Start today by briefly recapitulating where we were in the last class, we have seen that
even though the propositions of Newton. Where they laid the basic frame work for
deriving differential equations, there are a some practical difficulties and we identified
that difficulties. .
Let us enumerate them one by one, the difficult number 1 was that in a given system
mainly constraint system there will be constraint forces, just to recapitulate what the
constraints, where there were two types of constraints, holonomic constraints and non
holonomic constraints. And the holonomic constraints were those, which can be
expressed as a algebraic equation, why the non holonomic constraints are those, which
can be expressed as in equation. So, that was what we did and then we are said
that in order for the body to follow the constraint or constraint surface, the constraint
surface has to apply some kind of a force on it. So, there will always be some force
acting on the body the constraint force and in
.order to write the Newtonian equation for each body, you will have to take in account
all the force acting on it and therefore, the
constraint forces will have to be considered. So, number 1 was the constraint forces have
to be considered, so that was one difficulty, because the constraint forces are not always
easily quantifiable. Number 2, difficulty was
that in a inter connected system various body will interact with each other with forces
and these forces are in general produced by springs. If two bodies are connected by
springs, this body will be applying a force on the other body by means of this spring,
if two bodies move against each other by means
of some kind of frictional element, the one body will apply a force on the other body.
So, for each body this forces have to be considered and all this forces are vectors, so that
was another difficulty that we the formulation becomes very complicated. So, all the
given forces are vectors and we would like to simplify the matter by writing down the
differential equations in terms of things that are scalars, one would be comfortable
with that. The third problem was, that we have
seen that in case of a system with holonomic constraint for example, the pendulum, it is
completely stupid to write three differential equations, because it does not have it does
not move in all those directions. So, in general if there are n total number
of configuration coordinates no not n thrice n, if
there are n bodies there will be three times of n number of configuration coordinates.
But it is somewhat unintelligent to write all
the thrice n equations, rather we should be able to
write thrice n minus h number of equation, where h is the number of holonomic
constraints. So, the number of equations while the number actually required, where h is
the number of holonomic constraints. So, these where the three major difficulties with
the Newtonian formalism, that we need to overcome.
..
So, let us go one by one, first let us consider the situation of we consider the situation
of the constraint forces one, how to
and in countering this problem we argued that even
though the constraint forces are there in general they do not work. Because, the actual
motion is orthogonal to the constraint force, there are situations somewhat rare
situations, most situation the constraint forces do not do any work.
But, where the holonomic constraint is what we termed as rheonomic, where the
holonomic constraint is dependent on time. Say, there is a bowl in which you have
released a marble to move and the bowl is being moved around, so these are the
situations where you have a holonomic constraint all right, but that is dependent on time.
Now, when that happens we need to go into another stage of abstraction, we said that
we will not consider the actual motion, actual
velocity we will not multiply the force with the actual velocity.
Rather, we will multiply the force with a specific conceptually or emergent velocity,
emergent motion which is consistent with the constraint at every instant of time. That
means, if you have a pendulum with a oscillating support, then at every point the chord
has a particular angle and the bob is here. In that case, the admissible motion at that
position would be orthogonal to the direction of the chord.
So, the moment we conceptually imagine that kind of a possible motion admissible
motion. Then, we have a work the moment you multiply the forces with the, that
admissible motion we get a work, but that is what the actual work we done by this
.system. That is again admissible work, you might ask, what gain do you make by
considering this, yes we do will it will be clear to you later.
But, the what we did was we say that originally the equation was, originally the
Newtonian equation was j is equal to 1 to N, j is the number of mass points and it was
m j r double dot j that is mass into acceleration
term minus F j this is the applied force or the given force, this was equal to the constraint
forces. So, that was our logic that was the initial starting point, because this was the,
what Newton said. And then we said from here, that let us multiply
both these sides by the admissible displacement. There by we got this is a vector
equal to again the right hand side is equal to 1 to N F c this vector j times, now this
term will; obviously, vanish because for each of
this N bodies, the constraint force times the admissible motion is 0. So, is equal to
0 that is how we eliminated the constraint forces,
so keep this final equation written in your copy, we will refer to it later.
.
The second problem was the reduction of the
coordinates and in that what we did we have do, we said that after all in the thrice
n dimensional configuration space, each of the
holonomic constraints define a surface. And if there are two holonomic constraints
imagine for the sake of simplicity in visualization, the original configuration space is
three dimensional, in which you have defined one holonomic constraint that will define
a surface.
