The h Bar

Reading old math is hard to relate to modern math because old math is almost entirely in coordinates and with explicit equations

None of those abstractions

this is good, old-fashioned analysis

no bullshit

(some bullshit)

the book seems very respectable up to this point

I'm only 18 pages in, but hey

3:32 PM

what does the vector space V over a field F mean? I am quite confused.

@CaptainBohemian you come here talking about superconformal field theory but don't know what a vector space is?

I don't know what a vector space over a field mean, not vector space.

The multiplication of a vector is defined over the field $F$

@CaptainBohemian it's a module over a field $k$

$F$ is usually $\Bbb R$ or $\Bbb C$

For instance if you have a vector $v \in V$, $a v$ is also a vector for $a \in F$

($^*\Bbb R$ is the best field, of course)

Or $\Bbb R_\rho$

3:43 PM

so that field F is just the space of coefficients of vectors in V? like if there is a basis for V, we can expand any vector in V through the basis vectors with the expansion coefficients in the field F.

It's just an abelian 'group with operators' where the operators come from a field

Group with operators

In abstract algebra, a branch of pure mathematics, the algebraic structure group with operators or Ω-group can be viewed as a group with a set Ω that operates on the elements of the group in a special way. Groups with operators were extensively studied by Emmy Noether and her school in the 1920s. She employed the concept in her original formulation of the three Noether isomorphism theorems. == Definition == A group with operators (G, Ω {\displaystyle \Omega } ) can be defined as a group G together with an action of a set Ω ...

'a group with operators can be defined[3] as an object of a functor category GrpM where M is a monoid (i.e., a category with one object) and Grp denotes the category of groups.'

No problem

@CaptainBohemian kinda yes

Anonymous

@CaptainBohemian The "field" thingy is built in the very definition of vector space when you say that it is a set closed under finite addition and scalar multiplication. The scalars can belong to any field.

Oh no

When Grothendieck appears in your analysis, you done fucked up

help

3:59 PM

RIP

I think if it just calls that field F a coefficiet group, I would have understood what it means easier.

But a field isn't just coefficients

@CaptainBohemian the coefficients could be a bunch of matrices or a bunch of polynomials etc... but whatever they are, for a vector space they have to be 'nice' and behave like, say, the real numbers do, where you can do the usual easy algebra, and you formalize this by saying the coefficients live in a field. If you want to allow for more complicated coefficients, you generalize to module theory, or more generally a group with operators

@bolbteppa matrices generally do not commute, so they don’t form a field.

'could'

Apparently Cayley-Hamilton is less out-of-nowhere if you view it as part of the theory of polynomials with matrix coefficients solitaryroad.com/c152.html

What is a Smith Normal Form

ACuriousMind

4:25 PM

@JohnDoe if it is explicitly about the code it's likely off-topic here. You could try Computational Science.

@skullpatrol I have no idea why the one-box does that

@bolbteppa so only when the coefficients live in a field does the space form a vector space?

Yes, assuming your 'vectors' are vectors, i.e. they satisfy the obvious axioms of a vector space

5:18 PM

cayley-hamilton seems sensible to me when I note that $A-\lambda_i I$ kills the eigenvector $v_i$

so $\prod_{i=1}^n (A-\lambda_i I)$ should annihilate the subspace spanned by $\{v_i\}_{i=1}^n$

So if those eigenvectors are a basis for your vector space, then that matrix polynomial annihilates every vector

and so forth

(not going to assume that's a rigorous proof but the heuristic seems sound)

Swapnil Das

5:42 PM

Could anyone here recommend me some Undergraduate level thermodynamics books?

Anonymous

@SwapnilDas Almost all the BSc. Physics books have the same syllabus. If you want a book focused on thermodynamics you can check Reif.

Swapnil Das

Sure, thanks!

hurray

I found that paper on tensor distributions again

Swapnil Das

@Slereah I donno what you are talking about but please link the paper :P

@Semiclassical I want to think about it that way, but what if it doesn't have any eigenvalues, e.g. a real rotation, and you are not allowed complex numbers, seems like eigenvectors are missing the point and the whole Nakayama Lemma aspect is probably more important :(

5:51 PM

journals.aps.org/prd/abstract/10.1103/PhysRevD.36.1017

I think that tetrad action may be the Palatini action

Tetrad action isn't the Palatini action

Cartan formalism (physics)

I think there is a Palatini formalism and Palatini action which are different?

