The h Bar - 2016-06-07 (page 3 of 3)

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8:00 PM

String theory is nowhere near giving us predictions at that level of detail

@JohnRennie thought something like that. wonder if anyone has proposed any models that try to determine particle masses. (believe that a "emergent soliton" theory approach would yield something... eventually)

I'm not aware of any credible attempts to predict the mass spectrum

← starts working :P

@vzn All the masses in the Standard Model come from the Higgs mechanism. What you need to "explain" is the value of the Higgs couplings. These are input to the standard model, while in string theory the couplings of the effective gauge theories are produced by different configurations and angles of branes (together with the "one" input of string theory, the string tension/length).

But, to my knowledge, there's not yet known a reliable mechanism to actually produce the Standard Model nor how to select the "correct" brane configuration.

@ACuriousMind Stop your lies

8:03 PM

@vzn good luck, you're not the first to attempt it.

Neutrino masses don't come from the Higgs mechanism

@Slereah We don't know where the fuck neutrino masses come from :P

@JohnRennie understand what you mean. but actually there is very little research into emergent soliton dynamics.

and that's not even mentioning hadron masses

Well we know it ain't the Higgs field :p

I believe attempts were made using the early SU(5) and SO(10) GUTs, but they came to nothing.

@Slereah I thought lattice QCD could do a good job on hadron masses these days ...

8:04 PM

oh sure, but they don't come from the higgs mechanism, for the most part

It may well be an additional "inert Higgs", but could equally well be one of the variants of the seesaw mechanism

the existing theories seem rather complex when one counts all the separate "free parameters", generally particle masses, although maybe this is rarely pointed out. (epicycles, anyone?) :P

@vzn I think the usual number of free parameters in the Standard Model is something like 19. 19 inputs to predict everything is rather large predictive power.

Isn't it 26

@ACuriousMind ok 19 is not bad, but seems low, is standard model calculating mass of "all" the particles? eg quarks etc?

8:07 PM

@Slereah Might be 26, yes

Let's count 'em

123456789

coupling constants for SU(2) x U(1) : 2 (coupling constant and weak mixing angle)

@JohnRennie Those simulations often have additional "simulation parameters" that are adjusted to fit to experiment

← would like to see a chart of "all" the particles & how their masses are "calculated/ derived"...

8:09 PM

There's the SU(3) mixing constant

There's 9 elements for the Cabibbo matrix

4 for the neutrino mixing matrix

@vzn Getting the quark masses is not that difficult. Getting the mass of their bound states (i.e. protons, neutrons, etc.) from that is a tremendously difficult task due to QCD being strongly coupled in the low-energy regime

what do you mean by "strongly coupled"?

There's... 6 + 6 + 3 + 1 mass terms?

@Slereah A good number of those can be eliminated by gauge transformations, I seem to vaguely remember.

Does the VEV of the Higgs field count as a constant

technically it's dynamic

8:13 PM

@vzn The dimensionless coupling of QCD (its analogon to the fine structure constant) becomes much larger than 1 at everyday energy scales, so the usual perturbative methods don't work.

Well it works, it just means you have to do a lot of terms :p

@Slereah The parameters in the Higgs potential are the things you want to count as constants, probably

So yeah I think around 26 sounds okay

Plus c and hbar I guess

@Slereah which is interesting how computational complexity enters/ impinges on physics

@Slereah I don't think that's true - the perturbation series is asymptotic and for a coupling of $\frac{1}{n}$ usually doesn't yield good results beyond $n$th order.

8:17 PM

maybe

Which is why you have to do lattice methods, but then you get the problem that you don't really know how to take the correct continuum limit

What about $a \to 0$

let the map $\varphi : G \to H$ be an isomorphism. prove $|\varphi(x)| = |x|$ for all $x \in G$ i started with $(\varphi(x))^n = I_H$ since $(\varphi(x))^n = \varphi(x^n)$ we can say $I_H = \varphi(I_G)$ so now $x^m = I_G$ what do I do from here? How can I say $\varphi(I_G) = I_G$?

@Slereah Well, how do you take "$a\to 0$" in a necessarily finite simulation? Even abstractly, what do you do? First $a\to 0$, then $N\to\infty$? Both at the same time? Both at the same time, while keeping some ratios constant?