.So, that reduces the dimension by one, if you have another holonomic constraint that
will define another surface. So, ultimately the
body will be constrained to move in intersection of the two holonomic constraints
and that reduces the dimension by 2, so each holonomic constraint is reducing the
dimension by it is number. So, since ultimately the body is constrained to move on the intersection
of holonomic surfaces, there is no reason why we should still continue to define
the coordinates in terms of the old x, y, z
coordinates. We said, that now that we have identify a
subspace, where the body must lie the configuration point must lie not the body
really, because we are now talking in terms of
configuration space. If there are three bodies, there will conceptually thrice n three times
three number of coordinates, where still I would advise you to still visualize or extend
the visualization. We cannot visualize in nine dimensional space all right, but still
try extend the visualization, that it is some
kind of a space in which the moment you have defined a constraint it defines a conceptually
a surface. And the body this the configuration point
must lie on that surface, must move on that surface. And therefore, we said that now let
us define a new set of coordinates on that constrained surface and these are marked as
q, the generalized coordinates, so in case of
the spherical pendulum what would we do there is only one body. So, original set of
coordinates were x, y, z and then since the body is constrained to lie on the spherical
surface. Therefore, on the spherical surface the most
convenient coordinates are the spherical coordinates theta and phi. So, we define theta
and phi as the new set of coordinates consistent with the constraints and in a system
there could be some motions that are constrained to on a spherical surface, some
motion that are constrained to be on a linear side. In that case those motions will be we
would try to define the minimum number of coordinates consistent with the constraint
that completely define the positional status of
the system and that will be the ultimate our definition of the generalized coordinates.
So, we said that if we have a original thing like this and we know that the behavior is
constrained on this kind of a surface. Then, we would define a new set of coordinates
and then any point which was earlier expressed in terms of x, y, z, now would have to be
expressed in terms of q 1 and q 2. Now, extend that vision this x, y, z could be thrice n
.and this q 1 and q 2 could be again a large number of variables, just the idea is that
we have now defined a lesser number of variables
that are consistent with the constraints. So, q’s are, so suppose a body was there
at this position, then it is position is given by
this r vector in the original coordinate. If it is given by the r vector, then for this
body this r can r was earlier expressed in terms of
x, y, z, but now it has to be r has to be expressed
in terms of q 1 and q 2. New set of coordinates, similarly there was another body suppose
here it is coordinate say this would be r 1 and this one would be r 2. And all those
bodies’ coordinates now would have to be expressed in terms of the new set of
coordinates. And that can be easily done, because we can
say r 1 is equal to r 1 of the new set of coordinates q 1, q 2, q n and may be time,
it might also depend on time, because the surface itself might depend on time, similarly
that will go to r n capital N q 1. .
So, in short we can write then t j th mass point this is a vector as r j of q i and time,
where i goes from 1 to n. Now, once we have defined the old set of coordinates in terms
of the new set of the coordinates, we can also define the velocities, the velocity would
be r dot j or d r j d t. So, if r j is dependent
on all these, when you take the derivative it will
be expressed in terms . you will be able to write and all that plus
dived depends on time. So, this is nothing but, sigma i is equal
to 1 to n delta r j vector delta q i and this is q dot i
plus r j. So, this is how the velocity would be expressed velocity of each body fine,
.notice in the left hand side what we have, this body is moving .
bodies velocity expressed in terms of the new set of coordinate, we will use that. Now,
if I ask you how does the r dot j this is, so
this is r dot j how does this fellow depend on the
generalized coordinates. So, if i ask you del r dot j del q dot i what
you get, suppose you want to express this derivative, look at the right hand side this
is a generalized velocity in one direction i. So,
if we differentiate it with respect to another generalized coordinate you get 0, because
it general coordinate is a independent quantity.
So, the moment you want to do this derivative with respect to q dot i, then you
will find only that term will remain where this
and that are the same. All the others will vanish and this will also
vanish, because this tells how does the constraint surface vary with time and that
also does not depends on q dot i. So, ultimately what remains is the set clear, so we keep
this for future reference, we have argued that
this would be valid and we keep this for future reference fine. So, there are two things we
are keeping for future reference, one is this equation another is this equation we will
come back to that. Let us keep somewhere, so that we can again
refer to that fine . now let us get back to this picture, this was
the position r of a body, if I ask you what is it is
admissible motion it is admissible motion is on the surface.
