The h Bar

danielunderwood

02:34

For anyone interested in the question I asked about other number systems, the representation evidently doesn't change anything.

3

A: What would a base $\pi$ number system look like?

No problem will became easier to solve by choosing a different base, because they depend on the numbers, not on their representation. The solution of some numerical problems might look nicer, for some definition of nice, but that's as far as you'd go.

John Rennie

06:02

→ 1 message moved to trash

Slereah

08:24

morning

mathvc_

10:07

Hey. Can anyone explain it to me? It is claimed that every Einstein solution with cosmological constant in vacuum is locally isomorphic to Minkowski, deSitter, or adS?

David Z

@Axoren Did you get that edit? If not, I suggest including a link to the question here. It'll make it easier for people to see if they can meaningfully help you.

mathvc_

10:46

any help for my question?

bolbteppa

11:04

"The cases of spacetime of constant curvature are de Sitter space (positive), Minkowski space (zero), and anti-de Sitter space (negative). As such, they are exact solutions of Einstein's field equations for an empty universe with a positive, zero, or negative cosmological constant, respectively."

Anti-de Sitter space

In mathematics and physics, n-dimensional anti-de Sitter space (AdSn) is a maximally symmetric Lorentzian manifold with constant negative scalar curvature. Anti-de Sitter space and de Sitter space are named after Willem de Sitter (1872–1934), professor of astronomy at Leiden University and director of the Leiden Observatory. Willem de Sitter and Albert Einstein worked together closely in Leiden in the 1920s on the spacetime structure of the universe. Manifolds of constant curvature are most familiar in the case of two dimensions, where the surface of a sphere is a surface of constant positi...

@mathvc_ e.g. chapter 13, esp. section 13.3 here blau.itp.unibe.ch/newlecturesGR.pdf

@Slereah these notes cds.cern.ch/record/880599/files/CM-P00060426.pdf (the appendix) help get the Virasoro algebra commutators II.49 in Scherk the way he does it, am off by a factor if $1/2$ :(

mathvc_

@bolbteppa Thank you very much. That's nice! Actually it helps me to understand for maximally symmetric spaces, there are only three types of solution. But still not sure why all solutions in 2+1 dimension are from these three types?

bolbteppa

11:24

Slereah

12:08

@bolbteppa so anyway

What is a gauge transformation

why doesn't translation invariance not count as a gauge transformation

bolbteppa

I think because it's global not local

If you made translations local then you'd get extra terms when you differentiate and so a typical Lagrangian/action wouldn't be invariant, would represent shifting a system by different amounts at each point

Secret

what would a local translation invariance physically mean, like having the new action and the old action be differ by some fixed function say $e^x$...?

Slereah

@bolbteppa how do you express that mathematically

I mean gauge in the large sense

not just the principal bundle ones

bolbteppa

3

A: Local translations in curved spacetime

Ref. 1 defines a local translation on spacetime $M$ as a diffeomorphism. Note that the words local and global in this physics context mean point-dependent (=$x$-dependent) and point-independent (=$x$-independent), respectively. [Be aware that mathematicians in other contexts typically use the w...

Slereah

I mean like

For the simple system $L = (\dot x - y)^2$

$x \to x - \phi$ is a gauge transformation

But why isn't $x \to x + a$ for $\dot x^2$

Is it because it lacks constraint equations

bolbteppa

12:25

@Slereah $x \to x - \phi, y \to y - \dot{\phi}$ is the gauge transformation

Slereah

I know yes :p

I'm guessing the gauge constraint is more the defining characteristic than the gauge transformation

bolbteppa

The constraint videos give that example exactly

Slereah

I know!

bolbteppa

Because $\phi$ depends on $t$ it's local

Slereah

You could write it as $x + a(t)$ :p

bolbteppa

12:29

He also shows how the solutions are $x(t) = \alpha + \beta t + \frac{\gamma}{2} t^2 + \phi(t)$, $y(t) = \beta + \gamma t + \dot{\phi}(t)$ so if you shift $\phi$ by a constant that gets absorbed into the $\alpha$ constant so constant transformations are irrelevant I think

