@EmilioPisanty I'm sure you know more about attosecond pulses more than meow: are these pulses used for nuclear physics?

if the propagator is the transition amplitude $\langle x \vert U(t,t') \vert x' \rangle$ and the path integral is $\langle x',t' \vert x,t \rangle$ then i dont understand why the path integral is expressed in srednicki as an integral of the propagators? are these two not mathematically the same thing?

@Relativisticcucumber you are considering just one single electron's propagator, whereas we are interested in propagator of many particles

also, propagators chain properly, and that is pretty much the physical insight in path integrals too

@naturallyInconsistent what do you mean by "propagators chain properly"

@Relativisticcucumber $\int\langle x^\prime|U(t^\prime,t^{\prime\prime})|x^{\prime\prime}\rangle\mathrm dx^{\prime\prime}\langle x^{\prime\prime}|U(t^{\prime\prime},t)|x\rangle=\langle x^\prime|U(t^\prime,t)|x\rangle$

i see

@naturallyInconsistent wait sorry i think im still confused. so in the path integrals, we integrate over all paths which is integrating over times but in the propagator we do the same thing (im thinking of this picture in particular) so what do you mean when you say propagators of many particles

@Relativisticcucumber The path integral is integrating over time, i.e. over many time slices, all of space or phase space for each time slice. In QM, we only have one particle, so the propagator will only be about that one particle. It can actually be for multiple particles, but only fixed number of particles, that do not transform types. In QFT, there must be the ability to make new particles appear, and so the propagator must be found to be able to do that too

For example, consider an interacting spinless but charged Klein-Gordon field, we want to interpret $\left<0\right|\phi(y_1)\phi(y_2)\phi(y_3)\phi^\dagger(x_1)\phi^\dagger(x_2)\left|0\right>$

as the propagator Green's function for fields generated onto the vacuum at spacetime events x1 and x2, and later being detected at 3 detectors at spacetime events y1, y2 and y3

@naturallyInconsistent okay and is it that we calculate this via the path integral or am i misunderstanding? and are $\phi$ field operators in this case? which is like a sum of creation and annihilation operators? im not sure how to see why $\phi^{\dagger}(x_1)$ means unless this is the creation operator but :P

@Relativisticcucumber Yes, the conception is that it will be the creation of electrons, so that it can be annihilated later. We will be using the partition function trickery to get those from the path integration.

@naturallyInconsistent okay i see thank you

hi

Again, it is utterly unlikely that anybody would be using the path integral partition function trickery in QM, whereas in QFT it is the norm. Srednicki is thus bending over backwards to try to do it in QM to show some form of continuity in logic, but ends up with a presentation that is incredibly convoluted.

@naturallyInconsistent hello. it is more appropriate to say $\phi ^{\dagger} (x_1)$ is made of two summands, which create a particle and annihilate an antiparticle at $x_1$ respectively

sad

bah so i know that $\langle 0 \vert 0 \rangle = Z(J)$ but then what does this indexed notation of $Z_1(J)$ mean? and in this image i cant figure out where this definition of $Z_1(J)$ comes from or what it means? and then how does 9.11 even come to be? a taylor series...? i do not even understand how 9.11 can stem from 9.10 and where these P and V's come from and what they are

I think we will need the page before this (9.10); my Srednicki is at home

and i somewhat get why the integral of $J$ could be a source but what does it even mean to say $iZ_gg\int d^4x$ is a vertex?

@Relativisticcucumber the cubic in functional derivatives means that it takes 3 propagators and stitches them together at one point. In Feynman diagram form it will look like a vertex.

@naturallyInconsistent so this vertex is nothing physical and is purely pictorial ?

