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12:04 AM
0
Q: Which guidelines does this question violate?

OverLordGoldDragonThis question is one close vote away from closing, all four marking "Community-specific reason" - which is a new substitute for "Off-topic". The question clearly is on-topic. What guideline, then, is being violated? My guess for the 'real' reason is the opening statement, which some users found ...

 
The nuclear radius is $(r0)*a^(1/3) $ now if a=1 we are essentially talking about hydrogen nucleus . Then$r_0$ should be 0.8775fm but my book writes $r_0$=1.2 fm. Why?
 
12:34 AM
The eigenfunctions for a particle in a box are not eigenfunctions of the Hamiltonian of the harmonic oscillator
 
12:58 AM
@bolbteppa thank you for your help
 
 
5 hours later…
@skullpatrol it feels as though this debate is getting a bit tired now. People keep saying the same things, most of which are just their opinion and none of which can be proved.
 
We should go into neuroscience :-)
 
 
5 hours later…
11:50 AM
Argh, this meeting has been postponed by half an hour every half hour for the last two hours and now it's on Monday
It's fine, I didn't want to use the last two hours productively anyway, I guess :P
 
 
1 hour later…
12:57 PM
Hi, I am trying to compute the feynman rules for Gross Neveu as in this question : physics.stackexchange.com/questions/544358/…
I was hinted to compute $\langle\psi_\alpha(x)\bar\psi_\beta(y)\psi_\gamma(x')\bar\psi_\delta(y')\rangle$, which I proved to be $S_F(x-y)_{\alpha\beta}S_F(x'-y')_{\gamma\delta}$ minus the same term with indices permutated
I do not know how I could continue from here
Here $S_F$ is the Feynman propagator for fermions
 
1:44 PM
That post is not really asking about the Feynman rules, compare e.g. to QED the Feynman rules in momentum space are just things like the propagator and what you attach at a vertex
 
I'm having trouble understanding how to compute what to attach at a vertex
 
@ACuriousMind this paper (sec. 2) seems to say that the Henneaux view of primary first class constraints generating gauge transformations is wrong (even explicitly references the book)
In qed in momentum space it's like $-i e \gamma_{\mu}$ at a vertex, where the interaction term in the Lagrangian is $-e A_{\mu} j^{\mu} = -e A_{\mu} \overline{\Psi} \gamma^{\mu} \Psi$, so in GN it's probably $\pm ig^2$ right
Skimming the GN paper it seems this is more or less how they do it
 
Could you please link the paper you're talking about ?
 
1:59 PM
The Gross–Neveu model is a quantum field theory model of Dirac fermions interacting via four fermion interactions in 1 spatial and 1 time dimension. It was introduced in 1974 by David Gross and André Neveu as a toy model for quantum chromodynamics, the theory of strong interactions. It consists of N Dirac fermions, ψ1, ..., ψN. The Lagrangian density is L = ψ ¯ a...
First reference
 
Thank you
 
@bolbteppa A cursory reading does not convince me because the author demonstrates the "inconcistency" without considering the full extended action. The gauge transformation generated by the constraint acts not only on the field variables but also on the variables of the extended phase space. When you look to the reduced phase space you need to be careful. In particular eq. (1) is wrong, cf. eqs. (19.13) in H/T's book. Using (19.13), the variation of $F$ correctly works out to zero.
The author is essentially trying to use (19.11) for only $\epsilon^1$ on the reduced phase space where there are no longer two different $\epsilon^i$.
 
@ACuriousMind Why is a presheaf a functor on the opposite category?
rather than the category itself
 
This of course is inconsistent - in the passage to the reduced phase space, $\epsilon^1 = \epsilon^2$ needs to be imposed, as H/T correctly write.
@Slereah The arrows are the wrong direction
 
Wrong direction for what?
 
2:08 PM
For the categorical presheaf to coincide with the topological notion of presheaf when considered on the category of open subsets of a space with inclusions as morphisms.
Sheaves (at least for me) are much easier to understand if you first understand how they work in their topological incarnation
That is, forget all about categories and consider (pre-)sheaves as assignments of certain data (e.g. rings of functions, or vector spaces) to open subsets of a space
Then, since you already are familiar with bundles, understand that bundles and sheaves are the same idea of local data gluing together in some way - every bundle defines a sheaf by assigning to each open subset the space of bundle sections over it
Conversely, the gluing data that differentiates a pre-sheaf from a sheaf is exactly the gluing condition that glues together a global bundle from its local trivializations
 
What field are topological presheaves discussed in?
 
