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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 260 18 "Compton scattering" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 423 "We used \+
conservation laws for linear momentum and kinetic energy to solve one-
dimensional collision problems. In more dimensions the number of degre
es of freedom associated with the particle motion exceeds the number o
f constraints provided conservation laws. This means that we will not \+
be able to obtain complete solutions. Nevertheless, it is interesting \+
to investigate the constraints supplied by the conservation laws." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Suppose t
hat a particle of mass " }{TEXT 259 1 "m" }{TEXT -1 28 " is impinging \+
with velocity " }{TEXT 257 1 "v" }{TEXT -1 12 "0 along the " }{TEXT 
258 1 "x" }{TEXT -1 28 "-axis on a particle of mass " }{TEXT 256 1 "M
" }{TEXT -1 530 " that is initially at rest. The particles interact by
 a central force (force acts only along the mutual separation). To att
empt a solution based on conservation laws means to not take into acco
unt the details of the interaction. A complete dynamical approach invo
lves solving Newton's equations (or variants thereof, such as Lagrange
's, or Hamilton's forms), which takes into account the force law, and \+
which automatically obeys the conservation laws. An example of such a \+
complete solution was the problem of firing a cannonball." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 284 "1) The conservati
on of angular momentum for the motion of objects subject to central fo
rces implies that the motion is confined to a plane. This reduces the \+
three degrees of freedom in the position vectors of the masses to two.
 The number of degrees of freedom is reduced from 6 to 4." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "2) The conservatio
n of energy provides one scalar constraint, i.e., 3 degrees of freedom
 remain." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
86 "3) The conservation of linear momentum in the scattering plane pro
vides 2 constraints." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 282 "Thus, the problem cannot be completely determined from
 the conservation laws. This is illustrated in this worksheet, first i
n general, and then specifically for the scattering of photons off ato
mic electrons resulting in an energy transfer to the electron, i.e., C
ompton scattering." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 324 "We set up the equations for the constraints 2 and 3, i.e
., we begin in the scattering plane (having made use of angular moment
um conservation). We assume complete elasticity for the collision, alt
hough it is not difficult to incorporate in the energy balance a given
 amount of binding energy that has to be supplied to mass " }{TEXT 
264 1 "M" }{TEXT -1 37 " in order to free it from the target." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "En:=m/2*v0^2=m/2*v^2+M/2*V^2
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "To specify the momentum comp
onents after the collision we introduce two angles with respect to the
 " }{TEXT 261 1 "x" }{TEXT -1 42 " axis: theta for the scattering part
icle (" }{TEXT 262 1 "m" }{TEXT -1 46 ") and phi for the particle init
ially at rest (" }{TEXT 263 1 "M" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 39 "Momx:=m*v0=m*v*cos(theta)+M*V*cos(phi);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Momy:=0=m*v*sin(theta)+M*V*s
in(phi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "What are we intereste
d in?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "
The masses of the particles and the initial speed " }{TEXT 265 1 "v" }
{TEXT -1 20 "0 are usually known." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 29 "Unknown are four quantities: " }{TEXT 
266 1 "v" }{TEXT -1 2 ", " }{TEXT 267 1 "V" }{TEXT -1 105 ", and theta
 and phi. How can we use the three constraints to learn as much as pos
sible about the problem?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 85 "Before we jump into the general situation, let us \+
first consider the equal-mass case:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "En1:=subs(M=m,En);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Momx1:=subs(M=m,Momx);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Momy1:=subs(M=m,Momy);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 81 "Given that we have three equations we can ask for a solut
ion for three variables:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 
"sol:=solve(\{En1,Momx1,Momy1\},\{v,V,theta\});" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 8 "Maple's " }{TEXT 19 5 "solve" }{TEXT -1 77 " engine f
inds a set of four acceptable solutions. Let us see what they imply." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Set 1  \+
" }{XPPEDIT 18 0 "\{v = v0, V = 0, theta = 0\};" "6#<%/%\"vG%#v0G/%\"V
G\"\"!/%&thetaGF)" }{TEXT -1 46 " describes the situation before the c
ollision." }}{PARA 0 "" 0 "" {TEXT -1 6 "Set 2 " }{XPPEDIT 18 0 "\{v =
 -v0, V = 0, theta = Pi\};" "6#<%/%\"vG,$%#v0G!\"\"/%\"VG\"\"!/%&theta
G%#PiG" }{TEXT -1 54 " describes the mirror image of the situation in \+
set 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 
"Sets 3 and 4 assign the same speed to mass " }{TEXT 268 1 "M" }{TEXT 
