 6:03 AM
@ACuriousMind What if you had three electrons and one of the electron interacts with the other two. Can't there be a vertex with an incoming electron, two photons and an outgoing electron?

3 hours later… 8:38 AM
This part"It is clear that in taking the limit as the charge separation d goes to zero, the sum of electric fields will only contain terms of even order in d." 9:16 AM
Can Sigma0 decay into Lambda + photon without strong interactions? I'm not too fond of supermanifolds, but it's a neat coincidence that their 3 main types follow each other alphabetically
N-supermanifold, P-supermanifolds and Q-supermanifolds https://physics.stackexchange.com/questions/418043/most-trivial-neutral-pion-decay

I have the same question. So are these diagrams based on hadrons only outdated? Can a neutral pion directly decay into two photons with pair annihilation (as proposed in the question) with the quark model?

3 hours later… 12:00 PM
@Yashas there is no such vertex in QED @fqq what prevents it? what goes wrong with such vertices?
2  Why can there be only 3-particle vertices in Feynman diagrams? For example, when a particle and its anti-particle annihilate to form two photons, why is its Feynmann drawn the way it is in diagram 2 (3-particle vertices)? Why can't diagram 1 (4-particle vertices) be correct? Does diagram 1 viola...

I think I need to finish reading QED and I might self-answer the question. Doesn't not allowing more than three lines imply that an electron cannot interact with two other electrons in the same instant? I don't get what you mean 12:26 PM
I am trying to imagine a situation where an electron interacts with two other electrons at the same instant.
I can make the internal electron line go vertical, right? This would be like simultaneous interaction because verticle lines imply same time instant? nooo, the dreaded time axis on a Feynman diagram 12:42 PM
Kaon is made up of two quarks. Why can't the spins of the two quarks align to give a non-zero spin? is it the case that in a gravitational theory like general relativity, a tensor, like curvature, torsion, and nonmetricity, depends on the choice of frame?
like we can choose a frame to make the curvature vanish but in another frame which doesn't make the curvature vanish. 1:02 PM
You can't choose a frame in which curvature vanishes, if by curvature you mean the Riemann tensor. You can always choose a coordinate system (so-called "normal coordinates") in which the metric is the Minkowski metric at a point though.
The components of a tensor can depend on your choice of coordinates, but it is entirely possible to do differential geometry (and, by extension, GR) without choosing coordinates. Since choosing a coordinate system amounts to labelling the points on your manifold, you would expect that intrinsic properties like curvature shouldn't be altered by simply re-labelling the points on the manifold. @Bohemianrelativist @Yashas there is no such thing as an "instant" in a Feynman diagram
these graphs are representations of certain terms in a perturbation series, they do not represent any actual physical process "happening"
mathematically, Feynman diagrams are just graphs with labels "in" and "out" on the external legs, and it doesn't matter how you draw them - they do not have a notion of "time" or "space", particularly since we usually draw them for amplitudes in momentum space
the kind of vertices that are allowed or not allowed are just a feature of the specific theory, there is no universal reason why the Standard Model contains vertices of a certain kind and not another - it is simply an experimental fact that a theory with different vertices does not describe our world 1:28 PM
"From the Land of Mathematics our journey leads us back to physics. It’s nice to be home again"

1 hour later… 2:32 PM
I'm still kind of unsure about the distinction between $\phi(x)$, $\phi'(x)$ and $\phi'(x)$. Because these things are definitely not the same, and are definitely all used in my lecture notes. If $x\to x'$ is a diffeomorphism, $\phi(x)$ and $\phi(x')$ are just some field $\phi$ evaluated at two different points on the manifold. If $\phi'(x')$ is the pushforward of $\phi$ along the diffeomorphism that's also fine, the thing that's bothering me is $\phi'(x)$.
Would $\phi'(x)$ be the pushforward of the map $\phi$ evaluated at the point $x$?
Maybe this is way simpler than I am making it seem. I want to interpret the phrase "the transformed field" as meaning "the pushforward of the field", is this correct?
So $x$ and $x'$ are just two different points on the manifold and $\phi$ and $\phi'$ are the map (the field) and the pushforward of the map. Both of which can be evaluated at arbitrary points
actually now that I type it out, that sounds reasonable

