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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 322 14 "Work and Force" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 432 "We inves
tigate the relationship between force and energy (work). For energy-co
nserving systems one can formulate the equations of motion by making u
se of scalar quantities that represent the kinetic and potential energ
y for the particle in the force field. For non-conservative forces the
 work integral has to be calculated for the individual particle trajec
tories using a line integral. Both cases are considered in this worksh
eet." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 198 "
We begin with the conservative case. Take the example of a two-centre \+
electrostatic potential as experienced in a diatomic molecule by the v
alence electron (e.g., the H2+ molecule, cf. R.L. Greene, " }{TEXT 
257 30 "Classical Mechanics with Maple" }{TEXT -1 47 " (Springer 1994)
, pp. 88-93, and M.L. de Jong, " }{TEXT 256 37 "Introduction to Comput
ational Physics" }{TEXT -1 238 " (Addison-Wesley 1991), chapter 10). T
he problem of a spaceship, or asteroid, or small planet moving in the \+
field of two nearly stationary stars would be calculated in a complete
ly analogous way. Two protons are kept at a fixed distance 2" }{TEXT 
258 1 "d" }{TEXT -1 11 " along the " }{TEXT 259 1 "z" }{TEXT -1 28 " a
xis, one located at (0,0,-" }{TEXT 260 1 "d" }{TEXT -1 21 "), the othe
r at (0,0," }{TEXT 261 1 "d" }{TEXT -1 31 "). The electron is located \+
at (" }{TEXT 262 1 "x" }{TEXT -1 1 "," }{TEXT 263 1 "y" }{TEXT -1 1 ",
" }{TEXT 264 1 "z" }{TEXT -1 161 ") and being of opposite charge exper
iences an attractive force. It is convenient to express the electron d
istance from either proton in terms of its coordinate (" }{TEXT 267 1 
"x" }{TEXT -1 1 "," }{TEXT 268 1 "y" }{TEXT -1 1 "," }{TEXT 269 1 "z" 
}{TEXT -1 21 ") using coordinates, " }{TEXT 266 1 "r" }{TEXT -1 6 "1 a
nd " }{TEXT 265 1 "r" }{TEXT -1 15 "2 respectively." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r1:=sqrt(x^2+y^2+(z-d)^2)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r2:=sqrt(x^2+y^2+(z+d)
^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The electrotatic potentia
l energy in MKSA (SI) units " }{TEXT 19 3 "Usi" }{TEXT -1 155 " is the
n given in terms of the charges for the protons and the electron, Q an
d q respectively, the dielectric constant epsilon[0] and the two dista
nces as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "f[0]:=q*Q/(4*Pi
*epsilon[0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Usi:=-f[0]
*(1/r1+1/r2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 257 "For plotting as
 well as calculations it is more convenient to measure energy in units
 in which the dielectric constant, the charge units and a characterist
ic length scale are absorbed into a single constant. The characteristi
c length scale can be chosen as " }{TEXT 270 1 "d" }{TEXT -1 10 ", (or
 as 2" }{TEXT 271 1 "d" }{TEXT -1 97 "), i.e., the equilibrium separat
ion of the nuclei in the molecule. In these units we simply have:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "U:=unapply(subs(d=1,-1/r1-1/
r2),x,y,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 366 "There are various
 ways to visualize this function. It is possible to make cuts such tha
t a 2D graph conveys some information. Two of the coordinates have to \+
be fixed for this purpose, and keep in mind that all distances are mea
sured in terms of d which has been set to unity, i.e., all coordinates
 are measured as multiples of the half-separation between the nuclei.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "We ta
ke cuts along the " }{TEXT 275 1 "x" }{TEXT -1 7 " axis (" }{TEXT 274 
1 "x" }{TEXT -1 15 "=0) with fixed " }{TEXT 273 2 "y " }{TEXT -1 24 "(
treated as a parameter " }{TEXT 279 1 "y" }{TEXT -1 32 "0) and display
 as a function of " }{TEXT 272 1 "z" }{TEXT -1 111 ". We restrict the \+
range of the ordinate to avoid a huge scaling from the singularity in \+
the potential at (0,0,-" }{TEXT 277 1 "d" }{TEXT -1 11 ") and (0,0," }
{TEXT 276 1 "d" }{TEXT -1 15 ") respectively." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 44 "P1:=plot(U(0,0,z),z=-3..3,-5..0,color=blue):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "P2:=plot(U(0,0.5,z),z=-3..3,color=r
ed):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "P3:=plot(U(0,1,z),z=-3..3,c
olor=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots)
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{P1,P2,P3\});
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "It is straightforward to read
 off from this diagram the " }{TEXT 278 1 "z" }{TEXT -1 67 " component
 of the force for the electron when it is located at (0, " }{TEXT 280 
1 "y" }{TEXT -1 2 "0," }{TEXT 281 2 " z" }{TEXT -1 26 ") for the three
