2.3 Perturbation theory
is convenient in the case where ∂f /∂y = 0, because then
one immediately gets the first integral
The exact solution of the problems of mechanics is usually difficult, if not impossible. Often it is possible to get
∂f
= constant.
(9)
accurate enough solutions using a simpler perturbation
∂ ẏ
theory. It means that the solution is developed in Taylor
series with respect to some parameter, and only the lowest Often one encounters the case where ∂f /∂x = 0. For that
order terms in the expansion are calculated. This very ge- case it is convenient to change Euler’s equation to an alneral procedure is demonstrated by the following example. ternative form
d
∂f
∂f
Example We study the effect of air resistance on an
−
f − ẏ
= 0.
(10)
∂x dx
∂ ẏ
object thrown. For simplicity we assume the resistance
force is directly proportional to the velocity of the object, In the case that ∂f /∂x = 0 one gets the first integral
Fresistance = −kv. The equation of motion is
∂f
= vakio.
(11)
f − ẏ
mr̈ = −mgŷ − k ṙ,
(1)
∂ ẏ
where the time derivative is marked with a dot. The so- The proof of the second form (10) is obtained by direct
lution of this is of the form r(t, k). We assume that the calculation
effect of air resistance is relatively small, so we can expand
d
∂f
∂f
−
f − ẏ
r(t, k) as Taylor series with respect to k:
∂x dx
∂ ẏ
r(t, k) = r0 + r1 + r2 + . . . ,
(2)
∂f
∂f
∂f
∂f
∂f
d ∂f
=
−
−
ẏ −
ÿ + ÿ
+ ẏ
∂x ∂x
∂y
∂ ẏ
∂ ẏ
dx ∂ ẏ
where the subindex gives the order of the term in k (ri ∝
d
∂f
∂f
i
k ). We substitute this into the equation motion (1). This
= ẏ
−
= 0.
(12)
dx ∂ ẏ
∂y
has to be satisfied in all orders of k separately. In zeroth
order we get
Example 3. Minimal surface of revolution
mr̈0 = −mgŷ.
(3)
Let us find the surface of minimal area that is obtained
This is the equation of motion without air resistance. It can by rotation of a curve around the x axis; for example a
be integrated, and we get for appropriate initial conditions soap film between two rings. One slice of that has the area
p
(starting velocity V )
2πyds = 2πy 1 + ẏ 2 dx. For simplicity we leave the constant factor 2π. Therefore we have to minimize the integral
1
(4)
x0 (t) = Vx t, y0 (t) = Vy t − gt2 .
2
Zx2 p
A = y 1 + ẏ 2 dx.
(13)
Collecting the first order terms in (1) we get the equation
mr̈1 = −k ṙ0 .
x1
(5)
p
Because f = y 1 + ẏ 2 does not explicitly depend onx,
we get the first integral form the second form of Euler’s
equation (11):
Integrating this we get the corrections
k 1
1 3
k
2
2
Vx t , y1 (t) = −
Vy t − gt . (6)
x1 (t) = −
2m
m 2
6
C = f − ẏ
Based on this we can deduce that the object hits ground
(y = 0) at time
t=
2Vy
2
−
g
3
kVy2
mg 2
+ ...
=p
(7)
2Vx Vy
8 kVx Vy2
−
+ ...
g
3 mg 2
y
1 + ẏ 2
.
(14)
The general solution of this is (calculate as an exercise)
x−b
y(x) = a cosh
.
(15)
a
and the length of the throw
x=
p
∂f
y ẏ
= y 1 + ẏ 2 − ẏ p
∂ ẏ
1 + ẏ 2
(8)
y
a
3.0
All these are obtained as series expansion with respect to
k. Calculate the intermediate steps as an exercise.
2.5
4.2 Calculus of variations
2.0
1.5
Second form of Euler’s equation
1.0
Euler’s equation
0.5
∂f
d
−
∂y
dx
∂f
∂ ẏ
=0
x-b
(??)
-2
1
-1
0
1
2
a
1
− √ ln
2
√
p
2 − 2 − (y/a)2
√
.
( 2 − 1)y/a
(18)
y
a
1.0
0.8
0.6
The minimal surface of rotation realized by a soap film.
Source
http://www.funsci.com/fun3 en/exper2/exper2.htm
0.4
0.2
θ
x
Example 4. This example is slightly more demanding
a
0.5
1.0
1.5
and it uses variational notation, that is introduced in the
following section. Therefore read this example only after
The shape for the liquid surface for contact angles θ = 0
that.
and θ = 40◦ .
