The h Bar

6:37 AM

When you train a model to recognize if a prisoner will try to escape or not, you feed training data. And in the data, it turn out more black people have attempted to escape, therefore the AI will be biased. But there is nothing wrong with the algorithm, it does it's job. When it sees a black person, it immediately has a higher chance if classifying as "will attempt to escape" because the statistics tell it so.

I'm still looking for more detailed examples on how AI can be biased.

Korra

7:29 AM

Does anyone know what is the Laplace transform of $f(x)^2$ in a differential equation of df'/dx . Also, f is unknown

Captain Bohemian

7:43 AM

Maldacena didn't know what research is likeeven after he had completed his MSc! He said:“I really enjoyed taking the classes, but I didn’t know what research was like. It was still a big mystery to me,” he says. “In the end, I decided to take my chances.”

He hopes that other students in his position will not let that fear keep them from trying it. “Maybe they will find that they are better than they expected,” he says. “Or maybe they will love it more than they expected.”

I think he is worried too much about other students.

I had never had that kind of fear when I joined research community when I started my MSc though I didn't know what research is like then.

I found research is best job after being engaged in research.

before doing research, I had no idea of what a physics graduate can seek as a job.

Johan Liebert

10:08 AM

Can someone tell me what does this phrase: "I have served, I will be of service." mean? In the John Wick Chapter 3 movie I saw this phrase being used by Zero, the Director and John Wick Himself. Though one thing that I noticed was that this phrase was followed by some act of punishment (at least for two of them).

ACuriousMind

11:08 AM

@JohanLiebert It is not an idiom or anything.

So any meaning beyond its literal one must come from the context it's used in that movie.

JingleBells

When talking about socialism, why is it okay to treat cows different than humans? After all, it's all about equality right?

ACuriousMind

When talking about libertarianism, why is it okay to treat cows different from humans? After all, it's all about freedom right?

^similar non-sequitur

JingleBells

I will answer that, but please keep with my train

Forget socialism

I'm asking you, why is it okay to treat cows different than humans?

Faded Giant

When trolling a chat room, why is it okay to troll cows differently than humans?

JingleBells

I'm debating, not trolling

ACuriousMind

11:18 AM

@JingleBells It entirely depends on your moral framework. Utilitarians will argue about levels of the ability to suffer, Kantians about the ability to reason, etc.

JingleBells

@ACuriousMind I'm asking you.

John Rennie

In the interests of equality I am happy to cook and eat both cows and humans

JingleBells

Okay, here's my reason. I presume that socialists are okay with cows being treated differently than humans. Why? Because cows look different, they are less intelligent? Well, in today's world, people look different and have different levels of intelligence.

Faded Giant

The Simpsons - Meat & You - Partners In Freedom

►

ACuriousMind

@JingleBells I'm partial to the utilitarian argument that most animals do not possess levels of sentience and sapience that would make it possible for them to hold preferences about the future, so there is no direct moral reason to refrain from e.g. slaughtering them if you do so painlessly, but it is a difficult topic since we do not have a sapience-o-meter and are effectively just guessing here.

JingleBells

11:24 AM

@ACuriousMind So if you can't measure how much something suffers, then how do you determine if you should treat it equally or not?

ACuriousMind

@JingleBells We make an educated guess. What else could we do?

JingleBells

Alright, thanks for your time :P I'm watching Transformers: The Last Knight... i watch waaayy to many movies

ACuriousMind

E.g. there are animals that demonstrate long-term memory and planning and there are those that don't. We would tentatively rank the former as potentially more able to have preferences about the future than the latter

JingleBells

I'm not in favor of treating beings differently depending on their level of suffering. I'm in favor of treating beings differently by their ability to contribute to society and create value.

John Rennie

@JingleBells there is no rational argument for why it is OK or not OK to eat cows. There are only rationalisations.

