equations of motion for any holonomic system. Only a single scalar function of the
generalised coordinates and velocities, the kinetic energy and generalised force are
needed to use these n equations.
It is unnecessary to find the constraint forces. Newton’s laws were not used
explicitly but they are contained within equation (6.20). The advantage is that the
number of unknowns equals the number of degrees of freedom, unlike for Newtonian
vector mechanics in which the constraint forces are additional unknowns and more
equations must be solved.
If the forces are conservative forces we can eliminate the need to find FG explicitly.
6.4 Conservative forces and the definition of the Lagrangian
If the non-constraint forces are conservative (net work around some path is zero)
then by definition there exists a potential energy function V (q1 , ...., qn ).
Fi = −∇V(r1 , ..., ri , ..., rN )
(6.21)
∂V
for example the x component of this force is Fi,x = − ∂x
.
i
Equation (6.21) can be integrated to calculate the work done by a change in the
qa :
Z X
XZ
W =
Fi · dri = −
∇V(r1 , ..., ri , ..., rN ) · dri
i
=−
i
XZ
a
qa (1)
qa (0)
X
i
∂ri
∇V ·
∂qa
!
dqa = −
XZ
a
qa (1)
qa (0)
∂V
dqa
∂qa
(6.22)
In this last line we changed from viewing V as a function of the individual parts
to viewing the same potential as a function of the generalised coordinates. This is
only possible for holonomic constraints. The work is only a function of the end points
of the motion denoted by qa (0) and qa (1) in the above formula. This work does not
depend on the specific initial conditions or on the actual path qa (t), q̇a .
We can now integrate equations (6.11) and use (6.12) to obtain
Z X
W =
FG δqa .
(6.23)
a
Comparing this to (6.22) and noting that we can exchange the sum and integral signs
we can conclude that the generalised force is:
FG = −
∂V
∂qa
(6.24)
Therefore if there are only conservative forces and constraint forces we can bypass
the virtual work method and calculate the generalised force directly from the partial
– 30 –
derivative of the potential energy expressed in terms of the generalised coordinates:
V (q1 , ..., qn ).
Since we have assumed that the potential V is velocity independent, ∂∂Vq̇a = 0 we
can define the Lagrangian L ≡ T − V , substitute this into the general form of the
equation of motion to obtain the Euler-Lagrange equations in their final form:
d ∂L
∂L
−
= 0.
(6.25)
dt ∂ q̇a
∂qa
This is the second and most useful form of the equations of motion. The number of
these equations equals the number of degrees of freedom n. Thus we always have
exactly the minimum number of equations needed to solve the problem. Furthermore
we have reduced the entire problem of solving the motion to finding a single scalar
function. Of course, the equations of motion still need to be solved, but that is a
mathematics problem. The physics is entirely contained within equation (6.25).
6.4.1 Kinetic energy revisited
The required transformation of T and V from Cartesian coordinates is obtained by
applying the transformation equations (5.3,6.14)
T =
X1
i
2
X1
mi vi2 =
2
i
∂ri
q̇j +
∂qj
∂t
X ∂ri
mi
j
!2
.
(6.26)
If you carry out the expansion it is easy to find the general expression for T in
generalised coordinates:
T = M0 +
X
Mj q̇j +
j
1X
Mjk q̇j q̇k ,
2 j,k
(6.27)
where M0 , Mj , Mjk are functions of r’s and t and hence of the q 0 s and t.
T = M0 +
X
Mj q̇j +
j
1X
Mjk q̇j q̇k ,
2 j,k
(6.28)
where the constants M are definite functions of the r’s and t and hence of the q’s
and t:
X 1 ∂ri 2
M0 =
mi
(6.29)
2
∂t
i
Mj =
X
i
mi
∂ri ∂ri
·
∂t ∂qj
– 31 –
(6.30)
Mjk =
X
mi
i
∂ri ∂ri
·
.
∂qj ∂qk
(6.31)
Thus the kinetic energy can always be written as the sum of three homogeneous
functions of the generalised velocities T = T0 + T1 + T2 where T0 is independent
of the generalised velocities, T1 is linear in the velocities and T2 is quadratic in the
velocities. If the system is scleronomous (time independent) then T is always a
quadratic function of the generalised velocities since the first two terms in equation
(6.27) vanish.