@Relativisticcucumber From this page, we see that there is clearly an analogy being made, that free Lagrangian $\mathcal L_0$ corresponds with free partition function $Z_0(J)$ and intermediate with the full path integral $Z(J)$ and $Z_0(J)$ we are defining $Z_1(J)$ to be using just the first term of $\mathcal L_1$

@Relativisticcucumber This is why everybody who knows anything says that Feynman diagrams are nothing but pictorial representations of the perturbation series and have no physically sensible interpretations in their own right.

man the $Z$ is the root of all confusion i need to clear this up

@Relativisticcucumber It should not be confusing: The statement right at the end of page 59 said what it is.

i think my issue is more that i am seeing all of these things as mathematical statements, which is fine, but i am struggling to build that up into a cohesive picture that i can make sense of. usually i dont approach physics by remembering things -- i just learn and it builds a framework and i apply that but im getting lost because, while i agree with all of these statements mathematically, i feel i am reading long streams of random statements with no comprehensible meaning

i interpret $Z(J)$ as $\langle 0 \vert 0 \rangle$ which just doesnt seem to be useful and that trend carries on

as a result of the above, i cannot make sense of this content because i cannot just memorize these hundreds of lines of math

which tells me im doing smth v wrong

@Relativisticcucumber That is what I usually do too; sadly QFT is typically presented as this set of hundreds of lines of disconnected maths and no attempt is made to make a coherent whole.

I can only try to reassure you that you are doing something ok.

@naturallyInconsistent i dont see how one can learn it if this is true

@Relativisticcucumber But I just told you that the partition function trickery on Z(J) will be how we extract all the Green's functions propagators from it, and thus it is of fundamental importance.

@Relativisticcucumber That is why I was stuck for half a decade trying to read standard texts

okay stepping back for a moment, what im getting is that $Z$ is a generating function because when we functionally differentiate our path integral with respect to it, we bring down terms and this allows us to get the correlation functions. it is also called the partition function because we treat it similarly to the partition function in stat mech. is this correct? and we say this is the generating function of what? @naturallyInconsistent

or, i mean i assume generating functions should be called generators of something. is it the correlation functions?

consider the canonical ensemble. let $n^*_r$ denote the multiplicity of most probable state of the system-resevoir composite system.

@SillyGoose no thank you

in the equation on the bottom of this screenshot, the "for all r" would then not apply to $n^*_r$, right? or am i missing something. because i thought $n^*_r$ should not have a dependence on r, which labels the energy of the system (which then specifies the energy of the resevoir)

so in that case, $Z_1$ is the generator for the correlation functions for the system with lagrangian $\mathcal{L}_1$? i feel like this is wrong

@Relativisticcucumber Sorry, but I might be poisoning you on the terminology. Yes, this is exactly the same as the partition function in stat therm. I think generating function might be a more common name, but in maths the term generating function is also used, say, for the generating function of Legendre polynomials and so forth, and we do not use those generating functions the same way we do it in stat therm or in probability and statistics.

Yes, it generates the correlation functions, that we will interpret as exactly the same thing as propagators = Green's functions.

@Relativisticcucumber It is the generator for correlation functions that is missing the necessary counterterms, by definition

@SillyGoose Miao saw that: H O N K

Good luck

LOL

@naturallyInconsistent ok ok

@Relativisticcucumber the whole point is that $Z_1$ is a simpler system to study, so that you can temporarily ignore the annoyances coming from counterterms, and focus on the basic machinery of Feynman diagrams before having counterterms come in and make everything ugly.

@SillyGoose It is "for all indices/energies r", and yes, $n_r^*$ is not going to depend upon r per se; Really, it just means that for each energy level $E_r$ the value of $n_r^*$ is a different and depends upon $\alpha,\beta\ \&\ E_r$

i think this was a small progress. so then i just need to show that the taylor expansion does, indeed work. then i note that the vertices are due to the cube as you said and then the second sum over p contains the propagator and source. together these characterize the integral, so then we move to utilize feynman diagrams to organize this? so the feynman diagrams allow us to perform path integrals get the correlation functions? am i getting this right? @naturallyInconsistent

@naturallyInconsistent i see

@SillyGoose is sexy time a euphemism for physics with nI

shweg @Relativisticcucumber

@Relativisticcucumber You are starting to get it. The path integrals are already completely performed by getting the $Z_0(J)$ and writing the interaction as $e^{i\mathcal L(\partial/\partial J)}Z_0(J)$, and all that you are doing later on are just generating function trickery, organising them as Feynman diagrams, so that it helps you to organise the perturbation terms as integrals, etc.

@naturallyInconsistent ah sorry one more thing is that i keep seeing this mentioning of perturbation terms. where is perturbation happening?

is this the same perturbation theory as in QM?

