« first day (2635 days earlier)      last day (2286 days later) » 
00:00 - 14:0014:00 - 00:00

12:00 AM
Hm, I guess the Gupta Bleuler quantization is basically a primitive version of BRST quantization?
 
@Slereah it is
 
I got sucked down a real rabbithole trying to make sense of how GB is related to BRST, there is something crazy there alright
 
does it have spooky ghosts
IIRC GB is basically just the gauge condition written in QFT?
Like $\Box A |\Psi \rangle = 0$
Which I think is the $Q$ operator in BRST
 
Vaguely summarizing something I have been meaning to read up on
Gupta-Bleuler imposes the Lorentz gauge condition on the space of states, and the LG condition is related to gauge invariance of the action of course, and the LG condition being a scalar means it satisfies the Klein-Gordon equation, so that the space of states is invariant under time evolution. In the non-Abelian case,
trying to impose a scalar condition on the space of states will not satisfy KG and so the state space is not invariant under time evolution so we need a new symmetry to impose on the space of states to keep it invariant, and then the BRST current is like a Fourier transform of the Gupta-Bleuler conditions in the abelian case, and I'm lost...
 
"Private conversation 1957"
Haag is a chatty fellow
Maybe one day I can put the hbar in a bibliography
 
12:27 AM
ooooof
something's not right there
 
 
1 hour later…
1:54 AM
0
Q: What explanations are there for this strong spike in the use of 'angular momentum' in the 1960s?

Emilio PisantyA recent comment called my attention to the Google Ngram for 'angular momentum', which shows a very strong and rather sharp peak in the usage of the phrase shortly after 1960, followed by a steady decline with a loss of ~60% of the original frequency in today's usage. I find this extremely...

↑ extremely curious
 
@EmilioPisanty String theory angular momentum e.g. the Venenziano amplitude in dual models and the flurry of research trying to tie this to particle physics until QCD in the 1970's overtook this research and string theory waned, to be revived from a superstring perspective?
 
@bolbteppa and that's 20%-30% of the global usage of the phrase?
that'd be a lot of research in string theory
 
 
The Regge theory of complex angular momentum arose then and motivated it en.wikipedia.org/wiki/Regge_theory
 
My book says red light is deviated the least
But in previous section I read,it says” when minimum deviation happens (for i=e) light is paralllel to base
but in every photo of dispersion of light in prism red light is not parallel to base :/
 
2:06 AM
A similar strain would be that this Regge theory motivated (or was motivated by?) the research into axiomatic QFT from properties of the S matrix, with complex angular momentum underlying it in a big way apparently, too hard to make full sense of right now :p
 
In above picture (from Wikipedia) violet is parallel to base 🤔
 
@Fawad that is false in general
the angle between the red ray and the base depends on the initial incidence angle
that section of your book is likely describing a more restrictive geometry
the good thing is that you can just get a prism and check experimentally
 
@EmilioPisanty i will try today. I have prism
Should I do in sunlight or use white torch 🔦?
 
I was just pitching my idea
in this pic
I will detail what happened next in later
was talking in front of +300 people give or take
coding a product with some (supernaturally awesome) senior programmer/analyst
I think the person i am coding with right now is a genius
 
vzn
nice man thx for sharing :)
 
2:17 AM
@vzn no worries anytime hehehe
 
2:27 AM
@Cows awesome dude making the h bar proud :-)
 
 
3 hours later…
5:18 AM
Is there any difference between $\phi$ and $\{\phi\}$?
where $\phi$ stands for null set
 
Anonymous
@Yashas There's a symbolic difference, isn't it? The latter is a set actually having an element $\phi$ (which doesn't make sense till you define $\phi$).
 
Anonymous
The former is the good old null set.
 
What's the cardinality of a null set?
 
Anonymous
0
 
$|\phi| = 0$? and $|\{\phi\}|$ is 1?
 
Anonymous
5:22 AM
@Yashas Right. But $\{\phi\}$ doesn't really have any meaning unless you define $\phi$ as something.
 
Anonymous
The equivalent notation for $\phi$ is $\{\}$ if you insist.
 
Anonymous
Not $\{\phi\}$ (that one doesn't mean anything in general)
 
Anonymous
Some people call it the set containing the empty set. But that is again a definition. You can define it anyway you want.
 