 values of " }{TEXT 282 1 "y" }{TEXT -1 164 "0 (0, 0.5 and 1) represen
ted by the three curves above. Due to the superposition principle we c
an treat these potentials as 1D potentials for the motion with fixed \+
" }{TEXT 283 1 "x" }{TEXT -1 6 "0 and " }{TEXT 284 1 "y" }{TEXT -1 79 
"0, and thus the force is simply given as the negative of the derivati
ve w.r.t. " }{TEXT 285 1 "z" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 67 "P1f:=plot(-diff(U(0,0,z),z),z=-3..3,-5..5,color=bl
ue,discont=true):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P2f:=plot(-dif
f(U(0,0.5,z),z),z=-3..3,color=red):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
49 "P3f:=plot(-diff(U(0,1,z),z),z=-3..3,color=green):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(\{P1f,P2f,P3f\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "We added an option in the graph fo
r " }{TEXT 286 1 "x" }{TEXT -1 292 "0=0 to avoid a vertical line showi
ng up at the location of the singularity (incorrect joining of the asy
mptotic behaviour done by Maple without the option). Note that the sig
n of the force changes, and that the location of the equilibrium point
s (intercept with zero) depends on the value of " }{TEXT 287 1 "y" }
{TEXT -1 7 "0. For " }{TEXT 288 1 "y" }{TEXT -1 70 "0=0 the behaviour \+
is qualitatively different as a singularity appears." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 246 "How can the full 3D be
haviour be displayed? The correct approach would be to switch to cylin
drical (polar) coordinates and make use of axial symmetry w.r.t. the z
 axis. We prefer to keep Cartesian coordinates however. Thus we keep t
he constraint " }{TEXT 291 1 "x" }{TEXT -1 38 "0=0, and display the po
tential in the " }{TEXT 289 1 "y" }{TEXT -1 1 "-" }{TEXT 290 1 "z" }
{TEXT -1 50 " plane using a surface plot and a contour diagram." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Since we \+
called the " }{TEXT 19 5 "plots" }{TEXT -1 44 " package already, conto
urplot is available.." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "p
lot3d(U(0,y,z),y=-3..3,z=-3..3,view=-3..0,style=patchcontour,shading=z
hue,axes=boxed,numpoints=3000,orientation=[-30,15]);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 194 "The surface plot contains contours drawn at va
rious heights. We can turn it into a contour diagram by simply choosin
g the right projection (choice of theta/phi angles). Alternatively we \+
use the " }{TEXT 19 11 "contourplot" }{TEXT -1 18 " command from the \+
" }{TEXT 19 5 "plots" }{TEXT -1 57 " package. We avoid the singularity
 by picking a value of " }{TEXT 292 1 "x" }{TEXT -1 6 "0=0.1:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "contourplot(U(0.1,y,z),z=-3
..3,y=-3..3,axes=boxed,scaling=CONSTRAINED,grid=[40,40],contours=[-4,-
3,-2,-1.5,-1],filled=true);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "Th
e force acting on the particle is given as the negative of the gradien
t. We need to load the " }{TEXT 19 6 "linalg" }{TEXT -1 20 " package a
nd obtain:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "F:=grad(-U(x,y,z),[x,y,
z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 156 "Note that one could not \+
simply evaluate F:=-grad(U...) as this would amount to the multiplicat
ion of a vector by a scalar (namely by -1). Maple requires an " }
{TEXT 19 5 "evalm" }{TEXT -1 59 " wrapped around such a command to be \+
able to evaluate. The " }{TEXT 19 5 "evalm" }{TEXT -1 57 " procedure i
s used for matrix and vector multiplications." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "To confirm our previous r
esult for the z component of the force we graph:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 51 "plot(subs(x=0,y=0.5,F[3]),z=-3..3,-5..5,color
=red);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "We can demonstrate that
 the force field is curl-free:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 16 "curl(F,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "We c
an demonstrate by example that the line integral is independent of the
 chosen path. We choose two paths in the " }{TEXT 293 1 "y" }{TEXT -1 
1 "-" }{TEXT 294 1 "z" }{TEXT -1 257 " plane for which the potential w
as shown in the contour diagram. Let us calculate the work to reach th
e location [0,1,1] from the origin [0,0,0] in two different ways. Once
 through a straight-line connection, and once through an arc of a circ
le with radius " }{TEXT 19 9 "sqrt(2)/2" }{TEXT -1 30 " around the poi
nt [0,0.5,0.5]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 115 "To set up the line integral to calculate the work done b
y the force on the particle as it traverses the trajectory " }{TEXT 
295 1 "r" }{TEXT -1 1 "(" }{TEXT 296 1 "t" }{TEXT -1 203 ") we use a p
arameter representation of the path. If the path is differentiable wit
h respect to its parameter (e.g., time in physics), then the line inte