Slereah

Or is the point that the gauge function must be completely arbitrary

bolbteppa

The system being constrained leaves undetermined a time-dependent function in the solution which is completely arbitrary and picking one (that leaves the Lagrangian invariant) 'fixes a gauge' where we're free to pick anything leaving the Lagrangian invariant

Slereah

Alright

That does make sense

bolbteppa

So $A_{\mu}$ in EM is arbitrary up to $A_{\mu}' = A_{\mu} + \partial_{\mu} \Lambda(x)$ just as $x(t)$ is arbitrary up to $\phi(t)$ in the above example

Slereah

What about reparametrization invariance tho

That function isn't entirely arbitrary

bolbteppa

12:41

That is also a gauge symmetry I think

Basically, gauge transformations are local transformations that leave the action invariant, this unavoidably forces constraints on the action, but they have to be constraints that ensure the solutions are left undetermined while also leaving the action invariant, thus from Dirac's constraint theory these constraints simply must be first class constraints, because second class constraints allow us to fix the Lagrange multipliers,

while first class constraints do not, thus one can re-define gauge symmetries as a specific choice of Lagrange multipliers in the constrained Hamiltonian

Scherk makes a comment that choosing the multipliers is equivalent to the usual language of choosing a gauge

The videos also emphasize this in video 2 near the end

('undetermined up to leaving the action invariant' is probably a better way to say it)

Hmm:

"Here's an example: imagine you had a round bowl with a marble rolling around in it. You notice that if you rotate the bowl, the motion of the marble released in the same initial conditions is unchanged. From this, you conclude that as far as the physics is concerned, the problem is rotational symmetric. Now suppose the bowl is also brightly coloured with many paints.

You notice that these colours don't affect the marble's motion: you can repaint the bowl and still get the same marble motion. Is this the same as the rotational symmetry? No. You can paint the bowl arbitrarily: whereas for a rotation, you had to do the same thing to every point, you can repaint each point of the bowl how you like. As far as the physics is concerned, the bowl's colour is an arbitrary local symmetry: a gauge symmetry.

Whereas rotating the bowl gave us a new physical configuration that just reproduced the same physics, repainting the bowl actually leaves it the exact same—because the paint job is redundant information for describing the system, as far as the physics is concerned."

Turgon

what reparametrization invariance are you referring to?

bolbteppa

I guess the colors are like coordinates on the bowl, doesn't matter how you paint/label things to describe the motion in terms of colors

Slereah

Reparametrization invariance in general

Turgon

12:56

there's a specific one in hqet to ensure lorentz invariance so i was a little confused..

you're talking about field redefinition, right?

bolbteppa

Nambu-Goto action reparametrization invariance is a great example, action is invariant under reparametrizations, this unavoidably forces the existence of constraints on the action, we immediately expect that they will be first class, the canonical Hamiltonian is zero, this already eliminates the possibility of the primary constraints being second class, the only question is whether their PB algebra produces secondary constraints (which could then cause issues)

turns out they don't, and you get what is called the Virasoro PB algebra, which is said to represent this reparametrization invariance

(Wait the canonical Hamiltonian being zero does not necessarily guarantee it, actually doesn't matter, question is whether they are weakly zero or not)

bolbteppa

13:13

@Slereah 'Mathematically, a gauge is just a choice of a (local) section of some principal bundle. A gauge transformation is just a transformation between two such sections.' I think this is just a formal way of saying a gauge transformation is local since every point of the section need to be changed and each point along a section can change to a new section in it's own way

^to a point of the new section in it's own way^

Semiclassical

Here’s a question I don’t know the answer to: What is the smallest/narrowest aperture anyone has ever done in a diffraction experiment?

Probably I should be thinking not so much in terms of a double/single slit problem so much as, say, scattering from a lattice a la Braggs law

Secret

physicsworld.com/a/…

uh... cannot find anything about smallest aperture, only highest resolution

and this one is not really aperture limited

Semiclassical

13:29

Thinking in terms of “slit width” probably stops making sense once you get down to the atomic scale, heh

Secret

well, it will probably becomes more like "what is the distance between two cluster of atoms"

Semiclassical

Right

Distance between scattering centers

Secret

and it will no longer be rectangular

Secret

13:40

Hi @nitsua60, since you are mod, can you help me to unfroze my Rambles room so I can continue to investigate the $\pi +e$ as recently I accidentally flooded the math chat with my stuff and many users are annoyed of me as a result?