ACuriousMind

11:28 AM

There are plenty of rational arguments - there just isn't a single objectively correct one :P

That morality is subjective doesn't mean arguments about it cannot be rational.

user434058

Hey! Just letting you know, I have requested voluntarily deletion of all my SE accounts, just so that I can be more focused and have less distractions. It was fun hanging out with you guys, thank you for the good times! Bye :-)

JingleBells

:O

will miss ya :P

ACuriousMind

@FakeMod ...didn't you use to do this regularly and then come back anyway? :P

user434058

@ACuriousMind Yeah, it's kind of a habit :P

JingleBells

lol

bolbteppa

11:31 AM

"You notice that when we have passed over to the quantum theory, the distinction between primary constraints and secondary constraints ceases to be of any importance. The distinction between primary and secondary constraints is not a very fundamental one. It depends very much on the original Lagrangian which we start off with. Once we have gone to the Hamiltonian formalism, we can really forget about the distinction between primary and secondary constraints.

user434058

But all those times, I was at 101 rep. Right now I am at 6k rep. There's a slight difference...

bolbteppa

The distinction between first-class and second-class constraints is very important. ... The existence of second-class constraints means that there are some degrees of freedom which are not physically important. We have to pick out these degrees of freedom and set up new Poisson brackets referring only to the other degrees of freedom which are of physical importance Then in terms of those new Poisson brackets we can pass over to the quantum theory."

ACuriousMind

@bolbteppa It would seem to be important that we first define what we mean by "important" :P

bolbteppa

Just crazy how he modified the very Poisson brackets themselves, i.e. potentially affecting all of classical mechanics

The only motivation I think I can find for first-class and second-class is that in $\dot{\phi} = \{\phi,H_c \} + u_m \{\phi,\phi_m \} \approx 0$ a first class constraint reduces to $\dot{\phi} = \{\phi,H_c \} \approx 0$ (right?) while a second-class constraint is non-trivial $\dot{\phi} = \{\phi,H_c \} + u_m \{\phi,\phi_m \} \approx 0$ letting one fix the Lagrange multipliers $u_m$,

but then they are still indeterminate up to homogeneous solutions $v_m \{\phi,\phi_m \} \approx 0$ so a second-class constraint means unphysical extra variables lying around, and modifying the variables or equivalently PB's lets you get rid of them, though still seems like there's more going on

ACuriousMind

@bolbteppa It's not really a modification since it agrees with the normal Poisson bracket on the constraint surface for all first-class (=gauge-invariant) observables

I.e. the terms you're adding are not changing anything about the observable dynamics of the system

It's just a convenient expression that you get on the r.h.s. when you evaluate the evolution equations for the extended Hamiltonian

bolbteppa

11:46 AM

I don't think the Dirac bracket is weakly equal to the normal Poisson bracket in general if that's what you mean, if $A$ is promoted to $A^* = A - \{A,\chi_{\alpha} \} (\{\chi_{\alpha},\chi_{\beta} \})^{-1} \chi_{\beta}$ for $\chi_{\alpha}$ second-class, then $A^* \approx A$ and $\{A^*,\chi_{\gamma} \} = 0$ and the PB of two starred variables is

$\{A^*,B^* \} = \{ A,B \} - \{ A , \chi_{\delta} \} (\{\chi_{\delta} , \chi_{\gamma} \})^{-1} \{\chi_{\gamma},B\} \not \approx \{A,B \}$ (equivalently $\{A,B\}_D = \{ A,B \} - \{ A , \chi_{\delta} \} (\{\chi_{\delta} , \chi_{\gamma} \})^{-1} \{\chi_{\gamma},B\} \not \approx \{A,B \}$)

ACuriousMind

@bolbteppa It is, cf. e.g. eq. (1.51a) in H/T

Note that non-equality for gauge-variant observables doesn't mean anything because they're non-physical quantities anyway

bolbteppa

Oh yeah right because it's first class duh

Yeah that's a good point as well

ACuriousMind

Don't get me wrong - it's still a clever trick to have figured out, but it's not a change to the mechanics, just another change in our description of them just like the introduction of unphysical d.o.f. to begin with

Johan Liebert

@ACuriousMind Yes, I was asking regarding the context in that movie. Hope someone who watched that movie could clarify it.

BTW, I currently want to lose about 2000 reps. So if anyone wants a bounty on their non- homework type questions which adhere to site policy, please contact me. (With appropriate amount of reps that you want on that question.)

@FakeMod you too should give away your reps as bounty to help those who need some attention to their questions and then delete your account. That charity may also be helpful in increasing the content quality (due to competition for reps). :-)

bolbteppa

12:09 PM

Right, I think the motivation for second class constraints seems to be that they let you fix the Lagrange multipliers as you get a system of inhomogeneous linear equations for the multipliers, but you can't eliminate the possibility of being able to add homogeneous solutions as I mentioned above so you're still forced to deal with additional arbitrary terms in the theory and I think this just means they are non-physical variables in the theory that could be eliminated,

e.g. as in the HT example (1.45), hence why one defines the Dirac bracket to simply eliminate them. But the easiest example, a relativistic point particle, has only a first class primary constraint yet the Lagrange multiplier is fixed from Hamilton's equations and this has nothing to do with second class constraints fixing the Lagrange multipliers, so...

