Transcriber's Name: Priya
Theoretical Mechanics
Prof. Charudatt Kadolkar
Department of Physics
Indian Institute of Technology – Guwahati
Module No # 02
Lecture No # 06
Hamilton’s Principle
Okay welcome to the second week of this course
in the first week we learnt about haam we
learnt new ideas ahh like constraints generalize
coordinate configuration spaces and so on.
And then final we use the D’Alembert’s
principle to derive Lagrange’s equation.
So that basically summarizes the program of
classical mechanics what I will do is first
post the (refer time: 01:02) dynamical problem
here. See even if you are considering haa
single particle what is the that you wanted
to? If we know all the forces on the particle
then if we know it is initial position and
it is initial velocity then we want to predict
the path it is going to take okay. And then
of all the path that it is possibly choose
which path it will take? It will take one
of those paths which satisfies the Newton’s
Law’s okay which is given as a form of differential
equation which is mr double dot = force which
may be function of r, r dot and tn so on.
Even if you take more complicated system with
all the constraint thrown in with potentials
given and so on. Even then the problem really
reduces to now you are describe the system
in terms of a path in a configuration space
okay. So put everything together constraints
then you find out the accessible configuration
space and then system as a whole is just one
point in that configuration space. And we
know the initial condition that is if we know
q and q dot at some instant of time then ahh
we want to predict the path of the system
in the configuration space at some later time
t. And which path would it take again you
have this Lagrange’s equations which are
second order equations and you would say that
path take by the system is the one which satisfies
the Lagrange’s equations. Now you see for
us people who believe in nature it is slightly
you know uncomfortable feeling that it so
technical definition of which path the system
is going to take it is very mathematical definition.
It is does not seems the set of you know a
nice idea remember it is a feel it is a nice
idea about what nature is going to do and
that is where the Hamilton’s principle come
in. And the Hamilton’s principle this very
elegant principle says that the nature chooses
that path which has a least action. So of
course we may have to we will have to make
a technical definition of action but there
is a elegance in this statement it sounds
nice. But of course underneath there is mathematical
definition of course but it is such a simple
statement and then ahh in the course of this
week we will show that the Hamilton’s principle
is also equivalent to Lagrange’s equation
or Newtonian formulations that. So this is
our program for this week. Now I will start
the (refer time: 04:17) introduction with
classic puzzle okay look at this diagram here.
In this diagram it is like a big field which
is 100 meters by 100 meters and on the left
side of that field is the concrete and on
the right hand side on the right half there
is sand and there are these diagonal points
which have marked as A and B and the aim is
to go from A to B in shortest possible time
okay. What is the catch? Catches us. Concrete
is the nice hard surface you can run on it
with a speed which is 4 meters per second
but on sand you cannot run as fast as that.
So for example I have taken that to be 2 meters
per second okay now which path is the question?
The path that would take the least amount
of time now as far as the concrete is concerned
we can immediately figure out how to do this?
Because in the concrete region the time take
would be the length of the path divided by
the velocity which is fixed. So basically
we should be running in a straight lines on
concrete and in straight line on the sand
so ahah if you do not know the answer you
probably want to pass the video here think
about the answer and then ahh start again.
So here what we must do is run in the concrete
in straight line run in the sand in straight
line. How about a path like this? A path must
be somewhat ahhh like this this of course
is not a pretty good example. How about the
diagonal path? The diagonal path is the shortest
path but since we can run faster in concrete
it is common sense that we probably should
long longer distance or as much distance in
concrete as possible to get the shorter time.