1 hour later… 3:56 PM
∆ (∆G) = ∆V ∙ ∆P
I know dg=vdp how its derived
But how to derive the former Might help if you explain what you've written a bit more, what is the context and what are the variables? This was the question and solution 4:27 PM
@bolbteppa is $\mathbb{R}^{1|0}$ isomorphic to $\mathbb{R}$
I am starting to have doubts 4:43 PM
Or do you need to cut off the soul of the number for it to work
Algebra seems to work the same but otoh $1$ and $1 + \theta_1 \theta_2$ aren't the same number why are there grassman coordinates at all in $\mathbb{R}^{1|0}$ The commuting and anticommuting reals are just a split of the Grassmann algebra
ie if you have a Grassmann algebra of basis $1, \theta_1, \theta_2, ...$, then the even part of that algebra commutes and the odd part doesn't
Henceforth they are called the commuting and anticommuting real numbers
but otoh it's not 100% clear how I'm supposed to go from that to a physical spacetime where coordinates are regular numbers
Do I just truncate the Grassmann part of every coordinate? 4:59 PM
oh. I would have thought that $\mathbb{R}^{1|0}$ is defined just to have a single bosonic coordinate and nothing else that is why I hate super symmetry, yes
on the other hand, a lot of physicists also seem to assume that they are just $\mathbb{R}$
There does seem to be a mapping from $\mathbb{R}^n_c \times \mathbb{R}^m_a$ tp $\mathbb{R}^m$
But otoh the supercoordinates are usually defined as $(x, \xi)$
or somesuch
what is $x$ and $\xi$
Are we supposed to work with supernumbers for a while and then map it down to the reals in the end for actual values When two discs are rotating with let say angular velocity w, and then we bring one of the disc is brought in contact with the other disc, then $friction$ will act but I am not getting because of which $normal reaction$ is this friction acting and if we want to calculate the friction how can we? But otoh, the space of smooth functions on $\mathbb{R}^{n|m}$ is $$C^\infty(\mathbb{R}^{n|m}) = C^\infty(\mathbb{R}^{n}) \otimes \bigwedge^\bullet (\mathbb{R}^m)^*$$
Why is it just $\mathbb{R}^n$
mb I should make a topic what would you expect it to be if not $\mathbb{R}^n$ 5:20 PM
idk, what if I have a constant function $f(1) = \theta_1 \theta_2$
or something idk
or is that part of the right side?
idk
Hm, seems like it could be?
But why $\mathbb{R}^m$
What does the dimension of the anticommuting part has to do with it here @Charlie I don't mean Riemann geometry, but a more general geometry, like Riemann-Cartan geometry. In this kind of general geometry, if the manifold is parallelizable, you can choose a global frame with vanishing connection, then you have a vanishing curvature. This is the teleparallel gravity.
but my question is then the tensor, curvature, torsion, and nonmetricity, depends on the choice of frame. For example, if we choose a holonomic frame, the torsion always vanishes, but if we choose a frame which is not holonomic, the torsion can have a novanishing value. And if on a parallelizable manifold, we choose a global orthonormal frame with vanishing connection, the nonmetricity vanishes, but if we choose a frame which is not orthonormal, the nonmetricity can have a novanishing value.
isn't this contradictory because the tensor, curvature, torsion, and nonmetricity, shouldn't depend on the choice of frame. They should depend on the choice of connection only. Or am I wrong?
because connection, which is not a tensor, is also frame-dependent. 6:01 PM
I don't know anything about teleparallel gravity sorry 6:14 PM
