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1:03 AM
@ACuriousMind in general (not about obliv) you seem to operate under the assumption that people are tasked with things they have the competency to tackle
have you been formally introduced to the american education system ? XD
independently, i have a question about this image. i see that when we take the $t$ limit, we kill the exponentials, but i dont see how we are isolating the ground state? i thought it should be $\sum_{n = 0}^{\infty}\psi_n^*(q')\vert n \rangle$
 
1:31 AM
@Relativisticcucumber the image failed to send
 
oh no i failed to attach it. i cannot escape shameful events
 
and kitty already tried to give you a non-shameful exit...
 
LOL i accept my cringe and revel in it
 
@Relativisticcucumber $\varepsilon$ must be a tiny positive parameter that we take the limit to zero at the end. There is a lot of handwaviness in this: the exponential decay is done for all states but ignoring the ground state, and even though we should have loss of probability coming from the loss of all excited states, somehow we are only caring about the ground state and ignoring probability conservation. It really is just saying that the ground state is least attenuated compared to excited states.
and only finally are we saying that, since there is just one final state, the pure phase factor of that state does not matter, and thus the complex phase is thrown away.
 
"It really is just saying that the ground state is least attenuated compared to excited states." why is this true?
or what does that mean ?
 
1:54 AM
@Relativisticcucumber Oh, it is much easier to understand mathematically if they would just put in a bit more work. Take all the energies from the ground state $\left|q,t\right>=\sum_n\psi^*_n(q)e^{i(E_n-E_0)t}e^{\varepsilon(E_n-E_0)t}\left|n\right>$, then you can see that the ground state accrues no complex phase factor and no dampening compared to the rest.
 
@naturallyInconsistent but isnt it like we have $t \rightarrow -\infty$ then all the exponentials go to one anyways?
so no state has dampening?
 
@Relativisticcucumber when $\varepsilon>0$ as $t\to-\infty$ then $e^{\varepsilon(E_n-E_0)t}\to0$ except $n=0$
As for the meaning, it is kinda the same thing as when you shoot a beam of particles out in a ray, and after a long distance = long propagation time, the particles all decay to ground state.
 
@naturallyInconsistent how can an exponential ever go to zero :o
 
@Relativisticcucumber $e^{-\infty}=0$
gtg meeting
 
kms basic algebra. okay it makes sense now. for some reason i was thinking $e^{-\infty} = 1$
thank you and enjoy ur meeting
 
2:28 AM
i am also wondering about the equation of motion for a harmonic oscillator. in srednicki, he says that the EOM for a harmonic oscillator is $(\frac{\partial^2}{\partial t^2} + \omega^2)\psi(q,t) = f(t)$ but when i calculate this, i get $(\frac{\partial^2}{\partial t^2} - \omega^2)\psi(q,t) = 0$ using E-L equations. where does this $f(t)$ come from? he also says that we [...]
[...] can account for external forces in the lagrangian by adding terms $fq$ and $hp$ to the hamiltonian, but i thought that this means that $\frac{1}{2}m\omega^2Q^2$ is the $fq$ term that accounts for extra energy/force due to H.O., but it would seem this is a wrong understanding?
 
3:13 AM
and also i am a bit confused about what we can actually solve for in qft. so i was introduced to LSZ, but then told that we could not yet solve for the correlation function, so that would have to wait. then, introduced to the path integral formulation which writes probability amplitudes in terms of greens functions, which i was also told are very difficult to solve for. [...]
[...] finally, now we are writing the ground state to ground state transition amplitudes in terms of propagators, which it turns out are the correlation functions. however, when studying sakurai, i was told computing the propagators is extremely challenging. in fact, if we could compute the propagators, why would we even need path integrals?
and a final confusion, in srednicki, we set out to find $\langle 0 \vert 0 \rangle$ and then, we end up finding $\langle 0 \vert T Q(t_1)....Q(t_{2n}) \vert 0 \rangle$. I do not see why this is the result that we set out to get? one is an expectation of $\hat{Q}$ and the other is a ground state to ground state transition amplitude?
 
