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00:01
what changes is the training set and how much computing power has been invested into the model - there's a lot of variables there, but more or less the most powerful versions just have a variable called "free parameters" set much much higher than the initial versions, and have had much more computing power spent on their training
do you follow this AI stuff e.g. in the news ACM or is this just what you've picked up without deliberately seeking it out?
@qwerty I work in the tech industry and have no choice but to engage with it :P
my condolences lol
when everyone in upper management starts going crazy and demanding you build "AI" into everything, you need to at least try to understand what the heck they're talking about :P
mew
00:09
haha. if I didn't want to invest time into reading the actual papers and whatnot, are there actually any people who talk about it or provide sensible commentary without being part of the hype machine? maybe that's a tall ask
There is essentially no "neutral" commentary. You're either part of the hype or you're extremely critical of it :P
ok, I'll just maintain my subscription to the ACM hbar feed on this matter xD
its inescapable
pretty soon our fridges and toasters will have AI assistants
And it's a difficult line to draw because unlike with the blockchain/crypto hype there is something there, some genuinely interesting technology that has many potentially interesting applications
i hate its use in art
00:13
what interest do you personally see in it ACM?
tech bros and capitalists alike seeing absolutely no value in art
tbh there's just so much noise and fluff around it I tune it all out
same
But with few exceptions - such as the use in translation - it's mostly used to pretend it's "intelligent" without any cogent concept of what that even means
@qwerty Well, I think on a very basic level it's extremely interesting in and of itself that we can have a machine generate text that, if nothing else, at least is almost always grammatically perfectly correct without programming in the grammar itself (which isn't possible because natural languages are not captured by formal grammar)
Of course, this alone is not useful, but the problem is that it suffices to pretend that it's useful
But anyway, on a more practical level I do think that it is e.g. much more efficient for a translator to edit the pregenerated translation of some LLM rather than trying to start the translation from scratch. And if you have no other choice, the auto-translation is much better than nothing.
There are many specialized areas where these models, if tuned for that specific application, are genuinely better than any alternative
but what people are mostly trying to sell you as "AI" is some thin wrapper around GPT-4 and that's often worse than useless
00:30
interesting take. I'll think about it a bit more but my initial reaction would be that it seems, on the surface, that singling out language and translation seems to be related to that area having a level of inherent fuzziness and pattern matching. but I'm not sure.
@qwerty oh, certainly - pattern matching is what these algorithms conceptually do
It's not just language, there's also image recognition and generation as the second large application
to what extent is it rebranded machine learning in that case?
I mean in those applications
it used to be called machine learning before the hype :P
there isn't really any essential difference there, "AI" is just machine learning marketed differently
iirc machine learning had its own hype but I see
there's even some back-propagation where companies are sometimes re-defining perfectly traditional deterministic algorithms as "AI" just to be able to market their software as being "AI"
'(I mean literal "propagating backwards" here, not the backpropagation that's part of building neural networks :P)
 
4 hours later…
04:45
@ACuriousMind I expected that correction :P
 
1 hour later…
06:12
morning
 
2 hours later…
08:12
It's always hard to deal with like incredible geniuses that are also insane like Lawvere
Gotta figure out if he's talking nonsense or if he is just not bothering to explain a very genius move
 
3 hours later…
11:07
I just got an email that starts with "Certainly! Here's a revised and more detailed version of your message:"
I dare you to reply with "I'd like to speak with a human, please." :P
I'm pondering exactly how pissed off I want to sound in the reply :P
@ACuriousMind Something nefarious just happened. Looking up stuff about superconductivity I ended up on a paper and the name of the author was...
Rudolph Haag
I refuse to read it and declare my defeat
@SignorFeynman The guy from the theorem is Rudolf, not Rudolph
I copy-pasted someone who miswrote on physics forum :P
I can't edit it now but... It's him
11:23
neat
Bml
Bml
11:40
Hello everyone. I have a question. In this answer, it is said that rolling without slipping in the absence of friction is possible if a force is applied at a given distance $d$ from the center of mass of a rigid body.
