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John Rennie
John Rennie
5:37 AM
@Qmechanic under World Trade Organisation tariff rules European weather systems will be required to drop 30% of their rain on France before they cross the channel.
@akhilkrishnan I can't help I'm afraid. That's an area of physics that I don't know anything about.
 
skillpatrol
skillpatrol
5:57 AM
@JohnRennie i'm sure the french wouldn't have it any other way :-)
perhaps, germany will want a share?
 
Akash. B
Akash. B
hi
 
skillpatrol
skillpatrol
hi pal
 
Akash. B
Akash. B
got any interesting science article for me?
 
skillpatrol
skillpatrol
what are you interested in?
 
Akash. B
Akash. B
quantum mechanics
 
skillpatrol
skillpatrol
6:03 AM
have you checked out wikipedia?
 
Akash. B
Akash. B
yeah almost
 
skillpatrol
skillpatrol
i mean the references at the bottom are usually pretty good...
 
Akash. B
Akash. B
okay
 
John Rennie
John Rennie
@Akash.B the articles on the Quanta Magazine web site are very good.
 
Akash. B
Akash. B
@JohnRennie good
 
skillpatrol
skillpatrol
6:07 AM
very
 
 
1 hour later…
gaurang agarwal
gaurang agarwal
7:19 AM
Hello John Rennie remember about my surface tension problem
The Summery of the problem is:-
 
John Rennie
John Rennie
@gaurangagarwal hi
 
gaurang agarwal
gaurang agarwal
Remember for proving Young's equation we consider the boundary line or triple line(whatever it is) to be in equilibrium.
What exactly is this boundary line
Hi John Rennie
 
John Rennie
John Rennie
The boundary line is the edge of the drop i.e. the line along which all three phases, soil, liquid and air, all meet.
 
gaurang agarwal
gaurang agarwal
So the edge must be comprised of liquid atoms
 
John Rennie
John Rennie
You're over thinking this
 
gaurang agarwal
gaurang agarwal
7:27 AM
Yeah kind of
 
John Rennie
John Rennie
The surface tension is an emergent property like pressure i.e. it is the result of averaging out the interactions of all the atoms and moecules present.
 
gaurang agarwal
gaurang agarwal
But I can't digest that some sort of force is acting on a line and then we make an equilibrium over it
Because the equilibrium condition must hold on some material thing
 
John Rennie
John Rennie
An alternative way to look at this is that surface tension can be described as a force per unit length, but also as a surface energy i.e. an energy per unit area of the surface. The two definitions are equivalent.
I can explain why it's a surface energy as well as a force if you want.
 
gaurang agarwal
gaurang agarwal
Yeah I know that
Although I got an amazing proof of Young's law considering surface energies
 
John Rennie
John Rennie
OK, look at it this way:
 
gaurang agarwal
gaurang agarwal
7:32 AM
There is a book
SURFACE AND INTERFACIAL TENSION ~
Measurement, Theory, and Applications
An alternative derivation of Young’s equation follows the same route as
the derivation ofthe Laplace equation using a notional change ofthe location ofthe dividing surface. Consider the surface free energy ofthe system depicted in Fig. 4. Around the line of three-phase contact a cylinder is drawn with length L and radius R, and implicitly we assume that R and L approach
The figure is same drop over a solid figure that hui used in answering the question
 
John Rennie
John Rennie
Suppose we move the edge of the drop outwards a distance $dx$. Then for our line element $\ell$ the area of the solid air interface decreases by $dA = \ell dx$, the area of the solid liquid interface increases by $dA = \ell dx$ and the area of the liquid air interface increases by $dA = \ell dx/\cos\theta$
 
gaurang agarwal
gaurang agarwal
Yeah yeah I saw exactly this proof in the book
 
John Rennie
John Rennie
OK, so we're going to get the total energy change and differentiate it wrt $x$ to get the force.
So the energy approach and the force approach are equivalent.
The advantage of the energy approach is that you don't have to worry exactly what it means to apply a force on the edge of the drop.
 
gaurang agarwal
gaurang agarwal
Yes it got clear after this
But the question that remained is:- what it means to apply a force on the edge of the drop xD
 
John Rennie
John Rennie
Ultimately the force is being applied on the water molecules. That is, a water molecule at the point where the three phases are in contact experiences no net force i.e. it is in a minimum energy state.
 
gaurang agarwal
gaurang agarwal
7:41 AM
Yes but how can solid vapour surface tension act on water
 
John Rennie
John Rennie
I've never really thought about what goes on at the atomic scale.
 
gaurang agarwal
gaurang agarwal
May be the water molecules on the edge are pulled outward by adhesive forces but then we have to show that the adhesive forces are equal to surface tension which looks like something ....
 
