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12:19 AM
I'm trying to solve $(-\frac{1}{2}\nabla^2 -\frac{1}{r} + Fr\cos(\theta))\lvert\Psi\rangle$ using $\lvert\Psi\rangle = c_1\lvert 1s \rangle + c_2\lvert 2p_z \rangle$ hence it involves solving the equation $H c = E c$, using the secular determinant. However, I get to a weird result.
The determinant, yields: $(\frac{-1}{2} - \lambda)(\frac{-1}{8} - \lambda) - F^2 * C^2$, with $C$ being a number. I think this should not be too bad since -1/2, -1/8 are the solutions without magnetic field.
If I do baskhara, it wont solve unless given an $F$. I think this is why the problem says use $(1+x)^\frac{1}{2}\approx 1 + \frac{x}{2}$, and expand in taylor series.
1 hour later…
1:53 AM
asked here if anyone wants to help physics.stackexchange.com/questions/781257/…
2:12 AM
You are more correct than you think. We are not using n as the Landau level specification for the following:
$$\tag1e^{ikL}=1\qquad\implies\qquad\exists n\in\mathbb Z\ |\quad ikL=2\pi in$$
$$\tag2k=\frac{2\pi}Ln\qquad\implies\qquad\Delta k=\frac{2\pi}L\Delta n$$
The implied RHS in Equation (2) is the reason why we say that $k$ is quantised in units of $2\pi/L$. Now, if you want to count the number of states, you flip this around:
$$\tag3\mathcal N=\sum\Delta n=\sum\frac L{2\pi}\Delta k\approx\int\frac L{2\pi}\mathrm dk$$
4 hours later…
6:29 AM
Q: Who closed my question?

user23467I asked a question about quantum entanglement. I would expect that people commenting on the question would be familiar with this phenomenon and its philosophical issues. But someone named WillO made a comment that shows a complete lack of knowledge of what entanglement is. When I tried to clarify...

1 hour later…
7:55 AM
@ACuriousMind i think this is not correct. it doesnt have to do with the particles being identical. it's just that there exists no non-zero function $f(x_1,x_2)$ which gets multiplied by different complex numbers respectively when you exchange the arguments once and then once again
so there can be no state vector with this property. this is purely mathematical
@ACuriousMind in terms of operators, any operator that exchanges 1 and 2 also exchanges 2 and 1. so they both have to be the same operator mathematically
5 hours later…
12:44 PM
what is the scariest movie you have seen?
1 hour later…
2:13 PM
Newton's second law is sometimes stated as F=kma and k is later set to 1. What is the use of this approach? is this just how Newton phrased it originally?
it seems like a useless step
maybe it is to emphasize that we could've worked with other definitions of force too, with $k>0$ and $k\neq 1$
2:42 PM
hey @RyderRude
@GiorgiLagidze hey :)
hope you're doing good. i will ask the question now, but probably will be 3-4 paragraphs
okay, sure.
I was looking into the Landau's argument why L can't depend on x for free particle. and look, he doesn't like saying much and somehow, I got confused. so I got a good proof why L can't depend on x. - just exactly follow my logic and proof.
because the space is homogeneous in inertial frame, Lagrangians for position x+a and Lagrangian for position x must yield the same e.o.m. Because of this idea, it's easy to show the following: $L(x+a, x' , t) = L(x, x', t) + d/dt f(x,t)$. (adding total time derivative doesn't change the eom and this is what should be difference between our lagrangians of position x and x+a because we know due to space homogeneity in inertial frame, their eoms must be the same).
$L(x+a, x' , t) - L(x, x', t) = d/dt f(x,t)$ and from this, $a ∂L/∂x = x' ∂f/∂x + ∂f/∂t$. since on the left side, there's no proportion to $x'$, $∂f/∂x$ must be 0, hence f is only a function of t. and we got: $a ∂L/∂x = ∂f/∂t$. In order for this to work for any $a$, ∂L/∂x must be 0, i.e we showed that L can't depend on x.
before I move to my question, is it clear what I said all along ?
@GiorgiLagidze how did you get $a\frac{\partial L}{\partial x}$ on the LHS?
2:52 PM
well, $L(x+a, x', t) - L(x, x', t)$ is the difference between the lagrangians. since everything else is the same and they are just different by only a, that's what we get.
