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123
123
4:24 AM
GooD MoRning @Azmuth
 
Azmuth
Azmuth
Good Morning :)
 
123
123
what about corona situation in your region.
 
123
123
4:52 AM
What is the difference between projectile and circular motion is a sense that.
In circular motion velocity is constant it changes direction only. $\vec{a}_c = \frac{\vec{v}^2}{\vec{r}}$
In Projectile we also see direction is changing but acceleration does velocity change. Why?
 
 
1 hour later…
bolbteppa
bolbteppa
6:00 AM
You divided by a vector...
 
123
123
No $v^2$ is not vector
$|\vec{a}| = \frac{v^2}{r}$
 
bolbteppa
bolbteppa
You wrote $\frac{1}{\vec{r}}$
You can compare projectile and circular motion by asking when does projectile motion become circular motion i.e. when does a thrown object start orbiting the earth
 
123
123
my question was that in circular motion there is constant velocity. mean acceleration does two things change in velocity and change in direction.
In circular motion we can see that acceleration does only change in direction.
But in projectile motion acceleration does only change in velocity but we see that there is also changing direction. why this change in direction not in calculation.
All the acceleration we use in change in velocity but there is also direction changes continuously.
 
bolbteppa
bolbteppa
6:17 AM
The acceleration always points down in projectile motion, it always points towards the center of a circle in uniform circular motion
In circular motion, as the particle moves, the acceleration changes direction, this does not happen in projectile motion
 
123
123
@bolbteppa I know this frnd. I am confused there is also direction changes in projectile why we don't encounter this situation?
 
bolbteppa
bolbteppa
What does that mean
The acceleration does not change direction
 
123
123
In curvilinear motion we take acceleration as $a = a_{parallel} + a_{perp}$
$a_{parallel}$ changes velocity and $a_{perp}$ changes direction in terms of polar components.
 
bolbteppa
bolbteppa
In projectile motion, if you expand those vectors in terms of their x and y components, $\vec{a} = \vec{a}_{parallel} + \vec{a}_{perp} = (0,a_y)_{parallel} + (0,a_y)_{perp}$ you see they only have a y component that is non-zero
 
123
123
acceleration means Force here it is just multiplication of scalar mass. Consider it as 1
Means in projectile it can only changes direction? right
But physically when i see the projectile trajectory it is parabolic it changes direction. How we explain this? It conflict with our direction problem.
 
bolbteppa
bolbteppa
6:26 AM
In projectile motion, the acceleration always points down, but the initial velocity can point in any direction, so the y coordinate can go up and down because there is a velocity in the y direction also the acceleration has to make it change direction, while the particles e.g. moves right so you get parabolic shapes
 
123
123
Hi, @Azmuth
 
Azmuth
Azmuth
Hey :)
 
123
123
You also take a glance on my question. As @bolbteppa also helping me
 
bolbteppa
bolbteppa
 
Azmuth
Azmuth
$| \vec a | = v^2/r$
 
123
123
6:29 AM
@Azmuth Pls see my question above.
 
Azmuth
Azmuth
Velocity changes in both Projectile and circular motion.
@123 check these notes
 
123
123
@bolbteppa Hmm.. It means it is not necessary what we see in motion changes direction due to acceleration?
 
Azmuth
Azmuth
No.
 
123
123
@Azmuth I know all the calculation of projectile and circular motion. I was asking about what we see.
What i think now after discussion with @bolbteppa pls correct me if i am wrong
If motion is curved like in projectile it is not necessary which is due to acceleration. right or wrong
 
Azmuth
Azmuth
Motion is curved due to Force not acting in the direction of motion and that causes acceleration not being in the direction of motion and that causes curved path.
 
123
123
6:36 AM
This is my question. Whenever force is not in the direction of motion is always curved. right
 
Azmuth
Azmuth
yes
 
123
123
and we divide acceleration in two parts $a = a_{cetripetal} + a_{tangential}$
 
Azmuth
Azmuth
@123 wait, 1 sec... Someone else is asking me doubts in other room.
@123 yes, in case of circular motion
 
123
123
$a_c$ responsible of changes direction and $a_t$ responsible changes velocity
This could also be the case in projectile.
 
