What's the logic of Pauli-Lubanski? You want a unitary rep of Lorentz, do this by induced representations, which means fixing the non-compact generators, i.e. translations, and finding reps for only the compact generators, then boost. Since momentum vectors are eigenvectors of the $\hat{P}_{\mu}$'s, you want a rep of the group which fixes the eigenvalues, i.e. the $p_{\mu}$, i.e. those $L \in SO(3,1)^+$ s.t. $p = Lp$ i.e. $p_{\mu} = L_{\mu} \, ^{\nu} p_{\nu}$.
Now the Lie algebra of the little group is full of $\delta$'s s.t. $\delta p = 0$ i.e. $\delta p_{\mu} = \delta \omega_{\mu} \, ^{\nu} p_{\nu} = 0$ which implies $\delta \omega_{\mu \nu} = \varepsilon_{\mu \nu \rho \sigma} \delta V^{\rho} p^{\sigma}$ so that $\delta V^{\rho} = \frac{1}{2} \varepsilon_{\mu \nu \rho \sigma} M^{\mu \nu} p^{\sigma}$ which is Pauli-Lubanski and magically it's square is a casimir for massive reps, but, what's going on here in this witchcraft
And assuming that gravity IS present, when I leave the contact with the balloon, I know that my initial velocity wrt further cases would be taken as the velocity of the balloon.
But my question is, as soon as I leave the contact with the balloon do I start the downward motion at that same instant or do I first move up some more distance before the velocity that the balloon had given me becomes 0?
OK, you can use the first equation to find out how your velocity changes with time. Your velocity at the moment you let go of the balloon will be $+u$ (we'll call it u$ because it's your initial velocity).
So after a time $t$ your velocity will be $v = +u - g t $
I suspect that few of us end up living the life we expected we would when we were teenagers. The main thing is to enjoy what you're doing and try to have fun
@JohnRennie You should be proud of me. I complained about the lack of physics in my calculus of variations class so I volunteered to teach some physics on Thursday
@BalarkaSen if I'm taking the collared neighbourhood of some boundary obtained by removing some open set $U$ such that $\partial U$ has some level set function $f$, I can just take part of the tubular neighbourhood to define it, right?
Take some subset $U \subset M$, with boundary $\partial U$ defined by some function $f$ with $f^{-1}(0)$. If I take the tubular neighbourhood $N$ of $\partial U$, is the restriction $N \cap (M \setminus U)$ a collared neighbourhood?
Anyone knows if I can construct the stereographic projection of $S^2$ without using a bit of geometry(triangles etc)- a somehow purely algebraic construction. Thanks.
:) I know that. But how do I get there without pictures? If I can- if I cannot, OK. I would like to better understand how I could construct it without schemes, just using the relations given by the geometry of the sphere and the plain- it would feel more natural to generalize to higher dimensions. Maybe what I am saying doesn't make sense?? Tell if it's like that.
The stereographic projection is inherently geometric by definition. It's defined by pictures. You can use the pictures to generalize it to higher dimensions
AAAAAAAAAAhhhhhhhhhhhhh... :) Really? And so, why am I thinking this over so much? There is no way of thinking without pictures? @Slereah Yes, I know. Non-geometric I just meant to use for example the relation for the sphere(x^2 + y^2 +z^2 =1) without even needed to picture it. It's just a relation and with other relations then I could get to the point, no needing of much drawing. Thank, then, both.
For higher dimensions the map $f : S^n \setminus \{(0, \cdots, 0, 1)\} \to \Bbb R^{n-1}$ is just defined by $f(x_1, \cdots, x_n) = (x_1/(1 - x_n), \cdots, x_{n-1}/(1-x_n))$
Where $x_1, \cdots, x_n$ are the coordinates on $S^n$ coming from $\Bbb R^{n+1}$, i.e., $x_1^2 + \cdots + x_n^2 = 1$
The final formula can be defined totally symbolically
Thanks guys, I get what you mean, really. I'll try to rephrase just for the sake of saving any possible meaning from the question: at first, I thought that I could define the map as that, check that the two charts for N and S poles, do define a structure and then it would be OK. Then I thought, if there is a way to deduce the mapping from certain relations without the need of thinking about the line that passes through the sphere and the plane and certainly without ever thinking of triangles. But I get it, the whole point is that it is defined this way.