3:49 AM
@Relativisticcucumber sadly, it is the meeting that is never enjoyable. There were laughing moments, though.
@Relativisticcucumber In the totally completely isolated and free QHO, the RHS is zero. Later on, we insert a classical disturbing "source field" linear term $\psi J$, and it is added to the Lagrangian just so that functional differentiation by J brings down factors of $\psi$ into the expectation computation.
That is how you get the f(t) term on the RHS
 
4:05 AM
@Relativisticcucumber The correlation function / probability amplitudes / Green's functions are all difficult to solve for. What we are doing is to use the free-particle Green's functions, which are trivial, to perturbatively expand for the interacting ones.
@Relativisticcucumber we are using path integrals to reorganise and simplify the perturbation expansion. Originally it would be a doubly infinite sum of perturbation terms, and the path integral suggests Feynman diagram way to view what the perturbation scheme is doing, and resum it as one single infinite sum. That sum is still horrible, and we have to consider renormalisation, dealing with loops rather than the usual Taylor style expansions
@Relativisticcucumber The pure ground state to ground state transition amplitude is trivial and not very useful. Instead, the expectation of Q is the correlation function, which is interpreted as the Green's function
 
 
1 hour later…
5:20 AM
@naturallyInconsistent okay i think i can see this maybe
it seems feasible but i havent seen any perturbation theory mentioned yet so maybe its coming?
i think i might have the wrong starting point in mind. it seemed to me like we were trying to calculate probability amplitudes, which i thought were of the form $\langle n\vert m \rangle$ and that all of this formalism is to get theses sorts of amplitudes. is this wrong? and also is there a difference between scattering amplitudes and probability amplitudes ?
@naturallyInconsistent i c
 
@Relativisticcucumber all your $\left<n\vert m\right>$ will end up as $\left<0\vert\phi\cdots\phi\vert0\right>$
 
ffkhgggjl
 
from vacuum, create particles, propagate them a bit, annihilate them, go back to vacuum, what is the transition probability?
 
i guess i dont get what these time ordered operators are?
 
@Relativisticcucumber that is just the Hamiltonian evolution requires things to be time ordered?
the transition amplitudes only make sense if you created particles in the past and annihilated them in the future, not the other way around
 
5:34 AM
@Relativisticcucumber so in this message, the $\hat{Q}$ operators are creation and annihilation operators?
@naturallyInconsistent
 
@Relativisticcucumber you learnt that QFT field operators are made of momentum creation and annihilation operators, yeah? Yes
again, all those are really free particle delusions that we are insisting upon.
 
@naturallyInconsistent yes. so far i have seen creation and annihilation operators, counting operators, and position and momentum operators (though i am shaky on these two and what they mean), but i was unsure what $\hat{Q}$ was
 
@Relativisticcucumber you should not be seeing Q unless in old QM, and in that case they should be using Q to ease you into QFT later on, where it appears as field operators $\phi$
 
oh i see, so it is a field operator then?
 
it is meant to be a field operator in qft, yes
 
5:42 AM
okay i see
and why is $\langle 0 \vert 0 \rangle$ called $Z$?
 
It will be abused in the same way that partition functions are, as in statistical thermodynamics
 
what abuse are you refering to
 
@SillyGoose this aint for the best
 
The free particle $\left<0\vert0\right>$ is a pure number that we set to 1 and cannot be differentiated. We insert a classical source field J and forcibly make it a functional of J, so as to allow functional differentiation trickery to bring down a lot of $\phi$ in the middle, to get Green's functions of the free-particle propagators from this Z.
 
5:56 AM
i am confused why phase space here is 6N dimensional. if we are only considering points of the form $(q_i, p_i)$ for $i = 1, ..., 3N$, then aren't there only 3N points total? Why do we consider the other "ambient space"
oh wait
 
We do not proof, within physics, that such differentiations work / converge, that the partition function business works / converge, though maybe the mathematicians have already done it for us. But it is really just meant to be beaten into shape.
 
@naturallyInconsistent hm yeah this whole idea makes me very shaky
the $J$ business i mean
 
@SillyGoose 3N for q, 3N for p, so 6N
 
is it more like for each particle, we have $\mathbb{R}^3 \times \mathbb{R}^3$ from which the particle can take some ordered pair $(q, p)$
so phase space describes the unconstrained collection of all possible values of $(q_i, p_i)$ not the allowed values of $(q_i, p_i)$
 
@Relativisticcucumber What we can, however, definitively say, is that if such a partition function exists, then it must work out this way. And it might not work for a very big region, but it must work for some neighbourhood of zero, and that is more than enough goodness. (as long as the region of convergence is not itself of zero size. but then again asymptotic expansions have zero range of convergence yet still work fine.)
 
