@dmckee that's the funniest bone name I've ever seen =P

but that sounds extremely painful

Dunno if shock-suppressed pain is the best descriptor for that one, though. It's just that I'm relatively used to being bashed about (many years of martial arts), and this one was quite similar to other instances of fall-induced shock to the thorax, including inducing some heavy coughing. However, persistent sharp pain after the fall is another story.

What on earth are y'all talking about

@dmckee Look how fucking awesome salmanarif.bitbucket.io/visual

hey SE, where's my badge?

I want my badge

Epic @EmilioPisanty :

Nailed it.

@EmilioPisanty How can you see 18 edits? The edit approval queue wasn't ever so long on this site.

@EmilioPisanty :-) :-) :-)

@EmilioPisanty Now I can see your badge on your user page. You got it 34 mins ago.

But then, there's this waterfall analogy in one black hole website (to be refugees)

Hello everyone

@SBM hey

Hey @PhyMan

@EmilioPisanty Did you get your badge?

What does that mean @Secret?

It's an advertisement by the swedish government on printing a new type of bank note. The music later went viral because of the catchy tune and (at least as seen in facebook) is shared everywhere

Oh

And the kanji from yesterday isn't the congee on the Wikipedia but more like dal.

I don't know what you call it in English, yoghurt masala dal?

@SBM where?

Some chat discussion from yesterday with @blue @Kaumudi.H and @PhyMan

It's literally a category in itself

Not that either

One word with hundreds of meanings yikes

@BernardoMeurer Sometimes the body's reaction to a traumatic injury suppresses the pain involved and allows that person to continue to function at relatively high efficiency for a time.

This has its upsides, of course, but also its downside as you may worsen the injury by continuing to stress that part of the body.

Both of my "go straight to the emergency room, do not pass go, do pay $200+" event have gone that way.

If I take a syringe, suck some water in that is close to the boiling point at let's say 99°C. Then I shut the syringe at the front and pull a little on it, so some of the water inside the syringe starts boiling. Will at any point the boiling water inside the syringe start pushing the syringe further or do I have to keep pulling for more water to boil?

@pZombie When you pull on the syringe you reduce the pressure inside it.

The water boils until the pressure returns to ambient pressure, then it stops boiling.

If the water wasn't close to boiling point and I would pull on the syringe, then, since water isn't compressible, I would end up with water + some vacuum inside the syringe, right?

@JohnRennie But does that answer my question? Do I have to keep pulling the syringe, or will there at any point the steam push the syringe?

@pZombie No, because the water would boil in the reduced pressure and fill the syringe with water vapour.

@pZombie The water would boil until the pressure rises back up to ambient pressure, then it would stop boiling. So the pressure inside the syringe will never rise above ambient pressure i.e. the boiling water would never push on the syringe plunger.

I think I can see now what you mean. That makes sense.

Now if we imagine a syringe inside a vacuum chamber. Only the tip sticking out and being sealed. We dive the tip into a container of water, and open the seal or valve to allow some water to be sucked into the syringe.

sorry hit enter too early

The syringe has also a weight attacked on top of the plunger, which gains potential energy if the water pushes against it. I assume that there will be a certain height the weight will reach until the pressure from the water will stop pushing against the plunger. Am I correct so far?

What I am trying to understand is the dynamics of this whole process. From some videos I watched, some of the water seems to be turning into ice, while another part into steam. Yet, we have stored some of the energy in potential energy in the weight that has been pushed up. That energy is missing from the water+steam system or so I believe. So the water+steam system is colder than before or so I believe.

So if we were to choose the exactly right pressure inside the vacuum chamber, it seems to me that the water would end up as just colder water if we waited long enough

@Secret we were not talking bout that

Or let me put it differently. We have a syringe with some water inside. We heat the water up beyond 100°C but keep the plunger locked for some time until the water is at 120°C. Then we release the plunger. The syringe is very long and we ignore any friction. Eventually, when the syringe is extended far enough, the pressure inside should be equal to outside, and it would stop extending, correct?