Azmuth
Azmuth
@123 check these books, they are explained very well here: archive.org/details/FundamentalsOfPhysicsHalliday9thEdition
 
123
123
6:41 AM
I have read @Azmuth Halliday, Resnick
 
Azmuth
Azmuth
@123 I didn't say motion is curved due to 2 different acceleration components.
 
123
123
I know calculations are correct. I am asking question because of confusion.
 
Azmuth
Azmuth
It's just because direction of accelertation being not aligned to velocity vector
there is only centripetal acceleration and tangential accleration is = 0 (in uniform circular motion)
 
123
123
@Azmuth Yes. Also explain in projectile
I was thought whenever we see direction changes there is always two components of acceleration radial and tangential.
 
Azmuth
Azmuth
@123 In projectie there is only one component of acceleration, that's downwards, the sideways has no acceleration, only velocity
 
123
123
6:43 AM
In projectile this is not the case.
 
Azmuth
Azmuth
@123 No, not necessarily, the tangential components aren't necessarily required to be non zero to have change in direction.
 
123
123
@Azmuth yes. i didn't mean changes direction mean tangential is zero
If direction changes there is always a radial component of acceleration or not? which is responsible of changing direction.
 
Azmuth
Azmuth
@123 always radial component would be non zero.... (if the motion is on a plane)
 
123
123
OK let me clear one what in my mind.
In circular motion always radial component of acceleration which called centripetal acceleration $a_c = \frac{v^2}{r}$ at which velocity is constant. Take mass = 1 to take force and acceleration same.
In this case velocity is not changing acceleration due to force serves only changes direction.
 
Azmuth
Azmuth
@123 No, that's true only for *uniform circular motion)
 
123
123
6:49 AM
Is this correct or wrong
Yes yes in uniform circular motion we are taking ideal cases for both circular and projectile
 
Azmuth
Azmuth
@123 this i right
correct :)
 
123
123
If there is any tangential component of acceleration in circular motion it also changes tangential velocity and angular velocity.
With this idea i have in my mind whenever i see direction is changing (not necessarily circular) there is always radial component and tangential component. radial at which velocity is constant and tangential at which velocity changes.
Is this correct or wrong?
 
satan 29
satan 29
does anyone know when ACM comes online?
 
Azmuth
Azmuth
Uniform circular motion is defined as a a special case of circular motion where tangential direction is $0$. But it doesn't mean that it'd always remain $0$. There can be tangential components too, but then in such case, $a = v^2/r$ won't be applicable. You need velocity at an instant to make this true
@satan29 usually after 3 IST (my best guess, although, he doesn't has a regular time) @ACuriousMind
 
123
123
@Azmuth My confusion in projectile motion.
I see the direction changes in projectile. Why there is no radial component at which velocity constant which is responsible for changing direction of projectile. Using above analogy
 
Azmuth
Azmuth
6:56 AM
@123 $v^2/r$ will also work for projectile motion to find the instantaneous radius, but that's something radically different from the radius you see in circular motion
also, instantaneous radius is true for all kind of curved motion.
check this:
3
Q: Calculating force through ICOR (instantaneous centre of rotation)

Vasu Goyal A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ re...

newtonian-mechanics rotational-dynamics rotation
 
123
123
@Azmuth Yes this is what i wanted to know what happened in projectile it is actually changing direction
 
bolbteppa
bolbteppa
This is explained in the picture I sent above, I don't know what the problem is, the acceleration pointing down is making the velocity vector go from pointing up at an angle to pointing down at an angle until it hits the ground
 
123
123
@Azmuth Yes i agree with instantaneous. Because in curved we always need a support of calculus.
 