.
.So, your admissible motion delta r j this now has to be on the surface that now has
to be expressed in terms of the new coordinates
and how would you expressed that in terms of
the new coordinates. Exactly the same way we will say that this is delta r j del q i
times delta q i sum over i is equal to 1 to n correct
I am not writing the broken up form, because that will be the same for each body
j the derivative of that r j in terms of the ith
coordinate times the admissible motion along the ith coordinate.
So, this has to be summed up, so this is another thing that we keep for future reference,
getting too complicated know what you have essentially done is, that what are we
interested in we are interested in the transformation of the coordinates ((Refer Time:
22:37)) and who do we transfer. First this vector to the derivative of the vector; that
means, the velocity position and the velocity how do they transform into the new
coordinates. And then the admissible motion that is the three things we wanted to express
in terms of new coordinates, that is what we have written now fine ((Refer Time:
23:11)). Now, let us start with this equation that
we wrote, we had multiplied by the generalized by the admissible motion and obtained the
right hand side 0. Let us start from there . at this time our objective is number 2 all
the given forces were vectors. So, everything in the Newtonian framework
has to be expressed in terms of vectors, and then in any given system what
is the absolutely general scalars that we have
the energies energy is a scalar. So, we would like to express that was the
ingenuity of Lagrange, that he said that everything can be expressed in terms of energies.
So, let us try to frame the whole set of equations in terms of energies, which are
scalars and what are the energies kinetic energy
and the potential energy. So, that is what we are trying to arrive at, but you we will
have to go through a route, because the way Lagrange
himself did and the way you are to understand, they are a bit different. Because,
now we have the hindsight of how this thing can be derived far easily, that is why
we are going this way.
..
So, we start from here j there are n number of mass points, so N m j r double dot j minus
F j I am just copying this, this is a vector, this is a vector delta r j is equal to 0 that
is where we were. Now, we have obtained something
about how the generalized admissible motion transforms, we . that and that is what
we need to substitute here the first step. So, we substitute what
we have, you can also substitute on your copies, you will see N m j r double dot j
minus F j is a vector times all this stuff i is equal
to 1 to n I have just substituted this. Now, there are two summations here and is
immaterial in which order you sum them, so we bring this summation over i forward. So,
let us do that we bring this summation over i
forward, what remains at this term, so you have the summation over j, summation over
i just to recall what is summation over i the
number of generalized coordinates and summation over j is the number of bodies two
things. Then, we this term comes m j r double dot
j minus F j this and then this term delta r j
delta q i. So, this term and then you have to sum over fine, so this term is summed over
j and that term after that it is summed over
i let us put this term inside, so what do we have
is this part visible yes i is equal to 1 to n just put this inside you have sum j is equal
to 1 to N m j r vector double dot j times r j delta
q i I have multiplied it here minus F j vector r delta q i, just keep check that I am doing
it right fine done. So, we have this term we need to put a bracket
here, because the whole the summation is over the whole thing. Now, let us see what
each term means, here we have r j that is the
.position of the nth or jth body is double dot times this derivative, what does it means
physically just look at it. .
First suppose we have we take r dot j delta r j these are all vectors, suppose we start
with this term and take a d d t of this can you
do it, there are two terms here we will go by the
one the standard way of differentiation. So, we first differentiate this, so that would
be this is dot r dot here r double dot j times
this plus this remains now and this term is differentiated with respect to time.
So, we have r dot, so del r dot you can do it this way, so we have just written the
derivative like this. Now, you notice why did we do all this, because this term is this
term, so this term is expressible as r this is what we are interested in expressible as
this is not there r j del q i is equal to this minus
this r dot j derivative of j with respect to q i
minus r j is a vector dot. So, we have understood the meaning of the first term in terms of
two terms, which we are yet to understand leave it at this stage, but let us substitute
it here, let us substitute it here.
..