But what if while the plunger was extending, we had a weight attached on top of it, hence the weight gaining potential energy. Where did this energy come from then?

I assume it comes from the water, hence, if we lock the weight at the highest position. The water inside the syringe should be colder than if we just let it extend without the weight on it. Is this correct or am I thinking it wrong?

Walter Lewin, MIT professor: "All of you have now lost your virginity... in Physics!"

@user314159 just to check, you know that Lewin is a complicated figure, right?

I'm not saying don't watch the lectures, just be aware that they have a darker side from some perspectives

i don't like his lectures

He tries to entertain the audience much more than teach physics.

that's cos physics is boringh

What on earth did I just read...?

@Secret, @SwapnilDas can I share a game Idea with you

ok

It's a gambling game and you have to pick one of these categories

<1% : 100x

<10% : 10x

<20%: 5x

<50% : 2x

Now there are 100 players

each player selects one of the categories

Now if they select the <10% category and 10% or less peopel chose that category, they get a pay out of 10x

@Kenshin that's... way more info than I thought was publicly available

Payouts begin from the 1% down, (i.e. only 1 payout can occur per round, only the first category that satisfies the condition gets a payout)

How should one play the game?

oh and if there is no winning category, all players get their money back (minus commission)

But yeah, it's something to be aware of if one wants to go down Lewin lane

@EmilioPisanty D'you like The Yale guy? Ramamurti Shankar?

@peterh you mean the review counter on top right? That thing is irreparably broken

@SwapnilDas never heard of him

Kenshin do you know the answer?

Ooh ok.

@pZombie no is there an answer?

I invented the game

that's a very interesting game

ty

mind boggling

< 1% for 100 people means this category is only satisfied if there are exactly zero people choose it ?

<=

@Kenshin I would think that the best bet would, on average, be the 1%, as it's the first option on the list, so the other categories actually have a lower overall probability of winning than it first seems thanks to your 'only 1 payout can occur per round' rule. Just my thoughts

@Mithrandir24601 but if 2 or more poeple followed that strategy, you would always lose

because it will only payout if <=1 % choose that category

My thoughts are, is that you should choose the 1% category maybe about 1% of the time or just over (potentially jsut over given Mith's point that only 1 payout occurs and we start by checking the 1% category)

Ah, I get the game now - it's a game theory game, not a probability game :P

yea

that's why it is so mind boggling. If you follow what is supposed to be the best strategy, you end up losing, because given than people would not play random but also follow the best strategy, it becomes a loss, so there is no best strategy

so the answer must be that the best strategy for this game is to play random

@ACuriousMind Where are you when I need you?

but if playing random was the best strategy, and you assumed that everyone else was smart enough to figure that out, knowing that everyone would play random, you could actually then choose another category to max your profits

but again, that would be what everyone else would be thinking as well....

yes pZombie

@pZombie Hence, it's game theory :P

for instance let's say we assume you should pick the first category 1% of the time

then if everyone does that

you can calculate that this category will actually be vacent 65% of the time

(1 - 0.99^100

)

so then a smart player will chose this category 33% of the time

but then so will everyone else lol

so then they change

etc.

Love your game. You should post it as a question

on stack

I am not sure if this game can have a nash equilibrium, given that the strategy seemed to change on the fly

@Kenshin There'll probably be an ideal probability with which to pick the 1% option that means, on average, one person wins every time?

@pZombie maybe I don't want to release this just yet

@EmilioPisanty indeed I do know, hence the bold-facing on the word "virginity." :-)

His conviction came as shock to me.

@BernardoMeurer What ails you?

hello

@ACuriousMind What's the word for multimeter in german?

...*Multimeter*

@Slereah heyhey

.-.

What about voltimeter and amperimeter?

