# Froissart bound

Post-publication activity

Curator: Jean Zinn-Justin

The Froissart bound1 is the name of a very general property of the behaviour of total particle scattering cross sections at very high energy, e.g. at accelerators2.

## Intuitive reasoning Figure 1: The incident particle beam comes from the right. Within a circle of radius b0 in the plane of the target particle, the scattering probability cannot be larger than 1. Outside, it is non-negligible essentially in a ring of width 1/k. The total cross section is therefore bounded by π(b0 + 1/k)².

The nuclear strong interactions are typically mediated by bosons, the lightest of which is the pion. In classical mechanics, it is impossible for a particle to exchange a pion with another particle, because this would violate energy conservation at some time. In quantum mechanics, however, this is possible, during very small time intervals, because of Heisenberg's uncertainty principle. But this possibility, also called tunnel effect decreases exponentially with the distance. Its maximum is thus reached when the incident particle is closest to the target particle, at a distance – measured perpendicular to the incident line-of-flight – $$b$$ which is called impact parameter. One problem is that one does not know the intrinsic force of the interaction, just the dependence on the impact parameter.

Two cases have to be considered :

1. The impact parameter is small enough $$b\,<\,b_0\ ,$$ so that the intrinsic force of the interaction is not affected by the tunnel effect. The interaction probability is then less than or equal to 1.
2. For $$b\,>\,b_0\ ,$$ the tunnel effect wins, and the probability is bounded by the exponential $$\exp(kb_0-kb)\ ,$$ since it is bounded by 1 at $$b=b_0\ .$$

The total probability of case 1 is bounded by the area of the circle of radius $$b_0\ ,$$ i.e. $$\pi b_0^2\ .$$

The total probability of case 2 is bounded by:

$$2\pi\int_{b_0}^\infty \exp(kb_0 - kb)\, \mathrm{d}b = 2\pi\int_0^\infty \exp(- kb')(b'+b_0) \,\mathrm{d}b' = \pi /\,k^2 + 2\pi b_0/\,k$$

It turns out to be the area of a ring around the preceding circle, of width $$1/k\ .$$ The grand total probability is thus $$\pi(b_0+1/k)^2\ .$$

The quantity $$k$$ is given by the theory of tunnel effect, and it is related to the mass $$m$$ of the lightest meson that can be exchanged, in general the pion3, by $$k = m\,hc/2\pi$$4. The problem is to have an estimate of $$b_0\ .$$ By definition, $$b_0$$ is the impact parameter where a possible increase of the intrinsic coupling of the particles to the pion is cancelled by the damping exponential of the tunnel effect. It is generally accepted that intrinsic couplings of particles do not increase faster than some power of the center-of-mass energy $$E$$ as the latter becomes very large. $$b_0$$ is thus defined by the relation

$$C E^n\exp -k\,b_0 \,= \,1~;~b_0 = \,(n \,\ln \,E\, +\, \ln\, C)\,/\,k$$

where $$n$$ and $$C$$ are some constants, which we do not know. But the essential part of the reasoning is reached: the total cross section, which is bounded by $$\pi(b_0+1/k)^2$$ does not increase faster than $$\ln^2E$$ multiplied by some constant. The additional terms do not play any role, since they can always be absorbed into the overall constant at high enough $$E$$. This expresses the Froissart bound.

## Scheme of the proof according to the prevalent theories at the time of its formulation

In 1961, the French physicist Marcel Froissart proves2 that total scattering cross-sections among particles do not increase faster than the square of the logarithm of the energy as this energy increases. This is of some importance both for theorists and experimentalists.

For the theorists, it shows that one can write dispersion relations with at most 2 subtractions. These relations are used to relate the real part to the imaginary part of the scattering amplitude. In elastic scattering, for example, the imaginary part of the scattering amplitude in the forward direction is linked by the so-called optical theorem to the total cross-section. Putting a bound on the total cross-section allows to do calculations with a good confidence about the convergence of the integrals at infinity.

For the experimentalist, it is the big time to build ever more powerful accelerators. It is somewhat satisfactory to know that the experiments built around these accelerators would not be blinded by cross-sections increasing too fast.