6:02 AM
man okay
@naturallyInconsistent thank you for the help
 
@SillyGoose ACM had something to say about this. It turns out that it is better to consider a system not as the specific system that you have in your experiment, but rather to consider all possible systems with similar dynamics at once, so that all possible values of (q,p) are evolved forward, each as a single trajectory, but together the entire phase space evolution is handled
@Relativisticcucumber jigglypuffff
 
@naturallyInconsistent by all systems with similar dynamics, the precise meaning is same Hamiltonian but different initial conditions?
how nice it is to learn a bit of mech and s. mech at the same time >:D
 
@SillyGoose for this situation, definitely yes. Maybe even more, but yes for this.
 
also, when you recommended ian ford's book you said once one understand's the importance of entropy. by that, do you mean once one sees that (at least in textbook stat mech) all of the useful predictions of stat mech comes from entropy? @naturallyInconsistent
 
6:31 AM
@SillyGoose It is a combination of many books to get to a complete understanding. Firstly, one should learn from Feynman lectures Vol 1 on energy and heat to understand what the basic reasoning flows, such that we would be interested in the quantity called entropy. Ian Ford then takes over as a physics-motivated simple explanation of why entropy and partition functions work, and he defined a comically wrong thing to study, yet the microstate macrostate business still resembles reality.
If you want to be mathematically precise, you want to study some Callen. In there, Callen starts from assuming half of the 2nd law, and some other stuff, to generate the whole of classical thermodynamics, and then with a bit of postulating the form of entropy, it generates the whole of statistical thermodynamics.
Kittel & Kroemer is also a very recommended read.
I still think, however, that we should have a textbook that just does everything from the PoV of entropy; just about everybody does it from minimising some energy-like potential rather than always maximising entropy, but the pedagogically new thing that thermodynamics brings to us is entropy, not energy.
 
6:45 AM
@SillyGoose so you may understand this as just "when you understand all the crazy tricks we do with partition functions to get all the stuff we want", then you are already well versed in stat therm
in this, I'm defining stat mech as different from stat therm, even though they are obviously incredibly related, but just that stat therm side is easier, just play with partition function, stat mech you have to work out the physical kinetics and that is annoying.
 
7:08 AM
It is fun to think that when biologists originally told physicists about Brownian motions they were basically told "It's probably thermal motions you idiots"
 
 
2 hours later…
8:49 AM
@Slereah and it only took them about 80 years to come to that conclusion
 
Coffee and brownie motion!!!
 
physicists did not seem very interested by brownian motion for quite a while
but then again they can't just look at every time something weird happens in biology
 
Boltzmann was probably interested right?
Oh even that was rather late
But as we all know we were bad boys until yesterday, now we are much better
And besides everyone is hungry for a discovery, lol
I'll listen to a cat if he starts talking about HEP
 
EHEH :D
 
9:04 AM
but on HEP I'd want to ask, rather than talk
 
Socrates said all talking should be asking more or less, no? ^_^
 
@Amit How many biology paper have you read lately
 
@Slereah lol, nil as you might guess. In what respect are you pointing this out?
Oh I see, damn I am slow, didn't sleep enough
I'm not a physicist! I'm sure physicists go meowing like cats around biologists in their faculties to tell them something exotic.. no? :E I hope.
Hey, they took that ST theory about universe evolving with different fundamental constants from biology! Evolution! :D
 
like, I read that the infrared divergence is about how the photon has no mass, and thus a real world charged particle comes with infinitely many ultra-soft photons. Apparently the correct way to deal with this is via a concept known as infra-particles, and that changes the LSZ handling of that; im not sure how that goes
 
Academia is very open and multidisciplinary, don't say they are just as dogmatic as the people of the bygone days =[ I mean we can think it, but we better not say it. There may be kids here!!
 
9:10 AM
@Amit it is nice to exchange information: they taught meow meow about how their textbooks included how some structure in the brain does the auditory Fourier transform, and how another similar structure does the visual ones too. In return, miao miao taught them a basic understanding of Fourier transform.
 