Same but without the i, i.e. Voltmeter and Amperemeter

Well, that was simpler than expected

Thanks :P

If you want to be very German, you might call the multimeter a Vielfachmessgerät, the voltmeter a Spannungsmessgerät and the amperemeter a Strommessgerät

why would you want to be that German, though

@ACuriousMind Oh YES now that's what I was looking for

@ACuriousMind Do you want to help me understand a short computation in Milnor's book that I'm just blanking on currently?

@Danu Which one?

Morse theory; page 98, bottom. I can explain the context

@Danu Okay, I'm looking at it, what's the issue with it?

@ACuriousMind So I don't understand much about it---the first equality already weirds me out a little (what's up with that exchange of $u$ and $t$?---but I mostly just want to understand how to conclude that you get something nonzero in the end.

Especially since we only know $\partial_u v(u)=X\neq 0$, not its covariant derivative!

If you want some context let me know

@Danu Context plz

Not you guys.

(you should learn Riemannian geometry first; I can't give you a book's worth of background)

@Danu I'm not even sure what the $\frac{D}{\partial u} v(u)$ is supposed to be

I mean, $v$ is a path in $T_p M$, how is he taking a "covariant derivative" there?

Pull it back

Pull what along what?

So this $D/du$ stuff just means $\nabla^*_{\partial u}$ where $u$ is the parameter along the interval $I$

Along $V$?

I'm weirded out about this because in the first expression

Hey I understand what's going on!

Damn

I learned something

In uni

Amazed

we're dealing with a vector field along $\gamma_v$, i.e. the derivative there is $\nabla^*_{\partial_t}$ where $t$ is the parameter along $I$, the domain of definition of $W$.

@ACuriousMind Btw, let's denote this path by $\cal V$ from now on to avoid confusion with the tangent vector $v$

So yeah it is pretty weird to me---the last expression doesn't seem to be a covariant derivative of a vector field along $\gamma_v$ any more

@Danu What is $\nabla$ there?

LC connection on $M$

But $\mathcal{V}$ is not a path in $M$, nor is it defined along a path in M!

No, exactly

But the first thing is that

It's just a path in $T_p M$, i.e. a collection of tangent vectors at a single point, I don't see what the covariant derivative is supposed to be.

But yeah, it starts with the strange appearance of $\frac{D}{\partial u}$ one step earlier

@ACuriousMind You can still get one from the LC I guess, but I don't think that's what one is supposed to do

Should be $\frac{D}{\partial t}\frac{\partial}{\partial u}$

@ACuriousMind What if we start by just assuming that's a typo

Yeah, that's what I'd hope too

My problem is that even then I don't really know what to do

@Danu I.e. $u$ and $t$ are switched by accident? Then the second equality makes even less sense :P

Yeah... just put in $t$ instead of $u$ in the 3rd expression?

But how would $\mathcal{V}$ depend on $t$?

yeah, nope :P

So the whole book is a typo is you guy's assumption?

But this definitely makes sense: $$\frac{D}{\partial t}\frac{\partial}{\partial u} \exp(t\mathcal{V}(u))\bigg|_{(0,0)}=\frac{D}{\partial t}(d_v \exp tX)\bigg|_{t=0}$$

that's chain rule

But then...

Well, you know that $\partial_u \mathcal{V}(0) = X$

We need to use that $d_v\exp$ is the identity at the origin

@ACuriousMind Edited that in

But I don't know how to take cov. der's without passing to coordinates

Wait...isn't $d_v \exp(X) = 0$ by assumption?

Your $d_v\exp$ is Milnor's $\exp_\ast$, right?

Yeah, this is garbage :P

Watching this is great :P

Well, I have no idea what is happening there

0/10 Milnor notation

Milnor's style is so hard for me to comprehend. I had the same problem with Characteristic Classes.

(though I guess that was mostly Stasheff's work?)

I hope it's just a Milnor setback in your studies

That doesn't really work, but okay

@ACuriousMind I'll ask around in the math chat room sometime later, I guess.

@Danu I have a feeling it should be easier to see that $\exp_\ast(tX)$ is not constant.

