Comments on: Why is rapidity maximum at 0 ? http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/ private thoughts of a physicist and chessplayer Thu, 24 Dec 2009 08:50:26 +0000 http://wordpress.com/ hourly 1 By: dorigo http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97445 dorigo Tue, 20 May 2008 12:50:45 +0000 http://dorigo.wordpress.com/?p=1251#comment-97445 Hi Tim, thank you for pointing that out. The argument above is didactical in nature, but I should have probably mentioned it in the text... Cheers, T. PS I wrote about the W asymmetry in a dozen proceedings papers ;-) Hi Tim,

thank you for pointing that out. The argument above is didactical in nature, but I should have probably mentioned it in the text…

Cheers,
T.

PS I wrote about the W asymmetry in a dozen proceedings papers ;-)

]]> By: Tim Tait http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97443 Tim Tait Tue, 20 May 2008 12:33:25 +0000 http://dorigo.wordpress.com/?p=1251#comment-97443 Hi Tommaso, The rapidity distributions of inclusive W+ or W- at the Tevatron are neither maximum at zero not symmetric about it... the loophole in your argument is that it assumes that both of the PDF functions are the same function, but this need not be the case (and in fact is never true for W production because they are charged). In that case the monotonically decreasing up and (anti-)down quark PDFs balance to give a maximum at a finite value of y. Regards, Tim Hi Tommaso,

The rapidity distributions of inclusive W+ or W- at the Tevatron are neither maximum at zero not symmetric about it… the loophole in your argument is that it assumes that both of the PDF functions are the same function, but this need not be the case (and in fact is never true for W production because they are charged).

In that case the monotonically decreasing up and (anti-)down quark PDFs balance to give a maximum at a finite value of y.

Regards,
Tim

]]> By: carlbrannen http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97319 carlbrannen Fri, 16 May 2008 22:12:36 +0000 http://dorigo.wordpress.com/?p=1251#comment-97319 Tommaso, I'm surprised that I could read this so quickly and understand it. You're operating under your usual plane which is way over my head. Right now I'm getting ready for one of the lectures that the American Physical Society allows amateurs to give, tomorrow in Portland. The general idea will be to show how one can make exact calculations of bound states in QCD, and then find quantum corrections to them in line with how one does a Lamb shift correction to hydrogen bound states. As part of this, I wanted to write down a wave function that would depend on a finite number of points, say n=1,2,3. The idea is to treat those three points as "red", "green", and "blue" for QCD. My instinct was to write $latex \psi(x) = \Sigma a_n\delta(x-n)$ but it didn't work. Correct is $latex \psi(x) = \Sigma a_n\sqrt{\delta(x-n)} .$ Let's see if I got the LaTeX right... Tommaso, I’m surprised that I could read this so quickly and understand it. You’re operating under your usual plane which is way over my head.

Right now I’m getting ready for one of the lectures that the American Physical Society allows amateurs to give, tomorrow in Portland. The general idea will be to show how one can make exact calculations of bound states in QCD, and then find quantum corrections to them in line with how one does a Lamb shift correction to hydrogen bound states. As part of this, I wanted to write down a wave function that would depend on a finite number of points, say n=1,2,3. The idea is to treat those three points as “red”, “green”, and “blue” for QCD.

My instinct was to write \psi(x) = \Sigma a_n\delta(x-n) but it didn’t work. Correct is
\psi(x) = \Sigma a_n\sqrt{\delta(x-n)} . Let’s see if I got the LaTeX right…

]]> By: dorigo http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97296 dorigo Fri, 16 May 2008 12:56:20 +0000 http://dorigo.wordpress.com/?p=1251#comment-97296 Hi Andrea, in general yes, typical selections one applies to particles (say, a minimum Pt cut) make the rapidity distribution more peaked at zero. However, the camel plot was made at generation level. There, the only issue is the fact that PDF are not known at arbitrarily small values of x, where BFKL dynamics reign. Cheers, T. Hi Andrea,

in general yes, typical selections one applies to particles (say, a minimum Pt cut) make the rapidity distribution more peaked at zero.
However, the camel plot was made at generation level. There, the only issue is the fact that PDF are not known at arbitrarily small values of x, where BFKL dynamics reign.
Cheers,
T.