Imagine L1 = 10k(x+a)+x' while L2 = 10kx + x'. well use $a\frac{\partial L}{\partial x}$ and you will get $a 10k$ which is the same as if you subtract by hand
@GiorgiLagidze it is not generally true that $f(x+a)-f(x)=af'(x)$. This is only true for linear functions
@GiorgiLagidze this worked because you assumed L to be linear in $x$
@RyderRude you are missing that he is pointing out a much stronger criterion, namely that this relation is true for arbitrarily large a and not just the infinitesimal version. Then all higher derivatives of L by x must also vanish.
in particular, it is already sufficient to show that L is independent of x.
the logic i pointed out seems to work just fine in my opinion
I am not sure why you tried to consider the function f(x,t) and its total time derivative; I know that is the general prescription in Lagrangian mechanics, but it is awkward to have it in there for this context.
@naturallyInconsistent i am saying that $L(x+a,x',t)-L(x,x',t)=a\frac{\partial L}{\partial x}$ is a step in the derivation. this is not valid for arbitrary $L$
2:59 PM
@naturallyInconsistent reason is before Landau says anything in his book, he first mentions that if you add total time derivative of function f(x, t) to Lagrangian, it produces the same result. and we also know that space is homogeneous in inertial frame, which means that lagrangian for x+a and Lagrangian for x, must give the same results. the only way for these Lagrangians to give the exact same results(e.o.m), they must only be different by d/dt f(x,t)
okay i understood your argument now
@GiorgiLagidze I have read Landau myself. I know this derivation. I do not think he is meaning to use it here, but if you managed to give a good proof, power to you
well, he doesn't use this exact proof
one very good physicists gave me this proof
Again, I know
I just told you I've read Landau
ok, so you agree with the logic so far ?
3:01 PM
Yes, I do
okay, now my question :p
problem now is the logic and proof i showed didn't even account for free particle case. I mean, if you use the same proof for non-free particle, it's the same thing you will get. even if particle is not free, since we observe it in inertial frame, space is still homogeneous, so e.o.m's must still be the same at different positions. which means my proof just works fine for any case in inertial frame
Homogeneity of space immediately implies that $L(x+a,x',t) = L(x,x',t)$, i.e. that your $f(t) = 0$
BUT for non-free particle, we know L must depend on x, otherwise, it doesn't give newton's second law correctly. So in my logic proof, where am I making a mistake such as it fails for non-free particle ? i am only interested yet to find a flaw in my proof.
@bolbteppa dont you have to consider the general case where the Lagrangian changes by a total derivative
@RyderRude maybe he does, maybe he doesn't. with my proof, i show why f(t) must be 0
3:03 PM
Therefore $[L(x+a,x',t) - L(x,x',t)] = \frac{\partial L}{\partial x} a = 0$ so that $\frac{\partial L}{\partial x} = 0 $
but my question is something else now.
problem now is the logic and proof i showed didn't even account for free particle case. I mean, if you use the same proof for non-free particle, it's the same thing you will get. even if particle is not free, since we observe it in inertial frame, space is still homogeneous, so e.o.m's must still be the same at different positions. which means my proof just works fine for any case in inertial frame.
BUT for non-free particle, we know L must depend on x, otherwise, it doesn't give newton's second law correctly. So in my logic proof, where am I making a mistake such as it fails for non-free p
No you don't have to consider the general case where $L$ changes by a total time derivative because we are talking about the form of $L$ in any one frame, not comparing $L$'s at different points, we are saying at any point things have to be the same as regards $x$ translations therefore it can't depend on $x$
It is a very simple point, it doesn't need anything more than this
@GiorgiLagidze In the non-free case, really, it is just that it is two or more particles interacting. Then L=L(x,y,x',y',t), and L(x+a,y+a,x',y',t)=L; here, the system doesnt care that you have translated by a overall, but y-x is going to be a variable that ignores a but yet can affect the dynamics.
In particular, L can depend non-trivially upon y-x and thus have (relative) position dependence. This is what we really meant.
This fact shows that increasing the dimensionality of a space is a tremendously scary increase in potential complexity.
if i try approaching the same way, we got:
$L(x+a, y+a, x', y', t) - L(x, y, x', y', t) = \frac{d}{dt} f(x, t)$
$a\frac{\partial L}{\partial x} + a\frac{\partial L}{\partial y} = \dot x \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t}$
because of $\dot x$ on right side, left side doesn't contain the same proportion, i.e $\frac{\partial f}{\partial x}$ must be 0.