That kind of presheaves are often not very useful (precisely because they don't define a bundle, hence define no topological space), I don't think anyone discusses them standalone before jumping to sheaves
It's kinda like how much more people discuss groups than semigroups :P
In fact the only context off the top of my head in which I've ever met examples of presheaves is in constructions where you define some operation on sheaves and the naive outcome is only a presheaf, but then you sheafify and that's it.
 
This course on QG references the paper and "ABC before D: Issues with Dirac's view on first class constraints and gauge transformations" so maybe there is some technicality
I'd say it's fine to ignore to a first approximation anyway
@Slereah holy shi... I think I'm convinced Category theory is useful after reading section 1 of this, you're implicitly using category theory everytime you think about QM
A symmetric monoidal category is an obvious construction, what's happening...
 
2:24 PM
By that notion, you're "implicitly doing category theory" every time you do math (not saying that viewpoint is wrong, just saying it's not special to physics or QM at all :P)
 
That's the npov
but then you could say the same of any foundation of mathematics
you're implicitely using a Turing machine!
 
for instance, classical physics is also happening with a monoidal operation combining state spaces, just the monoidal operation is the direct product and not the tensor product
 
My hope was that category theory would turn subjects like algebra, topology etc... into one subject where one category proof covers some theorem in each subject (e.g. Cayley's theorem is just Yoneda's lemma) but I'm not smart enough to see how to do that yet or if it's even possible, but I don't see why it shouldn't be and why it wouldn't let you find new theorems in new subjects by just looking for analogues
 
I'd say it more makes it possible to view the subjects through a unified lens, e.g. many "existence and uniqueness" proofs are essentially proofs that the category under consideration "has" certain universal objects
 
Category theory could be the 'string theory' of mathematics :p
 
2:31 PM
I'm afraid you'll have to read some nlab for that!
also set theory already does that :p
or any other foundational system
 
I mean, I'm definitely a fan of e.g. describing constructions in term of universal properties. They capture the essence of what a "direct product" or "direct coproduct/sum" really is much better than specific definitions interior to a specific category, imo.
But that doesn't mean I find everything clearer or more obvious in categorical language. For instance, as I said, I find sheaves much easier to understand in their specific incarnation on topological spaces than in the abstract setting
 
Well
I guess I'd better read more algebraic topology!
@ACuriousMind I do wish the category people would more often attempt to do actual diagrams of categories
 
Well, everything except the most boring categories are infinite, unfortunately :P
I think the "nicest" at least mildly interesting ones are rigid ones like vector spaces where you have exactly one isomorphy class for each cardinality and everything is a sum of the one-dimensional vector space
But something like the category of groups I would have no idea how to even begin to draw
Also, if you've ever seen how algebraic geometers "draw" schemes, you really don't want them to draw categories :P
I've tried to read explanations of why the generic point is a fuzzy ball at least a dozen times and it never really clicked
 
2:48 PM
Not sure if this makes sense: so if the states A, B, C, ... of an electron and an associated collection of measurements $f, g, h, ...$ can be described as a category, then (trying to motivate a functor) a change of coordinate system can be described by a functor since it not only changes what the states A, .. look like but also what the f, ... operations look like. Then we can say any coordinate transformation is 'isomorphic' to any other using a 'natural transformation'
 
so intuitive
@bolbteppa usually people think of the state spaces of systems as the objects of the category, i.e. the category is that of (finite, if we're doing quantum information) Hilbert spaces
Then depending on what morphisms you allow (e.g. unitary, positive, POVM-type, etc.) you get different ideas of what a transformation/measurement might be
 
Hmm, re-reading the start of that pdf they are a bit fuzzy, I mean A and B don't seem to be different states, it seems like they could be saying A is the classical data of an electron before the measurement, then B is the classical data after the measurement, then C would be classical data after another measurement, and one could frame A, B, C... as the Hilbert space + different classical data to encode both the hilbert space but still make them look different...
Madness
 
@ACuriousMind You can attempt to sketch an infinite category!
 
3:05 PM
Sure, but as I said for anything less rigid than vector spaces I'd struggle
 
Hm
What are the objects for $\mathrm{Set}$
sets?
 
