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1:45 AM
is $[D_i, D_j]$ notation? or does that literally equal to the rhs of the equality sign?
 
2:21 AM
@imbAF A free particle of specific momentum is a stationary state but not bound. A time dependent Hamiltonian can have a bound state that is not stationary. It is not even clear that time dependent Hamiltonians allow for a sensible definition of what stationary states.
@Obliv You have already taken a step into semi-classical, because you are doing statistical thermodynamics. If you consider the microcanonical ensemble in classical mechanics, your integral over the entire phase space, even if constrained to a specific energy, is an infinite quantity. You need even just a bit of semi-classicalness to have that be finite, and then the usual mathematics would work.
@SillyGoose Minor nitpick: If you are assuming Boltzmann stats, then you are assuming canonical ensemble, not grand canonical.
@Obliv Do you begin to realise that your education in statistical thermodynamics is missing fundamental components? There are plenty of good textbooks that cover these conceptually important issues. I mean, what would a physics textbook that fails to cover the physics concepts relevant to a topic be of use for?
@SillyGoose Please don't invent stuff like mesostates; not least because things like that are almost certain to be used somewhere else in physics. Also, your definition is worded in a way that is trying to be general, but fails at it, because, imagine your AB to only be allowing thermal exchange. Then it cannot be in equilibrium by your definition, because it is not in mechanical and chemical equilibrium at the same time as thermal equilibrium.
@SillyGoose He isn't ready for postgraduate statistical thermodynamics yet...
@imbAF You can. The pathway is more sensible from MCE to CE to GCE. When going backwards, the results become much more approximate, with weird things like the fluctuation in E when you are meant to have already fixed E, say. It is a consequence of that fact that when we start directly with CE and GCE, things are assumed to be probabilistically distributed, so nothing is specified.
@imbAF The difference is negligible in the thermodynamic limit, by definition. However, it can be extremely visible when the quantum system being considered is small.
@SillyGoose It is very much encouraged. Not very useful to go back to sticks and stones when you already have rockets and lasers and drones.
@imbAF No, your conception of that is actually wrong. What we are doing is actually really first imagining that the system under consideration is infinitely large, that the boundary terms are gone. This is needed to throw away nuisances like surface tension. Then you mathematically and theoretically imagine a mathematical surface corresponding to the same size and shape as the experiment's volume, and try to match the experiment's results with that of this theoretical construct.
 
2:41 AM
@Mr.Feynman i mean the idea is right just was confused on some unimportant factor
 
@Obliv Which is why every good textbook on statistical thermodynamics make it a point to state that we don't really believe in the postulates going into MCE. Instead, at some point, we do as all working physicists do, and take CE and/or GCE as the basic assumption, and thereby get a much more theoretically believable distribution.
 
@SillyGoose you have to see it acting on something to but yeah
 
@Obliv That's the kind of disastrous pedagogy that would make meow have the urge to burn the whole institution down. So don't tell meow where you had the misfortune to be studying that at.
 
@lucabtz ah okay so if i write $[D_i, D_j](X, Y)$ i will see that $[D_i, D_j]$ is what witten wrote?
@naturallyInconsistent oh i see
 
@SillyGoose no acting on some function
 
2:46 AM
oh wait why is it acting on a function?
@naturallyInconsistent i think the isolated derivation is nice to see if that is what one is after since it is all calculus and not anything heavy in terms of conceptual physics (since i understood the concept to already be understood by the other person)
 
@SillyGoose oh nooooooo, that means you didn't understand what this is about. This is why I really really like the treatment in RHB, because the introduction of a scalar parameter converts the CoV problem into one that ostensibly looks exactly like the single variable calculus, so that familiar stuff reigns.
@SillyGoose Yes, but there are easier textbooks to follow than going straight to Pathria.
 
@SillyGoose yeah it's the same thing you would do with commutators in qm
 
by function in this case would we call $f$ a lie-algebra valued $0$ form? since earlier witten applies $D_i \epsilon$ where $\epsilon$ is a lie-algebra vlaued $0$ form
i am confused because to me it seems that the $D_i$ should be maps from $n$-forms to $n+1$-forms (maybe), both of which being Lie algebra valued
so the "test function" should be an appropriate $n$-form
if i think back to qm there is not really an ambiguity i guess because if we have [x, p], then the $x$ and $p$ both act on elements of Hilbert space by definition, so there is not a problem with me there in thinking of the test function as a wave function (if we are working in the position basis)
 
3:30 AM
hm this is my attempt so far to compute $[D_i, D_j]$. i am wondering how to proceed? (or maybe I am doing the manipulations totally wrong :P). I feel like we want to get all of the test functions to the right, so would rewriting some terms in terms of commutators and then using some derivative rules would obtain the final result?
 
