So, let us start today with slight digression,
a mathematical digression, and then I get
back to Lagrangian dynamics; I would like
to move over to Hamiltonian dynamics; and
this requires a mathematical transformation,
with which you are already familiar in other
context, but perhaps the name was not used
and so let me do that and point out what the
advantages are.
And this is the so called Legendre transformation;
and the idea and essence is the following.
If I give you a simple function of x, a single
variable x, and there are two ways of looking
at this function, one of them is to say that
for every value of x, I assigned a value of
f of x and therefore, if I assign this set
of numbers corresponding to each value of
x. Specify, numerical values of f of x for
each x, I have specified the function.
There is another way of doing it and that
is to say, I assign at each point the value
of the slope of the function. And at the next
point, I assign the slope once again and the
slope once again and so on and in this manner
I can build up the function if I specify,
at each point f prime of x at each point provided
of course, I also specify f of x at some reference
point otherwise of course you could shift
this entire graph parallel to itself and it
would be exactly the same as if I before just
gave you the slope but, if I also tell you
f of zero for instance then you know where
the intercept is and after that you can reconstruct
the function. This is the idea behind Legendre
transformation. Except that is not very useful
in 1 variable but when you have more than
one variable it becomes more exceedingly useful.
So, let me give you an example, suppose you
have a function f of x, y then of course,
you know that you can write its differential
df as delta f over delta x dx plus delta f
over delta y dy, and this quantity delta f
over delta x is some function of x and y so,
let me call this capital X of little x little
x let me use curly x little x in this fashion.
So, distinguish between these two quantities
and capital Y of x y dy by definition capital
X and capital Y are these partial derivatives.
And now, I might want to say that instead
of looking at little x and little y as the
independent variables I look at capital X
and little y as the independent variables,
in other words, the slope with respect to
x.
If I want to do that and I want to construct
the function g such that dg is equal to something
times d capital X plus something else times
d little y in recognition of fact that this
quantity this function g is a function of
capital X and little y the trick is very simple
all I have to do is to subtract from f minus
little x times capital X. Then of course,
if I take f lets define g as equal to this
dg is df minus X capital dx minus capital
X times little dx and of course, that term
cancels out and you are left with dg to be
a function of dx and dy combination of dx
and dy. So, this is the idea behind Legendre
transform.
Invariably you have to subtract out so lets
write down what dg is df minus x dX minus
X dx and that is equal to minus x dX plus
Y dy, because this term cancels out on both
sides. But, you must remember that this little
x must be now re-expressed in terms of capital
X and little y.
So, you need to take this equation X of x,
y equal to delta f over delta x and you have
to solve this equation, for little x and express
little x as a function of capital X and little
y. Once, you do that then you guarantee that
g equal to g of capital X and little y. You
could go on and say can I define a function
h which is a function of little x and capital
Y, then of course you can do that by defining
h is equal to f minus y times Y is equal to
h of little x and capital Y like one more
Legendre transformation. And finally, you
could transform the capital X and capital
Y all together so, you could define function
say p equal to p capital X capital Y this
would be f minus x X minus y Y. You can generalize
this to any number of independent variables
x y z etcetera.
So, the whole idea is simply to see what is
the most convenient function for your purpose
very often this fs and gs and so on. There
will be some kind of potentials or some kind
of functions like the laplacian, the lagrangian,
which I would like to transform to a set of
different set of independent variables for
some reason or the other. You are already
familiar with this in another context, what
is that?
Pardon me; the Gibb's potential in thermodynamics
used to this thing because if you recall we
are going to do thermodynamics but let me
recall to you what happens.
You start with the internal energy U and this
internal energy you transform and what is
this a function of what is dU a function of,
well dU is T dS minus P dV, for fixed number
of particles. So therefore, U is a function
of S and V. So, this in general dU, the first
law of thermodynamics can be written T dS
minus P dV plus mu dN actually if you change
the number of particles you also have a chemical
potential times the differential in the number
of particles. So, this of course implies that
U is a function of S, V and N as the independent
variable and it is obvious from this, that
T is delta U over delta S keeping V and N
constant, they are the partial derivatives.
Then you might want to consider a function
dF, which is equal to d of U minus TS and
that will now become a function of T V and
N because dF the T dS part is going to cancel
from here is going to be T dS and an S dT
so, it becomes dF equal to minus S dT minus
P dV plus mu dN. This implies, that of course
F is a function of T, V, N. So, what you have
done is Legendre transform going from the
internal energy U to Helmoltz energy F because,
you would like to have as controlled variables
the temperature volume and number rather than
the entropy volume and number.
