So, we are discussing the Maxwell’s equations,
in the relativistic case. Maxwell’s equations
written in the normal way in terms of the
electric and magnetic fields that itself,
it is a as a set Lorenz invariant, but we
could write yesterday in a more compact and
compact way the Maxwell’s equation; two
of the Maxwell’s equations, in fact. And
the way we had written that it is clear the
Lorenz invariance is clear in that this.
Now, today let us continue our discussion
on this and see how to take care of the other
2 equations. So, let us consider the object
that we had introduced yesterday now that
we have now that we are familiar with this.
We will have; I will call this as the field
tensor F mu nu, we did not introduce or call
it by any name yesterday. so, but it is usually
called the field tensor, it is a second rank
tensor and in the matrix form, we can write
it as 0 minus E x minus E y minus E z, E 1
0 minus B z B ym E 2 z B z 0 minus B x E 3
minus B by B x 0. So, this is what we had
written down yesterday.
And properties are that F mu nu is equal to
minus F nu mu; anti symmetric under the interchange
of these 2 indices mu and nu. And that we
said it is a second rank tensor. So, therefore,
it is actually going to all right, transform
in a like a second rank tensor, we will come
to that maybe at some other point of time.
It is not important here, let us see how do
we write down the lower components.
Let us look at the covariant form of this
tensor F mu nu. So, we have to lower 2 of
these indices, if you want to write it in
terms of the contra variant tensor. So, F
rho sigma; so, we will connect with F mu nu
covariant in this fashion g mu rho g nu sigma,
this is very similar to the raising and lowering
of the indices of vectors covariant to contra-variant
and contra-variant to covariant.
But then this let us see; what are the components
of F lower one. So, 0 0 or the diagonal elements
are all zeros. So, it is still anti symmetric
with respect to the interchange of these 2
indices. So, the diagonal elements are 0s.
Off diagonal elements F 0 1 is equal to g
0 mu is fixed to be 0, right and nu is fixed
to be 1. So, I can write it as g 0 rho g nu
is now 1 and sigma is anything F rho sigma.
rho sigma, rho and sigma are summed over all.
But we know that g is a symmetric diagonal
matrix. So, if one component is one index
is 0 the other index is also 0 for it to be
nonzero while. So, g 0 0 is 1 and g 0 with
any index. So, if rho is not equal to 0 the
first g the component of the first g is going
to be 0 then. So, therefore, the only non
vanishing term that you will get here is g
0 0 and sigma equal to 1. So, rho equal to
1 sigma equal to 1. So, this is rho equal
to 0 sigma equal to 1.
So, F 0 1 lower index indices is related to
F 0 1 upper index with the same indices upper
and lower cases and this is equal to g 0 0
is plus 1 g 1 1 is minus 1; therefore, altogether
there is A minus sign F 0 2 is similarly minus
F 0 2. F 0 3 is minus F 0 3.
We will now consider, we will now consider
mu equal to one and nu is either 0 or 1. So,
if it is 0 1 2 3. So, F 1 0 is equal to g
1 1 g 0 0 F 0 1 which is equal to again minus
F 0 1. In fact, since it is anti symmetric
in with the interchange of these 2 elements,
we do not really have to write this down at
all we can get it from F 0 1. So, we will
not write the rest of it at the 1 2 element
is new.
So, 1 2 is g 1 1, g 2 2, F 1 2 and both 1
1 g 1 1 and g 2 2 are minus 1s. Therefore,
together they will give up and therefore,
the sign is not changed in this case. Similarly
for F 1 3 which equal to F 1 3 and you can
see that F 2 3 is also equal to F 2 3 contra-variant.
So, altogether, we can write the matrix F
mu nu covariant as 0 there is a sign change
there. So, E x E y E z again, similarly for
the first column minus E x minus E y minus
E z diagonal elements are zeros.
In the case of off diagonal elements in the
second or belonging to corresponding to the
magnetic field there is no change in the sign
compared to the contra-variant case. So, it
is minus B z B y plus B z in the third row
minus B y in the fourth row and 0 diagonal
plus B x here minus B x here and 0 in the
diagonal term.
So, this is basically the contra-variant covariant
F mu nu. So, between contra-variant and covariant
F mu nu the electric field components change
sign and magnetic field components remain
the same that is easy to (remember). Consider
another object.
Which we will denote by F tilde mu nu equal
to epsilon mu nu rho sigma F rho sigma covariant
with a half here epsilon mu nu rho sigma is
defined in this fashion. Epsilon 0 1 2 3 is
equal to plus 1 and it is fully anti symmetric.