John Rennie
John Rennie
If you're going to start thinking about individual molecules you have to be careful. Adhesive force is an emergent force just like surface tension and pressure. All that matters are the forces between individual molecules.
 
gaurang agarwal
gaurang agarwal
Yeah and I am not a master blaster in handling this even if I am considering 3 molecules 1 of each state on the edge, the diagram is looking way messy
3 molecules are under a force of surface tension and adhesive forces by each other .There are many force triangles. Looks like illuminati xD
 
gaurang agarwal
gaurang agarwal
8:07 AM
Rennie can you please go back to where you have answered my question I think I've got something.
 
John Rennie
John Rennie
@gaurangagarwal you mean here:
0
A: Young's equation

John RennieTo try and make this clearer I've drawn a three dimensional picture of the drop: The surface tension is the force per unit length acting normal to a line. Consider the small segment of the perimeter of the drop I have marked as $\ell$. We'll assume $\ell$ is small enough that it can be conside...

 
gaurang agarwal
gaurang agarwal
8:20 AM
Yes I've posted it
 
John Rennie
John Rennie
@gaurangagarwal I'll have a look later. I need to work now for a bit.
 
PM 2Ring
PM 2Ring
@bolbteppa You may like to contribute an answer to this question: Doubt about mathematical construction underlying physical systems
 
gaurang agarwal
gaurang agarwal
8:42 AM
@JohnRennie okay nice to meet you and thanks for giving me your precious time
:-)
 
skillpatrol
skillpatrol
9:15 AM
\o @alarge
 
alarge
alarge
\o
 
skillpatrol
skillpatrol
are you still in shock over the warriors losing in 6?
i know i am
 
alarge
alarge
don't follow nba
 
skillpatrol
skillpatrol
st louis was a shocker also
10 straight road wins in the playoffs is unheard of until now
 
 
1 hour later…
Loong
Loong
10:42 AM
1969: 1.6021917(70) × 10−19 C
1973: 1.6021892(46) × 10−19 C
1986: 1.60217733(49) × 10−19 C
1998: 1.602176462(63) × 10−19 C
2002: 1.60217653(14) × 10−19 C
2006: 1.602176487(40) × 10−19 C
2010: 1.602176565(35) × 10−19 C
2014: 1.6021766208(98) × 10−19 C
2019: 1.602176634 x 10−19 C
 
skillpatrol
skillpatrol
10:56 AM
> as of 20 May 2019, its value is exact
 
akhil krishnan
akhil krishnan
11:37 AM
@JohnRennie . It is ok, Can we say that the presence of fractals implies chaos from a microscopic view point of the system?
 
Danu
Danu
o/ moin moin @ACuriousMind
Thank you for the reading material (funny coincidence: The guy who deleted his answer to that question of yours is a post-doc here in Hamburg!)
Those posts did not address my question(s) though
 
ACuriousMind
ACuriousMind
@Danu moin moin
 
Loong
Loong
A Curious Moind?
 
Danu
Danu
facepalm
Do you have time to talk at the moment?
 