Depending on your conditions, you might further
want to change from P dV to V dP because,
you might be able to control temperature,
pressure and number. Then what you do? You
go from this F by subtracting F minus PV plus
PV because, there is already minus sign here
and what would this give you dG here and that
is equal to minus S dT plus V dP plus mu dN
and that implies that G is a function of T,
P and N. Because, dG is proportional to a
combination of dT, dP and dN.
So, depending on what your external conditions
are, what your experimental conditions are?
If you can handle temperature, pressure and
the number as the control variables, then
you would like to have the Gibb's free energy
as a thermodynamic potential. These are all
thermodynamic potentials. And you go from
one to other depending on your convenience,
depending on the kind of processes you want
to look at and then, of course you know in
thermal equilibrium in a state of thermal
equilibrium, if you are a constant entropy,
volume and number then the internal energy
is at a minimum.
If you are in a stage, where the temperature,
volume and number are constant, then in thermal
equilibrium the Helmoltz free energy is at
the minimum and similarly, for the Gibb's
free energy. So, this is the reason for going
from 1 thermodynamic potential to another
simply so, that you can handle first write
the minimum principle and secondly you can,
decide on what your experimental control variables
are? Much the same thing is going to happen
in the Hamiltonian framework, going from the
lagrangian to the Hamiltonian, its going to
be a Legendre transform so, how do I do that?
Again, let me simplify notational little bit,
a Lagrangian L which is a function of q1 to
q n q1 dot to q n dot and possibly t, I have
written in abbreviated notation as L of q,
q dot and t, its actually a function of 2n
plus 1 variables in general, generalized coordinates,
generalized momentum and possibly time if
it is a non autonomous system. I would like
to get rid of these velocities and use the
partial derivatives of L with respect to these
velocities as the independent variable.
So, I would like to use delta L over delta
qi dot as the independent variable. And, let
me call this and define this, this is equal
to p sub i, and I call it a generalized momentum.
You can see why I am doing this? Because we
already saw from the Euler Lagrange equations
that, if you have delta L or delta qi equal
to d over dt of delta L over delta qi dot.
Then we know that if qi is a cyclic coordinate
or ignorable coordinate, this quantity here
is the constant of the motion.
So, every time you have a cyclic coordinate,
but corresponding momentum generalized momentum
is a constant of the motion. And this is how
invariance is start from again constants of
the motion will be identifiable. So, let me
define right now once and for all, delta L
over delta qi dot as the momentum the generalized
momentum conjugate to the corresponding generalized
coordinate and I will explain the meaning
of this word conjugate, but they go in pairs.
What happens if I do this? I would then like
to go from L to a new function H which is
a function of the qs and the ps collectively
even if i write as p and possibly t and how
should I define it? I should define this as
L minus the product of the q dots with the
corresponding ps but historically the reverse
sign it doesn't matter. So, let me define
this as summation over i equal to 1 to n qi
dot pi minus L. I have done to change the
sign and the reason is H is going to turn
out to be the total energy of systems, mechanical
systems in particular therefore, I would like
to retain a plus sign, no other reason.
Could have done it L minus q dot qi dot pi
dot let us leave it. However, H is a function
of q, p and t velocities do not appear in
the Hamiltonian, this quantity is called the
Hamiltonian and velocities do not appear in
it. So, how are you going to get rid of these
velocities? They are sitting here right here
so, what you are supposed to do is to solve
the set of equations and write qi dot as a
function of all the qs ts and possibly time.
So, you have to take this quantity differentiate
it and then of course you get a function of
q q dot and t and you are supposed to eliminate
if possible, the different qi dots and write
them express them explicitly as functions
of qs, ps and ts. And after that you pluck
that in here and where ever q dots appear
here, you again write them as functions of
qs, ps and ts and if you can do this, then
this is a function of q, p and t.
In principle, now when would you be able to
do this? When can you actually do this? Its
not obvious that you can always eliminate,
you have 1 variable its not so easy its not
so difficult you have single q, p and t, but
if you have a multiple scheme multiple set
of quantities, then its not so obvious, that
you can do this and I put it to you that I
will explain this, later on by example we
will see this.
That if this, matrix of quantities qi dot
delta qj dot, this is some function this is
labeled by the indices i and j and its obviously
written as 2 by 2 matrix, if that matrix is
non singular then you can always make this
change of variables. You can invert this set
of quantities, this is called the Hessian
matrix corresponding to this transformation
we will come back we will look at examples,
just want to point out here that you should
be able to make this transformation otherwise
there is no point in it.
Now, let us assume that we can do so and let
us ask what happens next. All right I pretend
I have done this, what happens next? What
should I do? I should try to find out that
H is indeed a function of qs, ps and ts, and
what it implies? Let us do that, see what
happens?
