The h Bar

Slereah

12:07 PM

Also are all causal curves actually Lipschitz

What about Rindler observers

they have unbounded derivatives

Ben Niehoff

@TheRaidersofLasVegas Hard to say. I enjoy Nakahara's book

@JohnRennie Isn't Part III essentially a masters? But I don't know if you get a degree for it

AccidentalFourierTransform

12:49 PM

so, is $\mathrm{Res}$ a vector field over $\mathbb C$?

Slereah

Res?

AccidentalFourierTransform

$\mathrm{Res}(f\mathrm dx+g\mathrm dx)=\mathrm{Res}(f\mathrm dx)+\mathrm{Res}(g\mathrm dx)$

and $\mathrm{Res}(\lambda f\mathrm dx)=\lambda\mathrm{Res}(f\mathrm dx)$

John Rennie

@BenNiehoff as far as I know the part III is unique to the maths department. Arguably it fulfils much the same role as a masters, but it isn't the same.

Slereah

Obviously not if it takes one-forms as arguments

Wait no

That is correct

Vector fields are linear functionals on dual vectors

AccidentalFourierTransform

$\mathrm{Res}\colon T^*\mathbb C\to\mathbb C$ linearly, right

Slereah

12:50 PM

So yes

AccidentalFourierTransform

cool

ACuriousMind

@AccidentalFourierTransform What is "Res"?

AccidentalFourierTransform

idue

ACuriousMind

Why does the residue act on 1-forms and not on functions, and why is there no point at which the residue is taken specified?

AccidentalFourierTransform

the point is a fixed parameter :-P

but yes, Res is defined on forms, not on functions

eg, what is the residue of $1/z$?

is it 1?

what if I take $\omega=2z$

what is the residue now?

ACuriousMind

12:54 PM

Ah, I see what you mean

Yeah, right, since it's defined through integration the invariant formulation should act on forms

AccidentalFourierTransform

^

Slereah

1:07 PM

"Milne and McCrea have obtained the remarkable result that Newtonian analogues exist for all the more important models considered by Robertson"

Man that's a 1934 paper

Pretty old

academic.oup.com/qjmath/article-abstract/os-5/1/64/1576447/…

"We can either modify our geometry in order to retain $\delta \int ds = 0$ as the paths of free particles, or retain Euclidian geometry and Minkowski spacetime and modify the variational principle by weighing the elements of path $ds$ with appropriate invariant weighting factors."

Mithrandir24601

1:25 PM

@JohnRennie that must have changed now as well :P to obtain a BA you have to have done a part I and a part II. Passing a part III in any subject entities you to the relevant masters degree

(In other words, a part III is some form of masters)

Slereah

1:42 PM

I'm putting Hoyle's steady state theory in the book because I feel a bit sorry for the poor guy

0celoñe7

@Slereah spacelike ones?

Slereah

There's two big definitions of causal curves

1) piecewise $C^1$ curves with matching time orientations

2) the one with the intersection of CNN possessing a $C^1$ curve connecting the two points

Are they equivalent?

The second one means that the curve is Lipschitz continuous

Which have a zero measure of non-differentiable points

Is that equivalent to piecewise $C^1$?

0celoñe7

The second is of course more general

AccidentalFourierTransform

wat

Slereah

@0celoñe7 Is there an example of a Lipschitz continuous causal curve that isn't piecewise $C^1$

Is it gonna be one of those garbage fractal curves

Also aren't Lipschitz continuous curves supposed to be of finite derivative

Oh wait

Rindler observers have unbounded acceleration, not derivative

Do they have unbounded derivatives, too???

I think so

0celoñe7

2:06 PM

@Slereah I don't have one

Slereah

or maybe Kriele meant locally Lipschitz continuous???

I think so since his definition is only the existence of a neighbourhood where that is true

0celoñe7

Same thing if the curve is defined on $[a,b]$

That should be an exercise for you...

@AccidentalFourierTransform what garbage are you reading?

Slereah

Hm

I know that path integrals are Holder continuous, and they're not piecewise continuous by any stretch of the imagination

I guess I just need slightly better behaved functions

0celoñe7

what the hell is piecewise continuous

Yashas

Tag Synonyms for Voting

not many people have voiced their opinion in that topic

Slereah

2:13 PM

Piecewise $C^1$*

0celoñe7

@Slereah what the hell is $C^{1*}$?

Slereah

Piecewise $C^1$, the * indicates a correction to my previous sentence

What is even the definition of a fractal curve

One that includes both fractal curves proper and stochastic processes

0celoñe7

fractals are meme math

I don't think serious people actually care about them

Slereah

Path integrals involve fractals

Kinda

0celoñe7

0

Q: Variational Principle - Extremum is Eigenvalue

Let $$A |a\rangle = a|a\rangle.\tag{1}$$ Then the expectation value is $$\langle A\rangle_{\phi}=\frac{\langle\phi|A|\phi\rangle}{\langle \phi|\phi \rangle}.\tag{2}$$ The differential is then $$\delta \langle A\rangle_{\phi} = \frac{1}{\langle \phi|\phi \rangle} \left( \langle\delta \phi |(A-\la...

ah Christ

@Slereah do you have to explicitly deal with the fractals

Slereah

2:21 PM

@0celoñe7 Define "explicitely deal"

The set of non-fractal curves is of measure $0$

And you have to use a special integral for the action

since you can't do Lebesgue integrals on non-Lipschitz continuous curves

0celoñe7

@Slereah What?

