00:00 - 17:0017:00 - 23:00

12:06 AM
That's true
What exactly is the Hermitian conjugate in tensor notation?
$\epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ doesn't reduce to $\psi^\dagger \psi$ in matrix notation does it?
It seems to be $\psi \epsilon \psi^\dagger$ right?
Does $\psi_\alpha$ correspond to $\psi_i$ in matrix notation and $\psi^\alpha$ correspond to $\psi_i^T$?
Hmm it can't be
@ACuriousMind

3 hours later…
3:40 AM
@DIRAC1930 if you use e.g. Srednicki's notation it does
you have to know which notation you are using

6 hours later…
9:17 AM
The $\epsilon$ is an inner product on the space of spinors
So yes that's about how it should be
$\psi^\dagger \varepsilon \psi$
I recommend learning spinor index notation, it does help to keep track of everything

Inner product for Weyl spinors in $D = 4$, what about other $D$

You're the expert here :p
I don't know a lot about spinors in arbitrary dimensions

I mean you could interpret $\overline{\psi} \chi = \psi^{\dagger\alpha} (\gamma^0)_{\alpha}^{\beta} \chi_{\beta}$ as an `inner product' in any D no?

Well, here it's a dual spinor acting on a spinor!
Not an inner product unless it's in the same space
You need some spinor musical isomorphism or whatever

9:49 AM
Trying to interpret a GR measurement as an holonomy
It is not a trivial idea
I have seen some papers hinting at it, though, so it's probably doable

You can interpret $\varepsilon_{\alpha \beta}$ as an 'invariant metric' which might be useful on the inner product point
In mathematics, a symmetric space is a Riemannian manifold (or more generally, a pseudo-Riemannian manifold) whose group of symmetries contains an inversion symmetry about every point. This can be studied with the tools of Riemannian geometry, leading to consequences in the theory of holonomy; or algebraically through Lie theory, which allowed Cartan to give a complete classification. Symmetric spaces commonly occur in differential geometry, representation theory and harmonic analysis. In geometric terms, a complete, simply connected Riemannian manifold is a symmetric space if and only if i...

Also there is that Thing where the G-structure of the manifold may influence the measurement scheme
Depending on the measurement you perform, I think they might be holonomies of different groups
I guess you can maybe construe the angles and other informations of your particle exchange as a parallel transport of a tetrad or something
but I'm not 100% sure if that works alright when one of the information is the proper time
but maybe that's just equivalent to the timelike part of the tetrad idk
since the original angular informations you get are spatial angles
But there is that whole idea that if there is no clock measurement, you lose part of the structure of your spacetime and instead measure its affine structure
So I guess instead of some $SO(n)$ holonomy you get naught but a $GL(n)$ holonomy
I should probably start with Minkowski space since that's the only case where this idea is developped

10:13 AM
Simultaneity connection form on a static spacetime is just a unit time translation vector
short enough
I guess things simplify a lot when your fiber is one dimensional
Also for the simultaneity bundle the interpretation of the vertical space is pretty easy

@imbAF It's not practical to reduce everything down to the QM level. It's ok for small systems, but the amount of computation required grows enormously as the number of elements in a system increases. So you need to use a hierarchical approach, where you work at the highest level that you can, and only go deeper when you need to.
To give a simple example, the rules of polymer chemistry tell you how a Lego brick behaves, but it'd be silly to work at the level of polymer chemistry when you're just trying to build something out of Lego.

10:29 AM
I do all my gardening from string theory

they say string theory is more like a jungle, than a garden
I guess one could create a garden within a jungle...

A jungle of jungles of jungles

I mean you know
I'm not gonna lie
Doing simple things from horribly complex theories is fun
but I wouldn't lose sleep over it if you can't do it
You're probably not gonna miss out on the meaning of life
Not gonna find the secret to immortality in quantum mechanics

The joy is in the journey

10:48 AM
the part that isn't fun is learning biochemistry
I hope you enjoy ion transport

Biochemistry is for people with photographic memories

That ain't me
You could ask me the pythagorean theorem and I will check it up just in case

Recall the 90 year old who got his PhD in physics?