### Kinematics and notations

The quoted article gives as an introduction the qualitative reasoning as presented in the preceding section. A more rigourous proof is then given, based on premises by now (XXIth century) completely outdated. Remember that at the time, pions were considered really elementary particles, with their own fields and the like. There was no consistent theory of these fields, and therefore it was tried to do completely without fields. It was hoped that the properties of the S-matrix (the matrix of all scattering amplitudes) would be strong enough to constrain it to a small family of acceptable solutions, among which a choice could be made, for example by fitting the existing masses, spins and other properties of observed particles. The two main ingredients that came into play to constrain the S-matrix were analyticity properties, some of which (in particular those giving rise to dispersion relation) had been proved in quantum field theory, and of course unitarity which should guarantee that in all situations, the total of probabilities of all possible outcomes is 15.

If we consider the simplest case, when there is scattering of particles $$1+2 \to 3+4\ ,$$ if we denote by $$p_i$$ the 4-momentum of particle $$i\ ,$$ we have of course $p_1^\mu + p_2^\mu = p_3^\mu + p_4^\mu$ (conservation of total energy-momentum), and moreover each particle has a well-defined mass $$m_i$$ given by $m_i^2= p_i^2 \equiv E_i^2 - \vec p_i~^2$
Combining all these equations gives $m_1^2=(p_2-p_3-p_4)^2=m_2^2+m_3^2+m_4^2-2p_2\cdot p_3-2p_2\cdot p_4 + 2p_3\cdot p_4$ $=(p_2-p_3)^2+(p_2-p_4)^2+(p_3+p_4)^2-m_2^2-m_3^2-m_4^2$

Let us denote \begin{align}s&=(p_1+p_2)^2=(p_3+p_4)^2\\ t&=(p_1-p_3)^2=(p_2-p_4)^2\\ u&=(p_1-p_4)^2=(p_2-p_3)^2\\ s+t+u&=m_1^2+m_2^2+m_3^2+m_4^2\end{align}

We can interpret $$s$$ as the square of the center-of-mass (c.m.s.) energy of the system $$p_1+p_2\to p_3+p_4\ ,$$ $$t$$ as the square of the c.m.s. energy of the scattering of particle 1 with antiparticle 3 of 4-momentum $$-p_3\ ,$$ giving particle 2 and antiparticle 4 of 4-momentum $$-p_4\ .$$ Similarly, $$u$$ can be interpreted as the square of the c.m.s. energy of the scattering of particle 1 and antiparticle 4 to particle 2 and antiparticle 3.

By virtue of the axioms of the S-matrix theory, the amplitudes of these 3 processes are one and the same analytic function of two independent variables among $$s,~t,~u\ .$$ Of course, not all values of these variables can be interpreted as physical amplitudes, since, for example, the scattering angles must be real. To simplify things, consider all masses equal and taken as unit. The threshold c.m.s. energy for reaction (12) is 2, i.e. $$s=4\ ,$$ and similarly for the other 2 reactions. Let us compute the scattering angle $$\theta\ ,$$ i.e. the angle between $$\vec{p_1}$$ and $$\vec{p_3}$$ in the c.m.s., for some value of $$s\ .$$ \begin{align}s&= 4(1+\vec p~^2)\\ t&=(p_1-p_3)^2 = - 2\vec p ~^2(1-\cos \theta)\\ \cos \theta &=1+\frac{2 t}{s-4}= -1 -\frac{2u}{s-4}\end{align} The physical region for scattering in the $$12 \to 34$$ channel is then limited by $$s>4~ ; ~ t<0~ ; ~ u<0 \ ,$$ all being real. We shall denote by $$A(\cos\theta)$$ the amplitude, since we shall for the most time work at constant $$s\ .$$

### Partial wave amplitudes and unitarity

It is interesting to decompose the scattering amplitude into partial waves, by integrating them with the Legendre polynomial $$P_\ell(z)\ ,$$ and multiplied by the kinematic factor $$\frac{\pi q}{2\sqrt{s}}$$ where $$q$$ is the length of the 3-momentum of one of the particles in the center-of-mass system (c.m.s.)$q = \frac{\sqrt{s-4}}{2}\ .$ $a_\ell(s) = \frac{\pi q}{2\sqrt{s}}\int_{-1}^1 A(z) P_\ell(z) \mathrm{d}z$

With this normalization, the partial wave can be written as a function of the phase shift $$\delta_\ell$$ as $$a_\ell\,=\,e^{i\delta_\ell}\ .$$

This decomposition is interesting to the extent that the value of $$\ell$$ is conserved whatever the changes of composition of the states: this expresses the rule of conservation of angular momentum in the c.m.s. If the only possibility is elastic scattering, that is, under the threshold for inelastic reactions, the probability of the state is conserved, that is, the amplitude can at most undergo a phase change.