Azmuth
Azmuth
@123 ICOR is a cheap and dirty method to approximate a very small/tiny section of any curved motion as a curved motion to use rotational dyanamics there.
@123 Even straight line paths require calculus (such as non uniform).
 
123
123
Yes you are right. I know the difference. Thanks
@Azmuth Yes it depend what parameters we are plotting whether it is behaving linear or not
 
Azmuth
Azmuth
7:00 AM
Unfortunately ICOR won't be any of any help in projectile path/motion calculus, you should look into the picture @bolbteppa sent above. The strategies to decode them into Newton's Laws are well pictured there.
@123 There are certain non linear cases which doesn't requires caculus either. (But as a physicist, a theoretical physicist, one should use calculus), god knows what complications a question has to offer,.
 
123
123
If we use radial and tangential component of acceleration due to gravity which is always pointing downward along y-axis. Can we see both components separately radial changes direction and tangential changes velocity?
 
Jack Rod
Jack Rod
@Azmuth hey!
Qm is gping great
 
123
123
@Azmuth Yes
 
Azmuth
Azmuth
@JackRod Hey, sup buddy?
 
Jack Rod
Jack Rod
Good
Wbu?
 
Azmuth
Azmuth
7:04 AM
@123 Hell no! That's why I told, ICOR is useless here, separating them as tangential and radial components, you are hitting yourself. Better separate them as 1. Direction of gravity and 2. Tangent to direction of gravity and not as radial and tangential ones.
@JackRod doing fine. Reading Algorithms, they are pretty hard. @PM2Ring suggested me a PhD level book and I'm still struck in first few pages.
Donald Knuth...! (God of Algorithm programming!)
 
123
123
@Azmuth OoKay ... That's why i am confusing. The idea of radial and tangential component did not work in projectile. right. It means it is not universal method for calculating the curves?
What is ICOR stand for?
 
Azmuth
Azmuth
@123 There's no single universal method for curves. Although there are some approximation ones (like I mentioned), areal velocity, coordinate rotation, etc, etc... but all are useless....
@123 Instantaneous Center of Rotation)
3
Q: Calculating force through ICOR (instantaneous centre of rotation)

Vasu Goyal A purely rolling wheel of radius $R$, has its ICOR at the point $P$ (as per the figure). If I calculate the centripetal force on a particle at the top most point of the rim, considering point $P$ as its center of rotation, I get the value of the force to be $F=2mR \omega^2$, where $\omega$ re...

newtonian-mechanics rotational-dynamics rotation
@JackRod Which branch you are in?
also called instantaneous radius method
 
123
123
Can we explain limitation of ICOR?
So, we can say what method is good for what type of motion.
 
Jack Rod
Jack Rod
@Azmuth did not I mentioned earlier?
 
Azmuth
Azmuth
@123 There are only limitations of ICOR.... only one advantage that it helps to visualize any curved motion as circular motion and helps us with circular dyanamics..
@JackRod I guess I forgot :P
@123 There's no single good method. Each situation can be resolved into multiple good ones.
@JackRod uhhhhhh!? Computer Science?!!!!
really?
 
123
123
7:12 AM
Is it possible to write most used method for calcuation. like cartesian, polar, ICOR etc.. what else
 
Azmuth
Azmuth
sorry, I'm involved in 2 rooms simultaneously....
@123 no
 
123
123
When you have time. list few of them. Or give me a link
 
Azmuth
Azmuth
@JackRod Have you all completed algorithms?
 
Jack Rod
Jack Rod
@Azmuth busy then we can talk later?
Mostly
 
Azmuth
Azmuth
Okay :)
@123 Rectangular coordinates, polar and circular motion method are the most common ones...
 
123
123
7:14 AM
Ookay. Thanks
 
Azmuth
Azmuth
I just realized that the series $\sum \limits_{\text{p primes}} \frac 1 p$ diverges... :O
DAMN! Editing
 
 
2 hours later…
123
123
8:59 AM
Is there book which explains energy with history. Like I have kimbell which explains electrostatic with history
 
Physics Meta
Physics Meta
9:43 AM
0
Q: Why are there close votes on this question?

