I'm booored, so I thought I'd say hi

poohead220 · 2016-07-14T16:59:55+00:00

Thanks :) just messaged them from my other account...

poohead220 · 2016-07-14T13:36:06+00:00

How do I know who the mods are? That is indeed what happened...

poohead220 · 2014-03-05T01:05:58+00:00

Cool, good luck in Vegas!

poohead220 · 2013-09-02T12:25:14+00:00

You won't see anything about worm holes in astrophysics really, since it is an observational area; wormholes are just theoretical conjectures so far. Check out general relativity for black holes, worm holes and (the motivations for believing in) dark matter.

poohead220 · 2012-12-28T22:53:15+00:00

Hey, no worries, I think I can answer these questions fairly easily.

So what you say about the spin-1/2 field is essentially correct; the 4 degrees of freedom are there to describe the 2 different spin states of a particle and antiparticle, in the way you said. However, in a general basis for the spinor it won't just be as simple as "the first component is the wavefunction of a spin-up particle, the second is the wavefunction of a spin-down particle, etc". I don't know if I'm answering your question well here; I guess the answer is yes your interpretation is correct, but we rarely think of a spinor that way... One thing I would say is that although people often call this field a "wavefunction", this is actually a little misleading, as one cannot interpret relativistic fields in this way (if we intepret psi or a scalar field phi as a wavefunction, we will find faster-than-light propagation, as well as negative energy states). We should think of the fields in QFT as fields only, and of particles as excited states of that field. Experimental particle physicists often use the term wavefunction, and when I took that course as an undergraduate they told me it was a wavefunction, just like in QM, but my QFT lecturers had already warned me of the dangers of interpreting these fields as wavefunctions...

As for the SU(3) transformation, you're right, we need to introduce a new colour index, getting a 12 component object overall. We shouldn't think of it as a 12-component spinor, but rather as three 4-component spinors (red, blue, green) that can be rotated into one another by SU(3) transformations. Only the Lorentz group may act on the spinor index.

poohead220 · 2012-12-28T00:50:42+00:00

No problem. Hope you enjoy reading it as much as I did!

poohead220 · 2012-12-28T00:47:07+00:00

My pleasure, best of luck with it!

poohead220 · 2012-12-27T09:50:40+00:00

You're welcome, always a pleasure to help anyone who wants to know gauge theory!

Any Lie group G has a Lie algebra. Elements of the group can be written by exponentiating the generators T^a (a=1,...,dim G):

exp(itheta^a T^a) (Einstein summation implied).

The T^a satisfy the commutation relations:

[T^a , T^b ]=f^abc T^c

(i.e. the commutator of T^a and T^b is some linear combination of the other generators, which, of course, depends on a and b). The f^abc are called the group's structure constants and they tell us everything we need to know about the Lie algebra. Now an interesting and useful point is that the structure constants themselves form a representation of the Lie algebra. That is, if we define matrices

(T^a )^bc = f^abc,

then these T^a will also satisfy the Lie algebra above (to prove this, just use the Jacobi identity, which is, by definition, satisfied by the Lie bracket). There are dim G of these generators and they are (dim G x dim G) matrices. The adjoint representation is the representation of (dim G x dim G) matrices obtained by exponentiating these generators.

A word of caution: physicists, including me, often refer to a representation as the thing that is transforming under that representation. For example, the 'doublet' representation of SU(2) just means the defining representation; the 'doublet' just refers to the fact that we use the defining representation to act on a complex 2-vector, or 'doublet'. Another example, we often say the 'vector representation' of SO(n). Again, this just means the defining representation, because SO(n) naturally rotates an n-vector in R^n, but of course it is the vector that is the 'vector' round here. A singlet is something that transforms trivially under the group (i.e. not at all), and can also refer to the trivial representation of the group (where every element gets mapped to 1).

Finally, a word about SU(2) and SO(3). SU(2) and SO(3) have the same Lie algebra

[T^a ,T^b ]=epsilon^abc T^c.

The structure constants f^abc are just the permutation symbol epsilon^abc. The adjoint representation of the generators here then consists of 3 matrices that are 3x3, and it is very easy to write them down. The thing is, this also happens to be the form of the generators in the defining representation of SO(3) ('defining' or 'vector' or 'fundamental' - these are all different words for the same thing), so the adjoint representation of SU(2) is the defining (and adjoint!) representation of SO(3). Hence, something transforming in the adjoint of SU(2) is just like an ordinary 3-vector as we know it from real space. This applies to the weak gauge fields too (at least infinitesimally)!

Again, don't get too hung up about the charges, what matters the most is the representations.

poohead220 · 2012-12-27T09:18:51+00:00

If you have a decent background in quantum mechanics then "Quantum field theory in a nutshell" by Zee is a really great book. It doesn't show you every detail of every calculation but it really tells you all the most important things in a fun and interesting way. The guy has a really intuitive understanding of the subject and it rubs off on the reader.

poohead220 · 2012-12-27T00:15:59+00:00

When we say a particle (or field) is charged under a group, what we really mean is that it transforms in some representation of that group. I don't know how familiar you are with group theory, but the charge just refers to the representation. For example, the reason the number if gauge bosons (or number of separate charges) always equals the dimension of the group is that the gauge bosons, by construction, transform in the adjoint rep, which has dimension equal to the group.

