The h Bar

123

6:04 AM

Hello Everyone...

qwerty

7:13 AM

do highly active users on SE write long-form notes or anything, and if so, where would you host them? personal website seems like the obvious place but something like centralised similar to arxiv or SE would make things easier to find. the ability to upload pdf files to a SE profile would be kinda cool...

Ryder Rude

7:29 AM

hi

@qwerty notes for what?

ACuriousMind

8:05 AM

@qwerty You seem to have some kind of idea that "highly active SE users" form a homogeneous group with broadly similar habits, but they're not. They might be currently professional academics, ex-academics or amateurs with a lot of time and so their habits will vary a lot. Obviously someone who is a professional academic physicist regularly writes "long-form notes" - either lecture notes for their teaching or research articles for publication -

but this has nothing to do with their SE activity, and I don't understand why someone who is not a professional academic might be caused by their SE activity to write such notes.

qwerty

9:05 AM

@ACuriousMind apologies if I caused offence. I did not mean to imply all active users have similar habits, but I did assume there would be an underlying broadly-defined goal of disseminating and discussing physics on some level. My point was that the Q and A format is only one short-form way of doing so.

ACuriousMind

@qwerty Oh, don't worry, you're weren't being offensive - I just think you underestimate how heterogeneous the user base is

qwerty

I am aware that there are professionals, ex-professionals and amateurs. I tend to write things down when reading, digesting or thinking about them and I assumed that other people did the same.

Slereah

@ACuriousMind They may even be John Duffield

ACuriousMind

@qwerty Well, for me personally, everything I know is already written down somewhere - that's how I learned it, after all. So I personally don't produce any texts except the answers I write. Over the years I have sometimes had the idea to write a textbook-like document in my personal style, but this would take a lot of time and effort I probably will never get around to.

one of my questions did trick mike stone into writing a 50 page paper once, though :P

Ryder Rude

i have thought about writin down my understanding in a note/book format

but i have been procrastinating it

it would be easier to look up for future reference

naturallyInconsistent

9:28 AM

It takes sooooo much effort. And now that I'm starting work on mine, it just keeps ballooning; to understand the unorthodox way im doing quantum, I would have to write the classical mechanics text first...

Bml

Hi everyone. Could you help me understanding the most voted answer to this question? There is something I cannot understand...

28

Q: Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$?

For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$ \frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$ From this, it follows there's also a corresponding scalar quantity conserved:$$\frac{d}{dt}\sum\limits_i^n m_i(v...

qwerty

@ACuriousMind interesting! i never feel I have understood something until I have done it in my own words/ own way. tbh, I dont know I'm going to make it in physics professionally (on the postdoc application cycle without serious prospects imo). actually i felt users (like yourself) on this site and wiki who I feel do a lot of good for physics/ people in general in that way, made consider me that SE would be a serious alternative way to contribute if I leave academia.

@naturallyInconsistent it really does! I have run into similar problems...

naturallyInconsistent

@Bml you should not state it as "most voted". You should just refer to the name, or just use the link that you get from sharing that particular answer, so that then we will all be on the same page

And you should formulate a question at least to a point whereby someone may guess what it is you are confused about.

ACuriousMind

@qwerty It is! physics.SE is definitely also a good way to "stay in touch" - knowledge you don't use tends to fade away, and the continuous engagement with physics here keeps it fresh

qwerty

@ACuriousMind haha, this is amazing!

Bml

9:37 AM

@naturallyInconsistent This is the link I get from sharing that particular answer:

31

A: Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$?

The factor of 1/2 is required from Galilean invariance, so that the energy mixes up with the momentum without a factor. This was understood before relativity, but it is largely conventional before relativity, since you could make the energy mix up with the momentum using some coefficient. Once yo...

Ryder Rude

@qwerty also, the site will be preserved as internet fossils

the answers will stay forever

and help people 100 years later

qwerty

indeed, hence my interest in uploading longform notes inappropriate for arxiv... long after personal domains are dead.

naturallyInconsistent

@Bml and what is your question?

Ryder Rude

there is an answer by joshphysics which is the best understanding of Newton's laws

it will help people forever

Bml

@naturallyInconsistent OK. The answerer says: "In Newtonian mechanics, the energy and momentum transform together after a Galilean boost. If you have a closed system with momenta $p_i$ that add up to zero (center of mass frame), the change in the kinetic energy after a boost, which shifts $p_i\rightarrow p_i - m_i v$ is $\sum_i {p_i^2\over 2m_i} \rightarrow \sum_i m_i~(v_i - v)^2 = \sum_i {p_i^2\over 2m_i} - \sum_i p_i \cdot v + \sum_i m_i {v^2\over 2}$.