"The uncountability of a set is closely related to its cardinal number: a set is uncountable if its cardinal number is larger than that of the set of all natural numbers."
the cardinal number of the set of all natural numbers is infinity, right?
What does it mean to say that the cardinal number of a set is larger than the cardinal number of the set of all natural numbers?
 
Anonymous
There are different kinds of infinities. That's a common misconception. For example there exists no bijection from $\Bbb N$ to $\Bbb R$.
 
5:28 AM
If the number of elements in a set is logically less than the number of elements in the set of all natural numbers, then it is countable?
 
Anonymous
@Yashas Yeah, roughly that's it.
 
This definition is freaking me out. Why does the definition specifically mention the set of natural numbers?
because countable numbers is the set of natural numbers?
 
Anonymous
@Yashas Because you count using natural numbers. :)
 
Anonymous
Right.
 
Anonymous
If you're interested in this stuff read the first few chapters of one two three infinity by george gamow. That's an excellent book and provides good intuition.
 
5:33 AM
the set of all even numbers is countable because it is a subset of the set of all natural numbers?
 
Anonymous
Yup
 
Anonymous
Or rather:
 
Anonymous
You can define a bijection from one to the other
 
Anonymous
$y=2x$ ($x\in \Bbb N$ and $y\in M$ where $M$ is the set of even natural numbers).
 
"X is nonempty and for every ω-sequence of elements of X, there exist at least one element of X not included in it. That is, X is nonempty and there is no surjective function from the natural numbers to X."
where X is an uncountable set
What does the first sentence mean?
Why can't I have an infinite $\omega$ sequence of elements of X? Basically, all the elements of X written in order.
"1. The cardinality of X is neither finite nor equal to |N|
2. The set X has cardinality strictly greater than |N|"
doesn't 2 imply 1?
Wikipedia states them as two separate points
 
Anonymous
5:49 AM
@Yashas You can do that only if there exists a notion of "order" in a set.
 
Anonymous
Do you know the ZFC axioms?
 
In any order.
I need to write down a sequence. I write down each element of the set to form the sequence.
@Blue hearing it for the first time
 
Anonymous
@Yashas That's simply not true. Consider a vector space or something
 
Anonymous
Check out the wiki article on well ordered sets maybe
 
Anonymous
I need to go now
 
5:54 AM
Everyone in the world should read that
 
6:04 AM
is it a good idea to apply for open positions in the same institute but in different fields?
 
Anonymous
6:47 AM
@Yashas Oh, I didn't notice that question before. Yes, 2 implies 1 and 1 implies 2 as well. They already wrote that any of the given conditions are sufficient.
 
Anonymous
But the proof of that is non-trivial
 
7:08 AM
but the cosmology group led by the professor with open position seems to be engaged with research regarding applications of fundamental thories.
too applicational to feel fundamental
 
Anonymous
@ACuriousMind When you come around could you please unfreeze this room: chat.stackexchange.com/rooms/63400/the-no-normie-zone
 
Anonymous
@BalarkaSen You there? I need some book recommendation
 
Go ahead
 
Anonymous
Scroll down to the First Year Second Semester here: jaduniv.edu.in/upload_files/course_file/1408616138-1.pdf
 
Anonymous
Under Mathematics IIIG have a look at the syllabus
 
Anonymous
7:20 AM
I'm not sure I know any good book which contains those stuff
 
Anonymous
And the professor is teaching everything randomly
 
Anonymous
Do you know any good book which covers all the topics in the probability portion ?
 
Anonymous
Mainly the last dozen topics
 
I'm afraid I don't
 
Anonymous
Ah :/
 
7:23 AM
I don't know anything about probability
 
Anonymous
It's okay. I'll probably pester the professor till he suggests me some books. :P Or else I'll just download some pdfs and refer those. Just that having things in one good book is always helpful.
 
7:59 AM
21
Q: Can we prove reality?

martinverdejoI heard someone make an assertion that 'We cannot really prove that there is reality.'. 'Reality' here would mean the universe and everything in it. You could look at an apple and think its an apple but it could really be something else, and we cannot really prove its "appleness". My initial an...