We study the shape of a liquid surface near a vertical
5 Two body problem
solid wall. We assume that the energies related to different
interfaces per area are: liquid-gas σ, liquid-solid σl , and
Closing of the orbits
gas-solid σg . We claim that this problem is described by
We obtained that in Kepler’s potential V (r) = −k/r all
the functional
orbits
are closed when E < 0, i.e. they return to the their
Z ∞ p
1
2
starting
point after one revolution. Next we show that this
2
E =
dx σ 1 + ẏ + gρy
2
is an exceptional case for a general potential V (r).
0
+ σl y(0) − σg y(0),
(16)
E
where y(x) is the surface height and E is the energy density
Veff
(the energy divided by the length that is perpendicular to
E>0
the x−y that we are considering here). The wall is at x = 0,
and as boundary conditions we have y(x → ∞) = 0. As an
0
E<0
r
exercise, justify that the energy (16) describes the problem
stated above and the second term in the integral descriV
bes the change of gravitational potential energy caused by
raising the liquid to height y(x) (ρ is the density of the
It is obvious from the figure that the system has a circuliquid). Show that by for the variation of E one gets f
lar orbit corresponding to the minimum of the effective
!
Z ∞
potential (??)
d
ẏ
δE =
dx −σ p
+ gρy δy
dx 1 + ẏ 2
l2
0
Veff (r) =
+ V (r).
(19)
!
2mr2
ẏ(0)
+ σl − σg − σ p
δy(0). (17)
The radius r0 and the energy E0 of the circular orbit are
1 + ẏ 2 (0)
obtained from the conditions
For the purpose of E having an extremal value with respect
dVeff
0
to δy(0), the expression in the parenthesis on the second
≡ Veff
(r0 ) = 0, E0 = Veff (r0 ).
(20)
dr
|r0
line should vanish. We define a contact angle θ so that
ẏ(0) = − cot θ, we get the condition σg = σl + σ cos θ. In addition we require that the second derivative V 00 (r ) >
eff 0
Interpret this as the equilibrium of three surface tension 0 in order for the orbit to be stable.
forces in the y direction. The integral term
p in (17)1 is of
We study a small perturbation to the circular orbit. This
the same type as above. Because f = σ 1 + ẏ 2 + 2 gρy 2
means
that E is slightly larger than E0 . We apply the orbit
does not depend explicitly on x, it is simplest to use the
formula
(??)
second form of the Euler’s
equation. Show that this gives
√
the relation y(0) = a 1 − sin θ between the contact
p angle
Zr
l
dr
and the height of the surface at the wall. Here a = 2σ/gρ.
√
φ = φ0 ± √
.
(??)
2
With some extra effort one can also make the integration
r E − Veff
2m
r0
and get
x − x0
a
=
1−
Based on this we determine the change in the polar angle
∆φ that takes the particle from the minimum radius rmin
p
2 − (y/a)2
2
to the maximum radius rmax :
l
∆φ = √
2m
rZmax
r2
rmin
√
dr
.
E − Veff
(21)
rmax
r0
∆φ
rmin
We expand Veff as Taylor’s series. The linear term vanishes and we get
1 00
E − Veff = E − E0 − Veff
(r − r0 )2 + . . .
2
(22)
From the zeros of this we get
s
rmax,min = r0 ±
2(E − E0 )
.
00
Veff
(23)
For small E − E0 we can approximate r2 ≈ r02 in the prefactor of the square root in (21). Making the change of
variables
s
2(E − E0 )
r − r0 = s
(24)
00
Veff
we get
l
∆φ = 2 p
00
r0 mVeff
Z1
−1
√
ds
πl
= 2p
.
00 (r )
2
r0 mVeff
1−s
0
(25)
00
Because for a general potential r0 and Veff
(r0 ) are independent of each other, in most cases ∆φ 6= π, in contrast to
Kepler’s potential.
Exercise: study the potential
V = arn+1 ⇔ F = −a(n + 1)rn r̂
and show
∆φ = √
π
.
3+n
(26)
(27)
It follows from this that ∆φ = π for Kepler’s potential (n =
−2), which result we know from above. Another special
case is harmonic potential (n = 1), for which ∆φ = π/2.
Also in this case the result is an ellipse, but now the center
of the force is at the center of the ellipse (not at a focus).
3