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Greetings, we will resume our discussion on
the Lagrangian formulation of equations of
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motion.
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I will quickly recapitulate the essential
consideration that we had undertaken. A mechanical
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system is considered to have evolved in such
a way that action would be an extremum. That
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is the principle of extremum action as it
is called. This is also referred to as the
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Hamilton's principle and this is a big departure
from the Newtonian formulation. In Newtonian
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formulation, we agreed that a mechanical system
evolves in such a way that it responds to
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the stimulus, which is imparted to the system
by a force. So, there is a cause-effect relationship,
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which is the principle of causality. It is
the important consideration and fundamental
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consideration in Newtonian mechanics.
The entire evolution of the mechanical system
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is described in terms of the Galilean principle
of relativity. Secondly, any departure from
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equilibrium is explained in terms of the cause
effect relationship, which is contained in
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Newton's second law; the principle of causality.
So that is the governing principle of Newtonian
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mechanics. The governing principle over here
has nothing to do with causality. It has nothing
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to do with force. It deals with the principle
of variation. It stipulates you to set up
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what is called as the Lagrangian for the system.
We have not yet mentioned as how to setup
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the Lagrangian. We have not gone there. We
only know that it is some function of position
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and velocity. In general, it is a function
of q, q dot and t.
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What kind of a function it is? We have not
described and never mind, we will work with
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it. At some point, we will have to answer
this question, what is the recipe to set up
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the Lagrangian? So, we will take it up at
that point. In the meantime, we carry forward
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the same starting point as in Newtonian mechanics.
The mechanical system is described by a point
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in phase, space by position and velocity,
rather than by position velocity. Directly,
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it is represented by a function of position
velocity, which is the Lagrangian. So, the
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Lagrangian is the function of q and q dot.
If you define an integral, it is known as
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the action. Action is the integral L dt from
t 1 to t 2. Let this action be an extremum.
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This is the governing principle, on which
the evolution of a mechanical system is described.
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Now, what this led us to? These are some of
the major steps that we discussed in the last
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class. So, I will not go through this in any
detail, but quickly recapitulate that. If
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this is an extremum, then any variation in
this would vanish and variation with respect
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to different paths in the phase, space in
the position velocity phase space that a system
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can take. So, these are like different world
lines or different paths that the system can
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take. Variations with respect to these parts
will lead to an action integral, which will
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remain stationary and it will not change.
If you go from one path to a neighboring path,
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which is somewhat close to the previous reference
path, along which, the system will evolve.
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This differential increment in action would
vanish, which we have restated in the equation
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at the middle of the screen here.
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Now, these are the alternative paths that
one can think of. So, this is a position,
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velocity phase space. We consider the evolution
of the system from time t 1, where the point
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with coordinates q and q dot describes the
state of the system. At a later time t 2,
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this system gets to this point along a certain
path. This is some wiggling arbitrary path.
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It could even cross itself, whatever it is.
It is some path that the system would take
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at later a time, t 2 and it gets to this point.
Now, I have show an alternative path over
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here in different color, which is slightly
different color and it is beige. I think I
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am not good at identifying colors. So, I suspect
this as beige. You can think of a slightly
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different path at some intermediate time between
t 1 and t 2. The system would be at different
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points at some specific points on each of
these two paths. So, let us say that if it
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were to evolve along the white path and if
the system is at this point at a certain time
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t, then at the same instant of time, if the
system has to evolve along the beige color
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path, the system would be here.
The horizontal distance between these two
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is the delta q at the time t and this delta
q at the time t. As you can see from this
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figure, it is 0 at the start and at the end,
it cannot be anything else. If it is non-zero,
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it can only be in between t 1 and t 2, but
not at the start and not at the finish. Those
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points are fixed and there is no variation
in delta q.
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Here, you have the term, in which delta q
vanishes and the remaining terms can be combined
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together. So, this term vanished and the remaining
terms have been put together in this expression
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here. Now, you are left with this term, which
is the first term amongst these three. The
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last term and these two terms are put together;
this is an integral del L over del q delta
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q dt. So, this del L over del q delta q dt
this is coming from the first term and from
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the second term, you have this expression
here.
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Now, in this, delta q is appearing in our
analysis because we are choosing different
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paths. There is no reason to suspect the beige
colored path that we had in our previous figure.
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It is the only alternative to the white colored
path that we have considered. We could think
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of n number of alternatives paths or any different
variation or anything, which is other than
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the path that the system would actually take.
So, any of those paths could be taken. Therefore,
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this delta q is completely arbitrary. There
is nothing special about that particular delta
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q or delta q dot that we had in the previous
figure. That was just one of the many infinite
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variations that one can think of. So, this
variation is being arbitrary.
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Now, you are confronted with a very simple
situation that you have a definite integral
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from t 1 to t 2. It is an integral over time,
so that is indicated by this dt at the end.
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What is being integrated out? It is the product
of two factors. One is a factor in this beautiful
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bracket and the other factor is delta q. The
integrand is expressed as a product of two
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factors of which, one is completely arbitrary.
No matter, what this arbitrary delta q is,
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the total integral must goes to 0. For arbitrary
delta q, this can happen if and only if, the
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factors inside this beautiful bracket is identically
0. So that comes as a necessary condition
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that the principle of variation holds. This
is the condition or the factor, which is inside
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this beautiful bracket. It must go to 0 and
this is what we call as a Lagrange's equation.