Try all combination, see which one converges best :p

8:20 PM

@Slereah Well, but what if more than one method converges, but to different things

many of those limits will just trivialize the theory

Well check with experimental results

@Slereah But then you put in the very thing you wanted to derive.

If it trivializes the theory my guess is that it's not the best!

Well we already know the proton mass gee

@Slereah But it could also mean that there is no well-defined non-trivial QFT of that type!

Could be, but can you trust a numerical simulation!

8:22 PM

Like, I think that the current tendency is to believe that $\phi^4$ doesn't actually exist as a non-trivial theory in the continuum limit

In 4D you mean?

@Slereah Uhhh...I think it's shown for $d>5$ and there is large evidence hinting at it for $d=4$.

Let's all throw QFT in the trash

@DanielSank scitation.aip.org/content/aip/journal/apl/108/21/10.1063/…

But I didn't really follow those developments, so take this with more grains of salt than what I usually say :P

8:24 PM

Strong field / ultrafast physics has a pretty bad track record with acronyms

particularly with animals, for some reason

but that one just pushes it way too far

"Virginia spiderwort" inside an acronym to get a V?

@ACuriousMind so how do you actually rigorously determine the orientation of [0,1]xM

user image

@0celo7 Go look up "orientation of product manifolds", I don't actually know any of these constructs by heart on the level of detail you usually want to know.

pretty sure MSE has a thread on this

@DanielSank It's a simpler way to make this nature.com/nphoton/journal/v8/n7/full/nphoton.2014.141.html

8:29 PM

Help

user image

The Hawking paper on chronology protection has Penrose Newman formalism

what do

@EmilioPisanty It's pretty...but what is it?

We call it a "bicircular" field

Well...I certainly can confirm that I see two circles on that graphic :P

8:32 PM

two laser fields focused collinearly, a fundamental with right-handed circular polarization, and its second harmonic at left-handed circular polarization

and they add together to make that nice three-lobed Lissajous figures

Ah...that's what that is

Then you focus them into a gas at some reasonable intensity, say, 10^15 W/cm2

and you look for emission of harmonics from nonlinear processes

user image

↑ and you get something like that

(axes are intensity vs harmonic order)

and you can get something like tens to hundreds of harmonics

Also

did anyone else get re-awarded any badges lately?

user image

... except I have it since October 2014

55

Q: Excavator and Marshall badges are being awarded over and over

The excavator has been awarded to... many users just some time ago. What a... bug? Certainly it is, too many unavailable posts :D Seems to affect Marshal and Excavator. Also looks like it's everywhere. SO's got a similar problem a couple of minutes before. Moreover, I can has THREE X mar...

bug status-completed badges

@EmilioPisanty Network-wide bug.

56

Q: Excavator and Marshall badges are being awarded over and over

The excavator has been awarded to... many users just some time ago. What a... bug? Certainly it is, too many unavailable posts :D Seems to affect Marshal and Excavator. Also looks like it's everywhere. SO's got a similar problem a couple of minutes before. Moreover, I can has THREE X mar...

bug status-completed badges

8:44 PM

cool, nice find

So it seems I have a few hours with seven gold badges, I should go get into a discussion and pretend I'm important.

@EmilioPisanty Maybe more like ten minutes. :P

@Tim The cache on the cards and badge list on the profile is 10 minutes; it will fall out shortly. — Nick Craver ♦ 1 min ago

@HDE226868 Gah. So much grandeur. gone.

what is an easy way to prove two groups aren't isomorphic (assuming their cardinality is the same) ?

Group isomorphism problem

In abstract algebra, the group isomorphism problem is the decision problem of determining whether two given finite group presentations present isomorphic groups. The isomorphism problem was identified by Max Dehn in 1911 as one of three fundamental decision problems in group theory; the other two being the word problem and the conjugacy problem. All three problems are undecidable: there does not exist a computer algorithm that correctly solves every instance of the isomorphism problem, or of the other two problems, regardless of how much time is allowed for the algorithm to run. == References... ==

heh

tough luck!

8:58 PM

so should I just look for examples where $\varphi(xy) \ne \varphi(x)\varphi(y)$?

