@barrycarter some further thought on this. look at elite/ leading edge/ "big science/ big physics" prjs eg say LHC or LIGO, and notice how much of it is actually highly sophisticated engineering... to me math/ physics/ engr all exist on a sort of continuum, interconnecting, synergistic, coevolving, none successfully advancing without the others. sort of a 3way yin yang? ... o_O

@Xylius It means someone starred it.

@barrycarter That is very much not true :-)

@DanielSank Why are things so complicated?

@DanielSank for what purpose though? One would flag a post for being inappropriate, but what does a star signify?

@BernardMeurer Because.

@Xylius It puts that post in the starred list so other people will see it for a while.

It's just a way to bring attention to a message.

I feel much better, you're basically the new Oprah Dan

It can also indicate that someone thought a message was funny and wishes to allow others to experience the humor.

@Xylius By the way, it's much easier to follow the flow of discussion if you reply to specific messages.

Mouse over a message from me and click on the arrow on the right side.

@BernardMeurer Is that good?

Sorry, I'm new to the mobile UI

@Xylius Didn't they recently upgrade the mobile UI?

@BernardMeurer what's your gf's name? I forgot already.

@DanielSank I wish she was my girlfriend

With that said, Kristina

@DanielSank yeah, they did, but I only just started using it yesterday, so still getting used to it

@BernardMeurer Grammar lesson: "I wish she were my girlfriend."

Even native English speakers do this wrong all the time though.

So maybe it's not "wrong" any more.

Wow, I had never noticed that sentence was wrong, thanks dan!

@BernardMeurer Sure. This is messed up all over the place so it may not really even be wrong any more.

I always refer to Fiddler on the Roof's famous line: "If I were a rich man...".

Why did you ask her name?

@BernardMeurer I was trying to tease you.

By calling her your girlfriend.

;P

I thought so! But then I thought "He's messing with me, I must not yield" :D

@BernardMeurer Nice countermeasures.

I appreciate a strong opponent.

Man, that song is so good.

I've been playing chess :p

I like the initial monologue

or dialogue I guess, depends on your religious views

It's pretty good yeah, was about to comment that, I have never watched that movie

It's a famous play.

My internet is so good I've been waiting to add a GPG key for some 10 minutes now

Nice.

Nice? What's wrong with you?!

:p

@BernardMeurer Nothing. My internetz are acceptably fast.

What's wrong with you?

@DanielSank In general? My right foot, my morals, my sleeping habits and my ideas on a healthy diet

@MAFIA36790 \o

Why do we always give high fives? I mean, I'm not saying we need a reason...

@BernardMeurer Hmmm, diet? Try more vegetables.

@DanielSank They're hard to make

What do you mean?

Vegetarian food is infinitely easier to cook that meats.

Here's my standard dinner when I think I need more vegetables:

It's delicious.

Step one I can do

step 3 is difficult

@MAFIA36790 I don't understand the question.

step 4 is just impossible

@BernardMeurer No it's not. You probably just haven't tried it.

To cook rice you put one part rice and 1.5 parts water in a small pot, turn the stove to low heat and come back in twenty minutes.

Or you buy a rice cooker and then you don't even have to set the heat or time.

You dump the rice and water into the cooker, push a button, and come back whenever you want.

@MAFIA36790 Yes. I don't have one any more because I got sick of having too many kitchen appliances.

I use one iron pan or a pyrex casserole dish for everything now.

BTW, I cook vegetarian at home.

I have good recipes.

Interested?

@MAFIA36790 I have only one good Indian dish?

Why?

Are you Indian?

Well well well. Let me tell you something.

I have a good friend from India. She was at my place for dinner one evening and I prepared this dish for her.

She liked it so much that she made it for her husband, who liked it so much that he asked her to make it again.

Also, just because you're Indian, does it mean you must eat Indian food?

@MAFIA36790 What... does that have to do with anything?

@MAFIA36790 I see.

The only Indian food I can make half-decently is Kabuli Chana.

@MAFIA36790 Seriously?

Where do you live?

Most of my American friends are very open in their food choices.

Some here in California have gardens and produce their own vegetables.

