displaystyle G} be a finitely generated linear group over a field. Then two following possibilities occur: either G {\displaystyle G} is virtually solvable (i.e. has a solvable subgroup of finite index) or it contains a nonabelian free group (i.e. it has a subgroup isomorphic to the free group...
2
 
There is a limit to how funny the word tits is, at least once one is out of kindergarten.
 
@JohnRennie you're probably one of those mythical people that no longer think Cockfosters is funny
 
@JohnRennie Feel free to move it to the trash.
np
 
I got banned (briefly) for saying I eat faggots.
 
I was just unaware that there was an "alternative." :-)
 
11:23 AM
@JohnRennie So would it be a bannable offense to mention any kind of liking for spotted dick?
 
As it happens I really like spotted dick with custard. One of those traditional English desserts for those of us who like that heavy feeling.
 
Yeah, it looks rich in calories.
 
:-)
 
@JohnRennie for the next time you get banned over faggots
 
11:30 AM
@skillpatrol spotted dick is traditionally made with a type of fat called suet, which is the fat round the intestines of cows and sheep. The resulting pudding is not light and fluffy - oh no. It is dense and stodgy and eating too much will leave you with the feeling that someone has pumped concrete into your stomach.
But, it is delicious.
 
interesting
 
@JohnRennie Is the '$\pi$ polarization' basically defined as polarization which is linearly polarized perpendicular to the $xy$ plane and '$\sigma$ polarization' which is linearly polarized parallel to the $xy$ plane?
 
@JohnRennie this triggers me
Please remove it
@JohnRennie He has a theorem on isometry groups of Rn.
 
12:05 PM
Security at OPM is insane
@Slereah do you know how to do a multipole expansion?
 
Is that the thing where u do a series of spherical harmonics
 
Maybe. I don't understand these things. If the problem is isotropic, why doesn't each x y or z coordinate look the same in the expansion
Instead it's supposed to be like x1x2+y1y2-2z1z2
Ah, the problem is not isotropic
Time to investigate
 
12:31 PM
Still wrong. And the google doesn't help because someone did a similar expansion and got something different but I can't reproduce what they got.
 
1:13 PM
damtp.cam.ac.uk/user/gold/pdfs/teaching/van_der_waals.pdf Can anyone tell me how they got equation (2)?
 
1:25 PM
put everything with the same denominator, expand out, factor a bit and then do $R + x_1 + ... \approx R$?
 
I'd have to check what combining denominators gives
 
Something ugly, no doubt
Product of 4 things
$$\frac{1}{R} + \frac{1}{R + x_1 - x_2} - \frac{1}{R + x_1} - \frac{1}{R - x_2}$$
 
Yeah. Expanding everything before combining denominators gives no cross terms, which is the issue.
 
$$\frac{(R + x_1 - x_2)(R + x_1)(R - x_2)}{R(R + x_1 - x_2)(R + x_1)(R - x_2)} + \frac{R(R + x_1)(R - x_2)}{R(R + x_1 - x_2)(R + x_1)} - \frac{R(R + x_1 - x_2)(R - x_2)}{R(R + x_1)(R + x_1 - x_2)(R - x_2)} - \frac{R(R + x_1 - x_2)(R + x_1)}{R(R + x_1)(R - x_2)(R + x_1 - x_2)}$$
$$\frac{(R + x_1 - x_2)(R + x_1)(R - x_2) + R(R + x_1)(R - x_2) - R(R + x_1 - x_2)(R - x_2) - R(R + x_1 - x_2)(R + x_1)}{R(R + x_1 - x_2)(R + x_1)(R - x_2)} $$
Since $x_i \ll R$ you can probably dismiss any term in $x_i / R^n$
Well, maybe not too much
Otherwise it's just $0$
 
1:41 PM
In mathematics, the Fibonorial n!F, also called the Fibonacci factorial, where n is a nonnegative integer, is defined as the product of the first n positive Fibonacci numbers, i.e. n ! F := ∏ i = 1 n F i , n ≥ 0 , {\displaystyle {n...
::facepalm::
 
::palmface::
 
2:03 PM
@EmilioPisanty what?
 
@0celo7 I was just intensely annoyed at the word "fibonorial"
like, this annoyed
 
I have this irrational hatred of Conan
 
That hair for starters
 
"Hate" is such a strong word.
It essentially consumes the hater.
 
