Download In this lecture we`ll discuss a very important concept or object

I.
THE PRINCIPLE OF VIRTUAL WORK
In this lecture we’ll discuss a very important concept or object, named generalized coordinates. This is essentially
what makes lagrangian dynamics so useful and great.
Basically, these generalized coordinates are coordinates that are in some sense the most natural ones for the system
in question. The choice of such coordinates is due to geometrical properties of the space where the system is free to
move, and this space is determined by what’s called constraints.
A.
Constraints
To understand what a constraint is, from the intuitive point of view, is really not very complicated. We’re in three
dimensional euclidean space. A free particle can move everywhere. Now, imagine that you impose a condition for the
movement of this particle: it has to move only in certain region of space, say, only on the surface of the sphere of
constant radius R. We’ve just constrained the movement of this particle. Before, it was possible to find this particle
at any point (x1 , x2 , x3 ). Now, you can find it only at points (x1 , x2 , x3 ) so that the coordinates are related by the
equation (x1 )2 + (x2 )2 + (x3 )2 = R2 . Essentially, we’re saying that instead of the three
coordinates
we need only two
of them, since we can write the third in terms of the others, e.g. (x3 )2 = R2 − (x1 )2 + (x2 )2 . We reduced the
number of degrees of freedom os the system from 3 to 2, by fixing one constraint equation1
f (x1 , x2 , x3 ) = (x1 )2 + (x2 )2 + (x3 )2 − R2 ≈ 0
Another example would be a particle constrained to move on a line. In this case tit’s pretty obvious that you’ll reduce
the number of degrees of freedom of the system from 3 to 1. Indeed, only one coordinate is needed to describe the
motion of a particle on a line: the distance from the origin.
Generally, this kind of constraint if given by a function that relates the coordinates of the system. If we have N
particles, and for each particle a we denote ~xa its position vector, then there could be K < N constraint equations
fI (~x1 , ~x2 , . . . , ~xN , t) = 0
where
I = 1, · · · , K < 3N
You can notice that we included a possible time dependence on the constraint equation. What does it means? It’s
simple: it says that the region where the particles are constrained to move is changing. For instance, if it’s a line, the
line is vibrating, or something like. If it’s a surface, it could be waving, like the surface of earth during an earthquake.
Or, if it’s our universe, it’s expanding.
Notice that this way the surface2 moves is independent of the way the particles on it are moving. In fact, we could
say that the surface movement depends on the way the particles move: this is more or less general relativity! It would
be a feedback reaction.
The constraints described in the way above are called holonomic3 .
You can wonder what are the other types of constraints. We’ll not get into details about them, since they’ll not be
studied. Just to mention: an non-holonomic constraint is given by a relation between not only the coordinates but
also between the velocities. There’re also the constraints that restrict the particles not to a specific place but encode
them into an volume (in the generic sense, so that it could be a surface, or something else). This kind of constraint
is given by an inequality, for instance fI (~x1 , ~x2 , . . . , ~xN , t) < 0.
1
2
3
Do not concern about the symbol used here; “≈ ”. This is the correct way to talk about constraints? , but here this is not needed.
Or line, or anything else. From now, we call surface any subspace of the 3 dimensional euclidean space. The correct name would be
hypersurface or hyperspace.
“It is remarkably hard to find the etymology of holonomic (or holonomy) on the web. I found the following (thanks to John Conway
of Princeton): ’I believe it was first used by Poinsot in his analysis of the motion of a rigid body. In this theory, a system is called
“holonomic” if, in a certain sense, one can recover global information from local information, so the meaning ”entire-law” is quite
appropriate. The rolling of a ball on a table is non-holonomic, because one rolling along different paths to the same point can put it
into different orientations. However, it is perhaps a bit too simplistic to say that “holonomy” means “entire-law”. The “nom” root has
many intertwined meanings in Greek, and perhaps more often refers to “counting”. It comes from the same Indo-European root as our
word ‘number”.”
S.Golwala
2
B.
The work done by a constraint force
As was mentioned in the lecture, constraint forces are responsible for work on the system. Now, we’ll see how it’s
done, and moreover, how this is related to the energy.
Consider a very simple situation: a single particle moving on a surface, like the sphere given before. The constraint
equation is f (~x, t) = 0. Let us write down the Newton equation but leaving explicitly who is external forces (like
~
friction of gravity, or something else) and who is the constraint force C.
