So, let us recapitulate where we ended last
time. The Maxwell equations implied that you
had two equations for scalar and vector potentials
del dot A plus del square phi equal to minus
row over epsilon naught; this is the first
equation. And the second one was one over
c square delta phi over delta t plus del dot
A, was there a gradient of this outside, gradient
of this plus instead of del square let me
write box A equal to the right hand side mu
naught j; is that a plus sign or a minus sign,
plus sign, where this box is l one over c
square d 2 minus del square. It is the four
dimensional analog, it is the analog and the
space time as we will see when we do relativity
of the Laplacian operator. This operator which
I just call a box is more properly called
the d'Alembertian analogous to the Laplacian
and as I said it is the wave equation essentially
the operator that governs the wave equation.
So, if you did not have this it would just
be the wave operator on A equal to mu naught
j. These were the two equations and then I
pointed out that this set of equations is
true for every pair of potentials A and phi.
So, the same equation is true if you had A
prime phi prime everywhere and so on and you
could choose the A and phi in such a way that
del dot A was equal to zero, could always
do this and that was gauge invariance. So,
in the Coulomb gauge 
del dot A is equal to zero. Of course that
does not fix uniquely A itself because you
could always add to this A something once
again something which satisfied del dot that
new vector field equal to zero and of course
you will still be in the Coulomb gauge. So,
it is a family of gauges really.
In the Coulomb gauge this thing goes away
and then you are left with Poisson's equation.
Now how do you solve Poisson's equation?
The only point you have to remember here is
that this will imply del square phi of r comma
t equal to minus rho of r t minus rho the
function of r and t over epsilon naught and
there is t dependence but then I am going
to argue that this t dependence is just a
spectator because whatever t appears on the
left appears on the right as well and this
is true for every t. You could treat this
t as a parameter in this equation and really
what you have to worry about is the r dependence
the fact that this is a partial derivative
with respect to space variables and that is
the leading a function of r on the right hand
side.
Now let us go right back and see what we know
from Coulomb's law for a single point charge.
I know that if you have a single point charge
at the origin and I ask what the potential
is at any point r in space, this is the scalar
function, this is the electrostatic solution
we are looking at just Coulomb's law and
what is this equal to, what is the potential
at any point? It is q divided by 4 pi epsilon
naught r. That is the potential due to a single
point charge at the origin and that is Coulomb's
law. Now of course if you had many charges
you would superpose the potential due to each
of these charges and that would be the answer.
So, once again I have for example some charge
density on the right hand side, I superpose
that each of these objects and then an equation
of the form del square phi of r is equal to
rho of r divided by epsilon naught minus.
This kind of thing is solved by saying that
phi of r at any point r is equal to an integral
over all the charges that you have and if
the charge is at some point r prime then it
is rho of r prime d v prime volume element
at the point r prime divided by 4 pi epsilon
naught modulus r minus r prime that is the
solution. If you have a distribution of charges
in space somewhere and this is a typical point
r prime and you are asking for the potential
at a point r, then all that matters is the
distance between r and r prime multiplied
by the volume element of charge here that
is rho of r prime d v prime integrated over
all r prime. So, the entire extent of the
charge distribution and divided by this kernel
mod r minus r prime. So, I know that the solution
to this equation is that that is the free
solution and the solution is the one that
is appropriate of saying phi of r vanishes
at r equal to infinity.
So, boundary conditions have been put in,
in writing that solution down and the boundary
condition that has been put in is that phi
vanishes when r tends to infinity when the
distance tends to infinity. So, with that
boundary condition, I know that this Poisson
equation has that solution. Of course, if
you specify finite boundaries and the potentials
on the finite boundaries and so on, you must
change the solution to Poisson's equation,
but this is a free solution, and I do the
same thing here. Now the fact that this is
a mathematical equation which has phi of r
comma t equal to rho of r comma t is completely
irrelevant as far as this free solution is
concerned and it is quite clear that for every
t this equation is true. So, I could certainly
write this in the right hand side
Sir second order equation, that is right,
how many boundary conditions do you need?
Good point, the second order differential
equation how many boundary conditions do you
need? Well, in the theory of partial differential
equations all second order partial differential
equations have been classified and such equations
you have to specify you have to find out first
whether it is well posed or not, whether the
solution exists or not, whether its unique
or not. It turns out that in this case this
kind of equation is called an elliptic equation
in this theory. It suffices to specify phi
at all points at infinity, so really it is
an infinite number of points and what I am
saying is that the boundary conditions specified
is that phi of r tends to zero as mod r tends
to infinity. And of course that is such spatial
infinity in all directions that is like a
big boundary. So, imagine the surface all
these charges are enclosed inside some finite
sphere and then you let the radius of the
sphere go to infinity impose the boundary
condition that on the surface of the sphere
the potential is zero, then it is a well posed
problem.