The Taylor's expansion fails. Radius of convergence = 0

@Relativisticcucumber way worse than QM.

oh no i have to go D: im sure i will be back tomorrow

thank you for the help @naturallyInconsistent

eek

@Relativisticcucumber We are treating interactions as perturbation (and counterterms). If we were mathematically powerful enough to just get $Z(J)$ from Equation (9.5) alone, we would not have had to work with $Z_0(J)$ at all. But we only know how to get Equation (9.7), so that we have to artificially break up the actual Lagrangian into Equations (9.8) and (9.9) and define weird shit like Equation (9.10)

@Relativisticcucumber lol, miao miao is much moar playful than normal kittens. All welcome red faced

@SillyGoose shweg

@Relativisticcucumber $Z(J)$ is not $\langle 0\vert 0\rangle$, that's $Z(0)$.

you seem to be desperate to find some sort of "physical meaning" of $Z(J)$, but there really isn't a straightforward one. In QFT, it's a convenient tool to generate correlation functions but much like LSZ there's a lot of math here until you arrive at the physical stuff like Feynman diagrams. That's just how it is.

Isn't J just a source term

@Slereah sure but "source" of what? :P

Eternal life?

that'd be $Z(☥)$

That would be the foundation of youth.

what

The foundation of youth is the source of eternal life.

uh, do you mean the fountain of youth?

@ACuriousMind no, it is $\left<0|0\right>_J$, the free field vacuum expectation for arbitrary source field J, normalised to have $Z(J=0)=1$. Of course, there is very little use of the $J\neq0$ bits

Yes, damn auto fill >8(

lol

@naturallyInconsistent yeah but I think the entire trouble here is that there's nothing physical about $J$ it's just a trick

@user726941 what is of use is the function Z(J), not just the value at 0

@ACuriousMind I had been telling Lorentzian vegetable and wacky fowl about this fact for quite a while by now, and with some moderate amounts of success.

u r required the neighborhood values to take the derivative at 0. so the function is useful

the meaningfulness of $\langle 0\vert 0\rangle$ is questionable anyway since it's usually divergent garbage we have to renormalize away :P

I mean isn't it pretty common in a variety of application to use a source field for various reasons

Like pretending that the source of your EM field is some background entity that you are not studying under the lens of QFT

it can represent interaction with a classical field

Does anybody else use the dark mode in chat?

Didn't even know it existed

Yup

on android

mobile

It's very easy on the eyes

@naturallyInconsistent for now, not really. Most nuclear transitions are at high (photon) energies that are very hard to reach optically. (There are a few exceptions, though.) There's nothing stopping you in principle, so long as it's an electromagnetic transition, but it's mostly not practical.

In the future, though... who knows! 🙂

I found it on the Opera browser @Slereah

@EmilioPisanty nice post about the attosecond.

I wanted to add the "time" tag, but five tags is the maximum.

technically isn't wrong lol

did life invent logic and sets or inherit them from the universe?

life perceives the universe in terms sets obeying the rules of logic, but is this how the universe exists, or is this life's delusion of the universe?

there is a strong argument for "this is how the universe exists"

"Brain power is applied biochemistry."

*just

🧪⚗️

@RyderRude I mean if you ask me, all these terminologies and ideas are just concepts that humans realized are useful

so i guess the question is the same as "do humans come up with concepts, or do concepts exist and humans just identify which are useful?"

granted though i don't really know if there's a real difference between those two

@SirCumference i was using "life" instead of "humans" because sets and logic are something very universal to life. different life forms can logically deduce and can group objects into sets

@user726941 thanks!

