Oborn: /* Deriving the Coulomb Force */

2009-05-11T16:27:38Z

Deriving the Coulomb Force

Oborn: New page: =The Nuclear Force= ==Deriving the Coulomb Force== TF_DerivationOfCoulombForce ;Poisson's Equation : $\del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})$ ...

2009-04-08T03:29:49Z

New page: =The Nuclear Force= ==Deriving the Coulomb Force== TF_DerivationOfCoulombForce ;Poisson's Equation : <math>\del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})</math> ...

New page

=The Nuclear Force=
==Deriving the Coulomb Force==

[[TF_DerivationOfCoulombForce]]
;Poisson's Equation
: <math>\del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})</math>

== The Deuteron==
== Nucleon- Nucleon scattering==
=== Cross section===
;Total cross section
:<math>\sigma</math> = <math>\equiv \frac{\# particles\; scattered} {\frac{ \# incident \; particles}{Area}} = \frac{j_{scattered} A}{j_{incident}}</math>

:<math>j =</math> current density = # scattered particles per Area.

Particles are scattered in all directions. Typically you measure the number of scattered particle with a detector of fixed surface area that is located a fixed distance away from the scattering point thereby subtending a solid angle as shown below.

; Solid Angle
:[[Image:SolidAngleDefinition.jpg|300 px]]
: <math>\Omega</math>= surface area of a sphere covered by the detector
: ie;the detectors area projected onto the surface of a sphere
:A= surface area of detector
:r=distance from interaction point to detector
:<math>\Omega = \frac{A}{r^2} </math>sterradians
: <math>A_{sphere} = 4 \pi r^2</math> if your detector was a hollow ball
:<math>\Omega_{max} = \frac{4 \pi r^2}{r^2} = 4\pi</math>sterradians

;Differential cross section
:<math>\sigma(\theta)</math> = <math> \frac{d \sigma}{d \Omega} \equiv \frac{\frac{\# particles\; scattered}{solid \; angle}} {\frac{ \# incident \; particles}{Area}}</math>

;Units
:Cross-sections have the units of Area
:1 barn = <math>10^{-28} m^2</math>
; [units of <math>\sigma(\theta)</math>] =<math>\frac{\frac{[particles]}{[sterradian]}} {\frac{ [ particles]}{[m^2]}} = m^2</math>

[[Image:FixedTargetScatteringCrossSection.jpg | 300 px]]
; Fixed target scattering
: <math>N_{in}</math>= # of particles in = <math>I \cdot A_{in}</math>
:: <math>A_{in}</math> is the area of the ring of incident particles
:<math>dN_{in} = I \cdot dA = I (2\pi b) db</math>= # particles in a ring of radius <math>b</math> and thickness <math>db</math>

==Scattering Length (a) ==

; Definition
: <math>a^2 = \frac{1}{4 \pi} \lim_{k \rightarrow 0} \sigma</math>

While scattering length has the dimension of length it really represents the strength of the scattering (probability of scattering). It effectively give the amplitude of the scattered wave.

;Note
: the above definition is essentially an expression of how the low energy cross section corresponds to the classical value of <math>4 \pi a^2</math>
:<math>\sigma</math> = scattering cross-section <math>\equiv \frac{\# particles\; scattered} {\frac{ \# incident \; particles}{Area}}</math>
:classically: the number of particles scattered = number of incident particles (the collision probability is unity)
: Area = <math> \pi a^2</math> = The area profile in which a collision occurs[[Image:ClassicalEffectiveScatteringArea.jpg | 200 px]]

<math>\sigma = \frac{{N}}{\frac{ N}{\pi a^2}} = \pi a^2</math>

To derive an expression for the scattering length lets start with a general expression for a scattered wave.

;A general scattered wave function has the form
:<math>\Psi_g = \frac{1}{(2 \pi)^{3/2}} \left [ e^{i \vec{k} \cdot \vec{r}} + f(q) \frac{e^{ikr}}{r}\right ]</math>

The first term represents a plane wave and the second term represent a modification of the plane wave due to the scattering in terms of the scattering probability <math>|f(q)|^2</math>.