FoundABetterNameI recently asked this question and noticed a close vote. So I simply commented on the post that maybe someone would share the reasoning for the close vote however no one did and today I noticed another close vote. I have the close vote privilege so I could see the reason is lack of clarity but I ...

discussion specific-question close-reasons vote-to-close
 
 
2 hours later…
Slereah
Slereah
11:22 AM
did I write anything objectionable here
 
ACuriousMind
ACuriousMind
I will read that after I have finished waking up :P
 
bolbteppa
bolbteppa
That question basically seems to be asking why is $e^{ipx}$ different to $u_p e^{ipx}$
 
Slereah
Slereah
@bolbteppa well they are isomorphic and all
you can do it
it's just not a particularly wise decision
 
ACuriousMind
ACuriousMind
@Slereah I think you're correct and likely wrote a lot of what Charlie needed to hear but I think you missed the core of the question
oh, no wait, you wrote the right thing
 
Slereah
Slereah
I probably should have written more on irreps I guess
 
bolbteppa
bolbteppa
11:29 AM
Are you saying the space of plane wave solutions of Klein-Gordon is isomorphic to the space of plane wave solutions of the Dirac equation
 
Slereah
Slereah
@bolbteppa Well the Hilbert spaces are, so... maybe?
The Hilbert space theorem is for the orthonormal basis, I don't know if it extends to the rigged Hilbert space
I'm not 100% sure what would be the definition of the "good Hilbert space"
 
ACuriousMind
ACuriousMind
The problem is that "isomorphic" has two different meaning in the QFT context: You can talk about the Hilbert spaces being isomorphic as Hilbert spaces, or you can talk about the physical state spaces being isomorphic as representations of the respective algebra of observables
 
Slereah
Slereah
something something faithful irrep of the Poincaré group
 
bolbteppa
bolbteppa
"The Fundamental Theorem of Infinite Dimensional Vector Spaces states that all Hilbert spaces with a countable infinite orthonormal basis are isomorphic"
 
ACuriousMind
ACuriousMind
when people say that two state spaces are "not the same", they are using the latter meaning, not the former
 
Slereah
Slereah
11:32 AM
but then again, you can just translate the symmetry operators to that weird Hilbert space, so
I don't know
 
Charlie
Charlie
well hello
 
ACuriousMind
ACuriousMind
you kind of danced around that when talking about spin and the mapping of basic vectors being nonsense
 
bolbteppa
bolbteppa
So even though a KG state with momentum $p$ is attached to two spinor states with momentum $p$ (different polarizations) they are isomorphic?
 
Slereah
Slereah
what makes $L^2 \otimes V_n$ a better group than $L^2$, as far as the representation goes
@bolbteppa In the sense that they can fundamentally be described as a big list of numbers (wrt to the eigenvalues), yes
they are big lists with the same number of eigenvalues
it's just not a very compelling space to put them in
 
bolbteppa
bolbteppa
As in a big infinite list
so everything is isomorphic as they are all countable lists
 
Slereah
Slereah
11:35 AM
Pretty much, yes
unless it's finite dimensional or not separable
But then that's getting into "µ-recursive functions, lambda calculus and set theory are the same" territory
Technically true but perhaps not very helpful
 
123
123
Is there any book which gives history of energy and work. Like i read kimbell which has history of electrostatic.
 
bolbteppa
bolbteppa
@Charlie I would really advise chapter 1 of Walecka Fetter's Many Particles QM book on second quantization
 
Charlie
Charlie
Are you just looking for the names of scientists that invented certain concepts @123?
 
123
123
I know few the scientist and history. but need to read what problem occurred to develop the idea energy KE , PE and work done.
 
Slereah
Slereah
Energy was mostly investigated by Mr. Hamilton, IIRC?
 
Charlie
Charlie
11:39 AM
I see @bolbteppa, thanks I'll have a read of that today
 
Slereah
Slereah
Like the modern notion of energy was invented by him
 
123
123
I learnt many things yesterday from you guys.
 