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How has it emerged? It comes as a necessary
condition that action is an extremum. So,
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we begin with the requirement. We demand that
action must be an extremum and then ask, if
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action is to be an extremum. What is the condition
that must be satisfied? A certain equation
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must hold the term, which is inside this beautiful
bracket. So, the equality of this term inside
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the beautiful bracket is the necessary equation,
which must be satisfied and this equation
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is called as the Lagrange's equation.
Now, we have the Lagrange's equation between
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do not have the Lagrangian itself we have
not discussed even as yet what is the recipe
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to construct the Lagrangian the only thing
we know about it is that it is a function
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of q and q dot so we will have to of course
address this because otherwise if we do not
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know what the Lagrangian is we cannot get
it is partial derivative with respect to q
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and we cannot proceed nor can we get its partial
derivative with respect to q dot because we
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have to know what is the mathematical dependence
on q of L and how does L depend on q dot so
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unless we know this we cannot really do anything
with this equation. So, let us see how we
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can get it.
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This is the Lagrange's equation. What we
do know is that space is homogeneous and isotropic.
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The space that we are dealing with is isotropic.
Its properties are same in all directions.
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Therefore, the Lagrangian cannot depend on
any specific direction because properties
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of the space are the same. That is what is
meant by isotropic. When Properties of the
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space is isotropic, the Lagrangian cannot
depend on the velocity. So, we are looking
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for a Lagrangian, which depends on velocity,
but not on the direction of velocity. So,
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what is the quantity that you can construct
from the velocity, which does not depend on
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the direction of the velocity? It can be V
dot; V being the velocity and the dot product
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of V with itself is V dot V. This does not
depend on the direction of V and it is a scalar.
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You can construct a quantity, if it depends
on velocity. The only function of velocity,
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which is independent of direction is the quadratic
function of velocity. The speed comes from
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the square root of V dot V. So, it must depend
quadratically on the velocity. Therefore,
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we pick a function f 1, which is a quadratic
function of q dot and q dot is our velocity.
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So, we expect the Lagrangian to be a quadratic
function of velocity.
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We take the simplest functions. As they say,
do not trouble trouble, unless trouble troubles
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you. Why do you look for more complicated
function? Look for the simplest function.
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What is the simplest function that you can
think of? Position, some function of q known
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as 2 q. It is the simplest function of velocity,
which does not depend on velocity; it will
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be a function of q dot square. That makes
the Lagrangian a sum of f 1 and f 2. Here,
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f 1 is a function of q dot square and f 2
is a function of q.
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Now, we have made some progress. We still
have not identified, what is the function
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f 2 of q. Is f 2 equal to alpha q? Is it alpha
q plus beta q square? Any polynomial function
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of q would meet the requirement that it is
a function of q. So, we have still not pinned
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down the exact functions and that is what
we will do. Now, we take the simplest function
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of q, which depends on the position. We know
that the potential energy depends on position.
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So that is a mechanical property that we are
familiar with. We expect it to be of great
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value in any discussion on mechanics. So,
we take the potential and we propose that
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the function f 2 q to be chosen. So that is
the negative potential energy. We make this
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proposal and we will still have to justify
this proposal. We have made this proposal
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with a certain hope that it will turn out
to be a good proposal. We have not provided
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reasons for it because we could think of other
functions as well
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At this point, we propose that the function
f 2 q is chosen, so that it is negative of
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the potential energy function minus V of q.
The simplest function of quadratic function
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of velocity is the kinetic energy, half mv
square. So, we propose n 2 q dot square, in
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which we have identified the inertia. What
we have done is we are choosing f 1 to be
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the simplest function of q dot square and
the simplest function of q dot square is just
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some constant times q dot square. The constant
that we choose is half the inertia half the
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mass. Why do we make that choice? We still
have to rationalize and so we make this proposal
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that f 2 q be chosen. So that it is minus
V of q. We propose that f 1 q dot square is
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proposed as m by 2 times q dot square. This
will identify the Lagrangian's T minus V
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and we hope to justify this choice.
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Now, what justifies this choice is a question.
We must now answer. Now, just ask yourself,
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what will this combination be? Here, m by
2 q dot square will be the kinetic energy.
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This is now giving us the difference between
the kinetic energy and the potential energy
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for the Lagrangian function. Now, you can
take the partial derivatives because you have
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written L as a function of q. So, the partial
derivative of L with respect to q will be
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simply minus del V by del q. You can also
take the partial derivative of the Lagrangian
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with respect to q dot because that will simply
be m by 2 times twice q dot. So, you can determine
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these derivatives. You can put these partial
derivatives in the Lagrange's equation and
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ask yourself, what do you get? The partial
derivatives of the Lagrangian with respect
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to q, which is the negative of the partial
derivatives of the potential energy function
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with q. The negative gradient of the potential
in Newtonian mechanics is the force. So, we
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get the force from this term and from the
other term, we get the momentum. In the Lagrange's
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equation, we have the time derivative of the
momentum. So, you will get dp by dt, which
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is the Newtonian force and essentially, what
you get is Newton's second law. This is what
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justifies our choice of f 1 and f 2.
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Let me go back to the previous slide one more
time that the proposal here was the Lagrangian.
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It can be expressed as a quadratic function
of q dot and a sum of another function of
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q.
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We choose these functions to be the kinetic
energy and the potential energy with the minus
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sign over here. This particular choice turns
out to be extremely profitable because it
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gives us results, which are completely consistent
with Newtonian formulation of mechanics. Now,
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Lagrangian formulation or a Hamilton's formulation
is an alternative formulation. It is not going
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to give us anything different. If it contradicts
with Newtonian mechanics, there will be trouble.
