Principles Of Least Action
Contents
• Fermat’s Principle of least time
• The Euler-Lagrange equations
• Light in vacuum and in media
• Lagrangian dynamics and Hamilton’s principle
• Symmetries and conservation laws
Learning Outcomes
• Know the Euler Lagrange equations and how to apply them
• Know Fermat’s Principle of least time
• Be able to formulate and determine the Euler Lagrange eqns for the path travelled by light in media
• Know Hamilton’s Principle
• Know the Lagrangian for a particle in a potential and be able to determine the
Euler Lagrange equations in simple problems
• Be able to explain the connection between ignorable coordinates and conserved
quantities
Reference Books
• PH211 Classical Mechanics notes.
• Analytical Mechanics, GR Fowles and GI Cassiday.
• Classical Mechanics, TL Chow.
• Perfect Form, DS Lemons.
1
We are going to explore an alternative formulation of classical mechanics which
at first sight appears very different from Newton’s Laws. It is a formalism that grew
out of optics and will allow us to study an area of mathematics called “calculus
of variation”. Of course it must turn out to be the same as Newton’s laws. This
alternative formalism makes some problems easier to solve but more importantly it
will give us new insights into conservation laws.
1
Optics
Our starting point will be to think about the path that light travels by. In these
enlightened times we might start from Maxwell’s equations and derive a wave equation
with light waves as solutions to determine how the light propagates. Before this
technology though Fermat proposed
Fermat’s Principle of Least Time: Light propagates between two points so as to
minimize its travel time
Thus for example in a uniform medium where the speed of light c is a constant
the minimum time of travel
d
(1)
c
is given by the path of shortest distance d ie a straight line. This is still a perfectly
good (if limited) description of light.
We can obtain more interesting results by thinking about media where the speed
of light changes.
t=
1.1
Snell’s Law
Consider two neighbouring regions of space in which light travels at different speeds
v1 , v2 - for example a glass air interface. We will be interested in the light that travels
from the point (x1 , y1 ) in the first medium to the point (x2 , y2 ) in the second
(x1, y1)
material 1 - v1
d1
θ1
material 2 - v2
(x, 0)
θ2
d2
(x2, y2)
2
In any one medium light travels in a straight line but in this case we have some
choice in where the light crosses between the media. Lets consider the arbitrary
crossing point (x, y = 0). The time of travel is
T [x] =
=
d1
v1
√
+
d2
v2
(x−x1 )2 +y12
v1
+
√
(2)
(x−x2 )2 +y22
v2
We now want to find the path (ie the value of x through which it passes) which
minimizes the time taken. Thus
dT
(x − x1 )
(x2 − x)
= q
− q
=0
dx
v1 (x − x1 )2 + y12 v2 (x2 − x)2 + y22
(3)
This equation though is just
sin θ1
v1
=
sin θ2
v2
(4)
which is Snell’s law.
In terms of index of refraction which is defined, relative to the vacuum, as
n1 =
c
v1
(5)
n1 sin θ1 = n2 sin θ2
1.2
(6)
Complicated Problems
We can imagine more complicated problems than that above where the index of
refraction is an arbitrary function of position. For example consider light moving in
a plane where the speed of the light is v(x, y)
y
xa
xb
x
Different paths are described by different functions y(x). The time to travel along
an arbitrary little piece of path is
3
distance
∆T =
=
velocity
√
dx2 + dy 2
v(x, y)
(7)
Summing such contributions up along a path gives the total time of travel
T [y(x)] =
Z
xb
xa
v
u
u
t
dy
1
1+
v(x, y)
dx
!2
dx
(8)
Now we want to find the path y(x) that gives the minimum time. This is the sort
of problem that Calculus of Variation is designed to address.
Exercise 1: To remind yourself about partial differentiation if a function is defined
by
T = a(t) b(t)3 ḃ(t) t10
where the dot indicates a derivative with respect to t. Give expressions for
∂T
,
∂t
2
∂T
,
∂b
∂T
,
∂ ḃ
dT
dt
Calculus of Variation
The mathematical method will be more widely applicable than the optics problems
so far discussed so we will use a more general notation in this section.
Consider a set of curves q(s) between two points (q1 , s1 ) and (q2 , s2 ) in the s, q
plane (we will only consider curves where the trajectory is single valued at each value
of s).
q
q2
q1
s1
s2
S
Imagine we are interested in one curve that minimizes the quantity
S[q(s)] =
Z
s2
s1
4
L(q, q̇, s)ds
(9)
L is just a number at each point on a given curve determined by the values of q and
s at that point and the gradient q̇ = dq/ds. The integral sums these numbers along
the line.