Especially since every source I can find that doesn't just cite Milnor for this theorem just declares it obvious this field is not constant :P

If the $W$ was constantly zero, then the "variation through geodesics" would not be a variation at all. That doesn't look as if it can happen.

Apparently $\Bbb R^4$ with an exotic smooth structure has completely different properties in GR and it's absolutely awful

although ironically its tangent bundle is the same as regular $\Bbb R^4$

@ACuriousMind Yeah, it is kind of a dumb thing. But I have to think about it for a bit.

Guys, I’m trying to understand why we can take the time-independent wave function to be real. I can follow what they do mathematically, but I don’t understand the last remark on energy. Why can we choose which wave function we want if they have the same energy? Griffiths says that there is a different wave function for each allowed energy. But does that mean that the ones that are associated with the same energy, must be linearly dependent?

@ShaVuklia it depends on the situation

if your energy level is non-degenerate, then $\psi$ and $\psi^*$ will be linearly dependent, and you can then make $\psi$ real by multiplying it with an appropriate complex phase.

if your energy level is degenerate, then it gets interesting

particularly if $\psi$ and $\psi^*$ are linearly independent

I have no idea what it means for the energy level to be (non-)degenerate.

but

this is the case e.g. for the $m=\pm 1$ states with $l=1$ in hydrogen

this question is like the second question in the first paragraph about the time-independent Schrödinger equation

so I'm proposing to keep things as simple as possible?

@ShaVuklia a degenerate energy level supports multiple linearly-independent eigenstates with the same eigenenergy

a non-degenerate level supports only a single independent eigenstate

yea right, well I haven't had eigenstates and eigenenergies and eigenfunctions and the like

I'm guessing then that

ok, fair enough

I should assume for now that they simply are linearly dependent?

$\psi$ and $\psi^*$

@ShaVuklia it's a reasonable assumption (just keep in mind that there are important situations where that fails)

where your quote says

> Since they have the same value for $E$ as $\psi$...

it just says that both $\psi$ and $\psi^*$ are solutions of the TISE with the same energy

yes I noticed that

and the two linear combinations are therefore also solutions of that same equation, so they also have the same energy

so for now I just assume that if solutions have the same $E$, then they are linearly dependent?

@ShaVuklia no

sorry

@ShaVuklia I think the point here is not that $\psi$ and $\psi^\ast$ are necessarily linearly dependent, but that $\psi + \psi^\ast$ and $\mathrm{i}(\psi - \psi^\ast)$ are real and span the exact same subspace as $\psi$ and $\psi^\ast$ no matter their dependence.

you're looking at the case where they are independent

@ACuriousMind @ShaVuklia Yeah, that can also work

@ACurious that was my first guess too

in fact, I also showed that an even potential can have even/odd solutions as their "basis"

in that case, I could write $\psi$ as a linear combination of an even an odd function

so I should be able to do that here too?

@ShaVuklia sort of, but not really

in one dimension, even functions and odd functions have different energies, pretty much always

a general wavefunction is always a linear combination of a linear and an odd function

but solutions of the TISE typically have definite parity

However, that structure normally breaks when you go to more general situations

right, I guess. But what ACurious said solved my problem. I'll stick to the spanning argument! (sorry, I wish I could think along with you, but I've only read chapter 1 and 2 from Griffiths, which is not the formal part of the book)

@ShaVuklia yeah, take it at your own pace ;-)

The big O notation doesn't make sense

Obviously we can't write $f(x) = \mathcal O(x)$

It should be $f(x) \in \mathcal O(x)$

guys

the Earth is flat

there's actually scientific evidence

I never knew

simple evidence

Take a level, put it on the ground

u will notice it is flat

unless it's not

Of course, I mean how hard is it to understand that we are all actually living in a giant pancake?

more importantly if he's on a plane, why not look out the window of the plane

That will be a slightly more informative experiment

@Slereah the notation makes sense, because it's simply convention.