@Obliv What? Do you want to prove that two groups are not isomorphic or that a given map is not an isomorphism?

or can i test for this to be true if they are isomorphic : $|x| = |\varphi(x)|$

prove that two groups aren't isomorphic

@Obliv that presupposes you know $\varphi$

which is sort of the point

@Obliv Then what is that $\varphi$ you are talking about?

Also are we talking finite groups

Finite groups would be easier

8:59 PM

↑ Slereah's point is crucial

uh it says the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$

Well it would be possible, for a start

Finite groups can be solved by enumeration, so they are computable

so infinite groups

Those groups are different

9:00 PM

So you've been tasked with showing that they're not isomorphic?

The first isn't even connected

@Slereah That's a topological concept, though

Wouldn't it matter for a lie group tho

@Obliv For that, you need to find an element of a certain order in one and then show that no element of that order exists in the other

... like, say $i^4=1$

with $i^2≠1$

(and also $i≠1≠i^3$)

which no real number satisfies

9:02 PM

@acuriousmind does that hold for all isomorphic groups?

an isomorphism $\varphi:\mathbb C^× \to \mathbb R^×$ would preserve the order of every element

i.e. $\varphi(i)^4=\varphi(i^4)=\varphi(1)=1$

and similarly $\varphi(i)^n≠1$ for $n=1,2,3$

that's impossible for a real number

hence no isomorphism

ah okay thank you

@FrancescoS Is the no-image-found on your profile picture intentional?

@EmilioPisanty No.. I see my image. Didn't see you anything?

user image

9:06 PM

@EmilioPisanty That might be an issue on your end, I can see his picture perfectly fine

@ACuriousMind Indeed it's probably my machine that's doing it

but it's still an issue @FrancescoS might want to look into

user image

ah

I know what's happening

you're loading from facebook

@ACuriousMind Coming back to Clebsch.gordan coefficients... If I have a tensor product of two tensors, each one with two indices (one for SO(3) transformations and the other one for SU(N) transformation) how can I decompose the bigger tensor in irreps? Should i have to symmetrize-antisymmetrize independently for each index?

I can see the picture here if I go there directly scontent.xx.fbcdn.net/v/t1.0-1/p200x200/…

but Privacy Badger is shooting it down

@EmilioPisanty I don't see the problem... the others can view my picture or not?

@FrancescoS They will unless they've got Privacy Badger installed

9:10 PM

@FrancescoS I'm not sure I understand what you want to do - C-G coefficients happen when you have a tensor product of irreps of a group and then want to decompose that as a direct sum of irreps. You can't decompose the tensor product of two rep of different groups in that way

But if Privacy Badger is shooting it down then it means that whenever someone sees your profile picture, Facebook knows about it and potentially what page you were seeing

which Facebook will quite happily jot down if they know your IP

So if you can re-host the image elsewhere, privacy-minded people will thank you

If not, just be aware that it's happening.

@Emilio Can I use that same strategy when showing that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic? But, aren't the orders of all non-identity elements infinite anyway? (for those groups)

@ACuriousMind Ok, I will give you an example. I have a massive fermion field which transforms under SU(N). I write a generic state with an upper index that is the fundamental of SU(N) and with a lower index, that is the 2 dimensional irrep of SO(3). Now, I want to make the tensor product and try to decompose it

@Obliv For positive numbers, yeah, but not for -1.

If you're comparing $\mathbb R^×$ and $\mathbb Q^×$, you simply note that the cardinalities are different and move on.

hmm. but the order of $-1$ is infinite too, no? ${-1}^n = (-1) + ... + (-1)$ $n$ times which never reaches $0$ I thought

9:15 PM

@Obliv These are the multiplicative groups we're talking about? where $(-1)^2=1$?

wait I misread

@FrancescoS What exactly do you mean by "make the tensor product"? Do you mean you're taking the tensor product of such a state with another state in the same rep?

no they are additive I edited the msg @emilio

@Obliv What Emilio said about the cardinalities still applies - they can't be isomorphic because there is not even a bijection between them to begin with.

I thought the cardinalities of the groups were the same

9:17 PM

@Obliv Then the orders game also fails, but the cardinalities is just the obvious nonstarter.