For example, my fiance was growing tomatoes and fava beans.

I like cooking for Indian people. You guys appreciate good vegetarian food more than most others.

@MAFIA36790 Butter gives me heartburn. I tend to use olive oil, and not too much.

@MAFIA36790 Thank you!

@DanielSank If I become indian will you cook for me?

@BernardMeurer Sure. I'll cook for you either way though.

@MAFIA36790 There was a short time when I thought I would have to marry an Indian. I was at a dance party with a bunch of physicists. There were many Indians. We were dancing somewhat.

Then someone asked for Indian music. Holy s---!

That got everyone dancing for real.

I never had so much fun at a dance party before or after that.

Apparently I was doing well because they said that I fit in and they would have to find me an Indian girl to marry o_O

@MAFIA36790 Really?

:: looks up butter paneer ::

@MAFIA36790 I see. It's a sort of vegetable and cream sauce.

@MAFIA36790 there was a period in which I wanted something like this for pasta.

I didn't know about butter paneer though, so I invented something.

It is essentially squash, tomatoes, and onions cooked to a sauce.

(and turmeric)

Turmeric is the best spice.

@MAFIA36790 I will describe it. It's so simple.

First you should fry the onions in spices until they are soft. Next, slice up the squash and the tomatoes. Then cook the squash in a pan for a long time until they wither (you can do this in the oven instead). Finally, slowly add the tomatoes so that there is never too much water in the pan.

That's it.

It does not need any salt.

I use this for pasta sauce. In fact, it is amazing to put this on a bread and toast the whole thing with some mozzarella cheese on top, like a pizza.

Next time I make it I'll take photographs and write it up like the other recipe.

One more thing I should mention. This is a general observation about vegetarian food.

If something seems a little bland, as if it needs a bit of salt or "something extra", instead get some yogurt and put a bit of it either with the food or right on top as if it were a sauce.

The tangy taste of yogurt is an excellent partner with most vegetarian food.

@MAFIA36790 What time?

Ok. I can make it to that.

Oh wait, 11 hours?

Damn.

I have to conduct an interview.

I actually might have time though.

What's wrong with this statement?

z is the direction along the string?

@MAFIA36790 ok

Hahaha

I was just going to suggest posting it on the site.

@MAFIA36790 I don't understand the notation ;_;

What is the $Z$ in $Z[d\Psi/dt]^2$?

Got it.

The square brackets are confusing. I thought Z was some kind of operator.

@MAFIA36790 I don't much understand the quoted text.

> turns out to be the force in the +z direction exerted on that part of the spring with the equilibrium position to the right of point z by that having equilibrium position to the left of point z, after the equilibrium value of the force, F0, has been subtracted out.

That whole sentence is just really hard to understand, and I think it has typos.

Should "spring" be "string"?

Also, what is $F_0$?

The whole thing is clouded in confusion.

@MAFIA36790 Oh, that's much clearer now.

I thought you said something about a "string" before.

I think you should include in the post the information you just told me.

@MAFIA36790 I understand now.

An edit will help.

I have to go for now though.

See you.

\o

Howdy

I haven't slept in 2 days

I think I got insomnia

fun...

Is that right? >_>

Aha?

K

Soo what's new?

?

I'm all alone... :(

Hello

So there might be a second reason why ff becomes $\text{f}\Gamma$

Apparently $\Gamma$ is the character 0 in latex

so in case of some error there could be a Gamma popping up

Utterly unrelated to physics, but it made me laugh. I particularly like the (current) top answer :-)

@JohnRennie Do you have cat on car problems too? I thought in the developed world, they didn't allow strays.

I have no car and no cat :(

i'd rather have a cat

@Slereah would it be at least slightly safe to assume you live in a place where public transport is extremely effecient and just good in general?

Because where I'm from, if you don't own a car, be prepared for hour long train/bus rides

Well I live in Paris, so no

@FaheemMitha My car gets washed when it rains, so I'm unconcerned about cats walking across it. I don't think there are many stray cats in Chester but there are lots of cats that seem to wander around wherever they like.