The de Broglie formula is for plane wavefunctions

ugh i don't even know what those are:l

Wavefunctions are $L^2$ functions, so they can be expressed as Fourier transforms

$$\int e^{ixp} \hat f(p) dp$$

In those circumstances, you can treat each $e^{ixp}$ individually

And for those "wavefunctions", you find that $p = h/\lambda$

This may not be apparent in the actual wavefunction, which is a mixture of many momentum states

What does the hat on $f(p)$ mean? And what is L^2?

Laplacian?

the hat means it's an operator

oh

and $L^2$ roughly means it's square integrable

Actually not here

It means it's a Fourier transform

oh really

interesting

well tbh, I didn't follow anything you said

haven't come across that in griffiths

i've only read chapter 1,2

Have you done wave mechanics first

Wave mechanics is pretty important to do before QM

i know a bit about waves

enough to follow griffiths in any case

(I'm not doing formal QM yet)

@ShaVuklia 1. The deBroglie formula/hypothesis is a statement prior to full development of quantum mechanics, you should not expect it to make straightforward sense when attempting to interpret it in actual QM. 2. In terms of QM, deBroglie just "claims" that the wavefunction of a particle with definite momentum $p$ is $\psi_p(x) = \mathrm{e}^{\mathrm{i}\hbar p x}$, where you may convince yourself that this works out to $p = h/\lambda$.

Waves and QM looks intriguing

@ACurious oh right, then. I'll try to verify that. Thanks

Now, no actual wavefunction has definite momentum, but due to Fourier transform, as @Slereah said, you may think of any wavefunction being the superposition of such plane waves - the integral is "just" a more general sum.

(Don't ask for mathematical rigor regarding the latter part :P)

yeah I didn't want to bring up the rigged Hilbert space with a beginner :p

The game is rigged

Is the superposition of the plane waves similar to the superposition of the stationary states?

because that's the only thing I've encountered so far

Well they are both superpositions of eigenstates

"superposition" is just a silly physicist word for "linear combination" :P

The first one is eigenstates of the momentum

The second one is eigenstates of energy

@ACurious lol:P

eigenstates ...

eigenstates are states with a definite value with respect to that operator

I can't wait until I read chapter 3 from Griffiths (but first my exams). I want to talk too about eigenstates and such :P

oh I was thinking linear algebra

@ShaVuklia It's funny 'cause it's true

@SBM It is just linear algebra, except when it's not but we physicists pretend it is, anyway :P

Must be fun, right?

@SBM It is :)

could you suggest a good resource?

Bunch of recommendations here, I can't personally vouch for any because I haven't read any

I don't know a whole lot of basic math books because during my degree we didn't have any books

It was just classes and xeroxed notes

Hey there. Does someone have any experience with the Creutz algorithm (also known as the Demon algorithm?)

@Mithrandir24601 Hi

Sorry this is a good christian channel

We do not deal with demon algorithms

Thank you @ACuriousMind

@JohnDoe Hi

@Mithrandir24601 How's it going?

@JohnDoe Let's see... Just back from Kung Fu. Currently having some hot chocolate and I've got my last ever exam on Wednesday. So not too bad :)

Thank you @ACuriousMind

@SBM I heard you the first time, you're welcome ;P

oh sorry, poor internet

@Mithrandir24601 Yeah that does sound good, especially the last exam part. Before I mention somethiing about what we discussed yesterday. Did you watch the boxing last night? Kell Brook vs Errol Spence?

@JohnDoe Nope. I do kickboxing (as part of KF), but I don't really watch boxing :P

What did you want to ask/mention?

@Mithrandir24601 Will type it now.

@Mithrandir24601 If you are busy don't worry about it.

@JohnDoe I'm not busy... No guarantee I'll know the answer though

@Mithrandir24601 Okay it's just an idea that you can comment on, not really a question.