3:46 AM
@SillyGoose yes I meant that. More in general it should be a section of some associated bundle
 
3:56 AM
is a wilson line a function from configuration space for the gauge field $W_R: P \to \mathbb{R}$ in the sense that it depends on both the closed curve in $M$ you integrate over and the gauge field $A_i$?
and so in this sense it is an observable in lagrangian formalism becuase it is a nice function from the configuration space of your system (here considering gauge fields) to numbers
 
4:51 AM
at a high level, is it accurate to think of a gauge theory as just your usual notion of physical theory in the following sense. We have a state space (a principle $G$-bundle), equivalent states (related by gauge transformations: special automorphisms on the $G$-principle bundle), and a dynamic relation (the lagrangian as a function from the $G$-principle bundle to numbers)
 
 
1 hour later…
123
6:04 AM
Hello Everyone...
 
 
2 hours later…
7:46 AM
is 10.1(a) proven by $\pi_*(A^\# f(u)) = \frac{d}{dt} \left. \left( \pi(f(u\exp(tA))) \right) \right \lvert_{t = 0} = 0$ since $\pi(f(u\exp(tA))) = p$ for all $t$ by construction? so its $t$-derivative is $0$?
@Mr.Feynman i have decided to try and learn some diffe g through nakahara hehe
 
I haven't used it but it should be good if you're interested in physical applications
@SillyGoose yes, one of the ways to understand the differential of a map is that it eats a vector tangent to a curve at a point and spit the vector tangent to the image curve at the image point
2
The image curve is what you wrote above
 
ah i see
 
Wait, let me say it better for this case
$A^\sharp$ is the vector tangent to $u\exp(tA)$ at $u$. That $f$ is there because vectors act on functions (germs, actually)
 
also so intuitively: we are constructing vertical subspaces. is this meant to be vertical in the sense that vectors $v \in V_uP$ move across fibers? And so, maybe vectors $v' \in H_uP$ will move only within a given fiber?
@Mr.Feynman germs XD
 
Vertical subspaces are by definition tangent to the fibers
 
7:59 AM
darn
 
What have I done
 
oh wait lol
okay so tangent corresponds to in the direction of
 
@SillyGoose your idea of vertical subspace is correct, those are the spaces tangent to the fibers (in this sense the "point" part is kept fixed). The horizontal subspace is not something that is naturally defined
Definining it means endowing your bundle with a connection
 
hm then i would have thought like gauge fields would have more to do with the vertical subspaces than the horizontals then :P bleb
 
I'm afraid I can't help with that, but the gist should be that a choice of horizontal subspaces is a connection on a principal $G$ bundle; via its representations you can consider the so-called associated bundles, which are vector bundles on which the group acts linearly (representations). The principal bundle connection induces connections on these associated bundles and these should be the gauge fields
Take what I said with a (or even two) grain of salt though
 
8:16 AM
@SillyGoose that's a good book
 
:D
why does the $u$ disappear in the $\omega$ subscript in hte final equality in $10.3b'$?
To my understanding the first equality is definition of a pullback, but I do not understand why the $u$ subscript disappears in the final equality
 
@SillyGoose do you mean $g$?
 
oh oops yes i mean g
also so can i define the image of the connection 1-form as the vertical subspace? $\text{im}\omega_u = T_uV$?
 
 
1 hour later…
9:53 AM
@SillyGoose technically the image is the lie algebra so no
But the lie algebra is isomorphic to the vertical space at any point
 
The business with DG is much worse now that you have connections: you also need to know bundles and connections :P
 
How can $X \in T_pM$ when $\omega_i$ is a $1$-form over $P$ and so should take argument in $T_uP$?
@Mr.Feynman the DG situation for me is already bad...
 