If I have a body of mass $M$ and radius $R$ on a rough horizontal plane and apply an impulse $\vec J$ to the body at a point at that same height $d$ from the center of mass, is the static friction force zero as in the previous case, or is it present? How can it be calculated explicitly?
12:03
@ACuriousMind is that the Reindeer then
12:18
sure, famous mathematical physicist Reindeer Haag :P
@Bml I don't really understand the question. The answer you linked makes no statement about whether or not a friction force occurs in any particular physical situation. It just points out that in some situations, for rolling without slipping to occur mathematically, you don't need a friction force.
if you say you have a "rough plane", you seem to be saying there's friction. So why would the be no friction?
Bml
Bml
12:44
@ACuriousMind In the answer I linked, if the force is applied at distance $2R/5$ above the center of a solid sphere, the sphere rolls without slipping in absence of friction, no? Or is the answer saying that motion is frictionless only if the force is applied at that distance?
@Bml Yes, it's saying that if there's no friction, then that force will lead to rolling without slipping.
Conversely the answer also directly implies that, if there is friction, then applying the force like that will not lead to rolling without slipping.
Bml
Bml
@ACuriousMind That is, $\text{absence of friction} \implies \text{force applied at a distance $d$ from the center of mass}$, yes?
I don't really understand what that implication is supposed to mean
there is really two implications inside of each other here: absense of friction -> (force applied like that -> object rolls without slipping)
likewise, presence of friction -> (force applied like that -> objects does not roll without slipping)
Bml
Bml
@ACuriousMind I am not sure about this, and it is also concerning my doubt. It can be shown that, for example, if an impulse is applied at a distance $2R/5$ above the center of mass of a rigid sphere on a rough plane, the motion results immediately from pure rolling. Thus, one of two is true: either pure rolling motion can occur with friction even if a force is exerted at that distance, or pure rolling motion occurs without friction even if the plane is rough. I am confused.
By "pure rolling" I mean "rolling without slipping."
@Bml I don't know what you mean. The answer you linked shows that if you apply a force at 2R/5 on a sphere, then if the resulting motion is rolling without slipping, then the friction force acting is zero
I don't understand your "It can be shown that...the motion results immediately from pure rolling"
Bml
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12:55
@ACuriousMind It makes sense, but I am not sure about this. See my previous message.
Yeah, I don't understand why you're not sure about this. You have an answer, and a proper mathematical derivation in the answer it links to. Where is the derivation that disagrees?
Bml
Bml
@ACuriousMind Wait, I'll try to send an image.
@SignorFeynman i am very surprised at just how many hep-th “big names” of the mid 1900s were also condensed matter big names
Bml
Bml
@ACuriousMind Sorry for the non-English description, but it is not essential for understanding. If you want I can translate.
@Bml How is the derivation here supposed to contradict anything? What your screenshots show is that if you have rolling without slipping with the only force acting on the sphere being that force at $h$ (i.e. no friction!), then the force is acting at 2/5 R from the center
Bml
Bml
13:11
@ACuriousMind Exactly what I said. Rolling without slipping motion occurs without friction even if the plane is rough if we apply a force at $2/5 R$ from the centre of mass...
@Bml Where did you get "even if the plane is rough" from?
the equations in your screenshot do not involve any potential friction force, i.e. to me they seem to assume no friction and derive the position of the attack point of the force from that
Bml
Bml
The problem statement is: "A homogeneous sphere of mass $M$ and radius $R$ is placed above a **rough** horizontal plane. Initially
the sphere is at rest. At a point $A$ on the surface and lying on a vertical $x-y$ plane passing through the centre of the sphere an impulse $\vec J$ directed horizontally to the right is applied.
1. determine at what height $h$ above the ground the impulse $\vec J$ must be applied for the sphere to move in rolling without slipping motion.