In which case can you do, say, a sequence
Starting from the empty set
and using the various set building items
 
Okay, that's right. Let's say it like this - it's easy for something as structureless as sets or as rigid as vector spaces. It's really hard for the interesting middle ground like groups, modules, manifolds, etc
 
So it seems like the reason GR is not a 'gauge theory' in the same way EM is a gauge theory, from a Lagrangian perspective, is that the EM Lagrangian has variables whose 'accelerations' are not determined, whereas for the GR Lagrangian the 'accelerations' are all determined, not sure if that's true haha
 
3:24 PM
Is it?
Isn't GR famously underdetermined
 
good evening everyone
https://imgur.com/a/BMbu2EH

Can anyone explain how does author makes the conclusion ?
"There is little chance of finding two parallel spin (triplet state) electrons together and the electrons act as if they repel each other"
9th line
in the image
 
I do know that GR has that weird thing where it is all constraints
 
The text describes that the wave function becomes zero when they are parallel so the probability of finding them parallel is 'practically zero'
 
@Slereah You mean because it has zero "real" Hamiltonian? Any system can be made (space)time-reparametrization invariant, i.e. made to consist only of constraints in the Hamiltonian, see e.g. physics.stackexchange.com/a/194925/50583
Eliminating the constraints gives you back a system with non-zero Hamiltonian. The Hamiltonian - zero or not - is not an intrinsic property of a physical system, but a matter of the choice of variables
 
3:47 PM
Seems like GR does actually have 'gauge' variables as described above (not fully sure yet), but:
"Relativistic theories of gravitation - such as Einstein's theory of general relativity - may also be considered as gauge theories. The bundle $P$ consists in this case of orthonormal linear frames (tetrads) of the space-time manifold $M$ and $G$ is the Lorentz group. Alternatively, one can take $P$ to be the bundle of orthonormal affine frames, in which case $G$ is the inhomogeneous Lorentz group.
There are, however, important differences between Einstein's theory and gauge theories such as electrodynamics or the Yang-Mills theory. First of all, the bundle of frames is soldered to the base $M$ whereas in other gauge theories the bundle is rather loosely connected to $M$...
The soldering results in the appearance, in theories of gravitation, of torsion, in addition to curvature, which occurs in any gauge theory. (Torsion is zero in Riemannian geometry, but being zero is different from not existing at all.) Moreover, the form of Einstein's equations of gravitation is different from the "generic" form of the field equations assumed in gauge theories.
The latter are derived from Lagrangian's quadratic in curvature, whereas the former are based on a linear Lagrangian. The possibility of constructing such a linear Lagrangian is also related to the existence of the soldering form on $P$."
 
@bolbteppa sir, you've seen this? writings.stephenwolfram.com/2020/04/…
^a different intrepretation...
 
Yes I made comments on it yesterday and the day before in here
 
okay...
@bolbteppa Sir, Is there any way to write electromagnetic tensor without $F_{\mu \nu}$ thing?
 
Not sure what you mean, but it's easy it's just $F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ and you can easily derive it 'from thin air' by looking for the equations of motion of a particle in an electromagnetic field if you want a natural example of where it could come from
 
@bolbteppa I totally didn't get that... :-O Where that $\partial_{\mu}A_{\nu} - \partial_{\nu} A_{\mu}$ thing came?
 
3:57 PM
What is $F_{\mu \nu}$?
 
In electromagnetism, the electromagnetic tensor or electromagnetic field tensor (sometimes called the field strength tensor, Faraday tensor or Maxwell bivector) is a mathematical object that describes the electromagnetic field in spacetime. The field tensor was first used after the four-dimensional tensor formulation of special relativity was introduced by Hermann Minkowski. The tensor allows related physical laws to be written very concisely. == Definition == The electromagnetic tensor, conventionally labelled F, is defined as the exterior derivative of the electromagnetic four-potential, A,...
^this one I mean
 
@bolbteppa If you search for "solder form" or "soldered" in the chat transcript you'll find me or Slereah saying pretty much this on various occasions :P
 
"A vielbein or solder form on a manifold $X$ is a linear identification of a tangent bundle with a vector bundle with explicit (pseudo-)orthogonal structure."
 
@dumbasarock when the the spin of the two electrons are parallel, the space part must be antisymmetric so the complete wave function of the electrons that is fermions remains antisymmetric.
 
it means it's a real vector and not a fake vector :p
 
4:01 PM
So using the word vielbein instead of solder form, the GR field is vielbeined to the base, while the others aren't, eh..
 
now the author assume that the two two electrons are almost near to each other such that their space coordinates are equal because of which the space wave function that was antisymmetric becomes zero due to the negative sign
if you calculate the probability density quantum mechanically, that would be almost zero, meaning the two electrons with parallel spin cant be near to each other because of which the probability of finding such a case becomes almost zero and the case here specific is is the case of parallel spin
 
@bolbteppa but even if the wavefunction are nearly equal the triplets wavefunction are still additive so they can't be zero as they are symmetric.
it's singlet which becomes zero.
but text says finding triplet close to each other is not probable
 
