The h Bar

6:26 AM

Hi All..

Hello @JohnRennie Sir

@JohnRennie Sir if you have time, can we discuss?

John Rennie

Hi :-)

I'm afraid I'm really busy this morning. I have a queue of question stacked up and some will be long ones. I can ping you when I'm free, but it won't be for a while. Sorry :-(

@JohnRennie Ok sir , i am waiting for you. Thanks

2 hours later…

8:04 AM

Hello @JohnRennie Sir, Are you available???

8:16 AM

@JohnRennie Can we say. We have a uniform rod and COM/COG is at the center of rod. If we apply force right side of the rod from COM then more mass is at left side of the force and less mass is at right side. Can we say the higher the mass lower the acceleration that's why left side stay behind due to more mass and right side move faster due to low mass. We say this phenomenon rotation??????

4 hours later…

Feynman_00

12:09 PM

@SillyGoose It turns out my grandma has got a goose

2 hours later…

1:45 PM

@SillyGoose 1. You can always purify a mixed state by changing the Hilbert space. 2. An arbitrary mixed state on a given Hilbert space is just any positive-semidefinite self-adjoint operator with trace 1.

2 hours later…

doublefelix

3:27 PM

Question about gluons and W/Z bosons - in practice these will decay before ever having a chance of touching a detector. But if their lifetime were longer, would we be able to observe them? Or are they somehow strictly virtual particles for some reason? I have background in QFT in general, but not much in QCD

@doublefelix neither are "strictly virtual particles", but gluons are confined due to their color charge - you can't have free color-charged particles. There is nothing in principle forbidding free W/Z bosons.

doublefelix

Thank you^^

6 hours later…

9:14 PM

@ACuriousMind Thanks for your answer here https://physics.stackexchange.com/questions/740676/are-spinors-intrinsically-nonlocal/740682?noredirect=1#comment1658516_740682. Your last comment read: "As my answer says, a spinor (without the "field" part) is just a vector in a certain representation of SO(p,q). This definition does not rely on there being any manifold, just like an abstract p+q-dimensional vector is not intrinsically tied to any manifold."

I'm trying to understand this. My representation theory is not great so that doesn't help. So fine, you can have a spinor which lives in "a …

@Jagerber48 it's not SO(p,q) that acts on coordinates

first, look at what happens to tangent vectors: A general coordinate transformation on the manifold acts via its Jacobian matrix on the tangent vectors

yes

now, the Jacobian matrix cannot directly act on spinors because the Jacobian will not in be in SO(p,q), just in GL(p+q)

as I also explain here, the notion of spin on a manifold (the spin structure) essentially means that we have a local basis ("tetrad", "vielbein") on which this Jacobian can act

after the transformation we have a "different" vielbein, and the two will be related by some transformation in SO(p,q) because both vielbeins are orthonormal bases and SO(p,q) are exactly the transformations between all possible orthonormal bases

this transformation then acts on the spinors

(generally this isn't something you really have to care about because usually we write down terms like $\bar\psi\gamma^\mu \psi$ where the spinors are contracted such that there's no free spinor index)

actually that's not quite right: If the Jacobian is not in SO, then it just doesn't have a defined action on the spinors - there is no action of a general coordinate transformation on spinor fields

the existence of the spin structure is essentially a compatibility condition that we indeed can globally restrict to transformations with SO Jacobians without becoming inconsistent

9:29 PM

Can we perform arbitrary coordinate transformation on a manifold that is supposed to have a spinor field on it? Is the spin structure a recipe to do that?

Or are the types of coordinate transformations we can perform restricted to those which have SO(p, q) Jacobians?

I mean, you can perform arbitrary coordinate transformations, they just don't act on the spinors :P

normal vectors transform under coordinate transformations because their basis are the derivatives w.r.t. coordinates and there is a natural way to transform them into the derivatives w.r.t. the new coordinates

but the spinor basis is not in terms of derivatives but in terms of the vielbein, which has to be an orthonormal frame

there is a natural action of coordinate transformations on frames, but if the Jacobian isn't in SO then the result of that action on the vielbein isn't orthonormal anymore and hence not a valid new basis for spinors

The vielbien is the matrix (in SO) that transforms a basis of tangent vectors in such a way that the metric becomes the standard diagonal metric

yes

Then I guess my next questions are (1) Is the spinor basis defined in terms of the vielbien, if so, how? and (2) Once we have the spinor basis, suppose we have a coordinate transform whose Jacobian IS in SO. Then how exactly does the spinor basis transform? I'm guessing it's something like it transforms under the "spin representation" of the Jacobian but I'm sort of just putting together words that sound right.

Silly Goose

a goose ! @Feynman_00

thank you i will look into that ! @ACuriousMind

9:42 PM

@Jagerber48 for an infinitesimal SO transformation with components $\lambda_{mn}$ in terms of the vielbein the action on a spinor (remember that spinor space is something on which $\gamma$ matrices act) is given by $\lambda_{mn}\gamma^m \gamma^n$ (the "so representation" part is that the $[\gamma_m,\gamma_n]$ are the infinitesimal generators of SO)

Ok that makes sense as an answer to my question (2). Still curious about (1). My question is basically this. I can see that tangent vector components transform under the Jacobian of a coordinate transform because the tangent vector basis is defined in terms of the coordinates. So there must be something that links the definition of the spinor basis to the manifold coordinates? I'm trying to understand that linkage.

it's not directly tied to the coordinates - the thing that ties the spinor to the manifold is the vielbein/spin structure

I guess I don't understand vielbein/spin structure well enough to tell if my question is answered.

Do coordinate transformations even induce changes in spinor components? Or are spinors in like, their own space (I have tensor products of QM Hilbert spaces in mind even though I don't really want to, I'd like to avoid QM...) and we need to apply transformations specifically targeted at the spinor space to transform their components?

the spinors do not transform under general coordinate transformations, no

Do they transform under some coordinate transformations? Such as those whose Jacobians are in SO?

9:54 PM

they're "scalars" in that sense

Do you have a recommended reference for learning about spinors in the way you've been talking about them?