This Dirac stuff is so confusing at times

Physics Meta

12:50 PM

0

Q: Do you want bounty on your question?

These days I ain't that active on this site and neither I use my privileges to help moderate this site. So I am thinking of giving away this reputation so that those who require attention to their questions can get it. So for getting a bounty on your question, do the following: Post the title of...

JingleBells

3:13 PM

Regarding college, you guys are talking as if college is the only place to meet people and talk ideas. I believe that's incorrect. You can always, after getting a job, talk to people, meet people, go to conferences, go to meetings. In fact, I believe the valuable learning of how the world actually works happens outside of college.

Charlie

3:32 PM

Of course, but in general going to college will increase the number and quality of jobs you are eligible for. No one is saying skipping college means you can't experience all of those things by yourself, but there are other benefits of going to college other than the social ones, specifically the qualification you're getting. @JingleBells

JingleBells

@Charlie Got it, thanks. If that's the case, college is definitely not for me.

What are you studying Charlie?

Charlie

well, physics :P

with varying levels of success

JingleBells

oh, lol, obviously

Any special topics ur interested in :P?

Charlie

theoretical particle

but I'm still a long way from even making a start on it

JingleBells

So bosons, quarks, leptons... these stuff? elementary particles?

Charlie

3:40 PM

those are types of particles yes

JingleBells

f*cking sentdex helped me on discord :O

@Charlie Is the existence of such elementary particles proven?

Seems really complicated stuff

Charlie

yes, as much as anything can be "proven" in physics

JingleBells

What do you want to do with that knowledge and understanding?

Charlie

research

JingleBells

I mean, after you research and everything. Do you learn it for self-satisfaction?

ACuriousMind

3:49 PM

"after research" usually means death or very high age if you stay in academia :P

Charlie

I mean research is a full time job that continues on

yeah

JingleBells

Don't you plan to do anything useful with the things you learn? Like maybe develop theories and do stuff like Einstein and so on?

Charlie

That would constitute "research"

JingleBells

got it

so you hope to discover cool new things in your realm of interest in physics?

Charlie

yeah that's research

JingleBells

3:51 PM

I believe in you Charlie, go be the next Einstein

ACuriousMind

just maybe hold off on the cousin-marrying thing

Charlie

clearly that's the secret ingredient

special relativity was clearly inspired by a special relationship with a relative

JingleBells

XXSDDD

ACuriousMind

not rejecting quantum mechanics might also be wise :P

JingleBells

good one charlie

Charlie

3:53 PM

thank you

satan 29

5:14 PM

hey all

is someone willing to discuss rotational dynamics?

Qmechanic

@satan29 : Don't ask about asking, just ask.

satan 29

@Qmechanic wow, I am honored to get a response from you!!

Tony Stark

Is it okay to post two answers on the same question?

satan 29

5

Q: Instantaneous axis of rotation and rolling cone motion

Suppose a cone is purely rolling (no slipping) around a fixed axis. I mean, it is revolving around a fixed axis perpendicular to the ground and passing through its vertex and also rotating, so the vertex is stationary. (sorry this might be a bit confusing but I hope you understand what I mean). S...

rotation rotational-kinematics

i Think i understand what selene routley is trying to say, but i have a minor confusion

ACuriousMind

@TonyStark If the two answers are fundamentally different, yes

JingleBells

5:18 PM

lol Tony Stark asking questions on SE. I thought u were a genius dude

satan 29

in that question, I dont see why the velocity of the center of the Base is Not wr

Tony Stark

@JingleBells A great Physicist once said what is there in the name.

John Rennie

Who needs Matthematica?

satan 29

suppose we are in the cone-frame

In the cone frame, The velocity of the bottom point should be simply $\Omega r$, no?

so the velocity of the bottom point in the cone frame should simply be $V -\Omega r$

although the final answer by selene routley does not satisfy this: so what is going on??

Safdar

5:40 PM

@satan29 what would be V here? (is it constant)?

satan 29

$V$ is the velocity of the center of the base of the cone

and i believe its going to be = $\omega d$

where $\omega$ is the angular velocity about the Z axis

ZeroTheHero

8:22 PM

@JohnRennie just more headaches for the math profs.