2:16 PM
The dark side has a good hold on me I'm afraid
But it feels good so 🤷🏻‍♂️
 
[Ramble no 1] If the Law of Attraction is a real thing, then I might be the first person to be able to utilise the dark side of it
 
@Slereah going up to second order seems to work. But I'm still missing terms when I try to reproduce what's in Sakurai
 
Add them by hand
Miracle factor
 
[ramble]
-> I am brimming with negativitiy, which by this "rule" means bad stuff is concentrated around me. The twist is that I am like a bomb, the more this negativity is concentrated, the further they will scatter. Thus I am left unharmed while the surrounding is screwed
(of course this is just a ramble. Now that the topic is changed, I should try to get my program to work or wasting another week of failing it)
 
Has anyone seen the movie The Raging Bull?
that is what hate will do to a person
 
@Slereah that's the usual proof of Gauss-Bonnet
Proof by magic/miracle
 
A friend of mine used to call it the Lourdes factor
When you just add a factor to make the equations work
It's a miracle
 
in The Periodic Table, 20 secs ago, by Secret
Ok nailed it: It's stable=opt, for some reason it needs a prior calculation
F*** you took me 2 weeks !
 
[Demonic voice] ~That is a complete underestimation of our power~
 
2:39 PM
::bows out::
 
3:09 PM
 
> [8] D. Freed and M. Hopkins, to appear
Great reference.
How am I supposed to find whatever they wanted to reference if they don't even give a working title?
> [23] To appear.
Arrrrrgh.
Why would you do this?
 
i'm pretty sure that's not how you're supposed to cite sources =P
 
Indeed, it is not
They just linked [23] every time they said "We hope to discuss this elsewhere"
I'm mystified as to what value they thought that added
 
Haha
@ACuriousMind I'm in DC, having flashbacks from FO3 in the metro
 
@0celo7 Heh
 
3:41 PM
@0celo7 which posts should I delete? The one you've replied to seems harmless enough.
 
On the lookout for Talon Company snipers
 
There's not much snow?
 
3:56 PM
@ACuriousMind That does look like a great reference. They did some interesting work... What are you reading?
 
@Danu That one's from here, see also my latest question
 
Just saw Trump's motorcade!
 
:-O
 
Also some Italian guy
 
@ACuriousMind Damn, are these different D. Freed and M. Hopkins'?!
The M. Hopkins I knew about does homotopy theory and stuff
 
3:59 PM
:36081637 Carlos the Jackal was Venezuelan not Italian
 
Ah, it is the same one. He has papers with Freed
 
@Danu Pretty sure that's the one
 
I wonder how that stuff is relevant to strings ^^
 
Thanks for the info @JohnRennie :-)
 
OBE
@0celo7 is this right?
for defining a tensor
or should it be element of instead of equals.
 
4:03 PM
@Danu Well, not all of it is relevant, but the differential character stuff is what the proper formalization of the C-field apparently should be, and provides a way to write down mathematically well-defined membrane amplitudes
 
That is $\mathcal T$ supposed to denote
 
OBE
oh right
that's the space
@Qmechanic imposter?
 
mister @JohnRennie how can i calculate the resistance between two points in an infinite resistor lattice?
@OBE no actually
nicknames are not unique in stackexchange
 
OBE
okay.
nice creativity though.
 
@ACuriousMind Why do you say, in the footnote, that global symmetries correspond to zeroth sheaf cohomology?
 
4:08 PM
i did not know he existed when i first signed up
 
@Qmechanic Google it! The calculation is the sort of thing you learn about, think "that's clever" then forget as soon as the exams are over. I certainly have no recollection of the method used. But it's such a standard question that a quick Google should easily find you the answer.
 
i just picked this name
@JohnRennie thanks anyway
 
@Danu That's not "sheaf" cohomoloy, just ordinary cohomology with values in $\mathrm{U}(1)$.
 
but how about an n dimensional lattice
how can i
 
haha
 
4:09 PM
calculate that
 
Which is just the constant $\mathrm{U}(1)$-valued functions, i.e. the usual global symmetry group of EM
 
Well, should still be the same as sheaf cohomology for locallyconstant maps hehe
 
nice
 
@ACuriousMind In hindsight, it's funny that I literally never saw $U(1)$ as coefficient group before. It's because it's a Lie group.
 
4:12 PM
But it's also Abelian, so it's a valid choice for a coefficient group
 
Also, a tiny subtlety in the notation that caused me to think it was sheaf cohomology was that I'm used to writing $H^k(X;G)$ for coefficients $G$ and $H^k(X,G)$ for sheaf cohomology with values in $G$
For sure @ACuriousMind. Just weird.
It's pretty useless for normal cohomological stuff because you don't have anything like a single generator
 
4:34 PM
Hey guys I think I found a simple and fascinating contradiction in physics
and was hoping someone could point out where I went wrong
 
Experience (many years of it) suggests you are unlikely to be the first to have spotted whatever your apparent contradiction is, but go ahead anyway ...
 