@naturallyInconsistent lol
It is a symbiotic fruition of knowledge, part of the mysteries of the timeline
The meow meow can never be fully described, it contains infinite information, like your photon buddies there
 
Yeah, Hawking thinks that the soft photons can contain all the entropy of black holes, so there
 
9:26 AM
Black furballs!! 🐈
 
You should have gone for 🐈‍⬛
 
I would if I knew I could
That doesn't make me a bad purrson
 
slow blinks
 
Feline energy saving mode engaged
 
9:41 AM
hi
@Amit there is a Kurzesagt video on this youtu.be/h6fcK_fRYaI?si=r2mtKR9WMy4LDbdM
 
@Amit It's actually a hypothesis first formally published by Smolin called cosmological natural selection and the parallels to biology were explicit from the beginning
it's also not explicitly string theoretical, though as Wiki says Susskind later ported it over to ST
 
it is formulated to explain fine tuning
 
Cool, yeah I heard it from Susskind so makes sense
 
i dont know why fine tuning needs to be explained the first place
probability theory only applies when there are other possibilities. if there r no other possibilities than this universe, how can u say this universe is unlikely
@ACuriousMind it is saying a collapsing black hole can create a new universe with different parameters
this doesnt seem like it's allowed by standard physics, as we first fix the parameters and then run the laws
 
10:15 AM
maybe it's like with fish. There's always a bigger probability space
 
i think something like this could b allowed by spontaneous symmetry breaking, changing the mass of particles
@Amit lol
yes, but this argument assumes its conclusion. it is only valid when u assume other possibilities in the first place
it is not needed when u consider this universe as the only possibility
 
As long as the anthropic principle doesn't lead to the misanthropic principle (due to physicists that use it for evil purposes, lol)
 
there is a creepy song with this name
what do u think about the liar paradox
is it just that "sentences need not be true or false"
 
The best physics related song that I know, lol
 
but liar paradox is about claims instead of general sentences, like "B is lying". Claims shud b true or false
 
10:26 AM
It's so good that I bet computer science guys try to say it's Shannon entropy!
Idk what's the big deal with the liar paradox, isn't it just another example of logical contradiction. Made more esthetically pleasing? lols
 
@Amit it is cool
@Amit two people are making claims and the claims r neither true or false
logical contradictions at least reveal the premise to be false usually
but this is more sinister. it's neither true or false
 
they're neither true or false, or rather you can't decide their truth value?
 
i'd say both?
i mean we cant decide because it's neither true nor false
not all sentences have truth values but claims shud
 
Huh? Why?
If it's a self-reference claim like "I always lie" it can't have a truth value
 
10:34 AM
then i think the paradox reveals that not all claims have truth values
how do we know the claims that we make about the universe are not self referential in a big circle structure?
i guess it's not much relevant for physics that not all claims hav truth values. but it became relevant for math in Russel's paradox
 
@RyderRude ah my example is bad btw, should be "This statement is a lie"
 
@Amit i mean that the circular structure can be bigger than this. there is a sequence of claims that are self referential in the end
what if the claims about the universe form a circular structure
some people propose "coherentism" as a self-referential model of the universe, but it is required to be internally consistent
 
@ACuriousMind lol, nice. But if mainstream'ness is any criteria I have a slight edge
@RyderRude We can't escape self-referentiality, seeing a bunch of atoms in the form of human beings trying to study another bunch of atoms that are called "inanimate", lol
That's why Plato wanted to be so... platonic. It feels good to convince yourself that some things transcend the natural world, in particular the ability to observe
 
@Amit It is a big deal because in classical logic there are supposedly only true or false claims, and one is supposed to be able to deduce them. The ability of self-referential claims to defy this dichotomy ends up delimiting the applicability of maths and computation.
 
10:50 AM
@naturallyInconsistent Yes it's a big deal when dealing with applying logic. When I said it isn't a big deal I meant that I never found it a surprising property.. that's as opposed to say, Russell's paradox, which I find to be much deeper and more interesting
 
@Amit But Russell's paradox is equivalent to Liar paradox...
 
Is it? I thought the Liar paradox is a rather very special case
Maybe you're right, and it's just because the Russell one always talks about sets, subsets, etc. that it looks a lot more complex. Maybe they can be shown to be equivalent...
 
@naturallyInconsistent I don't think it is; Russell's paradox is a problem with naive set theory that is fully resolved by axiomatic set theories like ZF(C); the Liar paradox on the other hand appears in any formal logical system where you can assert the truth or falsity of all statements of the logical language (e.g. by assigning Gödel numbers, then diagonalizing to find one that asserts its own falsity)
what is true is that one type of resolution to the Liar paradox looks very much like the resolution to Russell's paradox: Just restrict the rules of the logical system so that sentences that assert their own falsity are forbidden, just like ZF restricts the rules of what a well-formed set is so that Russell's paradoxical set is no longer a valid definition of a set
 
