L03:
Kepler problem &
Hamiltonian dynamics
18.354
Ptolemy
circa.85 (Egypt) -165 (Alexandria)
Greek geocentric view of the universe
Wednesday, February 5, 14
Tycho Brahe
1546 (Denmark) - 1601 (Prague)
"geo-heliocentric"
system
last of the major naked eye astronomers
Wednesday, February 5, 14
Johannes Kepler
1571 - 1630 (Germany)
Kepler’s 3 laws
Wednesday, February 5, 14
Isaac Newton
1643 - 1727 (England)
m1 m2
|F12 | = G 2
r
Wednesday, February 5, 14
Friedrich Bessel
1784 (Germany) - 1846 (Prussia)
Cygni 61
d~10.3 ly
gravitational many-body problem:
prediction of Sirius B, Bessel functions
Wednesday, February 5, 14
Friedrich Bessel
1784 (Germany) - 1846 (Prussia)
Cygni 61
d~10.3 ly
gravitational many-body problem:
prediction of Sirius B, Bessel functions
Wednesday, February 5, 14
Why do we study applied mathematics? Aside from the intellectual challenge, it is reason4 to Kepler’s
andanHamiltonian
dynamics
able
argue that weproblem
do so to obtain
understanding of physical
phenomena, and to be
able to make predictions about them. Possibly the greatest example of this, and the origin
4 Kepler’s
problem and
Hamiltonian
dynamics
applied mathematics?
Aside
from thedesire
intellectual
challenge,
is reasonof Why
muchdo
of we
thestudy
mathematics
we do, came from
Newton’s
to understand
theit motion
to arguewhich
thatdowe
so applied
to obtain
understanding
ofintellectual
physical challenge,
phenomena,
and to be
Why
wedo
study
Aside
from the
it is reasonof able
the planets,
were
known
to mathematics?
obeyanKepler’s
laws.
to argue that
we do
so to obtain
an understanding
physical phenomena,
and the
to beorigin
able to make able
predictions
about
them.
Possibly
the greatestof example
of this, and
able
to make predictions
about
them.from
Possibly
the greatest
example
of this, and thethe
origin
of
much
of
the
mathematics
we
do,
came
Newton’s
desire
to
understand
motion
4.1 Kepler’soflaws
motion
much ofoftheplanetary
mathematics we
do, came from Newton’s desire to understand the motion
of the planets,of which
were known to obey Kepler’s laws.
the planets, which were known to obey Kepler’s laws.
In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion
4.1TheKepler’s
laws
of planetary
motion
(i)
orbits of the
planets
are ellipses,
with the Sun’s centre of mass at one focus of
In the early seventeenth century (1609-1619) Kepler proposed three laws of planetary motion
theearly
ellipse.
In the
seventeenth century (1609-1619) Kepler proposed three laws of planetary motion
4.1
Kepler’s laws of planetary motion
(i) The orbits of the planets are ellipses, with the Sun’s centre of mass at one focus of
(ii) (i)
The
lineorbits
joining
planet
andare
theellipses,
Sun describes
equal
areas
in equal
intervals
of time.
The
oftheathe
planets
with the
Sun’s
centre
of mass
at one
focus of
ellipse.
the ellipse. The line joining a planet and the Sun describes equal areas in equal intervals of time.
(iii) The squares(ii)
of the periods of the planets are proportional to the cubes of their semimajor
The squares
of the
periods
of thedescribes
planets are equal
proportional
cubesintervals
of their semi(ii)
The axes.
line (iii)
joining
a planet
and
the Sun
areastointhe
equal
of time.
major axes.
These
based
on periods
detailedofobservations
by Tycho Brahe,
and put
to rest
(iii)laws
Thewere
squares
of the
the planets made
are proportional
to the cubes
of their
semiThese laws were based on detailed observations made by Tycho Brahe, and put to rest
any notion
thataxes.
planets move in perfectly circular orbits. However, it wasn’t until Newton
major
any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton
proposed his lawproposed
of gravitation
in 1687 that
thethat
origins
of this
motion
understood.
his law of gravitation
in 1687
the origins
of this
motionwere
were understood.