This page covers applications of the Cartan formalism. For the general concept see Cartan connection. The vierbein or tetrad theory much used in theoretical physics is a special case of the application of Cartan connection in four-dimensional manifolds. It applies to metrics of any signature. (See metric tensor.) This section is an approach to tetrads, but written in general terms. In dimensions other than 4, words like triad, pentad, zweibein, fünfbein, elfbein etc. have been used. Vielbein covers all dimensions. (In German, vier stands for four and viel stands for many.) For a basis-dependent...

Like 40% sure one side of my action is just that

0

Q: Tetrad Form of Einstein-Hilbert Action Help

I have seen it the claimed that the Einstein-Hilbert action can be written in terms of a tetrad $e_{\mu} \, ^a$: \begin{align} S &= \int d^n x \, e R(e_{\mu} \, ^a, \omega_{\mu a} \, ^b (e)) \\ &= \int d^n x \, e (T_{ca,} \, ^a T^{cb,} \, _{b} - \frac{1}{2} T_{ab,c} T^{ac,b} - \frac{1}{4} T_{a...

@bolbteppa Why would you not be allowed complex numbers?

That's an unnatural thing to do

How can there be SHM when there are two springs of different spring constants connected in parallel? It's counter intuitive because for even a small displacement both springs will exert different forces...

6:08 PM

You don't need $\mathbb{C}$ bruh

It's too complicated, or complex rather

Anonymous

@Abcd Are you worried about the rotational moment generated due to different values of the forces ?

@bolbteppa I don't think C is complicated

Why would you think that?

Anonymous

That doesn't matter. The horizontal motion is still SHM.

@Blue Yes, but what about the different values of forces?

You get a clean proof of Cayley-Hamilton over algebraically closed fields

6:10 PM

It's like one spring is pulling faster

and the other slowly

Hritik

Considering the planetory model of atom and disregarding any EM energy radiation, would atom still collapse ? It would not if electrons had a tangential velocity (same as earth) but how do they get that tangential velocity to begin with ?

Anonymous

@Abcd What about it? I don't understand your confusion. Due to the different values of the forces there will me a rotational moment, yes. But as the equations tell you, the restoring force is still proportional to the displacement $x$ from mean position.

Anonymous

Did you write down the equations?

@Blue Yeah, that's easy..$F_1 = k_1x$, $F_2= k_2x$

2 mins ago, by Abcd

It's like one spring is pulling faster

2 mins ago, by Abcd

and the other slowly

This is causing confusion^

Anonymous

What's the net restoring force?

6:12 PM

$(k_1+k_2)x$

Anonymous

Right. So that is still proportional to $x$

Anonymous

So it is SHM

Oh, I see.

Hritik

Anyone on my question please ?

@Blue My teacher gave this reason for no torque: "the block is a point mass". It wasn't convincing...was he even right?

Anonymous

6:16 PM

@Abcd Well, you can guess someone who says "block is a point mass" isn't in their right mind :P

lol, okay.

Anonymous

@Hritik How did electrons start revolving? I don't think that has any answer in classical physics (Bohr's Model)

6:51 PM

@Blue For SHM in series circuit can we say that there's basically SHM of "all the springs except the last spring+ block" system?

Anonymous

@Abcd What do you mean by "SHM of all the springs" ? Usually when there are some springs connected in series, there's a block attached at the end, which if you pull or push slightly, shows SHM motion

Okay, you are right.

Also, I don't think SHM is possible when the spring has mass, is it?

Anonymous

@Abcd Why not? All springs in real life have mass and almost all vibrations in nature are SHM (for small amplitudes and a small range of time)

Then why focus on "ideal (=massless)" springs only?

Anonymous

@Abcd That's just to make things simpler. If springs were considered to have mass then you would have to take into account additional factors. But over a small range, it would still be SHM

7:01 PM

@bolbteppa I guess the elementary case of interest would be when the characteristic polynomial is $p(\lambda)=\lambda^2+1$

Okay, it's given nowhere therefore I was unsure. If you have any article on that, please forward to me.

so take $A=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$

so a real rotation by 90 degrees, consistent with your suggestion

back in a bit

Anonymous

7:42 PM

@Abcd As such I don't have any article. But you can have a look at this

@Blue I solved a strange question in which the amplitude position was the centre of oscillation of the block...