If the curve that minimizes S is q̄(s) we can write the other curves as deviations
from it
q(s) = q̄(s) + δq(s)
(10)
subject to the boundary conditions
δq(s1 ) = δq(s2 ) = 0
(11)
The value of S for these curves varies from the value for q̄(s) by
δS = S[q̄ + δq] − S[q̄]
(12)
Since q̄(s) is the minimum though δS = 0 to lowest order in δq.
Let’s calculate S[q̄ + δq] to order δq
S[q̄ + δq] =
'
R s2
s1
L(q̄ + δq, q̄˙ + δ q̇, s)ds
R s2 s1
L(q̄, q̄˙ , s) + δ q̇ ∂L
+ δq ∂L
+ .... ds
∂ q̇
∂q
' S[q̄] +
R s2 s1
(13)
+ δq ∂L
ds + O(δq 2 )
δ q̇ ∂L
∂ q̇
∂q
Integrating the second term by parts (u = ∂L/∂ q̇, dv/ds = δ q̇ etc)
Z
s2
s1
"
∂L
∂L
= δq
δ q̇
∂ q̇
∂ q̇
# s2
s1
−
Z
s2
s1
d
δq
ds
!
∂L
ds
∂ q̇
(14)
The first term vanishes since δq vanishes at the ends of the path.
Thus
S[q̄ + δq] − S[q̄] = −
Z
s2
s1
!
!
d
δq
ds
∂L
∂L
−
ds + ...
∂ q̇
∂q
=0
(15)
This is only zero (at order δq) if
d
ds
∂L
∂ q̇
−
∂L
∂q
(16)
The Euler Lagrange equation.
Exercise 2: Work through the above derivation in the case where L depends on two
coordinates q and p. What two equations must then be satisfied by the minimizing
curve?
5
3
More Optics
Let’s return to the problem of the path light travels in a plane where the speed of
light depends on the position in the plane. We had found that the time taken to
traverse a path is
T [y] =
Z
xb
xa
L(y, ẏ, x) dx
(17)
where
1 q
1 + ẏ 2
L(y, ẏ, x) =
v(x, y)
(18)
This is of the form that leads to the Euler Lagrange equation
d
dx
!
∂L
∂L
=0
−
∂ ẏ
∂y
(19)
Lets look at a couple of examples
3.1
Light in Vacuum
In vacuum the speed of light is a constant so v(x, y) = c
The Euler Lagrange equation is
d
dx
ẏ
√
1 + ẏ 2
!
=0
(20)
ẏ
= constant
1 + ẏ 2
(21)
ẏ = constant, m
(22)
y = mx + c
(23)
Integrating this gives
√
The only solution of this is
or integrating
ie a straight line. This is our first example of the solution of the Euler Lagrange
equation giving the path that minimizes T . m and c are determined by the initial
and final position of the light.
6
3.2
Light in the Atmosphere
In the atmosphere the air temperature and density change with height resulting in
the speed of light depending on height - v(h). Equivalently we can write the refractive
index n(h) with
v(h) =
c
n(h)
(24)
Our result for the length of time light takes to travel some path h(x) can be
written as an optical path length
cT [h] =
Z
x2
x1
q
L = n(h) 1 + ḣ2
dxL,
(25)
We can use the fact that L is independent of x to simplify the Euler Lagrange
equation as follows. Note that
dL
∂L
∂L ∂L
ḣ +
ḧ
(26)
=
+
dx
∂x
∂h
∂ ḣ
The first term on the right is zero. Now replace ∂L
using the Euler Lagrange equation
∂h
∂L
d
=
∂h
dx
∂L
∂ ḣ
!
(27)
and we find
!
∂L
∂L
ḣ +
ḧ
∂ ḣ
∂ ḣ
dL
d
=
dx
dx
(28)
which is just
"
#
d
∂L
L − ḣ
=0
dx
∂ ḣ
(29)
which gives us
∂L
= constant, D
(30)
∂ ḣ
Note that this is only a first order equation rather than the second order Euler
Lagrange equation so is simpler to solve.
In our problem, using the explicit form for L above we have
L − ḣ
q
which simplifies to
n 1 + ḣ2 − q
n
q
ḣ2 n
1+
=D
=D
1 + ḣ2
7
(31)
ḣ2
(32)
Note that the physical meaning of D is the value of the index of refraction at the
point where the light ray becomes horizontal so that ḣ = 0.