So, dH therefore becomes equal to again let
me save myself a little bit of trouble by
assuming that I write this as q dot p forget
the i here you can put the summation and put
the indices in specified and so on. So, dH
becomes dq dot minus p plus q dot times dp
minus dL and what is dL, well this is equal
to p dq dot plus q dot dp minus delta L over
delta q dq minus delta L over delta q dot
dq dot minus delta L by delta t dt. I just
take the differential in the right hand side
L being a function of q q dot and p you can
write dL in that form. On the other hand,
dH itself should be writable as delta H over
delta q dq plus delta H by delta p dp plus
delta H by delta t dt.
Now lets compare both sides, we already know
that delta L over delta q dot is in fact p.
By definition, therefore, this term cancels
out against this term and now we are in good
shape because, p, q and t are independent
variables and if this is equal to that the
coefficients of dq, dp and dt must be equal
to each other; since is the completely independent
variables. What does that imply? This immediately
gives us set of consequences if I compare
for example, this with that it says q dot
equal to delta H over delta p. So, that has
matched this with that. Now, I match this
with that it hears by the physical input is
going to be, what do I get? I get delta L
over delta q equal to delta H minus delta
H over delta q. But remember, we are talking
about solutions of the Euler Lagrange equations.
We are going to use the equations of motion,
on these solutions trajectories delta L over
delta q by the Euler Lagrange equation is
known to delta L over delta q equal to d over
dt p, its delta L over delta q dot which is
p.
And therefore, this equation here becomes
t dot equal to minus delta H over delta q.
By erase that and of course if I compare this
with that for non autonomous systems it turns
out that delta L over delta t is minus delta
H over delta t. This therefore, is now the
content of Euler Lagrange equations that is
what we put in. And since, this is the set
that we arrive at incidentally notice that,
this definition p equal to delta L over delta
q dot has a conjugate has a corresponding
counter part which is this q dot is delta
H over delta p.
Whereas, p is delta L over delta q dot, it
is again the slope that is, just telling you
that you can always go from the Hamiltonian
to the Lagrangian back with the Legendre transform
and this is precisely the Legendre transform.
So, that is not very surprising, but this
equation here contains dynamics because, the
Euler Lagrange has the input in. This set
of equations is now exactly what we want what
is the advantage of this set of equations
over the Euler Lagrange equations.
They are first order equations, they are explicitly
first order equations in the 2n dynamical
variables. So, the great advantage is that
we reduce the second order set of equations
to first order set of differential equations
by going to the Hamiltonian framework. But
you are going to do lot more you are going
to get many more advantages what are they
and lets start writing them down.
So far I have not defined the Hamiltonian
anything more than p q dot minus L, I have
not said its the total energy of the system,
I have not identified it to the total energy,
this depends on what L would be. But as far
as I am concerned, you give me a system with
Lagrangian and provided I can make the Legendre
transform, I can always go to the Hamiltonian
what ever the function that be. And then,
once you have that function you end up with
this set of first order differential equations
and I now forget about the Euler Lagrange
equations and focus on solving this set of
equations.
But it has got magical properties, so again
lets write this down q dot is delta H over
delta p p dot is equal to minus delta H over
delta q. The minus sign is crucial the most
important minus sign in the universe is absolutely
crucial. You will see what it does has to
go along? The first point is that, if H is
independent of time explicitly then H is the
constant of the motion automatically, how
do we find out?
Well, if H is equal to H of q, p say this
is autonomous system then what is dH over
dt equal to delta H over delta q q dot plus
delta H over delta p dot by definition. But
on a solution trajectory, namely when the
Hamiltons equations, when Hamiltons equations
apply so you should call this Hamiltons eqations.
This is called to be a starting point in some
sense. When these equations apply, then I
replace q dot by delta H by delta p q. So,
this becomes delta H over delta q delta H
by delta p minus delta H by delta p delta
H by delta q and you see the role played by
this minus sign so this becomes identicaliy
0.
The Hamiltonian is therefore constant of a
motion as long as it is not explicitly dependent
on time. So, this is the way you discover
a constant of the motion this is always true
for every Hamiltonian system, which is autonomous
system, explicitly time independent Hamiltonian
you guaranteed that the Hamiltonian is a constant
of the motion. It is going to play a very
very fundamental role and what is that you
already begin to see that it has a favored
role over all other constants of the motion,
which we are going to now try to deduce? The
reason being that the way in which the free
space point changes its velocity is governed
by the Hamiltonian.
So, in that sense the Hamiltonian governs
the time evolution of a system. Therefore,
we call it the infinite decimal generation
of time translations, because its telling
if you give me qs and ps at any instant of
time p zero a t zero plus delta t the Hamiltonian
tells you where the system goes, where the
point go and so on. So, this is going to have
a very fundamental role to play in all that
they want to do. We will write down the Hamiltonian
for simple systems, but in general please
notice that the Hamiltonian as written here
this function very, very special role to play.