-1 129 ". If we assume that theta and phi are in the range  0..Pi/2, a
nd -Pi/2..0 respectively, i.e., both masses move to the right with " }
{TEXT 272 1 "m" }{TEXT -1 27 " deflected in the positive " }{TEXT 271 
1 "y" }{TEXT -1 15 " direction and " }{TEXT 270 1 "M" }{TEXT -1 41 " m
oving downward, it is clear that set 4 " }{XPPEDIT 18 0 "\{theta = phi
+1/2*Pi, V = cos(phi)*v0, v = -v0*sin(phi)\};" "6#<%/%&thetaG,&%$phiG
\"\"\"*(F(F(\"\"#!\"\"%#PiGF(F(/%\"VG*&-%$cosG6#F'F(%#v0GF(/%\"vG,$*&F
3F(-%$sinG6#F'F(F+" }{TEXT -1 36 " is the interesting case. The speed \+
" }{TEXT 269 1 "v" }{TEXT -1 20 " should be positive." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "An interesting conclus
ion can be reached:" }}{PARA 0 "" 0 "" {TEXT -1 307 "apparently the sc
attering angle phi is arbitrary (in the range -Pi/2..0), and the proje
ctile scattering angle theta is given in terms of phi. The well-known \+
textbook result for the equal-mass elastic scattering case is obtained
, namely that the two objects emerge at right angles with respect to e
ach other." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 230 "Furthermore, given a choice of angle phi, the distribution of \+
kinetic energies between incident and hit particle is fixed. For each \+
scattering angle phi (or theta) the kinetic energy of both the impingi
ng and hit object are fixed." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assign(sol[4]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "print(v,V);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 13 "KEm:=m*v^2/2;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "KEM:=m*V^2/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 13 "m:=1; v0:=10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "pl
ot([KEm,KEM],phi=-Pi/2..0,color=[blue,green],axes=BOXED);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 66 "Obviously the sum of the kinetic energies
 of the two masses (note " }{TEXT 274 1 "M" }{TEXT -1 1 "=" }{TEXT 
275 1 "m" }{TEXT -1 44 ") is independent of phi, as it has to equal " 
}{TEXT 276 1 "m" }{TEXT -1 3 "/2*" }{TEXT 277 1 "v" }{TEXT -1 4 "0^2.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 
57 "The extreme cases correspond to the following situations:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 17 "phi=0, th
eta=Pi/2" }{TEXT -1 37 ": the incident mass remains at rest (" }{TEXT 
19 5 "KEm=0" }{TEXT -1 56 "), and the angle theta is irrelevant. The h
it particle (" }{TEXT 273 1 "M" }{TEXT -1 151 ") moves in the forward \+
direction with maximum kinetic energy. This represents a head-on colli
sion that can be calculated in a one-dimensional geometry." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 18 "phi=-Pi/2, theta=
0" }{TEXT -1 57 ": the incident mass transfers no energy to the hit ma
ss (" }{TEXT 19 5 "KEM=0" }{TEXT -1 338 "), phi is irrelevant. This co
rresponds to the limit of a grazing collision when the particles inter
act only very weakly. More interesting is the observation that before \+
the limit is reached (small-angle projectile scattering) the hit parti
cle recoils with an angle of almost 90 degrees, and picks up a small a
mount of kinetic energy only." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 19 21 "phi=-Pi/4, theta=Pi/4" }{TEXT -1 46 ": eq
ual sharing of energy after the collision." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 "We are now ready to consider th
e general case." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "En:=m/2*v0^2=m/2*v^2+M/
2*V^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "Momx:=m*v0=m*v*co
s(theta)+M*V*cos(phi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "M
omy:=0=m*v*sin(theta)+M*V*sin(phi);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "sol:=solve(\{En,Momx,Momy\},\{v,V,theta\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 164 "We are now used to the fact that \+
the initial condition and its mirror image appear as solutions to the \+
system of equations. Let us analyze the interesting solution." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assign(sol[3]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "v;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 2 "V;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "theta;
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 281 "We notice that the relations
hip between theta and phi is more complicated. Let us first analyze th
e simple part. The kinetic energy of the hit particle is still a funct
ion of its scattering angle. Note that the arctan function can take tw
o arguments (look it up in the help pages)." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 13 "KEM:=M/2*V^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
102 "For fixed scattering angle phi we can explore the dependence of t
his kinetic energy on the mass ratio." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "phi:=-Pi/4; v0:=10; m:=1; M:=r*m;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 4 "KEM;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 16 "KEtot:=m/2*v0^2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "We p
erform a semilog-plot 'by hand' and cover the range of mass ratios fro
m 0.001 to 1000 on the abscissa." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "plot(subs(r=10^s,KEM)/KEtot,s=-3..3);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 110 "We learn from the graph that the equal-m