$a(\frac{\partial L}{\partial x} + \frac{\partial L}{\partial y}) = \frac{\partial f}{\partial t}$

to make this work for any $a$, $\frac{\partial L}{\partial x} + \frac{\partial L}{\partial y}$ must be 0, but that doe
@GiorgiLagidze The idea that "since on the left side, there's no proportion to $x'$" makes no sense - in general, $\partial L/\partial x$ has arbitrary dependence on $x'$
3:11 PM
would this logic be correct ?
@GiorgiLagidze no, because f can also depend upon y. The difficulty has spiked.
If you want to waffle in this manner, at best you can say that because $L(x)$ and $L(x+a)$ should give the same eom, we must have $L(x+a,x',t) = L(x,x',t) + \frac{\partial L}{\partial x}a + ... = L(x,x',t) + \frac{d}{dt} f$ so since the total derivative is irrelevant $L$ can always be chosen so that its Taylor expansion in $x$ can be set to zero after the first term i.e. $L(x+a) = L(x)$ i.e. $L$ does not depend on $x$
@GiorgiLagidze in the non-free case, you have more than one x variable. homogeneity implies $L(x_1+a, x_2+a, v_1+v_2)-L(x_1,x_2,v_1,v_2)=\frac{df}{dt}$. so u get $a(\frac{\partial L}{\partial x_1}+\frac{\partial L}{\partial x_2})=0$. this just implies that $\frac{\partial L}{\partial x_1}=-\frac{\partial L}{\partial x_2}$
yeah, and we just don't end up having $\frac{\partial L}{\partial x} = 0$ if we follow the same proof idea as I did. @naturallyInconsistent
@GiorgiLagidze notice that this doesnt rule out dependency on $x_1$ and $x_2$
3:14 PM
@GiorgiLagidze what would be the intended result from your attempted analysis here?
i think it makes sense now
@GiorgiLagidze to fulfill this condition, Lagrangians of more than one particle are of the form $f(x_1-x_2)$
@bolbteppa You don't even need to talk about a Taylor expansion, you can just say that obviously $L$ differs only by a total time derivative from a function that's constant in $x$ and so you can just choose $L$ to be constant
Right, really you don't need to do any of this
if we got 2 particles, total time derivative added to the Lagrangian that doesn't change the eom must be functions of both x and y ? are we sure about this ?
or d/dt f(x,t) would be enough
3:16 PM
And we have circled back to Landau's argument
the silly thing here is to expect that one could somehow derive constancy of $L$ while simultaneous considering the pseudo-symmetry case of changing by a total time derivative; of course you can't, but what is true is that among all these pseudo-symmetric Lagrangians there is one that is constant and that's the one we pick
@ACuriousMind yes, this is what Landau was doing; he could have helped by being more clear on this part, though. So dayum terse
terse-ty folks, yall
@naturallyInconsistent if we got L(x+a, y+a, x', y', t) = L(x, y, x', y', t) + d/dt f(x,t) - note f only function of x and t. In this case, eom for x coordinate won't change but for y, it will, right ?
@GiorgiLagidze I just told you that it will be wrong if you do that
f will have to be a function of y too
i know, but my last question is just out of curiosity
if it's just function of x
3:20 PM
There are many questions of curiosity that is really more a headache to consider than the potential illumination they can bring
@GiorgiLagidze since we are nitpicking, isn't the first step here wrong? Two Lagrangians giving the same EoMs does not imply they differ by a total derivative, they could be constant multiple of eachother
true but
d/dt f(x,t) doesn't mean it must be definitely something other than constant, it could be 5
or 10
d/dt f(x,t) includes cases when it's constant
That's not what I said
we've discussed the relationship between "same e.o.m." and "differ by a total time derivative" a while ago in this chat here
swiftly drops outta chat
3:29 PM
me ? :D
@GiorgiLagidze i watched that video with Brian and Dawkins. it was very intelligent
yes. lots of interesting things
yes, they interviewed each other about physics and biology
Brian Greene knows to ask interesting questions
true. and he is very enthusiastic
since 1985, he has been working on string theory
and he would love to see the theory at least fail so we can move on :D
there was this question by Dawkins that why do physicists even bother to talk about the singularity at the Big Bang. they can instead just begin the universe at a latter point in time :P
@GiorgiLagidze oh :P
3:36 PM
haha and brian's answer was pretty logical. also see the second video i sent
it is hard to fail it until we bring bigger colliders
that one is pretty interesting, but nothing related to physics or biology.
