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5:00 AM
oh no i have made a grave error in the past
u know how u can upvote a comment or flag it
well i didnt know you can like click the arrow
so when i hovered over the arrow and it was like "this was helpful"
i thought that was for the flag
so ive been flagging good comments
kms
hahahahaha
it is ok, the mods would just be rather confused for a while
how is the lorentzian vegetable?
i am eek how r u
what does eek mean?
kitty is having a ... blast ... hides red face
im unsure myself XD its a versatile word -- im at a very confusing point in my life lol
@naturallyInconsistent fkldjdslkfjdslkfjdslkfjdskl
@Relativisticcucumber sends clarity waves
@Relativisticcucumber asdlkjhsoidugekrtneritug indeed
5:07 AM
well at least someone here is having fun lol
jk i am also having fun
i am reading about path integrals
finally resuming qft after a week of none
oh no @JohnRennie flee
save your innocence
I was wondering that as well :-)
Oh, we had some long chats about path integration in both QM and QFT just not that long ago.
H O N K
i just thought that you are a nice elderly man and that you would want to be spared from the blast
now i feel awkward i have made many errors
5:09 AM
Nobody else knows that you made errors like that, Lorentzian vegetable
Let it be buried and never remembered
Did you say that the wacky fowl would go schwank?
@naturallyInconsistent ?? like the actual schwank?
i hope not dear god
What sound does the wacky fowl go for, then?
recently only heard the H O N K
@naturallyInconsistent sadly i learned that qft, at least the basics, are not physically interesting. many integrals
@naturallyInconsistent hehe
@naturallyInconsistent in what way do u mean "go for"
like their mating call?
5:12 AM
@Relativisticcucumber many initial stuff are just later straight up thrown away
@naturallyInconsistent kms
@Relativisticcucumber dunno. just some other sound of choice than H O N K
@Relativisticcucumber we can't help that QFT is a broken mishmash we do not really know how to coherently present to students
schwegg or squawk is their mating call
too bad i cant send my dolphin noise here idek how to type it
but i can really sound like a dolphin
@Relativisticcucumber uncomfortable with further mating calls, given the events that just passed
@Relativisticcucumber eek eek eek, eek eek
LOL
i become silent
5:16 AM
nuuu donttt
@naturallyInconsistent what is ur favourite hobby
dunno; maybe badminton, maybe games, too many to choose
:o badminton
unless, you consider nsfw stuff, in which case the list would explode
5:23 AM
what about badminton?
i like it but i am v bad
oh, im very ok playing with newbies
i used to have a friend that liked it so we played a lot
then they became my sworn enemy
so i have never played since
sigh
but here, meanwhile, the geographical displacement means no badminton for sooooo long
lets hope gym goes well
 
1 hour later…
6:47 AM
@Minsky hi. you are looking for this en.m.wikipedia.org/wiki/Heisenberg_picture . Look in the "Commutator relations" section for the exact analogy of Newton's laws in quantum mechanics
7:36 AM
can an electron electron collision produce jets, implying a qcd process among them?
8:35 AM
Can anyone give an example of a symmetric operator which is not essentially self adjoint but with finitely many self adjoint extensions?
 
2 hours later…
10:06 AM
Is it correct to say, in particle physics:
"It represents the characteristic momentum scale at which one is probing the structure of a hadron or nucleus." ?
10:23 AM
@Sanjana Such operators do not exist; if you're looking to the theory of extensions you should encounter a theorem about the form of all possible extensions (related to the Cayley transform) that explicitly shows operators either have no, one or infinitely many possible self-adjoint extensions.
one is finitely many
@Slereah but that's the "essentially self-adjoint" case she explicitly excluded
ah well
I recently learned that the extension of non-self adjoint operators in QM relates to the problem of classical systems that have an ill-defined Cauchy problem
which was neat to find
thank you @RyderRude
11:11 AM
@Slereah which relation is that?
via the boundary triplets?
Apparently the infinite extensions of the $\hat{x}^{-3}$ Hamiltonian correspond to the extensions of collision singularities classically?