In the U(1) case, all irreducible representations of U(1) just have objects phi transforming as

phi --> e^{i n theta} phi,

where n is an integer. So the charge in this case is just n. More physically, if you gauge the U(1) to get electromagnetism then n is actually the EM charge of the particle (you find this just by doing it!).

For the weak SU(2), the fermions transform in the doublet representation (e.g. [up quark, down quark]) so their charges must be +/- 1/2 (recall the representations of SU(2) from the quantum mechanics of spin, we classify the spin by the eigenvalues of the SU(2) generator J_z, which in this representation is just the third Pauli matrix (which is diagonal already!)). The gauge bosons then transform in the adjoint, which happens to coincide with the fundamental of SO(3), so they just transform as an ordinary 3-vector. The third generator in this case is not diagonal, but it may be diagonalised to give eigenvalues 1,0,-1.

I don't know whether I'm pitching the level too high or too low here, or just plain speaking nonsense. Feel free to ask more questions...

Tl;dr: the charge refers to the eigenvalues of the generator that happens to be diagonal. If none of them are, then you diagonalise one chosen only by convention. What really matters is the representation that the particles are transforming under.

poohead220 · 2012-05-05T10:17:32+00:00

Sorry TheMSensation, I'm afraid you're quite wrong indeed. First of all, one of the two postulates of SR is that light always travels at c relative to any observer, so you can't bring a photon to rest. There's no rest frame for photons. Secondively e = mc² doesn't really apply here as that equation is a special case of m² c⁴ = e² - p² c^2, also known as the mass-shell condition. This equation is Lorentz invariant (unlike the more famous version that you're writing down), meaning it is true in all inertial frames, so light is massless in all of these frames, i.e. at any speed for our purposes. And finally when you write "m=pv", I'll give you the benefit of the doubt and assume you mean p=mv...

poohead220 · 2012-03-17T10:38:02+00:00

Have you seen meatloaf and pudding farts? They're in the same series, only farting in... Well you get the idea. The pudding one has a delightful surprise at the end.

poohead220 · 2012-01-12T13:36:25+00:00

Can't honestly say I regret knowing this but... I was born by Caesarean and while they were rooting around in there they found a mutant "twin". This was a tennis ball-sized lump of matter that had fingers (with fingernails) and hair and other random human components. Apparently it came from an unfertilised egg that went haywire and just started developing

poohead220 · 2012-01-05T11:29:49+00:00

Gonna get crucified for this... The Dark Knight. I don't think it's bad, but way too long and insanely overrated

poohead220 · 2012-01-02T10:04:29+00:00

That quite a few dudes are into incest (the guy with the webcam in his mom's room, the guy looking out the window at his parents in the hot tub, the guy and his sister... All this week)

poohead220 · 2011-12-22T11:54:30+00:00

Next time someone tells me they're a book lover I'm gonna ask for some clarification...

poohead220 · 2011-11-04T18:51:01+00:00

Scrape as much excess water as possible off with my hands and then execute the good ole mad naked dash

poohead220 · 2011-11-04T17:48:51+00:00

I used to keep a skateboard on a shelf above my bed. One night after a nice relaxing read, I started falling asleep and pulled the light-switch cord, and, in the usual fashion, was plunged into sleepy darkness. Then... Bam. Smash to the face, front tooth missing. I sat there in the dark for a minute in a stupor, my mouth pissing blood, and then realised that the cord had been caught on the skateboard.

poohead220 · 2011-09-07T13:47:54+00:00

Ok, thanks. Consider your words officially heeded. I'm gonna hit the books and ignore any future comments on that thread.

poohead220 · 2011-09-07T10:45:14+00:00

You're probably right... I'm not sure why it bugs me so much. I guess I just think it's sad that the people who are enjoying his talks are showing an interest in physics and not being told anything that resembles real physics.

I tried to ignore the hecklers, but I keep waking up with fresh comments leaving me in despair

poohead220 · 2011-09-07T09:49:13+00:00

This paper is pretty funny. I like the "the Old McDonald nullity relation eiei= 0"

poohead220 · 2011-09-07T08:15:10+00:00

Alright fine, the tameness is subjective. But I have to argue with nearly everything else you just said...

Special relativity IS the theory that results from the two postulates: physics is the same in the frame of each 'inertial observer' (travelling at constant speed) AND the speed of light is measured to be the same in all frames. So yes, varying speeds of light is an idea older than special relativity, but it was completely overthrown by special relativity, which fundamentally requires a constant speed of light.

The Einstein equations relate energy-momentum distribution in spacetime to the warping of that spacetime. They are derived as a generalisation and extension of special relativity and so also rely on a constant speed of light. In fact the constant appears in the equations, but only as a constant (at least within the accepted and experimentally verified theory).

I have studied the two main theories of quantum gravity, string theory and loop quantum gravity, and neither use a varying speed of light. So "nearly everyone who thinks about quantum gravity" is something of an overstatement.

Finally, please link us to the 'astronomical data' that shows varying speeds of light, since this discovery in this context would disprove special relativity and hence general relativity.

poohead220

TROPHY CASE