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Towards
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the end of yesterday’s class, we learnt
on basic mechanism of deriving the first
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00:00:54,379 --> 00:00:59,690
order differential equation. And we said that
we will define the new set of coordinates
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as
p i which will be the derivative of the Lagrangian
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function with respect to q i dot.
.
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So, that is how we defined it and then if
we define this, then the derivative time
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00:01:16,620 --> 00:01:30,340
derivative of this term is nothing but, p
i dot, so your Lagrangian equation becomes,
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equation become p i dot minus derivative of
the Lagrangian with respect to q i plus
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00:01:39,150 --> 00:01:49,710
derivative of the Rayleigh with respect to
q i dot is equal to 0. We would normally
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00:01:49,710 --> 00:01:57,970
express it as p i dot is equal to derivative
of the Lagrangian with respect to q i minus
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00:01:57,970 --> 00:02:01,930
derivative of the Rayleigh with respect to
q i dot.
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So, you see, here is the dotted term and the
right hand side will have to be expressed
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in
terms of the un dotted terms. So, if you want
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to derive the first order differential
equations of any system, essentially you would
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need to write, q i dot is equal to
something and p i dot is equal to something.
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00:02:28,340 --> 00:02:29,340
..
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Now, p i dot is equal to something, that is
obtained from this equation and q i dot is
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equal
to how do you get it, how do you get it. You
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will actually get it from
this, let me illustrate how. Let us take,
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let us start with
you have already done the
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00:02:59,730 --> 00:03:07,250
spring pendulum system you have already the
Lagrangian written up, so let us start from
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00:03:07,250 --> 00:03:08,250
there.
.
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So, the spring pendulum system was, and we
had defined the…
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.
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.A plus r, where a was the constant and r
is the variable, and in it is term, we had
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written
already written, the Lagrangian as, this I
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will not derive all over again, I will simply
write, it was half m r dot square plus half
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m a plus r square theta dot square minus half
k
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00:04:02,830 --> 00:04:30,889
r plus m g by k square plus m g a plus r cos
theta minus m g a, that is how it was. So,
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this
is the Lagrangian function and then we know,
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that the p i is the derivative of the
Lagrangian with respect to q i dot, which
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in this case r and theta.
So, we will write p 1 is equal to derivative
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of the Lagrangian with respect to q 1 dot,
so q
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1 was r, so it was, it will be convenient
for us to in order to proceed, we will say,
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00:05:05,830 --> 00:05:11,409
q 1 is
equal to r and q 2 is equal to theta. So,
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00:05:11,409 --> 00:05:21,789
this is, let us write things in terms of q
1 and q 2,
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so this is,
.
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M.
.
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00:05:27,039 --> 00:05:34,259
Q 1 dot.
.
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So, that is the in the first coordinate the
momentum and in the second coordinate the
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momentum is, it has to be done with here,
so it is m,
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.
A plus q 1, a plus square into q 2 dot, so
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00:06:10,450 --> 00:06:13,999
notice that you have independently obtain
the
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00:06:13,999 --> 00:06:22,659
momentum in the two directions. Now, writing
the differential equation, so this equation
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00:06:22,659 --> 00:06:31,389
will give q 1 dot, the dotted term in the
left hand side is equal to,
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.
That is it, and this will give is, so you
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see, out of the four equation that you need
to
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00:06:55,759 --> 00:07:04,899
derive, two were simply obtained from this,
now to the other two. The other two would
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00:07:04,899 --> 00:07:10,539
be, where p 1 dot is in the left hand side
and p 2 dot will be in the left hand side,
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00:07:10,539 --> 00:07:14,250
so that
will have to be obtained as this equation
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.. This equation in this case
the R term is not there, so only this much.
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00:07:20,729 --> 00:07:21,729
..
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00:07:21,729 --> 00:07:28,520
So, we will have to write, P 1 dot is the
derivative of the Lagrangian with respect
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00:07:28,520 --> 00:07:34,020
to q 1,
which is something that we had already written
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up, so it is…
.
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00:07:39,979 --> 00:07:45,400
One thing, now, it has to be done in terms
of…
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.
In the right hand side, things have to be
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expressed in un dotted terms, in the right
hand
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side things will have to be, without the dots,
so just keep that in mind. So here, we have
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already written down the derivative of the
Lagrangian with respect to q 1, so write it
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in
the right hand side, so I am yet uncomfortable
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of this term, because here i have a dotted
term, which i do not want, in the right hand
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side, so that can easily be eliminated by
substituting this ..