Hopefully that's not me in the future
Apparently the holonomy of a spacetime measurement is indeed related to tetrad transport, I think
$$\delta t = \oint_\tau \sigma^* \omega d\tau$$
with $\omega$ the time translation form and $\sigma$ some choice of simultaneity surface, I think?
"So far we have considered stationary frames. In general, however, the trajectories of the observers which define the frame, are not generated by a Killing vector field.
This does not imply that the simultaneity is no more described by a gauge theory, but only that there is no connection preserving structure group."
" The curvature is the angular velocity of the observers at rest in the frame and the holonomy is the obstruction against the possibility of applying the Einstein synchronization in the large."
Damn holonomy

11:47 AM
Hi there, may I ask you about may idea of "mass is the source of space"?

Hi :-)
There are lots of vacuum solutions in GR. They even have a Wikipedia entry dedicated to them:
In general relativity, a vacuum solution is a Lorentzian manifold whose Einstein tensor vanishes identically. According to the Einstein field equation, this means that the stress–energy tensor also vanishes identically, so that no matter or non-gravitational fields are present. These are distinct from the electrovacuum solutions, which take into account the electromagnetic field in addition to the gravitational field. Vacuum solutions are also distinct from the lambdavacuum solutions, where the only term in the stress–energy tensor is the cosmological constant term (and thus, the lambdavacuums...
None of these contain any matter or energy, so they are all spacetimes that exist even though no matter or energy is present.

Yeah, ok, I know. But, look at the electrostatic equation div E = 4 pi rho. There's also the trivial solution that there is a constant electrostatic field everywhere without

Incidentally an electric field counts as matter/energy i.e. it curves spacetime like matter does.

Any souce

Well yes, but I don't see the relevance of that ...

11:51 AM
We are very used to flat spacetime around us. Not used to constant electric field around us.

True, but I don't see what point you are making. A constant electric field everywhere is physically unreasonable as it would have an infinite energy.

So it's understandable that we assume there is a constant flat spacetime in the absence of any source - however, it does not has to be necessarily. And that could explain dark matter.

No, not really
A spacetime without any matter is called Ricci flat

There are also boundary conditions in $\nabla \cdot E = 4 \pi \rho$ along with the other Maxwell equations, boundary conditions which assume the fields vanish at infinity, fixing the value of the free solution that you're adding to the above equation

No, we do not "assume there is a constant flat spacetime in the absence of any source"

11:55 AM
Dark matter doesn't have any characteristic of a Ricci flat sort of spacetime

Every field has infinite energy, that for what renormalisation is used...

Renormalization is a collection of techniques in quantum field theory, the statistical mechanics of fields, and the theory of self-similar geometric structures, that are used to treat infinities arising in calculated quantities by altering values of these quantities to compensate for effects of their self-interactions. But even if no infinities arose in loop diagrams in quantum field theory, it could be shown that it would be necessary to renormalize the mass and fields appearing in the original Lagrangian.For example, an electron theory may begin by postulating an electron with an initial mass...
The classical example there is an example of how jarring this is

@BarrierRemoval No.
In QFT a naive approach gives an infinite energy, but this is an artefact of the computational method. Fields do not eally have an infinitye energy.

Not?

No, of course not. Nothing in the real world has an infinite energy.

11:59 AM
It simply goes back to working with point particles

I thought they have because of infinite extent

Nope.

That's what renormalisation is for?

The infinite energy comes from the assumption that interactions take place at points.

Ok, sounds good

12:00 PM
It's a computational artefact.

Or evidence of strin....

This does not arise in GR as it's a completely different theory.

Okok, that renormalisation thing is good and cool and interesting, but I have to learn that before I can talk about it...
What I really want/need to talk is that "mass is the source os space" idea I have

Have a read of the free field example in the wiki above

Hm, I think the "connection form" if I try to go that route for the measurement problem is gonna just be the pushforward of the timefunction

12:05 PM
@BarrierRemoval OK, but that's clearly wrong as there exist spacetimes with no mass.

Also possibly I can involve the tetrads if I am lucky

Where do they exist?
Between galaxies?
Certainly NOT
Everything is moving in our universe
So there is no flat space
We assume it
Bit we do not measure it. Nowhere

Well we assume GR is the correct theory because it correctly describes experimental observations.
And GR tells us that vacuum solutions can exist, therefore GR tells us mass is not the source of spacetime.

