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WindowTitle->Lecture 16 MIT 3.016 (Fall 2009)   \251 W. Craig Carter  \
2003--2009
WindowFrame->Normal*)

(* Beginning of Notebook Content *)
Notebook[{
Cell[TextData[{
 "Generalizations of the Fundamental Theorem of Calculus (Part II)\n",
 StyleBox["(note: some of the symbolic integrations in this notebook will \
take a long time to evaluate)",
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Cell[TextData[{
 "Using Divergence theorem to find Hamaker Field (here modeled as \
point-wise",
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 ") for a Finite Body"
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Cell[TextData[{
 "The induced dipole\[LongDash]dipole interaction goes as -1/",
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 ", and it is this term which shows up in the Lennard-Jones potential.  This \
dipole-dipole interaction gives rise to the Hamaker forces between two \
bodies.  As a first approximation to the interaction between two bodies, one \
must intergrate 1/",
 Cell[BoxData[
  FormBox[
   SuperscriptBox["r", "6"], TraditionalForm]]],
 " from a single point in the first body to all points in the second body.  \
The net interaction energy is the sum over all such points and therefore this \
energy produces a six-dimensional integration. There are a few cases, such a \
two spheres, a sphere and a seminfinite half-space, that can be worked out \
exactly.  However, to model the interaction energy of more complex shapes, \
there are no closed-form solutions and one must resort to numerical \
integration.\n\nNumerical integration is a cpu-time-consuming numerical \
procedure\[LongDash]a six dimensional integration can be very expensive (time \
is money).  If there is a way to reduce the dimensionality of the \
integration, then we can reap rewards for our cleverness.  One trick is to \
use the divergence theorem to push the integration over a volume, to an \
integration over a surface. For example, we could use the divergence theorem:\
\n\[Integral]\[Integral]\[Integral]",
 Cell[BoxData[
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 " \[Del]\[CenterDot] ",
 Cell[BoxData[
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 " dV = \[Integral]\[Integral]",
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 "\nWe will work out such an example for a cylindrical volume interacting \
with a point in space, in other words, the London Interaction Potential for a \
finite cylinder. We will perform as many exact integrals as possible and \
reduce the cost of the numerical integration.  To do this, we must find a \
vector potential ",
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 " such that \[Del]\[CenterDot] ",
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 " = -1/|",
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 " where ",
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 " is a position in the integrated volume and  ",
 Cell[BoxData[
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 " is a point at which the potential is measured.  (A fairly general method \
to do this and similar problems can be found in ",
 StyleBox["Argento C; Jagota A; Carter WC",
  FontSize->12],
 StyleBox[" ``",
  FontSize->12,
  FontWeight->"Bold"],
 StyleBox["Surface formulation for molecular interactions of macroscopic \
bodies'' J.Mech. Physics Solids 1997,  pp 1161-1183 .",
  FontSize->12]
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Cell["\<\
The following is a ``guess'' at the vector potential; it will be verified as \
the correct one by checking its divergence.  We use (CX,CY,CZ) to represent \
points in the cylinder and (X,Y,Z) to be points in space\
\>", "Text",
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Cell["\<\
The following verifies that the correct vector potential is obtained, the \
divergence must be taked with respect to the coodinates of the integration:\
\>", "Text",
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Cell["\<\
We will integrate over a cylinder of radius R and length L along the z-axis, \
with its middle at the origin.  First, let's use the radius of the cylinder \
to scale all the length variables: Let (X,Y,Z)/R = (x,y,z); (CX,CY,CZ)/R = \
(cx,cy,cz), and L/R = \[Lambda] (the cylinder's aspect ratio).\
\>", "Text",
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Cell[TextData[{
 "Therefore, \[Phi](",
 Cell[BoxData[
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 ") =  \[Integral]\[Integral]\[Integral]",
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 " ",
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 "  = (\[Integral]\[Integral]",
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 " ",
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 ")/",
 Cell[BoxData[
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 "\n is the total interaction between a point an a cylinder.  We can exploit \
the symmetry of the cylinder: we know that the potential will be indendent of \
the angular coordinate in cylindrical coordinates; it will be a function of \
\[Rho] = ",
 Cell[BoxData[
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Cell[TextData[{
 "We will do three integrals over the cylindrical surfaces using this \
expression to define the cylinder: (cx,cy,cz) = (Cos[\[Theta]], \
Sin[\[Theta]], cz):\nThe cylindrical surface  is the domain  \[Theta] \
\[Element] (0, 2\[Pi]), cz \[Element] (-",
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 ")\nThe two caps r \[Element] (0,1),  \[Theta] \[Element] (0, 2\[Pi]), cz=\
\[PlusMinus]",
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Cell[CellGroupData[{

Cell["Cylinder Surface Normals and Integrands", "Subtitle",
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Cell["\<\
The following is a parametric representation of a cylinder surface that is \
coaxial with the z-axis (the cylinder ends will be included later)\
\>", "Text"],

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Cell[TextData[{
 "The infinitessimal surface vectors ",
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used to find the surface patch d",
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Cell[TextData[{
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 " \[Times] ",
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 " for the cylinder surface, there for the following (multiplied by d\[Theta] \
dz) is the infinitessimal oriented surface patch d",
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 "."
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Cell[BoxData[
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Cell["\<\
The integrand to be evaluated over the cylinder surface is the vector \
potential, dotted into the normal vector.  Because of the cylindrical \
symmetry of this model, we can convert to cylindrical coordinates.  One set \
of coordinates is for the cylinder surface (\[Xi] \[Rule] R Cos[\[Theta]], \
\[Eta] \[Rule] R Sin[\[Theta]]) for fixed radius R (which is a model \
parameter) and another set of coordinates for where we will be testing the \
potential (x \[Rule] \[Rho] Cos[\[Alpha]], y \[Rule] \[Rho] Sin[\[Alpha]]).  \
Because the potential must be independent of \[Alpha], we might as well set \
it to zero.\
\>", "Text",
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Cell[TextData[{
 "We have a choice whether to integrate over cz \[Element]  (-",
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the cylinder surface over cz and then the cylinder end over r, then we can \
integrate the sum of these over \[Theta] together."
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Cell[TextData[{
 "Try and see if we can do one of the two integrals---we have a choice of \
integrating over \[Theta] or (\[Zeta] for the cylinder sides, and  R) for the \
cylinder ends. We find a closed form for integrating  \[Zeta] for the sides \
and R for the top, and then subsequently numerically integrate \[Theta] for \
(0,2 \[Pi]).\n\n\n",
 StyleBox["This will take a while to finish, even on a very fast  machine (as \
of 2007, one minute to do the integral and about 6 hours minutes to simplify \
the expression).",
  FontColor->RGBColor[1, 0, 0]],
 "\nFor this reason, I' ve set up an option to read the result over the web, \
instead of having each of your computers spend time on a previously computed \
result:\n",
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 StyleBox[" 6.01 Oct. 2007,  If we first try the definite integral over \
\[Theta] \[Element] (0, 2 \[Pi]) , we get  ZERO which is  an incorrect result \
(I will include a discussion of why it is incorrect in the future))",
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Here we restrict z to the upper half-space. We will treat z=0 below.\
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Here is the limit of  the integral for z> 0 (CylinderIntegrandd\[Theta]) in \
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\>", "Text",
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Speed and Efficiency: When to Evaluate the Right-Hand-Side during a Function \
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Let's set a baseline to check efficiency. Here we check timing to integrate \
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The following is equivalent to the above (safer) definition---and will work \
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The following should take the *least* amount of time to perform, but as we \
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The following will not be rapid, because it has to do the symbolic \
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