@Obliv Nope. $\mathbb Q$ is countable, $\mathbb R$ isn't.

$\psi^a_{\sigma_1}$ is one state which transforms under $SU(N)\otimes SO(3)$. Then, I write $\psi^a_{\sigma_1}\psi^b_{\sigma_2}$ and I would like to reduce this tensor

what does that mean? $|\mathbb{Q}|$ is finite?

@Obliv It means the rationals are infinite but the reals are 'bigger'

There can never be a bijection between the two

If in doubt, it's in any 101 on infinite cardinalities

oh I see. the real numbers include the irrational numbers right?

@ACuriousMind why don't you know the gory details of diff top :(

9:19 PM

@FrancescoS Ah! Well, then you just separately decompose the $\mathrm{SO}(3)$ and the $\mathrm{SU}(N)$ parts. (By the way that really should be $\mathrm{SU}(2)$, not $\mathrm{SO}(3)$)

Maybe it's a vector field, you don't know

Cantor's diagonal argument

In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers; it was actually published much later than...

@Slereah He said "that is the 2 dimensional irrep of SO(3). ", so I know :P

Oh well :p

I hate the Newman Penrose formalism so much

@ACuriousMind what does that even mean

group theory is another reason why I don't like physics

9:21 PM

@ACuriousMind I see that if i symmetrize or anti-symm the upper indices is the same to symm or anti-symm the lower indices. So, I think it should bu sufficient to do the decomposition with just one index?

it wouldn't be so bad if physics books actually explained any of it

@FrancescoS I'm afraid you've lost me again :)

@ACuriousMind "again".. probably ;) sorry

You can't symmetrize indices of different groups

@Obliv yes

9:25 PM

They don't run over the same values

@Slereah Yes, I know.

I think wrt the quantum, that is called like

Selection rules

Or something

@Obliv What's interesting is that $\Bbb R=\Bbb Q\sqcup \Bbb I$, but $\Bbb Q$ is countable while $\Bbb I$ is not

Probably irrelevant

nvm

There are "more" irrationals than rationals

9:27 PM

Well yes

obviously

@Slereah Not obvious to me

what is the unclosed box symbol @0celo7

is that a union

Although

Most important subsets of the irrationals are countable

@Obliv $A=B\sqcup C\Leftrightarrow A=B\cup C\land B\cap C=\emptyset$

I use $\bigcup$

wait

9:28 PM

fuck, what is logical and

in TeX

$\And$

$\land$

\land

oh I see.. so it's a union of disjoint sets?

@ACuriousMind thanks sweet prince

@Obliv yes, it's called the "disjoint union"

It's a lie

disjoint union is more complicated

9:29 PM

?

user image

@Slereah I didn't know, please explain

You can do a disjoint union of non-disjoint sets

But then you have to index the sets

but then you're doing a disjoint union of the disjointed parts of the index of the sets? @slereah

@Slereah Just tell him the truth, then: It's the coproduct in the category of sets!

9:31 PM

@ACuriousMind Dammit

Well me I'm all about ZFC as a basis for mathematics

I'm not a category theory weirdo

@Slereah Right, you're a wormhole weirdo

Try saying that when I'm with hot alien babes

5

I'll say that when their eggs hatch from your flesh

I am strangely aroused

9:32 PM

P. hot

Remember the alien babe in Galaxy Quest?

Get me that mass of tentacles

You know, one thing I wonder is

If FTL Alcubierre drive don't work

so for the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ do I use the same idea, that one is larger than the other?

Is there any benefit to slow Alcubierre drives

Like technically you can have Alcubierre drives at any speed

@Obliv ...they are the same size.

how? 1/2 cannot be expressed in $\mathbb{Z}$

@Obliv what?

9:35 PM

"Same size" has a specific meaning in infinite sets, @Obliv

You can have a set as a subset of another set still being of the same size

Do lie groups help against sleeping disorders ?

we just said $|\mathbb{R}| > |\mathbb{Q}|$ because I thought the reals had irrational numbers

so $\mathbb{Q}$ has fractions, $\mathbb{Z}$ does not

@Obliv No, we said that because of the diagonal argument Emilio linked.

