We got a bird’s eye view into the, what
is called the standard model into the strong
sector of the standard model.
So, in the strong, we saw that the relevant
gauge group was s u 3 color and this was a
local gauge symmetry and so and this is sort of implied that, there were 8 particles spin one particles or 8 vector
particles called the gluons. And we saw that the hadrons, basically particles, which feel the
strong force were composed of quarks and we saw that the quarks, which were in the fundamental of s u 3 c.
So, these were the sort of things and but,
we also saw something very nice, it was using
a approximate symmetry, which was if you had
3 quarks, there is an approximate symmetry,
which was an s u 3 flavor. And we could we
found that this was able to organize mesons
and baryons but, it turns out that, there
are many more quarks more than just these
three.
And so, we will see, they are actually at
least we will explain that, there are 6 kinds
of quarks, which are organized into doublets
of some s u 2, which we will see in a short
while. But I am being little bit what do you
call, imprecise but, that is this is just
to give you charm and strange, and top and
bottom. So, there were 6 quarks and but, this
is the most massive guy and the so this in
some sense, these, these are very massive.
And so, if you want to talk of quarks, which
have low energies, which to construct the
mesons at some so, mesons were quark anti-quark
pairs. So, if you have a if you want to construct
something, which has a b b bar, just to create
a b b bar, you require twice the mass of that
thing. So, that is a very high energy object,
to create that so, it is much easier so, interms
of interms of masses these are lighter than
these are lighter than this, roughly speaking.
And in fact, s u 3 flavor is actually explicitly
broken by the factor, these do not have the
same masses for instance, in a obvious manner.
So but, but now, you could say that, you once
you able to track mesons, can we actually
be clever and try to use the factor. We know
some masses for these guys and work backwards
and work out the, so you pick an octet for
instance and you, you will find that because
the masses are different. You would expected
the, the mesons in an octet need not all have
the same masses, can we predict them in terms
of some three parameters.
And in fact, this is a very, very successful
way of preceding this thing but so the input
would be just a few numbers, like the masses
of these particles plus some knowledge of
the strong coupling strength. With these things,
you can actually have good estimates of your
masses and it is by I mean, you I would recommend
that, you look through it. It is a wonderful
idea of looking at something, for which we
do not have a complete understanding of things.
But nevertheless, you know from symmetry grounds
that, there are these constraint etcetera,
you can put them in and you find that, you
can do a good job. So, there is no field theory
necessary, in some sense to do that so, this
is, so this is one part of the standard model.
So, we looked at hadrons and this is the only
time, I am discussing fermions in this course.
So, what happens is that, there are leptons
and these are particles, which do not feel
the strong force. But, they feel other forces
so, for instance, we saw that, a neutron at
some lifetime of around 15, 15 minutes or
something like that. A free neutron it decays
into, you would have thought it should decay,
you know just based on charge conservation
it can decay into a proton, may be an electron.
But what people found is that there was some
missing energy and then, there was a puzzle,
should we think that, you know should we think
that, conservation of energy is violated or
something like that.
But, it turns out that does not true but so,
somebody was brave enough to predict new particle
called the, I guess it is a bar, anti electron
neutrino. So but, all these so, the thing
is that so, the thing is that, you find there
are more particles etcetera and this is actually
the forces involved in this are correspond
to something else and not these are not strong,
because these do not feel any strong interactions,
these two particles, this should there should
be something else, which is doing this.
And that is, that is sort of, let me something
called the weak force, it is a weak force
relative to say, you know electromagnetism
or something like that, that is why, we do
not say, also short ranged.
By the way, the strong sector if you look
at this, this does not explain, why it is
short ranged, just writing this thing, it
should be a I mean, it should be mass less,
etcetera. So, you might one way of doing it
is to say, may be there is some higgs mechanism,
which would become make it massive, that is
not the way it is going to happen. But, that
is exactly the way it is gonna happen out
here, for the weak force, it will be short
ranged because, the the vector bosons pick
up some mass, we will work out detail of this
mechanism in in the in this lecture.
So, it is short ranged and so, the idea is
to understand this force. But, the key point
here is that, they are these particles, a
whole bunch of particles like this. So, you
find that you have electrons, electron neutrinos
and the antiparticles you also have, what
you what people found is there was mu, nu
mu and tau nu tau. So, these are all leptons
so, these are exactly like the electron but,
it is more massive and this is also like this,
it is more massive.