Under the threshold for inelastic reactions, the phase shift is then pure real, and the modulus of the partial wave is 1. Above the threshold, the initial 2-body state can go into other states, so that the probability of finding it again after the reaction may be less than unity. This means that the phase shift $$\delta_\ell$$ can get an imaginary part, but only positive, in order that $$|a_\ell|^2\ ,$$ the probability that the initial state is found after the reaction, be less than unity.

### Analytic properties

In the physical region, saying that an interaction takes place implies that the phase shifts are non-zero. The amplitude thus takes an imaginary part for $$s>4\ ,$$ which means that there is a singularity along $$s=4\ .$$ Similarly, there will be singularities for $$t=4$$ and $$u=4\ .$$ One of the hypotheses of the S-matrix theory is that there are as few singularities as possible. But these singularities cannot be avoided, they come from unitarity itself. If one considers the plane of the complex $$z=\cos \theta\ ,$$ these singularities occur at $$z = 1+ \frac{8}{s-4}=\frac{s+4}{s-4}$$ for the $$t$$ singularity, and $$z = -\frac{s+4}{s-4}$$ for the $$u$$ singularity.

The effects of analyticity are much less trivial than those of unitarity. Since we postulate, as mentioned before, that the amplitude does not increase faster than some polynomial in $$s,~t,~u$$ it is bounded6 in the $$z$$ plane by some power $$z^k\ .$$ The function $\frac{A(z)}{z^{k+1}} \to 0 ~\mathrm{as}~z \to \infty\ .$

### Integral representation of the amplitude

It is then possible to express $$\frac{A(z)}{z^{k+1}}$$ as a Cauchy integral $\frac{A(z)}{z^{k+1}} = \frac{1}{2\pi i}\oint \frac{A(z')}{z'^{k+1}(z'-z)} \mathrm{d}z'\ ,$ the path of integration encircling once in the direct sense the point $$z\ .$$ But since the behavior at infinity of $$\frac{A(z')}{z'^{k+1}}$$ is good, nothing stops us from blowing up the contour to infinity, except for the singularities at $$z'=\plusmn \frac{s+4}{s-4}\ ,$$ and the multiple pole at $$z'=0\ .$$ We know little about the singularities, except that the amplitude will have there the continuation of an imaginary part, which will presumably extend to infinity in $$z\ .$$ The S-matrix axioms require that there should not be any unnecessary singularity, so the singularity at $$z=\plusmn \frac{s+4}{s-4}$$ is the end of a cut in the $$z$$ plane. We will denote the jump across this cut as $A(z+i0)-A(z-i0) = 2i \rho (z)\ :$

for $$z> \frac{s+4}{s-4}$$


$A(z-i0)-A(z+i0) = 2i \rho' (z)\ :$

for $$z< - \frac{s+4}{s-4}$$


where $$z\plusmn i0$$ is used to show on which side of the cut $$A$$ has to be considered.

Furthermore, deforming the Cauchy integral path across the multiple pole at $$z' = 0$$ will give the polynomial in $$\frac{1}{z}$$ cancelling the singularity of $$\frac{A(z)}{z^{k+1}}\ .$$

When the path of the Cauchy integral is moved as far away as possible, we are left with 3 terms: an integral over the jump of each of either cuts, and a polynomial coming out from the singularity of $$\frac{A(z)}{z^{k+1}}\ :$$