These proposed
laws were
based
on detailed
observations made by Tycho Brahe, and put to rest
Newton
that
Newton
proposed
that
any notion that planets move in perfectly circular orbits. However, it wasn’t until Newton
“Every object in the Universe attracts every other object with a force directed along a line
“Every
object
the of
Universe
attracts1687
everythat
other
object
with
force
directed
along a line
proposed
hisinlaw
gravitation
the
origins
of athis
motion
of centres
for the two in
objects that
is proportional
to the
product
of theirwere
massesunderstood.
and
of
centres
for
the
two
objects
that
is
proportional
to
the
product
of
their
masses
and
Newton proposed that
inversely proportional to the square of the separation of the two objects.”
inversely proportional to the square of the separation of the two objects.”
on Universe
this one statement,
it is
possible
to derive
Kepler’s
“Every objectBased
in the
attracts
every
other
object
withlaws.
a force directed along a line
Basedofon
this one
statement,
it is possible
to derive Kepler’s
centres
for
the two objects
that is proportional
to thelaws.
product of their masses and
4.1.1 Second law
inversely proportional to the square of the separation of the two objects.”
4.1.1 SecondKeplers
law second law is the simplest to derive, and is a statement that the angular momentum
a particle
moving it
under
a central force,
such as
gravity, islaws.
constant. By definition, the
Based on thisofone
statement,
is possible
to derive
Kepler’s
angular
momentum
L to
of aderive,
particle and
with is
mass
m and velocity
u isthe angular momentum
Keplers second law
is the
simplest
a statement
that
of 4.1.1
a particle
movinglaw
under a central force, such as gravity,
is constant. By definition, the
dr
Second
L=r^m ,
(53)
dt
angular momentum L of a particle with mass m and velocity
u is
Keplers second
lawrisisthe
to derive,
and is The
a statement
thatof the
angular
momentum
where
the simplest
vector position
of the particle.
rate of change
angular
momentum
is
dr
given by under a central force, such as gravity, is constant. By definition, the
of a particle moving
L=
(53)
dLr ^ m dt ,
= r ^ fm= and
r ^ f velocity
(r)r̂ = 0, u is
(54)
angular momentum L of a particle with
mass
dt
where r is the vector
the force,
particle.
The only
rateonofthechange
angular
momentum
where fposition
(r)r̂ is the of
central
depending
distanceof
r=
|r| and pointing
in the is
“Every object in the Universe attracts every other object with a force directed along a line
of centres for the two objects that is proportional to the product of their masses and
inversely proportional to the square of the separation of the two objects.”
Based on this one statement, it is possible to derive Kepler’s laws.
4.1.1
Second law
Keplers second law is the simplest to derive, and is a statement that the angular momentum
of a particle moving under a central force, such as gravity, is constant. By definition, the
angular momentum L of a particle with mass m and velocity u is
L=r^m
dr
,
dt
(53)
where r is the vector position of the particle. The rate of change of angular momentum is
given by
dL
= r ^ f = r ^ f (r)r̂ = 0,
(54)
dt
where f (r)r̂ is the central force, depending only on the distance r = |r| and pointing in the
direction r̂ = r/r. It can therefore be seen that the angular momentum of a particle moving
under a central force is constant, a consequence of this being that motion takes place in a
plane. The area swept out by the line joining
13 a planet and the sun is half the area of the
parallelogram formed by r and dr. Thus
1
1
dr
L
dA = |r ^ dr| = r ^
dt =
dt,
2
2
dt
2m
where L = |L| is a constant. The area swept out is therefore also constant.
4.1.2
First law
(55)
plane. The area swept out by the line joining a planet and the sun is half the area of the
where L = |L| is a constant. The area swept out is therefore also constant.
parallelogram formed by r and dr. Thus
1
1
dr
L
dA = |r ^ dr| = r ^
dt =
dt,
(55)
2
2
dt
2m
To prove Keplers first law consider the sun as being stationary (i.e., infinitely heavy), and
the
planets
it. The
The area
equation
motion
for a planet
is
where
L = in
|L|orbit
is a around
constant.
sweptofout
is therefore
also constant.
4.1.2
4.1.2
First law
First law
Toplane
provepolar
Keplers
first law consider
In
coordinates
the planets in orbit around it. The
dr
=
dt
d2 r
=
2
dt
In plane polar coordinates
d2 r
m 2 = f (r)r̂.