From our previous phase shift calculation, Schrodinger solutions tend to have the general form

:<math>\Psi_S = \sum_{\ell} A_{\ell} P_{\ell} [\cos(\theta)] \frac{U_{\ell}(r)}{kr}</math>
: <math>= \sum_{\ell} A_{\ell} P_{\ell} [\cos(\theta)] \frac{\sin(kr - \frac{\ell \pi}{2} + \delta_{\ell})}{kr}</math> (our previous solution was for <math>\ell=0</math>)

where

the angular part is given as
:<math> P_{\ell} \cos(\theta)</math>
and the radial part is
:<math>\frac{U_{\ell}(r)}{r}</math>

By comparing our general solution from the phase shift and our schrodinger solution we can cast f(q) in terms of the phase shift <math>\delta_{\ell}</math> and then define the cross section as

:<math>\sigma = \int |f(q)|^2 d \Omega</math>

in order to get a gereral expression for a scattering cross section which we then take the limit of the momentum going to zero in order to get a general expression for the scattering length.

: Math trick to recast plane waves
;<math>e^{i\vec{k} \cdot \vec{r}} = 4 \pi \sum_{\ell m} i^{\ell} Y^*_{\ell m}(\hat{k}) Y_{\ell m}(\hat{r}) J_{\ell m}(|k| |r|)</math>

where

:<math>\hat{k}</math> and <math>\hat{r}</math> are the <math>\theta</math> and <math>\phi</math> directions of <math>\vec{k}</math> and <math>\vec{r}</math> respectively.

To determine the scattering length we will be looking at <math>k \rightarrow 0</math> so let <math>\hat{k}=0 \Rightarrow \theta=0 , \phi=0</math> use the approximation

:<math>Y^*_{\ell m}(\theta=0,\phi=0) = \sqrt{\frac{2 \ell +1}{4 \pi}}\delta_{m,0}</math>

;using the above to recast <math>\Psi_g</math> to look more like <math>\Psi_S</math>
:<math>\Psi_g = \frac{1}{(2 \pi)^{3/2}} \left \{ 4 \pi \sum_{\ell} i^{\ell} \sqrt{\frac{2 \ell +1}{4 \pi}}\delta_{m,0} Y_{\ell m}(\hat{r}) J_{\ell m}(|k| |r|) + f(q) \frac{e^{ikr}}{r} \right \}</math>

:<math>\delta_{m,0} Y_{\ell m}(\hat{r}) =Y_{\ell 0}(\hat{r}) = \sqrt{\frac{2 \ell +1}{4 \pi}} P_{\ell} [\cos(\theta)]</math>

:<math>\Rightarrow \Psi_g = \frac{1}{(2 \pi)^{3/2}} \left \{ \sum_{\ell} i^{\ell} (2 \ell +1) P_{\ell} [\cos(\theta)] J_{\ell m}(|k| |r|) + f(q) \frac{e^{ikr}}{r} \right \}</math>

;which we want to compare to \Psi_S
:<math>\Psi_S = \sum_{\ell} A_{\ell} P_{\ell} [\cos(\theta)] \frac{\sin(kr - \frac{\ell \pi}{2} + \delta_{\ell})}{r}</math>

Using the identities:

:<math>\sin(kr - \frac{\ell \pi}{2} + \delta_{\ell}) = \frac{e^{i(kr - \frac{\ell \pi}{2} + \delta_{\ell})} - e^{-i(kr - \frac{\ell \pi}{2} + \delta_{\ell})}}{2i}</math>
:<math>J_{\ell m}(|k| |r|) = \frac{e^{i(kr - \frac{\ell \pi}{2} )} - e^{-i(kr - \frac{\ell \pi}{2} )}}{2ir}</math>

;By equating the two solutions

:<math>\Psi_S = \Psi_g</math>
:<math>e^{-ikr} \Rightarrow A_{\ell} e^{i\delta_{\ell}} = \frac{i^{\ell} (2 \ell +1)}{(2 \pi)^{3/2}}</math>

<math>e^{+ikr} \Rightarrow </math>
:<math>\sum_{\ell} \frac{A_{\ell}}{kr} P_{\ell} [\cos(\theta)]e^{i (kr - \frac{\ell \pi}{2} + \delta_{\ell})} = \sum_{\ell} \frac{i^{\ell} (2 \ell +1)}{(2 \pi)^{3/2} 2ikr} P_{\ell} [\cos(\theta)]e^{i(kr - \frac{\ell \pi}{2} )} + \frac{f(q) e^{ikr}}{(2 \pi)^{3/2}r}</math>

:<math>\Rightarrow \frac{f(q) e^{ikr}}{(2 \pi)^{3/2}r}= \sum_{\ell} \frac{ P_{\ell} [\cos(\theta)]}{2ikr} e^{- i \ell \pi/2} \left ( \frac{A_{\ell}}{kr} e^{i (kr - \frac{\ell \pi}{2} + \delta_{\ell})} - \frac{i^{\ell} (2 \ell +1)}{(2 \pi)^{3/2}} e^{i(kr - \frac{\ell \pi}{2} )} \right ) </math>
: <math>= \sum_{\ell} \frac{A_{\ell}}{kr} e^{i (kr - \frac{\ell \pi}{2} + \delta_{\ell})} - \frac{i^{\ell} (2 \ell +1)}{(2 \pi)^{3/2} 2ikr} P_{\ell} [\cos(\theta)]e^{i(kr - \frac{\ell \pi}{2} )}</math>