@dumbasarock the space probability, is zero,...
@dumbasarock means, the probability of finding the two electrons in such a case, with parallel spins, and that too close to each other, in unit volume of measurement
 
@bolbteppa it's part of the tensor bundle, and not the tangent bundle of the principal bundle!
 
look, the singlet state has a factor of root two...., i mean, that's a good probability than that of with almost zero
 
4:11 PM
triplet is symmetric and parallel case.
the approximation will not result in cancelling out of the overall wave-function
thus when we will find the probability density with it we'll get a value.
singlet case is antisymmetric and antiparallel case
equating the wace-function will result in the overall wave-function in value zero.
thus probability density will result in zero too as wavefucntion itself is zero
so how come author says that
"There is little chance of finding two parallel spin (triplet state) electrons together and the electrons act as if they repel each other"
 
@dumbasarock don't calculate the probability density from the complete wave function... i.e., don't consider spin part in it
 
finding the anti-parallel ones close should be less probable
I just want to know how being close results in triplet probability being less
 
@dumbasarock The text you quote explicitly says that the spatial wavefunction of triplets is anti-symmetric.
So when the two positions are close together, the spatial wavefunction - and hence the overall wavefunction - becomes close to zero.
What exactly is your problem with that statement?
 
but on previous page here it says the triplets are symmetric
 
It is probably talking about the spin wavefunction.
 
4:17 PM
@dumbasarock being close in a triplet probability is less because when a triplet is form both the spins are up that is is the spin part of the wave function is symmetric. show the electrons can have the overall wave function as antisymmetric, the space path must be antisymmetric
 
Even in your screenshot it says that at the top "there are three possible spin wave functions which are all symmetric"
 
The wave function is anti-symmetric, $\frac{1}{\sqrt{2}}[\psi_a(1) \psi_b(2) - \psi_b(1)\psi_a(2)]$, if $\psi_a(1) \approx \psi_a(2)$ and $\psi_b(1) \approx \psi_b(2)$ then $\psi_a(1) \psi_b(2) \approx \psi_a(2) \psi_b(1)$ which is just a complex number whose value is the same value you get from $\psi_b(1) \psi_a(2)$ so subtracting them is basically zero which means the wave function is basically zero
 
You need to pay close attention to the difference between "total wavefunction", "spatial wvefunction" and "spin wavefunction"
 
total = space . spin
total (anti for electrons, fermions) = spin (symm for triplet, because both up) . space (must be antisymmetric, i.e., negative sign, i.e., when electorns are close, ~ 0)
 
okay to make the total antisymmetric for triplet the space wave-function should be antisymmetric because the spin is symmetric already
 
4:20 PM
Unless I'm missing something, the point of that sentence is nothing more than explaining what happens when the electrons are close together so that the values in both parts of the anti-symmetric wave function are basically the same so you basically get zero on subtracting them
 
@dumbasarock you got it
and remember, probability density, what is it?
 
@bolbteppa yes that is the meaning but the wavefunction used for puttin values was of singlet while the less chances were told for triplet that was bugging me
 
probability density is the probability of finding the particle at a particular position or in a region of Space at a particular time
 
@Vivek absolute value of square of total wave-function
 
@dumbasarock I don't understand that sentence(?).
 
4:22 PM
so see, it has nothing to do with spin...
only space
 
@Vivek No, it's not. A probability density is not a probability, its integral is a probability.
 
on the other hand, there's a good factor of root 2, hmm, makes a good probability
 
I am definitely sure I am wrong but where ?
@Vivek what you are describing is something true but different
 
And you can perfectly well consider a probability density on the combined classical space-spin state space $\mathbb{R}^3 \times \{-1/2,1/2\}$ so that it is a proper probability density on states and not only on the spatial part.
 
@ACuriousMind yp, got confused
 
4:26 PM
@dumbasarock I can't tell you where you're wrong because what you said is not a sentence I could parse.
 
I am explaining the doubt once again from beginning.
Take a look at this image
https://imgur.com/a/cVWitON
 
Those are the spin wavefunctions, yes.
 
@dumbasarock what does that beta indicates? spin?
 
now going back to assumption, for close electron the ${\beta ^(+)}_1 = {\beta ^(-)}_1$ and ${\beta ^(+)}_2={\beta ^(2)}_2$
Those equal signs are approximately equal ( I forgot latex code for approximately equal)
so which of the state goes zero in above photo
it;s the singlets
 
No.
The $\beta$s are not the total wavefunctions, they are the spin wavefunctions
You can't plug a position into them
 
4:32 PM
now I got it
 
Your total wavefunction is a product $\beta \cdot \psi(x)$, where $\beta$ is one of the four expressions in that image and $\psi$ is the spatial wavefunction.
 
thank you very very much @Vivek @ACuriousMind
I just read the line "in order to have total wave-function antisymmetric...."
in the image I posted earlier
@Vivek Thank you very much for this @Vivek
 
5:01 PM
@dumbasarock wlcm...
 