(56)
dt
the sun as being stationary (i.e., infinitely heavy), and
equation of motion for a planet is
˙ˆ
ṙr̂ +
(57a)
d2rr✓✓,
m 2 = f (r)r̂.
(56)
dt ˙2
˙ ✓.
ˆ
(r̈ r✓ )r̂ + (r✓¨ + 2ṙ✓)
(57b)
In component form, equation (56)
dr therefore becomes
ˆ
= ṙr̂ + r✓˙✓,
dt m(r̈ r✓˙2 ) = f (r),
d2 r
¨(r̈
˙✓˙2 )r̂
˙ ✓.
ˆ
m(r
✓
+
2
ṙ
✓)
=+ 0.
=
r
(r✓¨ + 2ṙ✓)
2
dt
Putting
(57a) into
(53)
gives (56) therefore becomes
In component
form,
equation
dr
2
2˙
f (r),
L =m(r̈
r ^ mr✓˙ ) ==|mr
✓|.
dt
˙ = 0.
m(r✓¨ + 2ṙ✓)
Thus
Putting (57a) into (53) gives
r2 ✓˙ = l,
(57a)
(58a)
(58b)
(57b)
(58a)
(59)
(58b)
(60)
where l = L/m is the angular momentum perdrunit mass.
˙ Given a radial force f (r), equaL= r^m
= |mr2 ✓|.
(59)
dt r and ✓ as functions of t. A more practical
tions (58a) and (58b) can now be solved to obtain
result is to solve for r(✓), however, and this requires the definition of a new variable
Thus
r2 ✓˙ 1= l,
(60)
r= .
(61)
where l = L/m is the angular momentum peru unit mass. Given a radial force f (r), equaRewriting
equations
ofnow
motion
in terms
of the rnew
requiresofthe
tions (58a)the
and
(58b) can
be solved
to obtain
andvariable
✓ as functions
t. Aidentities
more practical
result is to solve for r(✓), however,1 and this1 requires
definition of a new variable
d✓ du the du
ṙ =
u̇ =
= l ,
(62a)
u2
u2 dt1 d✓
d✓
(61)
2
d du r =˙ u
d2.u
2 2d u
r̈ =
l
= l✓ 2 = l u
.
(62b)
dt d✓
d✓
d✓2
In component form, equation (56) therefore becomes
dr
2˙
Putting (57a) into (53) gives
Lm(r̈
= r ^r✓m
✓|.
(59)
2
˙ )dt == f|mr
(r),
(58a)
˙ = 0. 2
dr
m(r✓¨ + 2ṙ✓)
(58b)
˙
L
=
r
^
m
= |mr ✓|.
(59)
Thus
dt
r2 ✓˙ = l,
(60)
Putting (57a) into (53) gives
Thus
where l = L/m is the angular momentum2 per
Given a radial force f (r), equadr unit mass.
˙
2˙
r
✓
=
l,
(60)
= r ^tomobtain
= r|mr
✓|.
(59)
tions (58a) and (58b) can now beLsolved
and
✓
as
functions
of
t.
A
more
practical
dt
resultl =
is to
solve
for r(✓),
however,
and this
the definition
of a new
variable
where
L/m
is the
angular
momentum
perrequires
unit mass.
Given a radial
force
f (r), equaThus
tions
(58a) and (58b) can now be solved to obtain
2˙
1l, r and ✓ as functions of t. A more practical
r
✓
=
= . the definition of a new variable (60)
(61)
result is to solve for r(✓), however, and thisr requires
u
where l = L/m is the angular momentum per unit mass. Given a radial force f (r), equa1
Rewriting
motion
in terms
new✓ variable
requires
the
identities
tions (58a) the
andequations
(58b) can of
now
be solved
to robtain
as functions
of t. A
more
practical
=of the
. r and
(61)
u
result is to solve for r(✓), however, and
of a new variable
1 this requires
1 d✓ duthe definition
du
= in 2terms
u̇ = of 2the new=variable
l , requires the identities (62a)
Rewriting the equations of ṙmotion
u
u 1 dt d✓
d✓
. d2 u
(61)
2
d du r =
1
1
d✓
du
du
u
2 2d u
˙
l u̇ = = l✓ 2 =
= ll u , 2 .