 
8:23 AM
At first glance, I considered your ''reality" refers to ''real numbers" in contrast to ''imaginary numbers".
 
9:08 AM
Hello my duders
@CaptainBohemian I have a book on the topic of doing physics without numbers
It is possible but very unpleasant
 
@Slereah are you talking to me? I don't know much about number theory. I originally considered number theory has no application in physics until the recent, when I read from quanta magazine number theory has application in scatter amplitude.
 
I don't mean number theory
I just mean no numbers at all
 
I am not used to physics with numbers probably I have long touched physics formulated by symbols--I mean algebra.
 
Anonymous
9:29 AM
@Slereah What's the name?
 
Anonymous
BTW I'm not sure what physics with no numbers means
 
Anonymous
9:59 AM
@BalarkaSen Could you explain me the index of vector field thingy? I can't understand what's happening on page 5 of Nash Sen
 
Index = how much swirly is there
 
Anonymous
What is "swirly" ? :P
 
Anonymous
Lemme upload the picture for reference:
 
the loop-de-loops
its ok i have the book
what do you not understand?
 
Anonymous
10:01 AM
 
Anonymous
I'm elaborating my doubt:
 
Otherwise I have no idea how one can do physics without numbers. what does numbers mean, does it include algebraic symbols?
 
@Blue "Science without numbers"
Basically it uses Hilbert geometry axioms sort of laws
instead of numbers and functions
 
so... no coordinate geometry is involved, just axiomatics?
 
Anonymous
$f$ is a function from the disc $B^2$ into itself. Let $\mathbf{x}$ is an element of the boundary of $B^2$ and it is mapped to $f(\mathbf{x})$.
 
Anonymous
10:04 AM
So far so good
 
Anonymous
Then we construct a vector $v(\mathbf{x})=\mathbf{x}-f(\mathbf{x})$
 
Anonymous
OK
 
Anonymous
Now if we use this definition for all points on the boundary,
 
Anonymous
we get a vector field $v(\mathbf{x})$
 
Anonymous
Returning to point $\mathbf{x}$ we rotate $\mathbf{x}$ about the centre of $B^2$ by $2\pi$ once. While doing that, $v(\mathbf{x})$ also rotates through $2\pi$ once <----- what does this mean?
 
Anonymous
10:09 AM
Do they mean the vectors $v(\mathbf{x})$ have to cover all possible directions before we return to the same $\mathbf{x}$?
 
Anonymous
Is that called index $1$?
 
This is murkily written. $v$ is a vector field along the boundary of the ball $B$. Now look at the normalized map $f : \partial B \to S^1$ given by $f(x) = v(x)/\|x\|$, mapping to the unit circle in $\Bbb R^2$. OK?
 
Anonymous
@BalarkaSen Okaies
 
Anonymous
But I'm not sure I can relate this definition of index to the one given on Wiki: "The index of a vector field is an integer that helps to describe the behaviour of a vector field around an isolated zero (i.e., an isolated singularity of the field). In the plane, the index takes the value -1 at a saddle singularity but +1 at a source or sink singularity." Where's the source or sink singularity here?!
 
They're saying $f$ is a bijection.
In fact a homeomorphism but whatever.
@Blue The zero of the vector field is in the interior of the disk.
$v$ naturally extends to a vector field on the full disk, by defining $v(x) = x - f(x)$ for any point $x \in B$
The purported zero (which you haven't found yet in the book) lies inside $B$
The map $f$ detects how much $v$ swirls about that zero (in fact, it swirls around once)
 
Anonymous
10:15 AM
Wait a bit. Zero of vector field just means the point $\mathbf{z}$ at which $v(\mathbf{z})=\mathbf{z}-f(\mathbf{z})=0$ ?
 
Yes.
 
Anonymous
And that $\mathbf{z}$ need not necessarily be on the boundary of $B^2$. But how are we certain that it lies inside $B^2$ ?
 
