@Slereah if i stick to the action density and don't take a prefered time direction ... Do I still have to worry about diffeomorphism invariance?

@Mr.Feynman Near me there is a canal and the lockkeeper used to keep three geese. They used to wander along the canal path and would attack you and peck at your knees if you walked past. While hardly life threatening it was quite painful.

hehe

those things while funny can be dangerous if you're on a bike

there are birds here called magpies, they're related to crows. but leaner. well they will swoop you during nesting season, and sometimes even claw at your face, though that is fairly rare

(rare that they will claw, very common that they will swoop)

@antimony I'm confused, either you picked the wrong image or you're not talking about actual magpies. (The image is an Australian magpie, which is not a corvid and not especially related to crows unlike "true" magpies)

Didn't realize @ACuriousMind was a magpie racist

it's my darkest secret

@JohnRennie Well, not all corvids are crows, but all crows are corvids, no?

My knowledge of crow taxonomy extends only to what I could glean in 30s Googling, but in everyday life "crow" means any bird that's blackish, noisy and aggressive so I think we can forgive the Australians for calling Australian magpies "crows".

Though in the UK I can't think of any bird we colloquially refer to as a "crow" that isn't a member of Corvidae.

I mean, it's clear to me why they call it a magpie - it looks a lot like true magpies and seems to be rather intelligent

but that doesn't change that it is not genetically closely related to true magpies or corvids in general

convergent evolution can be weird, like the bunch of crabs that aren't actually true crabs

They are not completely unrelated. They are both passerines. I believe that's a fairly large grouping but it's still a small subset of all birds.

I love these essentially inconsequential arguments :-)

@JohnRennie I bet it was painful but nonetheless funny to imagine :P

@JohnRennie it's over half of all birds! (At least if the first sentence of the Wiki article is true)

@ACuriousMind I'm laughing because this is just the same discussion as the one about representations just replacing birds with groups

I had the corvid a few months ago

Not a fun disease

@Mr.Feynman It used to amuse me, but they attacked me one night when I was walking home after an evening drinking and I lost my temper and kicked one of them. After that they avoided me. Presumably this shows geese are capable of remembering humans.

@Slereah I already used that joke when talking about birds :P

@ACuriousMind That's within an order of magnitude of a small fraction of all birds.

Ah, the cosmologist approach to approximations

@JohnRennie Maybe now there some geese out there plotting to take you out

As for geese, there's a bunch of geese both right where I live and along the river through the city but I've never been attacked by them or seen them attach anyone (though they are pretty fearless when they think there's food)

maybe geese outside of continental Europe are meaner? :P

These were domestic geese - I wonder if that makes a difference.

@ACuriousMind I would expect British geese to be very polite if anything

Although you'd think domestic geese would be bred to be more placid.

Maybe it's just that they were used to humans so weren't as wary of them as wild geese. Or maybe they just hated me :-)

You have some fearful enemies out there

@JohnRennie not sure, some people use them as watch geese after all

True ...

We often eat goose for big family dinners. They have a lot more flavour than chicken and are a lot larger. Also goose eggs are very nice as they're richer than hens eggs. Though they're only available round about Easter.

I don't think I've ever had goose (not a lot of big family dinners here, we usually stay with the smaller chicken or duck)

They taste much like duck. I'm not sure I could tell the difference blindfold.

I'm afk 10 minutes and I find you talking about eating those geese :(

Well actually I'm laughing about the sudden change :P

Spin should appear also in the context of representations of the Galilei group, shouldn't it?

sure

Sometimes people talk about the Poincaré group as if spin appearead only there

there is a strange confusion around spin where a sizable fraction is convinced that spin appears only in relativistic QM

this is probably because historically it of course was considered in the context of the relativistic Dirac equation

you can come up with spin purely non-relativistically by thinking about Wigner's theorem and projective representations, but this rather algebraic approach to QM isn't as widespread

Oh, that makes sense

The same thing happened for the gyromagnetic factor of electron and Dirac equation now that I think about it

Or the same thing that happens for classical Schrödinger fields that people seem to loathe :P

If you read earlier textbooks on QM, spin in relation to projective reps is discussed in full detail. There is even a chapter in the Feynman lectures where he explicitly motivates it

A chapter in FLP about projective reps?