Liar liar, sets on fire
@ACuriousMind That's cool
 
however while everyone agrees that this is the correct resolution in the world of set theory, it's not so simple in logic: self-referentiality seems important to model certain ways in which we reason about the world and logic itself, and so this kind of restriction (e.g. by Tarski hierarchies of meta-languages that can only assert the falsity/truth of statements in the languages below them on the hierarchy) is far more controversial than the resolution of Russell's paradox
 
11:05 AM
While I do not disagree that the resolution of Russell's paradox by simply restricting the definitions so that the self-referentiability is no longer possible is a far easier solution than is wanted in general logic, I'd have to point out that 1) the reason why Russell's paradox became a problem is that the ability for self-referentiability reduces Russell's paradox to Liar's paradox, not the other way around 2) Before category theory came along, all of maths were built upon set theory,
 
the idea of the two paradoxes is the same - use self-referentiality - but they are not equivalent because self-referentiality of logical sentences is harder to give up than self-referentiality of sets (if you forbid self-referentiality of logical sentences completely, for instance, then what happens to sentences about Gödel numbers? Logic that completely forbids this kind of thing is likely far weaker than you'd like)
 
i.e. that includes mathematical logic.
(the "all of maths were built upon set theory" is very much Russell and friends' life work.)
 
@naturallyInconsistent huh? set theory is built on logic, not the other way around - the axioms of set theory are logical sentences
 
@ACuriousMind hi. the Godel sentence looks like a Liar paradox sentence only on the surface. however, it can be assigned a truth value (True). the Russel's paradox is however a liar paradox type sentence which cannot consistently be assigned a truth value
 
@ACuriousMind IIRC, it is the other way around. For example, Gödel encoded logical statements into numbers via sets, which is how computation became important for determining truth of statements.
 
11:09 AM
the Godel sentence is : "This sentence is non-demonstrable". If this is assigned the truth value "True", then it does not lead to any logical contradictions
 
@naturallyInconsistent Gödel's theorems are specifically about "logical systems that contain Peano arithmetic"; it's not that logic is built on sets, it's that his theorems are about logical systems which are at least powerful enough to prove the existence of natural numbers and the usual rules of arithmetic
 
the Godel sentence only shows the existence of true non-demonstrable sentences in any consistent axiomatic system at least as powerful as Penao arithmetic
the Godel sentence does not show that liar paradox type sentences can be constructed in such systems
 
@RyderRude I did not say that the Gödel sentence is an example of a liar paradox, I said that, in a language in which you can assert the falsity of arbitrary statements, you can use the same Gödel numbering and diagonalization scheme to find a sentence that asserts its own falsity.
 
@ACuriousMind Yes, I'm aware of that; but that is a proof regarding the nature of logic systems itself; it would be kinda difficult to claim that it is not saying something about logic itself when it is able to do something like that, as opposed to only being constructed out of logic.
(And also why I'm always pointing out that people who do not know about Gödel's completeness theorems are doomed to get the wrong interpretation about Gödel's incompleteness theorems)
 
@ACuriousMind the Godel sentence is "This sentence is non-demonstrable". It does not assert its own falsity. if it did, it would b a liar paradox type sentence
@ACuriousMind also, u did say that Godel numbering can construct the appearance of Liar paradox sentences
the Russel paradox, however, is a liar paradox type sentence
it shows up in set theories with unrestricted comprehension
 
11:15 AM
@RyderRude Again, I am saying that in a language where you can assert the falsity to arbitrary sentences you can use Gödel numbering and diagonalization to construct a sentence that is numbered N and says "The sentence with Gödel number N is false" to construct liar paradoxes, not that the usual Gödel sentence is a liar paradox.
what you're saying isn't false but you're not disagreeing with me
 
yeah, im with ACM on this one
 
@ACuriousMind oh. so this is Godel's construction applied to slightly different systems. i had never seen this result. sorry
is Peano arithmetic one such system
it cant be because it would imply Liar paradoxes are construct-ible there?
 
it is very well known and understood that Gödel knew his incompleteness theorems had important implications for truth, and simply decided to leave it for someone else to do it. Tarski is a lucky man in a lucky spacetime region to be able to cross the t and dot the i for Gödel.
@RyderRude he just stated it, i.e. yes. Why don't you read before asking stuff?
 
this is a great result showing explicit construction of liar paradoxes
 
well, no, you have to pay attention to the qualifier "in a language where you can assert the falsity of arbitrary sentences"
normal first-order logics cannot assert the truth of their own sentences due to Tarski undefinability
 