gral can be reduced to a definite Riemann integral." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "W =  int (" }{TEXT 315 
1 "F" }{TEXT -1 4 " . d" }{TEXT 316 1 "r" }{TEXT -1 9 ") =  int(" }
{TEXT 317 1 "F" }{TEXT -1 1 "(" }{TEXT 318 1 "r" }{TEXT -1 1 "(" }
{TEXT 319 1 "t" }{TEXT -1 6 ")) . d" }{TEXT 320 1 "r" }{TEXT -1 2 "/d
" }{TEXT 321 1 "t" }{TEXT -1 15 ", t = t0 .. t1)" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "For the straight-line con
nection we have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r_in:=[
0,0,0]; r_fin:=[0,1,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "
rSL:=evalm(r_in+t*(r_fin-r_in));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "rSL[1], rSL[2], rSL[3];" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 130 "The above does not work properly without the call to eva
lm, and the operation below does not work without the map construct, o
nce " }{TEXT 19 3 "rSL" }{TEXT -1 13 " is a vector." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 21 "vSL:=map(diff,rSL,t);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 14 "whattype(rSL);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 85 "In this parameter representation of the straight line the
 endpoints are realized for " }{TEXT 298 1 "t" }{TEXT -1 7 "=0 and " }
{TEXT 297 1 "t" }{TEXT -1 101 "=1 respectively. It is important to eva
luate the force along the trajectory, i.e., to substitute for " }
{TEXT 299 1 "x" }{TEXT -1 1 "," }{TEXT 300 1 "y" }{TEXT -1 1 "," }
{TEXT 301 1 "z" }{TEXT -1 39 " in the force assignment the values of \+
" }{TEXT 302 1 "x" }{TEXT -1 1 "(" }{TEXT 303 1 "t" }{TEXT -1 3 "), " 
}{TEXT 304 1 "y" }{TEXT -1 1 "(" }{TEXT 305 1 "t" }{TEXT -1 3 "), " }
{TEXT 306 1 "z" }{TEXT -1 1 "(" }{TEXT 307 1 "t" }{TEXT -1 124 "), so \+
that the definite integration makes sense, and the force evaluations h
appen at the correct positions as a function of " }{TEXT 308 1 "t" }
{TEXT -1 180 ", namely along the chosen trajectory. This is the most i
mportant aspect of the line integral to be understood by the beginning
 physicist (and also the most common source of error)." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 124 "The op(F) rather th
an F in the substitution is important. It is needed for the substituti
on to take place in all components:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "subs(x=rSL[1],y=rSL[2],z=rSL[3],op(F));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 11 "We use the " }{TEXT 19 7 "dotprod" }
{TEXT -1 20 " procedure from the " }{TEXT 19 6 "linalg" }{TEXT -1 46 "
 package to calculate the dot product between " }{TEXT 309 1 "F" }
{TEXT -1 1 "(" }{TEXT 310 1 "r" }{TEXT -1 1 "(" }{TEXT 311 1 "t" }
{TEXT -1 8 ")) and d" }{TEXT 312 1 "r" }{TEXT -1 2 "/d" }{TEXT 313 1 "
t" }{TEXT -1 26 ", and then integrate over " }{TEXT 314 1 "t" }{TEXT 
-1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "W1:=int(dotprod(
subs(x=rSL[1],y=rSL[2],z=rSL[3],op(F)),vSL),t=0..1);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 135 "We can check whether this result agrees with t
he difference between the potential energies evaluated at the two endp
oints respectively:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "U(op
(r_fin))-U(op(r_in));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "ev
alf(%,4); evalf(W1,4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "To see
 how the first of the two numbers came about look at the contributions
 to the potential difference:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 44 "evalf(U(op(r_fin)),4); evalf(U(op(r_in)),4);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 160 "The amount of work to move the electron from loc
ation [0,0,0] to [0,1,1] is negative, which means that on average the \+
force was acting more against the motion. " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "Try simpler trajectories for w
hich the force vector is aligned with the position vector, such as mov
ing from [0,0,0] to [0,0,1-" }{TEXT 324 1 "c" }{TEXT -1 98 "], i.e., f
rom the unstable equilibrium point towards one of the protons, e.g., t
he one located at " }{TEXT 323 1 "z" }{TEXT -1 70 "=1. Also see what h
appens as your path moves across the singularity..." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "The potential energy at
 the initial position " }{TEXT 19 12 "r_in=[0,0,0]" }{TEXT -1 28 " is \+
more attractive than at " }{TEXT 19 13 "r_fin=[0,1,1]" }{TEXT -1 366 "
. The difference in potential energy between the final and initial poi
nts of the trajectory is therefore positive. It costs energy to move f
rom [0,0,0] to [0,1,1]. This is consistent with the previous finding t
hat the work is negative: we have to invest energy to move the electro
n from [0,0,0] to [0,1,1]. The electron can trade kinetic energy to ac
hieve this goal." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 56 "Now we repeat the calculation for a circular trajectory:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "rC:=evalm(sqrt(2)/2*[0,
