« first day (4562 days earlier)      last day (377 days later) » 

1:10 AM
hm so why is category theory not something enjoyed in it of itself
i suppose i should read more about the intro theory to find out :P
^ what i mean by my first comment is that a category is a mathematical structure just as a group is or a vector space. and group theory and linear algebra are taught happily just as group theory or linear algebra. why does category theory need to be taught with respect to the knowledge of all other basic fields of mathematics
 
 
4 hours later…
5:33 AM
@DIRAC1930 how dare you scatter them
@ACuriousMind well, bookkeeping can be very important too
 
 
2 hours later…
7:28 AM
@SillyGoose Because it is meant to be a theory of mathematical structures. Sure, you can do category theory "abstractly" just with objects and arrows without ever giving them any meaning, but the main reason the notion of "objects and arrows" is interesting is because you are supposed to think about the objects as mathematical objects and the arrows as maps between them.
 
8:25 AM
> It has also become fashionable in the West to believe that all the laws of physics are already discovered. We assume we know all of them. While we have been pridefully crowing this tune, the Soviets have been steadily discovering new laws in secret, as well as new ways to circumvent the old laws.
Where is the secret soviet science
 
8:46 AM
> Don't believe everything they taught you in relativity; none of those guys ever had engineered a single general relativistic situation. & nothing they teach in GR is based on direct experiment.
 
9:21 AM
@DIRAC1930 by the way, that kind of scattering is what we call "mating" :P
 
 
3 hours later…
11:58 AM
@Slereah in Soviet Russia, universe obeys your laws
 
and time was invented by capitalists to sell more clocks :p
 
12:19 PM
@Slereah What about the standard GPS clock story
 
Paper is from the 80's, might be older idk
 
I wonder what consciousness is
Ok what is the evidence that nothing doesnt exist? I think maye nothing exists
Becuz if a conscious being dies, "nothing" starts existing from their perspective
So "nothing existing" is a consistent idea. It is a possibility
A dead conscious being really "thinks" nothing exists. But we would disagree with that. But how can anyone be objectively right in that debate?
So, "nothing exists" is really as good a viewpoint as "things exist". Becuz there's no way to settle the debate between the dead and the alive person
 
1:11 PM
consciousness was invented by capitalists to sell more philosophy books
 
2:07 PM
Consciousness main responsibility in humans is to create causality... you can see it in everything humans do, causality is the most important thing.
So thinking "nothing exists" has no meaning without a reference to "something used to exist" so it isn't real nothing anymore at all
Just a self induced concept trip lol
 
2:19 PM
@ACuriousMind hi can u give some link related to experiment and adventures related to cosmic rays?
 
2:44 PM
But to a dead person, nothing ever existed. There is no way to settle this debate with a dead person, which is y i think both viewpoints r equally valid @Amit
Also, statistically speaking, "nothing exists" is more true than "something exists"
This is becuz "something exists" is only true for an infinitesimally small portion of ur existence. "Nothing exists" is true infinitely more often. Like it is true before ur birth and after u r dead
 
3:08 PM
@Mr.Feynman wow a really late reply but i feel i am cursed to remember my dreams. i remember 2-3 dreams a night and i usually remember every detail. in fact i can remember even how aspects of the dream felt. for instance the other day in a nightmare i had my foot dragged through a green bean casserole and i remember exactly the feeling as if it happened to me. it is disorienting and very creepy and i do not like it. the dreams feel so real that they impact my moods XD
 
3:28 PM
I wish I could remember dreams like you @Relativisticcucumber
 
@Relativisticcucumber do you have something against beans or is the horror here that that ruined a perfectly good casserole? :D
 
3:49 PM
Today for the first time I corrected a wiki page :)
feeling proud
Probably didn't do a gr8 job
But atleast someone will take note of the importance of vectors
 
I think @Relativisticcucumber lives a dual life in dreamland hehe
 
 
1 hour later…
5:01 PM
@ACuriousMind he is Pythagorean
 
5:21 PM
is it accurate to say that the Maxwell equations permit wave-like solutions, but in order to actually find an electromagnetic wave, the electric (or magnetic) field must satisfy both the Maxwell equations and the wave equation?
my dreams only show their anxiety inducing faces around stressful times :P
 
@RyderRude "but to a dead person" - sorry i lost you there lol. There is no way to simulate non experience from within the experience
 
@SillyGoose I don't understand what you mean. The wave equations for the fields are what you get decoupling the Maxwell equations
But they are not equivalent to the Maxwell equations, in the sense that the latter contain more information. In fact, you derive the wave equations by taking the curl (i.e. differentiating, so you "lose information") of the curl equations
 
5:38 PM
0
Q: Question still attracting downvotes

Arunabh BhattacharyaIt is the same question which I have mentioned at On-topic question about wavelengths. It took several edits for the question to be reopened. Yet, my question has 6 downvotes and 1 upvote. What is the reason for the downvotes?

 
@Mr.Feynman I think that is an answer to a question equivalent to the question I asked
Exercise 1.1.xxx: Prove that Mr. Feynman's answer defined to be an operation on the equivalence class of Silly Goose's questions such that $f(x) = 1$ iff true and $= 0$ iff false is well-defined where $f$ is Mr. Feynman's answer operation and $x$ is an element of the equivalence class of Silly Goose's question.
lol i tried
to do a funny
is there a formal concept of a parameter dependent equivalent class?
 