@JohnRennie Yes, cats like to go whether their spirit takes them. Like in the Just So story.

@Slereah So Paris does not have good public transportation?

Interesting spread of questions you have on this site. Everything from beginner to (what looks like) research level.

I'm surprised of all places, a sophisticated nice city like Paris doesn't have good public transport.. I live in Australia, and we're just too it for our own good

eveyrthing is just too far away

it's alright

not a fun ride tho

so anyway

I am seeing that $[L^i, L^j] + [K^i, K^j] = 0$

But

$[L^i, L^j] = i \varepsilon^{ijk} L^k$

$[K^i, K^j] = \varepsilon^{ijk} J^k$

They differ by a factor of $i$

Did I forget another $i$ somewhere

Doesn't look like it..

Hm

Another place puts it as $[K^i, K^j] = -2i \varepsilon^{ijk} J^k$

also wrong but for totally different reasons!

Let's recalculate it

I now find $i \varepsilon^{ijk} L^k$

Close but wrong sign

$$[K^i, K^j] = [J^{0i}, J^{0j}] = i(g^{i0}J^{0j} - g^{00}J^{ij} - g^{ij}J^{00} + g^{0j}J^{i0}) $$

$$= -iJ^{ij}= -i \frac{1}{2}(J^{ij} - J^{ji}) = -i \frac{1}{2}(\delta^i_a \delta^j_b - \delta^j_a \delta^i_b)J^{ab}$$

Trying to read this without it auto rendering as math text is slowly killing my already dwindling neuron count

$$=-i\frac{1}{2} \varepsilon_{kab}\varepsilon^{kij}J^{ab} = i\frac{1}{2} \varepsilon_{kab}\varepsilon^{ijk}J^{ab} = i\varepsilon^{ijk}J^k $$

What did I do wrooong

Is it gonna be one of those $\varepsilon^{abc} = -\varepsilon_{abc}$ business

Use mathjax, @Xylius

Using the mobile version

so it doesn't render

Peskin does this annoying thing where he doesn't use Einstein notation for levi civitta symbols

So that issue could go unnoticed

@Slereah Hm..

No for euclidian space $\varepsilon^{abc} = \varepsilon_{abc}$

Hm

Does anyone else have an option on this? I'm stumped

@Slereah Maggiore puts it as $[J^i, J^j]=i\epsilon^{ijk}J^k$, $[J^i, K^j]=i\epsilon^{ijk}K^k$, $[K^i, K^j]=-i\epsilon^{ijk}J^k$.

This way you can define $\mathbf{J}^{\pm}=\frac{1}{2}(\mathbf{J}\pm i\mathbf{K})$ and get two copies of the Lie algebra of $SU(2)$.

Well yes, but what is wrong in the way i derived it

And got the sign wrong

@Xylius On Chrome and Safari mobile, Chatjax rendering should work. Just add it as a bookmark and invoke the bookmark when you're on the chat page.

@Slereah How do you get the second equation?

@Bass Thanks for the tip, much appreciated

@Bass That's the algebra of the Lorentz group

@Slereah Right.

@Slereah The second equation here is wrong. You need to switch the sign twice to move $\epsilon^{kij}$ to $\epsilon^{ijk}$.

hi

I have 2 question , physics subject

is here ok to ask them or should I go to another room ?

@Bass i did

That's why $i$ became $-i$

@Slereah You need to switch twice. So $i$ becomes $i$.

Oh

You can't do cyclic permutations

That's why

Any thoughts math.stackexchange.com/questions/1749434/… ?

Ahh, but the question is why does defining $\mathbf{J}^{\pm}=\frac{1}{2}(\mathbf{J}\pm i\mathbf{K})$ work?

Well that is what the exercize is about!

I mean, how would you know to do that before knowing that it would mathematically work?

I think it's just an overall common method?

Same way you get the Ashketar variables

Or that methodology where you do $E + i B$

for the EM field

I think the real reason you want to do this is that the exponential of anti-hermitian matrices is unitary (e.g. p9 isites.harvard.edu/fs/docs/icb.topic1288789.files/… )?

wouldn't it be antiunitary

Exponential of anti-hermitian matrices also unitary (P2(e) ma.utexas.edu/users/sadun/S06/Lie/Hwsol4.pdf ) right?