Look at equation 6.12 here feynmanlectures.caltech.edu/III_06.html

Oh, the spin one-half lecture

He has that phase on the LHS

I read that almost two years ago and I didn't know anything about representations back then

Spin doesn't appear much in classical mechanics due to the commutation relations

Since you can't have fermions in the same state you can't really have big fermion waves

As you can with EM for instance

there isn't much you can do with spinors that would have consequences in the classical limit

Sure, that's what happens with the polarization of EM waves but the problem was more about representations here

Those are mostly historical reasons I think

Spinors didn't appear in physics until Dirac I think?

So everything is about QM now for them

Even the name

even math people call them spinors :p

but you can use spinor reps in classical mechanics yeah

It is used in GR for instance

Also in computer graphics, I guess technically the quaternion rotation is done using a spinor rep?

Though then again, quaternion rotations are older than spinors proper

@Slereah Pauli

Was Pauli before Dirac?

Hm, one year before

People linguistically say that the group formed by exponentiating the 3 dim irrep of $\mathfrak{su}(2)$ is $SO(3)$ however we have $SO(3)\cong SU(2)/\mathbb{Z}_2$. Do they just mean locally they are the same?

I guess so, Pauli introduced spinors phenomenologically

@DIRAC1930 Did you mean $\mathrm{SO}(3)$ in the last isomorphism?

Yes sorry

@DIRAC1930 Lie's third theorem is about a specific Lie group yeah

$\mathrm{SU}(2)$ is the double-cover of $\mathrm{SO}(3)$, as such they are locally isomorphic

Which means $\mathfrak{su}(2)\cong\mathfrak{so}(3)$

There is a unique simply connected Lie group that integrates a given Lie algebra

But SU(2) isn't simply connected

@Slereah It is

Hm

$SO(3)$ isn't simply connected

It is the $\mathrm{S}^3$ sphere as a manifold

I guess

Regarding the point raised by @DIRAC1930 above, due to the fact the algebras are isomorphic, they have the same representations. The problem is that you can't always lift an algebra rep to a group rep, so the representations of $\mathrm{SU}(2)$ are not representation of $\mathrm{SO}(3)$ in general

But odd-dimensional irreps like the one you mentioned turn out to be

So there exists irreps $\rho$ and $h$ such that $\rho(SO(3)) \cong h(SU(2))$ for the odd dimensional case?

@DIRAC1930 I'm not sure what you mean. You have a group homomorphism $\Phi:\mathrm{SU}(2)\rightarrow\mathrm{SO}(3)$. Such homomorphism induces a homorphism (isomorphism, actually) of algebras $\phi:\mathfrak{su}(2)\rightarrow\mathfrak{su}(3)$. Now Given a $2\ell+1$ dimensional irrep $\Pi_\ell$ of $\mathrm{SU}(2)$ which induces a representation $\pi_\ell$ of $\mathfrak{su}(2)$, the map $\pi_\ell\circ\phi^{-1}$ gives a representation of $\mathfrak{so}(3)$

Only if $\ell$ is an integer this descends from a representation of the group

Are you sure $\Phi$ is an isomorphism? It's not invertible

@DIRAC1930 Not $\Phi$, but the induced algebra homomorphism $\phi$

Ah sorry I misread

That's the content of locally isomorphic

@DIRAC1930 they mean that the 3d representation of $\mathfrak{su}(2)$ induces a representation $\rho : \mathrm{SU}(2) \to \mathrm{GL}(3,\mathbb{R})$, and that this representation has $\rho(\mathbb{Z}_2) = \{1\}$ so that this descends to a representation $\rho' : \mathrm{SU}(2)/\mathbb{Z}_2 \cong \mathrm{SO}(3) \to \mathrm{GL}(3,\mathbb{R})$ and you have that $\rho(\mathrm{SU}(2)) = \rho'(\mathrm{SO}(3)) \cong \mathrm{SO}(3)$ (the image is just $\mathrm{SO}(3)$ as a matrix group)

this is a concrete example for what I was talking about yesterday:

Ah okay thanks. I calculated the symmetrized tensor product rep of SU(2) and showed $\rho(\mathbb{Z}_2) = {1}$.

I'm assuming the same thing happens for the 4 dim irrep of $SL(2,\mathbb{C})$ and the 4 dim irrep of $SO(1,3)$. We have the same $SO(1,3) \cong SL(2,\mathbb{C})/\mathbb{Z}_2$ thing going on there too.

