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DIRAC1930
DIRAC1930
12:01 AM
And one of the renormalization conditions is that Coloumbs law approaches the asympotic form i.e. "two classical (i.e. infinitely
heavy) particles at rest at a large distance apart must interact in accordance with
Coulomb's law"
 
 
7 hours later…
DIRAC1930
DIRAC1930
6:32 AM
@lucabtz The physical renormalization scheme (on shell renormalization scheme) is physical and intermediate quantities can be interpreted physically. E.g. one of the renormalization conditions is that $e$ is the classical quantity.
I think worrying about renormalization scheme independence and different renormalization schemes for this is completely superfluous and just overcomplicates a simple conceptual task. Everything can be interpreted without any confusion using the OS renormalization scheme
 
naturallyInconsistent
naturallyInconsistent
@DIRAC1930 which is what they had been trying to tell you all along.
 
lucabtz
lucabtz
6:56 AM
@DIRAC1930 yeah in that scheme the mass is the physical mass, however it doesn't really matter much for vacuum polarization
 
DIRAC1930
DIRAC1930
7:14 AM
So all in all $e$ is a fundamental constant and is defined through Millikan and other similar low energy experiments. Coloumb's law gets modified at distances substantially close to the electron. And what the electron actually carries is a complicated question
 
123
123
7:43 AM
Hello Everyone...
 
DIRAC1930
DIRAC1930
Is what I wrote last correct?
 
Sir Crackpot
Sir Crackpot
Oh no renormalization took over the hbar
Wilson, WILSON WHY?!
 
ACuriousMind
ACuriousMind
8:06 AM
@SirCrackpot Wilson didn't invent renormalization :P
 
Slereah
Slereah
He was just a beachball
 
DIRAC1930
DIRAC1930
Wilson ruined renormalization
 
Slereah
Slereah
Renormalization was already a thing in the 19th century
Big divergence problem in the 19th century was gravitational forces for a uniform universe
 
DIRAC1930
DIRAC1930
Everything in QED can be done properly using the OS renorm scheme. Once non-abelian gauge theory came into the picture, everything went downhill in physics
 
Slereah
Slereah
bit of a hard read tho
 
DIRAC1930
DIRAC1930
8:17 AM
Things that ruined physics I. Path integral methods II. Renormalization group methods III. Non abelian gauge theory
 
ACuriousMind
ACuriousMind
you don't have to do these things if you don't like them, but you also don't have to be so judgemental about your dislike of perfectly mainstream topics
physics is not "ruined" just because it isn't quite what you expected it to be
 
Slereah
Slereah
You can just become a crank and think physics was best before the 20th century
ideal solution
Everything since has been a lie
 
DIRAC1930
DIRAC1930
Ngl I'm probably half of a crank already
 
Slereah
Slereah
I think part of the problem rly is that people try to explain complicated mathematical things in the framework of even more complicated physical theories
Few people try to teach gauge theory or renormalization outside the context of QFT which is already pretty tough
Even though they're not fundamentally QFT concepts
 
DIRAC1930
DIRAC1930
I think the issue is that calculations in QFT are so long that people have tried to coin short sentences to describe what is going on which actually make no sense
The same thing kind've happens in the popular science communication promotion of QM but since calculations are so much simpler in QM, you can just dispense what you've been told and find out for yourself
Weinberg Vol 1 is actually underrated
Even if it's just to know what it important to learn
 
Slereah
Slereah
8:47 AM
^Classical renormalization stuff
 
Ryder Rude
Ryder Rude
9:14 AM
Von Neumann says "u dont understand math. u get used to it"
 
Sir Crackpot
Sir Crackpot
@ACuriousMind Wilson gave it a meaning
 
ACuriousMind
ACuriousMind
@DIRAC1930 What do you mean, "underrated"? It's probably the most frequent recommendation for what to read about QFT after the usual intro texts.
 
Sir Crackpot
Sir Crackpot
@DIRAC1930 I mean, Weinberg is considered one of the best books of QFT. How is it underrated
 
ACuriousMind
ACuriousMind
Everyone I know hates the notation - but that means they've all read it
"it's in Weinberg" is the answer to a lot of questions where the standard texts only give handwaving (or no) arguments
 
Slereah
Slereah
Truly the bible of physics
No need to read it, you just know the answers are there
 
PM 2Ring
PM 2Ring
9:17 AM
Amen
 
Sir Crackpot
Sir Crackpot
@ACuriousMind ::Weinberg proceeds to give a more elaborate handwaving
 
Ryder Rude
Ryder Rude
Von Neumann is right here. if math computations were simply understandable, we cud do everything in english
 
PM 2Ring
PM 2Ring
@RyderRude Except he would've used proper grammar and spelling. ;)
> "Young man, in mathematics you don't understand things. You just get used to them."
 