(62b)
ṙ r̈ ==
(62a)
2
2
dt
d✓
d✓
d✓
d✓ requires the identities
Rewriting the equations of motionuin terms uof dt
thed✓
new variable
2u
2u
d du
d
d
Equation (58a) becomes
2
2
˙ du =✓ l ◆
r̈ =
l1
= 1 l✓d✓
u
.
(62b)
du
2
2 = 1l
2
ṙ = d u
,
(62a)
1
dt2 u̇d✓=
d✓
d✓
u2 + u = u2 dt2d✓2 f
d✓.
(63)
d✓
ml
u
u
2
Equation (58a) becomes
d du
d2 u ✓ ◆
2 2d u
˙
r̈ = d2 ul
= l✓1 2 = 1l u
.
(62b)
2
dt
d✓
d✓
d✓
+ u = 142 2 f
.
(63)
2
d✓
ml
u
u
Equation (58a) becomes
✓ ◆
2
d u
1
1
+
u
=
f
.
(63)
14
d✓2
ml2 u2
u
14
ṙ =
r̈ =
1
1 d✓ du
du
u̇
=
=
l
,
u2
u2 dt d✓
d✓
2
d du
d2 u
2 2d u
˙
l
= l✓ 2 = l u
.
dt d✓
d✓
d✓2
Equation (58a) becomes
d2 u
+u=
d✓2
1
f
ml2 u2
✓ ◆
1
.
u
(62a)
(62b)
(63)
14
This is the di↵erential equation governing the motion of a particle under a central force.
Conversely, if one is given the polar equation of the orbit r = r(✓), the force function can be
derived by di↵erentiating and putting the result into the di↵erential equation. According
to Newtons law of gravitation f (r) = k/r2 , so that
This is the di↵erential equation governing
of a particle under a central force.
d2 u the motion
k
+ uof=the orbit
. r = r(✓), the force function can (64)
Conversely, if one is given the polar equation
be
d✓2
ml2
derived by di↵erentiating and putting the result into the di↵erential equation. According
to Newtons
of gravitation
This
has thelaw
general
solution f (r) = k/r2 , so that
k
d2 u
k
u = Acos(✓
✓
)
+
+ u = 0 2. 2,
2
d✓
ml ml
(65)
(64)
where
A the
andgeneral
✓0 are solution
constants of integration that encode information about the initial
This has
conditions. Choosing ✓0 =0 and replacing u by the original radial coordinate r = 1/u,
k
u = Acos(✓
✓0 ) + ◆ 21,
(65)
✓
k ml
r = Acos✓ +
(66)
2
ml
where A and ✓0 are constants of integration that encode information about the initial
conditions. Choosing ✓0 =0 and replacing u by the original radial coordinate r = 1/u,
which is the equation of a conic section with the origin at the focus. This can be rewritten
✓
◆ 1
in standard form
k
1+✏
r =r =
Acos✓
(66)
r0 + ml2 ,
(67)
1 + ✏cos✓
where
which is the equation of a conic section
with the origin at the
focus. This can be rewritten
2
2
Aml
ml
in standard form
✏=
,
r
=
.
(68)
0
k r = r 1 + ✏ , k(1 + ✏)
(67)
0
1 + ✏cos✓
✏ is called the eccentricity of the orbit:
where
2
2
u = Acos(✓2 ✓0 ) +
,
(65)
2
mlk
d u
+
u
=
.
(64)
2
2
d✓
ml
where A and ✓0 are constants of integration that encode information about the initial
conditions.
Choosing
✓0 =0solution
and replacing u by the original radial coordinate r = 1/u,
This has
the general
✓
◆ 1
k
k
= Acos(✓
✓
)
+
,
r = u Acos✓
+
(66) (65)
0
2
2
ml
ml
and ✓0 ofare
constants
of with
integration
that
information
about
the initial
whichwhere
is theAequation
a conic
section
the origin
at encode
the focus.