@Algebra2015 ask here! I can't guarantee to answer but I will try.

This drove me insane first time I seen it, asked a famous professor why we do this and he just said because it works, nobody seems to mention this but it seems like the real reason right?

The real reason is probably rooted in group theory shenanigans

Yeah I just think this might be the shenanigan haha

Showing that $SL(2,C)$ is a double cover of $SO(3,1)$ and all that

And $SL(2,C) \approx SU(2) \times SU(2)$

All that

@johm

john

may I ask here or via email ?

@Algebra2015 ask here ...

how to send file ?

[img]i.imgur.com/HDK8PhG.png?1[/img]

[img]i.imgur.com/HDK8PhG.png?1[/img]

imgur.com/HDK8PhG

@Algebra2015 you'll have to work this one through for yourself I'm afraid. It seems a fairly straightforward bit of electrostatics.

I tries

dd

In general we aren't keen on questions that are just calculation and are obviously homework. We prefer questions to be more conceptual.

@Algebra2015 you can't send files, in general, but you can have images show up inline in the chat. Just post the URL in a message by itself.

http://i.imgur.com/HDK8PhG.png

@Slereah nonononononono, that's not true! The l.h.s. is non-compact, the r.h.s. is compact

What is true is that $\mathfrak{su}(2)\times\mathfrak{su}(2)$ is the compact real form of $\mathfrak{sl}(2,\mathbb{C})\times\mathfrak{sl}(2,\mathbb{C})$, and that their complex representation theories are equivalent.

I don't know why the myth that the Lorentz group or its cover is somehow isomorphic to $\mathrm{SU}(2)\times\mathrm{SU}(2)$ is so pervasive in physics - we've known since Wigner that it is not, and that the non-compactness of the Lorentz group is essential for its representation theory and hence the classification of particles.

There's a lot of group theory that Wigner and Weyl did completely right, but that all standard physics presentations seem to get wrong.

@alarge I don't think many people sample from within the isolikeliihood contour with MCMC (though I've heard it discussed). It could spoil the posterior weights of the samples in a naive implementation, but mainly because the most popular code, MultiNest, uses ellipsoidal sampling

How would you know by looking at the Lorentz or Conformal group to split up the generators to get closed subalgebras? Makes no sense just looking at it to me :(

@ACuriousMind Well I used $\approx$ in a rather informal sense here :p

not to imply isomorphism or anything

hey @skillpatrol you want to discuss a written chat room etiquette during the chat session? cf. this:

no, but thanks for the offer :-)

@bolbteppa The idea of finding subalgebras is completely general, the maximal torus/Cartan subalgebra and the Cartan-Weyl basis for semi-simple algebras make sense in general. To see this as natural, however, requires the theory of roots.

@DavidZ: What is the agenda for today, anyway? Or is it secret until the session begins? ;)

@skillpatrol sure. But in case you ever wonder in the future why we don't have written chat room etiquette... this is why ;-)

@ACuriousMind not secret, there just isn't a really good way to post it, and also I don't have much of anything on the agenda. Just review of the replacement homework policy.

If we don't have specific things to talk about, it's just open discussion most of the time.

I've thought about making a meta post (or posts) to solicit topics for the chat session, but that might be too formal? I dunno. It seems like it could be overkill.

@ACuriousMind thanks, so I think you get a bunch of Lie Algebra elements ('tangent vectors') from your Lie Group which form a vector space basis, but because you have this extra product structure you can actually generate the basis from a subset of the basis generators, getting the Cartan subalgebra, therefore the lie algebra representation you started with must be reducible, right?