$\begin{array}{lcl}\displaystyle\frac{A(z)}{z^{k+1}}&= &\displaystyle\frac{1}{\pi}\int_{\frac{s+4}{s-4}}^\infty \frac{\rho(z')}{z'^{k+1}(z'-z)} \mathrm{d}z'\\&+&\displaystyle\frac{1}{\pi}\int_{-\frac{s+4}{s-4}}^{-\infty}\frac{\rho'(z')}{z'^{k+1}(z'-z)} \mathrm{d}z'\\&+&\displaystyle\sum_{n=0}^k \left.\frac{\mathrm{d}^n A(z)}{\mathrm{d}z^n} \right|_{z=0}\frac{z^{n-k-1}}{n!}\end{array}$

or, equivalently:

$\begin{array}{lcl}\displaystyle A(z)&=& \displaystyle \frac{1}{\pi}\int_{\frac{s+4}{s-4}}^\infty \frac{z^{k+1}\rho(z')}{z'^{k+1}(z'-z)} \mathrm{d}z'\\&+&\displaystyle\frac{1}{\pi}\int_{-\frac{s+4}{s-4}}^{-\infty}\frac{z^{k+1}\rho'(z')}{z'^{k+1}(z'-z)} \mathrm{d}z'\\&+&\displaystyle\sum_{n=0}^k \left.\frac{\mathrm{d}^n A(z)}{\mathrm{d}z^n}\right|_{z=0} \frac{z^n}{n!}\end{array}$

For $$\ell>k\ ,$$ the residual polynomial is orthogonal to $$P_\ell(z)\ ,$$ and: $\begin{array}{lcl}a_\ell(s) &=&\displaystyle \frac{q}{2\sqrt{s}}\int_{-1}^1 P_\ell(z)\left[ \int_{\frac{s+4}{s-4}}^\infty \frac{z^{k+1}\rho(z')}{z'^{k+1}(z'-z)}\mathrm{d}z'\right.\\&+&\displaystyle\left.\int_{-\frac{s+4}{s-4}}^{-\infty}\frac{z^{k+1}\rho'(z')}{z'^{k+1}(z'-z)}~ \mathrm{d}z' \right]~\mathrm{d}z\end{array}$

The integral over $$z$$ involves integrating the Legendre polynomial with a function that has a single pole. This leads by definition to the Legendre function of the 2nd kind: it is defined by: $Q_\ell(z') = \frac{1}{2} \int_{-1}^1 P_\ell(z) \frac{\mathrm{d}z}{z-z'}$ It is many-valued, since it has logarithmic singularities at $$z'=\pm 1\ .$$ We take the value that is real on the real axis outside of $$[-1,1]\ .$$ Let us note that an equivalent definition for $$\ell>k$$ is given by $Q_\ell(z') = \frac{1}{2} \int_{-1}^1 P_\ell(z)\frac{z^{k+1}}{z'^{k+1}(z-z')} \mathrm{d}z$ because $$\frac{z^{k+1}}{z'^{k+1}(z-z')} - \frac{1}{z-z'}$$ is a polynomial of degree $$k$$ of $$z\ ,$$ the integral of which with a Legendre polynomial of higher degree $$\ell$$ vanishes.

We can interchange the order of integrations, since the whole converges well. For $$\ell>k\ :$$ $a_\ell(s) = \frac{q}{\sqrt{s}}\int_{\frac{s+4}{s-4}}^\infty Q_\ell(z')\left[\rho(z')+(-1)^\ell\rho'(-z')\right]~\mathrm{d}z'$

### Use of the integral representation to get an exponential bound

Now the problem is to use the properties of the function $$Q_\ell$$ in order to get an exponential bound on $$a_\ell\ .$$