$Mathematics$ Rambles

For all my maths rambles that are not even qualified to fit in...

nitsua60

@Secret Aaaaaaas youuuuuuu wiiiiiiiish

Secret

thanks

Semiclassical

Lol

nitsua60

(sans the implicit declaration of love)

Secret

(Now to install a bot so it autobumps the room every 14 days. Seriously who is that smartpants in SE who think freezing rooms so that not even owners can unfroze them is a good idea)

Semiclassical

13:44

I don’t mind them being frozen, but a more obvious route to unfreezing than “beg a mod” would be nice

Secret

indeed, especially unlike most people, I do reuse old things alot, even if 5 years ago stuff

Part of the reason is ever since I read so many papers about time symmetry and Vaccaro's work, I stop thinking time as linear

old things are as relevant as the present day stuff if you preserve them well

Semiclassical

“The past isn’t prologue. It isn’t even past.”

(I think I’m stealing that from Babylon 5?)

Noooope

Secret

William Faulkner

William Cuthbert Faulkner (; September 25, 1897 – July 6, 1962) was an American writer and Nobel Prize laureate from Oxford, Mississippi. Faulkner wrote novels, short stories, a play, poetry, essays, and screenplays. He is primarily known for his novels and short stories set in the fictional Yoknapatawpha County, based on Lafayette County, Mississippi, where he spent most of his life.Faulkner is one of the most celebrated writers in American literature generally and Southern literature specifically. Though his work was published as early as 1919, and largely during the 1920s and 1930s, Faulkner...

this guy

Semiclassical

Yeah

“The past is never dead. It's not even past.”

Jake Rose

m.imgur.com/2vaVs3x Can anybody give me a hand with this? Maths people aren’t replying

Semiclassical

13:53

There is a line in B5 which riffs on that, but it’s not that one

Jake Rose

I’m having trouble understanding what would be in d let alone A

Semiclassical

14:08

first off, what will the sizes of A and d be? @JakeRose

(And why)

Emilio Pisanty

14:24

@Semiclassical halp

Jake Rose

A must have the size n x 3 and d will be n x 1

Emilio Pisanty

> For monochromatic light, rotational invariance is normally framed by requiring that the complex amplitude of the electric field obey an eigenvalue equation of the form A=B.

Jake Rose

But I don’t know what n will be

Emilio Pisanty

should that obey be obeys?

Semiclassical

@JakeRose That's a starting point. But you know more than that: Except in special cases, the intersection of a plane and a line is some unique point.

Jake Rose

14:26

Yes

Semiclassical

So you need Ax=d to have a unique solution.

if you translate Ax=d into equations, you'll have n equations in three unknowns

how many equations should you have if you want a unique answer?

@EmilioPisanty Hmm?

uh

Jake Rose

3.

?

Semiclassical

@EmilioPisanty how is that an eigenvalue equation :S

@JakeRose yep

Emilio Pisanty

@Semiclassical =P

Semiclassical

so you need A to be 3-by-3 and d to be 3-by-1

Emilio Pisanty

14:27

you'll be able to read the full thing shortly

Semiclassical

@EmilioPisanty I take it that's the question, then :P

ah, okay

Emilio Pisanty

@Semiclassical "by requiring that X obey an equation of the form Y"

vs

"by requiring that X obeys an equation of the form Y"

the former, right?

Semiclassical

ah

Jake Rose

so A are the equations, x the solution and d some constants or whatever?

Semiclassical

yeah, I think that's right

Emilio Pisanty

14:29

thx =)

Semiclassical

@JakeRose no. If you write out Ax=d, then the first component of each side will give the first equation

A isn't giving you the equations as such. It's telling you the coefficients of those equations.

Jake Rose

Oh yes sorry that’s what I meant

Semiclassical

Then yeah, you've got the right idea.