@BalarkaSen To put it differently, can I deduce fully symbolically the map from S^2 --> R^2 ?
What does "deduce" mean? Deduce from what? I wrote down the formula for the map in general
For a deduction you need to set premises and definitions. How do you define the stereographic projection, if not geometrically? I proposed a definition which does not involve geometry
Thanks. That's what I' saying. But, could I add a positive number to the above mapping, or multiply it by a positive number and still get a projection?
I often come across interesting questions, which I can not answer. As a consequence I can not verify the given answers while some seem to be helpful. Is it recommended to upvote such answers, where my knowledge doesn't reach put to verify them?
@Sid It's important to clarify that elected site moderators cannot impose ten-year suspensions; the longest they can apply (source, source) is one year. The ten-year suspensions are normally imposed by SE community managers and only in pretty exceptional circumstances.
I am writing for a friend who cannot ask a question here on Meta.
He has been suspended last week , asked to delete his account[s] and a message said that accounts cannot be removed before the end of the suspension.
Since he has been suspended for ten years, that seems tantamount to right being...
@Slereah Look, there are three things flying around. If $S \subset M$ is a submanifold there is (1) a tubular/$\epsilon$-neighborhood, which is a neighborhood of $S$ inside $M^n$ (if you have a Riemannian metric just define it to be points in $M$ which are $\epsilon$-away from $S$) (2) a normal bundle, which is a bundle $NS$ over $S$ such that $NS \oplus TS \cong \underline{\Bbb R^n}$ and (3) a function $\phi : M \to \Bbb R$ such that $\phi^{-1}(0) = S$.
(1) and (2) are equivalent objects; that's the content of the tubular neighborhood theorem
If you have a function $\phi$ in (3), $d\phi : TM \to T\Bbb R$ be the differential. Consider the kernel of $d\phi$. That's $NS$.
I understand that position eigenstates are delta functions but experimentally you will measure position with limited accuracy hence it follows that the actual quantum state you get is a superposition of these ideal position eigenstates or do you get a mixed state corresponding to these eigenstates?
@JohnRennie Nitpicky comment about your profile text: the internet was around in the eighties, the web was not. Not that many people care about the difference unless they actually used the thing in the pre-web era (I did, but only a little).
@EmilioPisanty As of right now we don't have any tool for looking at user information other than user pages (our just show some info and tools that aren't shown to most users).
But I did just notice that we can now see what questions are associated with a deleted account. That's new.
In looking at the question that went with the deleted account I'm guessing that user was young. Maybe they took the advice in that thread, negotiated with the team and got a reprieve?
@EmilioPisanty Yeah, I can see both through the user page, though in this instance I don't learn anything from the deleted one that you can't see from the others.
If I have the map f(x,y,z)=(x/1-z , y/1-z), defined to be the projection, can I just find the reverse function without having pictures? Do I have to exchange the z variable with something? Thanks anyway.
The little group of massless particles is $SO(2)$, which is abelian so irreps are 1-dimensional, and the generators is $J_z$, so we call the eigenvalue of $J_z$ the helicity of the particle, but why is it quantized? You can't use $J_x, J_y$ operators to prove it, need that $SL(2,C)$ is the double cover of Lorentz for some reason...
So you have irreps of abelian $\mathrm{SO}(2)$ being 1-d, and $J_z$ generating, it's eigenvalue $h$ being called helicity, only issue is you can't prove the eigenvalue is discrete $0,\frac{1}{2},1,\dots$ the usual way with $J_{\pm}$ operators because they don't exist, kick in the face basically
'there is a topological proof that $h$ is quantized based on $\mathrm{SL}(2,\mathbb{C})$'
A $\mathrm{U}(1)$ rotation of the little group is represented by $e^{i h \theta}$, and $\mathrm{SL}(2,\mathbb{C})$ is a double cover so 'a rotation by an angle $4 \pi$ around the momentum can be continuously deformed into no rotation at all, so the factor $\exp(4 \pi i h)$ must be unity, and hence $h$ must be an integer or half-integer.'