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So, that gives finally, the equation, so it
will be, this q 2 dot is p 1 by m, so 1 m
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00:09:10,731 --> 00:09:14,029
will
cancel off, it is p,
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00:09:14,029 --> 00:09:21,940
.
2 square here by m, and then it will continue
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a plus q 1 minus k, all the other terms are
without dots, so…
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..
Sorry, yes.
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00:09:42,410 --> 00:09:45,190
.
Cube.
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00:09:45,190 --> 00:09:50,750
.
No, no.
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00:09:50,750 --> 00:10:09,750
.
No, no, it will remain yes, so this is m g
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by k plus m g cos q 2, so we have obtain one
equation, one first order equation without
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any dotted term in the right hand side, have
you seen the technique. So, the first point
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was, that in this problem we had the
Lagrangian already derived, we say that let
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q 1 be r and q 2 be theta, and then your p
1 is
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derivative of the Lagrangian with respect
to q 1 dot, which yield at this, so that directly
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yields q 1 dot, as one first order differential
equation.
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P 2 was this, that directly yields the q 2
as the first order differential, q 2 dot as
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the first
order differential equation. P 1 dot, then
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is obtained from this equation and in this
specific problem, this term being 0, it is
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only this much, which we wrote down. We had
already derived this right hand side this
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one, but here a q 2 dot appeared, which should
not appear in the right hand side of a differential
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equation, so we substitute q 2 dot from
there and then we got the final equation.
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Similarly, p 2 dot is, it is the derivative
of the Lagrangian with respect to q 2, which
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is
you have already have it, it is minus m g.
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.
a plus q 1.
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.
Yes, it does not have a dotted term in the
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right hand side, so we are through. Do you
see
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the four equations, so this is how, the first
order equation need to be defined, if you
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are
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.obtaining it from the Lagrangian equation.
Let us do one problem, one circuit problem
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so
that you are little more comfortable with
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it.
.
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Let us see, let us circuit like this, notice
that if we did not derive or did not define
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the
. p i is this way, then all this will have
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to be obtained at hock. And,
getting rid of the dotted terms in the right
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hand side would not be, so simple as this,
that
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is why this is a systematic procedure of doing
that that is why the position and the
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momentum, these are taken as the conjugate
variables always, so here is E, here is C,
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L 1
L 2 and R.
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How . go about it, first we will define the
charges flowing in the
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loops, which are q 1 and q 2, these are the
position coordinates. Now, you would say,
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that my T is the energy stored in this one,
half L 1 q 1 dot minus q 2 dot square, plus
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the
energy stored in this one, which is plus half
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L 2 q 2 dot square, the potential here V is
the
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1 by twice C q 1 square, also due to this,
it will be minus since it is in the direction
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of q
1, E q 1.
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The this time there is a Rayleigh term, so
in the last problem it was not there, it is
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only in
q 2 direction, so it is half R q 2 dot square.
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So, from here, your Lagrangian is half L 1
q 1
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dot minus q 2 dot square plus half L 2 q 2
dot square minus 1 by twice C q 1 square plus
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E q 1, so you have it, you have it, so you
can easily write the Lagrangian equation,
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which
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.will now, take the form, can you just obtain
it fast, the Lagrangian way. Then I will
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obtain it, as a first order differential equation,
done so keep it in your copy.
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Now, notice what we have done on the sheet,
here we are writing, that the derivative of
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the Lagrangian with respect to the q 1 dot
is p 1, which is nothing but, this you have
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already definitely done it. Similarly, p 2
is derivative of the Lagrangian with respect
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to q
2 dot which is this, now in this case, the
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q 1 dot and q 2 dot are not directly given,
so
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they are mixed up. But you can see, these
are two equations and you can already very,
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very easily extract q 1 dot and q 2 dot from
these two, in terms of p 1 and p 2, cannot
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you, do that.
.
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So, you will get q 1 dot is equal to 1 by
L 1 plus 1 by L 2 p 1 plus 1 by L 2 p 2, and
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q 2
dot is equal to 1 by L 2 p 1 plus 1 by L 2
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p 2, check, from here sorry, from here ((Refer
Time: 18:33)) from these 2 equations, we extract
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q 1 dot and q 2 dot. Having done so we
will proceed to write down, the equations
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for p 1 dot and p 2 dot, so in this case,
they
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will have to be obtained from this equation,
because the R term is there. So, p 1 dot is
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equal to derivative of the Lagrangian with
respect to the q i q 1, that will come to
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be
. these two terms.