So, in some ways, it is sort of let me put
them all together like this, that is the sense
in which, they are paired up again and the
point is that, now you see, that there is
a beautiful symmetry with the with the hydronic
sector I mean, the quark sector and the the
lepton sector. There are exactly 3 such things,
3 out here so, these sometimes they are called
families so, this is called the u d e nu is
the first family, c s mu nu mu can be one
family, t b tau and mu tau and scientists
are very clever.
And they tried to look for some larger symmetries
like s u 5 or something, which we would mix
all of them together and that, and that has
some amount of success. But, it has also there,
I do not think there is a given model in in
some sense, as we will seen in this towards
the end of this lecture that, experiment lags
behind theory in some sense. And today for
instance, the l h c is verifying stuff, which
is known from now ages.
So, so what you find is, these are the kind
of leptons that you have and so, these are
again, I have written something as doublet.
But, there is a technicality involved I mean,
in the sense that, one usually breaks up electrons
into left movers and right movers and one
only one of them pairs up with the neutrinos
in the standard model. But, again things have
changed from from interesting sources for
instance, the sun is known be a good source
of neutrinos.
And we understand, how the reactions of the
of the in the core of the sun rather well
and so, we so, there is a prediction, theoretical
prediction based on, you know things like
normal reactions, in the lab, because, you
do not think that the reactions inside the
sun are going to be any different from what,
we doing in the lab. Because, it is the same
material, same substance so, cross sections
will be the same, degrades will be the same,
etcetera.
So, you can predict how much, how many neutrinos
you should see in the lab but, it turns out
that that, the number of neutrinos we see,
is very small. By the way, neutrinos are really
extremely weak I mean, they can just, it is
very easy for them to just go through the
whole hours without interacting with anything.
So, you can see, they are very very hard to
detect and so, they were really detected the
first time, around as a missing mass and you
you sort of say that, there is this particle.
So, these are very very light particles so,
the original guess was, they were mass less
and in the sense, they were only something
called left movers. But, there were no right
movers or vice versa, it depends on, I do
not know what the convention is, so, that
is not relevant. So, coming back to, what
I wanted to say is that so, neutrinos are
very hard to detect and I just lost the thread
of what I wanted to say. So, what we will
see today is, to understand a little bit more
about weak force and how, these become short
ranged.
So, this is called the electro weak symmetry
breaking 
and the group the relevant gauge group is
going to be s u 2 weak curl u 1 Y, sub-script
Y. So, this u 1 is not supposed to be the
electromagnetic u 1 so the so, these are all,
these are local gauge symmetry.
And now, I remember what I wanted to say,
what I wanted to say is that, this doublets
have to do with only one set of movers. If
you are exactly mass less but, the thing is
that the, today we know from experiments of
the solar neutrino part, we saw that the number
of neutrinos we observed was less. This is
also another thing called the atmospheric
neutrinos or the so, both of these things
they were some puzzles, which we knew, how
to resolve there is a I mean, there is illusion
involved giving masses to the neutrinos but,
rather small.
So, they have plenty of bounce on what, the
total masses of of these particles can be
and thereof I mean, it is rather small, it
is of the order of e v I think for instance,
the neutrino mass, etcetera. So, for instance,
the solar neutrino puzzle is resolved by saying
that, some of them get converted to some other
flavor, I do not work in this area so, I do
not know the litigated details of that, it
has been many years since I looked at it.
So, but, the key point is that, we as theories
we understand how to handle all these guys,
even if they have little masses, we know how
to do them. But, the reality is that, you
should know what nature chooses and so, this
is a very theoretical course so, we we will
just look at mechanisms and we need to prove
that, this mechanism is indeed.
Now, we see that, the so called standard model
has 3 local gauge symmetries one is s u 3
c, the other two are these s u 2 and u 1.
So, in principle, what one has in mind is,
if we go at very, very high energy energy
scales all these symmetries which are hidden
or broken will will become obvious.
So, what happens here is that, the symmetry
breaking mechanism is that, this combination
is broken down to u 1 E M. Now, let us just
do some counting s u 3 has 3 gauge bosons,
u 1 will have 4 and the so, what we will see
is that, this is so the higgs mechanism will
give masses to 3 of the 4 and will leave only
one thing, which is mass less. And that combination
is what, we should call the electromagnetism
because, that we know that, we are I mean,
even at lower energies, we do see a photon,
which to to a great accuracy has zero mass.