To do that, we use the generating function of the $$Q_\ell$$ as given in the standard textbooks7: $\sum_{\ell=0}^\infty z^\ell Q_\ell(x) = \int_{x+\sqrt{x^2-1}}^\infty\frac{\mathrm{d}\zeta}{\sqrt{1-2\zeta x+\zeta^2}(\zeta -z)}$ To be consistent with previous computations, we subtract  the terms up to $$\ell=k\ :$$ $\sum_{\ell=k+1}^\infty z^\ell Q_\ell(x) = \int_{x+\sqrt{x^2-1}}^\infty\frac{\mathrm{d}\zeta}{\sqrt{1-2\zeta x+\zeta^2}}\frac{z^{k+1}}{\zeta^{k+1}(\zeta -z)}$ We can infer from this expression the generating function for the $$a_\ell\ :$$ \begin{align} \sum_{\ell=k+1}^\infty z^\ell a_\ell &= \sum_{\ell=k+1}^\infty \frac{q}{\sqrt{s}} \int_{\frac{s+4}{s-4}}^\infty \left[\sum_{\ell=k+1}^\infty z^\ell Q_\ell(z')\rho(z')\right.\\ &+ \left.\sum_{\ell=k+1}^\infty z^\ell Q_\ell(z')(-1)^{\ell}\rho'(-z')\right]~ \mathrm{d}z'\\ &=\frac{q}{\sqrt{s}} \int_{\frac{s+4}{s-4}}^\infty \left[\rho(z')\int_{z'+\sqrt{z'^2-1}}^\infty\frac{z^{k+1}\mathrm{d}\zeta}{\zeta^{k+1}\sqrt{1-2\zeta z'+\zeta^2}(\zeta -z)}\right.\\ &+\left.\rho'(-z')\int_{z'+\sqrt{z'^2-1}}^\infty\frac{z^{k+1}\mathrm{d}\zeta}{\zeta^{k+1}\sqrt{1+2\zeta z'+\zeta^2}(\zeta -z)}\right]~\mathrm{d}z' \end{align}

We can again interchange the order of integrations and define: $\frac{q}{\sqrt{s}}\int_{\frac{s+4}{s-4}}^{\frac{\zeta^2+1}{2\zeta}} \frac{\rho(z')}{\sqrt{1-2\zeta z'+\zeta^2}}~\mathrm{d}z'=\sigma(\zeta)$ and $\frac{q}{\sqrt{s}}\int_{\frac{s+4}{s-4}}^{\frac{\zeta^2+1}{2\zeta}} \frac{\rho'(-z')}{\sqrt{1+2\zeta z'+\zeta^2}}~\mathrm{d}z'=\sigma'(\zeta)\ .$ These two integrals are taken over finite intervals, and therefore do not pose any convergence problems : they will be bounded by polynomials, just as their integrands.

In fact we can use the first integral mean value theorem to put a limit on the integral over $$z'\ :$$

$|\sigma(\zeta)| \leq \max(|\rho(z')|) \frac{q}{\sqrt{s}}\int_{\frac{s+4}{s-4}}^{\frac{\zeta^2+1}{2\zeta}} \frac{1}{\sqrt{1-2\zeta z'+\zeta^2}}~\mathrm{d}z'$

where the max has to be taken over the interval of integration. The integration is readily made and gives : $|\sigma(\zeta)| \leq \max(|\rho(z')|)\frac{q}{\sqrt{s}} \sqrt{(1-2\frac{\zeta(s+4)}{s-4} +\zeta^2}$

$|\sigma(\zeta)| \leq \max_{\frac{s+4}{s-4}}^{\frac{\zeta^2+1}{2\zeta}}(|\rho(z')|)\sqrt{\frac{(1+\zeta^2)(s-4)-2\zeta(s+4)}{2s\zeta}}$

We assume that $$\rho$$ and $$\rho'$$ are bounded by polynomials in $$z$$ and $$s\ .$$ We already have supposed that the degree of the polynomial in $$z$$ is $$k\ .$$ Let us assume that the degree of the polynomial in $$s$$ is $$A\ :$$ $|\rho(z,s)|+|\rho'(z,s)| < B |z|^k s^A$ The values of $$\zeta$$ correspond to values of $$\zeta = z + \sqrt{z^2 - 1}\ ,$$ that is, $$\zeta$$ lies between $$z$$ and $$2z\ .$$ The preceding bound thus gives automatically: $|\sigma(\zeta,s)|+|\sigma'(\zeta,s)| < B |\zeta|^k s^A$