What if the boundary is finite? Pardon me.
That is what I said if the boundary is finite
if for example somewhere you have a plate
and that plate is earthed and the potential
on that is zero then that is not the solution
to Poisson's equation. You have to ensure
that you add extra terms to it in order to
make the potential zero at this point but
we are talking about free space and we are
talking about free boundary conditions at
infinity. These are natural boundary conditions
at infinity the potential is zero. Then this
is the solution but this is only for illustration.
The point I am making is that, given a set
of boundary conditions appropriate set you
can solve Poisson's equation; that is all
I am trying to say and I wrote the solution
down without actually trying to solve this
equation. The simplest way of solving this
would be to go to Fourier transforms and then
this del square operator would just get multiplied
by k square and then you invert the Fourier
transform and so on. Since that would be a
course on Green's functions in partial differential
equations I do not want to do that right now.
Sir, is there a retarded potential? Pardon
me. Since it is a referent of time, now he
has brought up an interesting point which
is what I was going to do, he has anticipated
me. You see I have convinced you that if you
did not have this t dependence electrostatics
Coulomb's law we know is actually valid
and therefore that is the solution and now
I argue that for every t this is true because
t just acts as a spectator and then let me
paraphrase the question which I think is going
to bother him its bothering him and that is
the following. This potential leads to the
field because the physical field E depends
on this potential. Now the rho of r prime
t is over the entire charge distribution at
every point in space. Suppose I have a charge
distribution on the other side of environment
or galaxy and somebody there changes it. The
potential here is affected immediately because
you see no matter how far r prime is from
r, if this potential is changed a little bit
from the other side of the galaxy, then the
potential here is affected immediately and
this seems to violate relativity.
What would you say? Pardon me, but this is
what the equation says and that is the solution.
It is very easy to verify that that is the
solution to this equation. Exactly, E does
not depend on phi alone E also depends on
the partial derivative with respect to time
of the vector potential. So, remember this
formula; remember that E is equal to minus
delta A over delta t minus grad phi and we
have not yet solved for A. So, when you solve
for A you will discover that these A causal
effects are cancelled out. So, this is another
illustration of the fact that phi itself is
not physical. It contributes gradient of phi
contributes to the electric field but it is
not the whole story. You also have this piece
which you have to now solve for. So, that
is a good point good observation. In any case,
once that solved then the idea is that you
put in the solution for phi here and move
this to the right hand side and del dot A
is zero in the Coulomb gauge and then you
have to solve the wave function.
Now the next question you are going to ask
is what is the solution to the wave equation?
For this I have to write down the Green's
function, may be we will do it later in the
course but right now I do not want to get
into this because it is going to take me a
while to write down the solution to this equation.
Laplace equation was not so bad Poisson equation
because we already know Coulomb's law and
I just convince to do that; that is the solution
that is it, but this is going to take a little
more doing, solving an equation of this kind
it is not a elliptic partial differential
equation, this is called a hyperbolic partial
differential equation. This minus sign here
is very crucial and things are a little mess
here but you have to take it from me as an
article of faith that there is a well-defined
technique to solve this equation as well and
the problem is solved in principle.
But Maxwell's equation pi is not electrostatic
potential? No, it is not phi; phi is the scalar
potential. Pardon me. What are you trying
to do with the electrostatic potential? No
no no, I said let us go back instead of solving
Laplace Poisson's equation here, I have
an issue I said let us go back and recall
our knowledge of Coulomb's law and electrostatics
where I switch off all the time dependences
and I just have del square phi is equal to
minus rho of r and I said I know the solution
I know Coulomb's law. I put that down and
then my superposition that integral turned
out to be the solution and then I said t is
just a parameter in the exact equation here
the general case and I just put that down
there. So, that was just to show you that
that is the solution can be written down and
in this case I am not even doing that. I am
just saying that the solution can be written
down. We will not do that at the moment.
So, in the Coulomb gauge we know in principle
how to solve the Maxwell's equations. There
is the other gauge which is the Lorenz gauge.
In that Lorenz gauge this combination is zero
then you start by writing down the solution
to the wave equation with j as the source
term for the vector potential; put back in
here, this will no longer be zero here and
then del square phi is equal to the given
function rho plus a known function and therefore
once again you solve Poisson's equation
and the problem is solved in principle. Now
this is not my primary concern. We will come
back to this when we do relativity because
we would like to find out what is the meaning
of this gauge and I might as well say it here
right now.