$x^{-3}$?
in what case do you get a 3 up there?
Oh I guess it's $x^{-2}$ for the inverse cubic law
but yeah the inverse cubic law case
which is apparently pretty ill-defined
ah, it is the boundary triplets - the $c\leq 3/4$ case where you "have to choose what happens when the particle hits the singularity" is a special case of the general theory of self-adjoint extensions where you have to choose a very generic kind of "boundary condition" to pick a particular extension
 
3 hours later…
2:26 PM
hi
can a particle who has 3 components of kinetic energy only have 2 components in the potential energy part of a lagrangian
QM generalises the transformation of differential operators from the transformation of derivatives operators that we see in GR. This is because GR treats derivatives as vectors on a finite dimensional manifold while QM treats derivatives as operators on an infinite dimensional vector space
so like $L = T - V$ where $T$ has the 3 components, but $V$ only has the 2
@Obliv what do you mean by "components"?
like for polar coordinates you have the tangential and radial
2:32 PM
I mean in three dimensions you will generally have that $v^2$ has three summands regardless of what basis you write it down in
that's just a consequence of velocity being a 3-dimensional vector
yeah and you'd have potential energy in 3 dimensions too, the problem confuses me
why wouldn't the azimuthal angle matter
I mean the problem just told you to consider a situation where it doesn't
@Obliv consider the analogous problem in cartesian co ordinates. the kinetic energy can still depend on three variables while potential can be something like : V(x,y,z)=f(x)
then the azimuthal angle component of energy isn't conserved..
@Obliv there is no such thing as a "component of energy"
energy is a scalar
2:35 PM
then azimuthal component of momentum**
why would it not be conserved?
generally the momentum conjugate to a coordinate is conserved when the Lagrangian does not explicitly depend on that coordinate
what the book means here is $U(r, \theta, \phi)=f(r, \theta)$. $U$ is still defined on the entire configuration space. it's just independent of $\phi$ @Obliv
for e.g. you can consider the gravitational potential which is $U(r, \theta, \phi) \text{~} -\frac{1}{r}$
I get what you're saying but I'm trying to workout how it works if the object was spinning
why would the object be spinning?
the problem just talks about a particle moving in a certain potential
because the kinetic energy is specified in those 3 dimensions
2:39 PM
Newtonian point particles usually do not spin
OH
Ok..
I see
well ok even if the object wasn't spinning, if it had some other components of momentum that weren't in the direction of the potential, it still has some "energy" that isn't described by the potential energy function
which isn't terrible, because in reality it's fine to do that for certain cases, but in general that would be weird
I have no idea what you mean
the energy from momentum is fully described by the kinetic energy term
what does the potential have to do with momentum?
Ok imagine two spheres in space, keep one fixed. The one that is free can spin, move around in 3 dimensions and the potential energy function is only w.r.t the radius b/t the two
the potential depends on the position, not on momentum
@Obliv again, this is about point particles, not spheres
even at 0 potential energy, so when the spheres are gliding over each other, there will be some velocity of the free sphere racing around the other sphere
2:49 PM
they can't spin
you're just imagining a situation that has nothing to do with what your exercise is about :P
Ok then just the point is free, and still have a fixed sphere that provides the potential
the point can't spin, but it can still have components of velocity that aren't in the direction of the sphere
still not what the problem is about
but I'll bite: what does the direction of the velocity matter?
ok in the problem let's say the sphere provides the $U(r,\theta)$ so the point converts the energy along the radial axis and the longitudinal angle. If it's moving along the equator, that momentum doesn't change right?
why would a sphere "provide the $U(r,\theta)$"?
the problem literally doesn't say anything about where the potential comes from
Ok I'm not sure about that part
2:54 PM
and the usual example of a potential that is independent of the azimuth but not of $\theta$ would be that of a charged particle moving in a constant magnetic field in the z-direction
no spheres there
so how does the potential energy convert into KE
I have no idea what you mean by "convert"
like when the particle evolves in the system, lets say the potential energy goes to 0, so the energy is totally kinetic
The equations of motion are the E-L equations of that Lagrangian
@Obliv why would that happen
you seem to be postulating random kinds of motion without solving the equations of motion
in order to know what the particle does you have to solve the equations of motion
that's what they're for!