10:12 AM
What I mean is that all of your recent questions would find an answer or at least look easier learning some DG beforehand
@SillyGoose beware that you have $\sigma_i^*\omega_i$ acting on it
It's the pullback of the form via $\sigma_i$ and it's a form on $M$
 
 
1 hour later…
11:17 AM
hi
 
11:31 AM
Hey
 
123
Hello @RyderRude
 
12:11 PM
@123 hello
@123 hey
 
A 0 connection is just a connection over the 0th jet space of the manifold, which is the manifold itself
There's no curvature associated to it since the quantity it transports cannot change
 
yeah. i just read that lie derivative of two vector fields measures the difference in transporting the point by changing the order of transport
this sounded like an analogy with 1-connection, but now i dont see it
 
Zero connections are just connections on the manifold $M \times \{ \bullet \}$
Where the fiber is just a point
Can't really have a variation of a point
That's why you don't really have symbols associated to it
It's just directly the flow of the vector field
 
yeah. it seems like the notion of non commutativity of transport for lie derivatives is different from that of 1-connection transport
to talk about the lie derivative non commutativity of point transport, we need two vector fields, which would be two 0-connections
so there is no analogy
 
Another issue being that 0-connections don't take vectors as arguments, they are first order ODEs
They depend strictly on the position
There's only one possible tangent vector at that point
 
12:26 PM
yeah and u cant talk about change in a point after going thru a loop
the lie derivative instead talks about change in a point after transporting it along X then Y, and transporting it Y then X. so i thought it was an analogy
but one needs two 0-connections for this
 
@SillyGoose The principal bundle is not the state space; your state variables are fields like the gauge field or the matter fields that are sections of various bundles related to the principal bundle
 
the computations of $L_X Y$ and $[X,Y]$ give the same result. how to see that these computations shud b equal
 
@SillyGoose the whole point is that the notion of "vertical" is canonical and without choice, while there is no unique choice of "horizontal" - there's no structure that would pick out e.g. some "orthogonal" subspace to the vertical subspace. So in order to have a notion of parallel transport, we need to declare which complement of the vertical subspace is "the horizontal subspace". This is called an Ehresmann connection and equivalent to the data encoded in a connection form/gauge field
 
the definitions of $L_X Y$ and $[X,Y]$ r very different
i dont see their deep connections
shud we just say that these computations happen to be equal
 
Because locally the action of a vector field on a scalar function is a derivative
which is also what the Lie derivative is
 
12:38 PM
but
 
Hadamard Lemma
 
the lie derivative is directional derivative of another vector field after also considering the pushforward
and the lie bracket is like the commutator of two second order derivatives of a function
 
I don't see the problem, they are defined as different objects that turn out to be equal
 
yes, but how to see that these shud b equal, or is this a co incidence
 
Just like definite integrals and antiderivatives are completely unrelated ideas, untile the FTC connects them
@RyderRude what do you mean? There is a theorem proving it and it's not even that long
 
12:43 PM
FTC is understandable tho
 
It relies on Hadamard's lemma
 
@Mr.Feynman it's just explicit computation
@Mr.Feynman oh
my computation didnt use this lemma
 
@RyderRude And so is FTC
 
i just computed it
does this lemma help better understand the connection
 
12:45 PM
oh
so i should just accept this result
 
If you want to know more, you should he able to find somewhere on Spivak. There is a section discussing the geometric meaning of the Lie bracket
 
on a related note, how do u people decide when to pursue further understanding and when to accept the computation?
@Mr.Feynman thanks
 
I haven't checked it myself though
 
my book did give a geometric meaning but it's only for the bracket. it doesnt help connect the two
 
Vol 1, pag 159
 
12:47 PM
lemme see
 
Farewell now
 
on a related note, how do u people decide when to pursue further understanding and when to accept the computation?
everyone
 
on a related note, how do people decide when to pursue further asking the same question after a minute or when to stop? :P
2
 
123
Why potential energy difference is negative of the work done. Why is negative sign in potential energy?
 
1:28 PM
@123 The negative sign is just a convention to have minima correspond to stable equilibrium configurations and maxima to unstable configurations
Some, especially in mathematical physics use the opposite convention, i.e. $\vec{F}=\nabla U$
 
123
@Mr.Feynman Thanks for the good answer
physics.stackexchange.com/questions/315528/… answer by Mitchell. is fine or not in this thread?
 