11:22 AM
A lot of rather fundamental theorems are just various applications of Gödel's diagonalisation argument in different topics and guises to reduce a lot of important questions to the Liar paradox.
 
which you might see as a sort of "censorship theorem" that prevents logic from actually running into the liar paradox - you're just not allowed to talk about sentences being true from within the logic itself
and the proof of Tarski undefinability is essentially the diagonalization argument I alluded to above
 
The one we most have to mourn about these is poor old Hilbert. I also wanna know, dammit
 
it also mentions this : Tarski's theorem, on the other hand, is not directly about mathematics but about the inherent limitations of any formal language sufficiently expressive to be of real interest. Such languages are necessarily capable of enough self-reference for the diagonal lemma to apply to them. The broader philosophical import of Tarski's theorem is more strikingly evident.
so this theorem is for more practical languages rather than strong formal systems
i wonder if self-referencing capability brings more proving power than first order logic
it would be more interesting if mathematics is broader than first order logic. Humans did not use first order logic to do mathematics for long
 
@RyderRude the conversation above already provided an answer to this
@RyderRude and the answer to this is much closer than the answer to the above
 
mathematics has higher order logic too, but they dont have more proving power than first order logic
 
 
1 hour later…
12:35 PM
This also means that liar paradox cannot be relevant for claims of physics about the universe, as those claims are in the language of first order logic (as physics models are in ZFC)
 
12:56 PM
but some form of self reference is creeping in. like @Amit said, we are the universe talking about the universe and making claims
 
1:37 PM
Physics has plenty of worries about the status of logical sentences but it is usually whether or not a sentence is meaningful
I think any paradoxical sentence of that type about physics is likely to be put into the "not meaningful" bin
 
One can define canonically conjugate variables $A$ and $B$ via $\{A,B\}=1$.
There's another definition via the Lagrangian as $A=\frac{\partial L}{\partial \dot{B}}$

Are these definitions equivalent? [I cannot show them to be and think them not to be because the "Lagrangian" definition does not involve p or q]
 
@Sanjana The first definition is in the Hamiltonian context, the second in the Lagrangian, i.e. the first is an equation on cotangent space while the second is an equation on tangent space. In what sense do you want to show them to be "equivalent"?
 
Say I want to start with $\{A,B\}$ use the definition of Poisson bracket in terms of local Darboux coordinates $(q,p)$ i.e. $\frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial B}{\partial q}$ and substitute $A=\frac{\partial L}{\partial \dot{B}}$...This should after some algebraic manipulations give me $1$. Are my expectations incorrect?
 
The following is true: Since the Legendre transform is its own inverse, for any Darboux coordinates $q^i,p_i$ (i.e. a choice of coordinates on phase space such that $\{q^i, p_j\} = \delta^i_j$) you can take the corresponding Hamiltonian $H(q,p)$, Legendre transform it into some $L(q,\dot{q}^i)$ turning the $p_i$ into functions $p_i(q,\dot{q})$ in this process, and then you have $p_i(q,\dot{q}) = \frac{\partial L}{\partial \dot{q}}(q,\dot{q})$.
this is basically just the definition of how to pass between the Lagrangian and the Hamiltonian formalism
 
1:52 PM
@ACuriousMind But that is just for the coordinates themselves...What do I do for functions of the coordinates $A(q,p),B(q,p)$ (2D phase space for simplicity)---If I have such a pair with Poisson Bracket=1...Can I claim/show that $B=\frac{\partial L}{\partial \dot{A}}$?
Or the converse...if I assume $B=\frac{\partial L}{\partial \dot{A}}$ can I show $\{A,B\}=1$?
 
@Sanjana If you have such a pair then you should be able to use the theory of canonical transformations to get a Hamiltonian $H(A,B)$
you can't just randomly differentiate $L$ with respect to a function on phase space
$L$ is a function of $q$ and $\dot{q}$, so you first need to obtain $L$ as a function $L(A,\dot{A})$ via the Legendre transform of $H(A,B)$, otherwise what you're trying to do doesn't make any sense
 
What kind of geometry is used here?
To get $v = x_1^2 + x_2^2 + 2x_1x_2\cos\theta$
doesnt the law of cosines have a - not a +
 
It's a minus only for a side facing acute angle no?
 
ooh right
 
Oh, it's because theta is not the obtuse here
The cosine is negative if angle is obtuse
So it should be minus always
Look at where theta is
 
2:08 PM
u can also easily derive this using a dot product
 
Too easy!!
Vectors are cheating
Them physicists only dot and cross all the time!!!
 
cross product sucks big time
 
I'm pseudo insulted
 
convert to Clifford algebra
or tensors
 
Or quaternions?
 