ok ill do that and save my questions for after if any are left
Oh man my calculus knowledge is in shambles
how do I take $\frac{\partial L}{\partial \theta}(\sin^2 \theta)$
nvm its a product rule
just to be sure $\frac{d}{dt}(r^2\dot{\theta})$ is two product rules too
$2r\dot{r}\dot{\theta} + \ddot{\theta}r^2$
@RyderRude does that look right to u?
3:24 PM
ok this can't be right.. then I'd have to take $\frac{d}{dt}(mr^2\sin^2 \theta \dot{\phi})$
5 product rules wth
I don't know why you're acting as if differentiation is hard :P
the great thing about it, in contrast to integration, is that it's just a bunch of rules you have to stubbornly apply
yeah I'm sure i'll feel more confident the more I do it, I'm just nervous with chain rules
I feel like I'm doing something wrong
what exactly are you trying to do
it's like the same thing as this right$f(g(h(j(k(x)))))$
I have to find the momentum conjugates of the lagrangian expression
why does finding the conjugate momenta involve time derivatives?
the conjugate momenta are by definition $p_i = \frac{\partial L}{\partial \dot{q}^i}$, no?
3:32 PM
I'll be honest I have no idea what a momentum conjugate is, I've just been plugging in $\frac{d}{dt}\frac{\partial L}{\partial \dot{...}}=\frac{\partial L}{...}$
@ACuriousMind he did not go over that in class, but its in the homework oh well
why are you looking at an exercise that asks you to find conjugate momenta if you don't know what those are
sounds like you didn't do some reading you were supposed to :P#
He doesn't assign reading
D:
and why on earth would you just randomly try to compute derivatives instead of trying to find out what you're actually supposed to do???
3:34 PM
I was like 90% sure that was what it was asking
you obviously have access to the internet, you could've just searched what conjugate momenta are
It's kind of silly I have to even google what it's asking, I feel like this course should be a bit more self contained
But yeah you're right
I mean if your course genuinely just has this term appear for the first time on an exercise sheet without defining it then that sucks
I would link u the course site but I don't wanna get doxed, but in the syllabus it says a text isn't required but he recommends a couple. And the lecture slides don't mention conjugate momentum either
4:16 PM
What does "Which of the conjugate momenta are conserved?" mean in the context of a lagrangian mechanics question
I thought all momentum was conserved in classical mech
fqq
fqq
what about the momentum of a ball in a uniform gravitational field?
I'm not sure, a field cannot even have momentum can it
@Obliv hi. it is best to learn Lagrangian mechanics from a book in an organised way
they explain all this in the books
@Obliv the ball has momentum, the question is not about the field
you throw the ball, it goes up, it comes down again, obviously the momentum changes along that path of the ball, doesn't it?
I was under the assumption the gravitational body changes in momentum too (ever so slightly since the mass of the planet is so much larger)
as the ball moves further away and closer
4:23 PM
but the gravitational body is not part of your model here
you're just looking at the ball
the ball's momentum is not conserved
I should rephrase "isn't momentum conserved in a closed system in classical mech"
you might take that as the definition of a "closed system"
but no one said anything about the Lagrangian system being "closed" in this sense
@Obliv only the total momentum is conserved. the canonical momenta refer to individual momenta as there is one canonical momentum defined for each degree of freedom
What's the more common term, canonical or individual @RyderRude
canonical
4:26 PM
canonical momentum is a technical term
"individual" isn't
you will hear the term canonical momentum conjugate of a variable
Right, and it is conserved exactly when..
the potential energy function describes it?
just look at its equation of motion!
what's $\dot{p}_i$?
it is best to learn Lagrangian mechanics from a book
@ACuriousMind $\frac{\partial L}{\partial \dot{q^i}}$
4:28 PM
@RyderRude I really don't think just giving away the answers to the exercise here is the right thing to do.