The lower limit for the integration on $$\zeta$$ is given by $\zeta^2 -2\frac{\zeta (s+4)}{s-4} + 1 = 0$ $\zeta= \frac{\sqrt{s}+2}{\sqrt{s}-2}\ .$ Thus $\sum_{\ell=k+1}^\infty z^\ell a_\ell= \int_\frac{\sqrt{s}+2}{\sqrt{s}-2}^\infty [\sigma(\zeta)+(-1)^{\ell}\sigma'(\zeta)]\frac{z^{k+1}}{\zeta^{k+1}(\zeta -z)}\mathrm{d}\zeta$ If we expand the denominator in powers of $$z\ ,$$ we can identify the coefficients of the series in $$z^\ell$$ as: $\sum_{\ell=k+1}^\infty a_\ell z^\ell=\int_\frac{\sqrt{s}+2}{\sqrt{s}-2}^\infty [\sigma(\zeta)+(-1)^{\ell}\sigma'(\zeta)]\sum_{\ell=k+1}^\infty \frac{z^{\ell}}{\zeta^{\ell+1}}\mathrm{d}\zeta~,$ whence for $$\ell>k$$ $a_\ell = \int_\frac{\sqrt{s}+2}{\sqrt{s}-2}^\infty [\sigma(\zeta)+(-1)^{\ell}\sigma'(\zeta)]\frac{1}{\zeta^{\ell+1}}\mathrm{d}\zeta$ From the polynomial bound on $$\sigma$$ and $$\sigma'$$ $|a_\ell| < \left(\frac{\sqrt{s}-2}{\sqrt{s}+2}\right)^{\ell-k} \frac{B s^A }{\ell-k}$ for $$\ell>k\ .$$

### Putting together all the results

This exponential behavior in $$\ell$$ is the equivalent of the tunnel effect decrease of the first section. We can now choose between the stronger of the bounds for each partial wave amplitude: either 1 from unitarity, or this exponential. These will cross at some value $$L$$ of $$\ell\ ,$$ for which: $\left(\frac{\sqrt{s}-2}{\sqrt{s}+2}\right)^{\ell-k} \frac{B s^A }{\ell-k} < 1$ It is not easy to solve this equation, so let us make a drastic approximation, knowing that $$\ell-k>1\ :$$ let us solve $\left(\frac{\sqrt{s}-2}{\sqrt{s}+2}\right)^L B s^A < 1$ or for large $$s$$ $(1-\frac{4}{\sqrt{s}})^L B s^A < 1 \ .$ Taking the log of both sides: $\frac{4L}{\sqrt{s}} > \ln (B s^A)$ $L > \sqrt{s}\frac{A \ln s +\ln B}{4}$ This shows that we were justified in neglecting the constant terms with $$\ell \leq k\ ,$$ which get at some moment under this value.

The amplitude in the forward direction can be bounded by\begin{align} |A(1)|&< \frac{\pi q}{2 \sqrt s}\sum_0^L (2l+1) \,1\, |P_l(1)|\\&+\frac{\pi q}{2 \sqrt s}\sum_{L+1}^\infty (2l+1) |a_l(s)||P_l(1)|\\ &\sim \frac{\pi L^2}{2} + \pi \sum_{l'=0}^\infty (L +l') \left(1-\frac{4}{\sqrt{s}}\right)^{l'}\\ &\sim \frac{\pi L^2}{2} + \frac{\pi L \sqrt{s}}{4} + \frac{\pi s}{16}\\ &= \mathcal{O}\, (s \ln^2 s) \end{align}

We can now use the so-called optical theorem, which gives the total cross-section for all processes initiated by the scattering of two particles, in terms of the imaginary part of the forward elastic amplitude, which of course enters into the framework developed previously.

This theorem asserts that: $\sigma_\mathrm{tot} = \frac{4\pi}{k} \mathrm{Im} ~A(1)$ Here $$k$$ (different of the previous one, degree of the subtractions in $$z$$) is the 3-momentum of one of the particles in the rest system of the other one.

To relate it with $$s\ ,$$ note that the energy of the incident particle is $$\sqrt{k^2+1}\ .$$ The value of $$s$$ is the square of the 4-momentum of the system, that is, \begin{align}s&=(\sqrt{k^2+1}+1)^2 - k^2\\ &= k^2 + 1 + 2\sqrt{k^2+1}+1-k^2= \sim 2k\end{align}\ .

The total cross-section is thus bounded by some constant multiplied by $$\ln^2 s\ ,$$ which is exactly the result we got by the intuitive reasoning of the preceding section.