Anonymous
Why not outside ? (This might sound like an extremely silly question)
 
Let me finish dinner and I'll tell you the story in more detail. The book doesn't do it justice.
not dinner dammit lunch
 
Anonymous
Sure, have your dinner ;)
 
10:19 AM
My sleep schedule is so messed I can't be arsed to remember which is dinner and which is lunch
 
Anonymous
Sounds like me
 
Anonymous
:P
 
Ok, I'm done
 
Anonymous
I'm all ears, go on
 
@Blue This books is proving the Brouwer fixed point theorem. That says if you have a map $F : B^2 \to B^2$ from the closed 2-disk to itself, $F$ has to have a fixed point.
 
Anonymous
10:21 AM
Hope you didn't gobble down your food for me
 
I was almost finished. I don't eat a lot
live on biscuits, as a wise man once said
(nobody said that)
 
Anonymous
@BalarkaSen Yes, agreed
 
@Blue So outright "have a fixed point outside the disk" makes no sense, because the range of $F$ is the disk itself.
 
Anonymous
@BalarkaSen Yeah, so either the zero is in/on the disc or there's none at all. Right?
 
That is correct.
We'll in fact prove the third option is not possible
 
Anonymous
10:24 AM
Okay..interesting
 
Now, you can assume $F$ has no fixed points on the boundary, by cheating logic: if there was one on the boundary, you'd be done - you have found a fixed point of $F$!
So might as well assume $F$ has no fixed points on the boundary
i.e., for all $x \in \partial B$, $F(x) \neq x$
 
Anonymous
@BalarkaSen So we are just dividing it into two cases. (1) On the boundary (2) Inside. (1) is trivially true, so we look into (2), yes?
 
Anonymous
I mean if (1) was true, then it is trivial
 
(1) is trivially true? No. There are lots of self-maps of the disk which do not have fixed points on the boundary (consider $F$ to be shrinking radially by a factor of $\alpha > 1$ -- then $f(B) \subset B$, the image of the whole disks fits inside itself)
Right, that.
 
Anonymous
Hmm, so are we going to come back to case (1) later on? i.e prove that if there is no fixed point inside then there must be one on the boundary?
 
10:30 AM
No, look, there are two cases. (a) "There are no fixed points on the boundary" and (b) "There is a fixed point on the boundary"
 
Anonymous
@BalarkaSen Mmmhmm
 
We will show if (a) is true, $F$ has a fixed point on the interior. If (a) is false, then (b) is true, so (tautologically) $F$ has a fixed point on the boundary. In either case we get $F$ has a fixed point. QED
 
Anonymous
@BalarkaSen Oh, okhay. Makes sense now
 
Anonymous
Go on
 
Well, if $F$ has no fixed points on the interior, $v(x) = x - F(x)$ is a nonzero vector field along $\partial B$
Consider the map $g : \partial B \to S^1$ defined by $g(x) = v(x)/\|v(x)\|$ (this is well defined because $v$ is nonzero on $\partial B$).
 
Anonymous
10:37 AM
I forget the meaning of non-zero vector field
 
$v(x) \neq 0$ for all $x \in \partial B$
The vector is never the zero vector
 
Anonymous
You mean for no point on the boundary of the ball, the vector $\mathbf{x}-f(\mathbf{x})$ vanishes
 
Anonymous
Right?
 
Yes.
 
Anonymous
I'm not sure how that follows from that there is no $\mathbf{z}$ inside the ball, where $\mathbf{z}-f(\mathbf{z})=0$.
 
10:40 AM
I did not claim that anywhere
I just said $F$ has no fixed points on $\partial B$
That's hypothesis (a)
 
Anonymous
5 mins ago, by Balarka Sen
Well, if $F$ has no fixed points on the interior, $v(x) = x - F(x)$ is a nonzero vector field along $\partial B$
 
Anonymous
"on the interior"
 
Anonymous
Am I missing something
 
Oh that's a typo I did not notice.
Thanks, I meant the "boundary" not the "interior"
 
Anonymous
SO, you basically wanted to say that if there is no fixed point on boundary there is no $\mathbf{x}$ on the boundary, such that $\mathbf{x}-f(\mathbf{x})=0$.
 
10:42 AM
By definition, yes
 
Anonymous
I get till there
 
Anonymous
Okay
 
The consequence of it all is that the map $g$ defined above ^^^ is well-defined
 
Anonymous
7 mins ago, by Balarka Sen
Consider the map $g : \partial B \to S^1$ defined by $g(x) = v(x)/\|v(x)\|$ (this is well defined because $v$ is nonzero on $\partial B$).
 