Slereah
Slereah
Although there were plenty of precursors
 
123
123
Like K&K give one idea as physicist need force in term of position. i need more history
 
Slereah
Slereah
IIRC Hall's book has a whole section on why that Hilbert space is a good Hilbert space
I should read it in more details
 
bolbteppa
bolbteppa
11:41 AM
When they introduce creation and annihilation operators out of the blue, skip ahead to where it links up with the previous section they reanalyze what they got and express it in terms of creation operators that's basically the way to even motivate bringing them in as some operators in the occupation number formalism which do what's written there
 
Charlie
Charlie
@Slereah what does $\aleph$ and $\aleph_0$ represent in the answer you gave, I meant to ask. Does it represent a countable infinity?
 
123
123
why i dont see upload button
 
Charlie
Charlie
Perhaps you need over a certain reputation on the main site @123
to post images in the chat
 
Sir Cumference
Sir Cumference
@Charlie aleph null is countable infinity
 
Charlie
Charlie
I see, ty
 
123
123
11:43 AM
But it was there 2 days ago.
 
Sir Cumference
Sir Cumference
aleph 1 is the next highest infinity, aleph 2 is the next highest, etc
 
Charlie
Charlie
wait there more than two kinds of infinity?
 
Sir Cumference
Sir Cumference
wait nvm, i'm pretty sure it's aleph 1 that's the size of $\mathbb{R}$
 
John Rennie
John Rennie
@Charlie oh boy, are you in for a treat :-)
 
Charlie
Charlie
D:
 
Sir Cumference
Sir Cumference
11:44 AM
@Charlie yeah, some infinities are bigger than others. the smallest infinity has the size of the set of all integers
it's called countable infinity
 
Charlie
Charlie
I'm aware of only countable and uncountable infinities
 
Sir Cumference
Sir Cumference
but there are more numbers in $[0, 1]$ than there are integers in the whole number line. e.g. if i had infinite time, i could count all the integers, but i couldn't even begin to count all the tiny numbers in $[0, 1]$
 
John Rennie
John Rennie
@SirCumference yes, what is unknown (and undecidable) is if there is an infinity in between $\aleph_0$ and $\aleph_1$.
 
Charlie
Charlie
My assumption was that there was no way to interchange them by arithmetic operations. But you could for instance double the cardinality of the natural numbers and obtain a set with the same cardinality, countable infinity
 
ACuriousMind
ACuriousMind
@Slereah I've written another answer that focuses on that
 
Sir Cumference
Sir Cumference
11:46 AM
@Charlie you could consider the set of all subsets of $\mathbb{Z}$
for finite sets $S$, the set of all subsets has cardinality $2^{|S|}$. for infinite sets, the set of all subsets has a strictly bigger infinity
so you can always generate a higher one by looking at its set of subsets
 
Charlie
Charlie
But, $2^{|S|}$ is necessarily finite for finite $|S|$, right?
 
Sir Cumference
Sir Cumference
if an infinite set has cardinality aleph $n$, then its power set (i.e. set of all subsets)'s cardinality is named aleph $n+1$
 
Charlie
Charlie
So there is a "bigger" infinity than uncountable?
 
Sir Cumference
Sir Cumference
@Charlie yeah
 
Charlie
Charlie
:O
 
Sir Cumference
Sir Cumference
11:49 AM
@Charlie uncountable just refers to any infinity that's bigger than aleph 0 (i.e. the smallest infinity)
 
Charlie
Charlie
Do they exist for all $n$?
 