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So, it is important that the results which
come out of this are completely consistent
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with Newtonian mechanics. This consistency
is now explicitly manifest. What makes it
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possible for us to attain this consistency?
It is the proposal that the Lagrangian has
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written. It is m by 2 q dot square and we
did not provide any rationalization for the
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factor m by 2. We said that we would take
the simplest function of q dot square, which
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is just a linear function of q dot square
like alpha times k q dot square or k times
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q dot square. I suggested to choose the proportionality
to be half the inertia, m by 2.
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Now, we see that the particular choice m by
2 q dot square gives this del L by del q dot
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to be inertia times a velocity, which is the
momentum. That choice turns out to be a productive
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and a useful one. This is what enables us
identify the Lagrangian as T minus V. I will
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not say this as a very standard recipe to
construct the Lagrangian. When you go to more
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complicated situations, one has to keep track
of many other fairly subtle factors. You have
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to ask- is this Lagrangian invariant with
respect to Galilean transformations? If you
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are doing relativistic mechanics, you will
have to ask. Is Lagrangian, which is set up
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will be consistent with the Lorentz transformation?
So, there are many additional questions that
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you need to answer.
In this introductory discussion, there is
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no room for such involved issues. We do have
a basic foundation and this works for a good
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part of mechanics. You can define the Lagrangian
as T minus V. This also suggest us a recipe
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to do other problems because here after, we
will not recognize momentum as mass times
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velocity. It is a Newtonian definition of
momentum. We will rather use the Lagrangian
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definition of momentum and it is the partial
derivative of the Lagrangian with respect
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to the velocity. So, del L by del q dot is
our definition of momentum. Whenever we talk
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about momentum, we will talk about del L by
del q dot and not mass times velocity. This
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is what is called as the generalized momentum.
The definition of generalized momentum is
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that it is the partial derivative of the Lagrangian
with respect to the velocity. We can see a
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suggestion towards this and from this correspondence,
which we see over here.
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This is our generalized momentum, del L by
del q dot and that is the term we shall use
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in the future. What we have achieved by this
particular recognition of Lagrangian as T
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minus V? It is a complete equivalence with
Newtonian formulation. The foundations are
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completely different. The foundation in Newtonian
mechanics is the principle of causality, the
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cause effect relationship, the linear response
formalism.
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The foundation of Lagrangian mechanics is
the principle of variation. They have nothing
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to do with each other. These are completely
independent formulation and you cannot derive
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one from the other. These are independent
formulations, but they must converge and they
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must provide you with the same results as
long as you work within the domain of classical
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mechanics. Things change, when you go over
to quantum theory and that is a different
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story. In classical mechanics, Newtonian formulations
and the Lagrangian or Hamiltonian formulation
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must give you the same results.
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00:24:47,380 --> 00:24:54,380
Now, let us write a more general Lagrangian,
which is a function of q q dot and t. You
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will very soon see, when t must be included
and when it need not be included. If you have
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an isolated system, there are no missing degrees
of freedom. I discussed this in a different
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context earlier, but I think that sample is
a very useful one. If you are moving an object
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of a surface; you know that there is friction
at the surface and if you set up the equations
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of motions only for this bottle, but do not
include the particles of the table, which
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are also interacting with this, then you do
not have an isolated system. Then this system
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is interacting with something, which is not
taken into account in your equations of motion
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and it will show up as missing degrees of
freedom.
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When you do not have any missing degrees of
freedom, you will not have the Lagrangian
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to have any explicit time dependence. The
Lagrangian will always have implicit time
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dependence and the implicit time dependence
means that the Lagrangian depends on time
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by the virtue of its dependence on position
and also by the virtue of its dependence on
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velocity. It is because of that reason, the
Lagrangian depends on q and q depends on time.
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The Lagrangian becomes a function of time.
So, such a dependence is what is called as
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implicit dependence. If there are missing
degrees of freedom, there may be an explicit
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time dependence. So, this is the general expression
for the time derivative.
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The Lagrangian will come from its implicit
dependence on time, through its dependence
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on q, which in turn depends on time. So, there
is dq by dt over here. Dependence of the Lagrangian
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on the velocity depends on time. So, there
will be a d by dt of q dot, which is q double
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dot, which is the middle term that you see.
There will be a possible explicit time dependence,
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which is shown in the last term. So, this
is the complete expression for the total time
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derivative of a Lagrangian.
This is the Lagrange's equation. These two
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brackets must be equal to each other. Their
difference goes to 0. So, del L by del q is
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00:28:03,280 --> 00:28:10,280
d by dt of del L by del q dot and this del
l by del q is replaced by d by dt of del L
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00:28:12,620 --> 00:28:19,620
by del q dot in this step over here. You have
this term, which is del L by del q dot time
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the second derivative of q. You have the explicit
time dependence of the Lagrangian coming in
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the third term. If you look at these two terms,
they are algebraically completely equivalent
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to the derivative of a product of two functions.
The derivative of a product of two functions
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00:28:44,750 --> 00:28:51,750
is one function multiplied by the derivative
of the second plus the second function times
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the derivative of the first. It is q double
dot and so you have only rewritten this algebraically
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00:29:02,200 --> 00:29:09,200
as d by dt of a product of these two functions.
The last term shows up over here.
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Essentially, if you take this term, it is
a total derivative of the Lagrangian. This
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00:29:16,060 --> 00:29:22,070
is a total derivative of a product of these
two functions. So, I move this term to the
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left and it comes to the other side or rather
I move this del L by del t to the other side
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and move this to the right. So, just a rearrangement
of these terms gives us this relationship
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over here. The total time derivative is the
difference between the first term and the
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00:29:54,020 --> 00:30:01,020
Lagrangian. The first term is a product of
two functions.