@ACuriousMind If the impulse is applied at the height of the centre of mass, there is first a rolling with slipping motion, then a rolling without slipping motion in the absence of friction, because after a certain period of time the two linear and rotational velocities equalise, and the motion continues without friction. If the impulse is applied at a distance $2/5 R$ from the centre of mass, does the sphere move by rolling without slipping, but with friction or without friction?
sorry, I really don't understand what the problem is
Bml
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@ACuriousMind If the impulse is applied at a distance $2/5 R$ from the centre of mass, we know that the sphere moves by rolling without slipping (it is proved in the screenshots I sent), but with friction or without friction? You said "if you apply a force at $2R/5$ on a sphere, then if the resulting motion is rolling without slipping, then the friction force acting is zero", but by hypothesis the plane is rough, not frictionless.
In my opinion this is a contradiction...
13:28
@Bml That the plane is rough does not mean it has to always exert friction, only that it can.
both derivations clearly lead you to the result that in the case of the 2/5 R force, there is no friction force
Bml
Bml
@ACuriousMind OK, so is "pure rolling motion occurs without friction even if the plane is rough when we apply a force at distance $2R/5$" true?
I don't really know what that sentence means
Bml
Bml
@ACuriousMind Do we have these implications? 1) Absence of friction -> (force applied like that -> object rolls without slipping); 2) object rolls without slipping -> (force applied like that -> absence of friction); 3) Force applied like that -> (absence of friction -> object rolls without slipping)
@ACuriousMind How can they be brought together in a single implication?
Bml
Bml
14:15
@ACuriousMind Let me explain: if there's no friction, then that force will lead to rolling without slipping, but can we say that if we apply that force, then the object rolls without slipping in absence of friction, or that if the object rolls without slipping and we apply that force, there is no friction? I am confused about the premises and conclusions, I would appreciate it if you could help me out :-)
@Bml I think I started this whole conversation in the wrong direction. Let's take a step back: What's your definition of rolling without slipping?
Bml
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@ACuriousMind The object rolls (i.e. translates and rotates) but the point of contact is stationary instantaneously.
@ACuriousMind The condition for rolling without slipping is $v_{cm} = \omega R$, where $v_{cm}$ represents the velocity of the object's centre of mass.
@Bml Okay. If the point of contact is stationary, why would there ever be friction?
Bml
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@ACuriousMind If an external force is acting, a static friction force is required. The force of static friction plays an important role in the motion of rolling without slipping. Friction is only absent if the object continues at a constant speed.
I don't know what you mean by "The force of static friction plays an important role in the motion of rolling without slipping.". We've talked at lengths about several answers showing that specific forces can put specific bodies into "rolling without slipping" motions with the assumption of zero friction
if you just want to randomly assert stuff instead of arguing for it I don't know how to help you
Also, "Friction is only absent if the object continues at a constant speed." - yes, sure. The "rolling without slipping" motions in the answer we were talking about are all such idealized eternal motions. What's your point?
Bml
Bml
14:30
@ACuriousMind Sorry, I misspoke. I mean that "rolling without slipping" requires an angular velocity (and therefore angular acceleration) caused by a non-zero net torque with respect to the centre of mass (supplied by a force that does not act on the centre of mass, which may be a tension acting at the summit point or static friction acting at the point of contact) and static friction that prevents the point of contact from slipping.
Without non-zero torque with respect to the centre of mass and without static friction acting on the point of contact, how can rolling without slipping develop?
@Bml But that's just not true. If you start the motion via the force that is exactly at 2/5 R from the center of a sphere, the resulting motion is "rolling without slipping" with no friction force necessary
that's exactly what both the answer and your screenshots show
again, if you just refuse to believe that and assert the opposite I don't know what to say :P
Bml
Bml
@ACuriousMind OK, and my question is: how can we show that if a force is applied at distance $2/5 R$ from the centre of mass, then static friction is zero? I cannot understand this.
@Bml You can't show that because it's not true. In the real world, friction is obviously not zero, there are no eternally rolling spheres.
Bml
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@ACuriousMind No, I don't refuse to believe it, I'm just saying that I can't understand it (due to my limited abilities). Of course, I believe what you say, I just can't understand what I wrote earlier :-)
@ACuriousMind I mean, how do we prove it in an idealised world of an exercise?