Ok cool
 
let me guess
it is about de Broglie
 
nope, even more basic haha
 
If it's a twin paradox I shall scream :-)
 
So if we drop a weight above a table (from 1m) and it drops half a meter
then at the second it hits the half meter mark, its internal energy is the same
as it had before it was dropped at 1m
cause a portion of its potential energy was simply converted to kinetic energy
 
yes, the solution is the Heisenberg uncertainty principle
 
now same situtation but this time we drop the weight on a spring that is on the table
(the spring is 1m long lets say)
and the weight goes down to half the springs length where they are in momentary equilibrium
 
@Nova You mean total energy, not internal energy, right?
 
well, same thing right
 
4:41 PM
No, Internal energy has a specific meaning in thermodynamics.
 
@Nova No, internal energy is a thermodynamic quantity that would explicitly exclude the kinetic and potential energy you are talking about here
 
when I say "internal energy" im saying: the energy that the system (in this case, weight) has
 
Yeah, you should call that the total energy. But go ahead
 
@Nova it sounds to me as if you mean total energy
 
ok cool
anyways, the weight has fallen to the same place in the spring example as it did in the free fall
yet its total energy is different
because it has lost a portion of it to the spring (which gained it as potential energy)
 
4:44 PM
yep, so far so good
 
Yes, so the total energy of the whole system (weight+spring) has remained constant. What's the problem?
 
Ok, so here is the contradiction: the work done on the spring does not equal the energy that the weight has lost!
 
because the spring is also doing work on the weight
resistive work
 
That is impossible and does not, in fact, follow from anything you wrote.
 
4:46 PM
well
 
Yes ...
 
the weight does mgh work on spring
mgx*
no the spring does kx^2 work on the weight
 
and the weight does -kx^2 work on the spring :-P
and then you add mgh and use dimensional regularisation, and everything works out just fine
 
what? why doesnt the weight do mgx work on the spring?
that is its potential change
therefore all that energy must have gone into the spring
 
@Nova my mistake. I assume you mean the weight comes to a stop when it has compressed the spring by 0.5m so the change in PE of the weight is 0.5mg and the change in the KE is zero?
 
4:49 PM
yes
so we can (in a roundabout way) calculate the total energy of the weight
by figuring out how much work it does on the spring
and then figuring out how much work the spring does on the weight
and subtracting 2 from 1
cause the weight gains back some energy from the resistive work of the spring
 
If we measure the position of the weight as $x$ and it starts at $x=1$ and the spring is uncompressed, then the total energy of the system is $\frac{1}{2}m\dot{x}^2 + mgx + \frac{1}{2}k(1-x)^2$. I'm still not sure what the problem is.
 
Good afternoon
@ACuriousMind Oh, you're name's blue. Neat
 
@SirCumference ...it has been for a while now
 
@ACuriousMind Since when?
 
Since I was elected moderator
 
4:52 PM
We paint the moderators blue so we can spot them
 
Wait what
Huh, I never noticed...
 
you only just noticed this?
 
hmm,@ACuriousMind how do I turn on mathjax in chat?
 
@Nova Look at the link in the upper right corner of chat
 
Welp, that's a bit awkward.
 
4:53 PM
@Nova Just below the words "The h Bar"
 
@Nova Use chatjax++
It's better
Everyone Most people use it.
 
If the weight starts at rest and comes to rest at another point $x_0$, then we have that $mg = mgx_0 + \frac{1}{2}k(1-x_0)^2$. That's a quadratic equation whose one solution is obviously $x_0 = 1$, i.e. the thing didn't move and the other will be the equilibrium position. Absolutely no problem here.
 
Hmm, you know what would be cool?
Getting to pick your chat name color
2
 
I agree.
 
@SirCumference perhaps our resident Javascript hackers could put something together to do this ...
 
5:05 PM
...we can't even agree on a question for "The hbar tender" to ask :P
 
@ACuriousMind the formula you wrote doesn't make sense
left side mg has units for Newtons while the right side is units joules (cause those are formulas for work)
 
There is a $ 1\ \mathrm{m}$ on the l.h.s. that I didn't bother to write down
Since I set $x=1$
 
did you... did you write $\dot 1$ for some reason? or was it an $\mathrm i$?
 