ACuriousMind
ACuriousMind
@Danu Sure, go ahead
 
Danu
Danu
11:47 AM
Ok, so I've been running this seminar where we try to mathematically understand the notes by Figueroa O'Farrill called Electromagnetic duality for children
 
ACuriousMind
ACuriousMind
@Loong I'm tempted to change my name to that ;P
 
Loong
Loong
Hehe
 
Danu
Danu
We've now reached the critical point in the notes, where he starts discussing susy
 
ACuriousMind
ACuriousMind
@Danu Ah, I know of them (but haven't read them in detail)
 
Danu
Danu
this is where the notes start getting bad (imo)
and it's getting hard to understand exactly what's going on
anyways
I'm interested in his claim that N=1 SYM exists only in dimensions 3,4,6,10. He also says which types of spinors exist in these dimensions (Majorana, Majorana or Weyl, Weyl, Majorana-Weyl, in that order)
Now I'm pretty close to being content with my understanding, which is as follows
Let $M$ be an oriented, pseudo-Riemannian manifold with spin/spin^c/whatever structure. Then we can form the bundle of Clifford algebras Cl(TM) and take the fiberwise irreducible representation (let's not deal with annoyances about uniqueness of this rep in odd dimensions) to form the spinor bundle S
Clifford multiplication by the volume form gives a splitting $S=S^+\oplus S^-$ (where, for me, $S^+$ is actually the one with eigenvalue -1)
Now assume we have a $G$-principal bundle $P\to M$. Then a spinor in SYM is a section of $S\otimes \mathrm{ad}P$
A general section is called a Dirac spinor. A section of $S^\pm$ is a left/right-handed Weyl spinor, and if there happens to be a real structure on the irreducible Clifford module then a real section is called a Majorana spinor
So far, so good, right?
 
ACuriousMind
ACuriousMind
11:56 AM
Yup, seems fine to me
 
Danu
Danu
OK.
So now regarding existence of SYM in various dimensions
Obviously Weyl and Majorana spinors have half
$S$ is a complex vector bundle of rank $2^{d/2}$ or $2^{(d-1)/2}$ depending on the parity of $d$; We say that a Dirac spinor has this many degrees of freedom.
$d$ being the dimension of $M$, of course.
A gauge boson has $d-2$ degrees of freedom, and a necessary condition for existence of SYM is that this number matches the number of degrees of freedom of a spinor (since they're supposed to be related by a symmetry)
so we have to solve $d-2 = 2^{d/2}$ and analogously for Weyl/Majorana
this gives you the list $3,4,6,10$ that FOF has in his notes
 
ACuriousMind
ACuriousMind
I'm waiting for the "but" here ;)
 
Danu
Danu
However, he actually says that e.g. in $d=3$ there are no Weyl spinors
Why not?
The number of dof is fine. What other condition needs to be satisfied?
 
ACuriousMind
ACuriousMind
@Danu Because the splitting into Weyl spinors via the top-dimensional form works only in even dimensions
 
Danu
Danu
(similarly for $d=6$ no Majorana)
@ACuriousMind OK, that's simple enough.
 
ACuriousMind
ACuriousMind
12:00 PM
In odd dimensions it does not commute with the even subalgebra and so you can't use Schur's lemma to argue the rep splits
 
Danu
Danu
What about d=6?
@ACuriousMind What's the even subalgebra? even number of gamma's?
(I am woefully ignorant on the topic of Clifford algebras)
 
ACuriousMind
ACuriousMind
@Danu Yes - it's the Lorentz algebra, basically. Weyl spinors are representations of the Lorentz algebra but not of the full Clifford algebra.
 
Danu
Danu
The induced rep of the Lorentz algebra is this $\Sigma_{\mu\nu}=[\gamma_\mu,\gamma_\nu]$ right?
(up to factor)
 
ACuriousMind
ACuriousMind
@Danu Well, there are no Majoranas for Lorentzian signature in d=6, since there $p-q = 4 \mod 8$, but there are no Majoranas for that value of $p-q$, cf. physics.stackexchange.com/a/356339/50583
Funnily enough, there are Majoranas in Euclidean signature at d=6, which is why Wick rotation has never set well with me for fermions.
@Danu yup
 
Danu
Danu
@ACuriousMind Ah, a signature issue
Perfect.
Now, regarding that question, you didn't really get a satisfactory answer in the sense that nobody gave you a proof of existence of Majorana spinors. Did you ever end up figuring one out?
Is tehre s a simple description of the real structure?
 
