## Meson Theory on Strong Nuclear Force

The theory is very difficult, I only state the result.

There are 4 main mesons for strong nuclear force.

type mass [MeV] range symmetry force type
$\pi$  135  long  pseudo scalar  $-\sigma_{1}\cdot\sigma_{2},-S_{12}$
$\sigma$  400 ~ 2000  medium  scalar $-1, -L\cdot S$
$\rho$  775  short  vector  $-2\sigma_1\cdot\sigma_2, +S_{12}$
$\omega$  783  short vector  $+1, -3L\cdot S$

## Mean free path of a nucleon inside a nucleus

The mean free path is the average distance between 2 collisions. We simply copy the things in Bohr and Mottelson, Bertulani and Banielewicz, and John Lilley in here.

Since a nucleus has finite size, the mean free path can be view as transparency of nucleon scattering. Since the total cross section is smaller for higher energy, the mean free path is proportional to energy.

The wave number $K$ under a complex optical potential $V+iW$ is

$K=\sqrt{\frac{2m}{\hbar^2}(E-V-iW)}=k_r+ik_i$.

There are two solutions, one has imaginary $k_r$,

$k_r=\frac{\sqrt{m}}{\hbar}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}$

$k_i=\frac{m}{\hbar^2}\frac{W}{k_r}=\frac{1}{2\lambda}$

where $m$ is the nucleon mass, $E$ is the incident energy, $v$ is the velocity inside the nucleus, and $\lambda$ is the mean free path,

$\lambda=\frac{\hbar}{2W\sqrt{m}}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}\approx\frac{\hbar}{W\sqrt{2m}}\sqrt{(E-V)}$

Since the imaginary potential $W\approx-15-0.07E$, $V\approx-39+0.11E$ from 150 MeV to 400 MeV. The calculated mean free path for proton is shown in here. (Take $\hbar = 197.33 [MeV\cdot fm/c]$, proton mass $m=938.272 [MeV/c]$, the $\lambda$ is in fm.)

The blue line on the plot is exact calculation, the purple line is approximation $W << E-V$.

This plot is taken from S.S.M. Wong, showing the radial shapes of the volume term (simialr to central term) of proton-nucleus optical potential.

————–

in J. Killey, the $k_r$ is taken as,

$k_r = \frac{\sqrt{2m(E-V)}}{\hbar}$,

which is the wave number without $W$. The result only for weak $W << E-V$.

## Optical Model

In nuclear physics, the Optical Model means, we are treating the scattering problem is like optical wave problem. due to the incident beam can be treated as a wave-function. and this wave will be scattered by the target.

when the beam is far away from the target, the wave function of the incident beam should satisfy the Schrödinger equation in free space :

$\left( \frac {\hbar^2 } {2m} \nabla^2 + V(r) \right) \psi( \vec{r} ) = E \psi ( \vec{r} )$

and the plane wave solution is

$\psi ( \vec{r} ) \sim Exp ( \pm i \vec{k} \cdot \vec {r} )$

after the scattering, there will be some spherical wave come out. the spherical wave should also satisfy the free-space Schrödinger equation.

$\psi( \vec{r} ) \sim Y(\theta, \phi) \frac {Exp( \pm i \vec{k} \cdot \vec{r} ) }{r}$

Thus, the process of scattering can be think in this way:

$Exp( \pm i k z ) \rightarrow Exp( i k z ) + f ( \theta ) \frac { Exp ( i k r ) } {r}$

where f(θ) is a combination of spherical wave.

one consequence of using Optical Model is, we use complex potential to describe the nuclear potential terms in quantum mechanics.

when using a complex potential, the flux of the incident beam wave function can be non-zero. meanings that the particles in the beam are being absorbed or emitted. This corresponding to the inelastic scattering.

The reason for the “OPTICAL” is come form the permittivity and permeability of the EM field. for metallic matter, their permittivity or permeability may have a imaginary part. and this imaginary part corresponding to the absorption of the light. so, nuclear physics borrow the same idea.

the flux is defined as:

$J = \frac { \hbar }{ 2 i m} ( \psi^*(r) \nabla \psi(r) - \psi(r) \nabla \psi^* (r) )$

and the gradient of the flux, which is the absorption (sink) or emission ( source ) is:

$\nabla J = \frac {\hbar }{ 2 i m }( \psi^* \nabla^2 \psi - \psi \nabla^2 \psi^* )$

The Schrödinger equation gives the equation for the wave function:

$\nabla^2 \psi(r) = \frac { 2m} {\hbar^2} ( E - V(r)) \psi(r)$

when sub the Schrödinger equation in to the gradient of flux, we have:

$\nabla J = \frac {1} {i \hbar } ( V(r) - V^*(r) ) | \psi |^2 = \frac { 2} {\hbar } Im ( V) | \psi |^2$

we can see, if the source and the sink depend on the complex part of the potential. if the imaginary part is zero, the gradient of the flux is zero, and the wave function of the beam is conserved.