Squaring and rearranging we find
dh
=
dx
s
n2
−1
D2
(33)
Thus
x − x0 =
Rh
h0
q dh
(34)
n2
−1
D2
Explicit Example: Consider a ray of light that begins moving horizontally ( ḣ = 0)
at h = 0 in an atmosphere where
n(h) = n0 − λh
(35)
where λ is some constant. We must solve the integral
Z
x=
dh
q
(n0 −λh)2
D2
This can be done by changing variables to
(36)
−1
n0 − λh = D cosh φ
(37)
The integral becomes
x=−
Z
D
D
dφ = − φ + c
λ
λ
(38)
Returning to the original coordinates and requiring the boundary conditions ḣ(x =
0) = 0 and h(x = 0) = 0 gives the result
h=
n0
(1
λ
− cosh λx
)
n0
(39)
• When λ is positive n(h) decreases with altitude - this is what normally happens in
the atmosphere. Plotting the solution we find the form
h
(0,0)
apparent light path
to observer
x
Thus if we look up at the Empire State building it will appear taller than it
actually is.
8
• If there is a temperature inversion then λ is negative so n(h) increases with altitude.
Plotting the solution we find the form
h
x
(0,0)
We see “the sky on the ground” - a mirage.
Exercise 3:
(a) Consider a fibre optic cable lying in the z direction. The cable is made of glass with
index of refraction n(r), where r is the radial distance from the centre of the cable.
Working in cylindrical coordinates (r, θ, z) show that Fermat’s Principle implies light
travels on the path minimizing the quantity
Z
z2
z1
f (r(z), θ(z), r 0 (z), θ 0 (z)) dz =
Z
z2
z1
q
n(r) r0 2 + r2 θ0 2 + 1 dz.
where a prime indicates differentiation with respect to z. z1 and z2 are the zcoordinates of the end points of the path.
(b) If a light ray initially has θ 0 = 0 show, from the appropriate Euler Lagrange
equation, that the θ independence of f implies the path followed by the light is
described by a constant value of θ.
(c) Use the z independence of f to deduce that the first order differential equation
for rays travelling paths with constant θ is
f−
4
∂f 0
r = constant.
∂r0
First Integrals
We have seen two situations in which the Euler Lagrange equation simplifies from a
second order equation to a first order equation. These will be important when we
come on to Newtonian dynamics. Lets review them:
1) If L(q, q̇, s) is independent of the coordinate q
d
ds
∂L
∂ q̇
!
=0
(40)
So
∂L
∂ q̇
= constant
9
(41)
2) If L(q, q̇, s) is independent of the coordinate s
dL
∂L
∂L
=
q̇ +
q̈
ds
∂q
∂ q̇
(42)
using the Euler Lagrange equation gives
!
∂L
∂L
+
q̈
∂ q̇
∂ q̇
d
dL
=
ds
ds
(43)
which is just
"
#
d
∂L
L − q̇
=0
ds
∂ q̇
(44)
L − q̇ ∂L
∂ q̇ = constant
(45)
so that
Exercise 4: A smooth curved wire connects the origin to the lower point (x1 , y1 ).
A bead on the wire slides without friction from rest at the upper to the lower point
under the influence of gravity. It’s mechanical energy is conserved as it moves along
the wire. Choose down to be the positive y direction.
(a) Show that the time, T, required for the bead’s journey is
T =
√1
2g
R x2
0
r
0
(1+y 2 )
y
dx.
(b) Given that the integrand of the above integral is independent of x show that the
curve y(x) making T stationary satisfies the differential equation
dy
dx
=
r
(b−y)
y
(c) Change the dependent variable from y to φ where y = b sin2 φ/2 and show that
the above can be integrated to give the brachistochrone
x = b/2(φ − sin φ)
10
5
Newtonian Dynamics
We have seen that the motion of light can be described by a “principle of least time”.
Is there an equivalent rule that would describe the motion of a particle in Newtonian
dynamics? There is and it is enshrined as
Hamilton’s Principle: A particle travels by the path between two points that minimizes the Action.