You can Lebesgue integrate anything that is measurable.

I guess the derivatives need not be measurable...fuck

I've integrated derivatives of Lipschitz functions

Oh no, it's possiible

LocLip $\subset W^{1,\infty}$

Slereah

@0celoñe7 It needs a finite measure, yes

Which the fractals don't have

0celoñe7

2:36 PM

@Slereah they can fill $\Bbb R^3$?

Are they not defined on a compact interval?

John Rennie

@Mithrandir24601 perhaps I should stop making statements about Cambridge based on what it was like 35 years ago :-)

Slereah

@0celoñe7 They are contained in a finite volume but of infinite length

The Raiders of Las Vegas

@JohnRennie why didn't you go to Oxford?

Slereah

they wouldn't have him

Because he's a country bumpkin

0celoñe7

@Slereah if the integral is on [a,b] I have a hard time believing it could be infinite...

Slereah

2:43 PM

The integral is between two points of the manifold

The curve between which will be of infinite length

John Rennie

@TheRaidersofLasVegas I think Cambridge is a nicer town

0celoñe7

@Slereah did you ever continue reading Jost?

Slereah

Not so far

0celoñe7

$\Subset$

Aha

@Slereah What are the best Morrowind mods?

I think I'll do a playthrough...

AccidentalFourierTransform

2:59 PM

the one where you can kill kids

vzn

@0celoñe7 whoa! glib alert o_O :(

ACuriousMind

@0celoñe7 I think except for graphical stuff you shouldn't play with any mods - there are some good ones but they change the game considerably. It's broken enough without mods already ;)

AccidentalFourierTransform

$a,b,\cdots$ or $a,b,\dots$?

ACuriousMind

@AccidentalFourierTransform ...why would you ever consider the first?

AccidentalFourierTransform

wh$\cdots$ why not?

but yeah, \dots looks better

Secret

3:11 PM

ok... let's see if I can formulate this better... :

0celoñe7

@ACuriousMind I watched an "Ultimate Critique" of TW3 which was p. good

The Raiders of Las Vegas

@AccidentalFourierTransform do you know how to insert a start time into a YouTube video post in here?

0celoñe7

there was this moment where the guy was talking about Skyrim, and he concluded he had no reason to think Skyrim is a good game

AccidentalFourierTransform

yeah

right-click on the video

0celoñe7

It's interesting, when I think about Skyrim, I don't think it's that good

But as a whole, it's amazing

AccidentalFourierTransform

3:13 PM

"copy video URL at current time"

heather

hello

The Raiders of Las Vegas

hi

AccidentalFourierTransform

sup

0celoñe7

@TheRaidersofLasVegas Click, share, on the bottom

click the box

copy the link it gives you

The Raiders of Las Vegas

Right, thanks.

0celoñe7

3:15 PM

@ACuriousMind Broken in what sense?

AccidentalFourierTransform

the first non-trivial irreducible (projective) representation of the Lorentz group in $d=4$ dimensions has dimension $2$ (left/right handed Weyl). What is the dimension of this object for arbitrary $d$? is there a general formula, or is it erratic as in the Majorana case?

if I asked this on the main site, would people vote to migrate to math.SE? is it even a good question?

ACuriousMind

@0celoñe7 I think Skyrim is primarily made to be fun to play. The Witcher is primarily made to be what its makers wanted it to be.

@AccidentalFourierTransform Weyls exist in even dimension, and have half the Dirac spinor dimension.

I.e. $2^{\lfloor d/2 \rfloor - 1}$

AccidentalFourierTransform

nice

but is Weyl always the first non-trivial irreducible (projective) representation of the Lorentz group?

ACuriousMind

Do you mean "first" in the sense of "lowest-dimensional"?

AccidentalFourierTransform

yep

0celoñe7

3:19 PM

@ACuriousMind It's probably the leveling system that makes me enjoy Skyrim so much. I wish I could play SkyRe properly.

AccidentalFourierTransform

or, equivalently (?), the lowest spin representation

0celoñe7

The damn patches made the game unplayable though.

SkyRe conflics with basically every mod.

ACuriousMind

@AccidentalFourierTransform Well, the fundamental has Just dimension $d$, doesn't it?