5:17 PM
Imagine trying to work out the Hessian of the Einstein-Hilbert action...
 
Do it in $0+1$ dimensions
 
I really can't imagine directly working out the Hessian or examining the EOM directly for the metric variables to figure out if some of the accelerations are missing to show whether the EH action is really a constrained Hamiltonian system, but the way it seems to be done is:
"This action is invariant under general coordinate transformations (or passive spacetime diffeo-morphisms, $_P \mathrm{Diff}(M)$) and immediately, as the passive diffeomorphisms are a subset of the local Noether symmetries one can conclude, by Noether’s second theorem, that GR is a singular system and the equations of motion will have this symmetry. We therefore expect to find a constrained Hamiltonian system once we have completed the canonical analysis."
In mathematics and theoretical physics, Noether's second theorem relates symmetries of an action functional with a system of differential equations. The action S of a physical system is an integral of a so-called Lagrangian function L, from which the system's behavior can be determined by the principle of least action. Specifically, the theorem says that if the action has an infinite-dimensional Lie algebra of infinitesimal symmetries parameterized linearly by k arbitrary functions and their derivatives up to order m, then the functional derivatives of L satisfy a system of k differentia...
 
6:18 PM
btw @bolbteppa
Pretty good article
 
The way I understand (1.1) and (1.2) of that article is that (1.1) is derived by varying the coordinates $x'^{\mu} = x^{\mu} + \varepsilon$ directly in the Lagrangian and then requiring it to be invariant, while (1.2) is derived by asking how $g_{\mu \nu}(x)$ changes under $x'^{\mu} = x^{\mu} + \varepsilon$ and then requiring the variation to be invariant, so both really come from the same underlying translational invariance idea
I can't imagine carrying around the rest of that paper in my head when the derivation using what I said above is so simple :p
 
6:58 PM
@bolbteppa The interesting part is 1.3 though!
 
The (easy) proof of (1.2) shows why it reduces to (1.1) up to a total derivative and so takes the form (1.3)!
 
7:13 PM
@Slereah, how are you? I don't know what happened between us but I want to apologize and I want peace. I'm sorry for whatever I've done. This is my last attempt to make things right, if you decide to respond, it would be great, if not, I will not be trying to contact you anymore.
 
45
Q: Bourbaki's definition of the number 1

John BaezAccording to a polemical article by Adrian Mathias, Robert Solovay showed that Bourbaki's definition of the number 1, written out using the formalism in the 1970 edition of Théorie des Ensembles, requires 2,409,875,496,393,137,472,149,767,527,877,436,912,979,508,338,752,092,897 $\approx$ 2.4 $\...

"(I conjecture that Bourbaki's proof of 1+1=2, written on paper, would not fit inside the observable Universe.)"
 
 
2 hours later…
9:09 PM
@bolbteppa the OP has just posted an answer
There seems to be a rule of thumb in the logic community: if Solovay says it, it's correct. He seems to have garnered the reputation of being very, very careful. For example, he was the one who alerted Nash to a hole in his (Nash's) embedding theorem. — Todd Trimble ♦ 2 hours ago
 
Interesting reading about this stuff
37
Q: Were Bourbaki committed to set-theoretical reductionism?

Jeremy ShipleyA set-theoretical reductionist holds that sets are the only abstract objects, and that (e.g.) numbers are identical to sets. (Which sets? A reductionist is a relativist if she is (e.g.) indifferent among von Neumann, Zermelo, etc. ordinals, an absolutist if she makes an argument for a priviledge...

19
Q: Is the Bourbaki treatment of Set Theory outdated?

seraphOn page 5 of the following write up, the author asks why the Bourbaki did not notice that their system of Zermelo set theory with AC was inadequate for existing mathematics. Throughout the rest of the discussion, the author asserts that the Bourbaki group never acknowledged Godel's results on inc...

'I was able to confirm Matthias' calculations that the ordered pair (x,y) has a length of 4545 symbols' imagine trying to write this stuff out
 
9:29 PM
People got so carried away in the comments that the system created a chatroom
 
10:24 PM
lol thanks community
yesterday, by ACuriousMind
@JingleBells Other users are not required to explain themselves to you. If he doesn't respond, please don't ping him again.
 

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