WEBVTT
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In our last lecture we had started with giving
some equations which were equivalent of kinematic
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equations in the normal traditional classical
mechanics. In which we assumed that force
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is constant. We assumed that the motion is
along a straight line. So, that force and
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acceleration are in the same direction even
in relativity and then these equations could
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determine the velocity after a given time
and the distance travelled by the particle
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in a given time after that we evolved the
concept of force four vector. We had already
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defined what is a force? We defined in last
lecture what is force four vector? And then
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eventually came to the transformation of forces.
We realize that once we change the frames.
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We the force on the body also changes in relativity,
the force is also relative depending upon
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the frame from which you are looking their
force on the particle will also appear to
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be different.
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So, this is what we have mentioned here to
recapitulate we found out some equations relating
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to the motion of a particle under constant
force, and we discussed the force four vector
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and then derived the transformation of forces.
Now from today onwards will come to essentially
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the last topic of this particular series of
lectures on special theory of relativity,
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which is about the electric field and magnetic
field transformation. If you recall right
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in our first lecture when we are trying to
discuss the issues or the problems of the
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classical mechanics at that time we had mentioned
about the Lorentz force.
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So, this what I have written while beginning
our lectures, we pointed out an issue with
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the magnetic field in classical physics. We
had mentioned that in classical physics or
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this is to even relativistically.
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The force on a particle is given by if it
is a charge q, when the particle is charged
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with a charge q, the force is given by q E
plus V cross B where V is the speed of this
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particle in a given frame of reference. Now,
we had earlier discussed in the classical
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mechanics in the traditional classical non-relativistic
mechanics that acceleration is frame independent
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quantity. It means if you go to different
frames, the acceleration in the traditional
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classical mechanics, non-relativistic classical
mechanics will turn out to be same. But, the
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velocity of the particle may differ and in
general they will be different.
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So, we had said that if this force is a velocity
dependent force velocity being dependent on
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different frames. The force would also appear
to be different in different frames take for
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example, a frame of reference in which V is
0. It means the particle is at rest. In that
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particular frame of reference, the force on
this particular particle because of the magnetic
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field will be 0. Let us assume that E is 0.
So, it means there is no force but, on the
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other hand in any other frame of reference
when v is non-zero in principle the particle
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will experience a force which is v cross v
or the charge particle or experience a force
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which is v cross v. So, it looks funny of
course, we know that force is frame independent
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quantity, but we do not expect you know that
relationship to be obeyed when a force is
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zero E one frame of reference another frame
of reference it is a finite thing. At that
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time we also mentioned that some people can
always rise a question having our traditional
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classical mechanics knowledge that after all
V is also due to a motion of charge carriers
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and if I change the frame of reference. I
also expect that problem and if you do change
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and that time we had mentioned that see actually
the correct. The presentation or correct transformation
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of magnetic field and for that matter electric
field comes when we discuss special theory
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of relativity.
So, that is what we are going to discuss today.
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In fact, what we will be essentially telling
is that once we change the frame of reference
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both magnetic field and electric field would
change. So, in principle as far as relativity
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is concerned, there is no difference between
electric field and magnetic field depending
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upon the frame of reference. The field may
appear to be an electric field or may appear
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to be a magnetic field or a combination of
the two. So, let us start our discussion on
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this particular aspect today and then will
give eventually problem in some of the later
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lectures some examples we are we can discuss
this particular aspect little bit in more
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detail.
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So, this what we have had written. We realized
that the magnetic force is a velocity dependent
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force and velocity is frame dependent quantities.
I have just now discussed that. This is a
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Lorentz force which is F is equal to E multiplied
by E which is E is electric field plus V cross
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B where V is the velocity of the particle
and B is the magnetic field.
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So, this is the force which is being which
will be experienced by a particle which has
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a charge q or in this particular case I have
written e, so a charge e. So, now let us look
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for the transformation and for this particular
transformation we would essentially need the
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force transformation that we had earlier discussed
in our last lecture and we will see that how
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the electric field and magnetic field feels
to transform. So, that this particular force
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transformation is obeyed. So, we will take
some specific examples and using those examples,
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we will derive the electric field and magnetic
field transformations which appear to be somewhat
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more general.
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So, this is what I have written we discussed
at that time the electric and magnetic field
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transformations are actually found out from
relativity theory.
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So, that is what we are going to do today.
So, this is what I say electric electro-magnetic
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field transformation. So, EM essentially represents
electric field as well as a magnetic field.
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So, as we said that in relativity we do not
differentiate strictly between electric field
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and magnetic field. Because, what be appear
to one frame of reference observer in a frame
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electric field be appear to be mixture of
electric field and magnetic field to some
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other observer. So, we shall obtain it by
taking a specific case the results obtained
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are general.
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Now, let us consider the case, let us consider
that there is charge q which is instantaneously
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at rest in frame S prime. We can always assume
that that though this particular charge could
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be moving or could be accelerating or whatever
it is but, there is a frame of reference in
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which this particular particle or this particular
charge instantaneously at rest if you remember
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in our last lecture when we discussed the
force transformation also we considered this
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particular case. So, of course these frames
or initial frames, so if actually the particle
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is accelerating then if it is instantaneously
at rest in this particular frame of reference
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at related time it may not be. Because the
frames are actually initial frames and all
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are requiring that at instant of time when
I have discussed described in the situation
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this particular particle is at rest in this
particular frame at a later time it may not
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be. Now in this particular frame I assume
that there is a the particle is experiencing
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in electric field which I am writing as E
prime and it is also experiencing a magnetic
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field which I am writing as B prime. So, the
particle is instantaneously at rest in this
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particular frame and its experiencing electric
field and magnetic field which are given by
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E prime and B prime.