Anonymous
Right, yes
 
Anonymous
10:43 AM
Got it
 
Their claim that $v$ has index $1$ is equivalent to saying $f$ has "winding number $1$"
I wonder if I should tell you about that
It'd depart from the content of the book but it's the "right" proof
 
Anonymous
I heard that term in complex analysis I think
 
That's it
 
Anonymous
Looping around singularities and poles during contour integration
 
Anonymous
Okay, but I don't know about it in this context
 
10:48 AM
Consider simply the map $v : \partial B \to \Bbb R^2 - \{0\}$ (it has that range because $v$ is nowhere zero on the boundary)
 
Anonymous
@BalarkaSen Mhm
 
This is mapping a circle in the plane without the origin. So image of $v$ is a loop in $\Bbb R^2 - \{0\}$ going around the origin
 
Anonymous
@BalarkaSen Right, yes
 
Therefore you could look at $\oint_v dz/z$ after identifying $\Bbb R^2$ with $\Bbb C$
This number = 1
 
Anonymous
Wait a min
 
10:50 AM
That's what it all means
 
Anonymous
I've got a doubt :P
 
Go on
 
Anonymous
Say $f$ maps all the boundary elements $\mathbf{x}$ to the point $(\frac{1}{2}, \frac{1}{2})$ inside the unit circle. Then how is the image of $\mathbf{x}$ under transformation $f$, i.e. $f(\mathbf{x})$ going to form a closed loop around $(0,0)$? Or did you mean something else by "image of $v$" ?
 
It is possible that all of the boundary gets sent to a point. The image of $v$ is still a well-defined thing to integrate $dz/z$ on
The integral would be $0$ of course :)
So that scenario cannot happen
That's precisely what "winding number 1" means. The image of $v$ will go around the origin precisely once and not more, neither less
 
Anonymous
@BalarkaSen I don't understand why that scenario can't happen
 
10:55 AM
Oh and I meant $1/(2\pi i) \int_v dz/z$
@Blue The integral in our case is $1$
In your case it's $0$
Contradiction
 
Anonymous
I mean what is the problem with having winding number $0$? Did we initially assume that winding number of $v$ is $1$, during the proof, somewhere?
 
No I mean that's precisely what Nash-Sen's claim implies.
$v$ being index 1 implies the winding number of the image is $1$
I'm just reinterpreting the text in more precise language
The intuition is that if you move $x \in \partial B^2$ around the boundary circle back to itself, the vector $x - f(x)$ twists around $2\pi$, like the book says.
After you translate $x - f(x)$ to be a vector centered at the origin.
 
Anonymous
In the degenerate case, when all the points on boundary get sent to a point, even then when you return to the same $x$ after moving around origin by $2\pi$, even then $v(x)$ rotates by $2\pi$ (according the sense in which they wrote). They claim that for all such situations the index is $1$, when in fact it could be either $1$ or $0$ (albeit in the extreme case)
 
Anonymous
So it's NS being hand-wavy here
 
Anonymous
I like the winding number concept more, I guess
 
11:01 AM
No, you're wrong.... if $f$ sends all of $\partial B$ to a point, $v$ is the constant vector field along $\partial B$.
It means for all $x \in \partial B$, $f(x) = p$ for a specific point $p \in B$. Then $v(x) = x - p$
Therefore $v$ is the vector starting at $x \in \partial B$ pointing in a specific direction for all $x$
Maybe I am misunderstanding your thing
@Blue Wait, did you mean $F$ sends all of $\partial B$ to a single point in $B$?
There are way too many maps. $F, f$, etc
 
Anonymous
I think they are using "rotates" in a different sense than what I do. Let me draw a picture to illustrate my point
 
@Blue Done.
 
Anonymous
@ACuriousMind Thank you very much
 
@ACuriousMind Up early I see
:P
 
11:08 AM
@BalarkaSen Well, I have to be up early on Mondays, so if I used the weekend to return to my old sleep schedule, Mondays would be even less fun than usual :P
 
I had to wake up at 8 in the morning today so I sympathize
 
Anonymous
 
@Blue Ah, ok, so you did mean $F$.
Well, $v$ is a perfectly fine loop that goes around the origin then
 
Anonymous
@BalarkaSen Yeah, so that has winding number $0$ and index $0$, right?
 