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So, it will be minus q 1 by c plus E, now
we have to add the derivative with respect
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to q
1 dot, which is not there, so this ., P 2
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dot is derivative of the
Lagrangian with respect to the q 2 minus derivative
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of the Rayleigh with respect to q 1
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.dot, q 2 dot is, q 2 is not there, so this
terms derivative goes to 0 and this term derivative
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remains, so it is minus R q 2 dot. But, now
you have the q 2 dot in the right hand side,
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that will have to be substituted, so just
put it in here, so it is equal to minus R
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into all
these this stuff, R by L 2 you can write,
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is it pretty simple stuff.
There is no complication this business, only
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if you adopt the right technique, and the
. of the technique was to properly define
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the additional variable.
Now, let us ask this question, what did we
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do, we started from the Lagrangian
formulation where we had, the second order
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differential equations, then we additionally
defined, the conjugate momentum variables
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and then using that we derived the first order
equations.
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But ultimately if, the first order equations
is, our objective, why not derive them in
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00:21:36,800 --> 00:21:40,460
the
first order, why not go through the convoluted
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route of first deriving the second order
equation and then going to the first order
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equation. There exist, a technique to do that
applicable under most situations, so let us
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see that is exactly what is known as the
Hamiltonian formalism.
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.
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In the early 1830’s Hamilton proposed this
formalism and that is why it is known as the
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Hamiltonian formalism. In which, in the Lagrangian
formalism, we define the
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Lagrangian function, what was it, T minus
V, in the Hamiltonian form of function,
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formalism we will define the Hamiltonian function.
And in order to understand, what the
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00:22:42,860 --> 00:22:49,400
.Hamiltonian function is at this for more
systems, if you read the books in classical
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00:22:49,400 --> 00:22:57,500
mechanics it goes in a different way.
But, we will go through a route that is more
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intuitive for the engineers, because we
understand what a kinetic energy is, we understand
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what a potential energy is, and
therefore, we understand what the total energy
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in a system is, so the total energy system
is nothing but, T plus V. So, we will say
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that I define a function, I will write it
as script H
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is T plus V, and then we will try to formulate
the, Hamiltonian try to formulate the
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00:23:30,110 --> 00:23:34,550
equations in terms of the Hamiltonian function,
but presently we have define, what is
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known as the total energy function, as yet
I am not talking about the Hamiltonian
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function.
So, this is more intuitively clear to us,
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that I am talking in terms of the total energy
in the
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system T plus V. Now, there is a specific
property of the T term, T is the kinetic energy,
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it has a specific property, that it is in
mathematical term it is called the homogeneous
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00:24:06,190 --> 00:24:09,910
of
the degree 2. Now, what does it mean by the
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00:24:09,910 --> 00:24:31,830
homogeneous function of degree 2, what it
means is that if I ok, it was T would be a
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00:24:31,830 --> 00:24:38,910
function of what, suppose it is two dimensional
system, it is function of q 1 dot and q 2
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dot, it is a function of q 1 dot and q 2 dot.
So, it is normally and q 2 dot, so there are
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two variable velocity terms q 1 dot and q
2 dot
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and this fellow is a function of that. The
question concerns, what kind of a function
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is it,
it is that kind of a function, so that if
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I multiply the q 1 dot and q 2 dot both by
some
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number k that means, I just scale it up and
down, by some number k. Then the T,
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.
Yes, it will be reflected as k square terms
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times T, so any function that obeys these
property is called a homogeneous function
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00:25:38,790 --> 00:25:45,260
of degree 2. And there are some mathematical
rules, that this homogeneous functions of
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00:25:45,260 --> 00:25:50,840
degree 2 follow and we will use that so what
is
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00:25:50,840 --> 00:26:03,440
it, suppose I, now differentiate this equation
with respect to k, what do you have, we will
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00:26:03,440 --> 00:26:08,940
first say, I am differentiating this with
respect to k, then first I will have differentiate
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00:26:08,940 --> 00:26:15,030
with respect to k q 1 dot and then k q 1 dot
with respect to k, so what we have, we have
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00:26:15,030 --> 00:26:20,760
q
1 dot times derivative of T, with respect
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00:26:20,760 --> 00:26:27,100
to k q 1 dot, that is the first one chain
rule.