Yes, it does. I love GR. I do not wanna change a thing. Only the boundary conditions.

So if GR is correct mass is not the source of spacetime.

12:13 PM
But I told you, in electrostatics, div E = 4 pi rho also tells us that a constant electric field may exist without sources

Well so it could in a hypothetical universe.

Yes.
AND: in our universe there could it be that there wasn't any space left if you would take out all the mass.
It's about flatness of space between the galaxies. How do we know that the space is flat there????
We assume it
Because in the solar system, it fits the observations.

In your example we believe a constant electric field is possible because it is a solution to Maxwell's equations. In GR we believe a vacuum solution is possible because it's a solution to Einstein's equations.

But the solar system is in free fall in the gravitational field of the galaxy.

Spacetime isn't flat between the galaxies. Observations suggest spacetime everywhere is full of dark energy and that causes spacetiem to curve.

12:19 PM
Yes. But we do not calculate in electrostatics like "well, in the infinity, the field needs to approach the constant electrostatic background field". But in GR, we do!!!!

No we don't.

Yeah, ok, dark energy... But dark energy works on greater scales. You cannot explain dark matter with effects of dark energy. We can leave dark energy out from this discission.

The FLRW metric that (approximately) describes the universe does not assume asymptotic flatness.

Sure, we xo.
We do

Quite the reverse in fact as it assumes a roughly equal matter/energy density everywhere.

12:22 PM
Also "the source of spacetime" doesn't mean anything

Yes. I know that FLRW is not asymptotically flat. But that is not THAT kind of "not asymptotical flatness" I am looking for. FLRW is on universal scale. Doesn't matter between galaxies
"the source of space" is better.

Also no

It means a lot.

To you perhaps
but you may want to explain what you mean exactly

Time just does the opposite from what the space does as it has the opposite sign.

12:23 PM
mama mia

In the derivation of the Schwarzschild metric
Mama mia => hahaha. Please, help me :-)
Ok. Derivation of the Schwarzschild-metric: We assume there A approaches 1 in the infinity. Because then, the spacetime becomes Minkowski in the infinity.

@BarrierRemoval The time component has a different sign to the spatial components, but the statement "Time just does the opposite from what the space does as it has the opposite sign" is meaningless.

To quote Pauli
That's not right, that's not even wrong

@JohnRennie: No, it is not. From Ricci = 0 (ricci-flat space) you can derive B=1/A.
Where B is the coefficient in front of the time component
And A is the coefficient in front of the dr^2 component
In the static sperically symmetric metric
(general form)

If you want a counterexample just pick any gravitational wave spacetime
All of these spacetimes are Ricci flat and they are not at all like that

12:31 PM
Yes, there, the Weyl - tensor is not zero.
My image of "mass is the source of space" is perfectly fine with gravitational waves.

What does "mass is the source of space" mean

It means
1. That there won't be any space left if you would take out all mass of oir universe.

That is most certainly not true and not even really related to GR

2. (more practical) that A for a mass without any neighbours will approach 0, not 1.

unless you subscribe to relationism I guess, but then that's more of a discussion to take to philosophers
A isn't a generic feature of GR

12:35 PM
@BarrierRemoval the value of coefficients is tied to your choice of coordinate system, it doesn't actually mean anything to talk about " the time component" or "the dr^2" component for an arbitrary spacetime

Schwarzschild is a very specific spacetime

@BarrierRemoval Are you saying A = B = 1 applies only if we assume asymptotic flatness?

you need to fix your coordinates first, i.e. explain how you're choosing them for your particular spacetime, in order for that to make sense

And, because of meaning 2., the curvature inside galaxies and for the outer stars of galaxies, is steeper than with "A approaching 1". That exllains dark matter.
@JohnRennie: Yes, for the static case. Of course I know that with dark energy it isn't flat.

Aha, I think I get it. You're saying the spacetime between galaxies isn't really flat, therefore calculations for a galaxy based on asymptotic flatness are wrong.

12:37 PM
@JohnRennie: Yes
Not only "not really flat", but awfully unflat :-)

OK, but the spacetime between galaxies is experimentally well described by the FLRW metric.
And that predicts the spacetime is very nearly flat.