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So, let us see read this again we consider
a charge q which is instantaneously at rest
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in a frame S prime; however, it is experiencing
an electric field E prime and magnetic field
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b prime once I know this particular aspect
I can always write what will be the force
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on the particle which will be given by the
Lorentz force and now, let me use the relativistic
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notation because this is a particle which
is experiencing this particular direction.
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So in fact, this speed which will be given
will be u prime because it is S prime frame
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of reference remember the symbol v we have
reserved for the relative velocity between
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the frames and u and u primes are the speeds
of the velocities of the particles in respective
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frames.
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So, let us write the force on this charge
S the speed of the particle the velocity of
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the particle is 0 at that instant in this
particular frame. So, at that instant v cross
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b is 0, this was my force v cross b of course
here we have written u cross b.
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We write again F is equal to q E u
cross b prime. I am in frame S prime. So,
everything is been primed. This is the force
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which is being experienced by the particle
in S prime frame of reference. This was the
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electric field E prime and this is the magnetic
field B prime all I am assisting that is u
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prime is 0 at this particular instant of time.
So, this was 0. Therefore, the particle is
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experiencing only the force is due to only
electric field. The magnetic field would not
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cause a force on this particular particle,
because the instantaneous velocity happens
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to 0. So, the only force is due to the electric
field.
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So, this is what I have written S prime is
equal to q E prime. I can write this component
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wise because it will help me. So, I will have
to do this in components because, electric
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field I can always write as E x prime i plus
E y prime j plus E prime z prime k. I just
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take the components. So, I am writing F x
prime which is the x component of the force
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which should be given by q times x components
of the electric field y component of the force
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is q times the y component of the electric
filed and the z component of force is q times
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the z components of the electric field and
I am going to use this particular equation.
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So, what I have done. I have written this
particular thing in this particular piece
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of paper, so that I can keep on using it whenever
I want. So, now this is the equation which
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I have written here on this particular piece
of paper.
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F X prime is equal to q E x prime F y prime
is equal to q E y prime F z prime is equal
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to q E z prime then look at other equations
later. Now let us assume that there is another
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observer S in another initial frame S and
this is viewing this particular charge. So,
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there is another observer S which is viewing
this charge this is I am writing here.
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Let this charge be viewed by another observer
in frame S and let me assume that this S and
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S prime would be exactly the same condition
which we had been using right from beginning
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from Lorentz transformation. It means x axis’s
and the y axis’s parallel to y prime axis
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parallel z prime axis. The relative motion
is along the x direction and its along the
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x axis and x and x prime axis are always coinciding
this is the standard thing which I have written
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here.
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As usual we shall assume that the axis of
the frames are parallel and the origin of
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S prime moves on the x axis of S prime with
the speed V the relative speed between the
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frames we have always reserved the single
v. So, this is still I am using this same
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now when this particular observer looks this
particular particle at this particular particle
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at the same instant of time. It will assume
that if we find out that the velocity of this
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particular particle is not 0 but, because
this particle is at rest at that instant of
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time in S frame. So, therefore, this particular
particle velocity will turn out to be V which
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is the relative velocity with the frames.
This also one can use the velocity transformation
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we have taken done from similar examples earlier
this is very straight forward. We will always
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get if a particle is at rest in S prime. Its
velocity will turn out to be equal to v which
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is the relative velocity of the frames in
S prime frame of reference. So, the velocity
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of the particle is v in S frame of reference
that is what I had said according to S the
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charge moves with the speed v along the x
axis and of course this is velocity along
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x axis.
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Now we have said that in general I expect
that the force on the particle the electric
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field sorry, electric field on the particle
will turn out to be different and let us assume
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that the field that is seen by this particular
charge in this particular frame S will be
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E and B; electric field will be E and the
magnetic field is B. So, these feels in general
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will be different from the feels which have
seen by S prime observer used in the arguments
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which we have given in the beginning of the
lecture. So, what I can do what is our methodology
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is that I can always calculate what will be
the force on this particular particle in S
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frame. If I assume what will be the value
of if I take the electric field to be E and
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magnetic field to be B then I know how the
forces will transform from S prime to s. Because
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we have already derived the transformation
post transformation equation use these but
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the same equation and using those equations.
I will be able to find out the relationship
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between the electric field and magnetic field
as seen in S and as seen in S prime.
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So, this is my logic; this is the way I am
going to proceed. So, what I am going to do
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now is to calculate what will be the force
assuming the electric field to be E and magnetic
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field to be B in this particular frame and
then I know how the forces should transform
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using that I will calculate how the magnetic
fields and the electric fields use to transform.
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So, let us go to the next transparency as
I said according to S the charge particle
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is not at rest therefore, it would experience
a magnetic force also not just the electric
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field not just force because of electric field,
but will also experience the force because
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of the magnetic field and that particular
force will be given by F is equal to q E plus
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v cross B because and this v of course, is
along the S direction and this magnitude is
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going to the relative velocity between the
frames. So, I am written according to the
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S the force on the charge shall be given by
F is equal to E plus v cross B. here v is
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along the x direction.
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So, let us this particular equation in a proper
vector form and try to extract what will be
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f x what will be f y what will be f z. That
is why I am doing in the next transparency.
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Here I want to written out careful therefore;
because, there is a magnetic field terms also.
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Therefore, I am writing in explicit form F
is equal to q times E x i plus E y j plus
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E z k then it turn to magnetic field plus
q v is heavy component only in x direction
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therefore, I am writing as v i cross this
is the magnetic field which in general can
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be in any direction I write B x i plus B y
j plus B z k. But, as far as this first term
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is concerned it remains unchanged. I have
just copied it here. This is exactly identical
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to this but, now I can write this cross product
and simplify the second term. I know that
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i cross i is 0. So, when I take v i cross
B x i this will give me 0. So, that term will
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not can written then I take cross product
with the second term which is i cross j i
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cross j would give me k therefore, v i cross
B y j would give me v multiplied by B y in k.