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00:26:27,100 --> 00:26:43,280
So, the second term will be derivative of
T with respect to k q 2 dot, so I am
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00:26:43,280 --> 00:26:49,520
differentiating with respect to k using the
chain rule, it is a function of two things,
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00:26:49,520 --> 00:26:50,520
so I
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00:26:50,520 --> 00:26:55,800
.first differentiate with respect to this
and k q 1 dot with respect to k, that yields
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00:26:55,800 --> 00:27:01,350
q 1 dot.
And similarly, the next one is equal to in
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00:27:01,350 --> 00:27:09,640
the right hand side, what do you have, twice
k
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00:27:09,640 --> 00:27:24,660
T. Now, this is again a property of the homogeneous
function of degree 2 this equation,
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00:27:24,660 --> 00:27:29,730
and k being arbitrary, we could choose any
k still this equation will be valid, if we
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00:27:29,730 --> 00:27:36,470
choose k is equal to 1, then also this equation
will be valid, choose k is equal to 1.
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00:27:36,470 --> 00:27:37,470
.
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00:27:37,470 --> 00:27:51,760
So, for k is equal to 1 it will become q 1
dot, derivative of T with respect to q 1 dot,
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00:27:51,760 --> 00:27:56,240
plus
q 2 dot, derivative of T with respect to q
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00:27:56,240 --> 00:28:07,130
2 dot, will become equal to twice T q 1 dot
q 2
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00:28:07,130 --> 00:28:16,811
dot. So, if this is true for two variables,
we had assumed q 1 dot and q 2 and q 1 these
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00:28:16,811 --> 00:28:19,230
are
the two variables, if there are N number of
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00:28:19,230 --> 00:28:27,370
variables it will be the same, after all you
have to do is to, expand this in a longer
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00:28:27,370 --> 00:28:31,290
chain, so in general we can write, that this
left
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00:28:31,290 --> 00:28:48,320
hand side will be, summation of q i dot, how
many n, and this side is twice T.
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00:28:48,320 --> 00:29:08,360
Now, this result is known as the Euler’s
theorem that goes by the name of Euler, now
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00:29:08,360 --> 00:29:27,340
notice, that this term T, so we already know,
that the V term is normally independent of
207
00:29:27,340 --> 00:29:40,490
potentially it is not dependent on the velocities,
at least normally. We will come across
208
00:29:40,490 --> 00:29:48,150
situations, where it can be and in those cases
we will have to exercise some caution, in
209
00:29:48,150 --> 00:29:53,590
deriving the differential equations, but for
many system, most systems this is true. If
210
00:29:53,590 --> 00:30:00,850
this
is true then
211
00:30:00,850 --> 00:30:06,059
if this is so then we can write the Lagrangian
here, that is what we did earlier.
212
00:30:06,059 --> 00:30:07,059
..
213
00:30:07,059 --> 00:30:16,620
So, this yields summation over n, q i dot
derivative of the Lagrangian with respect
214
00:30:16,620 --> 00:30:28,300
to dot
is equal to twice T. So, this result we will
215
00:30:28,300 --> 00:30:33,990
use, just keep it in hand for now, now we
had
216
00:30:33,990 --> 00:30:41,660
said that we have we want to deal with, the
total energy function H, which was T plus
217
00:30:41,660 --> 00:30:52,180
V,
We will write it as twice T minus T minus
218
00:30:52,180 --> 00:30:57,620
V, twice T we know, what it is, T minus V
we
219
00:30:57,620 --> 00:31:09,430
know what it is. So, substitute you get, some
over I, q i dot derivative of the Lagrangian
220
00:31:09,430 --> 00:31:20,170
with respect to q i dot, that is what is the
twice T term minus the Lagrangian itself.
221
00:31:20,170 --> 00:31:32,840
And, this term also we know, what is this,
the pi, this term is nothing but, pi, so we
222
00:31:32,840 --> 00:31:42,110
can
write, q i dot p i minus L in the right hand
223
00:31:42,110 --> 00:31:50,170
side, now it is so happens, that this term,
this
224
00:31:50,170 --> 00:32:13,850
term in the right hand side is known as the
Hamiltonian function. Hamilton actually
225
00:32:13,850 --> 00:32:20,520
define this function and then showed, that
it is actually the same as the total energy
226
00:32:20,520 --> 00:32:27,120
function T plus V. But we had started from
the total energy, because for us it is a more
227
00:32:27,120 --> 00:32:31,850
intuitively appealing quantity and we showed
that the opposite, the total energy is
228
00:32:31,850 --> 00:32:34,950
nothing but, this function, but either be
it is a same thing.