What do we have after substitution we have, let us start from further left hand side,
because it will be rather large i is equal to 1 to small n, then we have a bracket then
it is in terms of j is equal to 1 to capital N,
this is what we are writing m j then this is to
substituted. So, it will be m j d d t of r dot j dou r j dou q i fine minus let us break
up the summations again j is equal to 1 to capital
N, the next term m j remains now comes this term r dot no this is vector dot j dou r dot
j dou q i. So, that substitution is over, but there was
another term remaining here, that has to be written. So, we will again write it as a separate
summation I will write here minus j is equal to 1 to N it remains as it is F j dou
r j dou q i, now this bracket will be closed times
your generalized coordinate generalize admissible motion is equal to 0. Now, after all
these business we have come to three terms, one term, second term and the third term
three terms. And these just by looking at these terms it
is very difficult to physically understand what
they are. So, let us try to physically understand what they are, we will come back to this,
so let us keep this here and let us write down individually, we said that we want to
express things in terms of energies.
..
A first energy is kinetic energy, what is the kinetic energy total kinetic energy, kinetic
energy expressed written as T. So, kinetic energy is half m v square for each body, so
half m 1 that is r dot 1 square plus half m 2 r 2 vector dot square plus, so that is
the total kinetic energy, that is half no I will keep
the summation outside half this is summed over
j is equal to 1 to N, N number of bodies half m j r dot j square fine do we have this term
around here no not really not exactly this term.
But, we have what could possibly be obtained from it is derivative, so let us take the
derivative of this term, this kinetic energy. So, this is the kinetic energy, so the derivative
of the kinetic energy, kinetic energy is after all dependent on the velocity, so we will
derive take the derivative in terms of the generalized velocity T del q dot i that is
the generalized velocity.
If you take what do you have in the right-hand side, we have the summation first j is
equal to 1 to capital N, when you take this derivative this two comes forward and gets
cancelled with the half. So, it will be m j r dot j and then the derivative of r j del
dot j dot q i, this is how would you express fine, so
we are better off because we can identify this
term somewhere here. So, we know that here this term was actually talking about the
derivative of the kinetic energy with respect to the generalized coordinate good at least
one thing is settled. Now, where did I keep the expressions here,
. see we have already obtained that the derivative of the velocities
would be equal to the derivative of the
.coordinates, we had already done this we had kept it for future reference that is where
we will use it. This term is then equal to m
j r dot j, but now we do not need to write the
velocities just the positions, so this is something that we will keep and then let us
also differentiate the kinetic energy with respect
to the position. .
So, we have derivative of the kinetic energy with respect to the position not q dot i it
is q i, then we will again have to write is as
the same summation of partial derivatives. So, it
will be summation of j is equal to 1 to capital N and in that case what will remain m j see
this is what we are trying to express or take a derivative of in terms of the q i. So, it
will be same way m j r dot j times the derivative
of r dot j times the derivative i, you can possibly identify that this is this term.
So, we are slowly homing on to what we wanted to express, . finally
if there is a force acting on this body, that force if now we expressed in terms of the
new coordinates how what shape will it take, in
the new coordinates suppose that generalized force is called Q i will be expressed as summation
of again j is equal to 1 to N F is the vector j r vector j. So, all we are doing
we had the actual forces applied F j, now we are
converting into transforming into the new set of coordinates Q i.
So, these are the forces generalized forces that are applied in the direction of the
generalized coordinates, so this will be the force. Now, suddenly we see that all the terms
that we wanted to know, . they are all clear what is this, this is
.actually the original forces acting on the bodies converted into the generalized forces,
what is this term, this term we have just indentified as this term.
So, this term is same as this term, so d d t of this is d d t of this clear, so and this
term is already we have derived. So, all we can now
express this equation in a much simpler and elegant form as i is equal to 1 to small n
and then we had written, now the all the summation go away, because this term summed
together is some simple term, this term summed together is some simple term, this
term summed is sum together is some simple term.
So, all these summation go away over j and we get d d t of del t q dot i, because that
is what we had written, this term were expressed
as this and that is what we have in that particular term. So, this is the first term
minus the second term here is what we wanted to
express, so we have del T q i dou Y dou q i that is the second term and the third term
is minus simply Q i, this whole thing simple
it has to be multiplied by the admissible motion is equal to 0.
