## Why is rapidity maximum at 0 ? May 15, 2008

Posted by dorigo in mathematics, physics, science.
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In a recent post where I shortly discussed a rapidity distribution of muons produced in low-energy proton-proton collisions at LHC, I formulated a conjecture: given parton densities in the colliding protons which are monotonously decreasing functions of the momentum fraction (and they are, for small fractions), the rapidity y of the parton-parton collision has a maximum for y=0.

Of course, I was guided by experience: I know for a fact that unless you bias your data sample with requirements such as trigger selections or analysis cuts, any time you plot a rapidity distribution you obtain something symmetrical around zero. But one thing is having experience of the rule, the other is proving there are no exceptions.

So I was triggered to demonstrate the rule. I was helped by a grad student in our group, Nicola Pozzobon… Without him, I would not have gone past the long-forgotten rules of the delta function!

Demonstration: Take parton distribution functions in the proton to be monotonously decreasing functions of the momentum fraction:

$f(x)$, $x \in [0,1[$ , $df/dx<0$.

The rapidity of the collision can be defined as

$y=0.5 \log (x_1/x_2)$

and so to obtain the distribution of rapidity arising from a given distribution of the partons in the proton we have to integrate on $x_1, x_2$ as follows:

$G(y) = \int dx_1 dx_2 f(x_1) f(x_2) \delta(y-0.5 \log(x_1/x_2))$.

The delta function, which is zero everywhere and equal to one only where its argument is zero, “picks up” the relevant values of $f(x_1), f(x_2)$ capable of contributing to a given value of y.

Now, if we solve $y=0.5 \log(x_1/x_2)$ for x_1 we find $x_1=x_2 e^{2y}$, so we can substitute in the expression for G and integrate in x_1, to find

$G(y) = \int dx_2 f(x_2) f(x_2 e^{2y}) /(2x_2 e^{2y})$,

where we have used the property of the delta function which says that

$\delta(h(x)) = \delta(x)/|h'(0)|$,

and the fact that

$h(x_1, x_2)=y-0.5 \log(x_1/x_2)$

from which

$h'(x_1)=dh/dx_1=-1/(2x_1)=-1/(2x_2 e^{2y})$,

or

$h'(x_2)=dh/dx_2=1/(2x_2)=1/(2x_1 e^{-2y})$.

If we had instead substituted $x_2$ we would have found

$G(y) = \int dx_1 f(x_1) f(x_1 e^{-2y}) / (2x_1 e^{-2y})$,

so we can generically write, using x with no subscripts:

$G(y) = \int dx f(x) f(x e^{\pm 2y}) e^{\mp 2y} /(2x)$,

with the agreement that both expressions implied by the sign swaps have to be true simultaneously.

Upon derivation with respect to y we find

$G'(y) = \int dx f(x)/x [ \pm x f'(x e^{\pm 2y}) \mp f(x e^{\pm 2y}) e^{\mp 2y}]$.

For y=0, the derivative of G is thus equal to

$G'(0) = \int dx f(x)/x [\pm x f'(x) \mp f(x)]$.

But these (the G'(0) with +- and the G'(0) with -+ signs chosen) are two expressions that must simultaneously be correct: and since they have opposite values for any x in the integration interval, they must be zero. If the derivative of a function is null, the function has an extremum there. CVD.

We have thus proved that the rapidity distribution has an extremum if the PDF f(x) are monotonous functions of x. It would be easy to show that this is indeed a maximum, but I will be content with the above result for this post.

So in conclusion, my conjecture is correct! If the parton distribution functions are taken to be monotonous, the rapidity distribution of the center-of-momentum of the collision is indeed maximum at zero… I prefer to use intuition usually, because the above demonstration took me more than an hour!

UPDATE: Hmmm, by thinking at the problem for a microsecond, rather than writing integrals and derivatives, I only now arrive at the following conclusion: Since the rapidity distribution is symmetrical around zero – from its very definition -, and since it must go to zero for y going to plus and minus infinity, it goes without saying that there is an extremum at zero, duh! However, it is still interesting to see how that arises mathematically. THis also proves what I know for a fact: a bit of analysis is worth more than a megabyte of programming – and the same can be said with thinking and computing formulas!

1. carlbrannen - May 15, 2008

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.

2. dorigo - May 16, 2008

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.

3. Andrea Giammanco - May 16, 2008

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?

4. dorigo - May 16, 2008

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.

5. carlbrannen - May 16, 2008

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…

6. Tim Tait - May 20, 2008

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

7. dorigo - May 20, 2008

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 đŸ˜‰

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