So, let us understand, how this works and
so it works in a in the in some ways, this
is the simplest way of doing it, there are
many other ways of doing it. So, so what we
do is, we consider a complex s u 2 doublet
so, here all these guys, you know all these
doublets, which I have written are doublets
of this s u 2. And roughly, it will look like
so, the generators should look like t 1 t
2 t 3, which we use. So, to the approximation,
this there was an s u 2 sub group sitting
inside, this s u 2 flavor that, roughly has
is related to this we will see little bit
more.
And then, instead of calling it phi 1 and
phi 2, I call it as phi plus and phi 0, it
is a doublet. So, it is a s u 2 doublet and
the u 1 is just the diagonal, u 1 it just
acts in on this thing so, I will just write
the transformation.
It is best written that way phi prime will
be the transforming as e power i theta a t
a plus I by 2, may be some angle alpha times
y at t alpha, where t a is just the poly matrices,
half of the poly matrices, it is just the
so, this theta is are the s u 2 transformations
and this is just alpha is the u 1 transformation,
y is just the identity matrix. So, that is
like saying that, the y charge of phi phi
and phi plus and phi 0 is 1, is plus 1.
And this half is so, this is just to fit certain
notations, y is sometimes called weak hyper
charge, I just call it the hyper charge. I
do not have freedom in in these definitions
because, these are standard so, I have given
you the transformation so, you can see this
is just the statement that it transforms like
a doublet, under this s u 2 and this just
says that, the diagonal. yes. No, no no they
will become the u one charges, u one under
electro magnetism. It is not this u 1 Y.
Yeah, it is not this u 1 Y that is what I
am saying, the u 1 chargers are plus 1 plus
1 if you want or if you want to put this half
into that, you can say half half. Now, we
just need to so, this is just you go ahead
and make this into a local gauge symmetry
and you write out this thing.
So, the Lagrangian would be 
and and then, so this is what, we will put
and so, let me just do one more thing, we
can also put kinetic energies for these two
things.
So, what I will do is, I will call the field
strength for this, I will call it g mu nu
and for this, I will call it f mu nu a.
So, minus one fourth 
and where, I need to define a whole bunch
of things, d mu of phi is the covariant derivative,
with respect to all these things so, it will
be D mu of phi let me pull the phi out minus
i. So, note that I have not put the coupling
constant here, I rescaled things so, I will
get it, I will get them out here.
So, I will introduce two coupling constants
little g for this and little g prime for this.
g times so, I need to use, if you remember
for the u 1 case, I mentioned that you can
see, that the charges, matrix of charges would
look like a T. So, this is just that so, Y
and T are assume in a row out here so, this
is just that.
So, W mu is the, these three are the gauge
bosons or gauge fields, for s u 2 and this
will be the gauge field for u 1 Y. So, they
are just local so, this is so far nothing,
just carry over whatever, we did for non-abelian
gauge theories out here. And this for the
abelian gauge theory, just putting them together,
next thing is to just define these things.
These you are worked out in an assignment
so, let me do the easy one first is there
any question, is there anybody raise a question.
So, this is the piece but, this is a non-abelian
gauge field so, we will get some extra pieces
out here so, f a b c which would be in this
case, epsilon a b c W nu mu nu. It does not
matter what is this, I am not 100 percent
sure about the sign here by this but, this
is what you should get.
So, just remember that, the g comes here because,
think about it, when you what did you do,
wherever you say W you put g W. So, you would
have written a g w here, g w here but, this
is W square so, we will get a g square. One
you can pull out and that, will cancel the
g square out here so but, there will be this
thing. This is a very important way of writing
it because, what we are doing here is, writing
the kinetic energy in standard form.
And this g, here is really the coupling so,
what you would do in perturbation theory is
to expand things out and work with g g as
a small object and you do computations. So,
this is actually much more trailer to computations,
etcetera so, if want to fix masses, etcetera
this is the way to do it. Because, otherwise
you would be missing factors of g in the mass
even though, we are in 3 plus 1 dimensions
and g is dimension less, is still a number.
and so, you need to keep track of that.
So, that covers all these things, I just need
to, what is it we need about you, we just
need you is something such that, it has an
minimum at a non-trivial value of phi which
is, phi naught equal to 0. We could choose
my favorite potential but, the only potential
that I seen to know to make that work but,
actually that is not very relevant. And so,
I am not going to say that, I will just say
that the thing is minimized by some value
where, I will write out what phi should be.
You just ask me questions; if there are any
terms in there you do not understand or go
straight ahead.