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We are looking at the total time derivative
of the Lagrangian. It comes from implicit
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dependence on time through the explicit dependence
on position and velocity, which are explicit
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functions of time. A possible explicit dependence
of the Lagrangian on time directly comes from
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the missing degrees of freedom. So, we make
use of the Lagrange's equation. From the
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00:30:50,860 --> 00:30:57,860
Lagrange's equation, we know that these
two terms are equal. The difference is 0.
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We express del L by del q as the time derivative
of this quantity over here. We already know
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the generalized momentum and that is our definition
of the generalized momentum. We combine the
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first two terms and write this as a time derivative
of a product of two functions.
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Now, if we rearrange the terms in this last
equation, we get the time derivative of this
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00:31:33,110 --> 00:31:39,630
difference equal to minus del L by del t.
So, you move this del L by del t to the other
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00:31:39,630 --> 00:31:46,630
side and move this d L by d t to this side.
So, you have only minus del L by del t on
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00:31:47,080 --> 00:31:54,080
one side. This tells us something very interesting.
In case, the Lagrangian does not have any
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00:31:58,990 --> 00:32:05,010
explicit time dependence, then the partial
derivatives of Lagrangian with respect to
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00:32:05,010 --> 00:32:12,010
time will vanish. So, we recognize the condition
under which, the total time derivative of
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00:32:18,120 --> 00:32:25,120
a quantity vanishes. If del L by del t is
equal to 0, which is the meaning of the Lagrangian
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00:32:29,490 --> 00:32:35,020
not having any explicit time dependence. If
the Lagrangian does not have any explicit
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00:32:35,020 --> 00:32:42,020
time dependence, del L by del t will go to
0 and d by dt of this bracket must vanish.
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00:32:45,310 --> 00:32:52,310
If the derivative of a certain quantity goes
to 0, the quantity must be a constant.
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00:32:53,820 --> 00:33:00,820
Here, del L by del t is 0. Now, it guaranties
that this particular function is the constant
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00:33:06,220 --> 00:33:11,280
and this function is called as the Hamilton's
principle function. This is the definition
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of Hamilton's principle function. It is
del L by del q dot times q dot minus the Lagrangian
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00:33:22,320 --> 00:33:29,320
and del L by del q dot It is our definition
of the generalized momentum, so the definition
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00:33:29,780 --> 00:33:36,780
of our Hamilton's principle function is
that it is p q dot minus the Lagrangian. This
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00:33:37,460 --> 00:33:42,860
is the definition and meaning of Hamilton's
principle function.
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00:33:42,860 --> 00:33:49,860
We know that it is a constant, if and only
if the Lagrangian does not have any explicit
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00:33:54,290 --> 00:34:01,290
time dependence. We have also seen the correspondent
with Newtonian mechanics through these partial
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00:34:02,710 --> 00:34:09,710
derivatives, which we have shown earlier.
We can ask ourselves- what is it that we can
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00:34:11,070 --> 00:34:18,070
learn further from this correspondence? If
you see this n p q dot minus L from the Newtonian
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00:34:23,349 --> 00:34:30,349
perspective by exploiting this correspondence,
you see that p q dot is mass times velocity.
261
00:34:33,319 --> 00:34:40,319
It is mv square and you have mv square minus
Lagrangian.
262
00:34:41,329 --> 00:34:48,329
The Lagrangian is kinetic energy minus the
potential energy and therefore, the Hamilton's
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00:34:49,250 --> 00:34:55,460
principle function is 2T minus the Lagrangian.
It is 2T minus the difference of kinetic energy
264
00:34:55,460 --> 00:35:02,460
and the potential energy, which essentially
gives a meaning to the Hamilton's principle
265
00:35:03,039 --> 00:35:09,829
function. What comes out of this is T plus
V, which is obviously the total energy of
266
00:35:09,829 --> 00:35:16,829
the system. The Hamiltonian, which we have
defined is the Hamilton's principle function.
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00:35:22,779 --> 00:35:29,779
It can be immediately identified with the
total energy. It is a conserved quantity,
268
00:35:31,710 --> 00:35:37,920
a constant quantity and the constant c is
not a matter of faith.
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00:35:37,920 --> 00:35:44,920
It is expressed by the previous result that
it has come from this d by dt of this Hamilton's
270
00:35:45,690 --> 00:35:49,190
principle function. It goes to 0, whenever
del L by del t is 0.
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00:35:49,190 --> 00:35:56,190
It is not just a matter of faith, it has come
as a result of this particular condition.
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00:35:56,400 --> 00:36:02,539
Now, what is the condition that the Lagrangian
must be? Whatever time dependency it has,
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00:36:02,539 --> 00:36:09,119
it must be only through q, q dot and not directly.
So, there should be no explicit time dependence
274
00:36:09,119 --> 00:36:16,119
that the Lagrangian can have.
If the Lagrangian does not have any explicit
275
00:36:18,559 --> 00:36:25,559
time dependence, will the Hamilton's principle
function correspond to the total energy? It
276
00:36:26,180 --> 00:36:33,180
is a conserved quantity. Do you see the Noether's
theorem talking to you in this result? The
277
00:36:40,880 --> 00:36:47,880
Lagrangian is independent of time is a symmetry
that the Lagrangian will remain the same,
278
00:36:47,980 --> 00:36:54,980
whether you formulate it yesterday or today
or tomorrow or even day after tomorrow. It
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00:36:59,190 --> 00:37:06,190
is the symmetry with respect to translation
along the time axis, any time in the past
280
00:37:07,990 --> 00:37:14,990
or present or future. The Lagrangian remains
the same because no information is missing;
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00:37:15,089 --> 00:37:22,089
no degrees of freedom are lost. You cannot
do that if you were to include friction because
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00:37:23,599 --> 00:37:30,599
every time you do this, there will be differences
in the interaction. They are not taken explicitly
283
00:37:33,559 --> 00:37:40,559
in your analysis, so they are left out.