It turns out that this quantity this set of
quantities phi over c and A together form
what is called a four vector a four dimensional
vector just as a vector that you have used
to in three dimensions is defined as a quantity
a set of three quantities which transforms
exactly like the coordinates themselves do
under rotations of the coordinate axis. In
exactly the same way this set of four quantities
will transform like the space time coordinates
do like time and the free space coordinates
do under Lorenz transformations and then this
would be called a Lorenz vector or a four
vector. So, this combination itself has a
specific transformation property just as ct
and r have a specific transformation property
under Lorenz transformations. I put a c here
because I would like to have all the components
of a given vector to have the same physical
dimensions and once I multiply t by velocity
fundamental velocity this has dimensions of
length.
So, this length and those three lengths here
x y z transform under Lorenz transformations
by the rules of Lorenz transformations and
my assertion is that this set of quantities
transforms in exactly the same way. They form
the four components of a four vector, the
time like component and the three special
components but you can find other such combinations.
For example c rho and j itself, so it turns
out that the charge density multiplied by
c and the current density have exactly the
same physical dimensions and that combination
also transforms like a four dimensional vector.
So, that is the reason why electromagnetism
looks so complicated when you write it in
terms of three dimensional vectors as we have
done here.
Once you write it in the four vector language
it is very very simple, it is very straight
forward and the equation were very elegant
indeed. You could also ask what the gradient
operator is; what is the analog of the gradient
operator four dimensions and just as we had
in d'Alembertian operator here, one would
like to define 1 over c delta over delta t
and of course the del operator but for a reason
it should become clear later it is minus the
del operator. This quantity would transform
like a four dimensional vector. That is the
appropriate four dimensional divergence four
dimensional del operator and it is not hard
to see that if you take the del operator and
dot it with itself you would get this d'Alembertian
and the dot product has to be defined in a
specific manner which we will do when we do
relativity.
So, the point I want to make is that this
combination here turns out to be the four
dimensional divergence of this four dimensional
vector and it would be written in the form
del mu A mu where mu runs over the values
this index runs over the values zero one two
three, zero for time, one two three for the
space variables and therefore setting this
equal to zero is equivalent to setting this
equal to zero and this is a scalar. So, this
means under Lorenz transformations the numerical
value of this quantity is the same in all
frames of reference all inertial references.
Therefore, once you set it equal to zero in
one frame its zero in all the frames. In other
words once you are in the Lorenz gauge in
one initial frame you are also in the Lorenz
gauge in all initial frames; that is the great
advantage of the Lorenz gauge. That advantage
is lost here because del dot A does not remain
unchanged when you go to another inertial
frame, it changes from frame to frame.
So, if it is zero in one frame it is non-zero
in another inertial frame and you can make
it zero once again by making a subsequent
gauge transformations but it could get tedious.
So depending on the problem that you have
you have to choose either the Lorenz gauge
or the Coulomb's gauge or any other gauge;
find your own gauge and solve your own problems.
So, its infinite amount of freedom that is
available; you have to observe some other
rules but otherwise the choice of gauge is
extremely useful and you saw here how useful
it was but I emphasize once again that physical
quantity measurable quantities in classical
electromagnetism would depend on the fields
on E and B.
If they depend on A and phi you must make
sure that they do not change under gauge principles;
so that is important as we saw. So, you can
use this freedom quite a lot but then the
fields should not change and indeed if you
recall what the energy density of the electromagnetic
field is it is epsilon naught e square plus
1 over mu naught B square multiplied by half
and it does not depend on A and phi; directly
it depends A and B and that does not change.
Similarly the point in vector which tells
you the momentum density is E cross B apart
from some constant and again depends only
on the physical fields. So, this is important
to remember and with this let us do the following.
Let us go back and ask what could possibly
be the Lagrangian for a charged particle in
an electromagnetic field. I am going to write
the Lagrangian down and then explain where
it came from. So, let me just first write
it down; I will just assert that this is a
Lagrangian and later we will justify, I will
justify it post factor based on the invariance
considerations.
So, for a charged particle in an electromagnetic
field the L for the particle and for the moment
I am focused only on the particle; I am not
worried about the field. The field is applied
from outside could change with space and time
but I am interested in the equation of the
motion of the particle. So, what can the Lagrangian
depend on? It depends on the coordinates of
the particle and the velocity of the particle
q and q dot but instead of q I really have
the three coordinates L r and the velocity
is v. It could depend on time explicitly because
the fields that I apply could change with
time. I could switch on an alternating current
or something like that an alternating voltage
in which case I must include this t explicitly
and this is equal to again the kinetic energy
one-half m v square and it is not t minus
v and the reason is the magnetic field produces
a force which is velocity dependent q v cross
b and this is not within the purview of the
normal t minus v kind of thing slightly different
and I will write the Lagrangian down and we
will see where it leads us.