So I have $\mathfrak{sl}(2,\mathbb{C})\cong \mathfrak{sl}(2,\mathbb{C} \oplus \mathfrak{sl}(2,\mathbb{C})$. I want to take the real form of the LHS so I can exponentiate it to $SL(2,\mathbb{C})$. What will the R.H.S. turn into? Will it just be $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$?

@DIRAC1930 did you forget a ${}_\mathbb{C}$ on the l.h.s.?

because no Lie algebra is isomorphic to the sum with itself :P

Oops yes I did lol

also, "the real form" is not unique

we have $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}\cong (\mathfrak{su}(2)\oplus\mathfrak{su}(2))_\mathbb{C}$, i.e. both $\mathfrak{sl}(2,\mathbb{C})$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ are real forms of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$

(they are the split and the compact real form, respectively)

so your question doesn't really make sense as written and I'm not sure what you actually mean

Can I write $\mathfrak{sl}(2,\mathbb{C}) \cong \mathbf{A}_\mathbb{R} \oplus \mathbf{B}_\mathbb{R}$ using the notation here en.wikipedia.org/wiki/…?

It seems legit but I always get something wrong

I mean, you just did, but it's wrong :P

the $A$ and $B$ are defined with an $\mathrm{i}$ and hence are not elements of $\mathfrak{sl}(2,\mathbb{C})$

Dammit lol

this is a common mistake and the reason one often sees claims that $\mathfrak{sl}(2,\mathbb{C})$ and $\mathrm{su}(2)\oplus\mathrm{su}(2)$ are isomorphic or whatever

but really they just have the same complexifications but are non-isomorphic real forms of it

Is this something that's easier to do at the level of representations rather than looking at the abstract algebra?

I don't understand the question, what does "this" refer to?

Well people write $X = x^\mu \sigma_\mu$ and then notice that a real $x^\mu$ corresponds to a Hermitian $X$

Then with this in mind I can write $X$ as an exterior product of two spinors and just see what restrictions that places on the $SL(2,\mathbb{C})$ tensors

Hi, in my college physics lab manual about measuring the phase velocity of transverse and longitudinal waves, the authors specifically mention that a rope with a SQUARE CROSS-SECTION must be used, without any reasoning. Is there any specific reasoning?

@DIRAC1930 Saying that $X$ is "Hermitian" already means you're in some representation

"Hermitian" doesn't mean anything for an abstract Lie algebra

Yes but I don't know how to progress at all just looking at the Lie algebra lol

what exactly are you trying to do?

@ACuriousMind Doesn't that come from the complexification being isomorphic to $\mathfrak{su}(2)\oplus i\mathfrak{su}(2)$? Do you mean that $\mathfrak{su}(2)\oplus i\mathfrak{su}(2)\not\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)$?

Well I have $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C} \cong \mathbf{A}_\mathbb{C} \oplus. \mathbf{B}_\mathbf{C}$. I have found a representation of the algebra given by $D^+$ such that $\{\mathbf{A},\mathbf{B}\}\rightarrow \{\imath \sigma_i,0\}$ and $D^-$ such that $\{\mathbf{A},\mathbf{B}\}\rightarrow \{0, \imath \sigma_i\}$. I can exponentiate this to a group but it will be the exponential of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$ and not $\mathfrak{sl}(2,\mathbb{C})$.

Is it best to assert that the $4$ dim vector representation acts on a real vector space and then see what conditions it produces on the spinor reps

@DIRAC1930 yes, of course, and $\mathfrak{sl}(2,\mathbb{C})\subset\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}$, so this is also a representation of $\mathfrak{sl}(2,\mathbb{C})$. What do you need?

I essentially want to write Lorentz transformations on $\mathbb{R}^{1,3}$ in terms of $SL(2,\mathbb{C})$ tensors on Weyl spinors

What do you mean? Weyl spinors are already a representation of Lorentz algebra and Lorentz transformations act on them in a definite way

Well I should be able to relate $\xi \chi^T$ to a vector of $\mathbb{R}^{1,3}$ with components $X^\mu$.