6.4.2 How to get rid of ignorable coordinates
If only the time derivative of a particular coordinate appears in the Lagrangian,
but not the coordinate itself, we say that this coordinate is ignorable or cyclic. For
example the position of a free particle is ignorable because it does not appear in the
Lagrangian which is simple 1/2mv 2 . Assuming only that the time derivative of a
coordinate q appears in the Lagrangian then the Euler-Lagrange equations are
d ∂L
= 0,
(6.32)
dt ∂ q̇a
which means that
p≡
∂L
∂ q̇
(6.33)
is a constant of the motion i.e. is constant in time as the motion of the system
proceeds. p is called the canonically conjugate momentum to the coordinate q.
(Quantum mechanics is founded on the notion of canonical conjugate momentum.)
It is possible to eliminate ignorable coordinates from the modified Lagrangian,
called the Routhian R ≡ L − pq̇ with n − 1 degrees of freedom.
6.5 Summary
The concepts of virtual displacement and virtual work were used to derive a general
equation of motion for holonomic systems of either type, scleronomic or rheonomic.
The Euler-Lagrange equations follow under the assumptions of conservative forces.
A single scalar function called the Lagrangian contains all of the information about
the possible motions of the system. The generalised coordinate system being chosen
to best fit the particular problem under study. Most importantly the system as a
whole is solved rather than the combined motion of the parts.
We have made some assumptions along the way:
• Constraints are holonomic: f (r1 , ..., rN , t) = 0
P constraint
• Constraint forces do no work:
fi
· δri = 0
– 32 –
• Applied forces are conservative: Fi = −∇V
• Potential V does not depend on q̇a such that
∂V
∂ q̇a
=0
Note also that within our new generalised coordinate system, the Lagrangian
L = L(q1 , ...qn ; q̇1 , ...q̇n ) where each of the generalised velocities are independent of
each other. This is one of the initially confusing features of Lagrangians, but it
works!
Euler found these equations in 1760 in a pure mathematics context. They were
first applied to physics by Lagrange 100 years after Newton. Lagrangian dynamics is
not a new theory, nor does it contain new physics. Whereas the Newtonian approach
is based on external forces acting on a body, the Lagrangian method utilises only
quantities associated with the body (the kinetic and potential energies) and the
concept of force does not appear. This is important because Energy is a scalar
quantity and thus the Lagrangian is invariant to coordinate transformations.
These equations contain Newtonian dynamics and more. We will see that conservation of energy is a direct consequence of having a time independent Lagrangian
function: Physical laws are invariant to time translations. Conservation laws such as
momentum or angular momentum follow from symmetries of the Lagrangian function.
– 33 –
6.5.1 Reference frames
The kinetic and potential energies must be evaluated in an inertial reference frame
where Newton’s laws hold. However, the variables that we choose to express T and
V can be variables with respect to any frame, even a non-inertial reference frame.
Consider a ball thrown upwards in a moving elevator accelerated with a = aelevator .
A person in an inertial frame will see the height of the ball obey the Lagrangian
1
L = mẏ 2 − mgy
2
(6.34)
with an equation of motion from Lagrange ÿ = −g with solution
1
y(t) = vo t − gt2 ,
2
(6.35)
where vo is the initial velocity of the ball. Now express the motion as seen from the
non-inertial frame of the person in the elevator. Let ỹ be the height of the ball then
the differential equation for the motion is
d2 ỹ
= −(g + a)
dt2
(6.36)
1
ỹ(t) = vo t − (g + a)t2 .
2
(6.37)
with a physical trajectory
If you simply calculate T and V using the non inertial frame variables the answer will
be incorrect since we derived the Lagrangian formalism from Newton’s laws, which
are only valid in inertial frames. Therefore we must continue to use the Lagrangian
from equation (6.34) and substitute into this expression the non inertial variables
T = 1/2m(ỹ˙ + at)2 = 1/2mẏ 2
V = mg(ỹ + 1/2at2 ) = mgy
(6.38)
which will give the correct equation of motion for either y(t) or ỹ(t). (Note that you
2
would get the wrong answer using the incorrect T = 1/2mỹ˙ and V = mg ỹ.
6.6 A simple example: Atwood’s machine
A massless string connects two weights over a frictionless pulley. This is a similar
system to that used in many funicular railways where a counterweight, or 2nd train
moves up the track whilst the other weight, or train, moves down.
6.6.1 Newtonian solution
– 34 –
The tension T is the same on both sides of the pulley.
The forces acting on the weights are F = T − mg = ma and
F = M g − T = M A.
P
The total force F = (M g − T ) + (T − mg) = g(M − m).
P
From Newton’s 2nd law
F = (M + m)ẍ, so we find
M −m
(M + m)ẍ = g(M − m) or ẍ = g
.
M +m
6.6.2 D’Alembert’s virtual work solution
Imagine a virtual displacement in which M moves downwards
a distance δx and m moves upwards the same displacement.