@Slereah I don't even know what $\mathbb{R}^{1|0}$ is :p page 1 here says it's completely legitimate to ignore all of these issues, at the end of the day the formal definitions are just a game to end up with whatever is done the normal way, I can't imagine the bundle/sheaf interpretation of local fermionic variables and black holes related to them, but the normal way will get there without thinking @bolbteppa Bit of a copout though! You're assuming smooth manifolds are the be-all end-all when they themselves are special cases I mean yeah, but also there is a methodology that uses that
I assume there is some bridge between the two A lot of it is a game to try to deal with the fact functions are single-valued by definition and we end up with a lot of multi-valued structures that end up using single-valued functions so sorting that out is a real mess but at the end of the day it's like an allegiance to writing computer codes in some old arcane language vs. something easy like C++ (where going back to axiomatic set theory for everything is going to the circuits 1's and 0's :p) I'm not sure C++ is the allegory you want for "something easy" 6:20 PM
Yeah I don't know a better way to say it :p Still, the supernumbers show up in enough papers and books that I'd like to know a bit more about them It's a 50:50 chance I'll be able to make sense of what super-Cartesian space is :p I'm not sure that "$\mathbb{R}^{p|q}$ is just the supermanifold of $\mathbb{R}^{p|q}$" is the insight that I need
I mean I guess that technically, you could probably work backward to show from the ring of smooth functions that the corresponding atlases are over whatever, but
not sure it's the most pedagogical way Super-Cartesian space begins from an algebra of functions on $\mathbb{R}^{p}$ rather than just $\mathbb{R}^p$, what the heck It doesn't
That's just how nlab does it
There are other ways
But they are also unpleasant
I'm not even 100% sure why the super algebra of functions is restricted only to Grassmann algebras of order $q$
Considering that a supernumber is of infinite order
I need a man
A man who understands this
some sort of SUPERMAN
I'm not even sure what $C^\infty(\mathbb{R}^{(p|q)})$ means
From the looks of it it doesn't map to $\mathbb{R}$, so what does it map to
or is that just how it's defined 6:41 PM
I can visualize a manifold, what does a supermanifold look like 6:51 PM
same but super
I don't think it's visualizable
$\Lambda_\infty$ is infinite dimensional
there's some process at some point where you're supposed to project all the weird coordinates onto $\mathbb{R}^n$ at some point I think
According to some paper the full Grassmann algebra is "an infinite dimensional Frechet-Grassmann algebra over C that is an associative, distributive, and non-commutative ring with degree, which is endowed with topology""
And the (anti)commuting numbers are its degrees
hm
"Setting up such theories in a proper geometric framework was a bit of a problem, because one was forced to work either on a space (like superspace) where no proper differential calculus was established, or on a "supermanifold" (like those of Konstant and Batchelor) where all the fields are commuting.
The definition by Rogers of $G_\infty$ manifolds seems able to bypass both these shortcomings in physical application, because these are actually Banach manifolds, and the natural fields on them are Grassmann valued. So, anticommuting variables and fields can be treated on the same ground as t
at least I guess I found the historical reason for this mess 7:11 PM
"Informally speaking, a supermanifold is a manifold in which the coordinate functions are smooth functions of the usual coordinates as well as the so-called odd variables.