ass case is the most effective for energy transfer. If the mass ratio \+
" }{TEXT 279 2 "r " }{TEXT -1 2 "= " }{TEXT 280 1 "M" }{TEXT -1 1 "/" 
}{TEXT 281 1 "m" }{TEXT -1 91 " is either small or large, then the fra
ction of the kinetic energy transferred to the mass " }{TEXT 282 1 "M
" }{TEXT -1 57 " in collisions with phi = -45 degrees becomes very sma
ll." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "T
he other variables that were determined looked somewhat clumsy. A squa
re root was not carried out to avoid case distinctions. With the speci
fication of " }{TEXT 283 1 "m" }{TEXT -1 5 " and " }{TEXT 284 1 "v" }
{TEXT -1 43 "0 they are, of course, somewhat simplified:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "v;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "simplify(v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 19 "evalf(subs(r=1,%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Mapl
e picks one of the roots, unfortunately the one with the negative sign
..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "Gi
ven that a simple square root is at issue we convert the RootOf to a r
adical expression:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "### W
ARNING: note that `I` is no longer of type `radical`\nv:=convert(v,rad
ical);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "To display the fraction
 of the total kinetic energy remaining with mass " }{TEXT 302 1 "m" }
{TEXT -1 108 " after the collision for the special case of phi = -45 d
egrees as a function of the mass ratio we calculate:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 36 "plot(subs(r=10^s,v)^2/v0^2,s=-3..3);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 141 "Obviously, this is the complement
 to the previous graph. Let us observe the corresponding scattering an
gle for the incident particle of mass " }{TEXT 285 1 "m" }{TEXT -1 1 "
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "### WARNING: note that
 `I` is no longer of type `radical`\ntheta:=convert(theta,radical);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "plot(subs(r=10^s,theta),s=
-3..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "The equal-mass case (
" }{TEXT 286 1 "s" }{TEXT -1 4 "=0, " }{TEXT 287 1 "r" }{TEXT -1 38 "=
1) results in theta=Pi/4 as expected." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 16 "In the limit of " }{TEXT 291 2 "r " }
{TEXT -1 2 "= " }{TEXT 289 1 "M" }{TEXT -1 1 "/" }{TEXT 290 1 "m" }
{TEXT -1 16 " going to zero (" }{TEXT 288 1 "s" }{TEXT -1 184 " large,
 negative), i.e., a massive incident particle as compared to the hit p
article, the scattering angle for the incident particle goes to zero (
note that phi is fixed at phi=-Pi/4)." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 16 "In the limit of " }{TEXT 294 2 "r " }
{TEXT -1 2 "= " }{TEXT 292 1 "M" }{TEXT -1 1 "/" }{TEXT 293 1 "m" }
{TEXT -1 20 " going to infinity (" }{TEXT 295 1 "s" }{TEXT -1 323 " la
rge, positive), i.e., a light incident particle as compared to the hit
 particle, the scattering angle theta approaches Pi, i.e., reflection \+
while the heavier object exits at phi=-45 degrees. This includes the l
imit of a head-on collision with perfect reflection, as the previous g
raphs showed that in the limit of large " }{TEXT 297 1 "r" }{TEXT -1 
44 " no energy is transferred to the heavy mass " }{TEXT 296 1 "M" }
{TEXT -1 41 " (its exit angle phi becomes irrelevant)." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "Another interesting
 case to investigate involves fixing the energy of the hit particle an
d asking for the allowed range of scattering angles phi." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Finally we
 are ready to discuss " }{TEXT 298 18 "Compton scattering" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 428 "
The discussion is slightly different to take into account the fact tha
t photons are not classical massive particles. It was observed in the \+
passage of light through matter that some light (X rays in particular)
 was emitted with an increased wavelength, i.e., reduced energy. The c
hange in wavelength depended on the scattering angle of the observed p
hotons. Compton's explanation is based on a corpuscular model for the \+
photons." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
102 "The problem can be set up as before, except that one needs expres
sions for photon energy and momentum." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 55 "The photon momentum before the collis
ion is labeled as " }{TEXT 303 1 "p" }{TEXT -1 88 "0 and given in term
s of Planck's constant h, the frequency nu, and the speed of light c:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "p0:=h*nu0/c;" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 49 "The momentum conservation law can be expressed as" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Momx:=p0=h*nu/c*cos(theta)
+P*cos(phi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Momy:=0=h*n
u/c*sin(theta)+P*sin(phi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "To
 formulate energy conservation we need to take into account the relati
vistic energy-momentum relationship (in the limit of high photon energ
ies the electron can be picking up a large amount of kinetic energy).