What do I mean by that? Meaning interchange
of any 2 indices keeping the others at their
positions will give you a minus sign. That
is, for example epsilon if I interchange the
first 2, I will get 1 0 keeping the third
and fourth at the same position 2 3 is equal
to minus 0 1 2 3 or it could be 0 2 1 3 where
I have interchanged 2 1 1 in epsilon 0 1 2
3. So, I get a relative minus n, etcetera.
Let us look at F tilde mu nu clearly, if mu
is equal to nu because of the anti symmetric
property of epsilon mu nu rho sigma if both
of them are the same, both if we need 2 indices
are the same that will give you a 0. That
is only way because one object will be say
for example, if I take epsilon 0 0 1 2 that
is going to be equal to epsilon interchanging
this to this and this to this.
Second; first and second will give me the
same thing at 0 0 1 no 3, but we already said
that interchanging any 2 should give you a
minus sign. So, that will give you consistent
solution as a 0. There is no other way that
you can get anything equal to minus one only
way is to set it to equal to 0. Therefore,
F mu nu is equal to 0 for mu is equal to nu
and it is anti symmetric with the interchange
of mu nu because epsilon mu nu rho sigma is
anti symmetric with the interchange of mu
nu.
Now, let us get the components F tilde 0 1
is equal to half epsilon 0 1 mu nu F all right.
So, F tilde 0 1 is equal to half epsilon 0
1 rho sigma F rho sigma. rho and sigma are
summed over. Now we see that since epsilon
already has fixed 2 of the indices 0 and 1
rho and sigma can take the rest of it rest
of the indices which are 2 or 3. So, it can
when I set rho equal to 2 sigma can take only
one value which is 3 because we cannot repeat
the indices.
So, here rho is equal to 2 sigma is equal
to 3. Add this to other possibilities. So,
that is what you mean by take summation over
rho and sigma. What are the other possible
rho values, we have taken one possibility
which is 2 we cannot take 0, we cannot take
one. So, the other possibility is 3 once you
take rho equal to 3 sigma is fixed to 2. F
3 2, but that is equal to epsilon 0 1 2 3
F 2 3 minus epsilon 0 1 2 3.
Because epsilon 0 1 3 2, if I interchange
3 and 2, it will pick up a minus sign. So,
that is equal to minus once F 3 2. But I will
do the same thing with epsilon sorry same
thing with F 3 2 as well 0 1 2 3 F 2 3 when
I change F 3 2 to F 2 3 that will pick up
another minus sign. So, making it plus epsilon
0 1 2 3, F 2 3.
Now, this were the identical terms and together
that will give me epsilon 0 1 2 3, F 2 3 which
is equal to 1 factor of one for epsilon 0
1 2 3 and F 2 3 is the covariant F 2 3. 2
3 is minus Bx. So, it is minus B x; now you
understand why we had taken a half factor
there it was in anticipation that we had put
a half there.
So, that we can get a minus B x without any
2 factors there or any other factor. In a
similar fashion, you can actually do your
homework and then get F tilde 0 2 as minus
B y F tilde 0 3 as minus B z ok.
And F 1 2 tilde is equal to half epsilon 1
2 rho sigma F rho sigma equal to half epsilon
1 2 are fixed what are available rho could
take 0 or 3. Let us take it to be 0 to start
then sigma is 3. So, F 0 3 plus epsilon 1
2 3 0 or rho equal to 3.
The other possible value F 3 0 and that gives
us epsilon 1 2 0 3 F 0 3 and epsilon 1 2 0
3 is minus epsilon 1 0 2 3 where I have interchanged
2 and 0 the second and third indices which
again to get into a form epsilon 0 1 2 3,
I have to interchange one and 0 first 2 indices.
So, that will give me a plus 0 1 2 3 F 0 3
which is equal to factor of 1 for epsilon
0 1 2 3 and F 0 3. F 0 3 is F 0 3 is E z.
So, this is equal to E z.
Similarly, we can get F tilde 1 3 as minus
E y and F tilde 2 3 as E x. The rest of the
terms are obtained by taking their interchange
of the indices from whatever we already have.
So, putting it together.
I have F tilde mu nu equal to 0 minus B x
minus B y minus B z; B x 0 E z; minus E y
B y minus E z; 0 E x B z E y minus E x 0.
Let me also write down F mu nu along side.
F mu nu, we had 0 minus E x minus E y minus
E z E x E y E z 0 minus B z B y B z 0 minus
B x minus B y B x 0. So, you can see that
we can get one from the other how do you get
one from the other F tilde mu nu is obtained
by taking.
Let us do the other way you start with F mu
nu change E to B all the components. So, wherever
you have an E x you replace it by B x. E y
goes to By E z goes to B z. And wherever you
have a B x you change to minus E x and E y
to minus E y, E z to minus E z. So, that will
give you F tilde mu nu or the other way around
with this.