The claim is just that in a scenario where you have the option to have no friction, then that specific force just causes rolling without slipping
it's part of your idealization that you just assume "no friction" can happen
there's nothing to prove because there is no claim being made
Bml
Bml
14:41
@ACuriousMind OK. But perhaps the heart of my doubt is another. If we apply a force at a certain distance from the centre of mass, there will be a rolling without slipping motion, in the absence of friction. If, on the other hand, we do not apply a force, but only an impulse at the same distance, after a certain time interval the impulsive force ceases its effect, so the object does not move under the effect of a force. Right?
I don't understand what you mean by "apply only an impulse". Impulse is just change in momentum, not something you can apply. Is this a mistranslation?
Bml
Bml
@ACuriousMind Yes. I mean ‘if we give the ball an impulse’. In any case, that force has a limited duration, and after a certain period of time the body proceeds at its own speed, without the presence of a force. That's what I meant.
@ACuriousMind That would change everything. If a force is acting, then static friction must be present, and that static friction will be zero at a certain distance from the centre of mass (in the case of a sphere, $2R/5$). If an impulse acts, the force has a limited duration, and after a certain period of time it ceases to exist. In this case, if the motion is pure rolling, there is no reason why there should be friction.
If friction is present, the motion is not rolling without slipping, but rolling with slipping. If there were a static friction force, the centre of mass would slow down and the motion would be uniformly decelerated instead of uniform. But if the CoM decelerated, since it is a motion of pure rolling, we would derive that its angular acceleration should be non-zero (negative) and the angular velocity of the sphere
should therefore decrease accordingly. Therefore the mechanical energy of the sphere (translational kinetic energy + rotational kinetic energy) would decrease in time. But in pure
@ACuriousMind Am I correct?
@SillyGoose I think that it's also because over the last century physics got so many new subfields, that you end up not knowing a thing about a subfield in your same field
And so you had people like Feynman who worked basically in almost every field of physics
@Bml The part that's correct is that there cannot be a friction force on a wheel rolling with constant velocity and no external force on it. I think you're focussing too much on this "if friction is present" part because it's not a binary thing. You can perfectly well use a model here where the surface "has friction", the kind that resists rolling with slipping, but does not have rolling friction, i.e. motions that roll without slipping go on forever at constant velocity
Did you people know this is a thing? :P
14:56
@SignorFeynman It's just that - there was barely QFT distinct from QM, let alone an actual split into hep-th and cond-mat QFT, really.
Bml
Bml
@ACuriousMind Yes, of course. I was only referring to one kind of friction, sliding friction, not rolling friction. Of course, as you say, there are rolling with slipping models where there is sliding friction but there is no rolling friction.
@ACuriousMind One can have an idea of that going through the third volume of the FLP. Of course, they are no standard treatment of QM even for 1965, but in a sophomore course you had Feynman discussing about (simplified models of) molecules, lattices and end it with a seminary on superconducitivity that is still used today to explain the Josephson effect. Even then, Feynman didn't mention stuff that today we consider standard for a QM class
Bml
Bml
@ACuriousMind If, on the other hand, a force is applied to the center of mass, what happens? If the plane is rough there will be friction on the wheel, whereas if the plane is smooth there is no friction on the wheel? Or is the cause that the force is applied on the center of mass and does not produce torque?
@Bml Sure, if the plane is rough, this force produces rolling-with-slipping and hence friction that opposes the "slipping" part.
i want to ask for permission to link the Theories of Everything podcast @ACuriousMind
15:13
Strike him down @ACuriousMind
@RyderRude It's not even been a week. If you can't even go a week without linking that one specific podcast you need to seriously widen your range of sources.
I mean, iirc you cannot link that podcast for specific reasons that still stand today, it was not a temporary situation, was it?
i liked one segment from the podcast that I would've liked to link. but I can't link it now. it was an articulation of how little humans know and why that is good
So an apology of ignorance? :P
yes... but in a good way
it was beautiful articulation
it goes on for a few minutes maybe
Bml
Bml
15:31
@ACuriousMind Only if it exceeds a certain value, right? That is, if it does not exceed a certain value the wheel will roll without slipping, if it exceeds that value it will slip. Right?