(and roots are the 'eigenvalues' of the elements of this subalgebra of operators acting on the same lie algebra operators in the space thought of as vectors)

@DavidZ point well taken ;)

@bolbteppa I'm not sure what you mean by "the Lie algebra representation must be reducible". It's not, the adjoint representation of semi-simple Lie algebras is irreducible. (The Cartan subalgebra is abelian, i.e. everything in it mutually commutes, but it is not central, it doesn't commute with things outside the subalgebra. Non-trivial center would indeed imply reducibility through Schur's lemma)

@ACuriousMind so the lie algebra of generators $J_i$, $K_j$ and their commutation relations before you form $\mathbf{J}^{\pm}=\frac{1}{2}(\mathbf{J}\pm i\mathbf{K})$ is already irreducible? Then why does it ("essentially", Ryder) split into two copies of $SU(2)$ upon forming these combinations, which constitute finding the Cartan subalgebra?

@bolbteppa Oh. I misunderstood what you meant by "finding subalgebras". What is happening is this: We take the Lie algebra $\mathfrak{so}(1,3)$, and complexify it. In the complexification, we choose the $J_i^\pm$ as basis. The Cartan subalgebra would be given by any pair $J_i^+,J_i^-$, as they commute. However, we can also observe that the $J^+$ and $J^-$ separately form subalgebras (not abelian ones).

These subalgebras are $\mathfrak{sl}(2,\mathbb{C})$, so $\mathfrak{o}(1,3)_\mathbb{C} = \mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$. And the real compact form of $\mathfrak{sl}(2,\mathbb{C})$ is $\mathfrak{su}(2)$.

Why do people even mention SU(2), though

Is it actually useful mathematically here

Because the representation theory of compact groups is very well-understood

Or is it just because they are used to SU(2) in non-relativistic QM

We know the reps of $\mathfrak{su}(2)$ - they are the spin representations labeled by half-integers.

Are they the same as the reps of SL(2,C)

Yes and no

The finite-dimensional representation theory of an algebra and its compact real form is the same

But?

But the representations which are unitary on the compact real form (and unitarity is the reason we want the compact form in the first place, as it simplifies the representation theory immensely) are not unitary for the other forms.

That's not too problematic since we do not expect it to be

@Slereah Oh, people are routinely puzzled by the fact that boosts are non-unitary on the target space of the fields :P

So at least some people do expect it

Well fields aren't states :p

@ACuriousMind Yeah, I think that the original $J_i$ and $K_j$ elements are anti-Hermitian right? So I guess in general the motivation for complexifying is to build more anti-Hermitian (or Hermitian) elements because the exponential of (anti)-Hermitian lie algebra elements gives unitary elements of a lie group right?

@bolbteppa 1. There is no notion of "Hermitian" on a Lie algebra. It doesn't have a conjugation operation. 2. The motivation for complexification is ultimately that the representation of the complexification are equivalent to representations of the original algebra, and that a complex semi-simple Lie algebra has a compact real form, whose representations are still equivalent, and we often understand the representation of the compact group very well - as in this case, we already know SU(2) reps.

3. You cannot expect an algebra of a non-compact group (such as $\mathrm{SL}(2,\mathbb{C})$) to be Hermitian in any representation, since it would exponentiate to a non-compact unitary subgroup, which is impossible in finite-dimensional spaces.

@ACuriousMind but page 2 (e) ma.utexas.edu/users/sadun/S06/Lie/Hwsol4.pdf says there is a notion of Hermitian, after all the lie algebra is just a collection of operators, also page 9/10 isites.harvard.edu/fs/docs/icb.topic1288789.files/… thus you need to exponentiate Hermitian/anti-Hermitian lie algebra elements to get unitary operators, which seems like an obvious reason to form them?

Just stick with nice compact lie groups first and see if this thinking makes sense

@bolbteppa Oh, but that is an argument that is particular to algebras of complex matrices, and their presentation as complex matrices. If you write e.g. $\mathfrak{su}(2)$ like it acts on itself, you get $\mathfrak{so}(3)$, and on the latter, you do not have a natural notion of conjugation, since it's just real matrices.

I would not want to motivate steps taken to deduce the representation theory by already choosing a particular representation.

So in the end the whole complexification is just a Clever Ruse

To get easily the representations

"The order created by God is on a foundation of uncertainty. The Book of Genesis explains that the world was an abyss of chaos at the moment of creation. Quantum mechanics is predicted in several additional respects by the Biblical scientific foreknowledge."

