Course summary
18.354
w
simplest Kepler’s
to derive, and
is
a
statement
tha
problem
er a central force, such as gravity, is con
a particle with mass m and velocity u is
dr
L=r⇥m ,
dt
ition of the particle. The rate of change
dL
= r ⇥ f = r ⇥ f (r)r̂ = 0,
dt
where we have ignored higher order derivatives in making the approximati
Now consider a single box with boundaries at x
/2 and x + /2. In a
Jx (x
/2, t)A⌧ particles will enter from the left and Jx (x + /2, t)A⌧ pa
through the right boundary. The number of particles in the box increases
Random walks & diffusion
N (x, t + ⌧ )
N (x, t) = [Jx (x
/2, t)
Jx (x + /2, t)] A
By dividing both sides by A ⌧ the number of particles per unit volume in
is seen to increase at the rate1
n(x, t + ⌧ )
⌧
In the limit ⌧ ! 0 and
n(x, t)
=
/2, t)]
@Jx
@2n
= D 2,
@x
@x
For strictly one-dimensional systems, the boundary area is just a point and, hence,
7
Mark Haw
David Walker
Jx (x
! 0, this becomes
@n
=
@t
1
[Jx (x + /2, t)
e V is
Z
Z d/dt inside the integral sign we must take
account
pndS =
rpdV.
(7) of this. The Reyno
⇢u · ndS,
⇢dV.so, and it can be shown that for a deforming,
(1)incompressible flu
does
V
Z
S
Z
S(t) are being deformed by the motion of the fluid, so if we want
to take the
d
Du
eaving
this volume through
the
bounding
surface
S
is
where u(x, t) is the velocity of the fluid
and=n is the
norm
⇢udV
⇢ outward
dV
ntegral sign we must take account of this. The Reynolds transport
theorem
dt V (t)
V (t) Dt
Z
Z
Z
can be shown that for a deforming, incompressibleZfluid
element
@⇢
⇢u
·
ndS,
Here
dV =
⇢u · ndS(2)
=
r · (⇢u)d
Z
Z
S
V @t
S
V
d
Du
Hydrodynamics
D
@ (8)
⇢udV =
⇢
dV
= have
+ (u · r)
Dt
(t)
V (t)
the velocity ofdttheV fluid
and
is
thehold
outward
normal.
Hence
we
Thisn must
for any
arbitrary
fluid
element
dV , thus
Dt
@t
Z
Z
Z
is called the convective derivative, and
@⇢we shall discuss it’s significance
@⇢
+ r · (⇢u)
dV =
⇢u · assuming
ndS = thatr
(3) = 0.
F ·is(⇢u)dV.
solely given by gravity,
@t
D S@
V @t
V
=
+ (u · r)
Z
Z(9)
Dt
@tThis is called the continuity equation.
Du
⇢
dV =
( rp + ⇢g)dV
for any arbitrary fluid element dV , thus
V (t) Dt
(t)
For fluids like water, the density
does not Vchange
very much and
vective derivative, and we shall discuss it’s significance in a moment. Hence,
@⇢ to neglect the density variations. If we make this approximation
F is solely given by gravity,
+ r Since
· (⇢u)this
= 0.must hold for any arbitrary fluid element
(4) we arrive at
to the incompressibility condition
@t reduces
Z
Z
Du
Du (10)
rp
⇢
dV
=
(
rp
+
⇢g)dV
=
+ g.
e continuity Vequation.
Dt
Dt r · u⇢ = 0.
(t)
V (t)
e water, the density does not change very much and we will often be tempted
This,
combined
with
equation
(4),
constitutes
Eu
hold forvariations.
any arbitrary
fluid
element
we
arrive
at the the
ensity
If we
make
this
approximation
thecontinuity
continuity
equation
Like
all
approximations,
this
one
is sometimes
very
good andthesom
can be tidied up a little if we realise that the gravitational force, being
ncompressibility condition
will
have to figure out where it fails.
Du
rp
= written
+ g.as the gradient of a scalar potential r
(11). It is therefore usual t
Dt
p⇢+
! p. This implies that gravity simply modifies
the pressure di
r6.1.2
·u
= 0.Momentum
(5)
equations
and does nothing to change
the velocity. However, we cannot do this
with the the continuity equation
(4),have
constitutes
the Euler
equations.
Things
if
we
a
free
surface
(as
we
shall
see
later with
water waves).
mations,
this
one
is
sometimes
very
good
and
sometimes
not
so
good.
We
So
far
we
have
more
unknowns
than
equations
(three
velocity co
p a little if we realise that the gravitational
force,
being
conservative,
can
be
Assuming the density is constant means we now have four equatio
Dimensional analysis
Elasticity
Calculus of variations
What is the di↵erential equation satisfied by Y (x)? To answer this question, we compute
What is
di↵erential
equation satisfied by Y (x)? To answer this question, we compute
thethe
functional
derivative
he functional derivative I[Y ]
1
= lim {I[f (x) + ✏ (x y)] I[f (x)]}
✏!0
I[Y ] Y
1 ✏
Z
= lim x2{I[f
I[f (x)]}
@f (x) + ✏ (x @f y)]
0
Y
=✏!0 ✏
Z x2x1 @Y (x y) + @Y 0 (x y) dx
@f 0
Z x2 @f
=
(x y) dx
@f (x d y)
@f+
0
= x1
(x y)dx.
@Y
@Y
0
Z x2x1 @Y dx @Y
@f
d @f
= the Euler-Lagrange equations
(x y)dx.