So, what we do is so, let say that, the the
symmetry breaking minimum is given by same
phi equal to 0 and some number a and which
was it to be real. So, this is what we do
and once, we had so, this is what we would
have done, if you writing the vacuum solution
for that particular user.
So, a is determine of course, by the minimum
the region where, you can see that, that implies
phi dragger a is a square.
So, the minimum occurs at say, phi dragger
phi equal to a square, this is real so, the
thing is that, once we turn on gauge fields,
the minimum this vacuum solution would be
a solution with W and A also being zero. This
is something we have seen again now, what
we have to do is, to to get the masses we
have to write out fluctuations and we could
be clever right.
So, write out to get masses, first thing is
that so, first let us understand the unbroken
symmetry. So, I have to show that, you get
a u 1, what is the unbroken symmetry so, to
the unbroken 
symmetry would be given by the one, which
where, act on g or a and give you back 0 a,
where g is some combination of these guys
yes, thank you, the reason I am not using
a mu is because, we will I i think for once,
I will use a mu for the electromagnetism.
So, so we need to do this but, this looks
very complicated to solve but, we can solve
it in a very simple manner.
We just need to see that the g is this kind
of element, this is a typical element, you
act on that and so, all you need to do is,
to ask what is the linear combinations of
of T and Y, which acts on this, which goes
to so, you just it is easy to see that.
So, suffices to, to consider this is the advantage
of to, find the elements of the lie algebra
which on, which annulate this. Because, g
is one plus something so, this is the one
so, you can say, this is g minus 1 equal to
0. So, you want to and you expand it out so,
it is better to think of this as, g minus
identity of 0 a equal to 0. So, we just need
to look at this thing and it is very easy
to see that T plus, T minus will not work
or T 1 T 2 will not do it.
But neither will T 3 neither will Y, but a
linear combination and that combination is
very easy to write. It is this combination,
interms of matrices T 3 is half minus half
so, this will become 1 and 0 and trivially,
it annulates this. So, this is the unbroken
linear combination interms of the lie algebra
generators and no other combination all the
others of this thing so this is the unbroken.
So, in other words, g e power whatever, you
put these things, you choose theta 3 to be
equal to so, what do I mean by that g, which
is equal to e power some I alpha into T 3
plus half Y, will do the job, will be an invariance.
So, this proves that so, we gets that there
is a u 1, which is unbroken so, you can see
that the advantage of having a lie algebra
is the symmetry breaking.
Even checking for symmetry breaking is reduced
to some actually finding out the it is just
a problem in linear algebra more likely, it
is enough to do for small element. And then,
you are done and by exponentiation, I get
the more general valuation so, from general
things, which we have seen so, we expect.
So, for global case, global global symmetry
what is it, we have how many goldstone boson
should we expect. So, we have s u 2 mod u
times u 1 broken to u 1, will have 3 because,
dimension. So, so 
this is 3 plus 1 is 4 and dimension of g h,
which is u 1 E and M is equal to 1. So, you
subtract these two, you get 3 goldstone bosons
and the higgs mechanism will tell you that,
these goldstone bosons will three goldstone
bosons will give masses two three of these
linear combinations. And one of them will
remain massless, that massless guy will identify
it with electromagnetic gauge field so, now,
we need to do, we need to put meet on to this.
We have to, we know what to expect but, we
have put numbers and see what we get so, just
me let let me remind you so, first step would
be to do fluctuations, to get the masses.
We will consider fluctuations and so, we can
see that, in the local so, you would write
phi prime, phi as some e power, write the
most general group element e power i theta
a t a plus i alpha.
Instead of that, I can actually say that,
I should get the other combination of something
like that acting on 0 a plus some eta. First
point is that, there are not 4 parameters
here because, one called linear combination
that is, this combination annulates everything.
So, I should remove infact, I do not need
to so, I can just get rid of in that sense,
I can just get rid of this Y, I can trade
and rewrite in this sense.
So, but, the point is that so these fluctuations
so, you can see are exactly 4 fluctuations
and in the local thing, we can shift this
and get rid of this. So, can choose a gauge
at this fluctuations goes away so, we will
just write so we will write the phi as 0 a
and eta is some real guy. So, it is eta is
a real fluctuation and of course, there is
the gauge fields so, there is W mu a and B
mu.
So, they were originally 0 so, I do not need
to 0 plus something, I just call whatever
I get as a fluctuations, fluctuation plus
this gauge transformation part, which to remove
the get rid of that, this is what we get.