5:55 PM
@SillyGoose isn't that just a function whose target is the quotient by the equiv. relation?
 
hmm
 
otherwise you'll have to explain what you mean by "parameter dependent", exactly
 
How do I prove this?^
 
well ideally one would be able to probe the relationship between equiv classes at different values of the parameter
 
@SillyGoose sure, but what does your parameter do? Does it change the equivalence relation or is the relation fixed?
do you have an example?
 
5:59 PM
@SillyGoose I'm not a math folk, you're gonna have to handwave to have me understand
 
i was thinking to be general so to allow the equiv relation to not be fixed (and also to allow the possibility that there may not exist a well defined "change in equiv class")
i can try to think of an example
well now so cosets are equivalence classes right
like in a quotient group
 
so i guess a simple example would be moving around a quotient group. you are fixing the equivalence relation and the equivalence class is parameterized by the representative
and so to define some sort of derivative w.r.t. to this parameter
which must exist i feel :P
 
I don't know what "moving around a quotient group" means
 
like starting at one representative of some equiv class
and then choosing another representative of potentially different equiv class
well hm i guess one should construct a situation where there is a clear parameterization of the group
 
6:04 PM
that sounds as if you're just describing a function into the original group
 
well let's say it is a function into the group
 
We have some group $G$ and apparently some normal subgroup $H$ and we're looking at ...what, exactly, in $G/H$?
 
i guess a function like you say $f: G \rightarrow G/H$ defined by $f(g) = g + H$ where $g \in G$ and $H$ is the normal subgroup
 
it is, it's just the normal projection onto the quotient that assigns to each element its equivalence class
 
oopsies hehe i was mixing up being a function with being injective again :P
so can there exist a derivative of $f$ which i guess in a simple case is the same for all representatives in the same coset
so $f'(k_1) = f'(k_2) = ...$ where $k_i$ is a representative of coset $k + H$
 
6:08 PM
what do you mean by "derivative"?
there is no notion of derivative on general groups
are $G$ and $H$ Lie groups?
 
hm so there is no notion of like "moving across a group" on an abstract group?
 
I don't know what "moving" across a group means or what it has to do with derivatives
 
well the extent of my knowledge has been reached :P
i guess i am thinking analytically when this is a purely group setting, so sure let us make them lie groups
i will be back ive got to do a korean speaking test :P
 
@SillyGoose in that case the derivative of $\pi : G \to G/H$ at the identity element is just the projection of algebras $\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$
 
Now shouldn’t there be groups that are also topologically spaces
In the most general sense
 
6:14 PM
@SillyGoose those are called topological groups
 
Groups that are topological spaces that are not manifolds
Does an analogue of a derivative not come along with a notion of distance?
 
@SillyGoose no, you need a smooth structure like on a manifold to define derivatives
also, general topologies don't even have a notion of distance, since there are plenty of non-metrizable topologies
 
Hm i see
So in the example you provided the derivative by construction sends representatives of the same coset in the quotient algebra to the same thing
 
6:30 PM
In the non-rel limit, does conservation of electron number and position number become a thing instead of conservation of position minus electron number?
 
 
1 hour later…
7:47 PM
What are your thoughts?
I'm just trying to get my intuition straight
If we state that we have 1 particle, are we by definition saying that it has to be governed by a free Hamiltonian because if it wasn't, we would have particle creation processes invalidating our original statement about having 1 particle
@ACuriousMind
 
In the Born oppenheimer approximation, the effective potential, whose graph looks like that of the harmonics oscillator but for increasing radiuses it has an extended side, which particles obtain this potential? The electrons? the particle that represents the relative motion of the nuclei?
 
8:38 PM
0
Q: Bug in MathJax rendering using Chrome?

ZeroTheHeroHere's a screenshot of part of a question as displayed by Chrome on my MacBook Air: You can see that some characters are rendered as boxes. Here's more or less the same bit as displayed on Safari: So it seems Chrome struggles to properly display some specialized fonts or characters. Is this n...

 
 
1 hour later…
9:49 PM
random thought but is there anything that really separates dog food from human food?
like i'm not ballsy enough to try it but if i ate dog food, is there any reason it wouldn't just taste like dried up meat?
 
you need to distinguish between dog food that is food that a wild dog or a wolf would eat and what we call dog food
 
yeah i'm thinking of domestic dog food
 
the latter is probably worse than the former, for both dogs and humans lol
but idk, probably someone did that experiment. sounds like a "Jackass" episode :)
 
@ACuriousMind Are we saying that the 1 in 1 out scattering amplitude approaches a value of 1 for low momentum?
Hence why QM works
 
@SirCumference a lot of it has more filler than meat. I'm no expert, but the dog food I tried tasted like cardboard. Presumably dog nutrition needs are different than human nutrition needs as well
I also tried a dog cookie once, which was much closer to normal cookies that people eat
Also probably depends on the exact dog food. The really cheap stuff is more filler and as you get more expensive, it's supposedly better and closer to real food (up to the point there are even some places with refrigerated dog food)
 
 
1 hour later…
11:06 PM
Hi All. I wonder if you guys talk about atoms behaviours here. I have a question related to compton scattering
 
11:24 PM
Just ask it. Someone will help if they can.
 
11:38 PM
I read the following: "Additionally, we call it Compton scattering if the electron is bound before the interaction and unbound after, if the binding energy is small compared to the photon energy".

I'm wondering why binding energy has to be small compared to photon energy. I know it doesn't say just less than, it means: "much, much less".

If you imagine photon energy is 300eV, electron's binding energy is 290eV, this would still result in ejection of the electron, but why would this be photoelectric and not compton ?
 

« first day (4562 days earlier)      last day (377 days later) »