Ryder Rude

9:41 AM

this answer physics.stackexchange.com/a/70188

Bml

@naturallyInconsistent Where did that $\sum_i {p_i^2\over 2m_i}$ come from if the goal is to show where $1/2$ comes from?

ACuriousMind

@Bml He's starting from that to show it results in no coefficient for the $(\sum p)\cdot v$ term. He could as well have started with an arbitrary prefactor $k$ with $\sum_i k \frac{p_i^2}{m_i}$ and found $2k (\sum p)\cdot v$ for that term, then concluded that $k = 1/2$ for the term to have no additional prefactor. Like many of Ron's answers, I don't find this argument convincing at all on closer inspection because there's no explanation here why we'd want that term to have no prefactor.

In the end I find this entire question somewhat pointless: Obviously the 1/2 is only a convention, since any multiple of energy is still a conserved quantity, and the useful thing about energy is that it is a conserved quantity, not its exact numerical value.

I prefer Qmechanic's answer, which is short and to the point: If you choose prefactors for all the other quantities as usual, then this 1/2 is what drops out of the work energy theorem as the natural conserved energy-like quantity to consider.

The relativity argument is just a red herring: We might just as well multiply the relativistic energy expression conventionally by 2 to get no 1/2 at first order, this is not a different explanation than "it's a convention" at all.

Bml

@ACuriousMind I guess he says we'd want that term to have no prefactor for reasons of simplicity, but in the end it is a matter of convention. As he says:

naturallyInconsistent

9:57 AM

@Bml The part of the answer that is tolerable, is the link to SR. Either you work with $$\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=m_0c^2+\frac12m_0v^2+\cdots$$ and convert that to momentum form as per usual, or you work with $$\sqrt{(m_0c^2)^2+(pc)^2}=m_0c^2+\frac{p^2}{2m_0}+\cdots$$ to obtain the $\frac{p^2}{2m_0}$ term.

Bml

"It is simplest if you take the coefficient $1/2$, ultimately because this is the natural nonrelativistic limit of relativity [but although it is the natural limit, this does not exclude it being a convention]. You would have a factor of $2$ in the $p$ mixing term if you didn't take the coefficient $1/2$. It is actually a testiment to the insight of the 19th century physicists that they adopted the most natural convention before relativity was discovered.

naturallyInconsistent

@Bml For centuries, people worked with $2E$ everywhere. It is cumbersome and irritating to have to remember where the 2 goes for all the integrals that would have to deal with the difference, but otherwise, the exact same physics happened. So, yes, it is purely a convention.

Bml

@naturallyInconsistent In what sense "it is cumbersome and irritating to have to remember where the 2 goes for all the integrals that would have to deal with the difference"?

naturallyInconsistent

@Bml Where you had some $\text{PE}=\int\vec F\cdot\vec{\mathrm dr}$, you will then have to do $2\text{PE}=2\int\vec F\cdot\vec{\mathrm dr}$. It is cumbersome, but people will make do

@ACuriousMind Actually, no, I disagree with this. The rest mass v.s. energy conversion rate would be altered; the factor of half will still appear to be the natural comparison between the rest mass energy contribution v.s. the non-relativistic kinetic energy contribution, and so I think Ron is correct.

Bml

@naturallyInconsistent But why would people have wanted to work with $2E$ or a factor of $2$?

ACuriousMind

10:12 AM

@naturallyInconsistent Sure, but the explicit 1/2 would vanish, to be counterbalanced by $E=2mc^2$, no? What you're saying is just a demonstration that "there has to be a 2 somewhere", but it is still convention whether you put it on a bunch of other stuff as a 2 or on kinetic energy as a 1/2. I would even agree the 1/2 is the better convention, but I don't think relativity completely removes this being a convention.

@Bml Because in the old times before classical physics was this neat assembly of well-trodden paths it is today, people simply noticed that a quantity proportional to $mv^2$ was useful because it was conserved. You don't need to put a 1/2 there until you have developed the rest of classical mechanics so far that you get stuff like the work-energy theorem that puts the 1/2 there as a consequence of all your other conventions.

naturallyInconsistent

@Bml they didnt know that they were making that choice. They just thought that they were eliminating the factor of half elsewhere

ACuriousMind

for a time, people weren't even really sure what "kinetic energy" really was supposed to mean - you can read up on the Vis viva debate where (bizarrely, from a modern viewpoint) people debated whether $mv^2$ or $mv$ was "the" conserved quantity to consider.

you can also see the interesting historical remark there that the 1/2 was "standardized" around 1820, pretty long after both energy and momentum had generally been established as useful

naturallyInconsistent

@ACuriousMind No, I'm saying that it would have been $2m_0c^2+\frac12(2m_0)v^2+\cdots$ and thus the factor of half will still be natural. Note that once we have SR, the natural thing to do is to set $c=1$, and then the $2m_0$ will really suggest that either we redefine energy units, or mass units, to get rid of this factor of 2. Because one object should have one object's worth of rest mass energy that is in agreement with each other.

Bml

@ACuriousMind OK, suppose we start with an arbitrary pre-factor $k$. with $\sum_i k \frac{p_i^2}{m_i}$. Why did he write \sum_i m_i~(v_i - v)^2?

ACuriousMind

@Bml ...because that's what a boost by $v$ does, it shifts all velocities by $v$?

Bml

10:23 AM

@ACuriousMind OK, but what exactly does this term ($\sum_i m_i~(v_i - v)^2$) represent?

ACuriousMind

oh, you perhaps mean the typo; yes, there should be a 1/2 there

The kinetic energy after the boost is $\sum_i \frac{m_i}{2}(v_i - v)^2$. This is then equal to the final expression for the energy in the answer, there's just a 1/2 in that intermediate step missing

which is pretty confusing since that's the entire point of the answer, but I guess 32 upvoters didn't care :P

Bml

@ACuriousMind Yes...

bolbteppa

@Bml That $1/2$ is there so that we get $F = ma$, not $F = 2ma$, the relativity argument is not correct, one fixes the relativity coefficient so that it reproduces the usual non-relativistic case, somewhere it amounts to a convention on the way that $m$ appears, and $F = ma$ is the standard convention

Ryder Rude

10:39 AM

@bolbteppa i think it is upto choice where one would take that constant if one removes it from KE

bolbteppa

If you send $m \to 2m$ then it goes away from the KE but arises in other places like $p = mv \to 2mv$ and $F = ma \to 2 ma$ and in the relativity formula etc

Ryder Rude

u r right. this constant gets transferred to the force

it's all definitions... this discussion is pointless

bolbteppa

It all reduces to the proportionality between F and a being some m, you can randomly call it 2m but why would you

Bml

11:02 AM

@ACuriousMind OK, suppose we don't know that factor $1/2$. The kinetic energy in the frame of reference of the centre of mass is $E_k = k \sum_i m_i (v_i - v)^2$. This is equal to: $E_k = k \sum_i m_i v_i^2 - 2k v \sum_i m_i v_i + k v^2 \sum_i m_i = k \sum_i \frac{p_i^2}{m_i} - 2k (\sum_i p_i) v + k M_{total} v^2$.

To make (conventionally) the energy mix up with the momentum using no coefficient, we have $2k = 1$, so $k = 1/2$.

From that: $E_k = 1/2 \sum_i \frac{p_i^2}{m_i} - (\sum_i p_i) v + 1/2 M_{total} v^2$, and since $(\sum_i p_i) v = 0$ by definition, we have $E_k = 1/2 \sum_i \frac{p_i^2}{m_i} + 1/2 M_{total} v^2$, where 1/2 \sum_i \frac{p_i^2}{m_i} is the kinetic energy with respect to the inertial frame of reference and 1/2 M_{total} v^2 is the kinetic energy of the centre of mass. Correct, or is there some flaw in the reasoning?

ACuriousMind

11:24 AM

yes

Ryder Rude

en.m.wikipedia.org/wiki/Many-worlds_interpretation physicists were extremely repulsive to Many worlds and disrespected Everett

> Everett recounted his meeting with Bohr as "that was a hell... doomed from the beginning". Léon Rosenfeld, a close collaborator of Bohr, said "With regard to Everett neither I nor even Niels Bohr could have any patience with him, when he visited us in Copenhagen more than 12 years ago in order to sell the hopelessly wrong ideas he had been encouraged, most unwisely, by Wheeler to develop. He was undescribably[sic] stupid and could not understand the simplest things in quantum mechanics.

Bml

@ACuriousMind So why does kinetic energy with respect to a fixed inertial system (not the CoM frame of reference) have the same expression? Shouldn't they differ by a minus sign?

ACuriousMind

I don't understand the question

Bml

@ACuriousMind Earlier I calculated the kinetic energy of a system of point masses with respect to the CoM frame, and it turns out to be the sum of the kinetic energy of the centre of mass and the kinetic energy with respect to a fixed inertial system. But shouldn't there be a minus sign instead of a plus sign?

If I choose to calculate kinetic energy with respect to a fixed inertial frame of reference (not CoM frame), I get the same result as if I calculate kinetic energy with respect to the CoM frame. This seems like a contradiction to me.

ACuriousMind

I'm not really following you, but now that I read what you wrote again, I don't really understand your third message

@Bml What do you mean "$(\sum p)v = 0$ by definition"? By definition of what?

Bml

11:35 AM

@ACuriousMind By definition of "CoM frame". The CoM frame is a frame in which momenta adds up to $0$.