@GiorgiLagidze yes. i saw the beginning of that :P
one guy kept saying there was evidence of God and the other guy kept saying there was none
in the end, tbh, both have points
i think Jordan was saying there is some evidence because of near death experiences and drugs
@GiorgiLagidze i havent seen it full yet
3:39 PM
he bases most of his logic on psychodelics
@GiorgiLagidze i also got some other videos of these people
shoot , maybe i know, maybe i don't
i thought in one video Dawkins was losing
it was a religion debate :P
but Dawkins is really good mostly
@GiorgiLagidze yes. Jordan's stance is hard to get
he likes to think and not only that, has a huge power in biology education. seems like a powerful guy to me
@GiorgiLagidze yes. He gives rational answers mostly
3:41 PM
Jordan thinks about it more like in a different way and he has points. if you think about it long enough, you will realize that he is good as well
he sometimes uses too high vocabulary for me :P
@GiorgiLagidze he defines truth as more than evidence based
same here. :/
i will see the other video too
thanks :)
thanks to you as well and everyone. time to get to Lagrangian and finally will close that chapter soon and move to Hamiltonian
4:08 PM
@GiorgiLagidze this is the correct version of this proof : $L(x+a,x',t)-L(x,x',t)=\frac{df}{dt}$ (The RHS can only be a function of $t$ for the EoM to remain unchanged here). Since the RHS only depends on $t$, u can immediately conclude that $\frac{\partial L}{\partial x}=0$ from the LHS
@GiorgiLagidze the original proof considers x dependency of the RHS, which is y u have to use some steps like "terms proportional to x' ", which r invalid because of what @ACuriousMind said
I don't particularly want to open this can of worms but I feel I have to: You must be aware that both Dawkins and Peterson are rather controversial figures for a variety of reasons (Peterson in particular holds views that would get him quickly banned from SE under the Code of Conduct were he to try and promote them here).
Randomly bringing them up in an unrelated chat and then saying nothing of more substance than "he likes to think" or "he gives rational answers mostly" is an extremely strange move (and exactly what trolls or astroturfing fans would do). I would like for this room to be a place where we can discuss even controversial things but this is far below the standard of discourse I would expect. Please do better.
@GiorgiLagidze there is also some lack of generality left because of what @fqq said. to take that into account, u will have to start with $L(x+a,x',t)-\lambda L(x,x',t)=\frac{df}{dt}$, for some $\lambda$. i dont know how the proof will proceed in this case. but this will make for the most general proof
4:42 PM
we have to recover $\lambda =1$ somehow from the general starting case
@GiorgiLagidze that question has actually left me wondering for long. it reminds me of that theory where the universe could've begun just now and we couldn't prove otherwise
so why do we want to begin the universe at the singularity anyway? we could just pick a later point as the beginning and it wouldnt make any difference to science (as in, all observables remain the same)
Most of the stuff scientists have observed so far will eventually lead us that far back. We don't have any direct evidence the universe started at any particular point in time, so we default to the most logical. Or at least, easiest to prove/disprove.
most people prefer to believe the past exists :P
@Someone yes. you could say that there are only two privileged time points to pick : "the universe began just now" and "the universe began at the singularity"
picking anything else would be arbitrary
but the choice really doesnt make any difference to science
because nothing can prove if the past happened
I once heard someone say that the past is an illusion made by memories we can never confirm, and that the future is an illusion created by our perception of time, which comes from memories.
but that's mildly terrifying to consider…
@Someone yes if we go by logical positivism, the present is the only thing that can be said to exist
4:52 PM
but then what's the point?
that just seems to lead to "do nothing because nothing's worth anything"
it is certainly seems more practical to think that the past really exists
@Someone Einstein's theory can be a life-saver here because the universe MAY not have a global present moment
this would make it meaningless to say "only the present exists". you would have to reduce to Solipsism and say "only my present exists"
Time isn't picky, but we are.
@RyderRude Couldn't this just devolve into the Boltzmann brain?