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 153 "The \+
kinetic energy before the collision is rest energy of the electron (it
 is assumed to be unbound, i.e., free, and at rest, we denote its rest
 mass as " }{TEXT 299 1 "M" }{TEXT -1 24 ") and the photon energy:" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "En:=h*nu0+M*c^2=h*nu+Eel;" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Before we can solve we need to
 express the electron energy after the knock-out in terms of its rest \+
mass " }{TEXT 300 1 "M" }{TEXT -1 24 " and its final momentum " }
{TEXT 301 1 "P" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 40 "En1:=subs(Eel=sqrt(M^2*c^4+P^2*c^2),En);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 39 "sol:=solve(\{Momx,Momy,En1\},\{nu,P,phi\});" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Matters are more complicated ob
viously..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 97 "We can eliminate the electron emission angle phi by squaring th
e momentum conservation equations:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "ex1:=solve(Momx,P*cos(phi));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 28 "ex2:=solve(Momy,P*sin(phi));" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 21 "eq1:=ex1^2+ex2^2=P^2;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 19 "eq1:=simplify(eq1);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "sol:=solve(\{En1,eq1\},\{P,nu\});" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(sol);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 3 "nu;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 226 "If
 we are just interested in the wavelength shift as a function of photo
n scattering angle, we do the following. First we make use of the rela
tionship  lambda*nu=c to obtain an expression for the scattered photon
's wavelength:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eq2:=nu=c
/lambda;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now we substitute the
 frequency nu0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "eq3:=sub
s(nu0=c/lambda0,eq2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq
3:=simplify(eq3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The wavelen
gth shift Delta is given by subtracting the original wavelength lambda
0 from lambda; the latter is obtained by isolating lambda in eq3." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "Delta:=simplify(solve(eq3,la
mbda)-lambda0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 331 "The wavelengt
h shift Delta does not depend on the original wavelength lambda0 and i
s given as a multiple of a natural length scale for the electron, name
ly the Compton wavelength of the electron  h/(M*c). If we factor out t
his unit, the wavelength shift as a function of the photon scattering \+
angle is given in the following graph:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "plot(1-cos(theta),theta=0..Pi);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 192 "The wavelength shift acquires a maximum in the ba
ckward direction. The maximum inelasticity equals twice the Compton wa
velength of the particle that has been hit (in our case a free electro
n)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "In
 the forward direction the photons pass through without energy transfe
r." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "It \+
is useful to obtain an idea about the length scale of the Compton wave
length for an electron:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "
lambda[C]:=h/(M*c);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "h:=6
.63*10^(-34)*_N*_m*_s; #Planck's constant in SI units" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "c:=3.*10^8*_m/_s; # speed of light \+
in SI units" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "M:=9.11*10^(
-31)*_kg; # electron mass in SI units" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "lambda[C];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "subs(_N=_kg*_m/_s^2,%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 409 "T
hus the maximum wavelength shift that a free electron can provide equa
ls approximately 5*10^(-12) m, or 0.05 Angstroms, or 5 picometers. To \+
put this in relation with the photon wavelengths we note that hard X r
ays have wavelengths that are thousands of times shorter than those of
 visible light (0.1 nm vs 450-700 nm, 1 Angstrom = 0.1 nm). For hard X
 rays the wavelength shift thus reaches the percent range." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 417 "We conclude that
 soft photons can backscatter from electrons without appreciable energ
y loss, while hard photons can transfer a sizable amount of energy in \+
a backscattering process. Of course, matters are in reality more compl
icated: soft photons can excite electrons to undergo transitions betwe
en bound states. It is just the consideration of the kinematics of ela
stic scattering that leads to the above conclusion." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 
1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }