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

Posted by dorigo in mathematics, physics, science.
Tags: , ,

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!