Slereah
Slereah
I'm not sure I'd trust the intuition of a weirdo like von Neumann on math
 
PM 2Ring
PM 2Ring
116
Q: What are some interpretations of Von Neumann's quote?

NebulousRevealJohn Von Neumann once said to Felix Smith, "Young man, in mathematics you don't understand things. You just get used to them." This was a response to Smith's fear about the method of characteristics. Did he mean that with experience and practice, one obtains understanding?

soft-question
 
Slereah
Slereah
9:23 AM
He may have different experiences from most
 
PM 2Ring
PM 2Ring
IMHO, that quote tends to be used by people who are having difficulties understanding new concepts. ;)
 
Slereah
Slereah
It may also be the case that some people have no trouble manipulating purely abstract systems and do not feel the need to gain an understanding of it
 
PM 2Ring
PM 2Ring
I think we can understand maths stuff, to an extent. I mean, I'm pretty comfortable proving the Pythagoras theorem, a few different ways. OTOH, if you also want me to give an iron-clad development of the axioms of Euclidean geometry before I prove Pythagoras, I'll have to decline.
 
Slereah
Slereah
Also you could make the argument that understanding is getting used to things :p
 
PM 2Ring
PM 2Ring
@Slereah That too.
 
Slereah
Slereah
9:28 AM
People aren't born with a natural concept of how space works, as far as we know
 
Ryder Rude
Ryder Rude
"getting used to" here means being able to tolerate not understanding the thing
 
PM 2Ring
PM 2Ring
There are some mathematical techniques I use that seem reasonable, and I kind-of know how & why they work, but that I wouldn't be able to prove from scratch, at least, not without heavily leaning on someone else's proof.
 
Ryder Rude
Ryder Rude
i dont understand repeated index summation when there r too many repeated indices
 
PM 2Ring
PM 2Ring
OTOH, can we honestly say we really understand the deep truth underlying even elementary stuff like 2 + 3 = 5
 
naturallyInconsistent
naturallyInconsistent
@Slereah No, I think they have discovered plenty of evolutionarily selected-for structures in the brain adapted to the 3+1D spacetime we live in.
 
Slereah
Slereah
9:32 AM
@naturallyInconsistent But to what degree
How much understanding do you start with
And do you start with that understanding, or is your brain simply capable of getting it from what it gets
 
Ryder Rude
Ryder Rude
@PM2Ring this stuff is taken for granted
many notions are taken for granted in life. natural numbers are one of those notions @PM2Ring
 
PM 2Ring
PM 2Ring
Someone once quipped, "If someone really understood Fourier analysis, they'd be able to teleport".
 
naturallyInconsistent
naturallyInconsistent
@Slereah Technically, to extremely varied degrees. The goodness of biological structures at getting the details correct can be very wrong and yet still function tolerably to procreate...
 
Slereah
Slereah
 
Ryder Rude
Ryder Rude
i dont understand this group theorem : G/ker(f)~=Im(f)
even tho it's a short proof
i think, with mathematics, u mostly just understand some axioms
u dont understand theorems
 
Sir Crackpot
Sir Crackpot
9:40 AM
Although your last statement is true, well it refers to more structured stuff
The one you mention is a basic theorem of algebra
It is also known as first isomorphism theorem
 
Ryder Rude
Ryder Rude
i dont get it. y shud it be true
it's really simple but...
can u explain this pls
 
Sir Crackpot
Sir Crackpot
First of all, given a map $f:G\to H$, if you consider as a map $f:G\to\mathrm{Im}(f)$ it is obviously surjective, ok?
In our case we want to preserve group structures, so maps are group homomorphisms
 
Ryder Rude
Ryder Rude
yeah
 
Sir Crackpot
Sir Crackpot
Well, in general $f$ is not injective
 
PM 2Ring
PM 2Ring
@RyderRude I think it's the other way around. You accept the axioms, because they seem like they're a reasonable basis for the theory, but without truly understanding them. But once you get used to those axioms, you can logically prove theorems using those axioms. And you can feel confident that your proof is correct.
 