cos(t),sin(t)]+[0,1/2,1/2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 6 "rC[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "plot([rC[2],
rC[3],t=Pi/4..5/4*Pi],scaling=CONSTRAINED);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 34 "map(simplify,subs(t=Pi/4,op(rC)));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "map(simplify,subs(t=5*Pi/4,op(rC)))
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "With the chosen parametrisa
tion the trajectory starts at 5/4 Pi, and ends at Pi/4. However, it pa
sses through the singularity at [0,0,1], and thus we integrate the oth
er way around:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 37 "map(simplify,subs(t=-3*Pi/4,op(rC)));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "t0:=-3*Pi/4: t1:=Pi/4:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "vC:=map(diff,rC,t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "W2:=int(expand(dotprod(subs(
x=rC[1],y=rC[2],z=rC[3],op(F)),vC)),t=t0..t1);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 199 "Maple has a hard time doing the integral in closed \+
form. We see that the integrand is non-trivial in the next command. Th
ere is no difficulty in integrating numerically if the singularity is \+
avoided." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "arg:=expand(dot
prod(subs(x=rC[1],y=rC[2],z=rC[3],op(F)),vC)):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 27 "evalf(Int(arg,t=t0..t1),4);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 186 "We are satisfied that the answer agrees with t
he floating-point evaluation of the previous work calculation for the \+
straight-line trajectory and the negative of the potential difference.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 748 "Now \+
that we established the fact that the potential energy allows to calcu
late the work done on particles by a curl-free force, we illustrate ho
w one uses the contour diagram to read off information about the force
. We fixed a plane when graphing the equipotential contours, and thus \+
we choose to illustrate only the force components in that plane. The f
orce is given by the gradient of the potential (up to a minus sign). T
he direction of the gradient vector originating at any point is orthog
onal to an equipotential contour passing through that point. The plots
 package in Maple contains a command that allows to draw a set of arro
ws on a discrete mesh for any given potential function in two dimensio
ns. To avoid the singularity we look at the " }{TEXT 326 1 "y" }{TEXT 
-1 1 "-" }{TEXT 325 1 "z" }{TEXT -1 18 " plane defined by " }{TEXT 
327 1 "x" }{TEXT -1 6 "0=0.5:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 90 "PL1:=gradplot(-U(0.5,y,z),z=-3..3,y=-2..2,axes=boxed,color=red,g
rid=[16,24],arrows=thick):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
150 "PL2:=contourplot(U(0.5,y,z),z=-3..3,y=-2..2,axes=boxed,scaling=CO
NSTRAINED,grid=[40,40],contours=[-2,-1.5,-1,-0.5],filled=true,coloring
=[white,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(
PL1,PL2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "The arrows give a go
od impression of the variation of the two force components in the " }
{TEXT 328 1 "y" }{TEXT -1 1 "-" }{TEXT 329 1 "z" }{TEXT -1 45 " plane.
 No information is conveyed about the " }{TEXT 330 1 "z" }{TEXT -1 24 
" component of the force." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 357 "The diagram illustrates the fact that it is not e
ntirely obvious that the work integral is path-independent: while walk
ing different trajectories from one endpoint to another one encounters
 force vectors pointing in different directions. Think about a path wh
ere the electron is first accelerated towards the proton located in th
e upper half-plane (at [0,0," }{TEXT 333 1 "d" }{TEXT -1 271 "]), and \+
then swings around towards the final location [0,1,1] working against \+
the force field. A very different experience as compared to the straig
ht-line or circular paths calculated before. Thus, the curl-free natur
e of conservative forces is a very powerful property." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "The potential in the \+
present case is given by a superposition of two central potentials wit
h respect to the positions of the nuclei [0,0," }{TEXT 332 1 "d" }
{TEXT -1 12 "] and [0,0,-" }{TEXT 331 1 "d" }{TEXT -1 134 "] respectiv
ely. Central potentials play an important role due to their spherical \+
symmetry (the potential depends only on the distance " }{TEXT 334 1 "r
" }{TEXT -1 146 ", not on the spherical polar angles theta and phi). T
his symmetry implies angular momentum conservation, both in classical \+
and in quantum physics." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{MARK "63" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 
1 2 33 1 1 }