## Differential Cross Section II

Last time, the differential cross section discussion is based on quantum mechanics. This time, i try to explain it will out any math. so, that my mum ask me, i can tell her and make her understand. :)

in a scattering experiment, think about a target, say, a proton fixed in the center, it is positive charged. if another proton coming with some energy. it will get repelled, due to the repulsive nature of Coulomb force of same charge. it should be easy to understand, if the proton coming with high energy, it will get closer to the target, or even enter inside the target.

the repelling angle of the proton is not just depend on the energy it carry, but also on the impact parameter ( we usually call it b , but i like to call it r). the impact parameter is the shortest distance between the target and the line of the moving direction of the proton at long long away.

if the impact parameter is large, the proton miss the target. it almost cannot feel the target affection. thus, it go straight and unaffected. when the r is zero, it will hit the target head on head. and due to the repulsion. it will return back.  so, we can understand. the smaller the impact parameter, the deflection will be larger. since it can feel the force stronger.

For same impact parameter, the higher energy proton will have less deflection, since it travel faster, spend less time by the force, and the deflection get less.

Thus, we have an idea that the angle of deflection is high for small impact parameter and high energy. And most important, it only depends on these 2 factors and the effect from the target.

since our detector can only detect some small angle over some small area. So, we can place out detector on some angle, get the yield, and this is the name –  differential cross section come from.

Now, we have a uniform flow of particle with energy E. they will be deflected by the target and go to some angles. If we detect at the deflection angle, see how many particles ( the yield ) we can get in each angle. we can calculate back the effect of the target. For example, for a small angle, the particle get little defected, and this means the particle is from large impact parameter. for a large angle, the particles are from small impact parameter.

In some cases, the number of particle detected will be very high at some particular angle then others angles and this means, the cross section is large. and this means something interesting.

Moreover, don’t forget we can change the energy of the beam. for some suitable energy, the particle will being absorbed or resonance with the target. that given us low or high cross section on the energy spectrum.

( the graph is an unauthorized from the link: http://www.astm.org/Standards/E496.htm )

The above diagram is the differential cross section obtained from a Deuteron to a Tritium ( an isotope of Hydrogen with 2 neutrons and 1 proton) target, and the reaction change the Tritium into Helium and a neutron get out.

The reaction notation is

$X(a,b)Y$

where a is incident particle, X is target, b is out come particle, and Y is the residual particle.

the horizontal axis is detector angle at lab-frame. and the vertical axis is energy of Tritium. we always neglect the angle 0 degree, because it means no deflection and the particle does not “see” the target. at low energy, the d.c.s. is just cause by Coulomb force. but when the energy gets higher and higher, there is a peak around 60 degree. this peak is interesting, because it penetrated into the Helium and reveal the internal structure of it. it tells us, beside of the Coulomb force, there are another force inside. that force make the particle deflects to angle 60 degree. for more detail analysis, we need mathematic. i wish someday, i can explain those mathematics in a very simple way.

Therefore, we can think that, for higher energy beam, the size we can “see” will be smaller. if we think a particle accelerator is a microscope. higher energy will have larger magnification power. That’s why we keep building large and larger machines.

## On mass deficit

The mass deficit is due to the potential energy loss.

But, why there are potential?

Coz there is a force associate with a potential.

Or in more correct way, the potential energy is due to the force. When 2 bodies in space has a attractive force, then it naturally go together. And the force accelerate the body and increase their kinetic energy.

Thus, we think the force created a field, which stored potential energy. When the force accelerates body, the potential energy converted into kinetic energy.

Imagine there are 2 bodies, A & B, Which at different location and have different force. If A is further away, when it comes at B, it has more kinetic energy than B, coz B is at rest at that position. Thus, further away, higher the potential.

Can thus potential be as high as infinite? In reality, because thing has surface, has size. The minimum distance between the 2 bodies is the sum of their radius. And if we sum up all potential, from the surface to infinity. We found that it is finite. And we like to set the potential at infinity is zero. Thus, it makes all attractive force has negative potential.

But, electric and positron do not has size, so, and electron and a positron can come together and the potential energy they can release is infinite! Since when they are apart infinite and the attractive force accelerate it, when they meet, their speed will be infinite!