Anonymous

@Abcd Let's talk tomorrow. I'm going to sleep now

8:11 PM

hmm. $A-\lambda I=\begin{pmatrix} -\lambda & 1 \\ -1 & -\lambda\end{pmatrix}$

So $(A-\lambda I)\begin{pmatrix} a \\ b\end{pmatrix}=\begin{pmatrix} b-\lambda a \\ -a-\lambda b\end{pmatrix}$

which to vanish needs $a=-\lambda b=-\lambda^2 a$ for some $\lambda$---no go

at the same time, one definitely has $A^2+I=0$

One can either explain that by allowing $\lambda=\pm i$, or by just verifying it

my thinking is that, if your characteristic polynomial contains complex roots, you can still factorize it into irreducibles over the reals of degree at most 2

so in addition to finding subspaces which $A-\lambda I$ kills, one also needs to consider subspaces killed by $(A-\alpha I)^2+\beta^2 I$

@bolbteppa so that makes things more complicated, but still within the spirit of what I was saying above

8:33 PM

@Semiclassical yes

the phrase to google for is the minimal polynomial of the matrix

right

Friedberg-Insel-Spence has a super nice section about it

stuff i'd have learned in a proper linear algebra course

Minimal polynomial (linear algebra)

right after the Jordan canonical forms

In linear algebra, the minimal polynomial μA of an n × n matrix A over a field F is the monic polynomial P over F of least degree such that P(A) = 0. Any other polynomial Q with Q(A) = 0 is a (polynomial) multiple of μA. The following three statements are equivalent: λ is a root of μA, λ is a root of the characteristic polynomial χA of A, λ is an eigenvalue of matrix A. The multiplicity of a root λ of μA is the largest power m such that Ker((A − λIn)m) strictly contains Ker((A − λIn)m−1). In other words, increasing the exponent up to m will give ever larger kernels, but further increasing t...

yeah. i know the phrase

I'm just used to hermitian matrices

8:35 PM

@Semiclassical yeah, well

Jordan blocks exist, too ;-)

ugh

@BalarkaSen crap I've put off this h-principle stuff long enough

yeah

try e.g. the matrix representation of the annihilation operator in the Fock basis when you truncate to a maximal photon number

@0celo7 just do it

8:36 PM

coherent states get gone

@BalarkaSen not EM

you get a single eigenvalue

with a single eigenvector

the insane paper about manifolds of metrics and h principles

I'm getting a weird thing where my numerical eigenvalues of a matrix $A$ aren't making sense when I shift it by $A+\lambda_0 I$

they should just shift by $\lambda_0$ as well

yes

A: AQFT: Can test functions obey the Klein-Gordon equation?

8:39 PM

4

No, they cannot. There is no non-vanishing smooth KG solution with compact support. Let $\psi$ be a compactly supported solution (I assume $m>0$, the massless case is a bit more complicated) and $T_{ab}$ the associated stress energy tensor. The integral of $T_{00}\geq 0$ over a spacelike Cauchy s...

How can you fix the hypersurface away from the support

I can't think of any foliation that would allow that

@0celo7 is the main theorem of any interest to me?

@BalarkaSen No.

Where is Moretti when you need him

what is the main theorem

@ValterMoretti please halp

8:41 PM

@BalarkaSen small deformations of metrics to negative scalar curvature measured in some topology generated by a background metric

you would not like it

oof

[Grl] M. GROMOV, Partial Differential Relations, Springer (1986).
[Gr2] , Stable mappings of foliations into manifolds, Math. USSR-Izvestija 3 (1969),
671-694.
[Gr3] , Structures Metriques pour les Varietes Riemanniennes, Cedic Nathan, Paris
(1981).

RIP

RIP in Gromov

@EmilioPisanty instead I’m finding that there’s an an anomalous eigenvalue of $A+\lambda I$ near zero, as though $A$ had an eigenvalue near $-\lambda_0$. But doing the spectrum of $A$ directly shows no such eigenvalue

Wtf

@Semiclassical ¯\ _(ツ)_/¯

8:45 PM

“Forget it, Dave, it’s Mathematica.”

9:18 PM

@Semiclassical my problem is "The Cayley-Hamilton theorem can be proved directly, and it is true not just for endomorphisms of a vector space, but for endomorphisms of any finitely-generated module over any commutative ring" math.lsu.edu/~madden/M7210fall2012/… i.e. it works without even real numbers let alone complex numbers saving examples like $x^2 \pm 1$, I'd say it's got to be like 'Euclid's Lemma' or 'congruences' or X abstracted

But I like the whole eigenvalue commutative factors thing

@0celo7 Do you get Moretti's proof here

although i'm thinking my issue might be that I'm thinking compact support on $\Sigma$ when maybe they're talking about $M$

9:46 PM

Ah yes apprently that was it

Although I am now wondering if KG solutions can be of compact support

Due to the whole Malament business

If the wavefunction is of local support then it is localizable and should be 0

10:06 PM

@Slereah where?