2:12 PM
yes!!
justice for Hamilton
according to a trilemma, explanations of the universe can either be coherentism, infinite regress or a dead end, none of which are satisfying to humans
 
Correction: to humans who want to explain the universe
 
yes, but can we make it well defined what do we mean by an explanation of the universe
 
No, it is always self reference because...
We are the world, we are the children..😂😂
 
lol
 
There's the axiom of choice that we're making
And messing up our lives
It's true we'll solve a paradox for you and me
Mic drop in the absence of air resistance
 
2:24 PM
philosophers look crazy
 
Wittginstein had to become one himself to be sure of that
 
is it fruitless to search for the ultimate understanding of everything
 
If it makes you happy, no, imho
 
i feel like we dont know anything
and it's very hopeless to understand everything
 
The socratic idea of knowledge is best imo. We can't know where truth is, only where it is not
 
2:30 PM
the ultimate understanding is a waste of time. scientists instead study specific stuff
@Amit how can we know where it is not
 
We make our ignorance more specific lol
 
we dont know what anything is, but we know how to work with whatever anything is
is that a good quote :P? @Amit
 
For me it is
 
great
 
Technology relies on reproducibility but science on refutation
 
2:37 PM
@Amit We may not even claim to know nothing.
 
The scientific facts are always socratic, but the theories give a feeling of knowledge because they have to generalize. That's why they're always ultimately wrong lol. But in tech we use them in a limited region so it is fine
@naturallyInconsistent Yes we don't know what we know seems quite right.. considering how memory works for example ;-)
 
it seems like people are satisfied with just being able to work with whatever reality is. this is y even physics is about only making predictions, and not about what anything is
2
 
@Amit If you are going to simply ignore what other people saying, im going to also start ignoring you.
 
But can't I just avoid this by using the chain rule?
I mean $\frac{\partial{L}}{\partial{\dot{A}}}=\frac{\partial L}{\partial q}\frac{\partial q}{\partial {\dot{A}}}+\frac{\partial L}{\partial \dot{q}}\frac{\partial \dot{q}}{\partial {\dot{A}}}$ where $L=L(q,\dot{q})$
 
@naturallyInconsistent Oh dear, why do you think ive ignored you?
 
2:42 PM
@Amit I'm making a rather particular point, and you seem to be reading the exact opposite, going on what you have been going on.
 
@naturallyInconsistent maybe I parsed your statement wrongly: "may not even claim to know nothing" I took as "we can't claim to even know what we know or not know"
 
ummm.. can we please just be nice even if there might be a misunderstanding? @naturallyInconsistent
 
It's okay Ryder he doesn't have to be nice to me
It helps me test the limits of my tolerance lol
 
yeah, naturallyinconsistent might be having a bad day.
 
We are the world...
 
2:48 PM
lol
 
what's the hardest part about doing lagrangian mechanics problems? I feel like setting up the lagrangian is my struggle, then solving the time derivatives
 
An analogy would be that two astrologers having a really nice time chatting about how the stars are foretelling a good day to go to the cinema and ignoring all the evidence and the people telling them something else.
@Obliv Lagrangian mechanics typically make mechanics problems an order of magnitude easier, so much so that problems have to be cooked up to be much more difficult than is sensible to attempt to solve via Newtonian mechanics. But you would first have to properly learn how to work with them, and as far as the chat logs go, you have been thrown into the deep end
 
Am I an astrologer wrt this analogy?
 
Eh, the complexity of these problems aren't THAT bad but I don't have the experience doing them yet.. and my swiss cheese knowledge doesn't help.
 
In particular, setting up the Lagrangian is usually the 2nd most simple part of the problem; the derivatives are the simplest.
@Obliv And many people in this chat have been telling you to fix the swiss cheese part first. Otherwise you will just find every step painful
 
2:52 PM
It's like years of swiss cheese Idk if it's even possible to fix at this point.
 