This can be
rewritten
conditions.
in standard
formChoosing ✓0 =0 and replacing u by the original radial coordinate r = 1/u,
1+✏
◆ 1
r = r0 ✓
,
(67)
k
1 + ✏cos✓
r = Acos✓ +
(66)
2
ml
where
2
Aml
ml2
,
r0 = origin at
.
(68)
which is the equation of✏ a=conic
k section with thek(1
+ ✏) the focus. This can be rewritten
in standard form
✏ is called the eccentricity of the orbit:
1+✏
r = r0
,
(67)
1 + ✏cos✓
• ✏ = 0 is a circle,
where
2
2
Aml
ml
• ✏ < 1 is an ellipse,
✏=
,
r0 =
.
(68)
k
k(1 + ✏)
• ✏ = 1 is a parabola and
✏ is called the eccentricity of the orbit:
• ✏ > 1 is a hyperbola.
• ✏ = 0 is a circle,
For an elliptical orbit, r0 is the distance of closest approach to the sun, and is called the
• ✏ < 1 is an ellipse,
perihelion. Similarly,
(69)
• ✏ = 1 is a parabola and r1 = r0 (1 + ✏)/(1 ✏)
is the furthest
from the sun and is called the aphelion. Orbital eccentricities are
• ✏ > 1distance
is a hyperbola.
small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though,
Halley’s For
comet
a veryorbit,
eccentric
but, according
toapproach
the definition
not the
an has
elliptical
r0 is orbit
the distance
of closest
to thejust
sun,given,
and isis called
a comet!
perihelion. Similarly,
r1 = r0 (1 + ✏)/(1 ✏)
(69)
is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are
• ✏ = 1 is a parabola and
• ✏ > 1 is a hyperbola.
For an elliptical orbit, r0 is the distance of closest approach to the sun, and is called the
perihelion. Similarly,
r1 = r0 (1 + ✏)/(1 ✏)
(69)
is the furthest distance from the sun and is called the aphelion. Orbital eccentricities are
small for planets, whereas comets have parabolic or hyperbolic orbits. Interestingly though,
Halley’s comet has a very eccentric orbit but, according to the definition just given, is not
a comet!
P
15
A
4.1.3
Third law
4.1.3
Third
lawthird law go back to equation (55). Integrating this area law over time
To prove
Keplers
gives
To
prove Keplers third law go back to equation
Z ⌧ (55). Integrating this area law over time
l⌧
4.1.3 Third law
gives
A(⌧ ) = Z dA = ,
(70)
⌧
2
0
l⌧
To prove Keplers third law go back
equation
A(⌧to) =
dA (55).
= ,Integrating this area law over time
(70)
where ⌧ is the period of the orbit. The area0A of an ellipse
can also be written as A = ⇡ab
2
gives
Z ⌧
where
a
and
b
are
the
semi-major
and
semi-minor
axes,
l⌧ yielding
where ⌧ is the period of the orbit. The area A of an
ellipse
can also be written as A = ⇡ab
A(⌧ ) =
dA = ,
(70)
2
where a and b are the semi-major and semi-minor
axes,
yielding
l⌧ 0
= ⇡ab
(71)
where ⌧ is the period of the orbit. The area
A of an ellipse can also be written as A = ⇡ab
2
l⌧
= ⇡ab axes, yielding
(71)
where a and b are the semi-major and semi-minor
The ratio of a and b can be expressed in 2terms of the eccentricity
l⌧ p
The ratio of a and b can be expressed in
of the eccentricity
= ⇡ab
(71)
b terms
=2 1 ✏2 .
(72)
a
b p
2 . eccentricity
The ratio of a and b can be expressed in=terms
1 of✏the
(72)
Using this expression to substitute b ina (71)
b p
= 2 1p ✏2 .
(72)
Using this expression to substitute b in (71)
2⇡a
a
2
⌧=
1 ✏
(73)
l 2p
2⇡a
Using this expression to substitute b in (71)
⌧=
1 ✏2
(73)
The length of the major axis is
l
2⇡a2 p
⌧=
1 2ml
✏22
(73)
The length of the major axis is
l
2a = r0 + r1 =
.
(74)
k(1 2✏2 )
2ml
The length of the major axis is
2a = r0 + r1 =
.