When the Lagrangian is independent of time,
284
00:37:42,369 --> 00:37:48,670
you have symmetry with respect to change in
time. Time is the parameter with respect to
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00:37:48,670 --> 00:37:55,670
which the Lagrangian is invariant. This invariance
is the symmetry principle. Associated with
286
00:37:57,230 --> 00:38:04,230
this symmetry, there is a conserved quantity,
which is the energy. The energy is a constant
287
00:38:08,220 --> 00:38:14,539
and you can see a manifestation of the Noether's
theorem associated with every symmetry principle.
288
00:38:14,539 --> 00:38:21,539
There is a conservation law and vice versa.
Now, this has come so neatly from the Lagrangian
289
00:38:23,940 --> 00:38:27,779
formulation.
290
00:38:27,779 --> 00:38:33,950
You can generalize this. If there are N degrees
of freedom, you can put a subscript p and
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00:38:33,950 --> 00:38:40,240
q dot and write the Hamilton's principle
function as the summation over i going from
292
00:38:40,240 --> 00:38:46,839
1 through N. So, this is a straightforward
extension. Let me remind you that the object
293
00:38:46,839 --> 00:38:53,839
of this course is not to get into the details
of Lagrangian formulation or into the details
294
00:38:54,500 --> 00:39:01,500
of Hamiltonian formulation. This is an introductory
course in classical mechanics. This is what
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00:39:02,099 --> 00:39:07,559
students would take as one of the first courses
in classical mechanics after high school or
296
00:39:07,559 --> 00:39:13,339
at some undergraduate level. At that level,
one would not take up any problems involved
297
00:39:13,339 --> 00:39:17,700
in classical mechanics in the Lagrangian formulation
or Hamiltonian formulation.
298
00:39:17,700 --> 00:39:21,990
We will not take up systems with many degrees
of freedom and so on. We will only illustrate
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00:39:21,990 --> 00:39:28,990
some examples. The focus of this course is
to reveal an alternative formulation of classical
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00:39:30,740 --> 00:39:37,740
mechanics as much as Newtonian mechanics.
It is inspired by the principle of inertia.
301
00:39:38,180 --> 00:39:45,180
Galileo's interpretation of equilibrium
goes into the first law. Newton's interpretation
302
00:39:45,299 --> 00:39:52,069
of departures from first law or from equilibrium
generated by interactions revealed by the
303
00:39:52,069 --> 00:39:57,200
cause effect relationships contained in the
equation f equal to m a, which is the second
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00:39:57,200 --> 00:40:04,200
law. An alternative formulation of mechanics
exists and it has a charm of its own. It has
305
00:40:06,759 --> 00:40:13,759
a beauty of its own, it is based on a completely
different principle. It is this alternative,
306
00:40:17,009 --> 00:40:24,009
we are introducing to the students in this
course.
307
00:40:25,430 --> 00:40:32,430
The quantities that we are going deal in this
formulation will be the generalized coordinate,
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00:40:33,529 --> 00:40:39,240
the generalized velocity and the generalized
momentum. So, we shall no longer talk about
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00:40:39,240 --> 00:40:46,240
coordinate, if it is just a position coordinate
in a Cartesian geometry. The definition of
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00:40:51,730 --> 00:40:56,900
the generalized momentum requires the Lagrangian
to be set up because unless you setup the
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00:40:56,900 --> 00:41:03,069
Lagrangian, you cannot determine its partial
derivative with respect to q dot. So, you
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00:41:03,069 --> 00:41:07,299
set up the Lagrangian and take its partial
derivative with respect to q dot. Now, you
313
00:41:07,299 --> 00:41:13,839
know what the generalized momentum, until
then it is unknown to us.
314
00:41:13,839 --> 00:41:17,960
We have seen that it reveals a connection
between symmetry and conservation law. We
315
00:41:17,960 --> 00:41:24,960
have already seen such an illustration. Because
of the fact that this formulation is so general,
316
00:41:28,220 --> 00:41:35,220
this can be easily extended to other expressions
of the Noether's theorem and I will illustrate
317
00:41:35,240 --> 00:41:39,400
another one.
318
00:41:39,400 --> 00:41:42,980
This is the one that we have already seen
that the Lagrangian of a closed system does
319
00:41:42,980 --> 00:41:48,940
not depend explicitly on time. Therefore,
del L by del t is 0 and the symmetry principle
320
00:41:48,940 --> 00:41:53,559
associated with this is the conservation principle
that the Hamilton's principle function is
321
00:41:53,559 --> 00:42:00,559
a constant and you call this as the total
energy. So, there is a connection between
322
00:42:01,420 --> 00:42:08,420
symmetry and conservation law that we have
seen over here. It is in this context that
323
00:42:09,180 --> 00:42:15,160
you call the Hamilton's principle function
as the energy. Otherwise, you simply call
324
00:42:15,160 --> 00:42:22,160
it as Hamilton's principle function, but
when it is a constant, you call it as the
325
00:42:22,690 --> 00:42:23,900
energy.