Jake Rose

I’m still quite confused about gett8ng said equations

Semiclassical

The main thing is that those three equations need to be linearly independent

Once you've got three linearly independent equations in the three unknowns, you can formulate that as a matrix equation and you're done

So, where can we get the three equations from?

There's not a lot of sources for equations in this problem, which helps

Jake Rose

14:35

The plane equation

And then each of the line equations?

Semiclassical

that's one

well, how many line equations do you have?

Jake Rose

3

So only 2?

Semiclassical

why only 2 of the 3?

(that's right, but why?)

Jake Rose

Because we only need 3 equations, and we already have another

That doesn’t sound as concrete as it did in my head

Semiclassical

Yeah, that's a bit too slippery

Jake Rose

14:38

Because by that logic the original line equation would be good enough

Semiclassical

Let's put it another way: Why wouldn't it be enough to use the three line equations on their own?

heh, sniped

Jake Rose

They’re all dependent on alpha

Semiclassical

Right. And alpha shouldn't appear in A

Jake Rose

Brb in like 10 min!

Semiclassical

np

Jake Rose

14:47

Why should alpha not appear in A?

Semiclassical

because the unique point should presumably not depend on alpha

alpha is just a parametrization

I guess another way to say it is that, once you know what alpha is, you know what the point is

so including it in there could only be redundant

Secret

I think that new site layout screw up all rss cause none of the new ones are working

Semiclassical

An unexpected change creating unanticipated problems? What a shock.

Secret

The xml code between the old and new version is slightly different as you can see here:

physics.stackexchange.com/feeds/…

physics.stackexchange.com/feeds/tag/quantum-mechanics

Secret

15:11

O nvm, it works now

(changed that to day) Now my room will be prevented from frozen foreverr

Jake Rose

@Semiclassical okay I can see that

Semiclassical

so, we need to have three linearly independent equations in the three unknowns x,y,z

As we noted before, the plane equation will work as one of those

Jake Rose

Yes

Then two of the line equations as in (stuff with x)=(stuff with y) and (stuff with y) =(stuff with z)

Semiclassical

you'll need to clarify that

can you be more precise? you should already have those equations at hand

Jake Rose

I’m on an iPad and can’t use latex you see

Semiclassical

15:24

fair enough

but, if you look at what the problem already had you do

Jake Rose

In the first part of the question it says show bla bla bla

Semiclassical

yeah

Jake Rose

The show that part is what I’m talking about when I say stuff with_

Semiclassical

right

So you've got X = Y = Z, as shorthand, and you want to do X=Y, Y=Z

Why not X=Z?

Jake Rose

I don’t know

I was pondering that myself

Semiclassical

15:28

Well, if you have X=Y and Y=Z, then you can deduce X=Z

So writing X=Z as a third equation is redundant

Jake Rose

Why not X=Y and X=Z but not the other?

Semiclassical

That would work as well

Jake Rose

Ohhh okay I see your point

Semiclassical

The point is that, while the three of them together are not independent

any two of the three implies the third

so you just need to pick two of them and move on :)

more generally, if X=Y and Y=Z then it's also true that aX+bZ = (a+b)Y

Axoren

@DavidZ Yeah, I got the edit. It seems though that the question may be best moved to an engineering SE. If that's the case, I'd rather it migrated than closed, if possible.

Semiclassical

15:31

so you can write out an infinity of other such equations just by picking various a,b

But they'll all be redundant, since they're implied by X=Y and Y=Z

Jake Rose

Okay I’ve got it now

Now how does this matrix help you solve anything?

Axoren

This question is more for my own curiosity than any practical application, because I was envisioning an island that would only appear during a storm when the tide rises significantly.

And wondering how such a thing could be mundanely possible.

Jake Rose

How do these equations tell you the poin t of intersection?