You could ask what about other constants of
motion? What does a general constant of motion
look like? And let us do that it can be related
to the Hamiltonian.
So, constant of the motion, suppose F is a
constant of the motion in general, of course
it is a function of all the qs ps and it could
be explicitly time dependent. Please notice
that the constant of the motion could be explicitly
time dependent, because the time dependence
which comes from the explicit time dependence
could be cancelled by the time dependence
of ps and qs.
So, in general of course the given system
dynamical system will have constants of the
motion which are time dependent some of them
would be time dependent. I will give you a
simple example in a minute, but what does
this imply says dF over dt equal to delta
F over delta q q dot, but q dot should be
replaced by delta H over delta p because,
you are on a solution trajectory plus delta
F over delta p times p dot but p dot is minus
delta H over delta q so, there is a minus
here plus of course delta F over delta t.
Now, this little structure here has very profound
implications have very deep meaning which
I will come to very shortly, this set of partial
derivatives please notice I abbreviated my
notation, if you have n degrees of freedom
it really stands for i equal to 1 to n return
i here. Really it stands for that and this
object which is constructed out of 2 functions
F and H is called the Poisson of F with H.
By definition, it is denoted as F curly bracket
H plus delta F by delta t. Its the Poisson
bracket of the F with H, it plays the same
role in classical mechanics, as the commutative
of matrices or operators would in quantum
mechanics and in fact the translation from
classical to quantum mechanics through the
Poisson bracket is very immediate we will
see that. So, you agree this is some function
of qs ps and ts after you finish this set
of complicated set of partial derivatives,
with some function of qs ps and ts and when
is F a constant of the motion if dF over dt
vanishes identically.
This means F is the constant of motion if
and only if the Poisson bracket of F, H plus
delta F by delta t is equal to zero. If F
should turn out to be explicitly time independent
then the last term goes away and then F is
the constant of the motion if and only if
its Poisson bracket with the Hamiltonian vanishes
identically. And I would then say Poisson
commutes with H, in analogy with what happens
in quantum mechanics whereas 2 operators commute
with each other in that sense I would say
this Poisson bracket is zero I would say Poisson
commute.
The other word that often use is F and H are
said to be in involution with each other,
I will write that down a little later when
to write down the criterion for integrability,
but at the moment I want you to focus on the
point that the way you test if some quantity
is a constant of motion now, is simply to
calculate the set of partial derivatives and
check if it is zero or not. If it is zero
identically, it is the constant of the motion,
if it is not then it is not a constant of
motion.
Now, of course this structure here has some
deep implications you could define the Poisson
bracket of any two functions of free space
variables. So, you could have some A which
is a function of all the qs and ps and ts
perhaps and some B and this Poisson bracket
here is defined as delta A over delta q dot
delta B over p dot minus the reversed partial
derivatives. It has some properties and the
first property is that this is equal to minus
B, A. That is immediately obvious you reverse
A into change A and B and you get a minus
sign.
By the way, there is an even more trivial
property which essentially says that A plus
B with C is A with C plus B with C that is
of course true, immediately all you have to
do is to plug it in there and this is at once
true. And similarly, if you took A and multiplied
it by some constant B this becomes alpha times
A, B. So, multiplication by a constant does
not do anything it just comes out of the partial
derivatives. So, this property here is anti
symmetry is crucial property of the Poisson
bracket.
There is another property, which is also a
crucial property and I will explain why that
so but for that we will need a preliminary,
you should ask what is A with BC, where A
B C are functions of the free space variables
qs ps and possibly t, what is this equal to?
Well put that in here you have a A here and
a BC here and then you expand this out. If
you do that you discover this is equal to
B times A with C plus A with B times C.
So, it has this chain rule kind of property
and of course, if these are functions this
is a function and that is a function of free
space variables everything commutes so, I
could have written this B on the right hand
side that C on the left hand side it will
make a difference. But I have in mind always
going over to quantum mechanics where these
would be operator matrices or operators then
of course A B is not equal to B A and this
is the correct order. Since, B appears on
the left here and C on the right that is,
exactly how it appears everywhere B always
appears on the left C on the right and that
is why I put this B on the left here and C
on the right here.
Now, I leave you to figure out what A B with
C D is you have to apply this formula repeatedly
so, this is the way you would find Poisson
brackets of various complicated functions
of the free space variables given a few elementary
Poisson brackets and the first of these, before
I do that let me, mention the property that
I was interested here this property is easy
to verify, first take the Poisson bracket
of B with C with some function take the Poisson
bracket with another function A and do this
in cyclic order B with C A plus C with A B
and answer turns out to be zero identically.