ACuriousMind
ACuriousMind
12:09 PM
@Danu The answer I accepted provides a proof - one only needs to know that the only irreducible representation of a full matrix algebra of size $n$ over a field $K$ is $K^n$, so whenever the Clifford algebra is isomorphic to a real matrix algebra, the irreducible complex representations is just the complexification of the standard rep on $\mathbb{R}^n$.
Conversely, it's relatively easy to convince oneself that there cannot be a real structure on this irrep of a complex or quaternionic matrix algebra
 
PM 2Ring
PM 2Ring
@Danu FWIW, the 3, 4, 6, 10 thing came up here the other day:
2 days ago, by bolbteppa
Another nuts comment - the dimensions of the division algebras 1, 2, 4, 8, are the dimensions of the massless Poincare little groups in dimensions 3, 4, 6, 10
 
Danu
Danu
that looks like a coincidence to me
 
ACuriousMind
ACuriousMind
This doesn't give the real structure explicitly, sadly, but it at least avoids any arguments about conjugation, duals or Hermiticity
 
Danu
Danu
@ACuriousMind Derp, I need to learn how to read.
I would like to know what the real structure is
(also I'd like something more explicit showing how the numbers come up)
also LOL -24 on that other answer, that's a pretty impressive score
 
ACuriousMind
ACuriousMind
@Danu You can bootstrap the isomorphisms between the Clifford algebras and the matrix algebras pretty easily by induction, see e.g. O'Farrils' maths.ed.ac.uk/~jmf/Teaching/Lectures/Majorana.pdf
(pages 7 & 8)
 
Danu
Danu
12:16 PM
excellent
That solves my problem. I have some more questions though.
 
ACuriousMind
ACuriousMind
:D
 
Danu
Danu
What FOF does in the notes is to first discuss N=1 in d=6, and then compactify in the sense of imposing the fields to be independent of the 5,6 coordinates
In d=6, $S^+$ has rank 4, and so the Weyl spinor he considers in the d=6 theory has 4 components, which also matches the number that a Dirac spinor has in d=4
He says that a Dirac spinor is indeed what you get. I'd like to understand what this means on a mathematical level.
My first naive guess is the following
I guess Cl(1,5) contains a copy of Cl(1,3) in the obvious way (simply taking only the first 4 gamma matrices). Then the statement that this Weyl spinor in d=6 becomes a d=4 Dirac spinor sounds, to me, like these 4 gamma matrices should preserve the Weyl spinor in d=6, which they don't. So now I'm confused.
What should I do in terms of representation theory to verify his claim?
 
ACuriousMind
ACuriousMind
Let's see
Abstractly, you need to show that the Weyl representation of $\mathrm{Cl}(1,5)^\text{even}$ (remember, Weyl spinors are not reps of the full Clifford algebra!) descends to the Dirac representation of $\mathrm{Cl}(1,3)$
For this, one needs a notion of how $\mathrm{Cl}(1,3)$ is embedded in $\mathrm{Cl}(1,5)^\text{even}$, which is not "in the obvious way"
 
Danu
Danu
12:31 PM
right
(but this "obvious way" of mine is an embedding right? or was even that already incorrect?)
 
ACuriousMind
ACuriousMind
Now, we have $\mathrm{Cl}(1,5)^\text{even} \cong \mathrm{Cl}(5,1)^\text{even} \cong (\mathrm{Cl}(3,1)\otimes\mathrm{Cl}(0,2))^\text{even}$
The last isomorphism is from lemma 2 of O'Farrill Majorana notes I linked above
And if you inspect it, you see that the even degree of $\mathrm{Cl}(3,1)\otimes\mathrm{Cl}(0,2)$ contains an entire copy of $\mathrm{Cl}(3,1)$ because you can get an even element as $\Gamma_i \Gamma_5 = \Gamma'_i \otimes \Gamma''_2$, so there is an inclusion $\mathrm{Cl}(3,1) \to (\mathrm{Cl}(3,1)\otimes\mathrm{Cl}(0,2))^\text{even}, \gamma_i \mapsto \gamma_i \otimes \Gamma''_2$
So the 6d Weyl rep indeed descends to a 4d Dirac rep, in a highly non-obvious way.
I made this up just now on the spot, so it might be wrong :P
 
Danu
Danu
@ACuriousMind What's the $\Gamma_5$ here?
 