18:09.330 --> 18:16.330
So, this what I have written v multiplied
by B y in k direction. Now i cross k is a
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minus a if you remember you have to go to
i j k i j k i. So, if you go i to i j, then
18:25.710 --> 18:30.900
I will get k if I go to j to k will get i,
but here I am getting and k to i then I must
18:30.900 --> 18:35.830
get j, but here I am going from i to k. So,
it will be minus j.
18:35.830 --> 18:42.830
So, this is what I have written here v multiplied
by B z i cross k is minus j which brings this
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minus sign and this is j. So, this is what
I have written here. Now like before I will
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pick up the x component of the force y component
of the force and z component of the force.
18:59.550 --> 19:05.490
So, these will be the force which we experienced
by the particle in S frame of reference. If
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the electric field was E and the magnetic
field was B both at this moment of time, I
19:10.080 --> 19:12.140
do not know what is E and B.
19:12.140 --> 19:17.170
I all am only writing the force component
and only if we which I know is the transformation
19:17.170 --> 19:24.170
of the forces. Same equation, which I have
written in the last transparency have copied
19:26.300 --> 19:33.300
it here and I picking up the different components.
So, I am writing x component of the force
19:36.460 --> 19:42.200
which will be the only i part. See i is appearing
only in this term nowhere else i is appearing.
19:42.200 --> 19:49.200
So, I just have picked up one term is just
q E x i. I am not writing because am just
19:49.520 --> 19:56.520
writing the x component of the force; x component
of the force is given by q E x. Let us write
19:57.910 --> 20:04.540
the y components of the force force for that
I have to pick up the terms containing j.
20:04.540 --> 20:11.540
This term contains j this terms contains j.
So, they are both contributing in a force
20:11.750 --> 20:18.750
in the y direction. So, this is E y minus
B v B z will be the y component of force multiplied
20:21.290 --> 20:28.290
by q. So, F y will be q E y minus v B z. Now
let us pick up the z component of the force.
20:31.730 --> 20:38.730
So, I have to look only the k terms. There
is a k here. So, this is E z k there is a
20:39.360 --> 20:46.360
k here there is a plus sign here. So, the
F z will be E z plus v B y of course, multiplied
20:46.720 --> 20:53.720
by q. So, F z will be q E z plus v B y this
is what I have written again here, because
20:58.730 --> 21:05.059
I am going to use immediately after that in
this particular piece of paper. F x is equal
21:05.059 --> 21:12.059
to q E x F y is equal to q E y minus v B z
and F z is equal to q E z plus v B y.
21:17.340 --> 21:24.340
Now, I know that what are the transformation
equations? Remember the particle is at rest
21:27.049 --> 21:33.700
in S prime frame of reference and in that
particular case our force transformation equations
21:33.700 --> 21:40.700
were somewhat simpler. So, those equations
I can write now. Because, the particle was
21:41.290 --> 21:46.720
at rest in a S prime frame of reference. So,
force F x of course, does not change will
21:46.720 --> 21:53.720
be explain. F x prime force in y direction
F y will become F prime y divided by gamma
21:54.230 --> 22:00.660
and force along the z direction F z will be
F z prime by gamma. This is a special case
22:00.660 --> 22:04.970
of the force equations which we have that
force transformation which I had used last
22:04.970 --> 22:11.110
time. In which I have said that if the particle
is happens to be at rest instantaneously in
22:11.110 --> 22:16.030
a given frame then the transformation equations
may not somewhat similar, and this precisely
22:16.030 --> 22:20.210
the case in this particular situation where
the particle is at rest in S prime frame of
22:20.210 --> 22:20.590
reference.
22:20.590 --> 22:25.660
So, I can write these equations in rather
simpler form and this is what I have written
22:25.660 --> 22:30.100
here and this is also I have written this
particular piece of paper. Here where I have
22:30.100 --> 22:36.470
written F x is equal to F x prime F y is equal
to F y prime by gamma and F z is equal to
22:36.470 --> 22:43.470
F z prime by gamma or let us look at this
particular paper carefully. I know that my
22:43.750 --> 22:47.520
force transformation equation must obey these
because, this is the way we have derived the
22:47.520 --> 22:53.420
force. If we have to be consistent these equations
have to be obeyed. I do not know what is the
22:53.420 --> 22:59.600
relationship as yet between E x and E x prime
and E y and E y prime and other fields other
22:59.600 --> 23:06.600
components of the fields. But I can always
use in this particular thing that effects
23:06.790 --> 23:13.790
must be equal to F x prime F y must be equal
to F y prime by gamma must be equal to F z
23:14.990 --> 23:21.900
prime by gamma because they have to obey these
force transformation equations. So, if I use
23:21.900 --> 23:26.740
this particular thing I can quickly find out
the relationships between the electric fields
23:26.740 --> 23:31.630
and the magnetic fields as we see in the next
transparency.
23:31.630 --> 23:38.630
So, this is what I have written. This is the
first equation which is q E x must be equal
23:40.179 --> 23:45.360
to q E x prime which because F x was equal
to F x prime. The second equation was F y
23:45.360 --> 23:52.360
is equal to F y prime by gamma; this was i
F y F y prime was just q E y prime divided
23:54.350 --> 24:01.350
by gamma. This is F z must be equal to F z
prime which is q E z prime divided by gamma.
24:02.390 --> 24:08.919
So, these are the three equations. First equations
is straight forward q will cancel gives you
24:08.919 --> 24:15.919
E x is equal to E x prime. Let us look at
two equations and see what what do they give
24:19.470 --> 24:26.470
us. This is just writing is equations simple
form as E x prime is equal to E x.