So probably we might think of a path like
this in a in this path you would run all the
way from A to the upper corner of the interface
and then run parallel to point B. Now here
is the comparisons of these paths so the green
path which is diagonal which is shortest about
141 meters it takes about 53 seconds but wait
the red path is about 162 meters but it takes
tiny bit less amount of time see we are having
much much longer distance but still you would
end up taking less time okay. So the question
is how do I find? Is there is path somewhere
in between the green path and red path which
probably is the shortest path haa shortest
time path. So how do we find this? One of
the ways you can do (refer time: 07:46) is
simply search for the path now we have already
figured out you know what is the nature of
the path so what I will do is there is a point
of crossing from concrete to sand and let
us say bottom of the picture to the crossing
point that distance actually defines the path
completely okay where you can do cross because
once you fix the crossing point A to that
crossing point is straight line and crossing
point to B is another straight line okay.
(video starts: 08:24) So here what I have
done in this one is you can actually change
the path as it function of crossing point
here so on the right hand of the graph you
have crossing point and we have just crossed
the haa midway mark and there you go. (video
ends: 08:42)The time taken which is plotted
on this graph on the right hand side immediately
shows that between that diagonal path and
the extreme path on the where you have run
all the way to the top of the figure somewhere
there at about 76.91 meters you have a best
time path and which takes about 50.5 seconds
and the distance run is of course larger than
the diagonal path but this is the one we wanted
to find okay. Now there is an (refer time:
09:21) other way use calculus see since ahh
we already have figured out the nature of
the path calculation the time for this path
is very easy. So the path can easily be given
by an integral of ds over v where ds is small
length element along the path and then because
the velocity is on concrete side is fixed
velocity on sand side is fixed the formula
can be immediately integrated to something
like this. So here is the time now all we
want to do is find a path or a processing
time where this particular time would be minimal
so all that we do is take a derivative of
this with respect to l that is the distance
of the crossing point and set it equal to
0 that immediately gives you ahhh value of
l as 76.91 meters okay. So this is the but
something more to this problem which I have
wanted to show you and that is. If I draw
a perpendicular the interface at the crossing
point and then I mark the angle here this
one as I and on the sand side I will mark
it as r and then you calculate sin I / sin
r that is equal to 2 right that is just the
speed in concrete divided by speed in sand
now this is just like the Snell’s law what
is the connection here? Oh the connection
is of course the famous Fermat’s principle.
Now Fermat’s principle (refer time: 11:15)
for the light says that this is geometrical
optics that the light ray would take such
a path between two point which takes the minimum
time or which takes the shortest amount of
time. So put it more formally what is do is
let us take one ahh say some region. In this
region I have these two points okay and I
want to find out which path if one of the
point emits ray which path this ray would
take to reach point B the catch is of course
that the refractive index here is probably
function of the space it is the inhomogeneous
medium. So in the inhomogeneous medium 
and then what happens to the velocity at each
point? At each point velocity of light would
be different and will be given by c / n okay.
So what we calculate is first of all you take
any arbitrary path between these two lines
and look at the small length element at some
point and your total time taken for this path
is integral from point A to point B of the
small line segment divided by the velocity
of light at that point. And we can of course
readjust this little and write this as 1 over
c integral from A to B n of course remember
this function of the point ds okay. At what
Fermat’s principle says is that of all path
that connects point A and point B so they
could be infinitely many paths which would
connects A and B only that path will be taken
by the light for which the time taken is minimum
okay. So this is what the Fermat principle
says. And this can be of course used to explain
the variety of phenomena there and one of
the common phenomena that people usually show
is that of the Mirage. What happens in Mirage
is that near the ground the air is warm so
the refractive index is higher and hence the
speed is speed of the light is faster near
the ground than above the air. So what would
happen is that the ray while it passes from
the source to the observer it actually goes
through slump like this because it tries to
travel more and more distance near the ground
rather than. And then of course the observer
would interpret as if the ahh the light rays
are coming from the ground and he sees the
reflection of the object even though there
is no water or there is no reflecting surface.
Okay the question that we want to ask is thus
now if there is a Fermat’s principle for
light and remember the light is in principle
that is not covered under the classical mechanism.
If there is the Fermat’s principle for light
why cannot I have such a principle or what
is that principle which would apply to the
classical systems or classical mechanics.
