meta user
network profile
| bio | website | igorivanov.blogspot.com |
|---|---|---|
| location | Liège, Belgium | |
| age | ||
| visits | member for | 2 years, 6 months |
| seen | May 5 at 13:17 | |
| stats | profile views | 415 |
postdoc in theoretical high-energy physics
|
Mar 28 |
awarded | Enlightened |
|
Mar 28 |
awarded | Nice Answer |
|
Nov 2 |
awarded | Yearling |
|
Nov 2 |
awarded | Yearling |
|
Oct 7 |
comment |
Why Are Even and Odd Regge Trajectories Degenerate? You might also want to take a look at the closely related concept of parity doubling in hadron spectrum, see e.g. this review: arxiv.org/abs/arXiv:0704.1639 |
|
Jul 18 |
awarded | Nice Answer |
|
Mar 15 |
awarded | Nice Answer |
|
Feb 2 |
answered | Half wave plate and angular momentum |
|
Jan 30 |
comment |
Does bunching reduce synchrotron radiation? @John — just to make sure: the term "bunching" means "the act of grouping particles in bunches", while "debunching" means "spreading out initially bunched particles into a more homogeneous distribution". You seem to be using "bunching" as an equivalent of "the number of bunches", which is not the correct usage. |
|
Jan 30 |
comment |
Does bunching reduce synchrotron radiation? Yes, you need bunches anyway for a proper functioning of a collider because particles are accelerated in RF cavities in bunches. |
|
Jan 30 |
awarded | Organizer |
|
Jan 30 |
comment |
Does bunching reduce synchrotron radiation? By the way, a remark to the title: it is actually debunching not bunching that reduces the SR. |
|
Jan 30 |
revised |
Does bunching reduce synchrotron radiation? removed the LHC from the text and from the tags |
|
Jan 30 |
answered | Does bunching reduce synchrotron radiation? |
|
Jan 5 |
awarded | Enlightened |
|
Dec 25 |
comment |
What specifically does the phrase “continuum limit” mean? @Marek — thanks for the effort, but still it does not sound persuasive enough to me. I don't really care about limit in the sense you mention: sure, Q is dense everywhere and its closure is R. I just wonder at which point we actually get continuity for the set of degrees of freedom, and as far as I see we still don't get it in the standard procedure. In my understanding continuity is introduced later: when we actually calculate something, we actually average whatever we get (Q or R, no matter) with some continuous weight functions. |
|
Dec 25 |
comment |
What specifically does the phrase “continuum limit” mean? Then at first step we add a node at 1/2, then we add a node at 1/4, 3/4, etc. You see that you never add irrational points, and at the end you get a subset of Q. In order to claim that you have a true continuum limit, you need to devise an explicit procedure and show that every real appears at some stage. |
|
Dec 25 |
comment |
What specifically does the phrase “continuum limit” mean? @Marek — yes, I know how rational are introduced, and btw the reals are introduced in an even more non-physical way. But I am not considering a limit staying in Q. I am considering your procedure and what I see at the end is Q, not R. Take 1D lattice of fixed size (=1). You take small spacing a and take the limit $a \to 0$. These words are not enough, you need to give a specific prescription of what you exactly do and then prove the result is the same for all prescriptions. Let's adopt the procedure when we divide a by 2 at each step. (cont.) |
|
Dec 24 |
comment |
What specifically does the phrase “continuum limit” mean? In this standard picture there is something which I never really understood. As far as I can see this limit leads to the set of rationals rather than the set of reals. In that sense you do not actually perform the continuum limit, only the "rational limit"! To get continuum, you'll need an extra "jump" at the end from Q to R, and there is no guarantee that everything valid for Q will be valid for R. |
|
Dec 17 |
answered | Can a multipass x-ray absorption cell be constructed? |