¨ = F~ + C
~
m~x
It’s obvious that there must be this constrain force, otherwise who would be keeping the particle on the surface? The
problem is that we don’t know which force is this. There’re lots of possibilities for it. Any two forces parallel to the
surface (at each point) can be summed up to give the constraint force such that the movement satisfies f (~x, t) = 0.
In order to take this arbitrariness out we fix that the constraint force is not given by a force parallel to the surface
but for a force orthogonal to it in each point. Precisely this force is obtained from the geometry of the surface by
~ (~x, t) as long as it doesn’t vanish on the surface. So, we have
taking its gradient ∇f
~ = λ∇f
~ (~x, t)
C
for λ a number or even a function of time.
Let us consider a conservative force - or, if you want to say it in a more impressive way: a force such that the
integrability condition is satisfied!
~ × F~ = 0
∇
⇒
~
F~ = −∇V
Then, Newton’s law becomes
¨ = −∇V
~ + λ∇f
~
m~x
It’s a well-known trick that when we have the acceleration, we can get the kinetic energy by taking the inner product
with the velocity. So, let us do it!
¨ · ~x˙ = −∇V
~ · ~x˙ + λ∇f
~ · ~x˙
m~x
d 1 ˙2
~ · ~x˙ + λ∇f
~ · ~x˙
m~x
= −∇V
dt 2
Now the R.H.S is trivial. It can be written in terms of total derivatives with respect to time of the scalar functions:
dV
~ · ẋ + ∂V
= ∇V
dt
∂t
∂f
df
~ · ẋ +
= ∇f
dt
∂t
Then
d 1 ˙ 2
dV
∂V
df
∂f
m~x = −
+
+λ −λ
dt 2
dt
∂t
dt
∂t
If f (~x, t) is satisfied, then
df
dt
= 0, which lead us to
dE
d 1 ˙2
∂V
∂f
m~x + V =
=
−λ
dt 2
dt
∂t
∂t
Now, as usual, let us consider potentials that do not depend on time. This means that the energy changes due to the
change of the surface:
dE
∂f
= −λ
dt
∂t
On the other hand, an energy change means that some work was done over the system. Of course, this work was done
here by the surface, or, by the constraint force.
3
C.
Generalized coordinates
As we saw, the inclusion of constraints in the system reduce its degrees of freedom. If there’re K constraints
equations we get 3N − K degrees of freedom. Let us associate to each of these degrees of freedom a coordinate q α ,
where α = 1, 2, . . . , 3N − K. These are not necessarily the usual xa coordinates, but eventually some of them can be
the same.
With that let us write down the xia coordinates in terms of the coordinates q α :
x1
x2
...
x3N
x1 (q 1 , . . . , q 3N −K , t)
x2 (q 1 , . . . , q 3N −K , t)
...
x3N (q 1 , . . . , q 3N −K , t)
=
=
=
=
One must notice that by doing this we’re in fact writing down xi so that the constraint equation is automatically
satisfied. That’s why we say that when xi are written in terms of the q α , we’re solving the constraints.
This becomes pretty clear when we, again, take the single particle on an sphere as the example. The constraint
equation is, as we saw, f1 (x1 , x2 , x3 ) = (x1 )2 + (x2 )2 + (x3 )2 − R2 . It’s just this single equation, so that K = 1. Then,
the number of degrees of freedom goes from 3 to 3 − 1 = 2. Then, we introduce the q α (α = 1, 2) coordinates named
generalized coordinates.
Now, the final step is to write x1 , x2 and x3 in terms of these generalized coordinates. How do we do it?
First of all, keep in mind the the constrain equation must be satisfied. Then the functions xi (q α ) are such that
xi xi − R2 vanishes. Notice that by taking
x1 (q 1 , q 2 ) = R sin q 2 cos q 1
x2 (q 1 , q 2 ) = R sin q 2 sin q 1
x3 (q 1 , q 2 ) = R cos q 2
we do satisfy the constraint equation. Commonly we call q 1 = θ and q 2 = φ.
Now, take the equations of motion
¨ = F~ + λ∇f
~
m~x
~ , this equation cannot be solved because we don’t know
The quantity λ is arbitrary, so although we have F~ and ∇f
λ. On the other hand we can eliminate it by considering the inner product between this equation and the so called
virtual displacement δ~x.