To do this, you have to verify this property
using that I expression, using this relationship,
using this formula you can verify that this
is actually branded and this is called Jacobi
identity. We will see what it uses in half
a second, but you see this tells you that
if you take 3 of these functions and you do
this in cyclic order repeatedly you add it
all up you get zero. Now, this property together
with this property for a set of elements,
any abstract set of elements A B C D etcetera,
elements of a class if it should so, turn
out that you can define the bilinear operation
between them.
In other words, something which involves two
of them at a time and produces another member
of this class and here there is free space
functions so, you take 2 free space functions
find the Poisson bracket you produce another
free space function, function of the free
space variables. If you can define a bilinear
operation among them which is anti symmetric
which has a minus sign when you interchange
2 numbers and for which the appropriate identity
is valid such a set of elements is said to
form a Lie algebra, this plus this together
imply that the set of elements is Lie algebra.
You are familiar with other examples, 1one
of them is immediately matrices, if you took
n by n matrices and define the bilinear operation
to be the commutator A B minus B A then of
course that is equal to minus of B A minus
A B and the triple commutator is zero identically.
So, matrices n by n matrices under the operation
of commutation form a Lie algebra. Functions
of free space variables for a system form
a Lie algebra under the Poisson bracket operation.
Can you give me another example, 1 which you
are very familiar with in three dimensions
very very familiar with A cross B ordinary
vectors would do exactly this, because you
know that if you took 2 vectors, when you
take the cross product you find another vector
this is certainly true. a cross b is equal
to minus b cross a, and you know that triple
product A cross B cross C plus B cross C cross
A plus C cross A cross B is also zero. So,
ordinary 3 dimensional vectors under the cross
product operation form a Lie algebra
There are numerous other examples but, these
are the simplest 1s and they are going to
have this is going to have very profound properties
we will see very, very shortly; so any questions.
So, you have seen what the constant of motion
is, something which in the autonomous case
is independent Poisson commutes with the Hamiltonian
and we have also seen that the Hamiltonian
itself is the constant of motion. Now, the
next question one should ask is, given these
properties can we write this down in more
compact form what is the real meaning of the
Poisson bracket? What is the geometrical meaning
of the Poisson bracket? This is worth asking.
Why does this particular combination of things
really turn out to be so fundamental, that
is the first question. The other question
is, is there something I can do to put this
whole business in the language of general
dynamical systems, what makes Hamiltonian
systems so special so very special? The answer
is the following.
Recall that our general dynamical system involve
variables x1 x2 up to x N and we had an equation
x1 dot is f1 of all the variables x N and
so on up to x N dot is f N of all the variable
x N. In our case, instead of a vector field
f1 to f N here a single scalar function namely
the Hamiltonian is determining the right hand
side.
So, that is a great simplification because
instead of having a vector field on the right
hand side you have q1 dot is delta H over
delta p1 right up to qn dot delta H by delta
pn and then you have p1 dot as minus delta
H over delta q1 up to pn dot with minus delta
H over delta qn. Of free space variables x
in this case are really q1 to qn p1 to pn
so, we have a situation where 2n equal to
N the free space is even dimensional for Hamiltonian
system.
And I write all the qs one after the other
and then write ps 1 after the other and I
call that a vector in free space a position
vector in free space. Then this quantity here,
tells you that the velocity of the free space
point is determined by certain partial derivative
with respect to a scalar function H. It almost
looks like a gradient it almost looks like
a gradient because, what would be the gradient
operator be in free space? This would be delta
over delta q1 delta over delta qn delta over
delta p1 delta over delta pn, this is my gradient
operator.
And it would be very nice, if I could write
x dot equal to the gradient of H. Is this
true? It is not true, because this is not
q1 but p1 unfortunately and this here is not
p1 but q1 and I have minus sign to both so,
it is not quite the gradient. However, we
can make it look like the gradient as follows,
lets define a matrix J to be equal to 0, the
unit matrix minus the unit matrix and 0, where
this thing is a n by n unit matrix. And this
here stands for the n by n null matrix.
Let me define matrix big matrix capital J,
we have 2n by 2n matrix in 4 blocks the top
block the top left hand corner block is just
a null matrix so, as the bottom right hand
corner block and then the identity matrix
in the off diagonal positions and minus the
identity the lower right hand side. Once I
do this, then it is a trivial matter to check
that this set of equations this entire set
of equations could be written as x dot equal
to J times the gradient of H.