This Lagrangian is q; this is the charge in
the particle. So, charge q and the particle
are assumed to be non-relativistic. So, this
entire formalism is valid non-relativistically.
So, even though the fields would obey Maxwell's
equations which have speed of light buried
in them, the particle itself is assumed to
move always at a speed much slower than the
speed of light; that is an assumption. Of
course we could write it down in the relativistic
case but we will do that after we study relativity
plus q times A dot v minus q times phi. That
happens to be the Lagrangian for a charged
particle in an electromagnetic field very
strange combination of this kind. Now let
me just try post factor to justify slowly
where it comes from. The Lagrangian is a scalar
as was pointed out. So, all terms on this
are scalars. We are trying to describe the
interaction of the particle charges charged
particle with the applied field. So, it is
reasonable that the some property of the charged
particle and some property of the field getting
coupled to each other.
This is the charge and that is the potential
scalar potential q times v is the current
due to this charge and the current is getting
dotted with the vector potential. You can
already see this is looking like starting
to look like a dot product of some property
of the particle and the dot product with some
property of the field and indeed if you recall
that j mu is zero and j current density and
the vector potential A mu is phi over c A,
the four dimensional analog of the current
density and the four dimensional analog of
the vector potential here the four vector
potential, then this combination here is nothing
but j mu A mu. It is the four dimensional
dot product of the current with the vector
potential; that is the reason for this Lagrangian.
But again at the moment you have to take this
Lagrangian from this given a matter of given
quantity and later we will justify it a little
more but I thought I should mention this here
as to where this comes from, what is the rationale
behind it and later on we will justify it
more rigorously. So, here is the Lagrangian
but this is not a very simple Lagrangian because
this quantity could depend on space and time
it could change. And similarly here plus q
of A r, t dot v and remember r is the coordinate
of the particle, v is the instantaneous velocity
of the particle and this is the Lagrangian.
Now what is the set of equations of motion?
Well, the Euler Lagrange equation say delta
L over delta each coordinate x y z, etc delta
L over delta x must be equal to d over dt
delta L over delta x dot; x dot is v sub x
by definition the x component of the velocity
and similarly for the y and z coordinates.
So, you have three equations of motion for
each. How are we sure that it is d by dt rather
than del by del dt; how are we sure that del
L by del x dot does not depend on time? It
would depend on time; it does depend. It does
not depend on other coordinates. It does depend
on other coordinates; that is why it is a
total derivative that is the whole point.
This is a total derivative; it is not differentiating
just the partial derivative.
In fact if you did not have time dependence
in the Lagrangian if you had an autonomous
system then this thing here would be zero
if you had a partial derivative and that is
not true. It is a total derivative; absolutely
it is very important. It is a good observation;
it is the total derivative with respect to
time. So, if the coordinate's changes with
time then you have to differentiate the time
derivative would the non-zero; it is crucial,
it is the total derivative. Now let us see
where this gives us and so on and similarly
for y and z. I could use the index notation
and do this but let us just do this with x
y z so you become familiar it looks familiar.
And let us compute what is delta L over delta
x dot equal to, what is this going to be?
Well, there is going to be a derivative here
because it is half m v x square and if you
differentiate it you get m x dot; that is
from this point here. Remember v y v z are
independent of v x as dynamical variables
so you differentiate with respect to v x alone
and you differentiate only half m v x square
here which produces a factor two cancels this
and gives you that but there is also dependency
here and this is A x v x and if you differentiate
it you get plus q A x. I will suppress the
r and t dependence inside, it is understood
and what is delta L over delta x equal to.
There is nothing from here because these are
velocities but there is dependence in here
on x y as well as z of every component.
So if I write that down you get q delta A
x over delta x v x plus delta A y over delta
x v y plus delta A z over delta x v z; that
is this term here if I differentiate with
respect to x and then there is derivatives
here; so you cannot forget that. So, it is
minus q delta phi over delta x this expression.
Now we plug it into the Euler Lagrange equation
and it says this quantity is equal to on this
side the time derivative of this the total
time derivative. So that is equal to m x double
dot plus q d A x over dt total derivative
and this is the equation of motion. So, now
permit me to erase this part then rewrite
it.