ACuriousMind
ACuriousMind
12:48 PM
@Danu I'm using the notation from page 8
 
Slereah
Slereah
Please
Write $\text{C}\ell$
It must be fancy
 
Danu
Danu
lmao
nevar
@ACuriousMind Btw he does this for (0,d) only right? But I trust you that signature is not an issue (?)
 
ACuriousMind
ACuriousMind
@Danu The even subalgebras of p,q and q,p are isomorphic, so it doesn't matter which way we do it
 
Danu
Danu
@ACuriousMind I mean he doesn't do 1,d-1
 
ACuriousMind
ACuriousMind
oh
 
Danu
Danu
12:52 PM
I didn't even consider the d,0 versus 0,d thing hahaha
but thanks for thinking for me :D
Also I'm very surprised at the aggressive tone in Bilateral's comment on your deleted post here. He's a super chill guy in real life
 
ACuriousMind
ACuriousMind
@Danu the internet changes people :P
 
Danu
Danu
One more question, for now
 
ACuriousMind
ACuriousMind
@Danu that might actually be a problem, I'm not sure
It definitely works to get an Euclidean Dirac spinor since $(5,1) = (4,0)\otimes (1,1)$ is definitely true :P
 
Danu
Danu
Now that we have a 4d Dirac spinor, one expects the supersymmetry variation parameter to also be a Dirac spinor. FOF now interjects "but in 4d the susy variation parameters are Majorana"?! [citation needed]
Is this a priori clear? Should this be obvious? Is this a rabbit out of a hat?
(this is crucial because it means we should interpret the 4d theory as N=2)
@ACuriousMind Close enough. The idea is clear
 
skullpetrol
skullpetrol
@ACuriousMind only if they let the internet change them :P
\o @Danu
 
ACuriousMind
ACuriousMind
1:04 PM
@Danu Isn't this just using the counting from earlier that in d=4 SYM only exists with Majoranas or Weyls?
 
Danu
Danu
Oh, right. Noice
hi @skullpetrol
I've said this before, and I sure as hell will say it again. You leaving academics is a serious loss for us @ACuriousMind
12
 
ACuriousMind
ACuriousMind
I'm not letting you guilt me into going back ;)
 
Danu
Danu
Just stay here and let me pass off your knowledge as my own in my seminars. That works
I'll be your Milli Vanilli
 
skullpetrol
skullpetrol
lol
 
ACuriousMind
ACuriousMind
Deal :D
 
Danu
Danu
1:09 PM
<3
Keeping the royalties though
 
ACuriousMind
ACuriousMind
You need them more than I do anyway :P
 
Ryan Unger
Ryan Unger
@ACuriousMind how the hell do you remember this after 2 year
 
Danu
Danu
That, I can imagine. The question is, how the hell did you manage to know all of that 2 years ago? :P
@ACuriousMind Feelsbadman
@ACuriousMind I have now double-checked this (things are hard when you're hung over...) and I can confirm that it checks out. I don't think there's a signature problem with the 1,d-1 vs 0,d.
 
Ryan Unger
Ryan Unger
You might find this funny...someone asked Fernando during his lectures what "adding chains" means
 
Danu
Danu
1:29 PM
I'm waiting for the punch line
or am I not getting it
 
bolbteppa
bolbteppa
1:46 PM
@PM2Ring would this be acceptable as an answer :p
Have to be super careful with this stuff
 
PM 2Ring
PM 2Ring
@bolbteppa LOL. This topic is way beyond my paygrade. I've read some of John Baez's stuff on octonions etc, but I get to the stage where I realise that I really didn't understand the last page or so. But he writes so nicely that you can easily fool yourself into thinking you do understand. :D
 
bolbteppa
bolbteppa
e.g. the division algebra little group link disparaged above, massless little groups in 3, 4, 6, 10 dimensions (these determine the actual degrees of freedom, particles being irreps of little groups) having dimensions 1, 2, 4, 8 is probably linked to the fact division algebras in these dimensions ensure that spheres $S^1, S^1, S^3, S&7$ are parallelizable rather than the octonions,
but but they are important mathematically anyway, there is an interesting big mystery here for sure
9
Q: Number of dimensions in string theory and possible link with number theory

user1355This question has led me to ask somewhat a more specific question. I have read somewhere about a coincidence. Numbers of the form $8k + 2$ appears to be relevant for string theory. For k = 0 one gets 2 dimensional string world sheet, For k = 1 one gets 10 spacetime dimensions and for k = 3 one ge...