Ryder Rude

Everett soon left academia

people ganged up on him

ACuriousMind

@Bml Sure, but in that frame the center-of-mass is at rest, so $v=0$, too

I don't really understand what you're trying to do

Ryder Rude

but today, it is seen as a good interpretation

Everett never got to see that

bolbteppa

> Gell-Mann and Hartle, along with a score of others, have been working to develop a more palatable interpretation of quantum mechanics that is free of the problems that plague all the interpretations we have considered so far.

Bml

@ACuriousMind OK, what does this change?

bolbteppa

11:37 AM

Basically once Bohr and Heisenberg could no longer properly defend proper QM, the casuists seen their chance and started bringing in their own desires again

Ryder Rude

@bolbteppa they defended it by disrespecting others...

they wanted to push an agenda, going by their conversation. i dont think they cared about what's true

bolbteppa

Yes, as I've pointed out multiple times, they said it was literally nonsense to think there were alternatives to standard (Copenhagen) QM

Ryder Rude

yes

this is a very bad way to do science. they turned into gatekeepers

Bml

@ACuriousMind Perhaps I have realised the error. The answerer starts from the CoM frame, with respect to which the velocity is $v_i$, then performs a Galilean boost assuming the velocity of the centre of mass is $-v$ (minus sign). Thus, $(v_i - v)$ is the velocity of a material point with respect to the fixed inertial frame of reference, not with respect to the CoM frame.

Here it all fits: I was not calculating kinetic energy in the frame of reference of the centre of mass, but in a generic fixed inertial system.

Bml

11:52 AM

Also, @ACuriousMind and @naturallyInconsistent, I cannot understand whether your arguments on special relativity are in disagreement or not. Could you clarify?

naturallyInconsistent

@Bml I made my case. ACM had not replied to that

ACuriousMind

I don't think we're disagreeing, I'm just willing to entertain much more stupid choices (i.e. $E = 2mc^2$ and refusing to redefine my mass unit) as "conventions" ;)

naturallyInconsistent

@ACuriousMind But even when you take that, you can see from my argument that the natural factor of half will still appear, and so I think Ron's SR argument is the only correct way to argue why the factor of half is natural to physics.

but alas, tada, miao miao gotta go to the gym

ACuriousMind

@naturallyInconsistent That's what I meant by "there's a 2 somewhere" - yes, the ratio you're talking about is real and "physical". But that does not force us to put the 1/2 into the expression for the kinetic energy necessarily, it's just the most efficient way to do this. You can say "the kinetic energy has a factor of 1/2 compared to the rest mass energy" or you can say " the rest mass energy has a factor of 2 compared to the kinetic energy", it's the same thing

Slereah

12:28 PM

@bolbteppa I'm not using any interpretation from that war criminal

Amit

@Slereah What do we really know about JASON?

Slereah

Not a whole lot though there's a few declassified JASON papers

mostly pretty boring technical stuff

Amit

So which of them is the criminal part?

Slereah

> In 1970, the Student Mobilization Committee published a set of secret minutes it had obtained on a 1967 Jason seminar on problems of counter-insurgency. The regular Jasonite participating was Dr. Murray Gell-Mann and the main thrust was to find ways of getting social scientists usefully involved in solving problems of interest to the military. Selected quotes:

“Gell-Mann: Can we find out what effect increasing police density or ear cutting, or other negatives have on villager attitudes?”

Things were certainly going on

bolbteppa

Slereah

12:37 PM

although as far as war crimes go, I probably wouldn't recommend Heisenberg either

bolbteppa

Who did you mean

Slereah

ocf.berkeley.edu/~schwrtz/SftP/Jasons.pdf

@bolbteppa He was working on the nazi atomic bomb

Amit

@Slereah lol... that's very little to go on but yes, something was certainly going on... but maybe Murray was actually being the heroic type and decided not to sell us off immediately to the secret race of aliens that the Pentagon aligned itself with... at least not until the end of the current millenium. He showed the aliens we still have potential by discovering quarks, so they decided to give us another shot for the next 1000 years. At least that's my theory but it's open to critique 🤣

Slereah

He was just your average stooge working for the military rly

it's just that this typically involves a lot of war crimes because that happens a lot in war

very naughty place

He probably fell for the usual trap that happens with all such war criminals

bolbteppa

He was just a simple high school teacher trying to take care of his family thanks to the lack of universal healthcare

Slereah

12:41 PM

"If we do these really horrendous things maybe the war will be shorter"

Happens a lot in history

Basically how chemical warfare came about

Amit

Probably Feynman had it easy because they suspected him to be a commie. Breaking into the safes that contained the A-bomb secret helped

Slereah

The US atomic program was certainly not above reproach

And some people involved were fucking nuts

Amit

Conclusion -- do wild shit to make the authorities suspect but not too much (e.g. Bohm being a full on commie)

Slereah

ie von Neumann

Amit

What about Bob Lazar?