And you get mass times the acceleration m
x double dot equal to or let us write m d
v x over d t equal to, move this to the right
hand side and let us see what happens. So,
you have minus q d A x over d t but that can
be written as minus q delta A x over delta
t and then delta A x over delta x d x over
d t and so on but let us bring that last term
here first. So, that is minus q delta phi
over delta x. So, that takes care of the last
term and then I have to keep track of all
the other terms. So, plus q times delta A
x over delta x v x plus delta A y over delta
x v y plus delta A z over delta x v z minus
the terms that were left out from here the
special derivatives and the first of these
was delta A x over delta x d x over d t that
is v x itself minus delta A x over delta y
d y over d t that is v y and minus delta A
x over delta z v z. So, that is the full set
of terms; painful doing it components, easier
to do it with index notation but this should
illustrate what is going on and of course
this term cancels and then you begin to see
what is going on.
This is equal to minus q rather plus q. What
is minus delta A x over delta t minus delta
phi over delta x. Remember that E was equal
to minus delta A over delta t minus grad phi;
so the x component of E is minus delta A x
over delta t minus the x component of the
grad of phi which is minus delta phi over
delta x. So, you see this is immediately emerged.
This is q times E sub x it has immediately
emerged plus q times and now let us take terms
together. So let us take the v y terms here
and you have v y times delta A y over delta
x minus delta A x over delta y minus v z times
delta z minus delta A z over delta x, so that
takes care of these two terms here. But what
is this equal to? Well remember B equal to
curl A. So B y B z is equal to delta over
delta x A y minus delta over delta y A x in
cyclic order of one two three.
So, this quantity is equal to B z and similarly
this quantity equal to B y. So, therefore
this says this is equal to q E x plus v, v
y B z minus v z B y that is equal to v cross
B the x component of it. So, it says the x
component of m d v vector over d t is the
x component of q E plus the x component of
q times v cross B and now it is true for every
component because I choose the x component
arbitrarily and its true component by component.
So, this immediately tells you that the equation
of motion is m d v over d t equal to q E plus
v cross B as vectors, this is the Lorenz force.
So, the Lagrangian has given us the correct
equation of motion. We work backwards I mean
shamelessly wrote this Lagrangian down in
order to produce the right equation of motion
but this is the way it is derived and as I
pointed out the Lagrangian itself is obtained
by invariance considerations.
So, there is something fundamental about the
whole thing, the fact that the four dimensional
dot product of the current with vector potential
is the Lagrangian. It is called minimal coupling,
it is the least you can do and it is been
experimentally verified to incredible accuracy,
ten decimal places or something like that
in quantum electrodynamics. So, we know this
is true and we have good reason to believe
that this is exact but the fact is that we
work backwards we really said we have the
Lorenz force as an experimental observation
and what is the Lagrangian that gives it,
well that complicated velocity dependent Lagrangian
happens to give it but now with your perceptive
you should ask the obvious question. I wrote
this Lagrangian down and that is a very very
deep point.
Remember that the Lagrangian that I wrote
down was one-half m v square plus q A dot
v minus q phi; two questions arise. You could
ask what about the electromagnetic field itself
that could change and it should also come
from Lagrangian. I should write a Lagrangian
density for the electromagnetic field; yes
indeed you can but we are not finding the
equations of motion of the electromagnetic
field. We assume we know them already, what
are they; the Maxwell equations. So, I have
not derived the Maxwell equations from a Lagrangian
density; that is a little more complicated
exercise. It requires classical field theory;
we have not done that. But the fact is for
the particle non-relativistic particle this
is the Lagrangian; does something strike you
funny about this Lagrangian, does it make
you uneasy a little bit? Pardon me. It depends
on the auxiliary variables, it depends on
A and phi; does not seem to depend on E and
B directly and yet we are producing a physical
equation here.
Please note the equation here is in terms
of E and B; these are physical quantities
measurable quantities on both sides. We started
with these auxiliary quantities which are
not unique. So it should bother you that the
Lagrangian is not unique not in the trivial
sense of adding a constant which does not
get differentiated but serious non-uniqueness
because it is in terms of A and phi which
I can change by gauge transformations and
the question is how are we sure that this
is going to remain as it is eventually. Well
the answer of course is trivial because if
I put any other A and phi exactly the same
thing would happen because E and B are gauge
invariant quantities and we know that under
gauge transformations E and B do not change.
So, just B as B is curl A, B is also curl
A prime and E is minus delta A over delta
t minus grad phi also minus delta A prime
over delta t minus grad phi prime. So, it
does not matter. On the other hand you could
ask what happens to the Lagrangian itself
if I make a gauge transformation. Now that
is a serious question, so what happens to
this L? And this leads us to the non-uniqueness
of the Lagrangian. Under a gauge transformation
L goes to L prime is equal to one-half m v
square plus q A prime dot v minus q phi prime
where A prime is related to A by addition
of a gradient of chi. So, this is equal to
one-half m v square plus q A dot v minus q
phi plus q grad chi dot v because that is
A prime plus q delta chi over delta t because
phi prime was phi minus delta chi over delta
t and this is the extra piece in the Lagrangian.