And for that (refer time: 15:20) I will take
one more example now look at this example
of a projectile motion. Now in this projectile
motion a particle is thrown from point A whose
coordinates are 0, 0 to the point B here whose
coordinates of point B are H 0 okay. Now I
want to post this problem slightly differently.
So your horizontal range is fixed you want
to throw a projectile from A to B not just
the horizontal range but the question is to
find path of projectile such that it reaches
B in given time and I will say capital T okay.
So we want to find out a path which haa so
so normally the problems are post slightly
differently see normally you would be given
the initial point and the initial velocity
which means the speed and the angle of inclination
or angle of projection and then you are ask
to find where is lands and after what time.
Here we are posing the problem as boundary
value problem I have already given you the
end points and I have also said find a path
which takes exactly that much amount of time.
Now we of course we already know the answer
to this you can ahh immediately put it in
we know the answer is parabolic path and so
on and the answer we already know and I will
call this path as x star t which is the times
t so this is and y star t will be equal to
half gt times capital T – t you can immediately
verify that this is the correct equation or
correct path for the projector and also gives
the time dependence. So at t = 0 the y coordinate
will be 0 at t = capital T also y coordinate
will be 0 and the ahh at small t = 0 x = 0
and at t = capital T x star will be equal
to H. So this is the correct path that we
know okay now what is special about this path
is the question we are trying to ask and that
is what is going to be your Hamilton’s principle.
So what is special what I am going to do is
first of all I will look at few other parts
which are nearby okay. So let me define new
paths which are x of t which is equal to H
/ capital T times t and y of t which is half
but instead of g I will put some other number
there a okay times t times T – t what is
this give you? This of course gives you lots
and lots of parabolic path all these paths
remember start at 0, 0 end at point B which
is h, 0 in time capital T okay. But they are
different parabola so basically the ahh heights
at the midpoint will be different depending
on the value of a. And we know only one of
those path is correct paths that is correct
path is this when a = g or a/g = 1 okay this
is the correct path. Now let me define since
we know the Lagrangian of this system. So
let me write first of all Lagrangian of the
system and the Lagrangian L will be equal
to 
square and minus so that would become + mg
times y so sorry – mg y. And I am of course
we can now apply Lagrange’s equation and
get that equation. But I will do I will define
a new quantity called as action so definition
of action is this so for every path from point
A to point B whatever that path may we have
chosen only a small family of those path which
are only parabolas but any possible between
A and B which travers says in time capital
T we will assign ahhh action A which of course
is function of the path of the path okay.
So r of t is nothing but x of t and y of t
will define this as integral from 0 to capital
T L. Remember L is function of x, y, x dot,
y dot and t integrate with it so will happen
is that this integral is of course easy to
do I have already given you x as a function
of t y as a function of T so we can calculate
x dot y dot also as functions of t. So the
whole of this can be expressed as a function
of T and then you can integrate okay and I
will leave that exercise to you people to
do. So after integrating what we get is action
A and I want to look at this action A for
different paths above the actual path that
we have found out which is x star y star.
So if I haa so these are the various path
that you would get if I change the value of
a for example the value of a here at the lowest
one here a would be 0 and somewhere in between
a = g path is there that is the correct path.
So what I want to do is this (refer time:
22:23) for each of this paths I will plot
action okay as a function of a / g you can
immediately see what happens here at a = g
or a / g = 1 the action is minimum see this
is what separate this path from the other
path and this is in fact the statement of
the Hamilton’s principle. So the correct
path here 
is the one for which action is minimum minimum
there you okay.So I will now post the statement
of the Hamilton’s (refer time: 23:24) ahh
principle the nature chooses the path of least
action but let us make this more accurate
okay. So let start by defining ahh consider
a system with N particles and N degrees of
freedom and you have looked at all the constraints
and everything. And then the system is described
by Lagrangian which is L and this is of course
function of q, q dot and t and remember each
of this q is q1 to qn because there are n
degrees of freedom. Now let me post the problem
just way I post a projectile problem so let
q1 be a point which is q at t1 and let q2
is another point at t2. So in the space in
the space okay I cannot draw the configuration
space of this schematic diagram in this one
I have this two points one point is q1 the
other point is q2. The aim is to start from
q1 at time t1 and reach q2 at time t2 and
then of course there are infinitely many paths
which connect the 2 and for each path we assign.