@SirCumference by "group objects into sets", I mean different life forms have the concept of distinguishable things, similar things, etc

yes, these concepts are useful for survival. here we can form a picture of early non-conscious life which is a bunch of molecules thriving in the environment. and the molecules, that survive after many generations, have these concepts of distinguishable things, logic and decisions inherited into their dna

but notice that our initial picture pre-supposed the existence of sets and logic in the outside universe. our picture was "a bunch of molecules thriving in the evironment and reproducing as time flows". This is from where life inherited these concepts into their dna

this means that the concepts of "a bunch of things" and "logic" exist in the universe

i think, to define concepts like "usefulness", we need to first assume that the objective universe comprises of " things obeying the laws of physics (i.e. sets and logic)". Then we can define "what's useful for a bunch of molecules to survive over time". so this viewpoint would mean that life inherited these concepts from their existence in the universe

Energy is the conserved quantity associated to time translation invariance. Momentum is the conserved quantity associated to spatial translation invariance. Which conserved quantity is associated to Lorentz boost invariance? (In SR and relativistic field theories)

@notmyrealname see e.g. physics.stackexchange.com/q/12559/50583

what is the deep connection between the generators of Lorentz algebra and the electromagnetic field tensor? I mean that both form an anti-symmetric 4x4 matrix. the generators of rotations occupy the same place in the matrix as the magneitc fields in $F^{\mu \nu}$

@ACuriousMind that's not what the post says!

yes, I think ACuriousMind meant to say the angular momentum tensor

@Slereah oops, you're right

but i cant edit it after this long

(removed)

I think i got it. this is because the generators transform like tensors under the action of the group

e.g. generators of translations form a four-vector

this fact shows up even in phase space mechanics. the group acts on generators too

i dont see yet any deep connection with $F^{\mu \nu}$ other than that they both are anti-symmetric

why would they be the same thing just because they're antisymmetric tensors

yes. i just realised that the only thing they have in common is anti-symmetry

idk. seeing $J_{ij}$ inside that matrix just reminded me of magnetic fields

All 2-forms are like that, doesn't mean they're all related

yes

@ACuriousMind this answer somehow also mentions this spooky analogy with the EM tensor :P

but the analogy is just anti-symmetry

You can always decompose such forms like that since that's just an identity of the levi-civita tensor

commutations relations of the rotation algebra can be written in terms of the levi civita tensor. i usually associate levi civita tensor with determinants and Hodge dual. why does the levi civita show up in the rotation algebra?

i think this appearance of $\epsilon$ is a co-incidence for the rotation algebra in 3D?

It just shows up in everything involving anticommutation

A lot of the weirder connections in geometry basically occur because they all involve anticommutation

but the commutator is always an-anti commutative operation? this means levi civita should show up in all Lie algebras?

i get it that it makes sense from a $[J_1, J_2]=-[J_2,J_1]$ perspective

but the rhs can specifically be written as a tensor multi of levi civita and $J_i$

@Slereah does the levi civita show up in 4D rotation lie algebra?

the rotation algebra in 3D also looks analogous to the 3D cross product algebra. the reason for the levi civita in at least the cross product is the Hodge dual operation

@EmilioPisanty im asking this because im myowself inside the nookooliar fooshun field. Those using lasers to do it want to be able to dump a lot of energy within a tiny pulse, so that the fusion can happen within the time between an initial pulse to dispel the electrons and other electrons coming back in to fill the holes thus left behind. Maybe I should be asking the laser fusion prof we are collaborating with.

Slereah, i'm sorry if i said said something stupid. thank u for ur time

@Relativisticcucumber The wikipedia says $\psi(x,t) = \int_{-\infty}^{\infty} K(x,t;x',t') dx'$, and that $K$ has the expression $K(x,t) = \int_{q(t') = x'}^{q(t)=x} D[q(t)] e^{i \int_{t'}^t L(t,q,\dot{q}) dt}$. In Dirac notation this is $\psi(x,t) = <x,t|\psi> = \int dx' <x,t|x',t'><x',t'|\psi> = \int dx' K(x,t;x',t')\psi(x',t')$. A slightly tricky exercise is to re-derive this in $\psi(x,t) = \sum_n c_n(t) \psi_n(x)$ notation which I'll leave to you.

Note $|x,t> = e^{-iHt}|x,0> = e^{-iHt}|x>$ so $K$ implicitly depends on the full $H$. Now in this notation we can nicely talk about an initial zero wave function $<x,t|0> = 0$ evolving into say another zero wave function, thus focusing on the energy of the fields involved in the Hamiltonian alone, in studying $Z = <0|0>$, so it's not just a bunch of random math

In other words, usually when we write $\psi(x,t)$ we're talking about say an electron moving in an electromagnetic field which is encoded in the definition of $H$, so studying $0$ is studying nothing moving through an electromagnetic field, thus we are implicitly studying the electromagnetic field (e.g. at two different times) through which anything could move in studying $<0|0>$

@bolbteppa i think u mean "studying $\langle 0|0\rangle _J$" . <0|0> is just 1 in ordinary QM

and what did u mean by a zero-wavefunction?