Let us define precisely what this virtual displacement is. We consider an instantaneous infinitesimal4 displacement
from the original position such that it’s consistent with the constraint. This means that the components xia are not all
independent from each other: they’re related by the constraint equation. In other words, this means that we solved
the constraint, and therefore we have xi in terms of the generalized coordinates xi (q 1 , . . . , q 3N −K ). With this, we can
write
δxi =
∂xi α
δq ≡ ∂α xi δq α
∂q α
(1)
Remember that there’s a sum over the index α from 1 to 3N − K For example, in considering the previous example
of the particle on the sphere, we have that the virtual displacements in directions x, y and z are related to the
infinitesimal variations of the angles θ and φ through the relations
δx = R cos φ cos θδφ − R sin φ sin θδθ
δy = R cos φ sin θδφ + R sin φ cos θδθ
δz = −R sin φδφ
Notice that θ and φ are the natural coordinates to describe things on the sphere with constant radius. These
coordinates live on the sphere, or, more precisely, on the tangent space of the sphere in each point. We shall come
back to this later. The important thing now is to realize that when working with these generalized coordinates the
bigger space (the 3-dimensional space) where the surface is embedded is not needed anymore. This is the key point
concerning generalized coordinates.
4
Remember that an infinitesimal is something that although non-vanishing is smaller than anything in the scale (of distance) of the
system you’re considering.
4
With a little effort you can write the displacement vector δ~r in terms of the unit vectors θ̂ and φ̂. Do it as an
exercise.
Going on, take the inner product of the equation of motion with the virtual displacement:
¨ − F~ − λ∇f
~
m~x
· δ~x = 0
Since the virtual displacement is on the tangent plane of the surface in each point, it will be always orthogonal to
~ , and consequently the last term vanishes, remaining
∇f
¨ − F~ · δ~x = 0
m~x
Now we have this sum, that in components can be written as
mẍi − F i δxi = 0
Here comes the crucial point: the term inside the parenthesis cannot be made equal to zero. Let us see why. The thing
is that δxi are not all independent. They’re constrained as explained above. For instance, in the case of the particle
on the sphere we have xδx + yδy + zδz = 0, which gives the relation between one coordinate with another. What
we’re trying to say is that we could make the term inside the parenthesis to be zero only if δxi were all independent
from one another. Then we would have separately that mẍi − F i = 0 for i = 1, 2, 3.
Why is this happening? One must remark that these vectors are not linearly independent because they’re too much.
The equation above is the projection of the Newton’s law onto the surface where the particle is constrained to move.
So, we should have a description of the dynamics in terms of the natural coordinates (generalized coordinates) of
this surface, and consequently, with the right number of degrees of freedom. The solution to this problem is however
simple: consider the virtual displacement in terms of the generalized coordinates, i.e., substitute (1) into the above
equation
mẍi − F i ∂α xi δq α = 0
The second term seems ok. In fact we can define Qα ≡ F i ∂α xi , which has dimension of force when q α stands for
length, or dimension of torque, when q α stands for some angle. To be brief, let us call it “generalized force”.
The first term must be written in a better way.
mẍi ∂α xi δq α = m
dẋi
d i
d
∂α xi δq α = m
ẋ ∂α xi − mẋi
∂α xi
dt
dt
dt
Using, in the last term, the fact that the time derivative and the derivative with respect to q α commute, which is
easily verified - and it was showed in the lecture - we get
mẍi ∂α xi δq α = m
dẋi
d i
∂α xi δq α = m
ẋ ∂α xi − mẋi ∂α ẋi
dt
dt
i i
Now we can write this last term simply as ∂α ( m
2 ẋ ẋ ) = ∂α T , where T stands for the kinetic energy. For the first
term we use one more trick.
The total time derivative of xi (q α (t)) is
dxi
∂xi
dq α
=
+ ∂α xi
dt
∂t
dt
from where we conclude that the functional dependence of the velocities is given by ẋi = ẋi (q α , q̇ α , t).
Now, consider the derivative of this velocity with respect to the generalized velocity q̇ α . We have
∂xi
∂ ẋi
= α
α
∂ q̇
∂q
This is the result we’ll use above to write
mẍi ∂α xi δq α = m
d i ∂ ẋi ẋ
− ∂α T
dt
∂ q̇ α
5
This can be once more simplified as
mẍi ∂α xi δq α =
d ∂T
d m ∂(ẋi ẋi ) − ∂α T =
− ∂α T
α
dt 2 ∂ q̇
dt ∂ q̇ α
Finally we get the desired result for the equation of motion
!
d ∂T
− ∂α T − Qα δq α = 0
dt ∂ q̇ α
Now, since δq α are independent from each other, the term multiplying it must be equal to zero:
d ∂T
− ∂α T = Qα
dt ∂ q̇ α
(2)