AccidentalFourierTransform

hmm good point

I think I got you

and for odd $d$, Dirac is irreducible, right?

unless Majorana exists, I guess

ACuriousMind

@AccidentalFourierTransform Well...not if there are Majoranas :P

AccidentalFourierTransform

3:21 PM

cool, thx

0celoñe7

daily reminder no one knows in which dimensions there are majoranas

AccidentalFourierTransform

:-P

The Raiders of Las Vegas

Stephen Hawking - A Brief History of Time

►

One of my favourite scenes.

@JohnRennie^

AccidentalFourierTransform

If the manifold is curved, partial derivatives have no intrinsic meaning and one must introduce a connection. Is there any sense in which functional derivatives may lose their intrinsic (?) meaning, and one is forced to introduce a connection on the space of functions?

something like $$\frac{\delta}{\delta f(x)}S[f]\to\frac{\delta}{\delta f(x)}S[f]+\mathcal A[f]S[f]$$

0celoñe7

@AccidentalFourierTransform Function spaces are vector spaces, Frechet/Gateaux derivatives are always defined

If you don't have a vector space, then you might have to do something strange

AccidentalFourierTransform

3:32 PM

well, partial derivatives are always defined too, but they may not be good for some purposes

@0celoñe7 and function spaces are always vector spaces?

hmm

0celoñe7

Lebesgue spaces, Sobolev spaces, Lipschitz spaces, Holder spaces, BMO, BV, ...

Well...you could have a functional on the space of metrics, for instance. That's not a vector space

But it's a cone, so it still works

AccidentalFourierTransform

the space of metrics is a cone? what does that even mean?

0celoñe7

Closed under addition and multiplication by positive numbers. A cone.

AccidentalFourierTransform

never heard of it :-P

0celoñe7

You need to do more functional analysis then.

AccidentalFourierTransform

3:41 PM

I don't need to do anything. I'm a free spirit.

5

I do that, with math.

0celoñe7

@AccidentalFourierTransform springer.com/us/book/9783540506256#otherversion=9783642080746 start reading

AccidentalFourierTransform

@0celoñe7 that, actually, doesnt look bad at all

some day, perhaps

is it good?

0celoñe7

@AccidentalFourierTransform It's the standard reference, but I haven't read it.

Giaquinta is a very good writer, for what it's worth.

I used his measure theory book for a class and am reading his elliptic regularity book right now.

Secret

Toy model for a probabilistic interaction:

Suppose there are two particles with initial pure states $\lvert A(0)\rangle$ and $\lvert B(0)\rangle$. We also have some known probability distribution $P : \{0,1,...,n\}\mapsto[0,1]$. Therefore $P(i) \in [0,1]$ and $\sum_{i=0}^nP(i)=1$.

Now, for each $P(i)$ there associates a unitary evolution operator $U_i(t_0,t_1)$, which describes the interaction the two particles will undergone in the time period $[t_0,t_1]$.

When the initial states of the particles $\lvert A(0)\rangle$, $\lvert B(0)\rangle$ is act by $U_i(0,t)$, it produces the correspond…

uh wait a sec.. I need to figure out how to take the time evolution of the density matrix correctly as otherwise I have two $P(i)$s in the final expression

AccidentalFourierTransform

who starred that? :-P

coincidentally, butts may be fun to look at, but when you are going to get your hands dirty, there are parts that are much more fun to play with than the butt

Slereah

3:52 PM

archive.is/j9EeV

heh

Sid

@AccidentalFourierTransform Heh. Let's star that too. :P

Secret

ok correction:

\begin{align}
\sum_{i=0}^nP(i)U_i(0,t)\lvert A(0)\rangle \langle A(0)\rvert U_i^{\dagger}(0,t) & =\sum_{i=0}^n P(i)\lvert A_i(t)\rangle \langle A_i(t)\rvert\\
\sum_{i=0}^nP(i)U_i(0,t)\lvert B(0)\rangle \langle B(0)\rvert U_i^{\dagger}(0,t) & =\sum_{i=0}^n P(i)\lvert B_i(t)\rangle \langle B_i(t)\rvert\\
\end{align}

AccidentalFourierTransform

@Sid mods are so not gonna like it

Sid

(I hope I don't get banned)

ACuriousMind

@Secret 1. Of all the things you need to explain, what a probability distribution is is not one of them. 2. Whatever you're doing here, at the point where you associate a different $U_i$ to each possible outcome, you're not doing quantum mechanics. 3. Your conditions for "distinctness" make no sense to me, That $X$ and $Y$ are "distinct" means usually that $X\neq Y$, not more, not less.

4. Since you are not doing quantum mechanics, it's completely unclear how you arrived at either of your final expressions.

@AccidentalFourierTransform So why did you post it?

Anonymous

3:57 PM

You really think AFT does everything for a reason ? :P

AccidentalFourierTransform

@ACuriousMind why do you think I only post things if I believe mods are gonna like it?

Im a rebel

Secret

Re 3: I am actually tried to express the idea that $A_i\neq A_j$ and $A_i \neq B_j$ and $B_i \neq B_j$ but I guess I stick too much on trying to express that via brackets

I think I should have wrote those $\neq$ straight away

Transcript for