No, the universe as a whole is described by FLRW

And the universe as a whole includes the bits in between galaxies.

As a sum, it is also flat in my image.
In my view, and this might be meaning number
3. Spacetime curvature means volume gain

If there were any substantial curvature of the spacetime between galaxies it would affect the dynmics of the expansion in a drastic way! It would be blindingly obvious from observations.

12:40 PM
How?

Any significant positive curvature would have recollapsed the universe long before humans evolved, and any significant negative curvature would have pushed all the other galaxies behind a cosmological horizon.

3. Is not perfectly correct, sorry. Of course, we all know, that the Ricci-Tensor means volume gain.
What you mean is dark energy.
I'm fine with dark energy.
3. Curvature means volume changes.
There is more volume inside the galaxies and less volume between the galaxies. And together they sum up to zero curvature.

I think you focus too much on colloquial notions like "more volume" and too little on the actual math

Ok, you clearly caught me :-)

if you have an actual hypothesis of how to explain dark matter, go write it down and submit it, plenty of journals would love to hear it
but if you can't write it down using the same math and formalism of GR other physicists use, then you don't have a hypothesis, you just have a misunderstanding of how GR/physics works

12:52 PM
But I'm not really able to put this all in maths. At least not so fast. Would have to learn so much... Even do not know what renormalisation means ;-)

it's easy to read some pop-sci account of quantum mechanics or GR and think one understands enough to reason about these topics, but...that's just not the case

@ACuriousMind: I tried that!!! But it's not scientifically hogh standard, I suppose.

I'm afraid physics is a lot like Karate Kid

if you don't know the math you can't do the physics, that's the unfortunate fact

You have to paint a lot of fences before you can do the karate kicks

12:53 PM
Kindest answer was from Nature astronomy :

Painting fences in this context is to do a lot of boring problems about throwing rocks or waves in a pond

@BarrierRemoval you don't need to do it fast, but in physics, like most modern natural sciences, it really is such that you have to learn to crawl and then to walk before you can run

Yes, ok. I know that. And: I really don't want to do that.
I have plenty if other tasks.... Argh

My advice is to get into politics instead

Any one volunteers? :-)

12:54 PM
There's no barrier to entry there
Any rube can become an expert there

Ok. Then lets talk about the maths.

Have a look at the maths in the classical electromagnetism renormalization example in the wiki article

What we measure is constant velocity at the edges of the galaxies.

Well something like $\epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ as an invariant doesn't seem to be the right one

@DIRAC1930 Errr what answer do you want
There are several things you could describe as "the non-relativistic limit of EM"

12:58 PM
@DIRAC1930 this is a good starting point

Oops edited the wrong comment

Landau Lifschitz has a whole chapter on the Pauli equation with EM fields if you want

In that paper, do they use $\delta_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$?

Constant velocities means partial derivative of the velocity (which is dependent on r, A and B) to r is 0.
With A=1/B this is a well defined equation for A. There, you got the metric.
Now we need to get the velocity out of the general static spherically symmetric metric. That is possible to derive.
Geodesics equations. Maybe I should ask a precise question about that in physics SE?
Please, do not close that one.

1:14 PM
I read the LL chapter briefly and it was interesting
He deals with the spinors as being wavefunctions right?
I'm trying to find out the correct spinor invariants (out of field operators) I can build to put in a non-rel Lagrangian
So $\epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ is one however it leads to $\psi \epsilon \psi^\dagger$ or something which means it can't be the right one because what always seems to appear in text books etc. is $\psi^\dagger \psi$

I believe you are having some issues with the notation
which is normal because spinor notation is bad
$\psi^\dagger \psi$ is in fact the very same thing
I would recommend uuuuuh
there's a few nice books on spinor notation but they tend to be weird books
Siegel has a nice section on it
Chapter II is all about spinors, including in full index notation

Ah okay thanks

physicists just decided to use neither a proper mathematician notation nor index notation for spinors

Why do people seem to skip spinor notation and go straight to matrix notation for non-rel?