Sir Cumference
Sir Cumference
so there's only one countable infinity, but the rest are uncountable infinities
 
Slereah
Slereah
Sure
 
Sir Cumference
Sir Cumference
@Charlie yeah
 
Slereah
Slereah
For any set, the power set is larger
 
Charlie
Charlie
11:50 AM
Well that is definitely some new information
 
Sir Cumference
Sir Cumference
iirc though, this infinity is bigger than any other:
Inaccessible cardinal
In set theory, an uncountable cardinal is inaccessible if it cannot be obtained from smaller cardinals by the usual operations of cardinal arithmetic. More precisely, a cardinal κ {\displaystyle \kappa } is strongly inaccessible if it is uncountable, it is not a sum of fewer than κ {\displaystyle \kappa } cardinals that are less than κ {\displaystyle \kappa } , and α < κ {\displaystyle \alpha <\kappa } implies...
 
Slereah
Slereah
It is, but it's not very often useful
 
ACuriousMind
ACuriousMind
@123 You need 100 reputation to be able to upload images to chat, cf. meta.stackexchange.com/q/317853/263383
 
Slereah
Slereah
Usually it stops being useful beyond $\aleph_2$
$\aleph_2$ is the cardinality of all the real valued functions
 
ACuriousMind
ACuriousMind
what about the cardinality of all functionals on real-valued functions? :P
 
Slereah
Slereah
11:51 AM
(Assuming the continuum hypothesis, don't at me)
 
123
123
@ACuriousMind Thanks, It means my reputation is not GooD :D
 
Slereah
Slereah
@ACuriousMind I mean $\aleph_2$ is already too big for most purpose, since that's counting every discontinuous function
I'm sure the cardinality of distributions isn't that large
 
Charlie
Charlie
@ACM ty for posting a second answer, if I've understood what you and Slereah are basically saying, it's that the Hilbert spaces of the free qfts are isomorphic at the Hilbert space level, but that this doesn't imply they are isomorphic in every sense?
 
Sir Cumference
Sir Cumference
@Charlie this also proves that a power set can't have a cardinality of aleph null, btw. bc any finite set's power set is finite, and any infinite set's power set is a bigger infinity (and therefore wouldn't be the smallest infinity)
 
ACuriousMind
ACuriousMind
@Charlie yes
 
Charlie
Charlie
11:53 AM
I could live with that, it would make sense
ok nice, ty for taking the time write those
 
Slereah
Slereah
one of the big issue of math is that there's a lot of meaning of "are the same"
 
123
123
What is the benefit of stackexcange. You guy earn money from to answer or any other?
 
ACuriousMind
ACuriousMind
hahahahaha...no, we're all volunteers here
we're exclusively paid in fake internet points
 
Charlie
Charlie
Yeah I guess my wording could have been more precise
 
Slereah
Slereah
You can send money, though
 
123
123
11:55 AM
@ACuriousMind It is no doubt you all guys gives lot of lot of help to me and others.
 
Charlie
Charlie
There should be a government scheme to trade in your internet points for money, like a casino
 
Sir Cumference
Sir Cumference
@Slereah i guess that's what equivalence relations are for tho
 
123
123
@Slereah I am also looking for money :D
 
Slereah
Slereah
@SirCumference there are many
Is $U(1)$ the same as $S$???
 
123
123
What is the charm and harm of being physicist?
 
bolbteppa
bolbteppa
11:59 AM
$U(1) = S^1 \neq S$
 
Charlie
Charlie
The Hilbert spaces in regular qm, once a basis is chosen, can the entries in their vectors (albeit infinite ones) be written countably?
 
Slereah
Slereah
@Charlie Yes, that is the basic idea of it being separable
 
123
123
How do we calculate the radial and tangential components of acceleration due to gravity in projectile motion?
 
Slereah
Slereah
The usual basis people use (the momentum basis) isn't technically a basis, but there are actual orthonormal basis that are indeed countable
for instance, Hermite polynomials
 
Charlie
Charlie
Ok that's kind of what I suspected, in that case how is it possible to establish an isomorphism between that and the uncountable real line(s)?
 