326
00:42:23,900 --> 00:42:30,900
Now, let us consider another symmetry. We
will now consider symmetry with respect to
327
00:42:30,999 --> 00:42:37,470
space. We considered symmetry with respect
to time, del L del t being 0. It is symmetry
328
00:42:37,470 --> 00:42:43,039
with respect to time that is it is the same
Lagrangian any point of time. So, its partial
329
00:42:43,039 --> 00:42:50,039
derivatives vanishes. Now, we consider changes
in the Lagrangian because of translational
330
00:42:50,880 --> 00:42:57,880
displacements of the system. Is there any
invariance coming because of this? This invariance
331
00:43:00,460 --> 00:43:07,460
would obviously come in homogeneous space.
If the space is homogeneous, you cannot have
332
00:43:08,319 --> 00:43:15,319
a different Lagrangian. So, delta L changes
in the Lagrangian because of changes in the
333
00:43:17,369 --> 00:43:21,900
coordinates.
Just to illustrate this, I will make use of
334
00:43:21,900 --> 00:43:27,369
the Cartesian coordinates, which is the simplest
or most familiar coordinate system that people
335
00:43:27,369 --> 00:43:34,369
use. So, this delta L will be determined by
partial derivative of L with respect to x
336
00:43:36,150 --> 00:43:43,150
times the displacement, delta x plus similar
term from the dependence of the Lagrangian,
337
00:43:44,160 --> 00:43:51,160
if any on y. Likewise, this term involves
the z coordinate. If this delta L is equal
338
00:43:54,029 --> 00:44:01,029
to 0, no matter what delta x is or no matter
what delta y is or no matter what delta z
339
00:44:03,180 --> 00:44:08,910
is. The sum of these three terms is not going
to 0 because delta x, delta y, delta z takes
340
00:44:08,910 --> 00:44:12,779
some special values, which adjusts with each
other. So that the sum of the three terms
341
00:44:12,779 --> 00:44:17,749
can go to 0.
You can always have 2 plus 1 minus 3 equal
342
00:44:17,749 --> 00:44:24,749
to 0 sum of three terms going to 0. Then,
none of the three terms is individually 0.
343
00:44:27,029 --> 00:44:32,930
This cannot be the case because delta x, delta
y, delta z are arbitrary displacements. So,
344
00:44:32,930 --> 00:44:38,319
no. matter what delta x is, no matter what
y is and no matter what delta z is, if delta
345
00:44:38,319 --> 00:44:45,319
L is 0. This is the statement of symmetry
and it is coming from the homogeneity of space
346
00:44:47,140 --> 00:44:54,140
and del L by del q must be 0, no matter what
delta x, delta y, delta z is. So, this result
347
00:45:01,170 --> 00:45:08,170
del L by del q equal to 0 comes from the consideration
of space being homogenous. Now, we makes use
348
00:45:10,240 --> 00:45:16,339
of the Lagrange's equation because del L
by del q minus time derivative of this term
349
00:45:16,339 --> 00:45:22,029
is equal to 0 and del L by del q is 0. It
means that this time derivative of this factor
350
00:45:22,029 --> 00:45:29,029
is 0 and this is of course the momentum. The
time derivative of the momentum is 0 and if
351
00:45:31,259 --> 00:45:38,259
the derivative of a function goes to 0, the
function must be a constant. So, we have exactly
352
00:45:38,890 --> 00:45:44,339
got a conservation principle that is absolutely
right.
353
00:45:44,339 --> 00:45:51,339
You see that the conservation of momentum
keeps coming every time, when there is symmetry.
354
00:45:52,140 --> 00:45:58,349
Earlier, we met the symmetry with respect
to time. Now, we meet symmetry with respect
355
00:45:58,349 --> 00:46:04,269
to translational displacements in homogeneous
space. We have discussed this earlier in the
356
00:46:04,269 --> 00:46:11,269
context of Newtonian mechanics. Here, you
see how it comes very naturally and easily
357
00:46:12,049 --> 00:46:19,049
out of Lagrange's equations. You know that
whenever the Lagrangian is independent of
358
00:46:23,980 --> 00:46:30,980
this coordinate q, is' What would happen,
if del L by del q goes to 0 every time? The
359
00:46:31,279 --> 00:46:38,279
Lagrangian is independent of a degree of freedom
q and the corresponding momentum will be constant.
360
00:46:39,269 --> 00:46:46,269
Now, this is a connection between symmetry
and conservation law. So, del L by del q equal
361
00:46:47,690 --> 00:46:54,690
to 0 guarantees that del L by del q dot equal
to p is conserved. The law of conservation
362
00:46:55,039 --> 00:47:02,039
of momentum arises from the homogeneity of
space. This is very nicely seated by this
363
00:47:03,069 --> 00:47:10,069
very well known law that the momentum, which
is canonically conjugate to a cyclic coordinate
364
00:47:12,349 --> 00:47:19,349
is conserved. Whenever, the Lagrangian is
independent of a coordinate, this coordinate
365
00:47:23,859 --> 00:47:28,730
is said to be a cyclic coordinate and that
is the meaning of a cyclic coordinate. Cyclic
366
00:47:28,730 --> 00:47:32,430
coordinate is a one, which does not appear
in the Lagrangian.
367
00:47:32,430 --> 00:47:39,430
If the Lagrangian is independent of the coordinate,
its partial derivative with respect to q vanishes.