Semiclassical

I guess the broader point is that an equation like X=Y is the equation of a plane, so X=Y and Y=Z is the intersection of two planes. But there's an infinity of ways to represent a line as the intersection of two planes, so you might as well just pick two and be done with it

Well, first off, that's not what they're asking you to do right now :)

All they're asking you to do is put it in the form of a matrix equation; they're not asking you to make that matrix equation useful

Jake Rose

Ah it says it in the question as if it is though

Can we go back to why alpha should not be in A

I’m still confus d by this

Semiclassical

15:34

Well, suppose it were. It should still be the case that Ax=d has a unique solution x, since it represents three equations in three unknowns

at least, it will be unique so long as we consider alpha as fixed

But if alpha appears in there, then there's really no reason to think of it as being fixed. You'd end up with your solution in x being a function of alpha, and therefore varying as alpha changes

Jake Rose

So alpha could be there as long as we consider it a constant

But since it’s just as unknown as the rest it’s just like adding a fourth variable

I see I see

Semiclassical

you can even say what that constant value is: it's alpha = (x-a)/l = (y-b)/m=(z-c)/n (assuming I've got the constants in the right order)

Jake Rose

Semi you’re my fave physicist

Semiclassical

lol

Jake Rose

I get it now that’s great

Semiclassical

15:37

One thing to note is that, if you wanted, you could express the matrix equation as A'v=d' where A' is 4-by-4, d' is 4-by-1, and v = (x,y,z,alpha)^t

Jake Rose

Would that not require a fourth equation?

Semiclassical

Yep. But now you can treat alpha as its own variable, and therefore take alpha = X as an additional equation

Jake Rose

Ohh I get you

Semiclassical

You could even just take the original line equations as such

Jake Rose

Yeah I see

Is that more useful?

Semiclassical

15:39

e.g. x-t*alpha = a

eh. In one respect, yes: you don't have to manipulate the equations as much in order to get into matrix form

The cost, of course, is that you've now got a 4-by-4 matrix A' instead of a 3-by-3 matrix A

So you really just end up shifting the work around

Jake Rose

With matrices I find I just don’t ‘understand’ the stuff. Determinants for example feels ridiculously abstract. Which is why I find this probably the hardest of all topics

Any tips?

Semiclassical

determinants are a bit goofy, yeah

The simplest perspective I know on determinants is geometric

Jake Rose

I get that in 2d but how do you think of it past say 3D?

Semiclassical

Eh, the same geometric idea works. A acts as a linear transformation on n-vectors; if you act A on a unit n-cube, then the image will have (hyper)volume det(A)

Jake Rose

Have you got any good web pages or places where I can read about this more thoroughly?

Semiclassical

15:45

Not off the top of my head.

And this geometric POV isn't really helpful when it comes to, say, the characteristic polynomial of a matrix being det(A-t I).

Jake Rose

No worries.

What would you say the most useful understanding is?

Semiclassical

not sure tbh. If I was learning from scratch, I'd probably say to look at Axler's presentation of it. But that might be too abstract

I mean, the basic idea when it comes to eigenvalue stuff is that you want Av=tv, therefore (A-tI)v=0, therefore A-tI is singular, therefore det(A-tI)=0

and then tedious manipulations :P

enumaris

dang, my RL algorithm is totally failing...now I gotta spend a bunch of time debugging what I did wrong -.-

Jake Rose

I get the maths behind edge values just not sure about intuition

Semiclassical

edge values?

Jake Rose

15:52

Eigen^

Autocorrect is unforgiving

Anonymous

@JakeRose You definitely should develop geometric intuition for eigenvalues: both complex and real. 3blue1brown is a good place to start.

Anonymous

As far as real eigenvalues are concerned, they simply denote the amount of stretching or squishing of eigenvectors (in linear algebra)

Anonymous

And eigenvectors are those which do not get knocked off their span due to the transformation

Anonymous

Complex eigenvalues correpond to rotations

enumaris

not fun...-.-

Jake Rose

15:59

I’ve seen his video, loved the essence of linear algebra series, but I can’t help but think my intuition isn’t complete because it’s more so a presentation of the ideas as to the full rigor of getting there

Semiclassical

Getting a geometric understanding of hermitian matrices is tougher

To the extent that I have intuition there, it's by saying "oh it's just QM"

Jake Rose

What about determinants? Again I think his vide9 is good but I would like a written source to get more rigor