@Bml if what exceeds a certain value?
Bml
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@ACuriousMind The applied force.
ah, yes, I suppose if the force isn't large enough to overcome static friction at all you get no slipping, sure
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@ACuriousMind Thanks :-)
Bml
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16:04
@ACuriousMind I was thinking, if we apply a force to the centre of mass of the wheel, how do we know if the wheel rolls (with or without slipping) or just translates?
@Bml You just compute what happens based on what you assume about the static friction
e.g. it will of course just translate if there's no friction at all
if you have friction it will roll in some form or another
Bml
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@ACuriousMind OK, and if there's no friction at all, if the force is not directly in line with the center of mass it will cause rotation and translation without rolling. Right?
I'm not sure what you mean by "without rolling"
rolling is just a mix of rotation and translation
Bml
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@ACuriousMind I mean that if the force is applied at the highest point of the wheel and there is no friction, then the wheel translates and rotates, but the relationship of pure rolling or rolling with slipping is not satisfied. If there was rolling with slipping or without slipping, the external force should be accompanied by a frictional force. No?
16:19
I don't see where the physics question is here. What's your definition of "rolling with slipping" where the answer to this is not obvious?
Bml
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@ACuriousMind Let me take a step back: if I apply a force at the highest point of a wheel on a surface where there is no friction, does the wheel roll with slipping or not?
@Bml Depends on how you define "roll with slipping"
Some definitions may explicitly involve the presence of friction, others may not
personally I think it's rather irrelevant how we define the terminology to apply to the completely unrealistic case of no friction at all :P
17:08
Are there non trivial solutions of GR which are periodic?
(Just wondering)
Is this dual interpretation of what it means to evaluate a field operator at some arbitrary spacetime point accurate. I am considering the field operator of a massless vector field:
1. $A^\mu(x)$ assigns a scalar and vector potential value at a spacetime point.
2. The field operator $A^\mu(x)$ assigns particle(s) to spacetime points
Is this a correct interpretation of what the output of a field operator for a value $x^\mu$ is ?
@MoreAnonymous Sure, gravitational waves
Also cyclical cosmologies
17:24
@imbAF Neither of those is correct in a quantum theory. $A^\mu(x)$ is an operator, it's not a "value" for the potential in the classical sense, and it does not really mean anything to say it "assigns particles to spacetime points" unless you're much more specific about what you mean by that.
@ACuriousMind I mean, for a given x^\mu value, what is the output ?
An operator
ok...I mean I am not sure what to make out of it. But I am asking because
In the lecture, we considered the scattering on an external static field
And it was said:
In this case quantum fluctuations of photons are not important (I don't understand what quantum fluctuations of photons are? Is he trying to say that we do not consider particles, in this case photons)
And also that $A^\mu$ is a classical function, a classical field that depends only on $\vec x$ and has no time dependency
So I though, I wanted to know:
What is the output of the classical field for an input $x^mu$. It must be a scalar and vector potential, no?
Yes
$A^0$ is what you call "scalar potential", the spatial $A^i$ are the vector potential
Classically (or if you are considering it "as an external field", which essentially means the same) $A^\mu(x)$ is just the value of those at $x$.
And what about quantum fluctuations of photons?
In this case
17:36
it's a not particularly precise phrase that just means you're treating the field classically, not as a quantum field so that photons could be generated or whatever
So quantum fluctuations have to do with particle creation and annihilation
which is the case for quantum fields
I suppose
it's a handwavy phrase that sometimes has a precise definition but generally people just mean that quantum stuff can happen :P
ahaa
I would not advise fixating on "quantum fluctuations" as carrying any particular insight :P
I understand
I have 2 more questions.
In the same topic, of static external field, we as I said above consider the scattering of a particle i.e electron to it.
And from my notes I see that we have a matrix element of the sort: $-ie\bar \psi \gamma^\mu A_\mu \psi$
And it seems like one can "mix" field operators of qft with what is the classical 4 potential $A^\mu$
Is this something normal?