24:26.840 --> 24:32.559
If you look at back equation the q will cancel
out here, E y prime this gamma will come here
24:32.559 --> 24:38.380
on this particular equation and this will
give me E y prime is equal to gamma E y minus
24:38.380 --> 24:44.419
v B z. Similarly, if you look at this particular
equation q will cancel this gamma I can take
24:44.419 --> 24:51.120
on this side. So, E z prime will be equal
to gamma times E z plus v B y. So, these are
24:51.120 --> 24:56.299
precise the equations that I have written
in the next transparency.
24:56.299 --> 25:03.030
E x prime is equal to E x E y prime is equal
to gamma E y minus v B z and E z prime is
25:03.030 --> 25:10.030
equal to gamma E z plus v B z v B y I am sorry.
So, what do this equations give us this equations
25:12.160 --> 25:17.110
tells us that if my force transformation equation
has to be obeyed, then the x component of
25:17.110 --> 25:24.110
the electric field must remain identical between
S and S prime frame of reference as far as
25:24.110 --> 25:28.049
the y component of the force is considered
if electric field is concerned y component
25:28.049 --> 25:33.880
of the electric field will be change. It will
become I mean in the y component in the x
25:33.880 --> 25:39.950
prime component it will be given by E y prime
if E y and B z happened to be the y component
25:39.950 --> 25:44.580
of the electric field and z component of the
magnetic field in S frame of reference then
25:44.580 --> 25:49.900
the electric field in the y component of the
electric field in S prime frame of reference
25:49.900 --> 25:56.040
will be given by this particular equation.
Similarly, the z component of the electric
25:56.040 --> 26:01.280
field will be given by this equation where
E z is the z component of the electric field
26:01.280 --> 26:08.280
in S frame and B y is the y component of the
magnetic field in S frame. So, it means if
26:08.370 --> 26:15.370
I know what is E x E y E z and if I know what
is B x B y B z. I will find out I will find
26:15.450 --> 26:21.299
that it is will be the electric field. So,
I get this equations will tell me electric
26:21.299 --> 26:27.460
field in a given frame if I know the electric
field and magnetic field in a different frame
26:27.460 --> 26:32.490
what is interested to realize then this electric
field thus depend on the magnetic field with
26:32.490 --> 26:35.880
the other frame. Similarly, this electric
field also depends not only the electric field
26:35.880 --> 26:41.070
but, also the magnetic field in the other
frame that is what we admen by saying that
26:41.070 --> 26:47.410
relativity electric field and magnetic fields
have they do not have really that sense separate
26:47.410 --> 26:52.620
context because what happened this particular
magnetic field is contributing. So, to see
26:52.620 --> 26:58.260
to electric field in different frame of reference.
So, electric field does depend on magnetic
26:58.260 --> 27:01.010
fields in a frame.
27:01.010 --> 27:06.559
So, this is what I have written above equations
give the electric field in frame S prime when
27:06.559 --> 27:13.299
electric and magnetic fields are known in
frame S always do a inverse transformation.
27:13.299 --> 27:16.110
If I know what is electric field what are
the electric field and magnetic field in S
27:16.110 --> 27:20.710
prime frame of reference. I can always find
out the electric I means at the moment electric
27:20.710 --> 27:26.960
field in S frame at that particular matter
all we have to do is to change prime to unprimed
27:26.960 --> 27:32.790
quantities change the sign of v the standard
prescription of going from a direct transformation
27:32.790 --> 27:38.600
to inverse transformation. So, this I have
written inverse transformation it is given
27:38.600 --> 27:42.850
by usual prescription. So, I have written
E x is equal to E x prime because that does
27:42.850 --> 27:49.429
not change. Here I have changed the sign here
E y is equal to gamma E y prime plus v B z
27:49.429 --> 27:55.260
prime. Here also I have changed the sign E
z is equal to gamma E z prime minus v B v
27:55.260 --> 28:02.260
B y prime. So, if I know electric and magnetic
field in S prime frame of reference. I can
28:02.450 --> 28:08.789
find calculate what will be the electric field
in S frame could there is a missing component
28:08.789 --> 28:15.789
you might have realized in these equations
done. This equation is to evaluate the electric
28:17.760 --> 28:20.500
field.
But I have not given the transformation; I
28:20.500 --> 28:24.650
have not given the equations what will happen
to the magnetic field. We also realized the
28:24.650 --> 28:29.280
magnetic field will also change that is what
we are saying. So, these equations only give
28:29.280 --> 28:34.410
me the values or the components of the electric
field in a frame. if I know electric field
28:34.410 --> 28:40.400
and magnetic field in a different frame. I
must also I have to calculate the components
28:40.400 --> 28:44.870
of the magnetic fields also in this particular
frame even happened to know electric field
28:44.870 --> 28:48.640
and magnetic field in a different frame. So,
for that I have to do little more exercise
28:48.640 --> 28:54.919
which is not that simple as the way we have
done it but, that is do it because that is
28:54.919 --> 28:58.780
also equally important. Because, in principle
I must be able to calculate both electric
28:58.780 --> 29:01.700
field and magnetic field in a given frame.
If I know electric field and magnetic field
29:01.700 --> 29:06.419
in a different frame that is what actually
the electric magnetic electro-magnetic field
29:06.419 --> 29:08.850
transformation would mean.
29:08.850 --> 29:14.750
So, this what I have written magnetic field;
I have to find out the magnetic field components.
29:14.750 --> 29:20.429
To find magnetic field in S prime when we
know the electric and magnetic field in s.
29:20.429 --> 29:25.690
We need another set of equations that we have
not yet found out. This is what I will do
29:25.690 --> 29:30.669
now to find out what are the components of
the magnetic field if I know electrical magnetic
29:30.669 --> 29:34.020
field in a different frame.
