WEBVTT
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In your last lecture we had discussed some
examples which involved momentum and energy,
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and the relationships how we change from one
frame to another frame. Specifically we discussed
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the center of mass frame, which we call C
frame. There is in contrast to C frame we
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have a l frame, which we call as a laboratory
frame, which need not be the frame in which
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the initial momentum 0, initial momentum of
the set of particle is 0, well in C frame
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it is 0.
We also said that some problems become easier
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to solve, if we go to center of mass frame.
We also discussed that some of the experiments
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high energy experiments are perform in the
center of mass frame in order to save energy.
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So, this what we have? Said about the last
lecture, we discussed some examples involving
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energy and momentum relationship. We also
discussed the concept of c frame and worked
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out some problems related to it. Today we
will like to go little ahead and introduce
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a new concept which of course in classical
mechanism is well-known, which is the concept
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of force.
In Newtonian mechanics, force is very important
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thing. We all know what is the force? So,
let us discuss what happens to the definitions
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of force once we comes to the special theory
of relativity, if you remember there two ways
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which we can talk of force in classical mechanics.
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One of course the force is the vector quantity.
So, always like with this, the vector sign
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F. we can write that F is equal to the rate
of change of momentum. When we say rate of
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change, it always means that time rate as
for its definition is consistent. So, force
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can be written as time rate of change of momentum
because in classical mechanics p is given
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by mass into velocity, where mass is suppose
to be costing.
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Therefore, equivalently we can write this
relationship in the form of mass into acceleration
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because rate of change of velocity with respect
to time is given as an acceleration. So, in
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classical mechanics we can use either of the
two ways, to express the force, either as
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the rate of change of momentum or mass into
acceleration.
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We will shall we shall just now see that in
a relativity, we have to choose one of the
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two and in peaceful in fact in relativity
we choose the first one. In relativity, we
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define force as the rate of change of the
momentum. So, we take the first definition
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not the second definition because as we shall
see that the two definitions are not same.
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Specifically as you know, that in momentum
relates to something, a in fact relates to
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the mass, in which the gamma u factor, we
just now discuss that.
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So, this what we said the force in special
theory of relativity is equated to time rate
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of change of momentum. So, the definition
that we have taken in relativity for the force
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is force is dp dt, were p is the momentum
of the particle. Let us see what does it mean,
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let us try to take this example forward, this
definition forward and try to get little bit
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more inside into, how our concept will change
with this particular definition of momentum
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this definition of force and also the wave
relativity defines momentum and other quantities.
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So, as we have said that, force in special
theory of relativity, we have said earlier
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is defined as gamma u m naught. So, u is a
particle is moving with a speed or with a
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velocity u in a given frame preference and
m naught is the rest mass of the particle,
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then gamma u m naught u is defined as the
momentum of the particle, this what we have
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discussed earlier in our lectures.
So, I must take the rate of change of this
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particular quantity, if I except this definition
of force which is force equal to rate of change
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of momentum, which essentially means that
force has to be written as d dt which is a
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time which is the rate of change of have that
time derivative. m naught which is here, u
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which is here and gamma u have express as
one upon under root 1 minus u square by c
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square.
So, force must be written in this particular
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fashion which essentially means that have
to take a time derivative of this particular
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quantity, where now you appear at two different
places, one here and another here. In classical
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mechanics in non relativistic mechanics this
was only place where you was appearing and
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must be took a time derivative just take give
a m into acceleration m into du dt which I
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can write an acceleration. But now if I take
this definition, I have to take this particular
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quantity take the derivative of this quantity
also. So, is a function in which, where you
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appears twice therefore, I must differentiate
by parts in order to get the actual expression
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for force applicable for special theory of
relativity, which have done the next transparency.
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So, we differentiate by parts, as we have
seen, the force that we have written was force
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is equal to d dt of m naught u under root
1 minus u square by c square.
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So, I am trying to differentiate by part.
First I take u, take the derivative of this
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quantity, plus I take this quantity multiplied
by derivative of u, this is what I have written
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this particular transparency. So, u multiplied
by d dt of m naught under root 1 minus u square
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by c square plus m naught divided by under
root 1 minus u square by c square multiplied
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by du dt is just simple differentiation by
parts. Now let us little more closely at this
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particular equation and try to see whether
I can make meaning out of it.