I guess I should say that $|0>$ is really the ground state $\psi_0(x,t)$

and the $Z$ thing is a bit subtle to fix

@bolbteppa yes

we are studying the transition amplitude from the ground state to the ground state (in asymptotic times), in the presence of an external source $J$

does anyone know what stochastic quantization is

@NiharKarve a bunch of nonsense

@RyderRude The only issue is really motivating this $J$ stuff and relating $<0|0>$ to what I wrote, it's not a big deal but there is a subtlety, I'll leave it for now and see if the main point gets through

@NiharKarve as you can see, the initial paper is one of these crazy 'derive QM from classical mechanics' papers

$U^i = \epsilon^{ijk} V^j W^k = (V \times W)^i$, he asked that back in 2010

yes. it has a great answer

i was just discussing why the the levi civita show up in the 3D rotation Lie algebra

@Amit that post is about cross product. u will b interested in it maybe

Hello folks

quick sanity check

can I ask someone here to access the PDF of this paper?

doi.org/10.3367/UFNe.2021.10.039078

(just checking that it is indeed publicly available, and not just within my institution)

@EmilioPisanty I can access it from home.

it's accessible (after checking the box saying I agree to the terms of use)

So I asked my professor this after class today but I didn't really understand the answer. In polar coordinates, $\rho = mv = m(\dot{r} + r^2\dot{\theta})$ but in cartesian coordinates $\rho = m(v_x + v_y)$ is this correct so far?

$\rho$???

who on earth denotes the momentum as $\rho$

I thought it was rho

it's $p$

for impulse

well anyway, I'm curious why in the classical newtonian definition of angular momentum (before the usage of lagrangian mech), we have $p_{\theta} = mv_{\theta} = mr^2\dot{\theta}$ instead of $p_{\theta} = mv_{\theta} = mr\dot{\theta}$

Oh yeah I remember researching the whole vis viva leibniz newton controversy

and how newton said impulse was more fundamental than leibniz' mv^2

since if we're to take the angular component of $v$, we just have $v = \dot{r} + r\dot{\theta}$

@ACuriousMind looks a little like a p

BTW we're covering noether's theorem in a couple of weeks so i can finally understand your answer to my question on SE from 2016 lol

@Obliv The Lagrangian $p_\theta$ is not the $\theta$-component of linear momentum

you have to be careful with the notation here

your second $p_\theta$ is the $\theta$-component of linear momentum, your first $p_\theta$ is an angular momentum (conjugate to $\theta$)

those just aren't the same thing

Okay, so "momentum" isn't defined as $mv_a$ for some a-component of v in any coordinates

but even before lagrangian mechanics (i think) there was the angular momentum equation?

"momentum" just isn't specific enough

there's linear (kinematic) momentum, there's canonical momentum and there's angular momentum

sometimes two of these coincide (i.e. one of them with the canonical momentum), but in general they don't

is canonical and conjugate the same thing

I just didn't want to type out "canonically conjugate" :P

what the heck does canonically mean? we're talking about canon physics not fanfic/non canon?

he hasn't mentioned canonical yet so that's why I ask

(tfw someone with english as a second language knows more about your own language) nvm i'll just google it

altho technically it's my 2nd language too, but at age 4

It follows the canon of the Catholic Church

Not to be confused with the cannon of the catholic church?

any set of generalized positions and their canonically conjugate momenta forms a nice (="canonical") set of coordinates for phase space is the origin, I think

$v_x + v_y$ is not the velocity, unless this is a vector equation?

so polar coordinates can be canonically conjugate or canonical as can cartesian?

yeah it's a vector equation

@Obliv a position coordinate is (canonically) conjugate to a momentum coordinate

$d \mathbf{r} = dx \hat{i} + d y \hat{j} = d r \hat{e}_r + r d \theta \hat{e}_{\theta}$ leads to your $\mathbf{v}$. maybe you need help deriving this

it makes no sense to speak of them being "canonically conjugate" in isolation, only together with their momentum: $x$ is canonically conjugate to $p_x = \frac{\partial L}{\partial \dot{x}}$ for any generalized position $x$

@ACuriousMind so the $\theta$ component of linear momentum $\ne$ the angular momentum? But the radial component of linear momentum $=$ the radial momentum?