@Someone what does that say?
4:56 PM
@RyderRude It's related to the idea that the simplest answers tend to be correct.
So the simplest explanation of the universe is a brain floating in space hallucinating everything that has ever happened, is happening, or will ever happen.
@Someone idk why i said "universe MAY not have a global present" i was thinking of spacelike foliations. The universe DEFINITELY doesnt have an objective present moment
because it depends on the observer. so the theory that only present exists is meaningless unless you go to Solipsism @Someone
well, it's about the same either way.
that's not what Boltzmann brains are about
they're an apparent paradox in statistical mechanics
did I misname something?
is it something else…?
@Someone wow. i recently saw a Kurzgesagt video on this
4:58 PM
you're just thinking about ordinary "brain in a wat" scenarios of which there are many
I was about to say that, actually, after checking a Wikipedia article.
Welp, not the same thing.
Just another brain in a vat.
it's true that Boltzmann brains are a very specific variant of this scenario in some sense, but that's it
the floating brain seems too ridiculous to be allowed by physics. i wouldnt call it the simplest explanation
Yeah, apparently it's closer to the idea of spontaneous formation than anything else.
it's also not about the "simplest" explanation but about probability
5:00 PM
So it's just a brain in a vat.
the simplest can be Solipsism depending on what u mean by simplest
a very particular brain in a vat that I misremembered the name of.
Some physicists also say that a mirror universe existed before the big bang. but how does that make any difference to science
the practice of science only requires u to assume that the present exists and that the future will exist
the past is useful as a concept but you are not required to believe in its existence
The Boltzmann brain idea is this: if you believe that the universe will end in heat death, then there is an eternity of heat death. No matter how unlikely a fluctuation, eventually it will occur. If you also believe that statistical fluctuations in heat death can include arbitarily complex structures, just exponentially more unlikely the more the structure is, then even our universe could arise, in an instant, as such a fluctuation.
And since there is an eternity in heat death, the present moment is more likely to be such a fluctuation than at any point before heat death
5:04 PM
the recent Kurzesagt video is about this in detail @Someone
shoulda done my research.
But that's not all: You only know that you exist, and so you could not distinguish being a brain inside such a universe fluctuation, or just being a brain imagining that it sees the universe. The system "brain hallucinating a universe" is much simpler than "brain inside an actual universe", and so much more likely to occur by our initial assumptions
it is fine. you just didnt remember jargon
so if you buy the premises of this argument, you must conclude that the most likely explanation for your existence is that you are such a hallucinating brain - a Boltzmann brain
@ACuriousMind but how can the brain live and function without the body
5:06 PM
so it's similar to brains in vats and the "simplest explanation" property
now, many people think this argument is very silly; the interesting thing is that different camps disagree about in what ways exactly it is silly
just in the context of heat death fluctuations
what do they say about its silliness?
@Someone one criticism can be that we are using the laws of physics that we have hallucinated according to this theory to deduce the fact that we are hallucinating
if these laws are also a dream, who's to say these are the laws of the true universe
this isn't meant to be an argument about philosophy of science, it's mostly an technical argument about how statistical mechanics works
this is the criticism i could make
5:08 PM
makes sense
you can, for instance, just reject the assertion that statistical fluctuations could create things as complex as brains in the first place
@RyderRude so it's a paradox that can only really be resolved in one of two ways—this is totally wrong, or our science is totally wrong.
@Someone no, not at all
well, brains need energy
and energy is hard to come by in heat death
@Someone no. it CAN be true that the laws of the true universe and the hallucinated universe are the same
5:10 PM
again, this isn't meant to be a philosophical argument (although you can start one), it's one about our construction of statistical mechanics
the core elements here are that the universe eventually spends an eternity in heat death and that arbitrarily complex fluctuations are unlikely but not impossible
in physics we often deal with weird probability distributions that assert things are unlikely but not impossible but actually everyone knows they're impossible
if we assume our universe to not be a hallucination, then this argument can be used to argue that there will be hallucinating brains in our far future
but using this to argue that our own universe is a hallucination is a circular argument
the simulation hypothesis hierarchy universe are very similar to this. it does not require the heat death, but it argues that intelligent life can create simulations of intelligent life, which leads to hierarchial universes @Someone
so it is likely that our universe is a simulation according to this argument. but it has never been proven if universes can be simulated
@ACuriousMind and im perfectly happy having the Big Bang just be the initiating point of such a fluctuation. Life just is; enjoy it and don't care too much about the rest.
yeah. It's difficult to represent true randomness without having it.