Anonymous

Axler develops determinants beautifully in his last few chapters

Anonymous

Ideally you shouldn't learn determinants until you get through a majority of linear algebra notions

Semiclassical

Does Axler talk about determinants geometrically?

enumaris

16:05

ughhhhhh nonworking algorithms bother me -____-

Anonymous

@Semiclassical He initially talks about it in terms of eigenvalue products and multiplicities but later goes on to prove that positive operators change volume by a factor of determinant, etc. It would be much more better with pictorial explanations but there are a lot of nice Math SE posts which deal with geometric interpretations of determinants...so that suffices

Semiclassical

Sounds right.

Anonymous

One of my most memorable days was when I could make sense of SVD geometrically XD

Semiclassical

lol, nice

enumaris

draw dem support vectors

Semiclassical

16:09

Something about mapping one ellipsoid to another

Anonymous

Yup

Anonymous

Wikipedia actually explains it well

Anonymous

I hadn't noticed it earlier

Anonymous

And iirc Jack Aurizio had a couple of nice posts on it

enumaris

oh wait SVD is singular value decomp...not support vector decomp lol

Semiclassical

16:13

this is a nice picture

sigh. I never understand why some image links work and others don't

Anonymous

Yeah that's a mystery, lol

Anonymous

.svg images behave strangely

enumaris

ain't no image links work

Anonymous

8

A: Change Chat Oneboxing to recognize SVGs

Fair point, and easy enough. Added to the next build. Of course (as with the other image types), this only looks at file extensions, and just assumes if the URL ends with .svg (or .jpg etc.) that it's an image.

Semiclassical

answered Sep 7 '12 at 14:09

...

Anonymous

16:19

Anonymous

Aha

Anonymous

You basically need to modify the URL a bit

Semiclassical

sounds right

Anonymous

You can get it directly from the "Full history" portion here

Anonymous

John Rennie

16:20

Wikipedia being sneaky - the link to the SVG isn't actually a link to the SVG file.

Anonymous

Click on the thumbnail to get the URL

Semiclassical

All of that said, I've yet to really run into a scenario where I've needed the SVD

I'm spoiled by the ubiquity of Hermitian operators in physics :P

Anonymous

I've seen it being used in machine learning contexts...

Anonymous

I guess it's there in Hastie's book

Semiclassical

I have heard of it in the context of QI stuff as well, e.g. Schmidt decomposition (c.f. Wikipedia claiming "The Schmidt decomposition is essentially a restatement of the singular value decomposition in a different context.")

Anonymous

16:25

(What exactly it was being used for....can't recall now)

Anonymous

I need to read the book really :P

Anonymous

I keep postponing

Anonymous

@Semiclassical Makes sense, makes sense

danielunderwood

17:17

asimovinstitute.org/neural-network-zoo

seems neat if you're interested in ML

Oh man I just found out the jetbrains ides have a right click option to search google for the highlighted text...this changes everything

enumaris

17:30

I don't think it's worth it to learn all of those architectures lol

maybe at a high level

danielunderwood

Well I was thinking of at least having an overview of each one. I'm certainly not going to read each paper to implement them lol

It is nice that the paper links are given though. Too many ML blog posts and such with no useful information or references

enumaris

yeah, like Hopfield nets were popular way back when...haven't seen them really used outside of academic curiosity tho

danielunderwood

17:51

You could make HNNs be the next big thing

enumaris

maybe lol

but I can't even get my RL algorithm to work yet

T_T

I'm like 90% sure I implemented it correctly tho...not sure why it doesn't work...

danielunderwood

RL is one of the things that initially got me interested into ML, but I still have no idea how to implement one. I've seen gym, but from the docs it seemed like it was just a library to be able to set up typical problems that could be solved by RL

enumaris

yeah

gym just gives you the environments

you have to implement your own RL algorithms to learn those environments

danielunderwood

Ahh the docs make sense then. When I first heard of it, I thought it was kind of like a RL version of keras

Also have you used TPOT for regression? It seems unhappy with the losses they list for regression in the docs

Rambles

Transcript for

$Mathematics$ Rambles