I admit I really do not understand why Google puts out stuff like this that obviously encourages cheating on assignments.

JMac

9:01 PM

@ZeroTheHero Yeah... I'm okay with the idea of general use math solvers, but gearing it towards answering homework seems odd

Thirsty for concepts

9:20 PM

0

Q: Superconductivity in ceramics

I understand that superconductivity mainly occurs due to the formation of the Cooper pairs in which electrons, instead of repelling each other, actually attract because one electron actually attracts the positive charges nearby which further attract the the other electron thus establishing a Coo...

superconductivity

Can anyone pls help me with this question

Charlie

9:37 PM

Is it common (in physics) to choose as a representation space a matrix vector space?

I'm not really sure what endomorphisms on a matrix vector space even look like, seems like it would be complicated

or maybe a tensor space, again sounds like it might be complicated

mostly just curious, I can't find reference to it online but then again I'm not entirely sure what I'd even be looking for

bolbteppa

I'm guessing you're thinking of the adjoint of a Lie algebra acting on the Lie algebra of matrices which we view as a vector space?

Charlie

hmm

bolbteppa

Yang-Mills is constructed in terms of the adjoints of su(2) and su(3) so pretty common

Charlie

just so I'm not confusing terminology, the "adjoint representation" s when we choose the representation space to be the Lie algebra itself right?

I've only just started covering this stuff

bolbteppa

9:53 PM

Yeah, $\mathrm{ad}(X) : \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$ such that $\mathrm{ad}(X)$ is a linear map $\mathrm{ad}(X) : \mathfrak{g} \to \mathfrak{g}$ defined by $\mathrm{ad}(X)(Y) = [X,Y]$

Charlie

ok nice

actually just because I haven't seen it before, what does the $\mathfrak{gl}$ in $\mathfrak{gl}(\mathfrak g)$ mean?

is that notation specifically used when we're talking about the adjoint representation?

bolbteppa

It's the Lie algebra of the general linear group acting on the vector space inside the $( )$'s, i.e. the lie algebra of the group of invertible matrices acting on that vector space

Remember in representation theory you map a group to a subgroup of the general linear group acting on some vector space

Charlie

oh I see

I hadn't actually given much thought to the general linear group itself being a lie group

bolbteppa

$\mathfrak{gl}(\mathfrak{g})$ is literally just a bunch of matrices acting on the vector space $\mathfrak{g}$ along with a Lie algebra commutator $[X,Y] = XY - YX$ defined on it

Charlie

actually that kind of raises a question for me, if the general linear group is just an abstract lie group, don't we have to choose a representation of it in order to associate it with the set of invertible matrices?

oh wait that's the point isn't it

ohhh I think I see actually

$GL(n,\Bbb K)$ is an abstract lie group, we choose it's representation on $\Bbb K^n$ and then other lie groups take as their representations a subset of that of $GL(n,\Bbb K)$'s

or rather as the codomain of their representation a subset of ...

Charlie

10:41 PM

So just so I'm clear, it's a bit imprecise to define Lie groups as subsets of the general linear group, it's only once we've chosen a particular representation that this really makes sense

they're just often defined that way in physics because the abstract group isn't particularly useful to us

this is what originally bothered me, it seemed strange to define the groups that way, but I guess it makes sense now

ACuriousMind

11:04 PM

@Charlie There are even Lie groups (e.g. the metaplectic group) that aren't subsets of any linear group!

bolbteppa

20

A: Are all Lie groups Matrix Lie groups?

As other answers mention, it is not true that any Lie group is a matrix group; counterexamples include the universal cover of $SL_2(\mathbb{R})$ and the metaplectic group. However it is true that all compact Lie groups are matrix groups, as a consequence of the Peter-Weyl theorem. It is also t...

$\mathrm{GL}(V)$ is just the group of non-singular linear operators mapping $V$ into itself. If $V$ is finite-dimensional then on choosing a basis for $V$ every element of $\mathrm{GL}(V)$ becomes a non-singular matrix. So it's all the same thing at the end of the day, since a matrix is really just a way of talking about a linear operator in a basis.

Sticking a bunch of matrices in a set and assuming they have an algebraic structure is fine mentally, but mathematically it's all sets, functions and relations so a matrix has to be justified in terms of functions

Charlie

physics math strikes again!

bolbteppa

Yeah, maybe Category theory makes it all easier :p

Transcript for