 
3 hours later…
4:18 PM
@ACuriousMind hm i see. i hope to tease out what the state space is :P. the language of some notes has been a little bit confusing because one note called the configuration space of fields the principle bundle itself. and almost all other notes call "the connection" the gauge field. but how can "the connection" be the gauge field since there are multiple configurations of the gauge field while there is only one connection
but maybe the nakahara reading is making this more precise. given a connection 1-form $\omega$ we pull it back through a local section $\sigma_i$ to obtain a $\mathfrak{g}$-valued 1-form $A_i$. so for a single choice of connection (equivalent to a choice of connection 1-form) we obtain many different gauge potentials $A_i$ via pulling back through different local sections $\sigma_i: U_i \to \pi^{-1}(U_i) \subseteq P$?
 
@SillyGoose It is correct that the $A_i$ are what we usually refer to as the gauge field, but I don't understand what you mean by "there is only one connection". The connection is additional data just like e.g. the metric in GR and when it is a dynamical variable in physics it can of course vary. There are many different connections on any given bundle, the field/state space is the space of all connections
 
oh i just mean that it seems like we choose to work with one connection
like in our physical theory we would fix $\omega$ that i described above (to my understanding)
or following nakahara's definition, we would fix our choice of horizontal subspaces
 
no, you would not
 
the connection/choice of horizontal subspace is the gauge field
just like the metric in GR - when we do the geometry we generally talk as if it's fixed, but in the physical theory this thing will be the solution to an equation of motion, i.e. determined dynamically
 
4:28 PM
by determined dynamically does that mean it will be constrained via the principle of extreme action
 
that's where the equation of motion usually comes from, sure
 
hm maybe i don't understand what the gauge field is. i am not understanding how the connection (the literal connection) falls under what i could conceive as a gauge field. Maybe I should ask: given a gauge field, what object is a configuration of this gauge field?
 
I don't understand the question. How would you say this for a simpler example: What kind of distinction do you make between "a scalar field" and "a configuration of this scalar field"?
 
well I guess "a scalar field" should contain all the data that its configurations contain. but, I would think that "a scalar field" (the field proper) "updates" to match what configuration it is in. for instance, if i am handed a particular configuration of this scalar field and asked what "the scalar field" is over each point in spacetime, i would tell them the configuration that it is currently in
 
I don't understand what you mean
A scalar field is a map $\phi : M\to \mathbb{R}$
 
4:36 PM
well maybe i can think about it in terms of particle mechanics :P. In particle mechanics instead of a field we work with a particle. a particle can roughly be defined as a collection of states it can occupy. analogously, a field might be defined as a collection of configurations it can take on
 
people sometimes talk about "field configurations", which is to fix a specific such map, or talk about the space of all such maps
but there isn't really any difference between "the scalar field" or "the configuration of a scalar field", it's just slightly different use of language
 
@SillyGoose who defines a particle as "as a collection of states it can occupy"?
A particle is what makes "blip" in my detector :P
and not every system that has "collections of states it can occupy" is a particle
 
okay wait i think maybe it is more conceivable now. so i have a lagrangian. I suppose there exists a gauge potential (as defined above) $A$. I write a term in the lagrangian with this $A$. I solve the equations of motion. Out I get a collection of $A$ that satisfy the equations of motion (among other things)? In other words, I get potentially a collection of connection 1-forms $\omega$ and potentially a collection of connections?
@ACuriousMind hehe it was a rough definition. maybe better to describe a system by the states it can occupy and then specialize to a system of a single particle. but there are still issues with the definition indeed
 
I feel you're going about this too abstractly, and I'm not known for making stuff too concrete :P
I have no idea what you mean by getting a "collection of $A$"
 
4:42 PM
well if i am doing normal electrostatics, i might be solving $\nabla^2 A^0 = 0$ with $A^i = 0$.
 
Be concrete: We have the EM action $S[A] = \int_M F[A]\wedge{\star}F[A]$, its equations of motion are the Maxwell equations (in terms of $A$)
these equations have solutions
 
this yields a collection of solutions $\{A^0 : \nabla^2 A^0 = 0\}$
 
Miao miao is watching on the sidelines and equally confused.
 
the nature of a gauge theory is that, in contrast to ordinary theories where the solutions to the e.o.m. are unique, they are only determined up to arbitrary functions of spacetime $f$, in that $A+\mathrm{d}f$ is a solution whenever $A$ is a solution
at what point does any talk about "collections" or "configurations" or whatever enter into this?
 
I think he wants to consider the space of all solutions to the eom, which are related by the gauge transformations
 
4:46 PM
It was confusing with the scalar field; not clear how talking about gauge fields would help clarify the discussion
 
When we say that there is a constant probability of finding a particle at any x (1D), does this imply that the wave function, that describes it, is a constant function ?
 