I guess so that only the worthy would enter
Also the Handbook of Spacetime has a very nice spinor section
Page 281
Pirate it, I am encouraging crime

1:22 PM
So it seems like in the Steigal book that all indices essentially anticommute

Spinors are slightly more annoying than tensors because there are four spinor spaces
as opposed to two
So you have indexes $\Psi^{A\dot{B}}_{C\dot{D}}$

$\epsilon_{ \alpha \beta } \psi^*{}^\alpha \psi^\beta$
$- \psi^*{}^\alpha \epsilon_{ \alpha \beta } \psi^\beta$
$-\psi^*{}^\alpha \psi_\alpha$
Am I doing things right with the signs?

I don't super remember the nitty gritty
things not commuting sounds okay certainly

Okay no worries

1:38 PM
although remember that there are two possibilities for the "classical" case : you can have things that are just spinors, and then you can have anticommuting spinors, with values in Grassmann algebras or something
so beware of what you're getting
The quantization of the first won't work due to various reasons, the quantization of the second is the Pauli equation

If only Dirac wrote a 2nd book on all of this

You can totally attempt to have a quantum theory of commuting spinors, but unfortunately you end up with a bogus theory with no lower bound on the Hamiltonian
which is a very bad sign

In Seigal, he says that $\overline{\psi}_\alpha = (\psi^\alpha)^*$ but how do I then sum over it if it transfers it to a down index?

Look p. 123, there is a better definition of all this

Is this done at the end of the calculation to convert to matrix notation? On pg. 88 there are some interesting comments about this for unitary, orthogonal groups etc.

1:48 PM
You can do everything in index notation, if you want

One thing I vaguely remember and should know better is 2-spinor notation vs 4-spinor

Honestly I would advise doing either everything in index notation or everything in proper math notation
What most physicists use is awful

What's proper math notation?

I’ve heard the former has some advantages but the latter remains more common

Treating everything as a complex vector space
And then just use the appropriate maps
If you're doing an inner product of two spinors, write it down
$\langle \psi, \xi \rangle$

1:50 PM
Ah okay
And then to check if it's invariant I would have $\langle U \psi , U\xi \rangle$?

Two-component spinor formalism is what I should have said

It should be
Although beware with spinors
Since it's a complex vector space things are a bit trickier

Man, I wish they taught us all this in physics
It seems like to do things properly you need to have done a theoretical physics or math degree

also check out the handbook of spacetime, too
I find it very nice
a little preview here

Im going to read the part on spinors now

2:03 PM
I don't think Geroch gets into Grassmann stuff, though

Should I be writing $\epsilon_{\alpha \dot{\alpha}}\overline{\psi}{}^\dot{\alpha} \psi^\alpha$ instead?

I mean you can write how you feel best about, as long as you keep track of everything properly

I'm just confused where the hermitian conjugate in matrix notation comes from in $\psi^*{}^\alpha \psi_\alpha$ where $\psi^*$ here is just the complex conjugate of the components

Hermitian conjugate?

Yeah
As in how to go from $\psi^*{}^\alpha \psi_\alpha$ to $\psi^\dagger \psi$

2:19 PM
Well that is essentially what that is
$\psi^a$ is a spinor, $\psi_a$ is a dual spinor
Therefore they form a linear mapping to the complex numbers
$\psi_a [\psi^a] \in \mathbb C$
Please note, though, so far we're entirely dealing with Weyl spinors
not Dirac spinors

Ah yes of course, i didn't think of it like that

So beware of any direct analogies

I think I've seen how to write Dirac spinors in terms of weyl spinors or something in SUSY lectures I have had

Yes
that does happen a lot
Dirac spinors are basically just a pair of two Weyl spinors

So in matrix notation $\psi^\dagger$ is a map such that $\psi^\dagger \varphi \rightarrow \mathbb{C}$

2:24 PM
Well be careful there, because as previously mentionned, there are 4 spinor spaces

Ah yes

Dual spinors act on spinors, dual conjugate spinors act on conjugate spinors
And the conjugation operator maps in the obvious way

what if I have a mixed rank 2 spinor such as $\epsilon_{\alpha \dot{\alpha}}$?
Like I had written here $\epsilon_{\alpha \dot{\alpha}}\overline{\psi}{}^\dot{\alpha} \psi^\alpha$

Well that one will map spinors to dual conjugate spinors

So that is a perfectly legal term?