229
00:32:34,950 --> 00:32:40,730
Now, notice this, we have this equation as
the total energy.
230
00:32:40,730 --> 00:32:47,710
.
No not yet, there has been no use of the Rayleigh
231
00:32:47,710 --> 00:32:49,990
function at all, because Rayleigh
function does not contribute to the total
232
00:32:49,990 --> 00:32:57,240
energy. Energy is nothing but, kinetic energy
plus potential energy, so Rayleigh function
233
00:32:57,240 --> 00:32:59,780
is entirely different affair, it does not
234
00:32:59,780 --> 00:33:06,860
.contribute to the total energy content of
the system, just two components. We will
235
00:33:06,860 --> 00:33:13,770
introduce the Rayleigh in the right place,
do not worry, we are not ignoring that so
236
00:33:13,770 --> 00:33:16,130
this is
the Hamiltonian function.
237
00:33:16,130 --> 00:33:24,130
Now, so we have two things in the left hand
side we have the H function, and in the right
238
00:33:24,130 --> 00:33:34,730
side we have the Hamiltonian function. And
in our nomenclature, we had called this term
239
00:33:34,730 --> 00:33:41,820
as . something like script H, because this
in concept represent the
240
00:33:41,820 --> 00:33:50,730
total energy and the Hamiltonian H is normal
H is this. We just showed that the script
241
00:33:50,730 --> 00:33:53,200
H
that we started with is equal to the Hamiltonian
242
00:33:53,200 --> 00:34:03,910
H, now what is this, this total energy
dependent on, total energy is dependent on
243
00:34:03,910 --> 00:34:08,150
the momentum the position and the time.
.
244
00:34:08,150 --> 00:34:27,580
So, we can write the total energy H is a function
of p i q i and time, that stands to reason,
245
00:34:27,580 --> 00:34:32,470
the momenta dependent on the, the kinetic
energy dependent on the momenta and the
246
00:34:32,470 --> 00:34:38,429
potential energy dependent on the positions,
and things can be dependent on time also.
247
00:34:38,429 --> 00:34:42,210
If
that is so then if you take a differential
248
00:34:42,210 --> 00:34:48,659
of it, we can write d H as, if it is a d,
it is
249
00:34:48,659 --> 00:34:57,950
dependent on three things, then we can expand
it by the chain rule to obtain some over i,
250
00:34:57,950 --> 00:35:06,230
it will be the first partial derivative of
H with respect to p i.
251
00:35:06,230 --> 00:35:23,780
Then d p i plus a partial derivative of H
with respect to q i d q i plus, this time
252
00:35:23,780 --> 00:35:25,700
it will not
be a summation, these two are summed over,
253
00:35:25,700 --> 00:35:32,440
this will be just the term, which is the
partial derivative of H with respect to t
254
00:35:32,440 --> 00:35:43,680
and this will be d t. So, this is how, this
term
255
00:35:43,680 --> 00:35:53,490
.expands, we have seen that this is also equal
to this term, so we can also expand this term
256
00:35:53,490 --> 00:35:58,780
in the same way, let us see, what it yields,
let us keep it here.
257
00:35:58,780 --> 00:35:59,780
.
258
00:35:59,780 --> 00:36:12,110
We have seen that the H is also i q i dot
p i minus the Lagrangian and now I want to
259
00:36:12,110 --> 00:36:26,260
write it as H, how will I write it. First,
you have this term here, it is dependent on
260
00:36:26,260 --> 00:36:30,000
q i,
dependent on p i, dependent on l, so we have
261
00:36:30,000 --> 00:36:36,890
to write, taking the derivative of this it
will
262
00:36:36,890 --> 00:37:03,360
be q i dot d pi and that will be summed over
i plus, again pi d q i dot, that is also summed
263
00:37:03,360 --> 00:37:16,190
over i. These two yield a derivative of here
and now this Lagrangian, Lagrangian we
264
00:37:16,190 --> 00:37:27,640
know is a function of q i, q i dot and time,
we have already seen that.
265
00:37:27,640 --> 00:37:39,530
So, it will have to be written as minus derivative
of the Lagrangian with respect to q i, it
266
00:37:39,530 --> 00:37:52,900
will be d q i also summed over i, minus derivative
of the Lagrangian with respect to q i
267
00:37:52,900 --> 00:38:03,340
dot d q i dot. Again summed over i and time,
it will not have to be sum summed over, it
268
00:38:03,340 --> 00:38:17,260
is derivative of the Lagrangian with respect
to t sorry, that completes the story then
269
00:38:17,260 --> 00:38:23,910
notice there are two terms involving d q i
dot and we can take them common.