Slereah
Slereah
12:03 PM
Laguerre polynomials
etc etc
@Charlie There is no such isomorphism
 
Charlie
Charlie
this question has probably been asked before, I've seen things resembling it slightly on se before
Yeah actually let me rephrase that
When we take $\langle x|\Psi\rangle$, this assigns to each point on (at least some subset of) the real line a value $\psi(x)$, which seems to require uncountable infiniteness
this has actually been bugging me for a while
 
bolbteppa
bolbteppa
@123 you can project the constant gravitational vertical acceleration vector onto each component but it's a silly thing to do this is not the way to think about it at all
It's very simple, see the picture I sent above, that's all there is to it
 
ACuriousMind
ACuriousMind
Jul 22 at 16:30, by ACuriousMind
@Charlie the game is rigged
or, if you just need to be reminded: $\lvert x\rangle$ does not actually lie inside the Hilbert space
 
Slereah
Slereah
 
Charlie
Charlie
Ah I see
 
bolbteppa
bolbteppa
12:15 PM
So the whole $\int F_{\mu \nu} \tilde{F}^{\mu \nu}$ thing being a total derivative vanishing at the endpoints for Maxwell, in the non-abelian case $\int F_{\mu \nu}^a \tilde{F}^{\mu \nu}_a$ can be expressed as a total derivative which does not have to vanish at the endpoints, and this actually leads to something that can be measured (but hasn't?)...
 
Charlie
Charlie
Once you have messed around with rigged Hilbert spaces for long enough, non-relativistic QM is mathematically rigorous in its entirety, right?
last question, before I go and get a sandwich :P
 
bolbteppa
bolbteppa
I doubt it, I'd guess it can't be because it's all equivalent to path integrals which are a bunch of nonsense rigorously
You can work that integral out explicitly for $\text{su}(2)$ and it relates/motivates instantons somehow... this is pretty cool
 
Charlie
Charlie
Oh I didn't realise the path integral wasn't well defined
 
ACuriousMind
ACuriousMind
@Charlie The rigorous versions usually avoid talking about rigged spaces entirely, but yes, QM with finitely many degrees of freedom (i.e. non-QFT) is entirely rigorous
@bolbteppa the path integral is only ill-defined for generic field theories, for non-field QM it is perfectly rigorous
 
Charlie
Charlie
ah
Ok, thanks, I can eat in peace now :P
 
ACuriousMind
ACuriousMind
12:19 PM
I should do that, too, actually
 
Ankit
Ankit
@JohnRennie in the example of a pendulum in an accelerating car as seen from an inertial frame , does the pendulum stay at its position due to inertia ?
 
John Rennie
John Rennie
@Ankit you mean if the pendulum is just hanging stationary i.e. not oscillating?
 
John Rennie
John Rennie
12:34 PM
@Ankit I have to go now. I'll be back tomorrow as usual.
 
bolbteppa
bolbteppa
I think the rigorous qft analogue is one of those 'lost causes' in that book
 
Slereah
Slereah
I believe in it
and nlab will do it
 
bolbteppa
bolbteppa
"the quantum propagator – in FQFT the value of the functor $U:\text{Cob} \to \text{Vect}$ on a certain cobordism"
Clearly if you're not doing functorial qft you're not doing qft
 
Slereah
Slereah
well you don't have to use path integrals
 
bolbteppa
bolbteppa
12:50 PM
I love the framing "A simple form of the path integral is realized in quantum mechanics, where it was originally dreamed up by Richard Feynman and then made precise using the Feynman-Kac formula" i.e. you're an idiot if you just get the 'simple' one and haven't done the “holonomy integrated against the Wiener measure” one "in the slice (infinity,1)-topos of smooth infinity-groupoids over the developing groupoid $\textbf{B } \text{U}(1)$"
 
Slereah
Slereah
tough but fair
 
bolbteppa
bolbteppa
It's so complicated it didn't even copy properly...
 
Ankit
Ankit
Can someone help me here .. the pendulum is not oscillating as John Rennie asked for ...
 