So, this is equal to one-half equal to L the
original L plus this quantity here now what
does one make of it. Well, please look at
it and realize that it is equal to q times
d over d t of chi because the total derivative
of chi which is a function by the way of r
and t in general as we know is equal to delta
chi over delta t plus delta chi over delta
x d x over d t and so on and that is just
this.
And I have written this simple thing down
that d over d t of a scalar function is plus
v dot del. When you act on a scalar function
on the right hand side this is exactly what
you get and you are familiar with the second
term in fluid dynamics, it is called the convective
derivative. So, the total time derivative
is the partial derivative with respect to
time plus the convective derivative. It is
just a rule of calculus here. Well let us
take this q in and write it as d over dt of
q chi. So, the lesson we have learnt over
here is that if you make a gauge transformation
the Lorenz force equation would not change.
It is in terms of physical quantities but
the Lagrangian changes by the total derivative
of a function of the coordinates and time
in general; that does not change your equations
of motion. This is a general statement not
restricted to electromagnetic fields or anything
like that. In general the Lagrangian is non-unique
through the extent that you can add to it
an arbitrary total time derivative of an arbitrary
function of the coordinates in time; that
is true for any system and if you go back
here and look at how we found the equations
of motion it will become obvious.
We started by saying I define an action t
1 and t 2 equal to integral t1 to t 2 L dt
and L was a function of q q dot and t in general.
I go back to the notation q for the coordinate;
I do not want to use it here because I used
it for the charge and now if I write A prime
is t 1 to t 2 L prime d t this is equal to
the original A plus an integral t 1 to t 2
d F of q and t dt. So, the original Lagrangian
to that if I add a d F over dt I get L prime
as L plus d F over dt for an arbitrary function
of the coordinates and time and that is what
A prime becomes. In the equations of motion
the Euler Lagrange equations were found by
varying A and claiming that the first order
variation is zero the principle of extremal
action. So, I varied it between time t 1 and
time t 2 by any path whatsoever no variation
of the q's at these points, delta q was
zero here and zero here.
But then what does this give you? This is
equal to A plus F of q of t 1 t 2 rather minus
F q t 1 t 1. I can integrate this out the
precisely because it is a total derivative
term and then it is the two points at the
ends. And now our variation to derive the
equations of motion permit all possible variations
inside with the end points kept fixed and
at the end points there is no delta q and
since this extra piece depends only on the
q's at the end points, these two terms do
not contribute to the variation at all. Therefore,
the equations of motion do not change. Now
of course you should ask what kind of transformation
is this, what does this mean, what kind of
transformation is this? It will turn out;
I will explain this as we go along, it will
turn out that this electromagnetic example
is a very, very useful one because it is very
instructive, it tells us what is going on
in the general case. Well, it was a gauge
transformation here.
So I could in some sense continue to call
that a gauge transformation. By that I mean
to the Lagrangian of a system, I add the total
time derivative of an arbitrary function of
the coordinates and time and the equations
do not change, not velocities also please
notice. I have not allowed for you to add
a function of the q dots here. This could
be done but then I have to start putting conditions
on this f but without any conditions this
is the general case. It is called profound
implications because when we do the Hamiltonian
framework it will turn out that this same
freedom of adding to the Lagrangian the total
derivative of such a function would be called
a canonical transformation in Hamiltonian
mechanics.
So, there is a close link between what was
on here and what we are going to study a little
later and then it will become much more natural
in that framework, but already you can see
that the Lagrangian is not unique, you can
add the total derivative and in the electromagnetic
case in this case physically a gauge transformation
is equivalent to adding the total time derivative
of an arbitrary gauge function to the Lagrangian;
does not change the equations of motion. That
is a lesson worth keeping in mind. Is there
somewhere we could have expected this? Well
you can put a q dot but you will have to put
further conditions on it. It would have to
be a suitable symmetric function and so on
but not getting into that right now, but this
is the simplest case. I am not sure if I 
have answered your question.
So, in this language it is obvious; I mean
in this language once I say there is no variation
here, it is quite clear you can add something
which depends only on these end points and
then the equations would not change. So, in
that sense we could have predicted this, but
it is going to have physical implications.
The implication of that in this context were
precisely gauge transformations but that would
come about from other transformations as well.
So, we are going to look at that. There could
be other symmetry transformations on the Lagrangian
which would leave the equations of motion
unchanged and the necessary condition for
that would be that the Lagrangian should change
only by the total time derivative of a function.