: <math>\Rightarrow f(q) = \frac{1}{k} \sum_{\ell} (2 \ell + 1 ) P_{\ell} [\cos(\theta)] \sin(\delta_{\ell}) e^{i \delta_{\ell}}</math>

:<math>\sigma = \int | f(q) |^2 d \Omega</math>
: <math>= \sum_{\ell \ell^{\prime}} \int \frac{2 \ell + 1 )(2 \ell^{\prime} + 1)}{k^2}P_{\ell} [\cos(\theta)] \sin(\delta_{\ell}) e^{i \delta_{\ell}} P_{\ell^{\prime}} [\cos(\theta)] \sin(\delta_{\ell^{\prime}}) e^{-i \delta_{\ell^{\prime}}}</math>

:<math> \int P_{\ell} P_{\ell^{\prime}} d \theta = \frac{2}{2 \ell + 1} \delta_{\ell \ell^{\prime}}</math>

:<math>\Rightarrow \sigma = \frac{4 \pi}{k^2} \sum_{\ell} (2 \ell +1 ) \sin^2(\delta_{\ell})</math>

:<math>a^2 = \frac{1}{4 \pi} \lim_{k \rightarrow 0} \sigma = \frac{1}{4 \pi} \lim_{k \rightarrow 0} \frac{4 \pi}{k^2} \sum_{\ell} (2 \ell +1 ) \sin^2(\delta_{\ell})</math>

For S-wave scattering <math>\ell =0</math>

:<math>\Rightarrow a = \pm \lim_{k \rightarrow 0} \frac{\sin(\delta_0)}{k}</math>

To keep "a" finite the phase shift <math>\delta_{\ell}</math> must approach zero at low energy

:<math>\Rightarrow a = \pm \lim_{k \rightarrow 0} \frac{\delta_0}{k}</math>

===Singlet and Triplet States===

Our previous calculation of the total cross section for nucleon nucleon scattering using a phase shift analysis gave

:<math>\sigma = \frac{4 \pi}{k_2^2}\sin^2(\delta_0)</math>

assuming <math>\ell</math> =0.

When solving Schrodinger's equation for a neutron scattering from a proton we were left with the transcendental equation from boundary conditions

: <math>\alpha \equiv k_1 cot(k_1R) = k_2 cot(k_2R + \delta)</math>

: <math>\sigma = \frac{4 \pi}{ k_2^2 + \alpha^2} \left [\cos(k_2 R) + \frac{\alpha}{k_2} \sin(k_2R)\right ]</math>

In the case of a Deuteron bound state

: <math>\ell =0</math> and <math>S=1</math>
:R = 2 fm
: <math>\alpha</math> = 0.2 /fm

when <math>k_2 \rightarrow 0</math>

: <math>\frac{\sin(k_2 R)}{k_2} = R</math>
: <math>\cos(k_2 R) = 1</math>

then

: <math>\lim_{k_2 \rightarrow 0} \sigma = \frac{4 \pi}{ 0 + (0.2)^2} \left [1 + (0.2) 2\right ] \left ( \frac{1 barn}{100 fm^2}\right)= 4.4 barns</math>

Experimentally the cross section is quite a bit larger

:<math>\sigma_{Exp}</math> = 20.3 b

Apparently our assumption that the dominant part of the cross section is S=1 is wrong. There also exists a spin singlet (S=0) contribution to the cross section.

When the neutron and proton interact (create a bound state or an intermediate state) their spins can couple to either a net value of S=0 or S=1. There is only one component along the quantization axis in the event that they couple to an S=0 state. There are 3 possible components (<math>S_z</math> = -1/2, 0 . + 1/2) in the event that they couple to an S=1 state.