We need to know what the “action” is. Let’s write it first for one dimensional
motion. The action is
S[path] =
R tb
ta
L(x, ẋ, t)dt
(46)
where the dot indicates differentiation with respect to the time, t. L is known as the
Lagrangian and is given by
L = kinetic energy − potential energy = T − V
(47)
From our calculus of variation result we know that the path that minimizes the
action satisfies the Euler Lagrange equation
d
dt
∂L
∂ ẋ
−
∂L
∂x
=0
(48)
We can now check to see if any of this makes sense (!). For a non-relativistic
particle in a one dimensional potential we have
1
L = T − V = mẋ2 − V (x)
2
The Euler Lagrange equation is therefore
∂V
d
(mẋ) +
=0
dt
∂x
which is Newton’s second law since
F =−
∂V
∂x
(49)
(50)
(51)
Note that the momentum of the particle is given by
p = mẋ =
11
∂L
∂ ẋ
(52)
5.1
Multiple Coordinates
We will want to solve problems in more than one dimension. The formalism is easily
adapted. The definition of the Action above in terms of the Lagrangian (L = T − V )
remains the same. If now we have several coordinates
qi
i = 1...n
(53)
(so for example we might call x = q1 , y = q2 z = q3 etc). In the derivation above of the
Euler Lagrange equation (section 2) we would have to take into account deviations
in the path in all of these coordinates. We would find that the change in the action
of a path close to the minimizing path would have the form
∆S = −
Z
s2
s1
X
δqi
i
d
ds
∂L
∂ q̇i
!
∂L
−
∂qi
!
ds
(54)
At the minimum the coefficients of each δqi must vanish independently so we get
a set of Euler Lagrange equations - one associated with each coordinate
d
ds
5.1.1
∂L
∂ q̇i
−
∂L
∂qi
=0
(55)
Generalized Coordinates
The reason that we have written the coordinates so generally as qi rather than for
example using x, y, z is that in some problems these are not the appropriate coordinates because of a constraint. A simple example to illustrate this is a ball on a wire
hoop
y
θ
x
The hoop stops the ball moving in the radial direction so the ball cannot be at
any arbitrary (x, y). The sensible coordinate to use is the angle θ.
Such a reduced set of coordinates are called generalized coordinates.
5.1.2
Generalized Momentum
A generalization of the idea of momentum can be defined in the spirit of (52). The
generalized momentum associated with a generalized coordinate is given by
pi =
12
∂L
∂ q̇i
(56)
5.2
Example: Projectile Motion
Consider the familiar problem of a projectile in a uniform gravitational field
y
x
We can obtain the normal Newtonian equations of motion from the Euler Lagrange
equations. We need expressions for the kinetic and potential energy of the system so
we can build the Lagrangian. The kinetic energy is just
1
1
T = mẋ2 + mẏ 2
2
2
(57)
V = mgy
(58)
and the potential energy
So the Lagrangian is just
1
1
L = T − V = mẋ2 + mẏ 2 − mgy
(59)
2
2
Now we find the two Euler Lagrange equations. The first associated with the x
coordinate is
d
dt
!
∂L
∂L
−
=0
∂ ẋ
∂x
(60)
which gives
mẍ = 0
(61)
The second equation associated with the y coordinate is
d
dt
!
∂L
∂L
−
=0
∂ ẏ
∂y
(62)
which gives
ÿ = −g
(63)
The two boxed equations are the standard Newtonian equations of motion.
Hopefully you’re starting to see the power of this technique now - the kinetic and
potential energies of a system are fairly easy to work out and then we just do some
maths. There’s not all that resolving forces business! The next problem is an example
that would be very hard by the standard methodology.
13
5.3
Example 2: Double Pendulum
Consider a double pendulum of the form
θ1
l1
v1
m1
l2
θ2
m2
v1 + v2
It would be pretty hard work to determine all the forces in play here. However,
the Lagrangian technique means we only have to calculate the energies of the two
masses to get to the equations of motion.
The first mass has a velocity ~v1 with magnitude l1 θ̇1 (v = ωr). The second
mass has both this motion plus a second contribution from the swing of the second
pendulum ~v2 with magnitude l2 θ̇2 . The total velocity of the second mass is therefore
~vtot = ~v1 + ~v2
so
(64)
2
vtot
= (~v1 + ~v2 ).(~v1 + ~v2 )
(65)
= (l1 θ̇1 )2 + (l2 θ̇2 )2 + 2l1 θ̇1 l2 θ̇2 cos(θ2 − θ1 )
where θ2 − θ1 is the angle between ~v1 and ~v2 .