Ryder Rude
Ryder Rude
9:45 AM
@SirCrackpot yeah
 
Sir Crackpot
Sir Crackpot
The degree of failure of injectivity is given by the kernel, similarly to linear algebra
 
Ryder Rude
Ryder Rude
oh
 
Sir Crackpot
Sir Crackpot
So identifying all elements multiplied by an element of the kernel you make it injective (not a proof, just explaining the gist)
 
Ryder Rude
Ryder Rude
@SirCrackpot yeah... i think linear algebra has a special case of this where u hav dim(V)=dim(Im)+dim(ker)
 
Sir Crackpot
Sir Crackpot
Algebdaic groups do not have dimensions but the conceptual idea is the same
if $a\in\mathrm{Ker}f$, then $f(a)=e_H$, where $e_H$ is the identity on $H$
Given $g\in G$ with $f(ag)=f(a)\cdot f(g)=f(g)$
That explains why injectivity fails. $ag\neq g$ in general, unless the kernel is trivial
 
Ryder Rude
Ryder Rude
9:52 AM
yeah
so non trivial kernel means a bunch of group elements of count= cardinality(kernel) get mapped to the same thing
so it is a homomorphism where these batches of elements get mapped to the same thing
so a quotient
 
Sir Crackpot
Sir Crackpot
@RyderRude yes, but if you know take the quotient by the kernel (which is normal, so the quotient is a group), you identify all of these
 
Ryder Rude
Ryder Rude
thanks
 
Sir Crackpot
Sir Crackpot
It may be useful to note the following: you may wonder if this removes all non-injectivity. Well, suppose you have $g,h \in G$ such that $f(k)=f(g)=f(h)$. Then $k:=hg^{-1}$ is in the kernel: $$f(hg^{-1})=f(h)f(g^{-1})=f(h)[f(g)]^{-1}$$
$$=f(g)[f(g)]^{-1}=e_H$$
But $h=kg=(hg^{-1})g$ so they are identified if we proceed as above
It'a the same proof one uses for linear algebra (categories loom over us) just replacing addition with group product, opposite with group inverse and linear map with group homomorphism
 
Ryder Rude
Ryder Rude
yeah...i realised they wer the same proof cuz vector spaces are groups
but they r infinite dimensional, so a bit different
in linear algebra, we hav dim(Im)=dim(V)-dim(Ker), which is like quotienting V
 
Ryder Rude
Ryder Rude
10:34 AM
@PM2Ring i meant to say that one mostly just understands some definitions and some simple theorems in math. definitions are often natural
 
Sir Crackpot
Sir Crackpot
10:52 AM
For a path integral non continuous functions are more "relevant" than smooth functions, right?
 
Slereah
Slereah
Depends on the path integral you're considering
From what I remember non-continuous functions are of "measure" zero in the usual QM case
 
Sir Crackpot
Sir Crackpot
Euclidean (?)
Many times I see people taking a discrete system, say XY model, and taking the continuum limit assuming that the direction field changes smoothly
Then the partition function is computed as a path integral
I wonder if this is dangerous
I'll give an example. Consider the Hamiltonian $$H=-\sum\cos(\theta_i-\theta_j)$$ where the sum is performed on a 2D lattice over nearest neighbours
 
Ryder Rude
Ryder Rude
@SirCrackpot i think i read non differentiable functions r more relevant
than differentiable ones
 
Sir Crackpot
Sir Crackpot
If you assume that directions vary smoothly, you can expand the cosine and in the continuum limit $(\theta_i-\theta_j)^2\sim(\nabla\theta)^2$
Eventually your Hamiltonian is of the form $$H\sim\int(\nabla\theta)^2d^2x+\text{const}$$
And then $Z=\int D\theta\exp\{-\beta H\}$
This doubt possibly applies to typical field theoretic stuff too: the classical action is a functional of smooth (differentiable) path, how can one integrate over non differentiable ones rjhrhshdjd
 
Ryder Rude
Ryder Rude
from i what know, the version of the path integral written in terms of $\dot {x}$ and stuff isnt the actual path integral. it's just pretty notation
one actual path integral can be written using discretisation of the action and limit lattice to 0
but ACuriousMind mentioned a more accurate "actual path integral" using measure stuff
i will try to find it
 
bolbteppa
bolbteppa
11:05 AM
> In the philosophy of science and particularly the philosophy of physics, the philosophical sentiment which expresses the following perspective on the description of physics by mathematics might deserve to be called exceptional naturalism or similar:

> Since nature (reality) is exceptional in that it has existence, it is plausible that it is the exceptional structures among all mathematical structures – such as the exceptional examples in the classification of simple Lie groups, the exceptional Lie groups – that play a role in the mathematical description of nature, hence in physics and s…
 
Sir Crackpot
Sir Crackpot
@RyderRude Wiener measures but I can't perform field theory computations thinking of those
 
Ryder Rude
Ryder Rude
yeah, that's the one
but u r right. books r using the imposter path integral in partition function calculations
 
naturallyInconsistent
naturallyInconsistent
@SirCrackpot Part of the niceness of replacing the usual time-slicing path integral with integrals over the amplitudes of the Fourier coefficients of the deviations from E-L equation solutions is that all these non-differentiable stuff will automatically be handled by the high order Fourier coefficients. The results are the same.
 
Ryder Rude
Ryder Rude
@bolbteppa Lisi's theory also uses the exceptional stuff tho :P
 
bolbteppa
bolbteppa
Ridiculously :p
 
lucabtz
lucabtz
11:14 AM
@RyderRude you will trigger him like this
 
Ryder Rude
Ryder Rude
my philosophy is that the mathematics of physics comes before the rest of the mathematics in the order of being, but not in the order of explanation
 
lucabtz
lucabtz
But I also agree that that theory is ridiculous except for the name that is so cool
 
Ryder Rude
Ryder Rude
i havent read his theory. he was talking about the emergence of time in a video
is his a quantum theory?
it says it's a unified field theory with the E8 group. so i guess he quantises it later
 
Sir Crackpot
Sir Crackpot
@naturallyInconsistent oh in fact they do that later
 
lucabtz
lucabtz
@SirCrackpot are non continuous functions even integrated over?
 
ACuriousMind
ACuriousMind
11:29 AM
@SirCrackpot the standard Wiener measure is over continuous functions; it's the differentiable functions that have measure zero and that's probably the "relevance" statement you remember
 
Sir Crackpot
Sir Crackpot
11:50 AM
@lucabtz @ACuriousMind oh okay, the path are non differentiable but continuous
In such case it makes sense to assume that the field doesn't change much for close spins
 
lucabtz
lucabtz
@SirCrackpot after all is called the continuum limit for a reason
 
Sir Crackpot
Sir Crackpot
This now makes it equivalent to the following question: how do we get away with having an action being defined on differentiable paths?
I mean, is there some reason why at the physics level of rigor we are okay with these non differentiable paths taking over?
 
ACuriousMind
ACuriousMind
@SirCrackpot I answered this question here
 
Slereah
Slereah
12:07 PM
The classical action in the path integral is a LIE
This is explained as far back as the original Feynman and Hibbs book
 
Sir Crackpot
Sir Crackpot
@Slereah I'm checking section 7.3 now
 
Slereah
Slereah
If you want the physicist handwaving
It's because in the path integral you have $dx^2\approx dt$
 
Sir Crackpot
Sir Crackpot
12:24 PM
Did you mean here? @Slereah
@ACuriousMind Thanks, I was hoping to understand it handwawingly for the time being but I'll read your answer
 
DIRAC1930
DIRAC1930
Does anyone know what Dirac's views on non-abelian gauge theories were? Or even internal symmeteries?
 
Slereah
Slereah
That's the one yeh
If you want the very handwavy version
Try to imagine what it would look like if you actually try to force a quantum particle to have a path
Take a measurement of $x$ and $p$ at infinitesimal time intervals
But from uncertainty you're gonna have that the variation from the expected path being $$\Delta x \Delta p \approx \hbar$$
 
lucabtz
lucabtz
@ACuriousMind idk how but your answers are always very enlightening
 
Slereah
Slereah
If you consider the measurement of the momentum by looking at it as some averaged speed, that's $$\Delta x \frac{\Delta x}{\Delta t} \approx \hbar$$
 
ACuriousMind
ACuriousMind
@lucabtz thanks, I try :)
 
Slereah
Slereah
12:33 PM
So your "derivative" will locally look like $$\dot{x} \approx \frac{\Delta x}{\Delta t} \approx \frac{\hbar}{\Delta x}$$
In the infinitesimal limit it's "too big"
But in your action what you're actually using is $$\dot{x}^2 dt \approx \frac{\hbar^2}{\Delta x^2} \Delta t \approx \hbar$$
That is the turbo handwaved version :p
if you want to do it properly you need to learn stochastic theory I fear
You do have some probability of your path actually having a smooth curve of course, but that probability is of measure zero
That is the "differentiable curves are of measure zero" part
 