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So, next transparency that should I am trying
to do. I am trying to look at this particular
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first term, which is d dt of under root 1
minus u square by c square, this m naught
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which are appearing outside I have brought
it inside, I multiplied by c square and also
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divided by c square. The idea is that, I can
express this in terms of the new definition
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of energy because I know this particular factor
multiplied by m naught, multiplied by c square
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can be written as the energy of the particle,
total relativistic energy of the particle
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which is defined as gamma u m naught c square.
So, this is my gamma u m naught c square.
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So, this whole quantity can be written as
the total energy of the particle divided by
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c square. So, the first term in that particular
integration by part, can be written as u multiplied
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by d dt of E by c square, where E is the total
relativistic energy of the particle.
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So, just expressing in a different form. Now
du dt we know that can be written as the acceleration,
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if you are given in a frame, if you see a
particle moving in the rate of change of velocity
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of that particular particle in this frame,
obviously defined as the acceleration. So,
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a has to be expressed is du dt which means
that the force now can be expressed as u multiplied
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by d dt of E by c square, which we have just
now discussed, plus du dt have written as
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acceleration of the particle. So, m naught
a divided by under root one minus u square
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by c square.
So, very strange we find that with this new
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definition of force, force is given in terms
of summation of two quantities in which only
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one of the quantity here depends on acceleration.
Here the first quantity depend actually on
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the time rate of the total energy of the particle,
which is in total contrast with the classical
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mechanics. So, let us look at some of these
things, let first discuss this at this particular
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equation and try to see what modification
I have to make in terms of our understanding
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in relativity, about the force which is quite
different from the classical mechanics, this
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is what I have summarize in the next three
points.
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First in that we realize that f naught is
not given, f is not given by gamma u m a.
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Had I taken other definition of force equal
to mass into acceleration, then probably I
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would have had a tendency of writing force
is equal to gamma u times m naught a, but
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as you have said that relativity we do not
take that particular definition. But take
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the definition, of rate of change of momentum
which has causes an additional term in the
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force. Therefore, it is very clear that in
this case force cannot be written as gamma
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u m m naught a in general, unless off course
of first term 0.
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If you remember the two terms, here this is
the gamma u m naught a. This term is an additional,
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if this term happens is to be 0; that is a
very different case. Otherwise it is always
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sum of two terms, and therefore of is not
return in terms of just gamma u m naught a,
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remember has ambition earlier just in the
beginning of this lecture, in classical mechanics
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these two definitions are equivalent. I could
have written force equal to mass into acceleration
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or I could have written force as rate of change
of momentum. Here I take only force as rate
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of change of momentum and not in general F
is equal to m a, which I might have a tendency
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see that, because I am depends on speed of
the particle relativity.
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So, just replace m by know gamma u m naught
is does an for there. The second thing which
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is also very interesting is that acceleration
may not be in same direction as force is look,
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here this force is the vector quantity. Acceleration
is the vector quantity. So, u is the velocity
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of the particle. Now we can always have a
situation, in which the particle is in of
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moving in a particular direction and acceleration
of the particle is in different direction
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or we apply a force which is in a different
direction. Therefore, which is not in the
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same direction as the direction of the velocity
of the particle.
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For example, a particle could be moving this
way, going this way and I may to decide to
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applying force in this particular direction.
Not classical mechanics could not made any
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difference. If the force was in this directions,
the acceleration could also be in the same
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direction because alpha is equal to m a. But
in this particular case, because the force
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also depends on the direction of u therefore,
the force will no longer be exactly in the
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direction of acceleration in will not be exactly
in the direction of F. It is also depend on
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u, which is an total contrast with the classical
mechanics, where force was just in the same
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direction as acceleration.