So, this says q1 dot on top would be the first
component delta H over delta q1 right up to
delta H over delta pn, but then you apply
the J matrix on the left hand side and you
would pick out just the right quantity. Because
out here would be a 1 top can be here that
1 would pick out for q1 dot precisely this
quantity here delta H over delta p1. So, this
makes it look almost almost not quite, but
almost like a gradient system. But the fact
is all the simplicity that you gain by saying
that your vector field on the right is derived
from a scalar field are going to be present
except it is a sort of twist at gradient,
it is not quite the gradient but it twist
the gradient.
Incidentally, this minus sign came from the
Euler Lagrange equations so, we cannot wish
it away, sitting there and it is a good thing
it was there because, it gave us the information
that the Hamiltonian is the constant of the
motion itself. As long as we have autonomous
systems so, this gradient here is called a
symplectic gradient. I will explain why this
word symplectic is used right here, but right
now it is terminology. It is called the symplectic
gradient of H.
This matrix J has interesting properties,
which will need J square is equal to minus
the identity matrix, that is the easy to verify
and this by the way stands for 2n by 2n identity
matrix. And that immediately tells us, that
J inverse equal to minus J. Because, J J inverse
is minus sign so, J inverse is minus J. So,
J has a inverse and its equal to minus itself
and incidentally, it is a trivial matter to
see, that this is also equal to J transpose
if I transpose, if interchange the rows and
columns this matrix. If I write the I J element
as the J I element then, that is just going
to be minus J very useful property. Can we
say something about the Poisson bracket now?
The answer is yes, because as you can now
see that what we wrote as A, B as summation
i equal to 1 to n delta A by delta qi delta
B over delta pi minus delta A over delta pi
delta B over delta qi this is what Poisson
bracket was by definition, can also be written
as well delta A over delta qi and then delta
A over delta pi suggests that you are taking
the gradient in free space of this quantity
A the scalar function A.
So, what is happening is that this quantity
is equal to gradient A transpose by gradient
A I mean a vector with elements delta A over
delta q1 delta A over delta q2 etcetera written
as a column vector, by transpose I mean the
corresponding row vector and that is getting
multiplied by gradient of B, which is a column
vector. But because you have this minus sign
not surprisingly you have a J in between.
Now what would you say if I wrote a column
vector and wrote a row vector on the left
hand side.
So, in ordinary space if I have a column vector
a1 a2 up to an and then I have p1 p2 pn and
this stands for the vector a and this stands
for the vector b, then of course once I write
it in this notation all I have done is to
take b dot a the dot product, because its
b1 a1 plus b2 a2 etcetera that is just the
dot product the ordinary dot product. That
has been replaced by a symplectic dot product,
if this is a symplectic scalar product and
its telling you that a and b these 2 functions
in free space, they are symplectic dot product
of the gradients is the Poisson bracket.
Now, go back to ordinary Euclidian space,
what happens if I tell you that you have 2
functions f of x f and g functions of the
space coordinate. Such that the gradient of
f dot product with the gradient of g is 0.
What would you say about the functions f and
g? So, gradient of f dotted with gradient
of g is equal to 0 or in matrix notation grad
f transpose grad g is equal to 0. What geometrical
information do you get by saying the gradient
of this function and the gradient of that
function dot product is zero, yeah their level
surfaces are perpendicular to each other.
So, this is telling you that if the Poisson
bracket of a with b is 0, then it says the
level surface of a and level surface of b
are orthogonal to each other in the symplectic
sense, with this metric, with this metric
put in. So, this is why I said that the people
often say that Hamiltonian dynamics is a study
of symplectic geometry. So, you have to pretend
that your free space the qs and ps have this
extra J sitting in their, when you want to
define their dot products, when u want to
define gradient and so on.
Once you do that, then you have what is called
symplectic geometry or a symplectic metric
and the whole of Hamiltonian dynamics has
very elegant geometrical interpretation. With
this metric, now I am going to come back to
this and explain when we do relativity what
I mean by orthogonality? When you have non
Euclidian metrics? So, this is something which
we will explain a little later. But right
now, you can see that once I put in this J
appropriately at various parts these things
start looking like Euclidian geometry, except
that I have to put in this J once in a while
you take care of all these minus signs.
So, this is in fact the significance of the
Poisson bracket, this is the way to look at
it. This is by the way an algorithm to calculate
the Poisson bracket but its real meaning is
here in this place. Now, let us come back
step back a little bit and do one more thing
which is absolutely crucial for a normal Hamiltonian
system and that is to define, what I meant
to this word conjugate momentum it did not
define that yet, so let us do that. Pardon
me what is meant by the order means?
This guy here by this, I mean a column vector
which is delta B over delta q1 up to delta
B over delta pn. Everything here includes
p you cannot talk about the qs and ps separately.
The whole free space is 2n dimensional always,
but by convention set I am going to write
the first n components as the qs and the remaining
n components as ps I could have inverted it
see later on but it does not matter about
I am going to stick to this convention.