Slereah
Slereah
nlab uses a lot of mathjax definitions, I think
 
bolbteppa
bolbteppa
Even the word developing didn't go right the first time :p
 
 
1 hour later…
bolbteppa
bolbteppa
2:00 PM
"I think it was Belinfante who first showed that the most general possibility, again, assuming that no orbital angular momentum is transferred in the fundamental decay process, was a mixture of five types of current: scalar, vector, tensor, axial vector, and pseudoscalar. This may be unfamiliar to today’s graduate students in physics, because the work of Marshak and Sudarshan made this kind of theoretical classification obsolete.
We now know what the right answer is, so who studies these other possibilities anymore? But, I find history fascinating, so allow me here to look back on this bit of history. It was clear that you had to have scalar or vector, because the other currents would not produce transitions between nuclear states of 0 spin, and there are transitions like the decay of Oxygen-14 into Nitrogen-14 where both nuclei have spin 0.
That is allowed by the scalar and vector current interactions, but would not be allowed by the other interactions. Then there are transitions like the one I mentioned when Boron-12 and Nitrogen-12 decay into the ground state of Carbon-12, which require the presence of tensor or axial vector currents.
So you have scalar and/or vector plus tensor and/or axial vector currents - in some mixture that no one knew. (The pseudoscalar interaction doesn’t produce allowed beta decay, so it is a somewhat tangential possibility.)"
That is very cool
 
Slereah
Slereah
2:42 PM
I remember that it was the big thing Feynman was known for for a while
Finding out that the weak interaction was a vector-pseudovector interaction
or whatever it was
 
Azmuth
Azmuth
Tell mah your halloween ideas!
now!
 
Slereah
Slereah
Stay at home
 
Azmuth
Azmuth
and?
 
ACuriousMind
ACuriousMind
^
 
Slereah
Slereah
that's about it
I mean I will do a variety of things, ie breathe, eat, metabolize
 
Azmuth
Azmuth
2:51 PM
I'm didn't step out of home for last 4 months.
 
Slereah
Slereah
I will perform a lot of Krebs cycles
 
Azmuth
Azmuth
I'd probably go to roof and capture #BlueMoon
 
Slereah
Slereah
Citric acid cycle
The citric acid cycle (CAC) – also known as the TCA cycle (tricarboxylic acid cycle) or the Krebs cycle – is a series of chemical reactions used by all aerobic organisms to release stored energy through the oxidation of acetyl-CoA derived from carbohydrates, fats, and proteins. In addition, the cycle provides precursors of certain amino acids, as well as the reducing agent NADH, that are used in numerous other reactions. Its central importance to many biochemical pathways suggests that it was one of the earliest components of metabolism and may have originated abiogenically. Even though it is branded...
I do that shit all day long
 
Azmuth
Azmuth
Cool!
You won't be capturing #BlueMoon?
Dun worry, If I did, I'll post my pictures here.
It happens in every 2.7 years
 
Slereah
Slereah
mb I will watch a spooky movie
 
Azmuth
Azmuth
2:52 PM
Me too!
 
Slereah
Slereah
If I am feeling particularly in the holiday mood
 
Azmuth
Azmuth
conjuring 2?
How's it?
 
Slereah
Slereah
Although Halloween isn't really a holiday here
 
Azmuth
Azmuth
Me too, holiday mood
@Slereah where?
 
Slereah
Slereah
The France
 
Azmuth
Azmuth
2:53 PM
@ACuriousMind you give ideas!
 
ACuriousMind
ACuriousMind
I concur with Slereah's plan of staying at home :P
 
Azmuth
Azmuth
@Slereah cool place, I'd dress as Iffel Towe
or somethign same name
@ACuriousMind no movies?
 
ACuriousMind
ACuriousMind
I've got an RPG session later actually, but we do that about every two weeks, so it's not special.
 
Azmuth
Azmuth
Role Playin Game?
Let's to a virtual party here
using Mozilla Hubs!
 
John Duffield
John Duffield
@bolbteppa : see Charles Galton Darwin's 1927 Nature paper "The Electron as a Vector Wave": nature.com/articles/119282a0
 
03:00 - 15:0015:00 - 22:00

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