2:26 PM
Sure
The classic one is the Pauli matrix
$\sigma^\mu$ is actually $\sigma^\mu_{A \dot{A}}$
That's why you have the bilinear $\bar{\psi} \sigma \psi$
Although beware because the Dirac spinors are a little weird
the "scalar" term mixes different spinors

So $\epsilon_{\alpha \dot{\alpha}}\overline{\psi}{}^\dot{\alpha} \psi^\alpha$ in matrix notation is $\psi^\dagger \psi$? But if you do it explicitly you will get something different.

You will get $\psi^*{}^1 \psi^2 - \psi^*{}^2 \psi^1$ or something instead of $|\psi^1|^2 + |\psi^2|^2$

Something like that
I'm not super big on Pauli spinors, unfortunately
People don't write much rigorous stuff about them
There is only love for spacetime spinors
7

The Pauli Hamiltonian is $$H= \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi$$ For the 1-component Schrodinger equation in absence of applied field we have the Lagrangian $$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{\hbar^2}{2m}\Big( \boldsymbol{\nabla}\psi^*\cdo... 2:43 PM Yeah I saw that just before you posted it lol 3:01 PM Hi again :-) Now, my question about maths and the toy metric is ready: physics.stackexchange.com/questions/680335/… Please, don't close. And please, calculate if you can :-) If I define an inner product as \langle \psi , \psi \rangle = \epsilon_{\alpha \beta} \psi^*{}^\alpha \psi^\beta, is this essentially saying that in matrix notation the Hermitian conjugate is defined as \psi^\dagger := (\epsilon \psi^*)^T? or something along those lines? 3:21 PM Apparently according to wikipedia \langle x, y \rangle = y^\dagger \mathbf{M} x which is known as the Hermitian form This would imply that the inner product for 2 spinor space is just the euclidian one. Something doesn't seem like it's adding up here 3:36 PM Somebody has a clue why my question was downvoted? I guess I should be happy that it wasn't closed. @BarrierRemoval it's not clear what you want to do - the Schwarzschild metric is the unique static, spherically symmetric metric. Assumptions about "A" or "B" don't make sense - if you change the stress-energy tensor from the Schwarzschild one to the non-Schwarzschild case you're apparently thinking of, then you have to just solve the EFE for that @DIRAC1930 it's just different notations, you should pick one/know what you are reading is using and stick to that it's pretty common in physics not to use "pure" matrix notation for spinors @fqq so when people write \psi^\dagger in matrix notation, what they actually mean is something like (\epsilon \psi^*)^T? 3:53 PM what people? to me they are the same. If we are talking about the same thing, \epsilon in matrix notation has one index raised -> it's the identity But then that wouldn't lead to \psi^*{}^1 \psi^2 - \psi^1 \psi^*{}^2  \epsilon_{\alpha \beta} \psi^*{}^\alpha \psi^\beta -\psi^*{}^\alpha\epsilon_{\alpha \beta} \psi^\beta  \psi^*{}_\alpha\epsilon^{\alpha}{}_{ \beta} \psi^\beta  What do I do now \psi^*{}_\alpha \psi^\alpha. We are back where we essentially started @ACuriousMind Schwarzschild isn't static uh...did I mean stationary? Also not! (the Killing field is spacelike inside the horizon) ah, I see what you mean 4:05 PM Not even to bring up the weird other versions of Schwarzschild with different topologies :p So it seems like in physics when they write \psi^\dagger \psi what they really mean is \langle \psi, \psi \rangle with some weird definition for \psi^\dagger in physics notation. I'm not 100% sure what does on with SU(2) spinors, really @DIRAC1930 I don't see how else you would define the adjoint Obviously they need the complex conjugate to have a real product, but idk what it looks like, component-wise as fqq says, that's the definition/purpose of \psi^\dagger - to produce the norm of \psi when applied to \psi we don't "really mean" \langle \psi,\psi\rangle, we defined \psi^\dagger so that that happens 4:09 PM so \psi^\dagger \psi is \epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta? there are so many spinor notations that I couldn't say off-hand :P By 'produce the norm' do you literally mean |\psi_1|^2 + |\psi_2|^2? While I don't know too much for the Pauli case, remember that the spinor metric is the "square root" of the metric Mapping it appropriately with Pauli matrices, you should get that it is equal to g So there is no reason here why the spinor metric should bz the same as the relativistic case 5 Consider the two components spinors (usually Weyl spinors): \psi = (c_1, c_2)^{\top}, where c_i are complex numbers. The metric in spinor space is usually defined with the second Pauli matrix \sigma_y (many authors are adding an i factor): \tag{1} \epsilon = i \, \sigma_y... I found this So it looks like something different is going on in the non-rel case In mathematics, the spinor concept as specialised to three dimensions can be treated by means of the traditional notions of dot product and cross product. This is part of the detailed algebraic discussion of the rotation group SO(3). == Formulation == The association of a spinor with a 2×2 complex Hermitian matrix was formulated by Élie Cartan.In detail, given a vector x = (x1, x2, x3) of real (or complex) numbers, one can associate the complex matrix x → → X... If you scroll down to isotropic vectors, the norm is the Euclidian norm is that's what is going on? Well as I said, your spinor metric should replicate the metric of space itself 4:21 PM In two dimensions, you can choose a basis of the Clifford algebra in which \gamma^0 = \epsilon, right. In any dimension, you can form a scalar from spinors as \overline{\psi} \chi = \psi^{\dagger} \gamma^0 \chi. However in 2D, \varepsilon also satisfies \varepsilon_{ab} = M_a^c M_b^d \varepsilon_{cd} when M has determinant +1 so it's an 'invariant tensor' and can be treated as a metric, this just agrees with the usual scalar invariant with the Pauli matrices, you can build vectors, ie$$\sigma_{A\dot{B}}^\mu \psi^A \xi^\dot{B} = V^\mu