270
00:38:23,910 --> 00:38:40,200
So, if you take them common you have, the
first term q i dot d p i, plus first let us
271
00:38:40,200 --> 00:38:45,570
take
them common, it is summed over i d q i dot
272
00:38:45,570 --> 00:38:56,230
and then inside the bracket it will be p i
minus delta Lagrangian of that is q i dot,
273
00:38:56,230 --> 00:39:21,000
then the rest of the terms remain. Now, notice
what is this?
274
00:39:21,000 --> 00:39:22,000
.
275
00:39:22,000 --> 00:39:31,720
.Will be 0, yes because this is nothing but,
this, so this whole term goes to 0, simplifies
276
00:39:31,720 --> 00:39:40,520
our affair, so it is only this, this and this,
so write it down.
277
00:39:40,520 --> 00:39:41,520
.
278
00:39:41,520 --> 00:40:17,300
D H, now yields i q i dot d p i minus, now
we already have . this
279
00:40:17,300 --> 00:40:31,119
equation, and from here we know this term.
We had already obtain this, and from here
280
00:40:31,119 --> 00:40:36,110
we know this term, so we let us write down,
your derivative of the Lagrangian with
281
00:40:36,110 --> 00:40:40,710
respect to q i is, now here we are including,
can you see that
282
00:40:40,710 --> 00:40:45,330
.
We are including this yes, so it is p i dot
283
00:40:45,330 --> 00:41:01,440
plus dot, so let us substitute it here, you
have d
284
00:41:01,440 --> 00:41:28,410
H minus this term, it will be, let us start
with plus and i, it will be minus p i dot,
285
00:41:28,410 --> 00:41:32,190
minus
derivative of the Rayleigh with respect to
286
00:41:32,190 --> 00:41:43,050
q i dot, it will be d q i minus Lagrangian
del t d
287
00:41:43,050 --> 00:41:57,330
t, well, well, well. Now, you see, we had
we have obtained this, from the right hand
288
00:41:57,330 --> 00:42:03,020
side
and this was another equivalent expression,
289
00:42:03,020 --> 00:42:10,530
we had obtained, in this case, we had stated
that H is dependent on p i q i t and we had
290
00:42:10,530 --> 00:42:14,380
expanded this.
And in this case, we had written that there
291
00:42:14,380 --> 00:42:20,110
is also the Lagrangian function and therefore,
we had expanded that in one case we had obtained
292
00:42:20,110 --> 00:42:24,180
this, another case we had obtained
that. Now notice, that they are almost the
293
00:42:24,180 --> 00:42:29,760
same, only here is a term, they summed over
i
294
00:42:29,760 --> 00:42:39,560
p i term, summed over i q i term, summed over
i t term, obviously this term and that term
295
00:42:39,560 --> 00:42:43,700
.should be the same, this term and that term
should be the same and this term and that
296
00:42:43,700 --> 00:42:50,880
term should be the same.
Do you see, how simply it becomes, all these
297
00:42:50,880 --> 00:43:05,640
mathematics ultimately then yields, three
very simple results, the first thing is q
298
00:43:05,640 --> 00:43:20,340
i dot will be equal to then this, minus p
i dot minus
299
00:43:20,340 --> 00:43:50,200
will be and
300
00:43:50,200 --> 00:43:53,780
normally in books you will find, these terms
written in the left hand side and
301
00:43:53,780 --> 00:43:58,840
these terms written in the right hand side.
But I have chosen to write it this way, because
302
00:43:58,840 --> 00:44:03,880
ultimately, what we want to extract are these
two, we will do that but let us understand
303
00:44:03,880 --> 00:44:09,660
this, this is the differential equation first
order, this is the differential equation first
304
00:44:09,660 --> 00:44:14,520
order,
this is not a differential equation, this
305
00:44:14,520 --> 00:44:20,940
only say that if the Lagrangian function is
independent of time, the Hamiltonian function
306
00:44:20,940 --> 00:44:26,320
will also be independent of time.
And, if they are dependent on time, they are
307
00:44:26,320 --> 00:44:29,710
dependent this way that is all, but this is
not
308
00:44:29,710 --> 00:44:34,170
a differential equation, therefore it does
not really define the dynamics. We are primarily
309
00:44:34,170 --> 00:44:40,480
then interested in these two equations.