No, 1!!!
Translate each of those vectors to have it's end at the origin, and trace out the arrow end of the vectors
You'll get a circular path around the origin in $\Bbb R^2 - 0$
That has winding number 1 m'dude
 
11:13 AM
@Blue You should post something in it else it'll freeze again in a few hours
 
Anonymous
Oh, let's continue our discussion in that room, @BalarkaSen
 
Anonymous

  the no-normie zone

no normies allowed
 
@ACuriousMind Done
 
@BalarkaSen Very good :)
 
2
Q: Why did the universe continue to expand from inertia after inflation stopped, if galaxies have no motion but space between them expands?

parkerI’ve read many answers on this site about space expansion and I will write from what I understand so please correct me if I’m wrong. The space between galaxies expands but the galaxies themselves are stationary and have no real motion due to the expansion. After inflation stopped, the u...

universe expansion is one of those strange things that we cannot comprehend it using newtonian like intuition
I think our current explanation for it is dark energy, but otherwise I am not sure why metrics can have time dependence
Since energy is not conserved at cosmological scales, I am not sure if one can think of some hamiltonian that drives the expansion
https://physics.stackexchange.com/questions/146739/what-caused-the-expansion-of-the-universe-to-slow-down-after-the-inflationary-ep
hmm...
 
11:27 AM
Are you interested in answering a complex analysis question over in the math room? @ACuriousMind :P
 
Halo. Can we make a material like this: en.wikipedia.org/wiki/Homeomorphism#/media/…
2
 
@skullpatrol Have I ever given the impression of liking analysis? :P
 
When you stretch it its mass increases at a rate which keeps it density constant.
 
user228700
Hello, everyone :-)
 
Or when you condense it loses mass
 
11:29 AM
Hey there @Kaumudi.H
 
user228700
@ACuriousMind How goes it?
 
Pretty well :) How about you?
 
user228700
Same here :-) I'm back home for a week again!
 
user228700
How's the new job?
 
@Kaumudi.H Going well, so far I love it
 
user228700
11:31 AM
Eyy, that's wonderful! :-)
 
user228700
Have you had a chance to make chocolate coffee yet?
 
@Kaumudi.H I've had chocolate tea before...
 
user228700
Hello! And did you like it?
 
@Mithrandir24601 Chocolate tea? Hmmm ...
 
user228700
@JohnR: Yello! :-) How's it going?
 
11:33 AM
@Kaumudi.H I actually bought a small tin of chocolate powder to take with me to work next week, so it's on the list! I've not really had a chance so far because the machine at work doesn't do chocolate, just coffee and tea.
 
@Kaumudi.H life goes on. Nothing of any great importance is happening at the moment.
 
user228700
Ah :-/ Well, as always, let me know if you ever do get a chance to try it.
 
user228700
@JohnRennie Hmm. You back at Wookey Hole, or was that scheduled for Feb.?
 
We're all squids eating dough in a polyethylene bag
2
 
user228700
@BalarkaSen New meme?
 
11:34 AM
No time to get chocolate coffee
 
@Kaumudi.H I'm off Friday after next.
 
@Kaumudi JR gets me, I think
 
user228700
@JohnRennie Cool :-)
 
user228700
 
@BalarkaSen you've been listening to too many Captain Beefheart albums! :-)
 
11:35 AM
On the contrary, just one!!
 
That's too many :-)
 
user228700
 
user228700
What.
 
user228700
OK, I have accidentally listened to the whole clip, and I am now afraid that my brain is fried.
 
11:37 AM
@Kaumudi.H It's nice, yeah :) It's a tea blended by a shop in Bristol that's kind of designed to go with chocolate, so they recommend putting some cocoa powder in as well :)
 
user228700
That's nice :-) I didn't think the combination would be very good, but you never know!
 
@JohnRennie It's combines chocolate and tea! The only way left to improve it (apart from bacon, obviously) is with marshmallows :)
@Kaumudi.H it wouldn't be so good for virtually all other teas I've got though :P
 
user228700
Hmm, right :-)
 
@Mithrandir24601 tea with chocolate and marshmallow. A work of emetic genius.
 
