@ACuriousMind OH, I see now. When we write the commutation relations in matrix notation and take linear combinations of the operators, we get the Lie algebra of $\mathfrak{su}(2)\times\mathfrak{su}(2)$, right?
@KyleKanos The result he mentions is zero, so I don't see how it is proportional to the vector. (I think, the question is vague enough that I may be wrong.)
@ACuriousMind Now does $\mathrm{O}(3,1)$ cover $\mathrm{SU}(2)\times \mathrm{SU}(2)$ or is it just isomorphic?
Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.
I) First recall the fact that
$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.
This follows partly because:
There is a bi...
There's a lot of complexification and compactification going on there
@0celo7 It is the same here, but one usually uses $\oplus$ for vector spaces (which Lie algebras are). Be careful, there is no $\oplus$ for the groups.
Nice, isn't it? :D (I see why this is confusing, but since, whenever there is an $\oplus$, it coincides with $\times$, you need not worry about this, I think)
@0celo7 A bit indirectly in the footnote, where he pokes fun at Zee's wrong statement (very similar to yours). $\mathrm{SO}(3,1)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$ are not directly related by an isomorphy or covering (which you can see by noting that one is compact, but not the other)
The representation theory of the Lorentz group is a very ugly thing
Hi, if I have a ball that is 2.0 kg and oscillating on a spring that has a force constant of 50 N/m, then what is the velocity of the ball at 1.5 m from equilibrium? I tried U = .5 kx^2 = 56.25, then 56.25 = .5 mv^2, solved for v and got 7.5, but this is wrong
So what does it mean, to the most anal pedant in the world, that given the generators of the proper orthochronous Lorentz group I can get two separate $\mathrm{SU}(2)$ copies?
@0celo7 Yes. And it's wrong. The representation theory allows you to classify representations of $\mathrm{SO}(3,1)$ by representations of $\mathrm{SU}(2)\times\mathrm{SU}(2)$, though, so I guess many make that wrong statement because they don't want to do delve into the details and think the reader will accept it if they wave their hands and say "isomorphy"
@0celo7 you have that the complexification of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is isomorphic to the complexification of $\mathfrak{so}(3,1)$, I think, since the complexification of $\mathfrak{su}(2)$ is $\mathfrak{sl}(2,\mathbb{C})$. (I'm not as sure as usual about what I've written here, but I think it's right)
But isomorphy of the complexifications does not imply isomorphy of the real forms, and does not imply isomorphy of the corresponding groups. It does imply that the reps of the groups are very closely related, however, and that is all we physicists want to use
@Pallas The original question you asked doesn't give a max displacement. You later said 2m. Is that from the question or is that something you came up with?
I went to a history class today. It is required to get a degree. I now remember why I hate history and love physics: historians never provide actual evidence.
@0celo7 Yeah, I don't immediately see it either, but I'm inclined to believe him ;) The anticommutator is probably just a compact way to write a term with order $\Sigma\gamma$ succintly (note that the order $\gamma\Sigma$ was already present)
Don't ask me what was added, though, I don't know anything about spin connections.
Be thankful that the arXiv is making a large part of current stuff available to everyone.
user54412
@0celo7 The counterargument would be that those not affiliated with a university are contributing a negligible amount to society's total research (or they are pharmaceuticals and so can buy any access they want).
@ACuriousMind Check out the typesetting in this book.
user54412
9:46 PM
Also, in the long run, it's not clear where savings would come from. My entire field (astrophysics) is by-and-large open access, but all that means is that we authors have to pay to publish. Thus to get the same amount of research we need comparatively more grant money.
It's $40 because non-affiliated researchers, i.e. companies, do pay.
In this day and age there is no good reason to have massive publishers sucking out the little money that academia has. Peer review doesn't cost much, and the quality is often dismal anyway. It merely serves to give the appearance of reliability or "novelness" even when the papers are in fact lacking both.
I find it difficult to wrap my brain around how energy/matter, chemical reactions and resultant entropy or waste heat exist in abstract spacetime. First, how can a clock measure "time" without energy? In addition, how can mass move through space without energy? I am thinking that energy may be...
@0celo7 No doubt, that's what it is. But in combination with the second sentence, I'm feeling a bit like my words have been stolen. Plus, my analogy is so much more apt than his
So it has nothing to do with $\nabla_\mu T^{\mu\nu}=0$?
>carts carry people
Lol
@Jimnosperm I thought that we cannot have local energy conservation in GR because the energy-momentum tensor of the gravitational field is not well-defined.
That's relevant. Energy and momentum are not conserved globally, but the energy-momentum tensor is conserved. Locally, we can construct a pseudotensor for energy and momentum that is always conserved, thereby meaning that energy and momentum are individually conserved locally
Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$.
I) First recall the fact that
$SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.
This follows partly because:
There is a bi...
I find it difficult to wrap my brain around how energy/matter, chemical reactions and resultant entropy or waste heat exist in abstract spacetime. First, how can a clock measure "time" without energy? In addition, how can mass move through space without energy? I am thinking that energy may be...