29:34.020 --> 29:39.780
Now, I look again to any special case and
then results happened to be general enough.
29:39.780 --> 29:46.780
Now the I alter this equation little bit which
makes things little more difficult mathematically
29:49.330 --> 29:55.630
little more difficult in earlier case I had
assume that the particle was at rest in S
29:55.630 --> 30:00.210
prime frame of reference. Now I will assume
that this particular particle is actually
30:00.210 --> 30:06.760
moving along the plus y prime direction of
S prime frame of reference. It is not at rest
30:06.760 --> 30:10.860
in a S prime frame of reference. But, its
moving with a velocity but, that velocity
30:10.860 --> 30:15.230
is directed along the y prime direction of
S prime frame of reference.
30:15.230 --> 30:22.230
So, we now consider a charge q which is moving
with a speed u naught prime along y prime
30:23.340 --> 30:30.340
axis in a frame S prime of course, like before
I have assumed that this experiencing if electric
30:32.410 --> 30:39.100
field E prime and magnetic field B prime in
this particular frame. Exactly this situation
30:39.100 --> 30:43.110
the only thing is which is different now that
I am assuming that the particle is not at
30:43.110 --> 30:48.570
rest and therefore, the Lorentz force also
have a term because of the magnetic field.
30:48.570 --> 30:54.280
So, let us write that particular term again
we do exercise the flow is exactly identical.
30:54.280 --> 31:01.280
I take this particular force write the components
of the force then I go to S frame of reference
31:01.289 --> 31:05.250
then I will wait what will be the velocity
of the particle which will not be little more
31:05.250 --> 31:11.750
complex assume what will be the electric field
and magnetic field in S frame and E and B
31:11.750 --> 31:15.960
then again like the forces then I know what
are the transformation of forces in the two
31:15.960 --> 31:20.549
frames and of course, in this case neither
of the frames you will find the particle to
31:20.549 --> 31:21.020
be at rest.
31:21.020 --> 31:27.150
So, I have to use generic most most general
form of transformation and using that I will
31:27.150 --> 31:34.150
be able to find out magnetic field equations.
So, let us do that. So, this is the force
31:35.630 --> 31:42.630
on this charge which has been given as F prime
is equal to q E prime plus u naught cross
31:42.990 --> 31:48.840
B prime this is S prime frame of reference.
In my earlier situations, this term was not
31:48.840 --> 31:53.520
there but, now it is there because I am assuming
because that the particle is moving along
31:53.520 --> 32:00.000
the z direction. This first electric field
term I have written just as a in the component
32:00.000 --> 32:07.000
form E x prime i plus E y prime j plus E z
prime k. u naught prime I realizes purely
32:08.539 --> 32:15.539
in the y direction. So, I have written u naught
prime j cross this B prime again I have written
32:15.780 --> 32:22.780
in the component form B x i plus B y B x prime
i plus B y prime j plus B z prime k then I
32:24.620 --> 32:28.470
will expand this particular thing I take this
particular cross product I will write all
32:28.470 --> 32:35.000
this terms just like before then take out
x component of F prime y component of F prime
32:35.000 --> 32:39.730
and z component of F prime just like before.
32:39.730 --> 32:45.270
So, this first term is exactly identical of
what I have written this first equation in
32:45.270 --> 32:51.280
exactly identical of what I have written in
my last transparency. Let us try to evaluate
32:51.280 --> 32:58.280
this if I take j cross i. This will be minus
k, Because, i cross j k j cross i will be
32:59.980 --> 33:06.330
minus k. So, you will have u naught prime
B x prime minus k. So, this what I have written
33:06.330 --> 33:13.330
here there is a negative sign u naught prime
B x prime; j cross j will give me 0 then j
33:15.159 --> 33:22.159
cross k will be i. So, u naught prime multiplied
by B z prime i and there is a plus sign here.
33:24.630 --> 33:31.630
Because, j cross k is plus i. So, plus u naught
prime B z prime i or using this equation I
33:33.460 --> 33:38.240
can resolve I can take out the x components
y component x component y component and z
33:38.240 --> 33:45.240
component of the force in S prime frame of
reference. Same equation which I has been
33:45.890 --> 33:52.890
written here without any change. Let us pick
up the x component it means F x prime.
33:54.490 --> 34:01.120
So, let us look at those terms which contain
i. There is E x prime which contains i. There
34:01.120 --> 34:07.710
is a u naught prime plus B z prime which contains
i. So, this becomes q E x prime plus u naught
34:07.710 --> 34:14.710
prime B z prime which is forms x components
of the force as for the y component of the
34:15.119 --> 34:21.049
force is concerned there is only one term
which contains j. There is no other term which
34:21.049 --> 34:28.049
contains j. So, F y prime is just equal to
q E y prime. Become the z component z component
34:32.809 --> 34:39.809
this term contains k this term contains k.
So, z component is q E z prime minus u naught
34:40.960 --> 34:47.960
prime B x prime. So, these are the x y and
z component of the forces as seen by an observer
34:49.079 --> 34:53.789
in S prime frame of reference. I have just
copied like before these
34:53.789 --> 34:59.279
equations in this particular paper because
I am going to use this things F x prime is
34:59.279 --> 35:05.430
equal to q E x prime plus u naught prime B
z prime F y prime is equal to q E y prime
35:05.430 --> 35:12.430
f z prime is equal to q E z prime minus u
naught prime B x prime these equations will
35:13.660 --> 35:15.309
see later.
35:15.309 --> 35:21.450
So, like before this charge is being viewed
by an observer in frame S satisfying the normal
35:21.450 --> 35:25.859
Lorentz transformation condition and of course,
I will assume that in this particular frame
35:25.859 --> 35:31.979
S the field seen by the particle or seen by
the charge is E electric field is E and the
35:31.979 --> 35:38.819
magnetic field is B and again write the force
and do the force transformation equations.