@Obliv no, that's $p_i$
note the dot
"conserved", after all, means that the time derivative is zero
I'm literally going step by step in this problem set. I don't even have equations of motion yet because that's part d)
@Obliv but it asks you which of the momenta are conserved!
that means you should look at the time derivative of the momenta
conserved means if its derivative is 0?
what do you think "conserved" means?
final minus initial is 0
4:30 PM
nope
$p_i = -p_f$?
if I run in a full circle and then stop again, my momentum at the start and stop is 0, but my momentum was not conserved along the motion
are you sure you're supposed to do this problem if you neither know what conjugate momenta are nor what conservation means?
like, this might just be accidentally the wrong exercise if you're having this much trouble
Wait no I just forgot this was week 2 we had $\dot{p_{tot}} = 0$ means total momentum is conserved
$\dot{p}_{tot} = \dot{p}_1 + \dot{p}_2 = F_{12} + F_{21}$
please also note that the total momentum is conserved only for the sum of the cartesian momenta. the sum of general canonical momenta is not conserved in general
general canonical momenta cannot even be summed because of different units
@Obliv sure, so when they ask you "which momenta are conserved" you're being asked to look at $\dot{p}_i$ and try to figure out whether that's zero or not
4:37 PM
@RyderRude I see, because of the change in units, it's like a totally new problem
@Obliv i meant that $\frac{\partial L}{\partial \dot{r}}$ and $\frac{\partial L}{\partial \dot {\theta}}$ have different units
your problem has nothing to do with total momentum anyway
Okay that makes sense I just have to find $\frac{\partial L}{\partial \dot{r}} = \dot{p}_r$ and see if it's 0 or not
@Obliv be careful: you again wrote down the equation for $p_r$, not $\dot{p}_r$
Oh right
also, a hint: while you can do this by actually computing all three time derivatives, if you use the E-L equations in a clever way you can do this without actually having to explicitly differentiate anything
4:41 PM
I want the clever way, I do not like taking 5 product rules :p
but dont tell me
I will try it first
then you have to stare at both the expression for $\dot{p}_i$ and the E-L equations until you see it :P
Oh dur I just take $\frac{\partial L}{\partial r}$
since it's equal to the left hand expression
wait that's not easier
oh right because it's a partial
that's a LOT easier
So i still have to do the time derivatives to get the equations of motion, but to check if can. momentum is conserved i just have to solve the right side of the E-L equation
5:05 PM
wait is $\frac{d}{dt}\sin\theta = \cos\dot{\theta}$?
@Obliv it is cos theta * dot theta. but why are you computing this?
i should say that the canonical momenta of any variable can always be summed over different particles, but only the total cartesian momenta and total angular momenta are conserved in most real world closed systems
but anyway, this isnt related to ur main question
ur question is about the conservation of individual canonical momenta. usually, books derive this result but u shud try this on ur own
5:22 PM
@Obliv in more detail, $\frac{\mathrm d\ }{\mathrm dt}\sin\vartheta=\frac{\mathrm d\vartheta}{\mathrm dt}\frac{\mathrm d\ }{\mathrm d\vartheta}\sin\vartheta=\dot\vartheta\,\cos\vartheta$
and I would point out that if your differentiation is shakey, the module itself will be a difficult sell.
@Obliv HINT: please dont bother computing any d/dt. just compute the rest of the EL eqn and stare at it
just do the partial derivatives
5:39 PM
sorry i think u were now solving the 4th question, which ask for the equations of motion
6:07 PM
yea I didn't realize that became $\dot{\theta}\cos\theta$ that changes things
so $\sin^2\theta$ becomes $2\dot{\theta}\sin\theta\cos\theta$
I got $$m(\ddot{\phi}r^2\sin^2{\theta}+\dot{\phi}(2r\dot{r}\sin^2{\theta}+2r^2\dot{\theta}\sin\theta\cos\theta))$$ for the $\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}$
 
5 hours later…
11:25 PM
how does one "set up" a problem like this? Do you still do the whole free body diagram business from newtonian mech

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