Just like modern medicine, if necessary pains are necessary, so be it; but doctors know to alleviate pain when they can
@Amit yes
@Obliv if you tell us some of your background, maybe we can give more useful advice.
 
well my geometry is basically nonexistent, algebra skills are hazy, all of the neat little tricks and shortcuts people use in solving things I seem to not know
Like in my diff eq class even partial fraction decomposition wasn't obvious to me lol
 
Off to the cinema then
 
It doesn't bother me because I will learn what I need to, i'm just worried that on exam days I will be poorly equipped
 
i'll watch constellations
 
2:56 PM
@Obliv are you under time pressure to finish uni or something? Or is this leisurely studies for fun?
 
time pressure
 
How long do you have; what year are you in; rough geographical location, etc, what will be on the exam, etc
 
2 years, junior, new jersey, i'd have to look it up
I do have an optics exam on friday. i'm guessing the best method is to just spam problems from the textbook and go over problem sets
 
holy, that is some intense time pressure
 
lol
 
2:59 PM
the scale of your problem is kinda outta what I can do for ya, and im already an expert on time crunched exam prep
 
pressure makes diamonds
 
ah, the americano
 
i could go for one of those. too bad I don't have an espresso maker
 
I'd suggest you look up the stuff that will be on exams, and tell us that whenever you post some question here, so that people can try to help you quickly
 
@Sanjana and how exactly did you get a function $\dot{q}(A,B)$ so that this makes sense?
again: Your $A$ and $B$ live on phase space, while the Lagrangian lives on tangent space with coordinates $q,\dot{q}$
in order for any of these formulae to make sense, you need to express the Lagrangian as a function $L(A,\dot{A})$, and the only way to do so is to first canonically transform your Hamiltonian into a $H(A,B)$, then Legendre transform that into a $L(A,\dot{A})$, then take the derivatives you want to take
if you want to be pedantic about the meaning of "conjugate" and self-adjoint operators etc. then you also need to be pedantic about the structure of classical mechanics
(this is not a slight: I love being pedantic about these things :P)
 
3:40 PM
If I'm working in some rectangular coordinate system, and I compute $\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}$ 1. should I not have any trig functions in it? 2. If I do, do i still get $\frac{d}{dt} \sin\theta \to \dot{\theta}\cos\theta$?
 
@Obliv 1. The form of the Lagrangian is generally arbitrary (but it is indeed rare that you'll find trig functions of rectilinear coordinates) 2. A time derivative is a time derivative and the chain rule is the chain rule.
 
oh wait I just realized the $\theta$ won't change because this is for an incline plane and a block sliding down it
I think I did something wrong. I'm not getting equivalent expressions of both sides of E-L equation. $$L = \frac{1}{2}(M+m)\dot{x}^2_1 + \frac{1}{2}m\dot{x}^2_2+m\dot{x}_1\dot{x}_2\cos\theta-mg(\ell-x_2)\sin\theta$$ I am getting on the left side, $(M+m)\ddot{x}_1 + m\ddot{x}_2\cos\theta$ but right side is 0?
$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1} = (M+m)\ddot{x}_1 + m\ddot{x}_2\cos\theta$
$\frac{\partial L}{\partial x_1} = 0$
Also this is an extra credit part of a problem so if it's really too difficult then I won't bother with it
 
what exactly is the problem here?
 
 
no, I don't mean the problem statement, I mean what's the problem with what you've done so far
why do you expect "equivalent expression" here
 
3:52 PM
i thought they were both supposed to be the same thing
 
@Obliv then what on earth would the equation tell you?
I don't think you understood what the E-L equations are for
 
you told me to see if con. momentum was conserved to compute d/dt of it. if i went thru and did the time derivatives I wouldn't have got 0 ?
like the easier way was to compute the right side of the e-l equations
but if i did it the hard way, would I have even got the answer 0?
 
the question is trying to tell you that something nice happens...
 
the E-L equations are true equations, not trivial identities - they hold for trajectories $x(t)$ that are the physical motions
that doesn't mean both "sides" have to be the same as functions of $x$ and $\dot{x}$
In your case, you've gotten $(M+m)\ddot{x}_1 + m\ddot{x_2}\cos(\theta) = 0$. That's a perfectly valid equation of motion
there is no problem here
 
If I didn't have the lagrangian and I wanted to see if momentum was conserved the newtonian way, what would I do?
nvm
I don't want to ask pointless questions
Ok the reason was because I was "checking" my work and found a different answer
 
3:58 PM
I see where the problem was. would do it like $\dot{A}=\dot{A}(B(q,p))$
But to express $\dot{A}$ as a function of $B$ I need $L(A,\dot{A})$...which is obtainable only from $H(A,B)$....so there's no escape from the canonical transformation you were talking about :(
 
this person didn't even compute the right side of e-l equation and just stated it was nonzero.
 
Can you send the full question, Obliv?
 
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