35:38.819 --> 35:45.549
According to S the charge moves with a velocity
u which now have to obtain by inverse velocity
35:45.549 --> 35:49.479
transformation. Because I know the velocity
in S prime frame of reference. I want to find
35:49.479 --> 35:54.369
the velocity in S frame of reference. So,
I must use inverse transformation and if I
35:54.369 --> 35:58.609
use inverse transformation equation. I can
find out what will be the value of u in the
35:58.609 --> 36:02.749
vector form or rather x y x y and z components
of u.
36:02.749 --> 36:09.749
Once I know what are the components in S prime
frame of reference. So, let us do that first.
36:10.989 --> 36:15.979
These are the inverse transformation equations
x component u x prime plus v divided by one
36:15.979 --> 36:22.950
plus v u x prime by c square. Because, the
charge is moving only along the y direction
36:22.950 --> 36:27.619
in S prime frame of reference.
Therefore u x prime is 0. Because it happens
36:27.619 --> 36:33.479
only a non-zero component along the y direction.
Therefore, u x prime is 0. So, denominator
36:33.479 --> 36:40.079
gives you only one this component is 0. So,
u x is equal to v it just picks some of the
36:40.079 --> 36:45.369
x component is the same as a relative velocity
in the frames as far as the y component is
36:45.369 --> 36:50.719
concerned of course, like before because this
term is 0. So, there is only one here u y
36:50.719 --> 36:55.549
prime is non-zero and there is a gamma. So,
this will be given by u naught prime divided
36:55.549 --> 37:00.729
by gamma. Because we have written the speed
along the y direction to be u naught prime.
37:00.729 --> 37:07.729
So, u naught prime divided by gamma is what
is the velocity along the y direction. u z
37:09.989 --> 37:16.509
is equal to u z prime divided by all this
things and I know that u z prime is 0. Because
37:16.509 --> 37:22.890
the particle is moving along only the y direction.
So, this gives me 0. So, I have velocity u
37:22.890 --> 37:28.739
components the x component is equal to the
v v is equal to v and the y component u naught
37:28.739 --> 37:35.140
prime by gamma. So, my u has two terms as
I have said or we using these two terms.
37:35.140 --> 37:42.140
I can again like the force. So, as I said
according to S the force on the charge shall
37:42.759 --> 37:49.089
be given by q E vector E is field seen in
this frame plus u which we have just now evaluated
37:49.089 --> 37:54.499
by inverse transformation cross B where B
is the magnetic field as will be seen by an
37:54.499 --> 38:00.170
observer in this particular frame of reference
set where the velocity components have just
38:00.170 --> 38:05.849
been given. I mean just given in obtained.
So, I will now write this particular equation
38:05.849 --> 38:10.969
vector form and then eventually get x y and
z components of the forces then compare with
38:10.969 --> 38:16.859
what I have obtained as F x prime F y prime
and F z prime and I must know that they must
38:16.859 --> 38:19.049
obey the force transformation equation.
38:19.049 --> 38:26.049
This rather Lorentz transparency. q is equal
to E plus u cross B. E like before I have
38:28.430 --> 38:35.430
written is E x i plus E y j plus E z k plus
q u I have written x component and y component
38:36.900 --> 38:42.319
x component was v I have written that as v
i y component of u naught component divided
38:42.319 --> 38:44.859
by gamma.
Also this I have written plus u naught prime
38:44.859 --> 38:51.859
divided by gamma j cross again B like before
B x i plus B y j plus B z k. First term we
38:55.650 --> 39:01.430
just write like that exactly the same way
here. Second term let us take the cross product
39:01.430 --> 39:08.400
and write it open it up. So, let us first
look at it v i cross B x i. This will give
39:08.400 --> 39:14.460
me 0. Because, i cross i is 0 then try take
the second term i cross j will give me k.
39:14.460 --> 39:21.460
So, first term will be E v B by k. So, this
what I have written v B y k then I take i
39:23.469 --> 39:30.469
cross k i cross k which will be equal to minus
u therefore, I write this as v B z with a
39:32.680 --> 39:39.680
negative sign here j. So, its minus v B z
j. Now I try to take cross product of this
39:41.549 --> 39:48.549
second term j cross i will be giving minus
k. So, this is minus there is a minus sign
39:49.089 --> 39:56.089
here u naught prime gamma B x k; j cross j
will give me 0 j cross k will give me i plus
39:59.390 --> 40:06.390
i therefore, plus u naught prime divided by
gamma multiplied by B z. I pick up the x y
40:14.229 --> 40:20.400
and z component of this particular equation
like before, let us look at the x component.
40:20.400 --> 40:26.319
x component will have E x term and will have
a term u naught prime divided by gamma B z
40:26.319 --> 40:33.319
which I have written here; y component will
have a term E y here it will have a term v
40:37.619 --> 40:38.410
B z here.
40:38.410 --> 40:44.200
There is no term other term which contains
j term. So, you will have E y minus v B z.
40:44.200 --> 40:51.200
This what I have written here E y minus v
B z. Let us look at the z component z component
40:53.539 --> 40:58.039
have three term which contain k this term
this term as well as this term.
40:58.039 --> 41:05.039
So, this will be E z plus v b y minus u naught
prime gamma B x. This what I have written
41:07.719 --> 41:14.719
here. So, the three components have been written
here which I am writing again here for the
41:15.059 --> 41:17.069
convenience here.
41:17.069 --> 41:24.069
F x is equal to q E x plus u naught prime
by gamma B z. F y is equal to q E y minus
41:26.130 --> 41:33.130
v B z. F z is equal to q E z plus v B y minus
u naught prime by gamma B x. Let us look at
41:36.069 --> 41:49.149
just this particular equation here and here
now when I am writing the relationship between
41:49.150 --> 41:53.619
F x and F x prime. I cannot use those simple
relationship which I could use in earlier
41:53.619 --> 41:58.420
case because the particle happened to be at
rest of mass frame of reference I will use
41:58.420 --> 42:03.969
the generating relationship. So, let us write
that we have to use that general inverse force.