Thus the total kinetic energy of the system is
h
i
1
1
T = m1 l12 θ̇12 + m2 (l1 θ̇1 )2 + (l2 θ̇2 )2 + 2l1 θ̇1 l2 θ̇2 cos(θ2 − θ1 )
2
2
The potential energy is determined by the heights of the masses
V = −m1 gl1 cos θ1 − m2 g(l1 cos θ1 + l2 cos θ2 )
(66)
(67)
and the Lagrangian is
L=T −V
(68)
There are two Euler lagrange equations - one associated with θ1
d
dt
h
i
m1 l12 θ̇1 + m2 l12 θ̇1 + m2 l1 l2 θ̇2 cos(θ2 − θ1 ) − m2 l1 l2 θ̇1 θ̇2 sin(θ2 − θ1 )
+(m1 + m2 )gl1 sin θ1 = 0
14
(69)
and one with θ2
i
d h
m2 l22 θ̇1 + m2 l1 l2 θ̇1 cos(θ2 − θ1 ) + m2 l1 l2 θ̇1 θ̇2 sin(θ2 − θ1 ) + m2 gl2 sin θ2 = 0 (70)
dt
These are pretty messy (but that was the point!). Things simplify a bit if we
assume that both θ1 and θ2 are small and expand to linear order. We then get
(m1 + m2 )l12 θ̈1 + m2 l1 l2 θ̈2 = −(m1 + m2 )gl1 θ1
m2 l22 θ̈2
(71)
+ m2 l1 l2 θ̈1 = −m2 gl2 θ2
These coupled equations in fact have normal mode solutions of the form
θ̈1 = −ω 2 θ1
(72)
2
θ̈2 = −ω θ2
ie the two pendulums oscillate with the same frequency.
To find ω you can try substituting in the form of the solution in (72) into (71).
You’ll find two simultaneous equations for θ1 and θ2 with two solutions. You’ll find in
one case θ1 /θ2 is positive and in the other it is negative. So in one case the pendulums
swing together and in the other case in opposite directions.
Exercise 5: If a system with generalized coordinate q has the Lagrangian
1
L = q̇ 2 − q 3
2
what is the Euler Lagrange equation describing the system?
Exercise 6: If a system with generalized coordinates ξ and ψ has the Lagrangian
1
L = ξ˙2 + cos ξ ψ̇ − ξeψ
2
what are the Euler Lagrange equations describing the system?
Exercise 7:
(a) Show that for a non-relativistic, free particle of mass m travelling with constant
velocity v the action S describing its motion reduces to
S = mvd/2
where d is the distance travelled. This was a form for the action proposed by Maupertuis who believed it reflected the simplicity and economy of the Creator-God....
(b) Consider such a particle rolling on a table in the x, y plane with speed v1 . Along
the y-axis there is a height discontinuity in the table which the particle can move over
15
at the cost of potential energy which reduces its velocity to v2 . If the particle starts
at (x1 , y1 ) to the left of the y axis and ends to the right at (x2 , y2 ) show that the
action for it passing across the y-axis at arbitrary y (assuming it travels in a straight
line except when it crosses the y axis) is given by
q
q
S = mv1 x21 + (y − y1 )2 + mv2 x22 + (y − y2 )2
By minimizing the action deduce the relation
v1 sin θ1 = v2 sin θ2
where the angles are the angles between the particle’s direction of motion and the x
axis before and after it crosses the y axis. Contrast this result with Snell’s Law for
light.
Exercise 8:Two blocks of equal mass M are connected by a flexible string of length
`. One block is placed on a smooth horizontal table and the other block hangs
over the edge. Using the length z of string hanging over the edge as a generalized
coordinate, write down the Lagrangian and use the Euler–Lagrange equation to find
the acceleration of the hanging mass in the following cases:
(i) the mass of the string is negligible,
(ii) the string is heavy with mass m distributed uniformly along it.
6
Conservation Laws
Finally lets look at one of the most surprising pieces of insight to come out of the
Lagrangian formalism - that is a deeper understanding of conservation laws. Mathematically this will just be to repeat the section previously on First Integrals but in
the dynamics arena we will see a new interpretation.
6.1
Ignorable Coordinates
If the Lagrangian does not depend on some coordinate qi it is called an ignorable
coordinate. Then ∂L/∂qi = 0 and it’s associated generalized momentum is conserved
as we can see from the Euler Lagrange equation
d
dt
∂L
∂ q̇i
!