Slereah

12:46 PM

I prefer

Rem Lezar "Higher"

►

Amit

@Slereah Didn't you grow up in France? This is a very Anglo-Saxish clip...

Slereah

I'm a Rem Lezar worshipper

Amit

Noted

Slereah

youtube.com/shorts/tX-vp0H643Q?si=O02qF6d0NeoDxjcj

Amit

That's a great way to sum up our in depth discussion

Ryder Rude

12:52 PM

"we were just obeying orders" is a common tactic for war criminals

Amit

Don't mention the war!

Ryder Rude

@Slereah also the atomic bomb

in the oppenheimer movie, oppenheimer suddenly grows a conscience when he thinks about the dead

or maybe because he thought about a future apocalypse

Amit

Most likely both

Silly Goose

is there a paper describing the basic mechanism of an atomic bomb? I am curious because it seems like quite a distinct example of a "micro- to extremely macroscopic" evolution of a system

Ryder Rude

@Amit yes

however, Truman never grew any conscience according to the movie

ACuriousMind

1:01 PM

@SillyGoose The simplest kind, pure fission bombs, are just bunches of subcritical radioactive material that you then bring together with some mechanism to achieve supercritical mass

Slereah

@SillyGoose Not a lot for obvious reason, but there is a famous one

I think it's the Teller paper or something

ACuriousMind

the Hiroshima bomb was essentially just a gun that shot one uranium mass at another

I'm not sure what kind of details you expect there from a paper that aren't engineering details :P

(and somehow I suspect you're not interested in engineering details)

Silly Goose

i think i am interested in the theory of how a (sustained) nuclear chain reaction is modeled (presumably using the framework of textbook qunatum mechanics). not necessarily how to start one

ACuriousMind

@SillyGoose What exactly is there to model? When some uranium atom splits, it releases on average X neutrons. Each neutron then has some probability per other uranium atom it encounters to induce fission of that atom. When this leads to each fission producing on average more than 1 neutron that hits another atom and splits it, you get a chain reaction - supercriticality. There isn't really anything particularly quantum about this except that there are probabilites :P

Silly Goose

i guess i am conceptually thinking that the time evolution of an atomic bomb (from pre-explosion to post-explosion) could be an interesting place to see a quantum-to-classical transition of dynamics

Slereah

1:11 PM

upload.wikimedia.org/wikipedia/commons/9/9c/…

that's the one

ACuriousMind

as I just said, you don't really need to understand much about quantum mechanics to understand an atomic bomb, you just need to know that one particular kind of uranium releases a lot of energy and more neutrons when you fire neutrons at it. Then you bring enough of that together to be dangerous and - boom

Silly Goose

hm so i mean is there no hamiltonian describing nuclear chain reactions

ACuriousMind

I don't think that would be a useful way of thinking about it, no

Ryder Rude

@Slereah are you also interested in philosophies of consciousness, or only philosophy of science?

Slereah

I'm not even conscious

Ryder Rude

1:14 PM

You've said that before :P I'm not believing it

Slereah

How do you know though

Ryder Rude

we don't have AI like u yet

Silly Goose

so it is more like a "phenomenological model"? it seems like one is just ignoring a microscopic picture

ACuriousMind

@SillyGoose What microscopic picture? The neutron hitting an atom is certainly microscopic! :P

Slereah

Trying to compute quantum calculations for an atom bomb is a bit overkill

Silly Goose

1:16 PM

hm okay

ACuriousMind

certainly there are some more detailed models of what exactly happens during fission, but they are entirely irrelevant for the workings of an atomic bomb

Slereah

I had nuclear physics class and I don't think we even used anything beyond the liquid drop model

Amit

@RyderRude How can he be conscious, being part of the eternal quantum whole, there is only a single unitary evolving consciousness we are taking part in... $|\text{OM}\rangle$....

Slereah

I think the fanciest model that's used in engineering circumstances is like "the nucleus is just a potential in which the protons and neutrons are"

Ryder Rude

@Amit but consciousness splits in MWI :P

Silly Goose

1:18 PM

separately, how does an atomic clock work? to my current understanding, we might have (ideally) a two state system $\lvert 1 \rangle$ and $\lvert 2 \rangle$. then we can pump in energy to the system to induce transitions $\lvert 1 \rangle \mapsto \lvert 2 \rangle$ in a precise way (e.g. after 2 seconds of energy pumping my $\lvert 1 \rangle$ will transition to $\lvert 2 \rangle$). Then, do we let the atom naturally decay $\lvert 2 \rangle \mapsto \lvert 1 \rangle$ at some natural rate

and then pump in energy again?