I think it's the use of the word "configuration" as distinct from the field itself which is causing some confusion too lol
 
@imbAF in which case it tends to be required to be a bounded region, yes
 
@ACuriousMind well all $A$ that solve the maxwell equations is a collection of $A$s, denote it $\mathcal{A}$, and i thought that a particular solution $A \in \mathcal{A}$ is what we call a configuration of the gauge field
 
@naturallyInconsistent yes, I am considering a bounded region but
One thing i find a bit confusing
you have the probability wave, the probability density, and finally the probability. So when you say that there is an equal probability of finding the particle at an arbitrary x positiion, you are making claims about the integral over a region of the pdf.
 
4:49 PM
@SillyGoose This sounds correct imo
 
@SillyGoose I don't know why you use "collection" instead of just saying set, and yes, in some contexts people might talk about a "configuration" as a specific solution. I don't understand how, given these interpretations of the words, I'm supposed to answer "Maybe I should ask: given a gauge field, what object is a configuration of this gauge field?"
 
But while the probability can be constant, the pdf or probability amplitude, can have expressions which differ from that of a constant expression. @naturallyInconsistent no?
 
@ACuriousMind im not sure when sets or collections are the appropriate object and i thought collections are more general :P
 
@imbAF const probability means constant pdf. But the probability amplitude can have any phase factor, i.e. it can have any momentum.
 
@imbAF no, it might have a space.dependent phase factor, i.e. $\psi(x)$ and $\mathrm{e}^{\mathrm{i}f(x)}\psi(x)$ have the same probability
 
4:51 PM
@ACuriousMind how do you define a gauge field and a configuration of it as?
 
@SillyGoose at no point outside of foundational set theory will this distinction ever be relevant
@SillyGoose that's the thing: I don't understand what that question means given our above understanding. Again: How would you answer this question for a scalar field?
 
@ACuriousMind @naturallyInconsistent Exactly. The probability amplitude might have a time dependent complex exponential. But in an exercise I am doing, for a particle in a box, it is said that the probability of finding the particle is constant. If I were to consider $\psi(x,t=0)$, how does the wave function look like. And in the exercise is taken as a constant i.e $\psi(x,t=0)=c$
which is a bold statement. Because the wave function itself can have phasors,which disappear when considering the pdf and calculating the probability
 
@imbAF I am quite certain that the question is only making the trivial statement that the particle is guaranteed to be found in the box, and that no shape is being specified of the PDF
Give us the full unadulterated question, and then we can discuss what it means
 
The particle is in a box. but I don't believe that means that the prob. is constant.
It is in the exercise. But it doesn't mean it is always the case
 
The probability that the particle is inside the box = 1
 
4:55 PM
Yeah overall
but that is always the case, even if the particle is not in a box, theres 1 for the particle being somewhere in the universe
There is the following potential $V(x)= \infty$ for $|x|\ge a/2$ and $V(x)=0$ for $|x|<a/2$
Consider a particle located exclusively in the left half of the potential at time t = 0, which is equally likely to appear at any location x < 0.
Which wave function $\psi(x; t = 0) = $\phi(x)$ describes this state at time t = 0? Note: Normalize $\psi(x; t = 0)$!
First of all I want to understand what it means for the probability of finding the particle at some x, is always constant
Now I know that the total probability of finding the particle in the box, is 1
 
Where did you get the "always" from?
they're just saying that at t=0, the probability of the particle is uniform in $[-a/2,0]$
 
@ACuriousMind what do you call a solution to the eom?
 
In this case, in the exercise
 
@SillyGoose ...I call it "a solution to the e.o.m."?
 
the probability of finding the particle in the box is 1
 
5:00 PM
why would I need any other word for it
 
So do you not use the word configuration
 
well, I might in the context of "configuration space", but generally not for "solution to the e.o.m.", no
 
What should I understand when it is said that the particle is equally likely to appear at any location x < 0. I don't believe it has to do with the probability of finding the particle in the box, which is 1.
 
2 mins ago, by ACuriousMind
they're just saying that at t=0, the probability of the particle is uniform in $[-a/2,0]$
and they're saying it's zero everywhere else
I don't know how else you could interpret this
 
It is also said, that it's eqally likely to apear at any location x<0. That's not the same as considering the probability of appearing somewhere, for x<0
 
5:04 PM
I don't know what you're saying
 
if the range is -a/2 to 0. And the probability is 1
I believe the probability from -a/2 to something less then 0, shouldn't be 1
 
look, all they're saying is that $\rho(x) = \frac{1}{a/2}$ in $[-a/2,0]$, and $\rho(x) = 0$ everywhere else.
 