The simplest example of this is $\mathbb{R}^p$, on which the coordinate functions form the algebra $\mathcal{C}^{\infty}(\mathbb{R}^p) \otimes \mathbb{R}[\theta_1,..,\theta_q]$ where $\theta_j$ are odd variables that are anticommuting...

Such a space is denoted by $\mathbb{R}^{p|q}$, and the general supermanifold is obtained by gluing spaces that locally look like $\mathbb{R}^{p|q}$..." I know the definition from locally ringed space
I just want to know the atlas definition and how it relates to it
this paper seems interesting for it "While this definition imitates that of smooth manifolds with obvious variants in the analytic and holomorphic categories, there is a striking difference: the odd variables are not numerical in the sense that they all have the value $0$. So they are more subtle, and a supermanifold is more like a scheme of Grothendieck on which the rings of the structure sheaf have nilpotent elements; indeed, any odd element in the structure sheaf of a supermanifold is nilpotent.
So a supermanifold is a generalization of a manifold at a fundamental level. However, the techniques for studying supermanifolds did not have to be freshly created; one could simply follow the ideas of Grothendieck's theory of schemes. Supermanifolds are more general than schemes because the coordinate rings are not commutative but supercommutative, a mildly noncommutative variant of commutative rings...
If we drop the smoothness requirement in a supermanifold, we obtain a superscheme that is the most general geometric object yet constructed."
So basically
If you want to follow formalism, the theory of Grothendieck schemes is probably the place to go possibly the original paper for supermanifolds
although apparently not, he already talks about it as though they were a thing
I think in the olden days they were called the "superspace" The first time superspace was set up was by Salam and Strathdee and in those papers they set up things like covariant derivatives etc... and commented that there should be some underlying geometry of their superspace and even comment on things like what the line element should look like etc... this was done the physicist way and so the Rogers paper is one of a few around that time trying to formalize it, the references in that paper even have 'to appear',
but even before these supergravity was set up the physicist way also
So this paper is probably a brilliant starting point for the formal approach doesn't help that there's apparently a variety of definitions for what a supermanifold is
some of which aren't equivalent
The definition as a manifold with a graded sheaf over it is due to Kostant, apparently 7:28 PM
Section 6 shows how her supermanifold reproduces the first definition of a superspace which is good So I guess that's the two things to connect "In curved superspace, as opposed to the homogeneous superspace of rigid supersymmetry, the $\theta$'s are anticommuting coordinates but not Lorentz spinors.

The action of the Lorentz group is defined on the tangent module at each point, and it is the odd part of each tangent vector which is a spinor." "Given an $(m,n)$-dimensional Kostant graded manifold $(X,A)$, it is possible to construct an $(m,n)$-dimensional $G_\infty$ supermanifold $Y$ over $B_L$ (which is in fact an $H^\infty$ supermanifold) such that $Y_B = X$ and the sheaf $A$ is isomorphic to the sheaf $H^\infty \circle \pi^{-1}$ "
Hm
Oh god That's obscene haha
It's good that you can trace these differences back to assumptions about the underlying topology Let's take a look at the original Salam paper 7:40 PM
He doesn't define superspace in the first one or two "In a recent article Wess and Zumino have invented an interesting new symmetry."
That should be close enough to the origin :p I don't know if he even uses the word superspace rather talks about superfields a few papers in I'll skim it a bit at least
Always interesting This one is like a summary of all the early ones thx 7:47 PM
The point of that table is (E) is the most general no I mean, it's the one the author wrote
Suspiscious
Also it was 50 years ago so who knows if that's the current model Yeah but it's taken as the main one or one of the two main ones usually (I'd say it's worth checking whether when people talk about it as one of the two main ones they ignore this table) also the modern versions I see tend to be sheaf based, so either that's wrong, or there is a sheaf version of that model Sheaf is part of definition A
I have to say a table like that is very convincing in terms of telling you the differences are usually small differences at the end of the day I dunno
According to the paper some of them aren't very useful
ie some of them don't allow fermions to exist 7:53 PM
"However, there is no obvious physical reason why superspace should be restricted to being a supermanifold of this type, or to being a DeWitt supermanifold. On the contrary, it seems very desirable to consider the full class of $G^{\infty}$ supermanifolds, admitting as it does the possibility of nontrivial topology in the fermion sector." could b
I will weigh in my fantastic opinion once I understand what the hell is going on I can see that if I read this then I'd be able to see why the other ones look superficially different e.g. "A crucial difference between the supermanifolds defined in this paper and those of Batchelor and DeWitt is that Batchelor and DeWitt both use a coarser, non-Hausdorff topology on $B_L^{m,n}$" probably makes things look different
"It is simpler than these definitions, but also much broader in scope, admitting a wider class of topologies, including a topologically interesting contribution from the "odd" part of the supermanifold, whereas all the other defi-nitions are essentially trivial in the $\theta$ sector. Also, it allows for the case where the Grassman algebra is infinite dimensional." The original Berezin paper isn't online >:|
oh wait
it is, but...