42:03.969 --> 42:10.969
Transformation to relate the forces in the
two frames see if you remember this was the
42:12.789 --> 42:19.789
x component transformation of the force the
force term of the four vector was work related
42:20.849 --> 42:27.849
to F prime dot u F x was equal to F x prime
plus v divided by c square F prime dot u prime
42:30.479 --> 42:34.519
if I take the dot product.
42:34.519 --> 42:41.519
I am sorry, this will be given by F x. Now
because, this is 0 this is 0 only one term
43:08.259 --> 43:15.259
will remain which is F y prime therefore,
I can write this particular term as just F
43:18.130 --> 43:25.039
prime multiplied by u y prime of course, denominator
because u x prime is 0.
43:25.039 --> 43:27.299
So, this denominator denominator there is
only 1.
43:27.299 --> 43:32.489
So, I can straight way write only the numerator
which gives me F x prime plus v divided by
43:32.489 --> 43:39.489
c square F y prime u naught prime. So, this
is relationship between x and x prime relationship
43:42.549 --> 43:48.369
between F y and F y prime cannot be. So, formed
found from the relationship; this term happens
43:48.369 --> 43:54.809
to be 1. Because, where x prime is 0. So,
this gives F y prime by gamma F z F z prime
43:54.809 --> 44:01.109
by gamma which is similar to what we have
done earlier because u x prime happens to
44:01.109 --> 44:05.709
be 0. So, these are the equations which I
have written here.
44:05.709 --> 44:12.709
So, what we have just now seen that F x is
equal to F x prime plus v by c square F y
44:13.680 --> 44:20.680
prime u naught prime F y is equal to F y prime
by gamma F z is equal to F z prime by gamma
44:21.650 --> 44:28.650
we already evaluated F x prime F x F y prime
F y F z prime F z all we have to do substitute
44:30.059 --> 44:33.989
in these equations.
So, let us first take only the x component
44:33.989 --> 44:39.819
of this equation substitute this value of
F x here this q E x plus u naught prime by
44:39.819 --> 44:46.819
gamma B z here F y prime is equal to q E x
prime plus u naught prime B z here. There
44:47.779 --> 44:54.779
is F y prime F y prime we put q E y E y prime.
So, I substitute things in this particular
44:56.329 --> 44:59.349
equation and let us see what we get.
44:59.349 --> 45:06.349
So, this what I have written using the equation
corresponding to the x component we get q
45:06.690 --> 45:13.690
E x plus u naught prime by gamma B z which
happens to be F x. This what is F x prime
45:15.109 --> 45:22.109
plus v by c square this is what was F y prime
using this particular equation you can see
45:22.279 --> 45:27.969
that q gets cancel out and I can write this
particular equation with E x prime is equal
45:27.969 --> 45:32.859
to E x because E x prime is equal to E x has
already been evaluated has already been found
45:32.859 --> 45:36.739
out when we did transformation obtaining E
x E y and E z components. So, E x prime we
45:36.739 --> 45:43.739
knew is equal to we know is equal to E x.
So, what I will do cancel this q substitute
45:44.259 --> 45:47.200
E x prime is equal to E x.
45:47.200 --> 45:52.549
So, using the following transformation that
is E x prime is equal to E x we can write
45:52.549 --> 45:58.680
the equation as follows which I am writing
here q has been cancelled out instead of E
45:58.680 --> 46:04.880
x prime I have written E x as you can see
that this E x would cancel out.
46:04.880 --> 46:10.910
Here, we have u naught prime divided by gamma
B z; once this E x has been cancelled out
46:10.910 --> 46:14.119
there is a u naught prime here there is u
naught prime here there is u naught prime
46:14.119 --> 46:21.119
here this u naught prime will also get cancelled
out. So, here I will only get B z by gamma
46:22.359 --> 46:29.209
just gamma I can take on the right hand side
and multiply it. So, B z will become equal
46:29.209 --> 46:35.239
to gamma which is this gamma which gets multiplied
here this term has already cancelled out in
46:35.239 --> 46:41.200
this particular term u naught prime has already
cancelled out. B z prime which is B z prime
46:41.200 --> 46:47.579
plus v by c square which I have written here
u naught prime has already cancelled out E
46:47.579 --> 46:53.819
y prime which I have written here. So, as
you can see I get a equation B z is equal
46:53.819 --> 47:00.819
to gamma B z prime plus v by c square E y
prime which tells me the z component of magnetic
47:01.200 --> 47:06.329
field in S frame of reference.
If I know the z component of the magnetic
47:06.329 --> 47:11.029
field in S prime frame of reference and electric
field y component of electric field in S prime
47:11.029 --> 47:17.359
frame of reference. So, this is one of the
equations you have obtained using this force
47:17.359 --> 47:22.799
equations which is part of the velocity magnetic
field transformation you mean magnetic field.
47:22.799 --> 47:28.759
So, as I have written this gives the inverse
transformation of the z component of the magnetic
47:28.759 --> 47:35.349
field. Now, when I look back at this particular
paper. This particular F equation F y is equal
47:35.349 --> 47:40.119
to F y prime by gamma. You can try it does
not lead to immediately to way new equation
47:40.119 --> 47:47.119
I will use this third equation. F z is equal
to F z prime by gamma. So, for this F z you
47:47.440 --> 47:54.440
know that there are three terms q E z plus
v B y minus u naught prime by gamma B x and
47:54.660 --> 48:01.660
for F z prime we have two terms q E z prime
minus u naught prime B x prime. So, in this
48:03.339 --> 48:07.739
equation I will substitute this value of F
z here I will substitute this value of F z
48:07.739 --> 48:10.690
prime this is what I am writing in the next
transparency.