who's working in 2D now?!?

This started off as a simple question lol

Nothing is ever simple with spinors

The Clifford algebra in 4D projects down to two copies of a 2D Clifford algebra

4:24 PM
I thought we were just talking about the fundamental repr of SU(2) but I see a lot of Lorentz stuff

Yeah, I don't think any Lorentz is needed for my question
for now at least

Right but for relativity, the spinors transform under $\mathrm{SL}(2,\mathbb{C})$

We're doing $SO(3)$ here though
Nothing so spicy

Okay, well the 3D clifford algebra is just the 2D clifford algebra and it's version of gamma 5 included, you still have the same thinking about invariants and using a $\gamma_0$

Okay so how can I build a normal SO(3) vector from SU(2) spinors?

4:34 PM
$1/2 \otimes 1/2 = 0 \oplus 1$

(by symmetrization and antisymmetrization of the product)

@ACuriousMind I have read there somewhere that Birkhoff's Theorem only holds for energies greater than zero.

A spin $j$ representation of $\mathrm{SU}(2)$ has $2j+1$ components. So for $j = 1$ you get a vector. How do you do this in practice. You need to realize that integer spin irreps of $\mathrm{SU}(2)$ are given by symmetric tensors, because contracting these against Levi-Civita vanishes thus they are irreducible, so $T^{ab}$ symmetric with $a,b = 1,2$ has 3 components $T^{11}, T^{12} = T^{21}, T^{22}$. This gives a vector, in fact you can associate those components to a $j = 1$ spherical Harmonic

Therefore, Schwarzschild metric is only the solution if all energies are greater than zero

You can do that for all higher integer spins, Landau talks about it. If this is all confusing, one first has to study irreps of angular momentum i.e. $\mathrm{SU}(2)$, it's all in that book

4:44 PM
Ah yes, but what about for field operators?

It's not going to change anything

Okay thanks
What is the inner product here?

2

The spinor inner product in particle physics is given by $\overline{\psi} \psi = \psi^{\dagger} \gamma_0 \psi$, where I take the convention that the zeroth gamma matrix is hermitian while the rest are anti-hermitian. This is invariant under spin group transformations, $\psi \rightarrow e^{\omega... @bolbteppa still the relativistic version :p Lol 4:56 PM It's still the same in 2D with$\gamma_0 = \epsilon\$ or something :p

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