Let us take F tilde mu nu. Take the for derivative
of this set it to equal to 0 to start with
and then that will give you as an exercise
this is an exercise divergence of B is equal
to 0 and minus curl of E is equal to dou B
by dou t. you will recognize this or these
2 as 2 of the Maxwell’s equations.
So, together with the earlier dou mu F mu
nu equal to j nu dou mu F tilde mu nu equal
to 0 for any nu give you divergence of E is
equal to 0; sorry divergence of E is equal
to rho divergence of B equal to 0; curl of
E equal to minus dou B by dou t; curl of B
which we got earlier is equal to j plus dou
E over dou t. This is a compact way of writing
the 4 Maxwell’s equation in a covariant
and explicitly or covariant fashion or Lorentz
invariant fashion.
I mean Lorenz covariant fashion. So, this
is what we have. And going on I am just wanting
to this here j nu is a four vector with first
component or 0 th component equal to rho,
and the charge density and 1 2 3 component
equal to the current density. all right. So,
let us come to expressing in the field now
in terms of the potential.
We know we can write electric field as gradient
of scalar potential and time derivative of
vector potential.
The first term is the electrostatic field
and second term corresponds to the dynamics;
electro dynamic term or which is due to changing
magnetic field. First term is curl free and
second term not necessarily curl free; curl
of the second term is going to give you the
variation in the magnetic field. Curl of A
is magnetic field B. So, that is what we have
for B; B is curl of A. So, if we have a time
changing the time varying electric field magnetic
field that will induce an electric field which
is given by this.
Let us focus on in the electric field expression,
let us denote the electrostatic field potential
phi and the vector potential, a together as
a for vector we can call it the 4 vector potential
and we already know dou mu is equal to dou
by dou t minus gradient dou covariant is dou
by dou t plus gradient.
So, now let us look at the E x. So, let me
write that again for the general. This one
is grad phi minus dou by dou t of E x is.
Let me first write the derivative term, minus
dou by dou t A 1 or A x minus grad; dou by
dou x A zero. So, this is equal to minus of
dou by dou t is dou 0 A 1 minus; so, now,
it should be plus dou by dou x is actually
minus dou a by dou x is dou 1.
So, we have dou mu equal to dou by dou t which
is dou 0 and minus grad. So, there is A minus
sign. So, there should be a minus sign here.
So, we have dou 0 A 1 minus dou 1 and A 0
which is equal to minus dou 0 A 1 minus dou
1 A 0. But if you look at the field tensor
F mu nu the zeroth mu equal to 0 nu equal
to one will give you minus E x.
So, taking F 0 1 as minus E x gives us dou
0 A 1 minus dou 1 A 0 as F 0 1. In a similar
fashion you can get 0 2 as dou 0 A 2 minus
dou 2 A 0. F 0 3 as dou 0 A 3 minus dou 3
A 0 now.
Let us switch over to B is equal to curl of
a B equal to curl of a B x is equal to dou
by dou y A z minus dou by dou z of a y which
is equal to minus dou 1 dou 2 A 3 minus of
minus plus dou 3 A 2. Since we know F 2 3
is equal to minus B x, we have that equal
to dou 2 A 3 minus dou 3 A 2. Similarly, F
1 3 equal to dou 1 A 3 minus dou 3 A 1 and
F 1 2 is dou 1 a 2 minus dou 2 A 1. Together,
We can write F mu nu as dou mu A nu minus
dou nu A mu, where dou mu is dou by dou t
minus grad gradient a mu is 0; sorry phi.
And what about the Maxwell’s equations dou
mu F mu nu is equal to j nu. So, that will
give, taking the derivative of F mu nu written
in terms of A nu and A mu, we have dou mu
dou mu A nu minus dou nu A mu equal to j nu
this is nothing, but dou mu dou mu acting
on A nu minus dou mu acting on dou mu A mu
is equal to j nu.
So, the Maxwell’s equation written in terms
of the potential in general looks like this.
Now let us come to something very interesting
partly something which we already know earlier
from our elementary electrodynamics discussions.
If you consider the expression for the electric
field when we consider the expression for
the electric field E equal to minus grad phi
minus dou by dou t of A.
If I take A mu and make it; change, make a
change in it to A prime mu which is actually
equal to the original A mu plus four derivative
of some scalar field. So, chi is some scalar
field 
and or scalar function. In this case its scalar,
chi is some scalar function. In that case
we have E prime equal to; So, essentially
this change in A mu will; sorry it should
have been A mu; phi will go to phi plus time
derivative of chi and A goes to original A
plus gradient of chi; A minus grad chi because
gradient has and relative minus sign compared
to the normal contra-variant vectors.