Ok, well how do I explain this:
$\int_{-a/2}^0 \rho(x)=1$
$\int_{-\a/2}^{-\a/4}\rho(x)=????$
This is what I am trying to get at
 
I don't know where your problem in parsing the exercise text here is, but I'm absolutely certain that's what it means
@imbAF ???? = 1/2, clearly
I don't understand the problem
 
Im also gonna have to point out that ACM specifically wrote down the entire PDF so that you can compute any required integral if you want
 
5:08 PM
TIL soccer is short for association football, in the same way that rugger is short for rugby football
 
In the exercise, it was deduced from the statement that the particle is equally likely to appear at any location x<0, that the wave function is of the form $\psi(x,t=0)=K$ K a const. Why? Can it not have some complex exponentials. Ofc time dependent ones vanish
because t=0
But you can have phasors of the form $ee^{ikx}$
 
@imbAF you are correct. There is no good reason why they have ignored the complex exponentials
 
So I guess for simplicity they take it as I wrote above
 
Yes, it is not a particularly well-defined question precisely because of the fact that it failed to specify the complex phase, but otherwise, what ACM wrote is necessarily what you should consider.
 
And one more thing, just for general knowledge, if it was said that the probability is not constant, then does this mean, that once you would calculate the probability $\int \rho(x)dx$, you'd get something that is time dependent?
Or no?
 
5:13 PM
If you are integrating over all space, that integral should always be 1. It just means that the PDF is not a flat line.
If it is not integrating to one, then the deviation from 1 signifies either particle creation or destruction
 
Well if I would integrate over half the box region, and I get a time dependent probability, I can argue that the particle, can "jump" to the other half, no ?
what it means the PDF is not a flat line? Constant ?
 
Yes, and it must
 
@imbAF why are we talking about time-dependency? Your entire problem statement is about t=0.
 
@imbAF not constant probability is almost always meaning that the PDF is not a flat line. Not that the total probability integral isn't 1
 
@ACuriousMind And one more thing, just for general knowledge...
@naturallyInconsistent what does it mean the PDF is not flat line? That it is a function of position and time ?
And not constant
 
5:17 PM
@imbAF non-trivial function of position. I dont think people much care that it depends upon time, but if it is a function of position, it might as well depend upon time, if you so wish
 
And is there a reason for not caring about the time dependency ?
 
The concept we are trying to get you to understand, is what humans mean with "(non-)constant probability"
Time is one way to make the thing non-constant, but the obvious centre stage is when it is non-constant in space
 
Aha, so essentially, you keep time fixed, and the pdf is position dependent, so it's not a flat line as you say
Or you can do the opposite, a fixed position and time varies.
 
I have no idea what the topic of this conversation is anymore
 
@imbAF yes, this is also non-constant, but we don't usually have a time-dependent PDF that is space constant. It makes no sense.
 
5:24 PM
I see
 
@ACuriousMind imbAF used confuse. It was very effective!
 
@ACuriousMind It was about a particle , that for t=0 was located in the left half of the box, and I needed to find the wave function describing it. Then, for extra-knowledge, I asked @naturallyInconsistent what it means when the particle is not equally likely to appear at any location x<0, in difference from what was said in the exercise, which was that it was equally likely
Additionally, I asked him, what would be the physical interpretation, of the total probability in the left half of the box, not being 1 or being time dependent (if it makes sense to say this).
To which, he confirmed, that one can interpret this situation as if the particle is "jumping" on the other half of the box
@naturallyInconsistent would it be correct to interpret $\psi(x,t=0)$ as the position dependent component, when doing separation of variables for the wavefunction of a system, ofc for a time independent potential ?
 
5:58 PM
"How much energy is dissipated as a function of time?" doesn't this just mean what is the lagrangian as a function of time?
 