17:41
What do you mean "seems like"? The classical $A^\mu$ are just numbers, why would you not be able to multiply an operator with a number?
to consider a situation where classical and quantum field interact with each other
@ACuriousMind Right
Ok thx for the clarification
And the other one, is a different topic
Symmetries
Does it make sense to take about gauge invariance? What I am trying to say is that, since gauge transformations are a subset of symmetry transformations, then invariance under them is a default thing
when you talk about gauge transformations
or symmetries
whichever of the two you'd use
I don't understand the question. Yes, gauge symmetries are symmetries. So invariance under gauge symmetries is gauge invariance. What's the problem?
But gauge invariance is a default thing when you talk about gauge symmetries
why is it necessary to emphasize on it?
what would you call it instead
don't get too bogged down in terminology
Ok. I just wanted to make sure that I was understanding things correctly
17:46
I don't know what you mean by "default thing"
Invariance under gauge transformations, is always the case, when the transformation is a gauge one
But I will follow Slereah suggestion and not focus to much on it
also like you don't necessarily know in advance that something is a gauge :p
sometimes you find it out
But the other reason why I asked is because, at some point during the lecture we mentioned things such as gauge boson, vector boson etc. And a term such as gauge/vector boson is strange in the sense that the first part is a mathematical concept while the 2nd term is a physical one
and then you can exclaim
Oh it's a gauge invariance
@Slereah Ok it makes sense now
17:51
@imbAF It will turn out that all massless vector bosons are gauge bosons (i.e. bosons of a theory with a gauge symmetry) and so the terms are interchangeable.
Ahaaa
And perhaps it's silly to ask but
if you had a theory with another type of symmetry
you'd call the corresponding bosons : (type of symmetry) bosons ?
what other types of symmetries are there? :P
rotational symmetry ?
and specifically you call the bosons of the gauge field the gauge bosons, not just any bosons
or translational symmetry
and so on
17:54
there is no analogy to a gauge field for "normal" symmetries
@imbAF A rotational symmetry could be a gauge symmetry, those are not really mutually exclusive.
One second cuz you are throwing new words
famously in GR all the spacetime transformations are gauge :P
gauge field = theory with a gauge symmetry ?
the gauge field is $A^\mu(x)$
what about in general?
17:56
I cannot teach you the whole abstract theory of gauge theories in chat, just accept that $A^\mu(x)$ is the gauge field and that effectively all massless vector bosons will come from a field that looks like $A^\mu(x)$
if you want the proof of this assertion you need to read something like Weinberg :P
@ACuriousMind aren't we mostly dealing with LT in GR?
why no?
why would you think that?
honestly
because
LT were introduced in SR. And therefore I thought that that would be the case in GR
18:02
I'm afraid GR has more symmetries than you want
Symmetries for days
I see
> Moreover, Engels sometimes anticipated scientific discoveries of the future. He pointed to the possibility of the existence of matter with no rest mass, and advanced the theory of the decisive role of labor in molding the physical and social forms of human existence.
hello folks =)
long time no see
just wanted to drop this here
0
Q: Definitions of cross products between tensors of rank 3 and higher

E.P.I have recently come across, in several papers, the notion of cross product between two rank-2 tensor, which I find useful for a project I am currently looking into. A useful formulation is given in L. Szabó. A note on cross product between two symmetric second-order tensors. J. Mech. Mater. Str...

in case any of you have any pointers
(I considered cross-posting to physics, but ultimately it's just math, so I'll keep it there at least for now.)
 
1 hour later…
19:18
Hello everyone :)
20:17
Ciao, Tobias
20:55
@EmilioPisanty welcome back pal
 
1 hour later…
22:13
HIIII
22:51
is the reason we don't(?) seem to usually hear the term "ambient space" in physics e.g. GR is because it usually means the immediate topological space rather than anything to do with the metric?
in what context would you expect to hear it? The spacetime manifold isn't embedded anywhere by default, there is no ambient space
oops, my eyes skipped over that part of the wiki definition
hm, actually, if it's about submanifolds, maybe if we were talking about ADM/3+1 foliation?

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