Why can't that prove all spin is quantized
$\mathrm{SU}(2)$ is the double cover of the little group $\mathrm{SO}(3)$ of massless particles, and so blah
For $SO(2)$ you only have the $J_3$ operator can you only write $e^{i J_3 \theta} |h> = e^{i h \theta}|h>$ and then the double cover argument means $e^{i4 \pi h} = 1$ so that $h \in \frac{1}{2} \mathbb{N}$, while for $SO(3)$ you have to work with $e^{iJ_{\pm} \cdot \chi_{\pm} + i J_3 \theta}$ stuff so you can't just assume it's a half-integer/integer
The $ISO(2)$ thing in Maggiore is not so bad, gets it in a few lines, then shows the non-$J_z$ lie algebra generators are Hermitian which means their lie group operators are unitary so they have continuous eigenvalues and this leads to continuous helicity, so it's like a choice to make it discrete in a sense, hmm
Today was my first time reviewing low-quality posts, and I accidentally pressed the "Looks OK" button a few times for posts that should clearly be deleted, as I was used to reviewing edits and "Looks OK" and "Approve Edit" are in the same place. Is there a way to reverse this decision?
So my question is, I understand the math behind the KW duality, no problem there, but I'm a bit confused about its role in Onsager's derivation of the 2d model
So my understanding of it now is that KW kills two birds with one stone, the two dimensional lattice is related to the one-dimensional chain through the geometrical argument of the dual lattice
@Slereah my mental calculus right now: on the one hand, I'm hungry and I forgot to grab my lunch this morning. on th eother hand, it's 12 degrees outside
But I don't see why it would be necessary to reexponentiate in terms of $\theta$ rather than say, just $\beta\epsilon$, given that $\tanh\theta = e^{-2\beta\epsilon}$ and the partition function has coefficients $\epsilon$ in front of the terms in the Hamiltonian:
Here's the tortured logic, so we know irreps of abelian $SO(2)$ are 1-d, we know casimirs provide irreps, but in the massless case $P^2$ and $W^2$ are no longer casimirs, however you can't just set $m = 0$ in $W^2$ to show it's $0$ you need to do the $ISO(2)$ proof, and then you still use $W^2 = 0$ to prove that $h = p \cdot J$, which makes sense to use since you're working with irreps, but god
So in the transfer matrix method this is reformulated to, say, $$ P = V_2 V_1 $$ where $$ V_2 = \prod \exp\left(\beta\epsilon \sigma^z_i \sigma^z_{i+1}\right) $$ and $$ V'_1 = \prod \exp\left(\beta\epsilon s_i s_i'\right) $$
iirc KW arises by considering two different ways of summing the partition function, and then using an argument about the critical point to link both expansions, giving the critical point pretty fast, e.g. topo-houches.pks.mpg.de/wp-content/uploads/2014/11/… but it's not really about the steps in either way of setting up the problem
The combinatorial steps are just a problem, and then the very end of the Landau solution with that crazy matrix is just too much, had to use mathematica for it
You can even derive $$\hat{V}_{ex} = - \frac{1}{2}\sum_{i<j} J_{ij}(1 - 4 \mathbf{S}_i \cdot \mathbf{S}_j) = - \sum_{i<j} J_{ij}(\frac{1}{2} + 2 \mathbf{S}_i \cdot \mathbf{S}_j) = - \sum_{i,j} J_{ij}(\frac{1}{4} + \mathbf{S}_i \cdot \mathbf{S}_j).$$ from general principles, 'exchange interaction', i.e. where the Ising hamiltonian comes from in the first place, in a way that motivates the other models
Eh, basically this is supposed to be an extension on what was already derived in class, we did the 1d ising model through a more tedious method than transfer matrices
So I wanted to do that, perhaps KW duality and a general scheme of solving 2D
I often come across interesting questions, which I can not answer. As a consequence I can not verify the given answers while some seem to be helpful. Is it recommended to upvote such answers, where my knowledge doesn't reach put to verify them?
Today was my first time reviewing low-quality posts, and I accidentally pressed the "Looks OK" button a few times for posts that should clearly be deleted, as I was used to reviewing edits and "Looks OK" and "Approve Edit" are in the same place. Is there a way to reverse this decision?