So, it will get linked to various other quantities;
we will see we will get back to this. This
brings me. Does Lagrangian exist for every
dynamical system? No, no, I am looking at
only that class of dynamical systems for which
you can write down Lagrangians for which the
equations of motion arise as Euler Lagrange
equations. So, that is a restricted class
of dynamical system. So, in between I also
went and did general dynamical systems for
which you cannot write any Lagrangian or anything
out. So, you can push that to some extent
but once you have a Lagrangian structure you
can push things much much more, so certainly
not all dynamical systems. This brings me
to a very very important observation and that
will take us back to general dynamical systems
and that is the following.
When I wrote the Euler Lagrange equations
down let us rewrite this thins. So I have
this L of q q dot and t and from this I had
equations of the form delta L over delta q
i equal to d over dt delta L over delta q
i dot and i was equal to 1, 2. We looked at
systems with a finite number of degrees of
freedom. So, these were Euler Lagrange equations.
And now one of the most crucial things in
solving general dynamical equation is to find
the constant of the motion. The moment you
find the constant of the motion, as we saw
in the case of harmonic oscillator, we had
just the energy as the constant of motion
and immediately that gave us the phase trajectory
as ellipses. So, the important lesson is every
time you have a constant of the motion you
have solved part of the problem. You would
like to find more and more constants of the
motion.
Now in this problem the face space is two
n dimensional, the trajectory is a two n dimensional
is a line in this 2 n-dimensional space. It
is like a piece of thread moving around in
this two n-dimensional space. To specify this
line in a two n-dimensional space, how many
constants of the motion should I have? I should
have 2 n minus 1 because if I have one constant
of the motion some function of the q q dot
equal to constant that gives me a surface
of 2 n minus 1 dimension. If I have another
independent constant of the motion another
2 n minus 1 dimensional surface, the mutual
intersection of these is generically 2 n minus
2 dimension. Just as in ordinary three dimensions
I have a sphere surface of a sphere that is
two dimensional; I have a plane that is two
dimensional. These two cut each other and
the result is a circle in general which is
one dimension. So, every time you have two
surfaces intersect the dimensionality of the
intersection is lowered.
Now you want to go from two n-dimensional
face space to a one dimensional object you
need to have 2 n minus 1 constants of the
motion. So, every time you have a constant
of the motion you solve the problem a little
bit. Sometimes you can use and they are going
to be closely linked to symmetries and that
is one of the things I want to emphasize in
this course. We would like to find the constants
of the motion; this is very very important.
In fact in a sense all dynamical systems of
this kind are governed by their constants
of the motion. Once you have done that job
is done. One of the things that happen in
Lagrangian mechanics is that sometimes you
may have a situation where the Lagrangian
does not depend on a coordinate on a particular
coordinate such a coordinate is called cyclic
coordinate; for reasons i would not go into
it right now. So, if L is independent of a
particular 
q j, let us say so I avoid confusion with
I, some particular q j then q j is an ignorable
or cyclic coordinate. What does that imply
ones from this equation?
This would imply d over dt delta l over delta
q j dot is equal to zero but that equation
is solved immediately and what does that imply?
I am going to use this C.O.M for constant
of the motion. So, it says delta L over delta
q dot that particular q dot is a constant
of the motion, does not change. Its numerical
value would be the same for a given initial
condition would remain constant in time. The
numerical value would depend on the initial
conditions and once you specify that it remains
exactly the same.
Therefore, a very very important thing is
to find cyclic coordinates because then you
find constants of the motion and every time
you do that you won the battle a little more,
you have gone a little further. Let us look
at motion of a particle in a plane two dimensions
and let us assume for example that the force
the particle sees is a conservative force
and that the force is always directed to the
center of the attraction or the repulsion
the central force, what the kinetic energy
would look like.
So, I have motion of a particle in a plane
and these are the x and y axis and I look
at polar coordinates plane polar coordinates
rho and phi. This is the square root of x
square plus y square and that is tan inverse
y over x. I use this rather than r and theta
because I want to reserve that for spherical
polar coordinates. The Lagrangian of this
particle is a function of the coordinates
x y, x dot, y dot but I could also write it
in terms of rho phi rho dot phi dot and let
us assume that it is a constant potential
in the sense that it is time independent and
always acts in a central fashion. It is a
function only of the distance from the origin
of rho. So, this is equal to the kinetic energy
half times m times v x square plus v y square
but I should write this in plane polar coordinates.
What is v square in plane polar coordinates?
Rho dot square because there is a radial velocity
rho dot square plus rho square phi dot square;
that is the kinetic energy minus the potential
energy v but now I am going to say this is
a function of rho alone central potential.