(74)
k(1 2✏2 )
Squaring (71) and replacing gives
2 m 2ml
4⇡
2a = ⌧r02 +
r
=
(74)
= 1 k(1
a3 , ✏ 2 ) .
(75)
Squaring (71) and replacing gives
k2
4⇡ m 3
2
confirming
Kepler’s
3rd
law.
⌧
=
a ,
(75)
Squaring (71) and replacing gives
k2
4⇡ m 3
⌧2 =
a ,
(75)
confirming
Kepler’s
3rd
law.
4.2 Hamiltonian dynamics of many-body
systems
k
confirming
law.
The
Kepler Kepler’s
problem 3rd
is essentially
a two-body problem. In the remainder of this course, we
4.2
Hamiltonian dynamics of many-body systems
will be interested in classical (non-quantum) systems that consist of N
2 particles. The
Squaring (71) and replacing gives
4⇡ 2 m 3
⌧ =
a ,
k
2
(75)
confirming Kepler’s 3rd law.
4.2
Hamiltonian dynamics of many-body systems
The Kepler problem is essentially a two-body problem. In the remainder of this course, we
will be interested in classical (non-quantum) systems that consist of N
2 particles. The
complete microscopic dynamics of such systems is encoded in their Hamiltonian
N
X
p2n
H=
+ U (x1 , . . . , xN ),
2mn
(76a)
n=1
where mn , pn (t) and xn (t) denote the mass, momentum and position of the nth particle.
The first contribution on the rhs. of Eq. (76a) is the kinetic energy, and U is the potential
energy. For our purposes, it is sufficient to assume that we can decompose (76a) into a sum
of pair interactions
1 X
U (x1 , . . . , xN ) =
2
(xn , xk ).
(76b)
n,k:n6=k
16
Given H, Newton’s equations can be compactly
rewritten as
ẋn = rpncan
H ,be compactly
ṗn = rrewritten
xn H.
Given H, Newton’s equations
as
(77)
That this so-called Hamiltonian
dynamics
ẋn =
rpn H , is indeed
ṗn = equivalent
rxn H. to Newton’s laws of motion
(77)
can be seen by direct insertion, which yields
That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion
pn
can be seen by direct ẋinsertion,
,whichṗyields
(78)
n =
n = mn ẍn = rxn U.
mn
pn
ẋ
=
,
ṗ = mn ẍn = rxn U. certain conservation laws.
(78)
n
An important observation is that
mn many nphysical systems obey
For instance, the Hamiltonian (76a) itself remains conserved under the time-evolution (77)
An important observation is that many physical systems obey certain conservation laws.
X ⇥ (76a) itself remains conserved
⇤ under the time-evolution (77)
For instance, dthe Hamiltonian
H =
(rpn H) · ṗn + (rxn H) · ẋn
dt
⇤
nX ⇥
d
⇤
H =X ⇥ (rpn H) · ṗn + (rxn H) · ẋn
dt =
(79)
n (rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0.
X
⇥
⇤
n
Given H, Newton’s equations can be compactly rewritten as
Given H, Newton’s equations can be compactly rewritten as
ẋn = rpn H ,
ṗn = rxn H.
(77)
Conservation laws
ẋn = r
rxn H.
(77)
pn H ,
n = equivalent
That this so-called Hamiltonian
dynamics
is ṗ
indeed
to Newton’s laws of motion
can be seen by direct insertion, which yields
That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion
can be seen by direct insertion,pwhich
yields
ẋn = n ,
ṗn = mn ẍn = rxn U.
(78)
m
n
p
ẋn = n ,
ṗn = mn ẍn = rxn U.
(78)
m
An important observation is nthat many physical systems obey certain conservation laws.
For instance, the Hamiltonian (76a) itself remains conserved under the time-evolution (77)
An important observation is that many physical systems obey certain conservation laws.
X ⇥(76a) itself remains conserved
For instance, the dHamiltonian
⇤ under the time-evolution (77)
H =
(rpn H) · ṗn + (rxn H) · ẋn
X
⇤
n ⇥
ddt
H = X ⇥(rpn H) · ṗn + (rxn H) · ẋn
⇤
dt
= n (rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0.
(79)
X
⇤
n ⇥
=
(rpn H) · ( rxn H) + (rxn H) · rpn H ⌘ 0.