Slereah

I looked into it once, for GR reasons

Amit

@RyderRude You know how I feel about MWI... 🤬 and when I say "I" , I mean the non-existing I

Slereah

Have you ever seen a rube goldberg machine

Atomic clocks are a huge mess of bits

There's a good PSE post about them but it's pretty complicated

Ryder Rude

@Amit u in the other world exists tho

Slereah

0

A: How atomic clock works?

I'm going to dump some notes of interest as I learn more and more about it here. https://www.youtube.com/watch?v=eOti3kKWX-c Inside the HP 5061A Cesium Clock by CuriousMarc has some good information. Launched in 1964, the HP 5061A is a very portable atomic clock similar to the ones that were take...

Ryder Rude

1:23 PM

@Amit do you prefer panpsychism or illusionism?

anyone seen Alien (1979)?

i turned it off midway recently. They were out to rescue people in dark mountains

maybe it gets good

Amit

2:00 PM

@RyderRude idk what is illusionism

Ryder Rude

2:13 PM

@Amit it says e.g. if u give a powerul ai an image and all the knowledge about how the human brain works, the ai can work out what the image looks like to a human subjective experience

Amit

Like the Talos principle!

Ryder Rude

this seems to be a game about AI

in the example i gave, the AI is "blind". it doesn't have visual experiences. But it can begin to have them after it has the knowledge

Amit

Yes, but the name is based on an argument by Straton that sounds quite similar

Ryder Rude

illusionism asserts that it can begin to have them

@Amit oh

Amit

and it's a good game

Ryder Rude

2:16 PM

It looks good

Amit

If you like spatial puzzles then yes

Ryder Rude

oh

i dont like puzzles :P

Amit

But you always puzzle about philosophy and physics... ^_^

Ryder Rude

i always stay in the non puzzle realm...

i just speculate and stuff

Amit

Ah I get it, you like the puzzles that don't have answers, lol

Ryder Rude

2:18 PM

i never do that conspiracy guy meme

@Amit yes.. :P

@Amit so what do u think about illusionism

Amit

I don't see exactly why it is in tension with panpsychism

Ryder Rude

yeah... Maybe. but most proponents would say panpsychism asserts that mental properties are a fundamental substance

maybe it is not in tension with some types of panpsychism. Rovelli's relational interpretation is one such panpsychism

if it is a fundamental substance, the ai cant simulate that substance without having it

Amit

if it's fundamental, perhaps that's exactly why we can in theory create a conscious machine, we just use the building blocks of nature after all

Ryder Rude

yes!

Chalmers always says that he doesn't see why silicon can't do it while biology can

Amit

I can only say that of the two, panpsychism sounds more far fetched, but there isn't an inherent contradiction

Ryder Rude

2:24 PM

Yes, but that's because how i explained illusionism. I have misled u!!

sorry

Relativisticcucumber

does whether something is a metal or not depend on temperature?

Amit

:O

Ryder Rude

it wasn't intentionl

Relativisticcucumber

or CAN it i mean

Ryder Rude

the point of the illusionism experiment was just to say that knowledge or physical information contains all the properties of an object

but i didn't manage to say that precisely

so, if the ai manages to see the object, but it sees it due to its access to "mental substances", then that is not illusionism

Amit

2:26 PM

is that what they used to call materialism?

Ryder Rude

yes... It is called eliminative materialism now

i think illusionism is a synonym

Relativisticcucumber

@Slereah so ive seen before conceptually saying that atoms have stable half lives and that this is used. going off of this post, it seems the time electron takes to reverse its spin direction would be a very "set" time, like that process takes that amount of time with certainty. is this so?

Ryder Rude

ooh I should've just given the Mary's room experiment to u...

Amit

@RyderRude But that's a point that logically seems to be untestable so at least not a valid scientific argument

Ryder Rude

@Amit yes. The Mary's room experiment is testable. I just gave u a bad experiment. Lemme explain

ACuriousMind

2:29 PM

@Relativisticcucumber Yes, if you just consider "metal" to describe a phase of matter with certain properties, see e.g. "metallic" hydrogen

Ryder Rude

In Mary's room, Mary's brain physically doens't allow her to experience colors

so the question is, is Mary able to see color by using knowledge of color ?

this is testable

if yes, then illusionism is confirmed

Because despite a physical handicap, she was able to see color

but it is "testable" in the very far fetched sense. It is not testable in any practical way

cuz if knowledge is the same as qualia, it is extremely complex knowledge for a real person in the experiment to process

Amit

@RyderRude You may find this talk interesting: youtube.com/watch?v=yrpO1eP4ecw

Ryder Rude

@Amit thanks!!