So for each path we assign action which is
defined as integral Ldt from t1 to t2 and
remember once I give you path I give you the
time dependence of all the q variables and
from there you can calculate q dot then the
Lagrangian is just merely function of t you
can do this integral. And then the Hamilton’s
principle says that the actual path of the
system 
the system is the one for which action as
defined there is minimum. Now this probable
is not the accurate statement in the following
sections I will refine this definition it
is actually considered as what is called as
the stationery point but this is the ahh Hamilton’s
principle. And what we are going to do in
the remaining section is this it is not only
this problem so Fermat’s problem is 1 which
is some sort of least time principle this
is a least action principle then I can probably
think of you know two points in this space
and then ask the question which is the path
which as the shortest distance such a least
distance path problem. So this is class of
problems which go under the name of variational
problems and the mathematics that underlies
all this is called as a variational calculus.
So in the next few sections that is going
is to be our focus to understand the ahh understand
how to solve these problems variational problems
of this kind and then we will proceed with
application to Hamilton’s principle. So
in this section we are going to focus on variational
calculus now there are set of problem which
we are interested in and these problems are
called as variational problems they all fall
in this category which is basically trying
to find paths which have something which is
minimized. So ahh in all these problems they
would be some fixed points and you would be
having paths between them with each path you
assign some quantity called as functional
and then trying to find a path for which this
functional becomes ahh minimum or rather extreme.
(refer time: 28:12) And the ahh history of
the variational calculus is also fascinating
you should definitely ahh try and read the
way it was developed. So it begins with ahh
Newton and he was trying work on a problem
where you have a solid surface of revolution
and the surface of revolution so this particular
solid is moving through viscous fluid and
what kind of surface would give you minimum
rays minimum resistance. Something looks like
a aerodynamics problem you have a vehicle
traveling through or Aero plane travelling
through air and you want to find what should
be shape of the nose so that it has the least
resistance in the air. The second ahh ahh
after that about 10 years later Bernoulli
post is famous Brachistochrone problem and
he post it as a open challenge and eventually
Newton also solved it ahh. This is ahh very
interesting reading and then finally Lagrangian
Euler also work extensively on this and Euler
is the one who actually coined the name variational
calculus. Okay let us start by defining the
notion of functional okay (refer time: 29:40)
so in this space I am going consider only
one dimensional case to begin with. So if
you have a space and I will mark this as a
y axis and this as x axis and you take these
two points here and this one is y1 at x1 and
this is y2 sorry for this and x2 okay. Now
there will be infinite many paths which join
the two points x1 y1 to x2 y2 and all these
paths basically are continuous or sufficiently
smooth functions of x. So y the path represented
by ahh function which is real valued function
it is continuous it is sufficiently smooth
this we will call it has this function is
called as is path between y1 and y2 remember
the x1 x2 is the domain which is fixed and
it is given and what we are going to do is
which each path we will assign a quantity
called as so ahhh scalar value J is basically
a function from set of all paths between y1
and y2 all paths between y1 and y2 to real
numbers okay. So we assign scalar real number
to each possible path this is in some sense
function function of functions. So the paths
themselves as functions and J is basically
assigning to each path some scalar value.