@Slereah exactly, the terminology was introduce in Pope Clement XIV's book on classical mechanics

@Obliv what is "the radial momentum"

"canonical" is just a math word to mean "the one you should care about"

the Official one

$\frac{\partial L}{\partial r} = \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}$?

but of course the $\theta$-component of linear momentum is not angular momentum, that doesn't even make sense unit-wise

@bolbteppa I can derive that, but what comes next for angular momentum I don't understand outside of the context of the e-l equation

but why then is the $\theta$-component of velocity the angular velocity?

$ds^2 = dx^2 + dy^2 = dr^2 + r^2 d \theta^2$, $L = \frac{1}{2} mv^2 = \frac{1}{2} m (ds/dt)^2 = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2)$, $p_r = \partial L/\partial \dot{r} = m \dot{r}$, $p_{\theta} = \partial L/\partial \dot{\theta} = m r^2 \dot{\theta}$

in Newton's time, I think momentum or impulse or vis viva whatever you want to call it, was the description of how the proportionality of mass*velocity gives you it

@Obliv I mean it just works out that way?

I'm not sure what the question here is, really

also I'm not even sure what "angular velocity" is supposed to mean for a general situation

If $(x,y)$ are your canonical coordinates, $(p_x = m \dot{x}, p_y = m \dot{y})$ are your canonical momenta, if $(r,\theta)$ are your canonical coordinates, $(p_r = m \dot{r},p_{\theta} = m r^2 \dot{\theta})$ are your canonical momenta, regardless of the choice, they satisfy $\{u^i,p_j\} = \delta^i_j$ in Poisson bracket language

Well I guess I should call $(x,p_x;y,p_y)$ the canonical coordinates on $(x,p)$ phase space

I know, and I get that $p_{\theta}$ is conserved if the potential energy function doesn't depend on $\theta$ @bolbteppa but .. is there some use of $mr\dot{\theta}$ ?

in the situation of potential energy function not depending on $\theta$, is $mr\dot{\theta}$ also conserved..?

Not sure what you mean, are you asking why is it $m r^2 \dot{\theta}$ and not $m \dot{\theta}$?

as in, $L = T - U(r)$ so $\frac{\partial L}{\partial \theta} = 0$

@Obliv I mean you already know that's a component of linear momentum, so it's conserved when linear momentum is conserved, at the very least (i.e. for a free particle)

ooh ok

just because it's not a canonical momentum in spherical coordinates doesn't mean it's no longer a linear momentum

it's also part of angular momentum, missing an $r$ proportionality

why would it be "part of angular momentum" if it's missing that $r$?

what is that supposed to mean?

in terms of factorization, $5$ is almost $10$ just missing a proportionality constant of $2$?

just with variables

In the case of $L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) - U(r)$, $\theta$ is a 'cyclic coordinate' and we see the E-L equation is $\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}} = 0$ so that $M = \frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta}$ is a conserved quantity

@Obliv I don't know what you're trying to tell me :P

$x^2$ is just $x$ times $x$, that doesn't mean you can say a lot about $x^2$ as a function by looking at $x$ as a function

what does this conservation imply.. that $\dot{\theta} = 0$?

we've been over this; you should remember what "conserved" means in this context

Thus you can set $r^2 \dot{\theta}^2 = M^2/m^2 r^2$

well yes, but then so is $mr\dot{\theta}$ @ACuriousmind so the angular component of linear momentum is also conserved

wait sorry

what

$\frac{d}{dt}(mr^2\dot{\theta}) = 0$ so $\frac{d(r^2)}{dt}$ could also be 0

or maybe only the product is constant!

you really can't say anything, in general, about $\dot{\theta)$ or $\dot{r}$ just from that equation

So either $\ddot{\theta} = 0, \dot{r}\dot{\theta} = 0$

@RyderRude Thanks, I can't resist a good cross product

with $\dot{r} = 0$ or $\dot{\theta} = 0$ ?

or is it more complicated than that

I'm not sure what you're doing :P

I'm looking at cases where $\frac{d}{dt}(mr^2\dot{\theta}) = 0 = mr\ddot{\theta} + 2mr\dot{r}\dot{\theta}$

is that differentiation correct..