Here force will produce an acceleration, which
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also depends on the direction of the speed
of the particle in that frame, this is what
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I have written. We also note that with this
definition of force the acceleration may not
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be in the same direction as the force. Force
is the another third thing which I would like
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to mention about this particular equation,
is that if force is constant it need not mean
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that acceleration is constant because it depends
the first term depends on u, in fact the second
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term also there is the gamma u. So, in peaceful
as speed of the particle changes, the same
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force may cause different acceleration while
in the classical mechanics, because mass for
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velocity independent, so if your force remain
same acceleration will always remain same.
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But here, acceleration will keep on changing
because first we apply force the velocity
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will keep on changing and depending upon the
velocity the acceleration velocity will keep
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on changing, this is totally contrast in the
classical mechanics, of course we can understand
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if you have to say that, our speed of light
is our ultimate limit and the particle will
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never able be able to cross the speed of light.
It means first we are applying the force and
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as the particle is trying to approach, the
speed of light, will obviously expect that
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it cannot accelerating this velocity cannot
change that much. So, the acceleration will
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not be that high must we are approaching to
the velocity of light. So, therefore, in that
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sense from redistrict concept we can understand
it, but from in classical mechanics it is
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total contrast.
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So, this what I have written even if the force
is constant, the acceleration is not, as the
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instantaneous velocity is also involved both
in the first term and the mass term. So, both
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the terms involve velocity. Therefore, you
not have in this. If the force remains constant
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as the function of time, it does not mean
that the acceleration also remain constant
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as a function of time.
Now let us look little more closely at the
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second term, the first term, the term which
involve u or which involve the rate of change
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of energy whether we can give a little more
physical meaning to the rate of change of
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energy on the term d dt.
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Now, we have said that this total relativistic
energy E can be written as sum of kinetic
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energy plus the rest mass energy m naught
c square. The rest mass energy we have discuss
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was a total relativistic concept. So, m naught
c square is the rest mass energy of the particle
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plus the kinetic energy of the particle will
give me the total relativistic energy of the
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particle.
So, if I just look at this time derivate of
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E upon c square, I can write this as time
derivative of m naught c square plus K just
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substituting for E, K plus m naught c square.
So, this what I have just substituted here,
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divided by c square. Of course the first term
will give me d dt of m naught and m naught
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is the rest mass of the particle, which of
course not expected to change as the function
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of time.
Therefore, this particular derivative will
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give me 0 therefore, this term can be simply
written as d dt of K upon c square, which
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is what I have written here. So, d dt of E
by c square can also be written as d dt of
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K upon c square because E and K differ only
by the rest mass energy and rest mass energy
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cannot change as the function of time.
Now, let us look go back to your classical
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mechanics and try to see, how this particular
K or rate of change of K is interpreted in
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classical mechanics. If one remembers the
work energy theorem, one would realize that
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kinetic energy is results because of a work
done by a given force.
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Let us look at the other transparency, as
you have said suppose there is a particle
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which is moving in a given direction and we
apply a force then if in a given time delta
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t, this particular particle moves by distance
delta r and we take a dot product of F and
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delta r. This is what is called work done
by the force and this by work energy theorem
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should be equal to the change in the kinetic
energy.
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So, if the particle was moving without any
force, of course is constant, the kinetic
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energy has to remain constant. It will not
change. Once we apply a force, once the force
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is applied, whatever is the displacement in
a given time delta t. If this displacement
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is delta r the F dot delta r F dot delta r
is called the work done by the force and this
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is equal to the change in the kinetic energy
because now under the application of force,
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the particles kinetic energy would change
and the change in the kinetic energy in the
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same time delta t will be given by F dot delta
r this is well known classical result which
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is called work energy theorem.
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This what I have written, if the displacement
of the particle in time delta t under the
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influence of a force F is delta r, often the
particle could be moving even for we are applying
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force, it does not change thing. As we have
said, in general assume, F need not be in
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the same direction, then by work energy theorem
the change in kinetic energy delta K is given
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as delta k is equal to F delta r. So, I can
divide this by delta t both the sides and
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assume that delta t tends to zero. So, these
deltas can be change to derivative, differentiates.
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So, that is what I will be moving next.
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The rate of kinetic change of kinetic energy
therefore, would be given by delta K divided
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by delta t, which I can write as, F dot delta
r by delta t. Delta r by delta t is the total
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displacement of the particle time delta t
divided by delta t therefore, this will give
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the velocity. So, this is the velocity of
the particle.