So that is why this one as special name so
this one is actually called as functional
okay and what we are interested in is not
any arbitrary functional but very specific
kind of functional which appear in all the
problems that we describer earlier or which
we will discuss later and that is a large
class of problems we are interested in. So
I will define this ahh functional which is
of very specific kind as a functional which
is written as y of x as integral from x1 to
x2 of some integral which I will call it as
f and f is of course function of y, y dot
here y dot is derivative of y with respect
to x commonly we of course write it as y prime
and x, dx. So these are the kind of functional
that we are interested in. Why does it not
depend on y double dot I will leave that question
to you that figure that out okay. Now here
as some examples in the first example (refer
time: 34:01) think of ahh plane in which you
are given two points and we want to find out
which path gives the shortest distance between
these two points if it is a Euclidian plane
we already know the answer it is a straight
path between the two but anyway we will post
this problem. So if take this some path between
these two points so this is x1 and this is
x2 if I take any path suppose y of x I immediately
can assign the length of the path as so length
of the given path as integral from x1 to x2
and integral over ds. What is ds? ds is the
small element here and assuming that this
is a Euclidean plane in which case this would
become integral from x1 to x2 of dx square
+ dy square to the power half and that I will
write as integral from x1 to x2 1 + y dot
square to the power half dx. And now we can
immediately identify the integral so your
length l is the functional of y and that is
basically integral of the function. So the
integral function which is function of y,
y dot and x here is in fact just 1 + y dot
square to the power half okay. And then the
second example us which we have already seen
(refer time: 36:26) which is if we have light
travelling from point x1 y1 to x2 y2 in some
inhomogeneous field then we assign with each
path for the light rays time 
of light between x1 y1 to x2 y2 again I will
write it as T which is function of path which
is given as integral and this is nothing but
ds divided by and from some x1 to x2 . Now
this again I will write it as so sorry I will
write this as 1 over c integral from x1 to
x2 of we already know how to write ds that
is nothing but 1 + y dot square to the power
half dx okay into n which is function of x
and y okay. So here again we can identify
the integral function the integrand function
which is function of y, y dot and x is n which
may depend on x and y into 1 + y dot square
to the power half okay. Now one more example
this is the famous Brachistochrone problem
(refer time: 38:42) so think of a situation
where you have the vertical direction and
horizontal and then you have 1 point here
and the other point here and you have some
wire which connects two points okay and then
there is a bead. So what we are going to do
is ahh so there is gravity pointing downwards
what we are going to do is this. From the
upper point you drop the bead with 0 velocity
and it will slide frictionlessly over the
wire the shape of the wire is fixed or is
given and then I would it would reach the
other point here and question that we want
to ask is this. So if this is y1 this one
here is y2 this x1 and this is x2 the question
we want to ask is this what is that shape
of the wire which will give me the minimum
time of slide okay. So let us post this the
time taken so if the shape of the wire is
given then the time taken is equal to again
similar to the previous problem it is integral
from x1 to x2 ds divided by velocity v but
we know velocity v at each point. So at each
point half mv square must be equal to so if
it as drop to this point which is y then clearly
this must be equal to mg times y1 – y and
that immediately gives you velocity v which
is nothing but square root of 2 times g times
y1 – y okay. And then I will put it back
into this equation so this becomes we already
know how to represent ds so this is x1 to
x2 ds is nothing but 1 + y dot square to the
power half and divided by the velocity now
is square root 2g times y1 – y dx and in
this case the integral or integral function
is 1 + y dot square to the power half y1 – y.
And lastly among the class of this problem
we already have the Hamilton’s principle.
So the Hamiltons principle (refer time: 42:16)
or Hamilton’s statement is in fact exactly
in the form of functional so here the action
in fact is already defined as t1 to t2 of
which depends on y, y dot and t dt remember
here the roll of x will be played by time
and y dot here is basically dy/ dt and y of
course it is posed as function of time t and
this is the path that we are assigning the
action to. So here this is how a functional
is assigned to paths and all these problems
what we want to find out is how to extremize?
How to minimize the functional or fondly physicist
always say action. For every kind of functional
that is the focus of our next session.