I'll sleep on this nvm

I gotta do optics hw and study anyway

nothing here has to be zero at all

you can just have $mr\ddot{\theta} = -2mr\dot{r}\dot{\theta}$

then $\ddot{\theta} = -2\dot{r}\dot{\theta}$

wait but $r\ddot{\theta} = a$ is linear acceleration right

so $r\ddot{\theta} = -2r\dot{r}\dot{\theta} = a$

but anyway just because $\frac{d}{dt}(a^2) = 2a\dot{a} = 0$ doesn't mean $\dot{a} = 0$ in general I guess

so $\frac{d}{dt}(mr^2\dot{\theta}) = 0$ doesn't mean $\frac{d}{dt}(mr\dot{\theta}) = 0$

@Loong thanks!

@Obliv No

$AB = 0$ implies $A=0$ or $B=0$, $AB + CD = 0$ does not imply any of them have to zero

thank you

does anyone know of any site where you can basically inpute lens,mirrors, etc with different parameters to simulate ray diagrams?

Can't seem to find anything on google

@ACuriousMind what i wanted was really just a definition. i see now that srednicki considers the definition to be $Z_0(J) \equiv \langle 0 \vert 0 \rangle_J$ which i understand to mean the path integral of the free field theory where we have added a source term ($J$). I think this is confusing because, to me, the subscript with 0 implies there should be other subscripts, and if we are just adding $J$ which is equal to zero, idk why we write as a function of $J$ but who am i to debate notation

not to mention before naturally explained the partition function relation, the choice of $Z$ was also not obviously reasonable to me :P

@bolbteppa i think i see what you are saying but i have one confusion. if i understand correctly, your main point is that we can study 0 particle states that start and end as zero particle states. if they stay zero particle the entire time, since our expressions depend on $H$, then we are studying the energy of whatever these particles would propagate through. is this correct? [..]

[...] the confusion i have is with the mathematics that you wrote. i feel $\vert 0 \rangle$ is qft notation and $\vert x,t \rangle$ is QM notation, so i was a bit confused about that. when you say zero wave function, i assume you mean vacuum state in qft?

$|n>$ is the level $n$ energy eigenfunction, so $|0>$ is the ground state zero energy level eigenfunction, it is different from $|x,t>$ and it's not qft notation it's qm notation for an eigenstate

@bolbteppa okay i misunderstood what you meant because usually we dont write $\langle x,t \vert n \rangle$ in the qm i took

but okay nonetheless is my conceptual understanding of what you are trying to say correct?

$\psi_n(x,t) = <x,t|n>$

Yes

okay i think that helps

@bolbteppa right i see, makes sense

okay thank you

i also have an opinion based question. in general, i feel like people have an impression that it is good to be able to work through things oneself and to not rely on others to learn things. however, i feel like if you go to a good uni and have great teachers, you don't learn anything yourself -- you just show up and are taught everything in a sensible way. [...]

[...] it seems nobody is criticizing this system, so im confused about why there is such a push for learning things on ones own and why this is valuable? i feel like if you know something, you know it, and i dont see why it matters at all how that information is obtained?

this is not specific to stack exchange -- i just have this impression for people in general

Learning physics is not "obtaining information" and you don't "just show up and are taught"

> Spoon feeding, in the long run, teaches nothing to the learner other than the shape of the spoon.

is it wrong to read $U$ here as the expectation value of the energy $\langle E \rangle$?

where by expectation i mean the expectation value w.r.t. to the distribution of energy eigenstates, i.e. $\langle E \rangle = \sum_r p_r E_r$. this is mathematically the same as teh second statement of (1) in the screenshot

but im not sure why we'd give it this ambiguous name $U$ "the average energy per system in the ensemble"

also in eq (2), do we identify the term being differentiated on the right hand side of the equation as the moment generating function for the energy (random variable) in the canonical ensemble?