48:10.690 --> 48:17.690
So, let us use the equation corresponding
to the z component of the force. So, this
48:18.509 --> 48:24.709
is this gamma I have multiplied because on
the both sides. So, I get gamma times q E
48:24.709 --> 48:30.949
z plus v B y minus u naught prime by gamma
B x. This quantity was equal to was equal
48:30.949 --> 48:36.499
to F z and this was F z prime which was actually
divided by gamma; gamma I have taken on this
48:36.499 --> 48:43.499
side. So, this becomes q E z prime minus u
naught prime B x prime this E z prime I know
48:43.619 --> 48:48.769
how to convert in terms of E z by using the
transformation of electric field which we
48:48.769 --> 48:50.509
have just now obtained.
48:50.509 --> 48:55.569
So, this what I will be substituting in the
next thing using the following transformation
48:55.569 --> 49:01.749
E z prime is equal to gamma E z plus v B y
which we have just now derived I will substitute
49:01.749 --> 49:07.569
in this equation and I will get this equation
left hand side remains exactly identical.
49:07.569 --> 49:14.569
u has been cancelled out for E z prime I have
written gamma E z plus v B y and this term
49:16.190 --> 49:23.190
remains as it is or if you can see very clearly
this gamma E z will cancel of this E z there
49:24.369 --> 49:31.369
is a gamma v B y which cancels of this gamma
v.
49:32.059 --> 49:39.059
If you look at this term this gamma will cancel
with this gamma you have minus u naught prime
49:39.390 --> 49:44.839
B x on the right hand side you will have minus
u naught prime B x prime u naught prime and
49:44.839 --> 49:51.009
u naught prime will cancel but, this minus
sign is you cannot be cancel this gives me
49:51.009 --> 49:56.619
B x is equal to B x prime. So, like we had
E x is equal to E x prime I get B x is equal
49:56.619 --> 50:03.619
to B x prime. So, this gives us the transformation
of the x component of magnetic field now I
50:04.229 --> 50:09.059
know how to find out B z and how I know how
to find out B x.
50:09.059 --> 50:16.059
I have not still would find in equation corresponding
to y which is quite little but, more afford
50:16.949 --> 50:22.130
for obtaining that transformation of y component
of the field we have to assume that the charged
50:22.130 --> 50:27.069
is moving in the plus z direction plus z prime
direction like we have assumed in plus y prime
50:27.069 --> 50:32.499
direction in S prime frame of reference. So,
you do exactly the same calculation again
50:32.499 --> 50:37.769
but, now we assume that the charge is moving
in plus z prime direction in S prime frame
50:37.769 --> 50:38.930
of reference.
50:38.930 --> 50:45.930
Otherwise we do all the calculations exactly
identically. We can carry out exactly similar
50:47.829 --> 50:52.709
analysis as we have done just now the first
equation would give the transformation of
50:52.709 --> 50:57.309
y component of the magnetic field just like
in this particular case it had given me the
50:57.309 --> 51:02.410
transformation of the z component of the field
if I would have assumed that the charge is
51:02.410 --> 51:06.459
moving in a plus z prime direction in S prime
frame of reference.
51:06.459 --> 51:11.390
The first equation will give me the transformation
correspondent to the y component which we
51:11.390 --> 51:18.390
will not work out here. One can try to do
it yourself and the result that we will obtain
51:18.440 --> 51:23.789
will be equal to B y is equal to gamma B y
prime minus v by c square E z prime. Now,
51:23.789 --> 51:30.309
I have found out all the three inverse transformation
corresponding to the magnetic field; I can
51:30.309 --> 51:35.119
always find out the direct transformation
by doing the same prescription replace v by
51:35.119 --> 51:39.499
minus v prime by unprimed unprimed by prime.
51:39.499 --> 51:46.499
So, these are all by equations which are relating
to the magnetic field this is what we call
51:47.369 --> 51:54.369
as the direct transformation. I know electric
field and magnetic field in S frame of reference.
51:55.489 --> 52:01.359
I will find out the magnetic field in y S
prime frame of reference. Here, I know the
52:01.359 --> 52:06.160
electric and magnetic field in S prime frame
of reference. I can find out the magnetic
52:06.160 --> 52:11.199
field components in S frame of reference.
We have already found out the relationship
52:11.199 --> 52:18.199
which gives me E x E y E z E x prime E y prime
E z prime essentially means that for example,
52:18.779 --> 52:23.769
if I know what are the magnetic field and
electric field in S prime frame of reference;
52:23.769 --> 52:29.440
I can always find out what will be the electric
field and magnetic field in S frame of reference
52:29.440 --> 52:35.499
or vice versa using these equations.
So, essentially we have got three sets of
52:35.499 --> 52:39.979
equations corresponding to the electric field
components and three equations set of equations
52:39.979 --> 52:46.440
corresponding to three equations corresponding
to the magnetic field components. So, if I
52:46.440 --> 52:50.609
know electric field and magnetic field; we
I can find out electric field and magnetic
52:50.609 --> 52:54.599
field components in any other frame of reference
and as you can see very easily from these
52:54.599 --> 52:59.359
equations they are all interdependent for
evaluation of electric field I also need of
52:59.359 --> 53:04.979
the magnetic field for evaluation of magnetic
field in general I also need to the electric field.
53:05.789 --> 53:11.150
So, here I come to end of this particular
lecture. I have just summarize this particular
53:11.150 --> 53:17.329
lecture we have done lot of special calculations
to find out the transformation related to
53:17.329 --> 53:20.249
electric and magnetic field.
Thank you.