This is my definition of free space point;
this is my definition of gradient the del
operator in free space obviously. Very shortly,
we are going to see that it does not matter
which variables you call the qs and which
variable you call the ps you can actually
exchange. These are generalized coordinates
in generalized momentum and there is a deep
symmetry between them. I am not saying the
physical dimensions of a length and the linear
momentum are the same, but apart from physical
dimensionality we can always interchange variables.
So, the beauty of it is although we started
out from very very humble beginnings this
is telling us something extremely deep going
on in major, the fact that dynamics is happening
in free space is not a accident; it is got
a very deep geometric structure.
Now, let me define what is meant by conjugate
we call that our independent variables q1
to qn p1 up to pn. And I could ask, what is
the Poisson bracket of say qk with ql? What
is this equal to? Certainly, qk and ql are
dynamical variables, free space variables
and they could ask, what is the Poisson bracket
of this with that?
So, this is i equal to 1 to n delta qk over
delta qi delta ql over delta qi minus in the
reverse sorry, pi minus the reverse term.
But you see that, qs and ps are independent
variables completely independent variables
so this is zero therefore zero this is completely
zero so this is equal to 0, identically 0.
In exactly the same way, pk with pl is identically
zero and the reason is at some stage when
you do this partial derivative you are going
to take this function and differentiate with
respect to q and that is identically zero.
But it is another matter, if you ask what
is qk with pl? This is not going to be zero,
what is this going to be? Summation i equal
to 1 to n delta qk over delta qi delta pk
over delta pi pl minus delta qk over delta
pi delta pl over delta qi. And this term is
0, because the ps do not get differentiated
with qs nor do the qs with respect to the
ps so, that is gone. When is this term equal
to non-zero quantity, when k equal to i otherwise
its 0, because these are independent variables.
Therefore, this is equal to summation over
i delta ik the chronicle delta ik and this
of course is chronicle delta il and you have
to sum over all possible i s and this is going
to be 0 unless k is equal to i, this is going
to be zero unless l equal to i, the free indices
here are k and l therefore, this is equal
to delta kl, just the chronicle delta kl.
So, that defines for me, the canonical Poisson
bracket relations this set of relations are
called the standard or canonical Poisson bracket
relations.
Now, once these relations are satisfied, then
I say that the momentum pk is conjugate to
the momentum the generalized coordinate qk
or qk and pk form the conjugate pair. So,
all the free space variables get broken up
into pairs and notice here that each qk Poisson
commutes with all the other ps except its
conjugate, when the Poisson bracket is equal
to 1 on the right hand side.
This is very familiar in some sense, because
if you look at positions and momentum in quantum
mechanics, the commutator of position coordinates
1 component with another is zero, momentums
similarly is zero, but position with momentum
the right hand side would give you an ih cross
and the units operator if these are conjugate
variables. This is the germ of that, its starting
with that. So, that relation in quantum mechanics
has actually arrived at from here at some
sense follows from in this place.
So, there is in fact a correspondence which
will tell you how to go from Poisson brackets
to commutator does in quantum mechanics? And
the Hamiltonian framework lays the groundwork
for everything. Now of course, once you are
given these relations and the Poisson bracket
formulas you can find the Poisson brackets
of any functions what so ever. And lets write
down, what a Hamiltonian is for a simple system
like the simple harmonic oscillator and see
how this works?
So, I have for instance you could do this
in general, you could do this for a set of
particles so, if I have a system for which
the lagrangian as a function of q q dot is
equal to I will look at a autonomous system
now, is equal to 1 half the kinetic energy
minus the potential energy and the kinetic
energy is equal to it will take like a general
function, but let me take 1 half mi qi square
summed over i over all the particles I to
n minus V of q1 to qn and this is an example.
So, I have in mind the Lagrangian which is
just, whose independent generalized coordinates
are just the Cartesian coordinates of the
set of particles, call them q1 up to q n and
the potential which is function only of the
coordinates, no explicit time dependence here.
What is the Lagrangian in this case? First
of all, what is pi? This is delta L over delta
pk delta qk dot and this is equal to mk qk
dot, because only that qk is going to get
differentiated and it gives this mk qk dot.
And indeed in this simple example, it is clear
that the momentum is mass times the velocity
and vice versa, the velocity is 1 over the
mass times the momentum does it always have
to be true, the real definition is sitting
here in this place.
And therefore, what is the Hamiltonian? The
Hamiltonian is equal to summation over i pi
qi dot minus L, but that is equal to summation
over i pi but for qi dot I have to put in
this here and invert this expression here
write the q dot in terms of the p, because
remember the Hamiltonian is the function of
qs ts and time possibly.