Fair enough

@Mithrandir24601 I just want to find you think of the reasoning of the following. My understanding of a text that I amusing So the proabbility of function of knowing nothing about say using is that. For a say case of measuring the position of an object in space, when having no knowledge, the probability is given as a constant function $P(x) = c$. But for the case where $x > 0$ we consider the function as described yesterday $$P(x) = \frac{c}{x}.$$The reason for this is as follows:

@Mithrandir24601 We consider that certain probability functions should be invariant under transformations which belong to transitive groups. Hence for the case of the positive quantity we have that it is not invariant under the translation group hence we can't consider a constant probability function. Does this make any sense?

@JohnDoe I'd say (but I'm no expert) that it depends on what you want to do with the probability distribution - under translation/addition, the above isn't a Haar measure over the positive reals, but it is under multiplication

@Mithrandir24601 Yeah I understand. The book I'm reading is Quauntum Measurment Theory with Applications. The idea seems to be that you consider the situation, like an object in space (where the probability is with respect to position) or say a lake (where the probability is with respect to volume) and then you consider which transitive group you require it to be invariant under tranformations. So it the object case it should be invariant under translations...

@Mithrandir24601 For the lake we want it to be invariant under the group of scaling tranformations...and so on.

@Mithrandir24601 The purpose is just to motivate a probability function. I think the idea is much more complicated because they give references to other books which deal with it in detail.

@JohnDoe So if you want it invariant under scaling, then $$\frac{1}{x.y.z}dx\, dy\, dz$$, should be reasonable (in 3D)...

If I haven't mentioned it already, a Haar measure is a measure on a group, so that's why it's different for multiplication and addition - is that what you're asking?

(although there is a way to make it invariant under both addition and multiplication)

@Mithrandir24601 I don't know. Since for the one dimensional case you have as you stated yesterday that $\int_{c}^{d}p(bx)dx = \int_{c}^{d}P(x)dx$, so in the one dimensional case it is invariant in that sense I guess...

Is there any way of getting latex on The mobile website?

@Mithrandir24601 No I'm not asking anything. I just wanted to see what you though of that idea of defining probability functions based on their invariance with respect to groupd.

Looks around for the nearest field theorist Anyone here with some experience solving loop integrals using Mathematica? I doubt my question is worth an actual post on the site, so I figured I'd ask here...

@JohnDoe Oh, OK - it's a good idea. It's what I'm working on at the minute though, so of course I'm going to say that :P

@Mithrandir24601 Yeah it sounds interesting, unfortunately it's not the point of this book so it's going to move on to other stuff. But I want to learn more about group theory, and their applications in QM.

@JohnDoe Fair enough. So you want to know my thoughts on why it's useful?

@Mithrandir24601 Yeah

@Mithrandir24601 I read further now they mention the Haar measure as an alternative to defining the probability in terms of coordinates...

@JohnDoe Ah, OK... I don't know what it's use is in maths, but I'll go with what I'm currently using it for, keeping things as simple as possible (otherwise it might get messy): Say you've got a single qubit that you can do arbitrary operations (gates) on to get a single qubit circuit. This circuit is represented [not sure if this is the right word or not] by SU(2). So, if you want to integrate over the group, you use the Haar measure

(i.e. if you want to integrate over all possible circuits)

@Kaumudi.H Thanks, yes, I'm now safely in Somerset :-)

@Mithrandir24601 Oh okay. Yeah I want to read up on that since I'm doing a project on two level systems later this year. Does 'gates' refer to any operator acting on a quibit?

good night

@JohnDoe It can also be used to integrate over a subset of that group

@JohnDoe Any unitary with determinant 1

@Mithrandir24601 Kewl very interesting

@SBM Night!

Open question: does anyone know of any mathematica scripts that compute Feynman parametrizations other than this one? I'm almost certain that they must exist, but I can't for the life of me find them anywhere on the internet...

@JohnDoe so, yes that would be any ideal evolution of the state (things like decoherence probably messes this up a bit though, so you might have to use U(2) instead of SU(2) or something, depending on the processes involved, but I'd have to look at this further to be sure)

@blue results tomorrow?

All the best!

::minor lightbulb moment:: Is the Haar measure also related to the whole intensive properties thing? :O

@blue English showed me the finger. 91%