Also is the correction term necessary for a rather loose demonstration?

Rovelli doesn't use it and he still gets correct results for the free field

@FrancescoS As an algebra, the Clifford algebra is already generated just by the 4 $\gamma^\mu$. As a vector space, it is $2^4 = 16$ dimensional with basis $1,\gamma^\mu,\sigma_{\mu\nu},\epsilon_{\mu\nu\rho\sigma}\gamma^{\nu}\gamma^\rho‌​\gamma^\sigma,-\mathrm{i}\gamma^5$.

Strictly speaking, only $\mathrm{i}\gamma^5$ is an element of the Clifford algebra, not $\gamma^5$ itself (the algebra is a real), so neither $L$ nor $R$ lie in it

@Slereah it is necessary for the proof (not a loose one); and to see the relation between the Wiener measure and the corrective term you can see Reed Simon II, page 229

it's pretty clear ;-)

(Also, my$\sigma_{\mu\nu}$ are $\mathrm{i}$ times yours, for the same reason)

@ACuriousMind Ok, sorry… i didn't know the difference. I means the vector space of the matrices

So anyway

Schwartze is a pretty good book so far

Goes into topics not covered by Peskin or Weinberg

So a nice addition

@FrancescoS What is "the vector space of the matrices"

@ACuriousMind Like $M_n(\mathbb{C})$

(the space of all $n\times n$ matrices with complex entries)

I am just guessing...

@yuggib you are right

I hoped to do more physics during my week off

But between friend visits and family and tidying up the place

Not much time

Orphans have it pretty good

@yuggib Uhhhh...

@FrancescoS Abstractly, the objects $\gamma^\mu$ are not matrices

I imagine they're isomorphic to matrices

You may represent them as matrices, but the Clifford algebra is not naturally written as matrices in any particular way

Those math people

They never want to represent anything

@ACuriousMind I use them as matrices in this case.. Anyway, even if they are not matrices and they are basis of the vector space (as you said), the operation $\gamma^\mu \gamma^\nu$ is defined not?

@FrancescoS the product is defined only if you have an algebra

Yes, the Clifford algebra is an algebra, it has a multiplication

an algebra is in particular a vector space, and that may be a part of your answer; since the clifford algebra is generated by the gammas and it is not isomorphic to the whole $M_n(\mathbb{C})$ but only embedded in it as a subspace, then the gammas alone are not a basis of $M_n(\mathbb{C})$

however, it may be that adding some other matrices you get a basis of $M_n(\mathbb{C})$

actually I may say that the Clifford algebra is embedded in a subalgebra of $M_n(\mathbb{C})$...

right ACM?

Yes - for $\mathrm{Cl}_{3,1}(\mathbb{R})$ it is exactly $M_4(\mathbb{R})$

nice to know...then here you go with your basis @FrancescoS ... not of $M_4(\mathbb{C})$ but of $M_4(\mathbb{R})$

ook..

@FrancescoS I get the feeling it might be better if you say why you want a "basis" of the algebra ;)

but my questions was if I can decompose any product as a linear combination of the basis elements.. This is just what I need to know in this moment..

@FrancescoS No. $\gamma^\mu\gamma^\nu$ is linearly independent from all $\gamma^\mu$.

@ACuriousMind Because I need to write all possible lorentz invariant pairs of bilinears

Ugh...have fun with that ;P

ahaha yes, this is a good answer

@ACuriousMind Look at here web.hep.uiuc.edu/hepg/Theses/Bouchard_2011.pdf pag 14,15

@ACuriousMind there is something I need. But I don't understand why he does not consider for example the operator $ (\sigma_{\rho\sigma}\sigma_{\mu\nu})[\sigma^{\mu\nu}\sigma^{\rho\sigma}] $

@FrancescoS Aha! So you were talking about the space $M_4(\mathbb{C})$!

Yes, I am working with complex matrices

sorry for this misunderstanding

Wait

Yeah, the statement that $R,L,\gamma_\mu R,\gamma_\mu L,\sigma_{\mu\nu}$ span $M_4(\mathbb{C})$ as a complex basis is correct

If fields of different mass are unitarily inequivalent

Why did it take so long for Haag's theorem to get discovered

Aren't interacting fields generally of different mass from the free fields

@FrancescoS $\sigma_{\rho\sigma}\sigma_{\mu\nu}$ contains just sums of products of four gamma matrices, but these are all proportional to $\gamma^0\gamma^1\gamma^2\gamma^3$, and that's something like $\mathrm{i}(L+R)$

Is that why Lubos Motl is really flippant when talking about Haag's theorem

@Slereah What kind of question is that? Do you know a proof that the CCR representations of different masses are inequivalent?

No I don't!

That the example is obvious doesn't mean that it is straightforward to see that they are inequivalent.

@ACuriousMind and what about for example $\sigma_{\rho\sigma}\sigma_{\alpha\beta} \sigma_{\mu\nu} .... \sigma_{\epsilon\lambda}$ for example?? I would like to understand if I can stop the research of 4-fermions interactions taking just the basis elements

Is that what Haag proved?

Or was it proven beforehand

@ACuriousMind I know a proof...but they're inequivalent representations of the CCR algebra over $L^2(\mathbb{R}^d)$

@FrancescoS What? You just multiplied more $\sigma$s to what is a multiple of $\mathrm{i}(L+R)$. As a general rule: Products of $n$ gamma matrices with $n>4$ can always be written as products of $n-4$ gamma matrices multiplied with some combination of $L$ and $R$

@ACuriousMind thank you!