6:55 PM
howdy
how do you guys like pita chips
 
7:16 PM
consider a gauge theory defined by a principle bundle $P \xrightarrow{\pi} M$ with gauge group $G$, gauge field $A$, and action $S[A]$.
Define a connection as a choice of horizontal subspace $H_uP \subseteq T_uP$ at every $u \in P$ (satisfying certain axioms).
Is it accurate to say: (1) in solving the equations of motion obtained by stationizing $S[A]$ one obtains a set of solutions $\{A\} \equiv \mathcal{A}$; (2) each $A \in \mathcal{A}$ corresponds to a connection $1$-form $\omega$, which corresponds to a connection; (3) some pairs $A, A' \in \mathcal{A}$ correspond to the same connection 1-form and so the same connection; (4) it is an abuse of language to call a gauge field the connection, but a gauge field is in correspondence to a connection
(5) if we denote by $A \sim A'$ the equivalence relation "$A$ and $A'$ correspond to the same $\omega$", then $\mathcal{A} / \sim$ is in bijection with some subset of all connection $1$-forms, which is in bijection with a subset of all connections.
 
 
1 hour later…
8:29 PM
pita chips are great
 
8:44 PM
@SillyGoose theyre good with hummus
 
9:05 PM
the human brain can hold complex ideas now and manipulate them
this is an incomprehensible level of thinking to other smart organisms like dolphins. dolphins cant even imagine basic algebra, maybe they can imagine some arithmetic
so there might be some other level of thinking that outshines humans in the same way
 
9:28 PM
anyone wanna help with a basic physics problem
 
10:19 PM
@SillyGoose no, not really
it's a bit difficult to talk about this because there's the issue of different posssible trivializations. but let's talk about $\mathbb{R}^n$ where this isn't an issue
your (3) doesn't make any sense: An Ehresmann connection specifying a horizontal subbundle corresponds uniquely with a connection form $\omega$, which in turn corresponds uniquely with connection forms $A_i$ in trivializations. Hence "a connection", "a connection form" and "a gauge field" are all in bijection.
so no to (4), a gauge field is a connection is a connection form, these things are completely and utterly equivalent
and also no to (5) - it is not the case that "different $A$" would correspond to the same connection, but rather it is the case that different connections (corresponding to different $A$) may be related by a gauge transformation, and hence have the same action (since the action is gauge invariant); the space of "physical states" is the space of all connections/gauge fields modulo gauge transformations
 
okay i think i see. so $A \leftrightarrow \omega \leftrightarrow \text{connection}$. Furthermore, the action induces an equivalence relation on the set of one of these objects (which then induces the equivalence relation on all of them)?
and is it because $A$ actually appears in the action that we talk about the "gauge transformations" as acting on the space of these gauge fields $A$?
 
10:35 PM
the action is invariant under gauge transformations
and more specific arguments show that these transformations relate physically equivalent fields (or "configurations"), i.e. no observable quantity is changed by these transformations
hence they are a redundancy in the description, and we should consider the "true" space of physical configurations the space of fields modulo gauge transformations
this idea has nothing directly to do with the notion of connections, in principle you can have gauge transformations you have to quotient out without a connection formalism, it just turns out that the most relevant gauge theories are of this Yang-Mills type that can be phrased in terms of connections
 
10:59 PM
I am almost 100% overcomplicating this problem
The current induced is $I = \frac{vB_0\ell}{R}$ and the force on the bar by the magnetic field is directed opposite to its motion so $F_B = IB\ell=\frac{B_0^2\ell^2}{R}v$. So $$F_{net}=\frac{B_0^2\ell^2}{R}\frac{dx}{dt}-m\left(9.8\frac{\text{m}}{\text{s}^2}\right)=m\frac{d^2x}{dt^2} \implies m\frac{d^2x}{dt^2}-\frac{B_0^2\ell^2}{R}\frac{dx}{dt}+m\left(9.8\frac{\text{m}}{\text{s}^2}\right)=0$$
I made this into a 2nd order ODE and used the auxiliary equation
I don't see how I'm wrong in my reasoning but I'm also not sure if he expects 3 hypothetical solutions..
using the auxillary equation $am^2+bm+c=0$ we can discern that our equation fits as $a=\text{mass}$, $b=\frac{B_0^2\ell^2}{R}$, $c = 9.8\cdot\text{mass}$ so $b^2-4ac = \left(\frac{B_0^2\ell^2}{R}\right)^2-4m^2\left(9.8\frac{\text{m}}{\text{s}^2}\right)$ The solutions depend on whether this quantity is greater than $0$, equal to $0$, or less than $0$.
I was also considering just using a lagrangian but I can't think of what the potential energy term would be for the damping force
 
11:23 PM
yeah nvm I can just use a basic kinematics definition lol
 

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