I guess this would be the vector $X\rightarrow X^\mu\sigma_\mu$

Which leads $\xi \chi^T$ to be a Hermitian matrix

However, what happens to the $D^+ \otimes D^-$ acting on it I have no idea

so you want to show that $(1/2,0)\otimes(0,1/2)\cong (1/2,1/2)$?

duck is so very delicious

I think so

hi, in AP French's book Vibrations and Waves, pg 168-170, he derives the formula for forced vibration for a stretched string. But he states that "the important part of the above result is that a large forcing response with a small driving amplitude by having the forcing take place at a point which is close to being a node of one of the natural vibrations". How does this matter? Why would forcing near a node produce larger responses?

@DIRAC1930 or is the question how to show that the usual vector representation on $\mathbb{R}^{3,1}$ is the one labeled by $(1/2,1/2)$?

because the first isomorphism is essentially true by definition of what the labels $(s_1,s_2)$ mean, at least if you defined them nicely :P

Here's a link I found for reference: wiac.info/docviewerf for my last question

I guess maybe this problem occurs more simply in non-rel physics. If I have the fundamental rep of $SU(2)$ and I take the tensor product with itself and symmeterize it to form an irrep, I will have a cset of complex matrix acting on $\mathbb{C}^3$. The fund irrep of $SO(3)$ is however a set of real matrices acting on $\mathbb{R}^3$

@insipidintegrator that's a link to web PDF standards?

also, I think your question would be a perfectly fine question for the main site

(you should quote what the "above result" refers to, though, not everyone has access to every book)

@DIRAC1930 that's...not really true, if we're talking complex representations then the fundamental irrep of $\mathrm{SO}(3)$ is a $\mathbb{C}^3$, too

what is true is that the fundamental rep has an invariant real form

but I don't see why that would be relevant unless we have some explicit reason to go looking for real representations (as when looking for Majorana spinors)

that is to say: I still haven't really understood what you're trying to do

If I understand it, they're trying to relate the four-vector rep to the Weyl spinors rep explicitly and they might be uncomfortable with having complex 4-vectors

okay, so the correct statement here is that $\mathbb{R}^{3,1}$ is the real version of the representation on $\mathbb{C}^4$ that we label by $(1/2,1/2)$

Why is every conversation here about spinors

My least favorite topic

Slereah, you crack me up :P

at least so far I haven't had to talk about Majoranas, but this is uncomfortably close :P

What do you mean by real version of the representation on $\mathbb{C}^4$?

I feel I keep repeating myself :P

@ACuriousMind If I were you, I would phrase that like "I'm using myself as a bibliography" in this case :P

source: me

That reminded me of Hachiman

wth is a Hachiman

And before you ask, Hachiman is an anime character

I was too slow

He repeats "source: me" a bunch of times

also, "an anime character" is about as precise as "a character from a book" :P

A character from this anime

Italian detected!

I should change my settings lol. I keep googling stuff in English and get italian wiki first

ACM, if you don't mind me asking this again, what did you mean here? I thought $\mathfrak{su}(2)_\mathbb{C}\cong\mathfrak{su}(2)\oplus i\mathfrak{su}(2)$

how does what you say contradict what I said?

because $\mathfrak{su}(2)_\mathbb{C}\cong\mathfrak{sl}(2,\mathbb{C})$

Unless you meant this

ah, careful: $\mathfrak{su}(2)_\mathbb{C}\cong \mathfrak{su}(2)\oplus\mathrm{i}\mathfrak{su}(2)$ is an isomorphism of vector spaces not of Lie algebras

Oh, that makes sense

a proof that $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is not isomorphic as a Lie algebra to $\mathfrak{sl}(2,\mathbb{C})$ is that $\mathrm{SU}(2)\times\mathrm{SU}(2)$ is compact, but $\mathrm{SL}(2,\mathbb{C})$ is not compact, so these are two non-isomorphic simply-connected groups, hence their Lie algebras cannot be isomorphic

Oh, that's a nice straightforward proof

Thanks

The condition is $D_+^\dagger=D_-$ it seems

The next step is to figure these equations out for $x^\mu x_\mu=0$ (on the light cone) so I can write $X=x^\mu \sigma_\mu$ as an outer product $\xi \otimes \xi^*$

lol

I will probably be here for a year lol

Actually I'm going to try mathematica

I have more than 4 equations for 4 unknowns and I know what the Hermitian matrix should look like

I think

Actually I give up lol

I love that we're still talking about spinors :P

Spinors are maybe my favorite mathematical object right now

Also, in the last year the question "wth is spin?!" turned into a more elaborate "wth is really a spinor?!"

@Mr.Feynman The question never ceases