Isn't that just cocoa?
Dressed up :P
 
11:42 AM
@JohnRennie you can also add vanilla essence and honey (it sounds like an odd combination, I know, but for this specific tea, it's quite good)
The tea is brewed at about 4 times the normal strength as well
 
Whoa! Heavy on the caffeine?
 
Anonymous
I don't know why people like marshmallow....it tastes sooo bland!
 
Anonymous
Or have I never had good marshmallow
 
Go full Indian and try marshmallow with curry
 
lol
sweet & spicy
 
Anonymous
11:47 AM
As for tea I simply love masala tea (although very few people can make it well)
 
I have found cold tea to be quite refreshing.
 
Anonymous
Yeah, cold tea/coffee are good too.
 
cold but not iced?
 
Anonymous
I hate the Starbucks cold coffee though
 
Anonymous
They just ruin it
 
Anonymous
11:49 AM
I can make better ones at home
 
Anonymous
@skullpatrol Well, iced too, sometimes
 
Anonymous
Cold by default means iced here
 
Right I think it's known more popularly as iced tea
Anyway, must get back to eating dough inside my polyethylene bag
 
Also known as "Ice tea"
 
@Blue masala chai is also very nice :)
(yes, I know chai means tea, I'm just agreeing with you)
 
Anonymous
11:54 AM
Yep...do you get it in the uk?
 
@Blue yeah, I've got a tin of it. I've also got recipes (although I'm not sure if 'recipe' is the right word :P) for how to make it :)
 
Anonymous
Ah, very nice
 
Perhaps, "instructions"?
:P
 
Also, the above comment about chai is as a result of people in the UK often calling it 'chai tea' instead of 'masala chai' or 'spiced tea' (or some variant thereof) making me facepalm a fair bit
 
We all know British people are expert in the art of drinking tea
user image
4
 
11:57 AM
@skullpatrol yeah, sounds about right :P
 
Anonymous
Lol..."chai tea" ;D
 
Anonymous
People would give you strange looks if you say that here
 
@BalarkaSen Is that Asterix and Obelix?
 
Yes of course :)
 
IT IS!! :D
@Blue I seem to be the only person who does that here :P
 
12:00 PM
Redundancy is often perfectly acceptable, as in "last and final call" :P
 
12:34 PM
Is the mass gap issue just the decomposition of the Hilbert space as $$\mathcal{H} = \{ 1 \} \oplus \mathcal H_{p^2 = m} \oplus \mathcal H_{p^2 \geq M}$$
 
What "mass gap issue"?
 
0
Q: Does $\Theta ^\mu _\mu =0$ only imply on-shell conformal invariance?

pppqqqThis is in part a reiteration of this old phys.SE question, which did not receive much attention. It is usually stated (see e.g. Ref. 1, §4.2.2) that a traceless energy momentum tensor $\Theta {^\mu} {^\nu}$ implies the invariance of an action $S[\Phi]$ under conformal transformations (here and...

and then I thought I am seeing hodge theaters...
 
12:52 PM
@ACuriousMind The presence of a mass gap in interacting theories
I guess that should be more $$\mathcal{H} = \mathbb C \oplus \mathcal H_{p^2 = m} \oplus \mathcal H_{p^2 \geq M}$$
It's hard to follow Haag because he uses a lot of vague arguments and then say "Oh by the way that argument is total garbage"
I don't know what's real anymore!
 
@Slereah The mass gap is the claim that there isn't a continuum of e.g. glueball states with arbitrarily low mass.
I'm not sure what you intend to do with your formula there
 
Ah alright
So mostly unrelated
 
1:07 PM
While studying dynamics I found out that instantaneous acceleration is perpendicular to the velocity vector at that instant,plz explain how?
 
@ffahim That's not true in general, you'll have to be more specific for us to explain anything.
 
I am thinking if there is any conservation law not corresponding to symmetry?
 
@ACuriousMind thanks! exactly what I was looking for
 
1:29 PM
damn. basically I don't understand a damn from the answers.
 
00:00 - 14:0014:00 - 00:00

« first day (2635 days earlier)      last day (2286 days later) »