So, the first thing is to realize is that
now, so, I can I can drop this piece in the
gauge. So, I just need, you can see that to
leading order, this phi here is just 0 a plus
eta, what I will do is, I will do it in two
steps, I will forget about eta. We will do
the eta masses later so, we will just put
0 a so, we will compute.
So, then this term, if you look out here d
mu on some constant, a is a constant so, this
vanishes. So, this will just give you these
kind of terms, let just go ahead and play
with it a bit, I can erase this.
So, what I am going to do now is, I want to
work out what is d mu of 0 a, in principle,
I can do it with a plus eta but, I do not
want to do it for now. So, what is this we
need to work out but, this is nothing but,
the derivative acting on this is 0. If I put
the eta, they will give me a D mu of eta so,
that will be minus i g and t x where, half
sigma s.
So, we can just I can work out what g W mu
a T a acting on 0 a will be, this just half
of poly sigma matrices. So, I can just pull
that out, half g into 
again this was this expression was there in
your assignment, previous assignment so, you
get something like that. And let me just call
this combination W W mu plus, if I wanted
to do things, I should put root two’s etcetera
so when I look at the mass term I will remember
that.
At this point I will just leave it as this
so, this is just what I will call, W mu plus
so, this is equal to half g into a so, I can
even pull out this a, I get W mu plus here.
And this term will give me minus that is,
this piece, first piece I am pulling out,
a minus I and next one will be, this is not
easier. It is just g prime over 2 into Y,
which is just into 0, I am just jumping steps
here that is, this piece because, Y is a diagonal
matrix. So, any diagonal matrix multiplying
0 a at this, I pull out the a, this is what
I get so, you can see that, if I add, I just
need to add these two things and multiply
by minus i, I get what I need.
So, I get D mu of 0 a is equal to, you can
keep the halves outside, g a, even the a I
can pull out, g so, this is what I get. You
can see that, only one linear combination
of W mu 3 and D mu is coming out here, there
is an i minus i, which will disappear. So,
what I have to do is, mod square of this minus
I will go off, important thing is W mu, the
plus will become star of that is, W mu minus
so, we just work out, what we get for this.
So, now we see so, what we get is a square
by 8 into that is, this piece minus why did
I do this 8, there was half from this, is
there any half I have forgotten anywhere?
No, I have not, good. Now, I just do not want
to mess up that is all, minus 
what is there a half or I mean, I cannot see
what I have there, half half half, I am I
am perfectly fine so, this is what I got.
So, this I will equate to the following, why
am I using capital m, I have no idea. Let
us use small m, m W square, it does not matter,
you correct for that, that is true, legally
speaking that is required. So, I am correcting
it out here and then, here what we will see
so, this I would like to be half, I will define
something called m z z square.
I have not defined, the whole bunch of things
I have to define, only thing I have defined
is W mu plus and W mu minus. I do not put
a half here because, of the reason that, where
is that because, this I should have a root
2 in the definition, to get it correct. And
so, what is z, is its actually nice it is
just more or less that combination but, I
have to sort of make it. So, just comparing
things we see that, m so, z is so, what we
have is 3 bosons call them, W mu plus W mu
minus and z, these pickup masses.
The mass for the W’s square is just, just
by comparing is upon 8 and then, m z square
so, m z square will have this thing now. I
am little bit unsure about this, this could
be 4 and 2 depends on, the there are some
conventions. So, which I which are often,
I do not remember but, what you can see here
is that, this is a concrete thing that you
see.
And I forgot to mention that there was a combination,
which the unbroken combination, these are
the charges so, we will define Q, we will
define it to be T 3 plus half Y, do you see
anything familiar in this, what was that gellman
E C M relation. Now, going back to the original
definition of phi, you can check that the
first one so, this may, this is exactly, which
was 1 and 0. We says that, the guy below has
0 charge and the upper guy has charge 1 so,
this so this Q is nothing but, the electromagnetic
charge. So, we just need to work out, what
is the combination, which is which becomes
A mu and that is easy to see.
A mu is just the orthogonal combination to
this and I think, you just should exchange
g and g prime and change a sign here. So,
g prime, the angle is nothing but, now, this
also you can it is easy to see that, the W
mu plus and minus will have no Y charge. So,
T 3 eigen values, which is what, they are
plus and minus so, the plus fellow will have
electromagnetic charge plus 1 and this will
have charge plus minus 1 and this will have
charge 0 because, it commutes it comes from
the T 3 and y sector.