Equating this to zero, yields
0
@Y
dx @Y
x1
@f
d @f
Equating this to zero, yields the Euler-Lagrange
equations
@Y
dx
@Y 0
0 =
(17)
(17)
(18)
It should be noted that the condition I/
@f Y = d0 alone
@f is not a sufficient condition for
0 might
= even indicate 0 a maximum. It is often possible, (18)
a minimum. In fact, the relation
@Y
dx @Y
however, to convince oneself that no maximum exists for the integral (e.g., the distance
t should
noted that
thebecondition
I/asYwe=like),
0 alone
is not
a sufficient
condition for
alongbea smooth
path can
made as long
and that
our solution
is a minimum.
To be rigorous,
however,
one should
also consider
the possibility
that the minimum
is merely
minimum.
In fact,
the relation
might
even indicate
a maximum.
It is often
possible,
a local
minimum, oneself
or perhaps
the no
relation
I/ Y =
0 indicates
a point
of inflexion.
owever,
to convince
that
maximum
exists
for the
integral
(e.g., the distance
It is easy
to check
thatmade
Eq. (18)
yieldsasthe
equations
long a smooth
path
can be
as long
weNewton
like), and
that our solution is a minimum.
Surface tension
Singular
perturbations
tary di↵erential equation (Acheson, pp.
2
↵erential equation
2
d u
du
✏ 2+
= 1.
dx
dx
might argue that we can neglect this term, the solut
u = x + C.
onsider the full problem the solution is
e,the
using
the Cauchy-Riemann
we see that
Z-plane
uniform flow pastequations
a more wing-like
shape? (Note that we have
d some technical details here,
such2as the requirement that dF/dz 6= 0 at any
2
@
@
@
@
would cause+a blow-up
of the +
velocity).= 0.
=
(6)
2
2
@x
@y
@x@y @y@x
Conformal mappings
le
conformal
maps
e used
for . We
can therefore consider any analytic function (e.g.,
he real
map
is and imaginary parts and both of them satisfy Laplace’s equaZ = F (z) = z + b,
(12)
onents u and v are directly related to dw/dz, which is conveniently
onds to a translation. Then there is
dw
@
@
=
+Zi = F=
u= ze
iv.i↵ ,
(7) (13)
(z)
dz
@x
@x
consider
at anangle
angle↵.↵ to
The
corresponding
onds to auniform
rotationflow
through
In the
this x-axis.
case, the
complex
potential for
i↵ . Using the above relation,
wast
= au0cylinder
ze i↵ . In
this
case
dw/dz
=
u
e
0
making angle ↵ with the stream is
u0 cos ↵ and v = u0 sin✓↵.
◆
2
R past
i
mine the complex
potential
for
since
we
know
that
i↵ flow
i↵ a cylinder
W (Z) = u0 Ze
+
e
ln Z.
(14)
Z
2⇡
✓
◆
R2
=
u
r
+
cos
✓,
(8)
i↵
0
s expression could also include
the term ln e = i↵ which I have neglected.
r
constant however and doesn’t change the velocity.
al
the complex
potential
re part
is theofnon-trivial
Joukowski
transformation,
✓
◆
2
2
completely controls the dynamics. In the process of deriving this result we
rn about a rather remarkable phenomenon in rotating fluid dynamics.
Rotating flows
The Taylor-Proudman theorem
r a fluid rotating with angular velocity ⌦. The equation of motion in the fram
e rotating with the fluid is
@u
1
+ u · ru + ⌦ ⇥ (⌦ ⇥ r) =
rp⌦ + ⌫r2 u
@t
⇢
r · u = 0.
2⌦ ⇥ u,
re two additional terms: the first ⌦ ⇥ (⌦ ⇥ r) is the centrifugal acceleration, w
e discussed before. This can be thought of as an augmentation to the pre
tion, using the identity
Taylor columns,
1 etc
⌦ ⇥ (⌦ ⇥ r) =
2
r(⌦ ⇥ r)2 .
rth, we will simply absorb this into the pressure by writing
p = p⌦
⇢
r(⌦ ⇥ r)2 .
Solitons
KdV equation
Active matter
FIG. 12: (a) Hedgehog defect. Magnets have no line defects (you can’t lasso a basketball), but do have point defects.
⃗ (x) = M0 x̂. You can’t
Here is shown the hedgehog defect, M
surround a point defect in three dimensions with a loop, but
you can enclose it in a sphere. The order parameter space, remember, is also a sphere. The order parameter field takes the
enclosing sphere and maps it onto the order parameter space,
wrapping it exactly once. The point defects in magnets are
categorized by this wrapping number: the second Homotopy
group of the sphere is Z, the integers.
(b) Defect line in a nematic liquid crystal. You can’t
lasso the sphere, but you can lasso a hemisphere! Here is the
defect corresponding to the path shown in figure 5. As you
pass clockwise around the defect line, the order parameter ro-
Two parallel defects can coalesce
each one individually is stable: ea
the sphere, and the whole loop ca
Π1 (RP 2 ) =
than the 1-D curve we used in th
You might get the impression t
is really just seven strength 1 de
You’d be quite right: occasional
but usually big ones decompose
doesn’t mean, though, that addi
gives a bigger one. In nematic li
defects are as good as none! M
line defects: a loop in real space
thing it can’t smooth out. Forma
group of the sphere is zero: you c
For a nematic liquid crystal, tho
ter space was a hemisphere (figu
on the hemisphere in figure 5 th
by twisting and stretching. It d
but you have to remember that t
on the equater really represent t
tation. The corresponding defect
which rotates 180◦ as the defect
shows one typical configuration (
fect). Now, if you put two of thes
cancel. (I can’t draw the pictures,
lenging exercise in geometric visua
defects add modulo 2, like clock ar
school:
Topological defects