.
310
00:44:40,480 --> 00:44:51,180
And these two equations immediately yield,
q i dot is equal to derivative of the
311
00:44:51,180 --> 00:45:01,660
Hamiltonian with respect to p i and p i dot
is from here, I want I am trying to extract
312
00:45:01,660 --> 00:45:18,550
it, it
is this minus, that is it, that is what the
313
00:45:18,550 --> 00:45:28,500
Hamiltonian equations are first order nice
shape.
314
00:45:28,500 --> 00:45:32,590
We will illustrate, how to apply this two
physical systems, definitely that is what
315
00:45:32,590 --> 00:45:36,600
we are
doing always, we derive a formalism and illustrate
316
00:45:36,600 --> 00:45:39,270
with it large number of examples, do
not worry about it.
317
00:45:39,270 --> 00:45:44,960
.But notice one thing, that if this term is
ignored that means, if some system does not
318
00:45:44,960 --> 00:45:47,180
have
this term for example, this spring pendulum
319
00:45:47,180 --> 00:45:55,350
and stuff. Then it takes a very nice
symmetrical form, can you see that very nice
320
00:45:55,350 --> 00:45:58,960
form, q i dot is nothing but, the
Hamiltonian derivative of the Hamiltonian
321
00:45:58,960 --> 00:46:02,210
with respect to p i, and pi dot is nothing
but,
322
00:46:02,210 --> 00:46:06,810
the derivative of the Hamiltonian with respect
to q i with a negative sign, that is all.
323
00:46:06,810 --> 00:46:09,910
Very
nice set of equation that is why, this has
324
00:46:09,910 --> 00:46:16,860
intrigued physicists for a long time and many
physicists had devoted their life time studying
325
00:46:16,860 --> 00:46:22,590
this, nice set of equations.
And, if this is ignored, what you have that
326
00:46:22,590 --> 00:46:26,050
kind of systems where, such equations will
be
327
00:46:26,050 --> 00:46:44,170
valid are called the Hamiltonian systems.
So, the Hamiltonian systems are that is it,
328
00:46:44,170 --> 00:46:53,310
whether the Rayleigh term does not exist,
which means there is no dissipation or friction,
329
00:46:53,310 --> 00:47:00,170
then you have got Hamiltonian system. Remember
such systems are actually very
330
00:47:00,170 --> 00:47:06,430
important, supposing one a scientist is studying
the motion of mars obviously, that is a
331
00:47:06,430 --> 00:47:11,000
Hamiltonian system, the friction dissipation
is very small.
332
00:47:11,000 --> 00:47:16,520
If one is studying the motion of electrons
in a atom, Hamiltonian system, we do not have
333
00:47:16,520 --> 00:47:25,110
any concept of friction there. So, these things
are not really imagination, this being
334
00:47:25,110 --> 00:47:31,570
ignored actually happens, but for our purpose,
since we are primarily interested in
335
00:47:31,570 --> 00:47:38,400
engineering systems, we will include this,
except for a few exceptions, which are also
336
00:47:38,400 --> 00:47:43,880
you know not really coming from engineering,
but our exercise to understand such
337
00:47:43,880 --> 00:47:48,440
systems, this spring pendulum and all.
But for most engineering system there will
338
00:47:48,440 --> 00:47:52,580
be some kind of friction or dissipation, so
we
339
00:47:52,580 --> 00:47:58,110
will include this. Remember in other books
you will not find the equations with this
340
00:47:58,110 --> 00:48:08,390
term, only I am including it, so that you
can derive this, equations using this formalism,
341
00:48:08,390 --> 00:48:15,690
so we have this equations and then we need
to understand, how to obtain the equations
342
00:48:15,690 --> 00:48:22,859
with this. Let us go back to the circuit example,
that we take took today, where was it,
343
00:48:22,859 --> 00:48:37,700
circuit example was here, here was the circuit
example we took...
344
00:48:37,700 --> 00:48:38,700
..
345
00:48:38,700 --> 00:48:45,050
So, here we have already written down the
T, we have written down the V and we have
346
00:48:45,050 --> 00:48:52,220
written down the R. Notice, that in the Hamiltonian
formalism, we need to write down
347
00:48:52,220 --> 00:48:59,300
the right hand side, and right hand side means,
H as derived in as has taken derivative in
348
00:48:59,300 --> 00:49:05,099
terms of p i, H has taken derivative in terms
of q i.