Is there a cyclic coordinate? Phi is the cyclic
coordinate immediately it follows that phi
is a cyclic coordinate.
So, you are absolutely guaranteed phi is a
cyclic coordinate 
and that implies that delta L over delta phi
dot is a constant of the motion and what is
delta L over delta phi dot. Now we are coming
to that, it is angular momentum. So, if I
differentiate it here I get a phi dot here
and if I differentiate the two goes away.
So, it is equal to m rho square phi dot and
of course you recognize that this is the moment
of inertia of the particle about the origin
and that is the angular speed angular velocity.
So, this is equal to the angular momentum
of the particle about the origin and we guaranteed
it is constant. Of course we know that in
a central force there is no torque and therefore
the angular momentum is constant but it is
nice to see that it comes out automatically
as a trivial statement. Phi is a cyclic coordinates,
so the angular momentum is constant. Incidentally
let me call this in anticipation of the notation
p sub phi to show that it is the momentum
conjugate to phi. You are going to use this
word conjugate repeatedly; I will explain
what it means but let us anticipate this a
little bit.
So, it says this is a constant of the motion;
that is a big help. In this problem how many
degrees of freedom are there; two. What is
the dimensionality of the face space; four.
Now you like this problem to be solvable.
How many constants of the motion do you need?
You need three. How many do you have? We have
found one but we also know in the back of
our minds that the total energy is going to
be constant, actually the total Hamiltonian
is going to be a constant. So, we have a second
one and we need one more. We need one more
and that is not always easy to find in this
problem but let me anticipate things again
by saying that in problems of this kind if
you have a two n-dimensional face space sometimes
if you have n constants of the motion that
is enough, you can find the rest. They are
called integral but I am jumping ahead little
bit but this is an example of such a system.
These two are enough; they are independent,
the angular momentum and the total energy
are independent. You can specify each of them
independently and therefore, this problem
is in some sense solvable.
Let us do the same thing for three dimensional
motion we will look at what happens. In three
dimensions I have to use spherical polar coordinates
and I am again going to assume 
why do I always put x here y there and z there;
we could have done it the other way. Why do
I do that? Pardon me. Well, why do not I choose
x from right hand side and why outside the
board and put setup there. I would like to
choose the right handed coordinate system,
like to choose it just. No special reason
but I would like to make sure that when I
twist from x to y I go upwards and there is
no special reason. It is a left-handed coordinate
systems are just as good and this reminds
me of this famous TV scene about George Bush
because every time in class I explain right
handed coordinates I always use the left hand
because the right hand is chalk but then George
Bush does this when he says well the right
hand does not know what the left hand is doing;
this is very, very famous scene.
So, here we are going to have L of r theta
phi r dot theta dot phi dot and that is equal
to one-half m and now there is a r dot square
plus r square theta dot square plus r square
sine square theta phi dot square minus v of
r alone; that is a central potential, is there
a cyclic coordinate? Yes, it is a cyclic coordinate.
So, once again the azimuthal angle phi is
a cyclic coordinate. This implies that the
derivative with respect to phi dot is a constant.
So, it implies that m r square sine square
theta phi dot equal to p phi equal to a constant
of the motion. What is that quantity? It is
in fact the angular momentum but in three
dimensions angular momentum has three components
unlike in two dimensions where it has just
one component
In three dimensions angular momentum has three
components and therefore this is just one
component the azimuthal component of the angular
momentum and buried here although we do not
see it as cyclic coordinates, there are other
constants of motion. There is actually the
total angular momentum that too is a constant
but that is going to involve theta, it is
going to involve other components as well.
So, this is just one component and in this
problem by the way how many constants of the
motion do you need to solve this problem?
You need five but like I said these are systems
where three would do. We have the total energy
the Hamiltonian, we have this P phi and we
need one more to solve it and it will turn
out to be the square of the total angular
momentum.
Now once you have that the problem is solved.
Incidentally that is the reason why when you
go to quantum mechanics the hydrogen atom
is specified by a principle quantum number
and arbitral angular momentum quantum number
and a magnetic quantum number because constants
of the motion translate into quantum numbers
in quantum mechanics. This is the connection
between these two which we will see. So, once
again I will stop here. We see that the existence
of cyclic coordinates is extremely helpful.
We discovered the cyclic coordinate here by
changing to spherical polar coordinates; had
we looked at it in Cartesian coordinates,
this would not have been apparent. So, you
also begin to see the advantage of using symmetry
to choose coordinate systems appropriately.
So, let me stop here and then we take it up
from this point; we are ready to do the Hamiltonian
problem.