That's not the definition of cardinality, though

For instance, $\Bbb N$ has the same cardinality as $\Bbb P$

The set of prime numbers

what about two uncountable sets? Can you tell which one is larger by some other reason then?

are there only 2 categories of infinite sets, countable and uncountable?

9:38 PM

No

@Obliv Sets have "the same size" if and only if there is a bijection between them.

There's a whole hierarchy

@Obliv if you're smart it's obvious that they have the same size

@0celo7 kys

just like it's obvious that there are more irrationals than rationals

@Obliv just ask @Slereah

9:39 PM

In general, $\vert \mathcal P(X) \vert > \vert X\vert$

so if you could theoretically show that a bijection existed between $\mathbb{R}$ and ,say $\mathbb{Z}$, then they are of the same size?

@Obliv Yes, but Cantor's diagonal argument shows there is no such bijection.

but how can you show a bijection exists between $\mathbb{Q}$ and $\mathbb{Z}$?

i guess i'll take a look at that argument

I wonder, is there a topological argument for that?

@ACuriousMind ?

Those are sets

9:41 PM

@0celo7 what

That works even with no topology

@ACuriousMind that $\Bbb Q\cong \Bbb Z$

@0celo7 They carry no topologies if you're just asking for bijections!

ah I see $\mathbb{Q}$ has infinite representations for fractions so it would fall under that cantor argument

@ACuriousMind so?

give them topologies and see what happens

9:42 PM

IIRC the basic proof for that is that you can just assign an ordinal for every fraction

@0celo7 what?

Either that or

the zig zag

wait no you guys are saying they're the same size. Shit

user image

$\Bbb N \to \Bbb Q$

There's a version for negative fractions, too

user image

That helps a lot, thanks @slereah

9:45 PM

That's not quite $\Bbb Q$ since the equivalence relation isn't applied, but $\Bbb Q$ is a subset of that set

leads me to wonder if such a bijection could be produced with $\mathbb{Q}$ on the outsides to produce $\mathbb{R}$

So obviously the cardinality will be smaller than $N$

So in the end, the cardinality is between the cardinality of N and the cardinality of N

So it is the same

Quite a lot of subsets of R have this cardinality

The natural, the integers, the rationals, constructible numbers, algebraic numbers, computable numbers

@ACuriousMind given $\Bbb R$ the standard topology and see what happens!

See if another proof works!

user image

I don't read french

What does that say

9:50 PM

It says "zwischen"

@0celo7 How is endowing these sets with a topology going to make something "happen" that tells me something about mere bijections between them?

Maybe they are homeomorphic

@0celo7 In the subspace topology of the reals, they aren't, because one is discrete and the other isn't.

I know they're not homeomorphic

Then what is your point?

9:53 PM

I don't have one, forget it

@Slereah How very French

"One can adapt the lemma in [7], p. 295"

My god

Even Hawking just says to look up the proof in HE

2

what?

HAHA

what are you reading

The chronology protection conjecture

10:12 PM

""The quantity h is rather like the surface gravity of a black hole. It measures the rate at which the null cones tip over near $\gamma$. As in the black-hole case, it gives rise to quantum efFects. However, in this case, they will have imaginary temperature, corresponding to periodicity in real time, rather than in imaginary time, as in the black-hole case"

10:24 PM

So the chronology protection paper was basically just saying that the stress energy tensor diverges in Misner space

@ACuriousMind Ah, GP has a full tutorial on how to orient $[0,1]\times M$

Now to read the full Visser paper on chronology protection

"It should be emphasised that the universe appears to exhibit a “defense in depth” strategy in this regard. For appropriate ranges of parameters describing the wormhole (such as masses, relative velocities, distances, and time shifts) Casimir effects (geometry induced vacuum polarization effects), wormhole disruption effects, and gravitational back reaction effects all contribute to the fight against time travel."

Bloody universe

10:52 PM

Gee Visser says that making a time machine with the Morris method seems easy

But step 0 is "acquire a wormhole"

seems a tad hard

11:35 PM

@ACuriousMind , @HDE226868 loving this query, though

data.stackexchange.com/stackoverflow/query/471647/…

"Badges people shouldn't have"

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