Sir Crackpot
Sir Crackpot
1:22 PM
@lucabtz There is this feature that I'll dub "ACMness" for which he write things in the way you need to know :P
Thanks @Slereah I needed this kind of hand angular momentum
 
Slereah
Slereah
It is fairly handwavy but you'd be surprised at how much you can do with it :p
 
Ryder Rude
Ryder Rude
i hav a vector field $v$ and a tensor field $T$. i take the diffeomorphism generated by $-v$ to be $\phi (p,t):M\to M$. Then the lie derivative of $T$ at point $p$ is $ \lim _{t\to 0}\frac{JT(\phi ^{-1} (p,t))-T(p)}{t}$. $J$ is the Jacobian of $\phi(p,t): M\to M$ at $p$
is this correct
sorry $J$ is a product of the Jacobian and the inverse Jacobian according to the index structure of $T$
 
nickbros123
nickbros123
1:36 PM
apart from callen's view point, is there any other interesting axiomatisations of classical thermodynamics?
 
Slereah
Slereah
What's Callen's view point
 
nickbros123
nickbros123
callen develops stuff with entropy as the very first thing
Its probably a popular way to approach, so forgive if i attribute it completely to Callen :p
 
Ryder Rude
Ryder Rude
for lie derivative visualisation, shud i think of diffeomorphisms generated by the negative of the vector field with respect to which we r taking the lie derivative?
suppose the diffeomorphism of $v$ is $\phi$ and that of $-v$ is $\phi ^{-1}$
 
nickbros123
nickbros123
I dont know anything about differential forms, but can in-exact differentials be considered as differential forms?
like heat, for eg
 
Ryder Rude
Ryder Rude
so under the diffeomorphism $\phi ^{-1}$, we get $f'(p)=f(\phi(p))$ like we wanted
so im imagining the river of $-v$ under which the contents of the manifold are flowing
 
Sir Crackpot
Sir Crackpot
2:09 PM
@nickbros123 yes, of course
Exact differential forms are only a subset of all differential forms
 
ACuriousMind
ACuriousMind
Jun 29, 2022 at 11:24, by ACuriousMind
@ManasDogra for "heat is a co-chain", see this excellent answer by joshphysics
 
lucabtz
lucabtz
@RyderRude it looks correct, the compact way to write what you are writing with $J$ is the via the pullback of tensor fields
 
Ryder Rude
Ryder Rude
yeah
@lucabtz what about the visualisation? should i think of the lie derivative wrt v in terms of the diffeomorphisms generates by -v?
 
lucabtz
lucabtz
2:24 PM
@RyderRude the visualization is very similar as the definition of the usual directional derivative
 
Ryder Rude
Ryder Rude
ooh just take the direction derivative but with a twist
 
lucabtz
lucabtz
you are essential constructing some incremental ratio
 
Ryder Rude
Ryder Rude
twist being the pullback
 
lucabtz
lucabtz
the point of the pullback is just that you cant compare tangent spaces at different points
 
Ryder Rude
Ryder Rude
yeah
it is co ordinate dependent that way
so i should think of the lie algebra commutator as this directional derivative of left invariant vector fields
earlier i was thinking of river flows
 
bolbteppa
bolbteppa
2:29 PM
@DIRAC1930 I would like to read anything he says on this stuff
 
bolbteppa
bolbteppa
2:44 PM
 
More Anonymous
More Anonymous
3:22 PM
@bolbteppa O_O
 
nickbros123
nickbros123
4:04 PM
@ACuriousMind ive saved it, for my consumption 2 yrs down the line
 
Bml
Bml
Hello everyone... Do you know how to make an animation of a torsion pendulum in Manim?
 