If you sum up the two possible cross-section,<math> \sigma_S</math> and <math>\sigma_T</math>, then you must weight them according to the possible psin compinations such that

:<math>\sigma_{tot} = \frac{3}{4} \sigma_T + \frac{1}{4} \sigma_S</math>

; Solving for <math>\sigma_S \Rightarrow</math>

:<math>\sigma_S = 68</math> barns

Because the cross sections depend on the spin date we can conclude that

; The Nuclear Force is SPIN DEPENDENT

;Also
: Using the spatial wave functions for the singlet and triplet state one can deduce that

:<math>a_T</math> = + 6.1 fm <math>\Rightarrow</math> there is a triplet np bound state
:<math>a_S</math> = - 23.2 fm <math>\Rightarrow</math> there is NO singlet np bound state

Doing similar but more complicated calculations for p-p and n-n scattering results in

:<math>a_{p-p}</math> = -7.82 fm <math>\Rightarrow</math> there is NO pp bound state
:<math>a_{n-n}</math> = -16.6 fm <math>\Rightarrow</math> there is NO nn bound state

==The Nuclear Potential==

From the above we have found information on the range of the nuclear force, it's spin dependence, and it's ability to create non-spherically disrtibuted systems (quadrupole moments).

=== The Central Potential ===

No matter what potential Well geometry we choose for the nucleon, we consistently find a term which is purely radial in nature (a Central term).

:<math><V_c(r)> = \int U^*(r) V_c(r) U(r) dr = \frac{\hbar (\ell + 1) \ell}{2m} \int |U(r)|^2G(\delta) \frac{dr}{r^2} </math>

where

:<math>G(\delta)</math> = a parameterization of<math> V_c</math> which is constrained by scattering phase shift information.

=== The Spin Potential ===

We know from the lack of a p-p <math>(^2NHe)</math> or n-n bound system that the nuclear force is strongly spin dependent. This is reenforced even more based on our observations of the S=1 n-p bound state (the Deuteron).

Experiments also indicate that parity is conserved to the <math>10^{-7}</math> level. <math>\Rightarrow</math> experiments with an relative precision of <math>\frac{\Delta X}{X} > 10^{-7}</math> have yet to find a parity violation.

:<math>\Rightarrow</math> The spin potential would have terms involving spin scalar quantities because a spin potential with terms that are linear combinations of spin would violate parity.

Consider a spin potential function such that the total spin is given by
:<math>\vec{S} = \vec{s}_1 + \vec{2}_2</math>

The scalar spin quantity <math>S^2</math> would be given by

:<math>\vec{S} \cdot \vec{S} = s_1^2 + 2\vec{s}_1 \cdot \vec{s}_2 + s_2^2</math>

or

:<math>\vec{s}_1 \cdot \vec{s}_2 = S^2 -s_1^2-s_2^2</math>

====Spin Singlet ====
if
: S=0

then
:<math><\vec{s}_1 \cdot \vec{s}_2 >= <S^2> -<s_1^2>-<s_2^2></math>
: <math>= \left [ S(S+1) - s_1(s_1+1) - s_2(s_2+1) \right ] \frac{\hbar^2}{2}</math>
: <math>= \left [ 0(0+1) - \frac{1}{2} (\frac{1}{2}+1)- \frac{1}{2} (\frac{1}{2}+1)\right ] \frac{\hbar^2}{2}</math>
: <math>= - \frac{3}{4} \hbar^2</math>

====Spin Triplet ====
if
: S=1

then
:<math><\vec{s}_1 \cdot \vec{s}_2 >= <S^2> -<s_1^2>-<s_2^2></math>
: <math>= \left [ S(S+1) - s_1(s_1+1) - s_2(s_2+1) \right ] \frac{\hbar^2}{2}</math>
: <math>= \left [ 1(1+1) - \frac{1}{2} (\frac{1}{2}+1)- \frac{1}{2} (\frac{1}{2}+1)\right ] \frac{\hbar^2}{2}</math>
: <math>= + \frac{1}{4} \hbar^2</math>

====Construct the Spin Potential ====

Let
:V_1(r) = spin singlet parameterized potential
:V_3(r) = spin triplet parameterized potential

Then

:V_s(r) = - (

==Yukawa Potential ==

[[Forest_NucPhys_I]]

@@ Line 4: / Line 4: @@
 [[TF_DerivationOfCoulombForce]]
 ;Poisson's Equation
-: <math>\del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})</math>
+: <math>\nabla^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})</math>
 == The Deuteron==

NuclearForce Forest NucPhys I - Revision history

Oborn: /* Deriving the Coulomb Force */

Oborn: New page: =The Nuclear Force= ==Deriving the Coulomb Force== TF_DerivationOfCoulombForce ;Poisson's Equation : \del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta}) ...

Oborn: New page: =The Nuclear Force= ==Deriving the Coulomb Force== TF_DerivationOfCoulombForce ;Poisson's Equation : $\del^2 \phi(\vec{\eta}) = - \frac{e}{\epsilon_0} \delta(\vec{\eta})$ ...