A: AQFT: Can test functions obey the Klein-Gordon equation?

4

No, they cannot. There is no non-vanishing smooth KG solution with compact support. Let $\psi$ be a compactly supported solution (I assume $m>0$, the massless case is a bit more complicated) and $T_{ab}$ the associated stress energy tensor. The integral of $T_{00}\geq 0$ over a spacelike Cauchy s...

This one

But he answered

Turns out I mistook what he was saying

Sir Cumference

Hi

A: Lower limit of the size of the Universe? (WMAP)

Been trying to find out if anything special happens for compact support solutions of Klein Gordon

Alas not much info

Although maybe the issue doesn't lie with the solution but with the inner product

Sir Cumference

Ah, I finally understand what Wikipedia might be referring to when it says the Universe must be at least 156-554 Gly in diameter

According to this post

7

From the Friedmann equations, you can derive that $$ \dot{R}^2 - \frac{8\pi}{3}G\rho R^2 = -k c^2, $$ where $\rho$ is the total density of the universe and $k$ is a constant that determines the shape of the universe: $k=-1,0,1$ for an open, flat and closed universe, respectively. If the universe ...

~145-538 Gly would be the diameter of the Universe assuming there is a minimal positive curvature, below a measurement precision

Which is still a possibility given our measurements from Planck and WMAP

10:36 PM

@Slereah this is well known energy inequality stuff

@0celo7 what is

@Slereah you can't have compactly supported solutions for the wave equation

@0celo7 ah yes, that makes sense

Malament is safe for another day

Why not?

@Slereah because of what Valter said

also because of uniqueness theorems for such things

I mean initial data that is compactly supported

10:40 PM

oh, that's fine

you can evolve any Schwarz class data

Is it, though

@Slereah maybe not for Klein-Gordon, but for sure for the wave equation

For a relativistic point particle it means the probability presence is 1 in the compact support

Which might be verbotten

Although I need tp check what the actual probability is

Since KG has that weird inner product

I'm not sure what you're asking for

Well, for let's say the Klein-Gordon theory of a single relativistic quantum particle, the wavefunction is just gonna be a solution to the KG equation

Which can be a function of compact support on $\Sigma$

And since the wavefunction doesn't disperse FTL, it will always be of compact support

Which means (possibly) that if you perform a measurement $E_\Delta$ of the presence of the particle in a compact region $\Delta$, then $E_\Delta \Psi = 1$

But

10:47 PM

Moretti is saying that you can't have $\psi\in C_c(\Bbb R\times\Sigma)$

Yes, obviously

in particular, $\psi$ is nonzero on each $\Sigma_t$

it can have finite spatial extent

But that's not what worries me

I don't know what's worrying you then

Because according to Malament's theorem, you should have $E_\Delta = 0$ for all regions

10:49 PM

...what?

But maybe that's a subtlety of the theorem because I'm not 100% sure if this is all related to such a system

loocsieulb

sup my dudes

wait @0celo7 did you want me to take a picture of him

of course

Hm

loocsieulb

how badly do you want it

10:50 PM

Maybe the issue I'm having is

loocsieulb

because i'd have to sprint AF

One of malament's requirement is localizability

But

i'm not sure if this is for some states, or for every state

I should look up the original proof

It's a pretty broad theorem so he doesn't use Hilbert space language much

since it has to apply to a lot of quantum theories

link.springer.com/chapter/10.1007/978-94-015-8656-6_1

that's the one

it's one of those theorems that just keep bothering me

god, is that a book typed in word?

"An assignment to each spatial set $\Delta$ of a projection operator $P_\Delta$ on $\mathcal H$"

And the localizability condition is $P_{\Delta_1} P_{\Delta_2} = P_{\Delta_2} P_{\Delta_1} = 0$

"The "localizability condition" captures the requirement that the particle cannot be detected in two disjoint spatial sets at a given time"

And Malament's theorem is that given reasonable conditions, $P_\Delta = 0$, the zero operator

Which doesn't sound right

I think I need to post that question tomorrow

For an initial condition $\psi(x, 0) = f(x)$, the wave equation solution is $f(x - t)$, right?

or something similar

@Slereah in 1-D, the solution is given by en.wikipedia.org/wiki/D%27Alembert%27s_formula