]]> By: Andrea Giammanco http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97295 Andrea Giammanco Fri, 16 May 2008 12:36:22 +0000 http://dorigo.wordpress.com/?p=1251#comment-97295 Hmmmm, ok, but the whole argument for using rapidity as a physics variable in hadron colliders, isn't that (in some approximation) QCD distributions are flat in eta? (My source: Barger & Phillips, "Collider physics".) This flatness is exploited for example in UE-subtraction for jets (and indeed it's the rationale, as far as I know, for using cones in eta-phi instead of theta-phi, as would be more natural for the signal alone). Of course this flatness cannot be absolute, because this would mean infinite integral between -inf and +inf (and indeed I believe the result of your mathematical demonstration, because it relies on the assumption of finite area under the distribution). But I guess it holds for a very large range of eta. (Here I'm referring to eta because this is what we really use, I should refer to y.) So, I was wondering: maybe also your empirical observation of a dromedar shape is an artifact of selection, and not only the camel! (The camel would be a second-order artifact.) I mean that maybe the maximum would be at 0 even without the artifact, but the artifact makes it more visible. The artifact could come from the fact that y and pT are not independent, and low-pT particles are detected less easily. What do you think about that? Hmmmm, ok, but the whole argument for using rapidity as a physics variable in hadron colliders, isn’t that (in some approximation) QCD distributions are flat in eta?
(My source: Barger & Phillips, “Collider physics”.)
This flatness is exploited for example in UE-subtraction for jets (and indeed it’s the rationale, as far as I know, for using cones in eta-phi instead of theta-phi, as would be more natural for the signal alone).
Of course this flatness cannot be absolute, because this would mean infinite integral between -inf and +inf (and indeed I believe the result of your mathematical demonstration, because it relies on the assumption of finite area under the distribution). But I guess it holds for a very large range of eta.
(Here I’m referring to eta because this is what we really use, I should refer to y.)
So, I was wondering: maybe also your empirical observation of a dromedar shape is an artifact of selection, and not only the camel! (The camel would be a second-order artifact.)
I mean that maybe the maximum would be at 0 even without the artifact, but the artifact makes it more visible.
The artifact could come from the fact that y and pT are not independent, and low-pT particles are detected less easily.
What do you think about that?

]]> By: dorigo http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97285 dorigo Fri, 16 May 2008 08:39:11 +0000 http://dorigo.wordpress.com/?p=1251#comment-97285 Carl, of course you are right on all three counts. The "demonstration" was not useless, however, because it allowed me to ponder on a few issues with the rapidity distribution. For me, computing integrals is not an everyday occupation, so it is always refreshing... Cheers, T. Carl, of course you are right on all three counts.
The “demonstration” was not useless, however, because it allowed me to ponder on a few issues with the rapidity distribution. For me, computing integrals is not an everyday occupation, so it is always refreshing…

Cheers,
T.

]]> By: carlbrannen http://dorigo.wordpress.com/2008/05/15/why-is-rapidity-maximum-at-0/#comment-97255 carlbrannen Thu, 15 May 2008 22:48:56 +0000 http://dorigo.wordpress.com/?p=1251#comment-97255 In English, to my ear, the word "monotonically" is preferred to "monotonously", which is more suitable for describing most politician's speech rather than a function. Also, the mathematicians are going to disagree with your definition of the delta function in that it's infinite at the selected point rather than 1. The 1 is what you get when you integrate over the selected point. Also, the whole argument that there is an extremum at 0 seems to me to be automatic with an even function. This does not prove that there is a maximum; it could also be a minimum. In other words, this does not at all exclude the camel function. In English, to my ear, the word “monotonically” is preferred to “monotonously”, which is more suitable for describing most politician’s speech rather than a function. Also, the mathematicians are going to disagree with your definition of the delta function in that it’s infinite at the selected point rather than 1. The 1 is what you get when you integrate over the selected point.

Also, the whole argument that there is an extremum at 0 seems to me to be automatic with an even function. This does not prove that there is a maximum; it could also be a minimum. In other words, this does not at all exclude the camel function.

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