So there has to be an end to the simulation chain somewhere with a true random system, totally unsimulated.
And we haven't detected a pattern to observed randomness yet.
5:21 PM
@ACuriousMind somewhere out there in the multiverse, there is an Earth that had the doomsday scenario in Oppenheimer, i.e. atmosphere destroyed by nukes.
@naturallyInconsistent I mean that part isn't the weird part about the Boltzmann brains, you're right that if we just look at the universe fluctuations this is just a multiverse theory where the universes are fluctuations in a heat-dead larger universe rather than any of the other weird constructions
spoilers (people may not have seen it)
@ACuriousMind you know im not into multiverses, but i dont mind entertaining possibilities that are not ruled out by physics and common sense.
in Everything Everywhere All At Once, there are many absurd universes shown
5:28 PM
@Someone do you think that it could be possible to simulate a universe
i think we are making a categorical error when we say that
@naturallyInconsistent It's the kind of speculation which I think is pretty fun as long as we all agree the answers are ultimately completely irrelevant
There can certainly be theories which allow for interaction of the realities. it is not always unscientific
for e.g. some theories link universes using a black hole
but is there reason to give these theories any thought? i dont know the details to say that
5:49 PM
@ACuriousMind is this the moment we agree that we have a fundamental disagreement and prepare for a duel to the death?
only if someone throws gas on the fire :P
uncomfortable flashbacks to a certain German historical issue
@user726941 hi :)
hi pal
@naturallyInconsistent eh
in another universe maybe
5:59 PM
@ACuriousMind i suppose we have learnt from Galois
Enlightenment principles shine again
illuminati approves
i want to discuss what the relevance of these other realities can be. can you share your thoughts @naturallyInconsistent
Yes, I can, but I will not do it when the conversation partner is you, RR. We have been through this way too many times.
@naturallyInconsistent it's surprisingly unclear what exactly happened with Galois' duel
@ACuriousMind unfortunateness happened, leaving behind that one gem of goodness.
I just meant it's very unclear, what, if anything, we should learn from Galois' death we couldn't from a million others: Don't point weapons at each other :P
6:07 PM
Should I ask on the history of math and science.SE what exactly happened with the duel?
it's extremely impressive that he came up with his theory at like age 20. Nothing short of inventing calculus.
Weren't they fighting over a woman?
it is said that most mathematical fields are ultimately linear algebra. can the same be said about calculus?
6:13 PM
in the limit?
what do you mean?
i think calculus is its own thing because of limits @user726941
That is what I meant.
no way to reduce that operation to matrices
@user726941 oh :)
@RyderRude this is false
6:22 PM
@naturallyInconsistent can you explain?
I was very shocked when I saw someone do those symbolic manipulations to get derivatives using matrices; when $\varepsilon$ is a matrix that squares to zero. Turns out to be really helpful for symbolic calculus.
it would not have any meaning of limits as per usual calculus presentations, purely just an algebraic trick, but it obviously works
@naturallyInconsistent that's probably some variant of non-standard analysis, it doesn't matter that you can choose the $\epsilon$ there to be a matrix, you just need the algebraic property to square to 0
@ACuriousMind hence why i said i was shocked, yet respected it to work
There was a recent question about the delta distribution, so I opened it to see if ACM had interacted with the post :P
6:25 PM
narrator: ACM did interact with the post.
The narrator speak Japanese in my headcanon (more precisely Kaguya-sama narrator for those who know it)
ACM, are you subscribed to the Dirac delta tag or something?
i havent noticed ACuriousMind having delta distribution fetish yet
it is mostly gauge theory fetish
@naturallyInconsistent not exactly, but:
gruppenpest alert
6:28 PM
@RyderRude no fetish, indeed
@ACuriousMind Why did that take up so much screen space.
@ACuriousMind this answers my questions that u like QFT more than GR
Windows' screenshotting functionality sucks, I only screenshot the lower white area
but for some reason it decided to save it with a bunch of transparency above it
ok, np
good night yall
6:33 PM
good night :)
i feel that my SE reputation score keeps increasing on its own
are they giving points for something other than upvotes
that sounds like a marketing question
it says "access moderator tools" at 10000
I think you can see chat flags in other rooms, etc.