So, this becomes pi dot and another pi dot,
so divide it by mi minus that takes care of
this part minus 1 half summation over i mi,
this is q dot square qi dot square but I have
to write qi dot as pi dot divided by mi and
then I prove that pi square over mi square
plus the potential that gives you the familiar
expression 1 half pi square over 2m plus V
equal to T plus V. This is how you identify
for such systems the Hamiltonian and the total
energy of the system. So, the minus sign which
we used in defining L has disappeared and
it is become a plus sign.
And this is the reason I chose to define Hamiltonian
as pq dot minus L rather than L minus pq dot
so, that I would get the total energy. Now,
since we know the Hamiltonian as the constant
of the motion for autonomous systems, we know
immediately that this combination here provides
you with the first constant of motion. And
in many cases, we simply write down the Hamiltonian
as T plus V, but you must express in terms
of moment.
Now, let us look at the other example very
quickly of the particle in a charged particle
in an electromagnetic field, what happened
there? What happened to the momentum? Remember
the Lagrangian as a function of r v and possibly
t was equal to 1 half mv square plus qA dot
v minus q phi. So, what is the momentum p
conjugate to the position r? This is delta
L over delta v, I am sorry to abuse notation,
by this I mean I do not mean to differentiate
with respect to vector, by this I mean a vector
formed by differentiating L with respect to
each component separately.
So, short hand for it slight the gradient
operator and the velocities you will write
it in casually in this form and this is equal
to mv, but here you have to differentiate
with respect to v and it turns out to be qA.
So, that gives us a very, very important relationship,
which says that the momentum conjugate to
the position for a charged particle in an
electromagnetic field is not the mechanical
momentum, which would be mass times the velocity.
But it also involves the field explicitly,
there are many many names given for this but
I would like to simply call it the conjugate
momentum 
or canonically conjugate momentum and this
is sometimes called the mechanical momentum
also sometimes called the kinematic momentum.
Which is what it is? For normal motion without
presence of field. But it is important to
remember, that p is not equal to mv, that
it is not in most cases it is not if particles
moving sufficiently fast certainly not equal
to mv itself, but here you also have the field
appearing potential appearing. Now, you should
ask a very deep question immediately, once
I tell you p equal to mv plus qA what would
that be? Pardon me say this slowly A is not
unique, A is not unique it is not gauge invariant,
it looks like we ended up with trouble. Because,
you have a momentum which I said is a free
space variable in physical, and now I am telling
you it involves A, its seems to be gauge dependent
this is going to lead us to some trouble,
but let see what happens? Go ahead with this.
What is the corresponding Hamiltonian and
then I stop here I take it up from that point,
H of r and p and may be t in this problem
is equal to, you plug this back in here you
can solve for this of course put it back in
here and then it will turn out that this becomes
p minus qA whole square over 2m plus q phi.
That is the Hamiltonian of a charged particle
in a electromagnetic field. p is just replaced
by p minus q, the p of the free particle is
replaced by p minus q.
And you guarantee absolutely that absolutely
guaranteed that the Poisson bracket of px
with x or x with px is 1, x with y with py
is 1 and so on. Why did we use this symbol
p for this combination in this case? I leave
it to you as an exercise now to write down
the Hamiltonian equations of motion and verify
that its exactly the same as Euler Lagrange
equation the Lorenz force equation that you
got, but notice the plus sign appeared here.
And A is appeared here in a strange way, although
it appeared linearly there, there is a quadratic
term here. The term which depends on A square
so, it is almost as if A is inducing itself.
This is related to the phenomenon of diamagnetism,
the fact that you have his A square term it
is going to be non trivial term, there are
many other non trivial terms going to happen.
If i square this as a p dot A term, there
is a A dot p term that does not matter in
classical mechanics.
But in quantum mechanics, A is going to depend
on r therefore, p dot A is different from
A dot p, you have to keep track of them. But
this is going to give us the right answers,
this is the correct Hamiltonian for the charged
particle in the electromagnetic field and
we will see, what happens if I make a gauge
transformation? What is going to happen this
stage? And it is going to be related back
to the fact, that you can add a total time
derivative to the Lagrangian nothing happens.
But we have to ask, what is the meaning of
that statement in the Hamiltonian framework?
What is the corresponding thing? And it will
turn out it corresponds to something called
a canonical transformation which does not
change the physics.
So, again all these things are closely linked
to each other and this is a very intricate
machinery, but I will start tomorrow next
time, I will start with this point and we
will look at a few problems in Hamiltonian
mechanics especially, if we have stability
and so on. Good question let me give the detailed
answer to that next time, says if the Hamiltonian
more physical than the Lagrangian, no I do
not say that at all. I will list the advantages
and the disadvantages of each of the two formalisms
then we will see just as physical.