368
00:47:41,670 --> 00:47:45,910
Therefore, the partial derivative of the Lagrangian
with respect to q dot is the corresponding
369
00:47:45,910 --> 00:47:52,910
velocity, which gives you the corresponding
momentum. This correspondence is implied by
370
00:47:54,410 --> 00:48:01,410
the term of canonical conjugation. This canonical
conjugation refers to this particular explicit
371
00:48:04,519 --> 00:48:11,519
correspondence between momentum and position.
So, the momentum p is said to be canonically
372
00:48:13,609 --> 00:48:20,609
conjugate to a coordinate. It is given by
the partial derivative of the Lagrangian with
373
00:48:20,650 --> 00:48:27,650
respect to the corresponding velocity del
L by del q dot. So that is the momentum, which
374
00:48:27,700 --> 00:48:32,099
is canonically conjugate to a cyclic coordinate
is conserved.
375
00:48:32,099 --> 00:48:39,099
Like I said, extend this for many degrees
of freedom. You will have to sum over all
376
00:48:41,559 --> 00:48:46,960
the degrees of freedom. So, it is essentially
the same original expression for the Hamiltonian,
377
00:48:46,960 --> 00:48:53,960
which you wrote for all degrees of freedom.
Now, you make demands of the properties of
378
00:48:55,039 --> 00:49:02,039
the Hamiltonian. What would be a differential
increment in the Hamilton's principle function?
379
00:49:02,519 --> 00:49:09,519
So, this dH will get it and it is the increment
from changes in q k dots, which are changes
380
00:49:12,220 --> 00:49:18,940
in the velocities, changes in the momenta
as the product is coming in the Hamiltonian,
381
00:49:18,940 --> 00:49:25,200
because of changes in the Lagrangian.
So, the increment in dH of the Hamilton's
382
00:49:25,200 --> 00:49:32,200
principle function comes from the change in
this product. This product can change because
383
00:49:36,220 --> 00:49:40,859
of either of these two terms. So, it will
be the first function times increment in the
384
00:49:40,859 --> 00:49:47,859
second and the second function times the increment
in the first, so they both are contributors.
385
00:49:50,359 --> 00:49:57,359
So, the change in the product, q k dot p k
will be a sum of first two terms. You must
386
00:50:02,529 --> 00:50:08,989
subtract from this because of this minus sign.
The increment or the change in the Lagrangian
387
00:50:08,989 --> 00:50:15,989
is due to the dependence of the Lagrangian
on q, a change in q and also the dependence
388
00:50:16,900 --> 00:50:23,549
of the Lagrangian on the velocity and the
change in velocity. Of course, you must sum
389
00:50:23,549 --> 00:50:30,549
over all the degrees of freedom.
Each term has been summed over k, but there
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is something we ought to notice. Some of you
would have already noticed that if you look
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00:50:39,789 --> 00:50:46,789
in the first and the last term, both involve
increments in the velocity dq dot. For each
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00:50:53,279 --> 00:51:00,279
degree of freedom, a degree of freedom is
necessarily independent. The corresponding
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00:51:00,319 --> 00:51:06,660
coefficients must be equal because this one
comes with the plus sign and this one comes
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00:51:06,660 --> 00:51:13,660
with the minus sign. So, these two terms must
cancel each other. They must cancel each other
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00:51:18,859 --> 00:51:25,859
because del L by del q dot is nothing but
the momentum itself.
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00:51:27,470 --> 00:51:31,730
What is p? It is the partial derivative of
the Lagrangian with respect to the velocity.
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00:51:31,730 --> 00:51:38,730
So, those two terms cancel and you are left
with the middle two terms. In the middle two
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00:51:46,619 --> 00:51:53,619
terms, the coefficient of dq in the second
term is del L by del q. It is equal to the
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00:51:54,599 --> 00:52:01,480
time derivative of the momentum. As we know
from Lagrange's equation that del L by del
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00:52:01,480 --> 00:52:08,480
q must equal to d by dt of the corresponding
moment. So, del L by del q is now identified
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00:52:11,819 --> 00:52:18,819
as p dot. The subscript k must be kept off
the track and you must sum over all the terms.
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00:52:19,890 --> 00:52:26,890
Now, this is our expression for the increment
in the Hamilton's principle function. Hamilton's
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00:52:28,920 --> 00:52:35,920
principle function's in a way as Hamiltonian
mechanics is done. It must be written as a
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00:52:43,009 --> 00:52:50,009
function of position and momentum and not
as a function of position or velocity. Now,
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00:52:50,700 --> 00:52:55,170
these are two different formulations - the
Lagrangian formulation and the Hamiltonian
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00:52:55,170 --> 00:52:58,730
formulation. I am going to rub this point
and I am going to hyphen this any number of
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00:52:58,730 --> 00:53:05,730
times, even at the cost of repetition.
The Lagrangian formulation must be used in
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00:53:05,999 --> 00:53:12,509
terms of position and velocity. The Hamilton's
formulation must be used in terms of position
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00:53:12,509 --> 00:53:19,509
and momentum. So, we express the Hamiltonian
as a function of position q and momentum p.
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00:53:27,730 --> 00:53:31,460
It is either position or momentum or both
momentum and position. It is the same thing
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00:53:31,460 --> 00:53:38,460
and does not matter which one you write before.
These are classical dynamical variables and
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00:53:38,859 --> 00:53:43,670
the order in which you write them does not
matter.
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00:53:43,670 --> 00:53:50,670
This dependence on position and momentum gives
the expression for the differential increment
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00:53:54,249 --> 00:53:59,980
in the Hamilton's principle function as
dH equal to the partial derivatives of the
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00:53:59,980 --> 00:54:06,980
Hamiltonian with respect to p times the increment
in p. Where can this increment come from?