The applied force, gravity, does work M gδx on M and
−mgδx on m.
Figure 5: Atwood’s
The acceleration of M is ẍ downwards and of m is ẍ up- machine
wards.
The inertial force on M is ẍ upwards and the inertial work done on M is −M ẍδx.
The inertial force on m is ẍ downwards and the inertial work done on m is
−mẍδx.
D’Alembert’s principle gives (M − m)gδx − (M + m)ẍδx = 0.
M −m
Solving this for the acceleration gives ẍ = g M
.
+m
6.6.3 Euler-Lagrange solution
Atwood’s machine is a conservative system with holonomic, scleronomous constraints.
There is one independent coordinate, x, the position of the other weight given by
the constraint that the distance between the weights is l, the length of the string.
The potential energy is V = −M gx − mg(l − x).
The kinetic energy is T = 12 (M + m)ẋ2 .
The Lagrangian has the form L = T − V = 12 (M + m)ẋ2 + M gx + mg(l − x).
Therefore there is only one equation of motion involving the derivatives
∂L
= (M − m)g
(6.39)
∂x
∂L
= (M + m)ẋ
∂ ẋ
which we can substitute into the Euler-Lagrange equation
d ∂L
∂L
−
= 0.
dt ∂ q̇a
∂qa
(6.40)
(6.41)
to get (M + m)ẍ − (M − m)g = 0 and thus
M −m
ẍ = g
.
(6.42)
M +m
Note that the tension T does not appear in the non-Newtonian analyses of this
problem.
– 35 –
6.7 Changing Coordinate Systems
We shall now show that Lagrange’s equations hold in any coordinate system. In fact,
this follows immediately from the action principle, which is a statement about paths
and not about coordinates. But here we shall be a little more pedestrian in order to
explain exactly what we mean by changing coordinates, and why it’s useful.
Let’s rewrite the positions of N particles with coordinates ri as xA where A =
1, . . . 3N . Then Newton’s equations read
ṗA = −
∂V
∂xA
(6.43)
where pA = mA ẋA . Recall that the number of degrees of freedom of the system is said
to be 3N . These parameterise a 3N -dimensional space known as the configuration
space C. Each point in C specifies a configuration of the system (i.e. the positions
of all N particles). Time evolution gives rise to a curve in C.
Let
qa = qa (x1 , . . . , x3N , t)
(6.44)
where we’ve included the possibility of using a coordinate system which changes with
time t. Then, by the chain rule, we can write
q̇a =
dqa
∂qa A ∂qa
=
ẋ +
dt
∂xA
∂t
(6.45)
In this equation, and for the rest of this course, we’re using the “summation convention” in which repeated indices are summed over. Note also that we won’t be too
careful about whether indices are up or down - it won’t matter for the purposes of
this course. To be a good coordinate system, we should be able to invert the relationship so that xA = xA (qa , t) which we can do as long as we have det(∂xA /∂qa ) 6= 0.
Then we have,
ẋA =
∂xA
∂xA
q̇a +
∂qa
∂t
(6.46)
Now we can examine L(xA , ẋA ) when we substitute in xA (qa , t). Using (6.46) we
have
2 A ∂L
∂L ∂xA
∂L
∂ x
=
+
(6.47)
∂qa
∂xA ∂qa
∂ ẋA ∂qa ∂qb
while
∂L ∂ ẋA
∂L
=
∂ q̇a
∂ ẋA ∂ q̇a
– 36 –
(6.48)
We now use the fact that we can “cancel the dots” and ∂ ẋA /∂ q̇a = ∂xA /∂qa which
we can prove by substituting the expression for ẋA into the LHS. Taking the time
derivative of (6.48) gives us
2 A
d ∂L
d
∂L
∂xA
∂L
∂ x
∂ 2 xA
=
+ A
q̇b +
(6.49)
dt ∂ q̇a
dt ∂ ẋA ∂qa
∂ ẋ
∂qa ∂qb
∂qa ∂t
So combining (6.47) with (6.49) we find
∂L
d
∂L
∂L ∂xA
q̇a −
=
− A
∂qa
dt ∂ ẋA
∂x
∂qa
(6.50)
Equation (6.50) is our final result. We see that if Lagrange’s equation is solved in
the xA coordinate system (so that [. . .] on the RHS vanishes) then it is also solved in
the qa coordinate system. (Conversely, if it is satisfied in the qa coordinate system,
so the LHS vanishes, then it is also satisfied in the xA coordinate system as long as
our choice of coordinates is invertible: i.e det(∂xA /∂qa ) 6= 0).
– 37 –