So, this, this, this whole thing says that,
you should find two, two charge charged and
u 1 bosons with masses given by this and you
also get some, you get another vector boson,
which is charge less, which has this. So,
usually it is, this is written z 0 and this
could be W plus and minus.
So, the the angle tan theta W is defined to
be g prime over g, let me just check this,
I do not want to write some wrong formula.
So, you can see that we can write, this would
imply that, sin theta w is so, the sin theta
upon cos theta so also same.
So, you can see that, this is this for instance
is minus cosine theta W, W mu 3 and this thing.
So, it is just a rotation by an angle theta,
W in this phase of D mu and, and this is the
photon, which does not appear in in the masses
so, this is massless. Now, you see that, this
relation is what I had remind when I said,
you know it looks so, there is nothing wrong
with thinking of this T 3 as related to the
other the the T 3, we saw in the flavor thing,
etcetera and it does not mess up anything
because, u and D sit in a doublet obviously.
So, we can look at numbers and the numbers
are nice M W is a 80 G e V, M Z is around
91 point something, let me leave it at that
91 G e V and sin square theta W is around
0.23. So, these are experimentally measured
things so, you could turn things around and
see that, these are also, you can see that
what are, what are all the observables? Observables
are the masses of M W and M Z these are things,
which we observed and then, you can, you can
see that for instance, what can we see M W
by M Z square, the ratio of them will be g
square by this thing. So, that would be sin
square theta W so, now you can see that, the
experimentally observed things, really fit
into this thing, and it really fits this pattern.
So, this goes by the name of Elashow Weinberg
Salam model and I must tell you these models
were proposed the the s u 2 cross u 1. Certain
parts of it was proposed well before the higg
mechanism was known and it took, I suspect
it took, it will early 60’s I think, took
enormous courage on their part. And people
did not even know that, the quantum theory
was good so, very brave people. So but, now
there is still one more guy, whose mass we
have to figure out and that is, the eta.
Now, I will look at so so, this will give
you one bunch of term, this will give you
half of that, will give you D mu of eta whole
square plus, maybe a interactions, etcetera
plus plus more terms higher order pieces.
But, this will give the kinetic energy piece
and then, there will be pieces with eta, with
the gauge bosons, these are interactions in
the field theory sense.
But now, what does this give you so, here
we will see that, you will get u, what is,
what would phi dagger phi b, it would be a
plus eta whole square. Now, we have to expand
this to the second order in eta so, this will
be equal to some u at a plus, the first order
piece will go out and then, you will end up.
So, we need to take a second order derivative
of this of evaluated at whatever you know
phi equal to with a half may be. Now, so,
this is a part so far, I mean all these things
did not involve, it only involve the value
of the vacuum expectation value, this thing.
And we did not need anything, any details
of the potential but, now, we need the curvature
of the potential at that thing and that, tells
you the mass of the eta square.
So, the prediction is of this, if this model
is correct the prediction is, there is a massless
scalar particle corresponding to eta, this
particle is called the higgs, and a lot of
money is being spent at LHC to actually find
this particle. But you can see that, there
is no way the, this data that we have the
fact that we have observed all these particles.
We know these things, I am giving you precise
numbers, that will give a, you can work things
around and you can get for me what is…
But that is not going to help you, fix the
mass of this particle because, this depends
on something else. So, you can go ahead and
put that lambda phi 4 kind of theory that
thing, which I write there lambda mod phi
square minus a square whole square, expand
it and you will find that, it depends on lambda
and lambda is not fixed by anything. So, there
is no prediction, you can make for this mass
now, people have other ways of putting constraints,
observance, these things and so, that comes
to my experiments.
But theoretically, there is no way, we can
predict it this, this is a serious problem,
I mean you do not know, is it low, is it high,
I mean these are the kind of things. And in
fact, recently in the news, I saw a sort of
possible masses where, in fact I think familiar
people ruled out some segment and you have
bank in the middle of somewhere. So, there
were range of masses I think, 100 to 200 GB
where, you know somewhere in between it is.
So, that the open question is, is it a light,
is it below that range, is it above that range,
these means different things.
And so, we so far, we have never observed
a fundamental particle, which is scalar we
have observed lot of scalar particles are
examples of that. But, they are not fundamental,
they, we know that they are made up of quarks
but, we have not seen a single scalar particle,
fundamental particle. And so, in some sense,
this is a holy gift and may be, may be that
is why, it is called the god particle so,
I answered my own question.