Wait! The speed cannot be infinite, the highest speed is the speed of light and for an object has mass, it never move at the speed of light!

So, when electron and positron hit each other from infinity. However, relativity does not limit the energy and the kinetic energy. The result is, it can release infinite energy.

But at first, infinite exist. In reality, we can just approximate the “infinite” by far far away. When the potential change very little. But far far away is still very short compare to infinite. Thus, we don’t have infinite energy source.

In laboratory, we can accelerate electron and position at very high speed and contain several GeV. Thus when those particles annihilate, they release GeV energy due to the kinetic energy and their mass, which is just 0.000511GeV. Does not help much.

## Mass of particles and nucleus

in Nuclear physics, the particle we deal with are so small and so light, if we use standard unit, then there will be many zero and we will lost in the zeros. for example, the electron has mass:

Mass( electron ) = 9.11 × 10-31 kilograms
Mass( proton ) = 1.67 × 10-27 kilograms

see? as the special relativity give us a translation tool – E = m c^2, thus, we can use MeV to talk about mass.

Mass ( electron ) = 0.511 MeV
Mass ( proton ) = 938.3 MeV

thus, we can see, Proton is roughtly 2000 times heavier then electron ( 1000 : 0.5 ).

Mass( neutron ) = 939.6 MeV

neutron is just 1.3 MeV heavier then proton.

The nucleus is formed by proton and neutron. so, in simple thought, an nucleus with Z proton and ( A-Z ) neutron should have mass

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) = Mass ( A, Z )

where A is the atomic mass number, which is equal the number of nucleons in the nucleus, and Z is the proton number.

However, scientists found that it is not true.

Z x Mass( proton ) + ( A – Z ) x Mass ( neutron ) > Mass ( A, Z )

Some of the mass is missing! But that is explained why nucleus will not break down automatically. since it need extra energy to break it down.

we called the mass different is Mass Deficit. or Binding energy.

Mass Deficit = Mass( A, Z) – Mass ( proton + neutron )

some one may think that the binding energy is the energy for holding the nucleus together. in order to hold the nucleus, some mass was converted into the energy to holding it. this is INCORRECT. the correct argument is, the binding energy is th energy require to break it down.

think about a simple 2 bodies system, like sun and earth. at far far away, when both of them are at rest, the total energy is Mass( sun) + Mass ( earth ) + Potential energy.

when the earth moves toward to sun, the potential energy converted to the Kinetic energy, so the earth moving faster and faster. but, in order to stay in the orbit, some K.E. must be lost so that it does not have enough ( or the same) energy to run away. Thus, the total energy of the system is lesser then the total mass.

another analogy is electron orbit. when an electron was captured by an atom, it radiate energy in order to stay in some energy level. thus, the total energy of the system again less then the total mass.

any any case, the mass of the sun and earth and electron does not change, but the potential changes to negative, thus it makes to total energy lesser.

similar idea hold for nucleus, but the potential of it are great different, because there are a Coulomb Barrier. Thus, in order to make a nucleus. we have to put so many K.E. to again this barrier, then the resultant nucleus release the Mass Deficit energy and also the input K.E..

 a scratch on the nuclear potential. there are a Coulomb Barrie. ( by wolframalpha.com)
When the nucleus is radioactive and undergoes decay. this mean, it Mass deficit is positive. thus, it will automatically break down to another nucleus until it mass deficit is negative again. during this process, the emitted particle carry K.E. which is from the potential. Not the mass for one nucleons.
Remember, Mass( nucleus ) = Mass ( protons + neutrons ) + Potential

## Scattering phase shift

for a central potential, the angular momentum is a conserved quantity. Thus, we can expand the wave function by the angular momentum wave function:

$\sum a_l Y_{l , m=0} R_l(k, r)$

the m=0 is because the spherical symmetry. the R is the radial part of the wave function. and a is a constant. k is the linear momentum and r is the radial distance.

$R_l(k,r) \rightarrow J_{Bessel} (l, kr )$

which is reasonable when r is infinite and the nuclear potential is very short distance. when r goes to infinity,

$J_{Bessel} (l,kr) \rightarrow \frac {1}{kr} sin( k r - \frac{1}{2} l \pi )$

for elastic scattering, the probability of the current density is conserved in each angular wave function, thus,

the effect of the nuclear potential can only change the phase inside the sin function:

$\frac{1}{kr} sin( k r - \frac {1}{2} l \pi +\delta_l )$

with further treatment, the total cross section is proportional to $sin^2(\delta_l)$.

thus, by knowing the scattering phase shift, we can know the properties of the nuclear potential.

for more detail : check this website