(79)
which is just the statementnof energy conservation. Other important examples of conserved
quantities are total linear momentum and angular momentum,
which is just the statement of energy conservation. Other important examples of conserved
X
X
quantities are total linear momentum
and
angular
P =
pn ,
L = momentum,
xn ^ pn
(80)
X
X
n
n
P =
pn ,
L=
xn ^ pn
(80)
if the pair potentials only depend
on the distance
n
n between particles.
There exists a deep mathematical connection between such invariants and symmetries of
if the pair potentials only depend on the distance between particles.
the underlying Hamiltonian, known as Noether’s theorem. For example, energy conservation
There exists a deep mathematical connection between such invariants and symmetries of
is a consequence of the fact that the Hamiltonian (76a) is note explicitly time-dependent
the underlying Hamiltonian, known as Noether’s theorem. For example, energy conservation
and, hence, invariant under time translations. Similarly, conservation of linear momentum is
islinked
a consequence
of the fact invariance
that the Hamiltonian
(76a)ofisangular
note explicitly
time-dependent
to spatial translation
and conservation
momentum
to rotational
and,
hence,
invariant
under
time
translations.
Similarly,
conservation
of
linear
momentum
is
invariance.
linked
spatial
translation
invariance
of to
angular
momentum
rotational
Fortothe
remainder
of this
course, itand
willconservation
be important
keep in
mind thatto
microscopic
invariance.
symmetries and conservation laws must be preserved in coarse-grained macroscopic continFordescriptions.
the remainder of this course, it will be important to keep in mind that microscopic
uum
symmetries and conservation laws must be preserved in coarse-grained macroscopic continuum descriptions.
Reflection of underlying symmetry !
(Noether’s theorem)
4.3
Practical limitations
Deriving
Kepler’s limitations
laws required us to solve a second-order linear ordinary di↵erential equa4.3
Practical
tion, which was obtained by considering the idealised case in which a single planet is orbiting
Deriving Kepler’s laws required us to solve a second-order linear ordinary di↵erential equa-
uum descriptions.
4.3
Practical limitations
Deriving Kepler’s laws required us to solve a second-ord
relevant problem in the world contains many more than three degrees of freedom. Indeed a
tion,contains
which
was obtained
by systems
considering
the idealised ca
physical problem …
typically
1023 interacting
particles (Avogadro’s
number), which
too many
particles
in interesting
theitsun.
If we
consider
a more
realistic
problem
is so great a number that
is unclear
if the
mathematical
techniques
described
above arein which
of any use. The central aim of this course is to make theoretical progress towards underteracting with each other via gravity, the problem becom
standing systems with many degrees of freedom. To do so we shall invoke the “continuum
two planets
orbiting
theofsun
one encounters
the
hypothesis”, imaginingfor
thatjust
the discrete
variable (e.g.
the velocity
a particular
molecule
Solution:
continuum
theory
…
leads to(e.g.
new
challenges
of fluid)
can be replaced
with
a
continuum
the
velocity
field v(x, t)).solution.
There are many
which there is no general
analytical
Lagrange s
subtleties that arise in trying to implement this idea, among them;
to this problem if we restrict the planets to move in th
(i) How does one write down macroscopic descriptions in terms of microscopic constants
mass of one of them is so small as to be negligible. In
in a systematic way? It would be terrible to have to solve 1023 coupled di↵erential
to the ‘three body problem’ one must use ideas from 1
equations!
equations
investigate
their
stability.
However,
now
(ii) Forces and e↵ects
that a prioriand
appear
to be small are
not always
negligible.
This
turns out to be of
fundamental
was complicated
not recognised universally
number
of importance,
equationsbutfor
things until
to happen
the 1920’s.
(iii) The mathematics of how to solve ‘macroscopic equations’, which are nonlinear partial
di↵erential equations, is non-trivial. We will need to introduce many new ideas.17
In tackling these problems we will spend a lot of time doing fluid mechanics, the reason being
that it is by far the most developed field for the study of these questions. Experiments are
readily available and the equations of motion are very well known (and not really debated!).
Furthermore, fluid dynamics is an important subject in its own right, being relevant to many