 
4 hours later…
fiziks
fiziks
8:06 PM
Hi Folks, I have a question about this question (physics.stackexchange.com/questions/295867/…) and some separate reading I am doing.
How does \nabla (\epsilon(\vec{x}) \vec{E}(\vec{x}))=0 become \nabla \vec{E} = \nabla( \log{\epsilon} ). Specifically, where does the log arise from? I think i am a little confused about the tensorial nature of \epsilon since I don't normally deal with inhomogeneous media
 
Sir Crackpot
Sir Crackpot
$$\nabla (\epsilon(\vec{x}) \vec{E}(\vec{x}))=0$$ become $$\nabla \vec{E} = \nabla( \log{\epsilon})$$
ftfy
 
Semiclassical
Semiclassical
@fiziks it's usually helpful to switch to index notation for that. so the $i$th component of $\nabla\cdot (\epsilon \vec{E})=0$ would be $\partial_i (\epsilon_{ij}E_j)$
tho actually that looks pretty weird to me. dimensionally it seems suspicious
 
fiziks
fiziks
I'm not talking about divergence, i am talking about the gradient @Semiclassical
 
Semiclassical
Semiclassical
hmm
 
ACuriousMind
ACuriousMind
the notation is very weird here, I can't really figure out what the $\nabla$ on both sides of the equation is supposed to be doing
 
Semiclassical
Semiclassical
8:16 PM
yeah, doing gradient on a vector field is confusing
 
fiziks
fiziks
Text I am reading also states grad(div(E)) = - grad(ln epsilon) . E
 
Semiclassical
Semiclassical
@fiziks is there an epsilon missing from the LHS of that?
 
ACuriousMind
ACuriousMind
so Maxwell's equation in vacuum I'd usually write as $\nabla \cdot \vec E = 0$
but what's $\nabla(\mathrm{ln}(\epsilon))$ supposed to be, it can't be a gradient
 
fiziks
fiziks
@Semiclassical I don't think so. If you consider div(E) = rho/epsilon I can kind of see it come from a product rule? but wasnt able to derive it exactly
@ACuriousMind Epsilon is a tensor field in this case
 
Semiclassical
Semiclassical
this does feel like it calls for index notation rather than vectors
 
ACuriousMind
ACuriousMind
8:19 PM
so this is, as Semiclassical said, $\partial_i (\epsilon_{ij} E_j) = 0$
 
fiziks
fiziks
Yes hashing it out in index is probably a good idea. I will take some notes and then come back and see if i can get some results
 
Semiclassical
Semiclassical
it's not especially obvious how $\ln \epsilon$ is supposed to be interpreted here
is $\ln$ just entry-wise here?
 
ACuriousMind
ACuriousMind
product rule yields $\partial_i \epsilon_{ij} E_j + \epsilon_{ij} \partial_i E_j = 0$, but you can't just divide by $\epsilon_{ij}$ to get something ln-like since it's summed over
 
Semiclassical
Semiclassical
@ACuriousMind yeah
 
ACuriousMind
ACuriousMind
that question is pretty baffling because the answer seems to perfectly know what the question is talking about, but after reading the answer I'm none the wiser :P
 
Semiclassical
Semiclassical
8:22 PM
if it's just a scalar then one can divide by $\epsilon$ to get $\partial_i (\ln \epsilon)E_j +\partial_i E_j=0$
so maybe this is one of those cases that only makes sense in the scalar case?
 
ACuriousMind
ACuriousMind
I mean if the tensor is diagonalizable you could go to the diagonal frame
 
Semiclassical
Semiclassical
yeah
 
ACuriousMind
ACuriousMind
still there's an $E_j$ in your equation behind the ln that doesn't fit with the equation in the question
 
user 85795
user 85795
9:15 PM
 
More Anonymous
More Anonymous
9:40 PM
Does this make sense?

Probability of a System: Probability of reaching a consensus for a paticular region of space x Probability of performing a negative entropy (in the classical mechanics sense) at that instant of time
(heyyy)
 
lucabtz
lucabtz
10:02 PM
@MoreAnonymous i dont know if it is only me but it looks like a bunch of words put together to me
kinda like when people talk about science in movies and the people writing the script have no idea about science
 
Sir Crackpot
Sir Crackpot
Yo, I'm back
I'm back now!
Not yet, damn cache
Now I am
Thanks, ACM
 
ACuriousMind
ACuriousMind
10:36 PM
@Mr.Feynman uh, what for?
are you malfunctioning?
3
 
Mr. Feynman
Mr. Feynman
@ACuriousMind I thought you did that magic with the cache button
 
ACuriousMind
ACuriousMind
nope
 
DIRAC1930
DIRAC1930
@bolbteppa youtube.com/watch?v=xJzrU38pGWc at 17:05 , Friedrich Hund asks Dirac about his views on modern developments in theoretical physics
2
 

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