I like talks about this

Amit

np.. it mentions Chalmers among other things...

Ryder Rude

nice

i will see it

@Amit have you seen paranormal activity

It is a found footage horror

Amit

2:33 PM

:O no... Oliver Lodge etc? I read about this stuff but wasn't too convinced to say the least

Ryder Rude

No, idk about Oliver Lodge

Amit

so i'm not sure what you're referring to

Ryder Rude

in the beginning, it says "we would like to thank the families of the victims " :P

Amit

why would i see something like that

Ryder Rude

it is fictional

but shot like real found footage

and they try to lie in the beginning

Amit

2:35 PM

only now I realized you're talking about a film. No I don't watch that kind of genre.. I thought you were asking me about my experiences of the spirit world, lol

Ryder Rude

lolol

ACuriousMind

@Amit if you want to learn actual philosophy of consciousness, most sources like Wikipedia would be better than trusting what Ryder Rude says. Mary's room is not an argument about "illusionism", it's an argument about physicalism. Eliminative materialism and illusionism are not synonyms, and while all eliminative materialists tend to be physicalists, physicalists do not need to be eliminative materialists.

Ryder Rude

@Amit it is really good tho. it makes u afraid of small noises

@Amit there are maybe some subtleties with the stuff i said like ACuriousMind says

the problem is that these philosophies are extremely close, so it becomes hard to take care of all the small differences

Amit

@ACuriousMind Thanks. Honestly I'm just going along for the Ryde 😉 and I won't try to learn anything seriously from a chat conversation with anyone. Nonetheless your care regarding accurate info is duly appreciated and commendable

@RyderRude It's fine as I said, it's not really a subject I'm very much into apart from a casual interest

Ryder Rude

i believe that dividing philosophies into discrete categories is supposed to serve as a guideline in discussions. Too much accuracy is a red herring

knowledge can't be divided by discrete boundaries. This is especially true in philosophy

@Amit ok :)

ACuriousMind

2:43 PM

@RyderRude Then why are you using those discrete categories? This is exactly the same thing as with mathematical technical terms: If you do not want to talk at the level of accuracy these terms imply, don't use them.

Ryder Rude

@ACuriousMind i was just about to give the example of Mathematics as something where discrete categories are perfectly well defined and accuracy is good

ACuriousMind

But using technical terms and then, when called out on their inaccurate use, claiming you didn't want to be accurate anyway is just irresponsible.

Ryder Rude

@ACuriousMind i was using the terms as guidelines...

it helps to have some terminology, or one would have to build discussions from scratch

but too much accuracy is a red herring

but also, I'm not experienced in philosophy, so maybe ur point holds to some extent

ACuriousMind

That's what someone says who does not want to invest the actual effort required to understand the nuances of a topic, but wants to sounds knowledgeable about the topic anyway. The difference in this case matters enormously, since a physicalist may believe that every mental state has a direct physical counterpart: For instance, they may think that there is a specific neural configuration of the brain that corresponds to "imagining the color red".

An eliminative materialist would claim that this mental state is too vague of a concept and does not actually exist; that most of the "mental states" we commonly talk about exist only as turns of phrase and have no direct reflection in any physical substrate.

Ryder Rude

yes, i was using eliminative materialism in exactly the sense u wrote

Suppose Mary's room were about physicalism, them a physically handicapped Mary would merely get to know the neural configuration counterpart, but not the experience

she would get to know the experience if the experience were the same thing as the neural countepart,

and that is eliminative materialism

@ACuriousMind also, this is not physicalism. Allowing for mental and physical counterparts that are correlated is not enough to make it physicalism. It could be dualism or dual aspect monism

ACuriousMind

2:54 PM

Well, the problem is that you've also incorrectly recounted the standard formulation of Mary's room, she is not physically incapable of perceiving color, the point is that she's trapped in a black-and-white room (hence the name, see Wiki).

Ryder Rude

@ACuriousMind i am aware...

I am taking liberties in framing the experiment

also, for it to be physicalism, one would have to say that all that exists is physical. This would depend on one's definition of physical

ACuriousMind

@RyderRude So "In Mary's room, Mary's brain physically doens't allow her to experience colors" is just a straight-up lie on your part, then?

Slereah

you can see colors stuck in a white room, just press on your eyes

ACuriousMind

If you are aware that Mary's room is something different, and you said this anyway being fully aware of that, that's just a lie

Slereah

it's not rocket science

Ryder Rude

2:58 PM

@ACuriousMind it's not a lie. I was trying to express the same idea!