349
00:49:05,099 --> 00:49:06,099
.
350
00:49:06,099 --> 00:49:20,620
So H has to be expressed as, it has to be
expressed as p i, the problem is, that if
351
00:49:20,620 --> 00:49:24,180
I simply
write it as T plus V, it is expressed in terms
352
00:49:24,180 --> 00:49:28,450
of q i dot not as pi, so the first task will
be, to
353
00:49:28,450 --> 00:49:37,340
express it in terms of p i, so that is what
you have to do first. In this case, in this
354
00:49:37,340 --> 00:49:47,500
.particular example H is T plus V, and let
us write down T plus V first, using the
355
00:49:47,500 --> 00:49:53,430
expression for the T and the V ., that we
have already written. We
356
00:49:53,430 --> 00:50:09,770
will say, half L q 1 dot minus q 2 dot square,
plus half L sorry, this is L 1 this is L 2,
357
00:50:09,770 --> 00:50:14,240
q 2
dot square and the other terms will be now
358
00:50:14,240 --> 00:50:25,230
plus, remember T plus V, 1 by twice C, q 1
square minus E q 1.
359
00:50:25,230 --> 00:50:35,950
Now, we need to express this in terms of the
pi’s, now we had already obtained this,
360
00:50:35,950 --> 00:50:41,200
and
this, so we can substitute, so if you substitute
361
00:50:41,200 --> 00:50:53,830
this here, we get, where is it. This term
is p
362
00:50:53,830 --> 00:51:12,940
1 by L 1, this term, so we can easily express,
half L 1 p 1 by L 1 square, plus half L 2,
363
00:51:12,940 --> 00:51:26,500
this is p 1 plus p 2 by L 2 square plus the
rest remains the same, twice C q 1 square
364
00:51:26,500 --> 00:51:36,869
minus, we had done, so we will just simplify
it, to twice L 1, because. L 1 goes square
365
00:51:36,869 --> 00:51:53,870
here, P 1 square plus 1 by twice L 2, p 1
plus p 2 square plus 1 by twice C q 1 square
366
00:51:53,870 --> 00:52:07,520
minus E q 1. So, that is how we have derived
H in terms of the p’s and q’s, the rest
367
00:52:07,520 --> 00:52:11,320
is
algorithmic, the rest is algorithmic.
368
00:52:11,320 --> 00:52:12,320
.
369
00:52:12,320 --> 00:52:22,720
We will say simply that the derivative of
H with respect to p 1 is nothing but, q 1
370
00:52:22,720 --> 00:52:30,120
dot
simple, and this immediately yields p 1 by
371
00:52:30,120 --> 00:52:41,930
L 1 plus p 1 plus p 2 by L 2. Similarly, q
2 dot
372
00:52:41,930 --> 00:52:53,420
is derivative of the Hamiltonian function
with respect to p 2, which is p 1 plus p 2
373
00:52:53,420 --> 00:53:06,560
by L
2, just check. Now, p 1 dot is minus derivative
374
00:53:06,560 --> 00:53:18,330
of the Hamiltonian with respect to q 1,
which is these two terms appear, it will be
375
00:53:18,330 --> 00:53:31,590
minus q 1 by c plus E done, P 2 dot is well,
in
376
00:53:31,590 --> 00:53:39,619
.the sorry, I did not include the Rayleigh
term, but in any case the Rayleigh term, does
377
00:53:39,619 --> 00:53:44,400
not
depend on q 1 dot, so even if we should have
378
00:53:44,400 --> 00:53:49,080
written it here, we are not wrong.
And this is minus derivative of the Hamiltonian
379
00:53:49,080 --> 00:53:59,650
with respect to q 2 minus Rayleigh with
respect to q 2 dot, that becomes, this is
380
00:53:59,650 --> 00:54:18,030
0 this remains minus R q 2 dot and then you
substitute minus R by L 2 p 1 plus, so that
381
00:54:18,030 --> 00:54:25,530
is how you write down the differential
equations. Simple procedure can be applied
382
00:54:25,530 --> 00:54:31,520
and can be applied to most systems, most in
the sense that we had made some assumptions,
383
00:54:31,520 --> 00:54:36,421
while deriving the Hamiltonian
equations. If some system violates them, then
384
00:54:36,421 --> 00:54:42,630
we have to exercise caution, otherwise this
simple procedure works, we will continue in
385
00:54:42,630 --> 00:54:43,630
the next class.
386
00:54:43,630 --> 00:54:43,630
.