So, this will tell me that I have a gradient
of phi plus time derivative of a scalar function
minus time derivative of original A, 3 vector
A minus gradient of chi and this is equal
to minus grad phi minus derivative of grad
chi minus dou by dou t of A minus of minus
plus dou by dou t of grad chi, which is equal
to the original E, because the second and
the fourth terms are exactly the same excepting
that their signs are different. So, bottom
line is that if we change A mu to A mu prime
which will which picks up derivative of a
scalar field in addition to the original A
mu, then the electric field remains the same.
How about the magnetic field? Magnetic field
B is equal to curl of A which is equal to
curl of; or B prime, now is equal to curl
of A minus grad chi which is equal to curl
of A minus curl of gradient of chi. And curl
of gradient of any function, any scalar function
is going to be equal to 0 is equal to 0. It
is a property of the curl of a gradient. So,
this is equal to curl of A which is nothing
but the original B. So, B and A are invariant
as A goes to A plus 4 dimensional gradient
of a scalar function chi. Electric field is
not changed magnetic field is not changed.
So, there is a freedom in choosing the potential
A mu you cannot change the zeroth different
components of A mu independent of each other
arbitrarily. If the change in A mu is in this
fashion that A mu goes to originally A mu
plus gradient or dou mu of a scalar function
any scalar function denoted here by chi, then
the physical electric and magnetic fields
are not changed. So, if we consider the case
where electric and magnetic fields are measurable
and therefore, physical quantities, then A
is not uniquely fixed for a given E and B.
Or we cannot directly attribute therefore,
any physical meaning to a mu actually that
is not exactly the way we interpret.
We will say that there is some arbitrariness
in A mu. So, to physically interpret it as
anything we have to remove these unwanted
degrees of freedom. But this particular freedom
or arbitrariness in the potential is very
important in particle dynamics. We call this
the gauge transformation. This particular
way of changing A mu to A mu is called gauge
transformation. And we see that the electric
and magnetic fields are invariant under such
gauge transformation.
So, we can say physics is invariant under
such gauge transformations. And this is going
to be a very important point note when we
come to the dynamics of the particles in quantum
field theory. So, what we will do is to see
how to remove the unwanted degrees of freedom
in A mu and try to fix this. So, that later
on when we give physical meaning to this A
mu we have physical quantities without any
arbitrariness in it.
There are different ways to do that.
%%%%
So, let us look at one way of doing it consider
A mu. So, it has 4 components phi A x, A y,
A z when we say these are not uniquely fixed,
we can actually think about some relation
between these quantities that will try to
fix this. Let us consider the 4 divergence
of A mu, for any arbitrary choice of a mu
it is not guaranteed that dou mu A mu is equal
to 0.
But let me start with some potential which
I denote actually by A mu prime or A prime.
Divergence of that is not equal to 0, but
then I make a transformation to A mu which
is equal to a prime mu plus dou mu chi where
chi is some scalar field scalar potential
scalar function. Take the derivative of A
mu that is a derivative of A prime mu and
dou mu dou mu of chi. Now we can choose chi,
so that dou mu dou mu chi is equal to minus
dou mu A prime mu.
Since this transformation is allowed for any
scalar field let us say we get to some field
there is a scalar function kind let us choose
a scalar function which satisfies the condition
dou mu dou mu chi is equal to minus dou mu
A mu prime. That will then give us dou mu
A mu is equal to 0. So, we can always choose
some scalar function to set the potential
which is divergence less. To start with if
it is no divergence less then we can always
consider the scalar field which will give
us; which will set change the A mu so that
the new A mu is divergence less.
We have an A mu which satisfies this condition
is that the only arbitrariness in that that
is not if we further make any change to some
A mu plus divergence; A mu plus dou mu lambda
where lambda is some other scalar fields.
This scalar field cannot be completely arbitrary,
because in that case the for divergence of
A mu may not be equal to 0. But if we take
dou mu a mu of lambda to be equal to 0, or
if we choose lambda, so that dou mu dou mu
a mu lambda equal to 0 then; So, this is by
choice; so, that this is equal to 0.
This will give us dou mu A mu going to dou
mu A mu plus dou mu lambda which is equal
to dou mu A mu plus dou mu dou mu lambda,
which is equal to dou mu A mu, because the
second term is equal to 0 that is the property
of the chosen lambda. So, we have we can fix
this in this fashion. So, after fixing A mu;
so that therefore, divergence equal to 0 there
is still some arbitrariness in that and that
arbitrariness is that we can again and dou
mu lambda, so that dou mu dou mu of lambda
is equal to 0 without changing the electric
and magnetic field.
So, the electric and magnetic field or the
physics remains the same still. So, we will
come to this gauge freedom and how the Maxwell’s
equations are written within this and tell
a little more about the gauge conditions that
we have just now mentioned or discussed; in
the next discussion.