@AccidentalFourierTransform I did write $\dot 1$ at first. No idea why my fingers decided to do that
 
yeah, no spiders here
we're all humans
 
5:15 PM
::nods in a human fashion::
 
@ACuriousMind Isn't it fair to say that the spring is doing work on the weight?
 
Sure, the difference in potential energies $mg(1-x_0)$, or equivalently the difference in the spring energies $\frac{1}{2}k(1-x_0)^2$ is the work done by the spring on the weight (or by the weight on the spring, depending on your point of view).
 
Which then means it's fair to say that the weight is gaining energy from that aforementioned work?
 
Why would the weight be gaining energy?
 
5:23 PM
cause the spring is doing work on the weight
even though it's net losing energy
if we look at just one side of it
we can say the spring does work on thte weight and therefore must add some energy into it
 
No, I don't want to say that.
The change in energy is given by the total work done on the weight. This is computed not merely from the force from the spring but also from the force of gravity.
The spring is not "adding energy".
As long as the force from the spring is smaller in magnitude than the force of gravity, the spring is gaining energy, not the weight.
And the equilibrium position is precisely where the two forces are equal
 
Sure, but can't I break it into pieces conceptually
the weight loses a bunch of energy cause it does work
then it gains a little of that energy back because something else does work on it
so the net is still energy loss
 
If you wish you may say that. I just don't think that's a particularly helpful way of putting it.
 
ok, so if we are on the same page
then a free fallign weight which has no opposing forces to do work on it
should lose more potential energy than a weight which has a spring doing work on it (if they fall the same distance)
yet, they end up losing the same potential energy
which means my conceptual breakdown must be flawed somehow
 
I don't see how that would follow at all.
The free-falling weight does not do any work.
Its energy is simply converted from potential to kinetic energy, but in this simple model, the weight itself exerts no force on anything and hence does no work.
Note that the total energy of the free-falling weight at half height is still the same as at the beginning, while the total energy of the weight on the spring is much less, since part of it is now in the spring.
 
5:36 PM
In QM, to show the spin computation relations, $[\hat{S_i}, \hat{S_j}] = i \epsilon_{ijk}\hbar \hat{S_k}$ holds, is it necessary to prove for all permutations?
 
prove using what?
that is pretty much a definition
 
@ACuriousMind Is gravity not doing work on the weight?
during freefall?
 
@Nova Of course. But that's different from the weight doing work.
The falling weight does no work - why should it lose any energy at all? (And, in fact, it does not - shortly before impact its total energy is the same as when it began its fall)
 
@AccidentalFourierTransform Using the definitions of the operators $S_{x}, S_{y}$ and $S_{z}$
 
and what is the definition of those operators?
 
5:39 PM
@AccidentalFourierTransform $S_{x} = \frac{\hbar}{2}(| + \rangle \langle - | + | - \rangle \lange - |
)$ and so for $S_{y}$ and $S_{z}$.
 
The weight on the spring does work - on the spring. That's why it has less total energy at half height than when it started.
 
@ACuriousMind gotchya, I think my understanding of work was a bit wonky then
 
@JohnDoe hmm that is a very poor and unhelpful definition tbh
but if you are willing to take that expression as a definition, then yes, you have to check all permutations
 
@AccidentalFourierTransform If it states that you need to use the orthonormilty of $| + \rangle$ and $| - \rangle$.
God damnit
So boring.
 
@ACuriousMind Thanks for your help
 
5:41 PM
@JohnDoe you can use the Pauli matrices
it is essentially the same thing
but matrix algebra is usually faster because we are used to it
 
@Nova can I suggest a different way of looking at the problem?
 
@AccidentalFourierTransform Yeah true, could do that.
 
@JohnRennie Please
 
If you actually did this experiment you'd find the weight bounces up and down. In fact it undergoes simple harmonic motion. From 1m the weight falls down to 0.5m then bounces back up to 1m and so on. The centre of the motion would be at 0.75m.
From 1m down to 0.75m the weight is accelerating downwards i.e. the net force on the weight is downwards. It reaches maximum velocity at 0.75m, then from 0.75m to the lowest point 0.5m the weight is decelerating i.e. the net force on the weight is upwards.
So the motion divides into two parts: the bit above the midpoint at 0.75m and the bit below the midpoint. Above the midpoint work is being done on the weight, so its KE increases, and below the midpoint the weight is doing work so its KE decreases.
 
5:59 PM
@ACuriousMind When I'm doing quantum mechanics with two particles, I have to do the integrations over all six variables, right?
 
To get expectation values? Yes.
 

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