It's an equivalent set up

the reason i framed it this way was to avoid loopholes and imprecision

i previously described an AI experiment, and u can see i wasn't able to be precise enough

ACuriousMind

@Slereah I guess we have to only put the disembodied head of Mary into the room then - can't rub your eyes without hands

Slereah

She's having a rough time

Phosphene

A phosphene is the phenomenon of seeing light without light entering the eye. The word phosphene comes from the Greek words phos (light) and phainein (to show). Phosphenes that are induced by movement or sound may be associated with optic neuritis. Phosphenes can be induced by mechanical, electrical, or magnetic stimulation of the retina or visual cortex, or by random firing of cells in the visual system. Phosphenes have also been reported by meditators (called nimitta), people who endure long periods without visual stimulation (the prisoner's cinema), or those who ingest psychedelic drugs....

It's like LSD for poor people

Ryder Rude

@Slereah u can see, my version doesn't have this loophole

ACuriousMind

@RyderRude Then the intellectually honest way to say this would be e.g. "This isn't exactly the original Mary's room, but I think the following setup captures the same essence with fewer issues:". But you didn't do that, you just told someone else that this was Mary's room. Which is simply factually wrong.

Slereah

"They were first reported in Western literature by Alcmaeon of Croton in the fifth century B.C. "

ACuriousMind

3:02 PM

I don't know how you expect anyone to take you seriously when the factual accuracy of your statements seems to fluctuate wildly depending on what you currently find most convenient

Slereah

"The eye obviously has fire within, for when the eye is struck fire flashes out."

Ryder Rude

@ACuriousMind i was trying to keep it short..... and i have expressed the exact same idea. The content is the same

Slereah

I guess that's why some people thought that vision started at the eye radiating outward

I guess it makes more sense in that context

Ryder Rude

problem variants are a thing!

ACuriousMind

@Slereah I wasn't aware LSD was expensive :P

Slereah

3:04 PM

@ACuriousMind This isn't the 60's anymore daddy-o

you ain't getting your tab for a dollar

Ryder Rude

why do u not want to focus on the content of the argument ,but on irrelevant details

ACuriousMind

@RyderRude I find it very revealing that you consider the literal truth of your statements to be an irrelevant detail, and I'm done with this conversation.

@Slereah true, I pay in euros

Slereah

Bette times

naturallyInconsistent

@ACuriousMind I'd vehemently disagree, but it is clear that you aren't paying anywhere near as much attention to that than would be warranted for the amount of disagreement I'm putting in. I am in no way arguing that "KE has the half, compared to rest mass energy" v.s. "rest mass energy has the 2, compared to KE". I am arguing that after SR is known, you can no longer argue that missing this factor of 2 or half is natural any more.

ACuriousMind

3:31 PM

@naturallyInconsistent I understand that you think we disagree, but I don't understand what exactly the disagreement is. Maybe I've expressed myself poorly. I agree that the relativity argument is a good way to argue that calling $\frac{1}{2}mv^2$ instead of $mv^2$ "kinetic energy" is "natural".

I still maintain, as natural as this is, it's still a convention, and an alternative would be to e.g. call $mv^2$ kinetic energy and $2m$ the rest mass energy (and perhaps introduce this factor of 2 in other places, too). This would be consistent, just unnecessarily complicated.

naturallyInconsistent

Ah, you think natural looking notation suggestions are still conventions. Maybe that is why

bolbteppa

In relativity, you have an action $S = - \alpha \int ds$, you fix $\alpha$ by the non-relativistic limit, which means invoking $T = \frac{1}{2} m v^2$. Alternatively it comes from generalizing $F = ma$ to a relativistic equation, but this also assumes $F = ma$ and $T = \frac{1}{2} mv^2$.

ACuriousMind

@naturallyInconsistent see - we disagree, but only about the meaning of the word "convention" ;)

naturallyInconsistent

lol

Ryder Rude

3:46 PM

plato.stanford.edu/entries/qualia-knowledge

see the above link. Apparently, the loophole Slereah gave is a known one

and "Mary is monochromatic from birth" is a known way to modify the experiment

Slereah

6:09 PM

Really as far as thought experiments about qualia go, I prefer the Bridgman version, where he mentions the case of kittens who do not open their eyes for weeks

and calls it a kitten crisis

Mr. Feynman

6:48 PM

@ACuriousMind the meaning of convention is conventional

Claudio

10:36 PM

when we write $\mathbf{L}^2 = \hbar^2\left[N(N+1)+a_k^{\dagger}a_k^{\dagger}a_ja_j \right]$

the number operator defined as $N = a_ja_j^{\dagger}$ is the sum of the three number of operators $N = N_x+N_y+N_z$, which yields $N|n_x,n_y,n_z\rangle = (n_x+n_y+n_z)|n_x,n_y,n_z\rangle = n_{\text{tot}}|n_x,n_y,n_z\rangle$

ok yeah it must be I forgot about repeated indices lol :P

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