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00:54:07,319 --> 00:54:14,319
It can come only from increments in p and
q because that is what the Hamiltonian depends
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00:54:15,430 --> 00:54:22,430
on. The corresponding coefficients will obviously
be the partial derivatives of the Hamiltonian
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00:54:23,400 --> 00:54:28,769
with respect to the corresponding Hamiltonian
parameters. So, del H by del p times dp plus
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00:54:28,769 --> 00:54:34,509
del H by del q times dq. I think it is always
good to see this physics behind the partial
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00:54:34,509 --> 00:54:41,509
derivative, so that you do not look at it
as just mechanical algebraic mathematics or
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00:54:41,519 --> 00:54:48,519
calculus through each term. It is physics,
which is talking to us that there is an increment
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00:54:52,989 --> 00:54:59,989
in Hamiltonian. This increment can only come
from the increments in momenta and from the
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00:54:59,999 --> 00:55:05,900
increments in the coordinates because those
are the quantities on which, the Hamiltonian
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00:55:05,900 --> 00:55:12,900
depends on. So, dH becomes this and this must
hold good for each degree of freedom, independently.
425
00:55:20,089 --> 00:55:27,089
This is how we have two expressions for the
differential increment in the Hamiltonian.
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00:55:30,309 --> 00:55:37,309
One is this and the other is this. Both involve
arbitrary increments in the momenta and arbitrary
427
00:55:43,869 --> 00:55:50,869
increments in the coordinates. They all add
up to the same increment in the Hamilton's
428
00:55:51,609 --> 00:55:57,170
principle function. So, the corresponding
coefficients must be equal and they better
429
00:55:57,170 --> 00:56:04,170
be right for each degree of freedom.
The coefficient of dp in this is q k dot must
430
00:56:07,180 --> 00:56:12,529
be equal to the coefficient of dp in this
term. It is del H by del p so del H by del
431
00:56:12,529 --> 00:56:19,529
p. It becomes equal to q dot and the coefficient
of this dq, which is p dot must be equal to
432
00:56:21,529 --> 00:56:28,349
this coefficient of dq over here. There is
a minus sign over here, so this is carried
433
00:56:28,349 --> 00:56:35,349
over here. So, these equations that you see
at the bottom, del H by del p equal to q dot
434
00:56:36,859 --> 00:56:43,859
and del H by del q equal to minus p dot are
known as Hamilton's equations of motion.
435
00:56:44,170 --> 00:56:50,950
What do they tell us? They tell us how q and
p evolve with time. That describes how a mechanical
436
00:56:50,950 --> 00:56:57,599
system evolves with time. That is the fundamental
problem in mechanics. How do you describe?
437
00:56:57,599 --> 00:57:03,059
How do you characterize the state of a mechanical
system? How does this evolve with time? It
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00:57:03,059 --> 00:57:10,059
is time dependence and it is time derivative.
What mathematical equation provides this time
439
00:57:12,470 --> 00:57:19,470
evolution? It is an equation of motion; it
connects the position, velocities and accelerations.
440
00:57:21,109 --> 00:57:28,109
It provides an equation of motion and these
are known as Hamilton's equations of motion.
441
00:57:34,190 --> 00:57:41,190
There are some issues, which I have not gone
into a great depth. I mentioned them for those
442
00:57:44,989 --> 00:57:51,989
who are interested, intelligent and who would
like to take proactive steps to read further,
443
00:57:53,930 --> 00:58:00,930
but I am not going to discuss that in this
course. I have emphasized that this alternative
444
00:58:01,380 --> 00:58:08,380
formulation is based on the principle of variation.
It requires the action integral to be an extremum.
445
00:58:10,930 --> 00:58:16,960
I pointed out this extremum could be a minimum,
it could be a maximum, it could be a saddle
446
00:58:16,960 --> 00:58:23,829
point. In more general terms, it is called
as the stationary point.
447
00:58:23,829 --> 00:58:30,829
There are complex questions that one can ask.
Is the stationary, local or global? These
448
00:58:32,049 --> 00:58:37,390
are fairly complex issues, fairly advanced
concept. They go well beyond the scope of
449
00:58:37,390 --> 00:58:44,390
this course. I will not get into those details.
Usually, we know that even from elementary
450
00:58:47,109 --> 00:58:52,769
calculus, we distinguish between a minimum
and maximum, only by looking at the second
451
00:58:52,769 --> 00:58:59,769
derivatives. So, there are additional differentials,
which come into play. I will not discuss that
452
00:59:03,160 --> 00:59:09,450
and with this, we are ready to conclude this
particular class.
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00:59:09,450 --> 00:59:14,190
I will mention a few references. There are
some excellent papers in the American journal
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00:59:14,190 --> 00:59:21,190
of physics, which you will find very instructive.
In particular, I like this title very much
455
00:59:21,869 --> 00:59:28,869
- Getting the most action out of least action.
This is really a very catchy title for a paper
456
00:59:31,680 --> 00:59:34,710
because the least action that he is talking
is obviously the principle of least action
457
00:59:34,710 --> 00:59:41,710
and not the least action. We always like to
do, which is do nothing. So, the least action
458
00:59:42,319 --> 00:59:49,319
is what Thomas Moore refers to. He gets a
lot of it and the title is very nice - Getting
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00:59:52,190 --> 00:59:59,190
the most action out of least action. There
are other papers by Taylor and his collaborators,
460
01:00:00,109 --> 01:00:07,109
Hanca and so on. You might find these papers
quite instructive and with that we will conclude
461
01:00:10,450 --> 01:00:11,150
this class.