Obviously @JohnDuffield it is people who are responsible for behaving badly.

But why the sudden change in the last 20 years @JohnDuffield was my question?

@bolbteppa I should clarify that whereever I said "complexification", I meant of the algebra, not of the group (the two notions are different)

Also, for a real Lie group, $\exp(\mathrm{i}X)$ is not an element of the Lie group (unless you're a physicist and defined the exponential with that $\mathrm{i}$ in there in the first place)

Yeah, so right now it's like you have lie group elements of the form $e^{iH}$ then you find the lie algebra, note it's 'reducible' to a cartan subalgebra, complexify the cartan subalgebra elements because you can then do the lie algebra Cartan/polar decompositon $e^{iA}$ which are a representation of your lie group elements? If you started from a real lie group just set $H = -iH'$ then you'd still get some complex quantity very similar to what you started with.

I don't understand what you just said. The Cartan algebra has nothing to do with the complexification, and the complexification of the algebra has nothing to do with polar decomposition

@MAFIA36790 Yes. This is something they've been talking about for about a month.

The link goes to the meta.stackexchange.com post where they brought it up.

@ACuriousMind Getting confused by the signs in this answer of yours. If $[M_1,M_2]=iM_3$ and cyclic, then $M_\pm:=M_1\pm iM_2$ results in $[M_+,M_-]=2M_3$, which fits into your equation, but $[M_+,M_3]=-M_+$, does not fit your equation. Maybe it has to do with my edit (which you approved), but it does not seem to fit before the edit, too.

Ugh. Signs. ;P

Yep :)

But I'm not able to get it right.

Tried adding a sign or a $i$ into the definition of the $M_\pm$'s, but that does not affect the algebra.

I seem to recall that the signs I use in that answer mean that the $L_{-n}$ are the raising operators.

The equation $(1)$ in the question seems to be right.

Ah, no, it's right. The $L_{-n}$ create the states from the highest weight, and are the lowering operators

@ACuriousMind So $L_{-1}$ is $M_-=M_1-iM_2$ in $SU(2)$? But then the commutator of the ladders with $M_3$ (or $L_0$) does not fit.

Yes, I see your problem

If you reverse the ladders like saying that $M_+$ is $L_{-1}$, then it's the commutator $[M_+,M_-]$ that's wrong.

Maybe your equation defines the Witt algebra, but it's a different story for $SU(2)$?

@BernardMeurer Where can I find the program? guessing github? link please :p

@Bass I am absolutely sure the three things are the $\mathfrak{sl}(2,\mathbb{C})$ algebra, i.e. complexified $\mathfrak{su}(2)$.

The Wikipedia article on the Witt algebra agrees, fwiw

@ACuriousMind Yep, just saw that. It does seem to fit naturally into the Witt algebra equation, before you notice that sign..

@MAFIA36790 How was your JEE exam?

@ACuriousMind Hmm, the same article says that "over the reals", it's $sl(2,\mathbb R)=su(1,1)$. Shouldn't that be $su(2)$?

@Bass No, it means the real span with "over the reals". $\mathfrak{su}(2)$ is the compact real form, but it is not spanned by $L_1,L_0,L_{-1}$ as a real vector space - It's $L_1 + L_2$ and $\mathrm{i}(L_1+L_2)$ or something like that that span it.

what are stackexhange chat bots? and what are they used for?

@ACuriousMind Couldn't that be the issue at hand? Limiting the Witt algebra to $n\in\{-1,0,1\}$ gets $\mathfrak{sl}(2\mathbb C)$ or $\mathfrak{sl}(2,\mathbb R)$, which is not the same as $\mathfrak{su}(2)$.

@Bass So what? The commutation relations in the question are also not those of $\mathfrak{su}(2)$ - neither $L_+$ nor $L_-$ are in it

They are those of $\mathfrak{sl}(2,\mathbb{C})$.

Btw, what's a compact real form in this context?

A real algebra that, when complexified, gives back the original one, and on which the Killing form is negative-definite.

@ACuriousMind Doesn't the question deal with the Lie algebra $\mathfrak{so}(3)=\mathfrak{su}(2)$?