−
dpi
∂L
=0
=
∂qi
dt
(73)
so
pi = constant
(74)
This is clearly a mathematical fact but there is a deeper interpretation. If L only
depends on q̇i not qi itself then we can shift
16
qi → qi + const
(75)
and leave the Lagrangian, L, (and hence the physics) invariant. This is a symmetry
- translation invariance in the qi direction.
Thus we learn that the true relation is
symmetry
conserved momentum
→
This is a new insight we have not seen before in Newtonian mechanics.
6.2
Energy Conservation
The second example of a first integral above was the case where L does not depend
explicitly on t. This implies that a quantity known as the Hamiltonian is conserved
H=
X ∂L
i
∂ q̇i
q̇i − L
(76)
To prove that it is conserved we explicitly calculate
dH X d
=
dt
i dt
!
X ∂L
X ∂L
X ∂L
X ∂L
∂L
q̇i +
q̈i −
−
q̇i −
q̈i = 0
∂ q̇i
i ∂ q̇i
i ∂t
i ∂qi
i ∂ q̇i
(77)
= 0.
using the Euler Lagrange equations and ∂L
∂t
In simple systems the Hamiltonian is just the total energy of the system as we
can see for example in one dimension where
1
L = mẋ2 − V (x)
2
(78)
so using the definition above
1
(79)
H = mẋ2 + V (x)
2
In conclusion here we have learnt that time translation invariance implies energy
conservation.
6.3
Example - Central Forces
Consider a particle moving subject to a central force ie in a potential V (r)
r
F
θ
Ο
17
The kinetic energy of the particle is
1
1
1
T = m(ẋ2 + y˙2 ) = mṙ2 + mr2 θ̇2
2
2
2
(80)
thus
1
1
L = mṙ2 + mr2 θ̇2 − V (r)
2
2
There is a Euler Lagrange equation associated with the r coordinate
d
dt
!
∂L
∂L
−
=0
∂ ṙ
∂r
(81)
(82)
giving
∂V
∂r
Plus a second equation for θ, which since L is independent of θ, is just
mr̈ = mrθ̇2 −
d
(mr2 θ̇) = 0
dt
which tells us that angular momentum is conserved.
The Hamiltonian is also conserved and is given here by
1
1
H = mṙ2 + mr2 θ̇2 + V
2
2
(83)
(84)
(85)
which is the total energy.
Exercise 9: If a system with generalized coordinates x and y has the Action
Z 1 2 1 2
ẋ + ẏ + cos ẏ − x dt
S=
2
2
what quantities are conserved?
6.4
Hamiltonian and Energy
Finally it is worth stressing that the Hamiltonian is not always the energy of the
system. As an example consider a bead on a hoop that is being rotated, as shown,
by a torque
ω
a
θ
18
m
θ and t describe the position of the bead so are good generalized coordinates. The
kinetic energy is given by
1
T = m(a2 θ̇2 + a2 sin2 θω 2 )
2
(86)
V = −mga cos θ
(87)
and the potential energy by
Thus
1
L = m(a2 θ̇2 + a2 sin2 θω 2 ) + mga cos θ
2
Since L does not depend on t the Hamiltonian is conserved. In particular
(88)
1
1
H = ma2 θ̇2 − ma2 sin2 θω 2 − mga cos θ
(89)
2
2
Although H is conserved the total energy of the system is not since to keep the hoop
rotating the torque must be applied doing work on the system.
19
Section C
This question is both a mock exam question and part of your course
work. You should hand in your answer to be marked to my pigeon hole
by friday the 24th February 2006.
A uniform rod of mass m1 hangs horizontally from two fixed supports by equal
lengths l of string. A bead of mass m2 can slide freely on the rod. The system, shown
below, only moves in the vertical plane.
θ
l
m2
m1
(a) Show that the Lagrangian describing the system is given by
1
1
L = m1 l2 θ̇2 + m2 (ẋ2 + 2lẋθ̇ cos θ + l2 θ̇2 ) + (m1 + m2 )gl cos θ
2
2
where θ is the angle of the strings to the vertical and x the distance of the bead
from an end of the rod.
[5]
(b) Write down the Euler-Lagrange equations for the system.
[4]
(c) Which coordinate is ignorable and why? What is the corresponding conserved
quantity?
[4]
(d) If the system is released from rest (θ̇ = ẋ = 0) at an angle θ = α then the
resulting motion is described by
l(m1 + m2 sin2 θ)θ̇2 = 2g(m1 + m2 )(cos θ − cos α)
Show, by differentiating with respect to t or otherwise, that this solution is
consistent with the Euler-Lagrange equations.
[7]
20