string-theory mathematics
 
PM 2Ring
PM 2Ring
2:05 PM
@bolbteppa I think you're right, but I don't understand this stuff well enough to have an informed opinion. It's a bit like hearing poetry in a foreign language. The patterns are pretty, but I don't know what it means. ;)
 
bolbteppa
bolbteppa
2:23 PM
Another interesting thing is that you get the division algebras by adding e.g. to R, C, etc.... a 'root of minus 1' (Cayley-Dickson) and just wanting a division algebra at every step which is only possible for a few steps, the norm forcing the first few added roots to anti-commute. Clifford algebras can also be thought of as a field along with a bunch of anti-commuting roots of minus one but you just sacrifice needing it to be a division algebra
For a Euclidean Clifford algebra over R the first three (in terms of dimension) are literally R, C, H
 
PM 2Ring
PM 2Ring
Argh. We have a new numerologist on Astronomy. astronomy.stackexchange.com/a/32308/16685
@bolbteppa Anti-commutation makes sense physically, given how 3D rotation works. Losing associativity is a bit harder to cope with.
 
PM 2Ring
PM 2Ring
2:56 PM
What do you do when your homework question gets closed? Repost it, of course, with two more questions. :facepalm: physics.stackexchange.com/questions/486217/…
 
danielunderwood
danielunderwood
3:09 PM
But they need to get their homework done for next week!
 
Emilio Pisanty
Emilio Pisanty
3:56 PM
@ACuriousMind ..... for now, anyway .....
 
 
7 hours later…
danielunderwood
danielunderwood
10:56 PM
So it's not clear to me why a minimized action consists of the path of minimized infinitesimal steps. Is there a good answer to that on the site? I feel like I've seen one before, but can't seem to find anything with the search terms I'm using
 
ACuriousMind
ACuriousMind
@danielunderwood Why does the shortest path from A to B consist of the shortest infinitesimal paths between the points on it? ;)
 
danielunderwood
danielunderwood
11:17 PM
That seems obvious in Euclidean space, but could you not cook up some fancy geometry where it isn't the case? Like if you had the Poincare half-plane, wouldn't the shortest step always be up? (though then you'd never reach the final point if it weren't straight up)
Though something something ant on an apple
 
ACuriousMind
ACuriousMind
@danielunderwood But a "step up" doesn't always respect the boundary conditions. Note that both "shortest path" and "extremal action" are always relative to some boundary and/or initial conditions.
Forget about "infinitesimal" steps: The following is true for any functional $F$ on "paths" that behaves additively (like distance and action do): Given two paths $\gamma_1$ and $\gamma_2$ that share begin/end points so they can be concatenated into a path $\gamma$: If $\gamma$ extremizes $F$ among all paths with its start/End point, then $\gamma_1$ and $\gamma_2$ also extremize $F$ w.r.t to their start/end points.
The proof is simple by contradiction: Since $F$ is additive, if there was a $\gamma'_1$ that had a (say, lower) value instead, then the concatenation of $\gamma'_1$ and $\gamma_2$ would have a lower overall value of $F$ than that of $\gamma_1$ and $\gamma_2$, so $\gamma$ cannot have been extremal.
Note also that there is no single principle of extremal action - there are different ones, with different boundary conditions and variables being varied. Qmechanic has written quite a few answers on these, iirc
 
danielunderwood
danielunderwood
It does seem rather sensible when you describe it in that way. I guess I just find it difficult to think about overall boundary conditions in the same context as infinitesimal steps
 
Ryan Unger
Ryan Unger
11:40 PM
@ACuriousMind damn proof by contradiction
 
ACuriousMind
ACuriousMind
the best kind of proof :)
 

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