@Slereah the paper where the theorem is proved is by Wightman; the non-normality of ground states can be found in bratteli-robinson

@FrancescoS Another way to see this is that you can reduce arbitrary products of gamma matrices to products of the four gamma matrices - they anticommute with each other and if they hit themselves, they just square to the corresponding entry of the metric.

now I gotta go...see ya later/tomorrow

@ACuriousMind Good morning!

@MAFIA36790 Possibly:

amazon.com/Group-Theory-Nutshell-Physicists-Zee/dp/0691162697

They mess everything up

I haven't read it so this isn't a recommendation, but it's been getting some good reviews.

@BernardMeurer Good evening

@ACuriousMind Tell me something I don't know

Good old Wightman

@BernardMeurer The magic system of Dungeons and Dragons is based on a series of novels by Jack Vance, and thus called "Vancian magic" in RPG circles.

@ACuriousMind That's interesting, I always wanted to play RPG

I much prefer the World of Darkness magic system

Where everything runs according to the...

Fuck I forget the term

General belief kind of thing

ah yes, the paradigm

Or the consensus

@MAFIA36790 It's a ritual now :)

@ACuriousMind thanks as always :)

And of course

Science is just the current Consensus

Meaning that it is all magic

@Slereah I always say, it's just a trick

@MAFIA36790 He got banned for a month

One thing I found rather disappointing in the Technocratic book is that the school of Correspondance is described as using the space bending powers of general relativity

But in practice most of the spells are done by cameras and remote control

Playing a technocrat team with people from here sounds like it would be dangerous really

Since they have to cast spells by justifying what they do with sciency things

Sounds like it could lead to endless nitpicking

I don't want to spend hours checking if your CTCs really work as you say they work! ;)

"Wait, have you thought to take into account the vacuum back reaction!"

And then 3 hours of checking calculations

Although time travel is a pretty risky gambit in Mage

Since it's pretty hard to justify according to the Consensus

Hiding magic from the Consensus is usually more along the lines of "Hm this wall seems a bit worn down, I bet it could collapse ANY MINUTE"

wink wink

my orbifolds book arrived!!! but it's inaccessible to me for now. :(

@3075 How come?

I don't know enough math.

Ah i see, painful I know

need to read Lee's manifold books first.

the author of the book recommended I read the book and learn the background as I progress.

but that seems kind of impossible.

How would I learn about what would be the relativistic mass of a spinning uniform ball with a surface electric charge that orbits a centre point and how would it be distributed?

"Relativistic mass" is an outdated concept, and I don't know what you mean by it being "distributed"

@StevenStewart-Gallus your question is a bit unclear. Perhaps you could give us some of the background to it to help us understand what you are getting at.

@ACuriousMind Alternatively, one could just ask where the kinetic energy of the object is distributed.

@StevenStewart-Gallus I don't know what that is supposed to mean, either. You can either conceive of the ball as one object, then it just has kinetic energy, no "distribution", or you can conceive of it has being constituted of its, well, constituents, then each of them has some energy

What is the kinetic energy of each point on the ball?

Are you trying to ask for the stress-energy tensor associated to a rotating fluid?

Each "point" really doesn't have energy, but you can associate energy density.

Yeah I meant energy density of the spinning orbiting ball.

@StevenStewart-Gallus Have a look at this calculation of the stress-energy of a slowly rotating star.

May I ask why you are interested in this specific energy density?

@ACuriousMind I'm looking at the energy of atoms and particles from a classical perspective and comparing them with the energy stuff from quantum equations.

@StevenStewart-Gallus why?

@AccidentalFourierTransform Your profile attributes my utterance to the wrong user :P

@StevenStewart-Gallus Also, there is no classical perspective of atoms. The atom is a fundamentally quantum mechanical thing, treating it classically mostly produces garbage.

@ACuriousMind Nah, just perhaps a less... abrupt sound?

I dunno, it's just so in-your-face.

Actually screw that. Yes, I want a fanfare.

@DanielSank What time were you thinking of for Saturday

@JohnRennie im in there

@user507974 Evening.

Would have to be.

@DanielSank so like 7?

@user507974 Not sure. I'll email you and the others and see when people want. The issue is that I have folks from out of town visiting who probably aren't interested in Smash Bros., so we might have to do it later.

@user507974 Shoot me a quick email to make sure I have the right one for you.

@DanielSank done

@DanielSank i dont mind later much, im nocturnal and i was just a bit worried about it being earlier because im busy early evening

@user507974 You go to Berkeley?

@BernardMeurer nope, what made you guess that

Too broad?

@ACuriousMind yep, I know, it was on purpose :-P

do you like my profile better now?

@DavidZ hi :-) about my comment earlier this morning, I just wanted to say that I agree with you about the quality of the post. When I first read the question, before it was edited and the questions split, it seemed to me that there was actually only one question: "what does E=mc2 mean?"

so to me it was not really too broad. Im not saying the question should not be closed (because for example if Im right and the actual question was "what does E=mc2 mean" then the question would be a duplicate of e.g., physics.stackexchange.com/questions/147058/…)

Im not saying the question is a good one - it is not - but too broad didnt seem quite right to me. Not that it really matters though...

Sometimes I realize that some of my SE questions were dumb

But then I keep it secret, so nobody knows!

is @0celo7 unbanned yet

@slereah

Probably not

Still a bit over a week I think

:[

Is the displacement along an arc length a vector?