6:41 PM
oh these tools are very mid. it displayes their list if u click on it
but idk what kind of cool tools i possibly expected :P
it says : review reviews, delete questions and view reports
these are the tools you get at 10000
7:13 PM
@ACuriousMind Oh, I forgot about watcher tag. Now I think of you as some sort of guardian/vigilante defending those tags
@Mr.Feynman do you like group theory more or differential geometry?
@Mr.Feynman u love cheese, right?
my grandparents are going to italy this week and are seeking cheese recommendations XD
7:30 PM
why is the juxtaposition of those two messages so funny
@Relativisticcucumber I like cheese but I don't know shit
My knowledge limits to the name of a some kinds of cheese and whether I like them or no
Sep 1 at 18:42, by Mr. Feynman
@Relativisticcucumber difficult to say. I very much like Primosale, which a type of Pecorino. Then, I love Grana Padano, Parmigiano Reggiano (very similar, but don't tell that to people from Emilia-Romagna). Last but not least, my beloved Gorgonzola
why did you write that if you wanted to delete it afterwards :P
@ACuriousMind I mean, there is no contradiction :P
i thought it was embarrassing somewhat
2 mins ago, by Mr. Feynman
My knowledge limits to the name of a some kinds of cheese and whether I like them or no
didn't say there was
7:33 PM
I hate that I can't add a "t" to my message
your old message is indeed perfectly described by naming cheese and liking it
@RyderRude lie groups are differential geometry
@Mr.Feynman yes, but still... do you like any subject more? lets count lie groups in group theory
I like DG more, I guess
7:35 PM
Differential forms and connections are too exciting
yes. but solving the equations just comes down to co-ordinates again :P
Before topology and de Rham comes in :P
how dare you besmirch de rham
i have to check out whats de Rham
seems very deep
@ACuriousMind predictable :P
Jan 29 at 16:48, by Slereah
they hate sunlight and the smiles of children
7:39 PM
sunlight sucks for sure
chlidren's smiles are ok I guess?
@ACuriousMind now that's a true videogamer hep-th
what is a comprehensive diff geom book, covering topology and De Rham too? or is it okay to rely on physics books ? @Mr.Feynman
just kidding, I remember our discussion about whether
I'm not serious about de Rham by the way, just sour grapes :P
i think if you know group theory and diff geom, you know pretty much all of the math for physics. is this accurate
7:42 PM
@RyderRude I don't know, I don't like relying on physics books for math. As to de Rham and topology I can't advise any book
Also, I'm in my physics period of the year now :D
does that mean u get to take all physics courses?
No, it means I'm free from the need for math
next semester i get to take two quantum courses and ive got to choose two math courses~~
Until I fill the tank with crimes against math, I'll keep surviving with physics only
maybe i can try to swing an independent study in topology applied to physics as a math elective
7:46 PM
And I guess that it won't take long to fill it with the Standard Model
i feel quite surfeited with scrutinizing proofs
@SillyGoose I had two electives. The first one I chose sucked. Hard.
what was it?
The subject is fine
The course wasn't
7:48 PM
Now I have to decide what to do about the other elective course. I've put an experimental particle physics course as a placeholder but I already have another one. I'm not doing two hep-exp courses lol
i am thinking of diffe g and maybe a lighter course for the math electives but idk maybe the topics in an intro diffe g course are not so relevant
what's an experimental course like? you learn about experiments done in the field?
@SillyGoose life is sucked out of you
Seriously, experimental setups are typically discussed
Along with methods of measurements and stuff like that
7:51 PM
hm i could see takign one such course in one's field of interest being somewhat helpful
although perhaps an "eat your vegetables" sort of situation
Then you might have some exp courses writing down the SM lagrangian 💀
@SillyGoose I like veggies
🥦 is good
is there a way of describing the normalization condition on fully bosonic (fermionic) states in representation theoretic terms?
what do you mean by "the normalization condition"?
that a state is normalized just means it has norm 1 :P
7:55 PM
perhaps he means the normalization factor after symmetrization of states
ooh there's something else too
there's the relativistic normalisation
p|p'=2omega delta
is this what u meant @SillyGoose
i mean this normalization condition (4.1.7.), which takes into account the indistinguishability of certain states
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