So, true that we went through a lot of you know partial derivates and messy stuff, but
ultimately the objective was to simplify stuff and here is the simple thing. And you
notice, that here is something that is multiplied by the admissible motion and the
admissible motion along the first coordinate and the admissible motion under second
coordinate are all independent quantities. And this term summed over i would be 0 only
if this term is 0 always can you see that. So, here is the term for the ith thing it
is multiplied by the ith admissible motion and all
that summed over will has to be 0. And that can be 0 only if since this cannot vanish,
this is something that we have assumed that it
can be just any motion, therefore this term must vanish. Not only that this term must
vanish for all the generalized coordinates, not
the summation, just individually along all the generalized coordinates this term must
vanish.
..
So, that gives a very simple equation d d t of dou T dou q i dot minus doh T doh q i
minus the generalized forces equal to 0. Now, this is the Lagrangian equation in the most
generalized form, in the most general form, you might wonder that, so far we were
initially talking about energies, energies would be kinetic as well as potential. But,
so far we have talked out only kinetic energy, some
where does the potential energy come or where does the potential energy go.
Now, you notice that here we have got a generalized force, what does that mean, it
means the forces on the bodies in the direction of the generalized coordinates. So, in case
of the pendulum, if you apply a force in the direction of the theta, so that will be the
generalized force. Now, what does this force consist of what could be the different types
of forces acting on a body just think, these are the forces, so these are the forces that
we need to consider.
What kind of forces could it be, it could be one forces
like the gravitational force it could be force applied by one fellow on another
fellow by means of a spring in between. So, a
force could be also externally applied somebody pushes a body, so externally applied
force
and there could be also the force due to friction. So, there could be all these types
of forces have we taken it into account everything is there any other type of force that
you can imagine, that can be applied on a body, no we have taken everything in.
So, if you drop this the last one, then all the other types of forces are actually derivable
from a suitable defined potential energy I will show that. But, . this
.will have to be separately treated, but other forces gravitational force we know that there
is something called a gravitational potential and the gravitational force is the gradient
of that potential you learned in school. But,
yes you may not know how that can be done for
a force applied through a spring or the externally applied forces.
But, for now take it for granted I will illustrate all of these separately, take it for granted
that all these forces can be treated the same way we treat the gravitational force. That
means, in the gravitational force we said that there is a gravitational potential and
the force that is applied is actually the derivative
or the gradient of the potential. .
So, the potential energy is expressed as v, so the generalized force can be obtained as
the gradient or negative gradient clear and that
satisfies your concept of the gravitational force. But, in general in order to visualize
imagine in this way, the potential is like a
potential surface, say suppose you have got a potential surface like this and the body.
So, this is the q q i coordinate and here is the
body sail, so body’s position in q i is this much
from here to here, what will be the force acting on the body the gradient.
And you might imagine a three dimensional surface and in that if the body position is
such, then the force acting on the body is nothing but, the gradient of the potential
surface. So, potential is conceptually a surface, you have got the q 1 coordinate, q 2
coordinate generalized coordinate and in that it is a surface and the position of the body,
if you know then depending on the position it applies a force which is a gradient.
.And how it also applies to other kind of forces I will come to, but that at least satisfies
your concept of the gravitational potential does it not. Gravitational potential concept
was exactly this, even electro static potential’s concept is exactly the same, what was the
concept of electrostatic potential, that if you have a particular body at a particular
place how much force is applied and the force is
obtained as the gradient of that potential. So,
always the potential’s concept is the same that the force is a gradient of the potential
negative of the gradient like here, how it applies I will to the other kinds of forces
I will come to that later.
.
But, once you assume this, then this can be substituted from there we will substitute
and write d d t i dot minus here we substitute
minus dou v dou Q i, so this will be plus. Now,
Lagrange defined a new term called, now it is called Lagrangian we write it as script
L which is the kinetic energy minus the potential
energy. So, this can easily be seen that this is a derivative of the Lagrangian with
respect to q i, can you see that this term the
whole term. But, this term well here you were you are
taking a derivative of the kinetic energy with
respect to the velocity and the potential is normally not depend on the velocity. So,
we can very simply write it as d d t of derivative
of the Lagrangian with respect to minus derivative of the Lagrangian with respect
to, so simple after all this day’s labour, we
have obtained such a simple equation.
.That is the actual form of the Lagrangian equation for a non dissipative system,
because we have yet not considered this term, how to consider this term
we will come to that later. But, this is the Lagrangian equation
for non dissipative system for conservative system. Let us stop here
and we will illustrate how to apply this to physical system in the next class.
.