18:54.710 --> 18:59.650
So, delta r delta t can be written as the
velocity of the particle and if I make delta
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t tends to 0, I can take this as instantaneous
velocity of the particle. This is why I am
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writing, F dot u and this deltas, I am replacing
by d because I am assuming that delta t is
19:11.540 --> 19:17.620
tending to 0. I can write this as dK dt is
equal to F dot u.
19:17.620 --> 19:24.620
Now, therefore, the expression of force can
be written more clearly, the way we wanted
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to write it, containing two term, the first
term has been reinterpretated in terms of
19:32.460 --> 19:39.460
F dot u. So, it will be written as F is equal
to u multiplied by F dot u divided by c square
19:41.530 --> 19:48.060
plus m into a, where of course m is gamma
u m naught that is a member. So, this m is
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also speed dependent or the velocity dependent
mass, it is gamma u m naught.
19:53.030 --> 19:58.140
So, this is the way we write, we can thus
write the expression of force in terms of
19:58.140 --> 20:05.140
acceleration, as F is equal to u F dot u divided
by c square plus m a. Of course, one can see
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that in non relativistic limit, this expression
reduces to standard expression. In non- relativistic
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limit, this m can be assumed as m naught because
gamma u will be very close to 1, because u
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is very small in comparison to c, gamma u
is very, very close to 1. Therefore, m will
20:24.390 --> 20:30.050
be equal to m naught and if u is very small
in comparison to c there is a c square in
20:30.050 --> 20:33.630
denominator, these values will be very small
in comparison to c square.
20:33.630 --> 20:38.770
Therefore, even the first term will be neglect
and therefore, F can be written as F is equal
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to m naught a, the way we are used to writing
classical mechanics. Therefore, we can see
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that it really reduces so its standard expression,
in the non relativistic limit. Now let us
20:52.690 --> 20:59.690
look at some special cases, in which the force
happens to be in the same direction as acceleration,
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which we have just now said. Because of this
complicated equation that we have I would
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have called a complicated, in which there
u was involved in there is a u involved here.
21:08.490 --> 21:13.070
There is u involved that means there is work
done by the force involved, then only we get
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force equal to mass into acceleration.
So, you have two terms, now what are the cases
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in which force and acceleration could be in
the same direction that can happen in two
21:24.170 --> 21:31.170
cases. One of the cases will be, where this
term 0. This term happens to be 0 then F dot
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u for example, when F dot u 0 it means force
is applied perpendicular to the direction
21:37.310 --> 21:41.840
of this speed, discontinues velocity then
this term will be 0. Then F will be given
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equal to mass into acceleration. F and a will
be in the same direction.
21:46.290 --> 21:52.440
There is a second special case, where we have
only one dimensional motion, it means the
21:52.440 --> 21:56.130
particle is moving in the particular direction
you are applying assoils force in the same
21:56.130 --> 22:00.960
direction. So, particle is speed changes also
in the same direction, all the motion is confined
22:00.960 --> 22:01.660
in the x direction.
22:01.660 --> 22:08.040
So, instead of for example, in this particular
case which have written here I take u direction
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to be this and force direction is also in
the same way. So, the motion of the particle
22:14.240 --> 22:19.590
is in the same direction, is only a one dimension
motion. The particle layout motion will keeps
22:19.590 --> 22:23.190
on moving in the straight line. If its keeps
on moving in straight line then in that case,
22:23.190 --> 22:27.700
this vector u and this vector a is in the
same direction. Therefore, force will also
22:27.700 --> 22:32.990
be in the same direction as acceleration.
So, the two special cases where I will see
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that force and the acceleration are having
the same direction. Let us discuss these two
22:37.490 --> 22:43.710
cases these, two cases are interesting and
are of course, comparitively.
22:43.710 --> 22:50.710
So, what I have said we can see that the two
special cases when force is parallel to the
22:51.910 --> 22:52.810
acceleration.
22:52.810 --> 22:58.570
Let us take case one, if the force is always
perpendicular to the instantaneous direction
22:58.570 --> 23:05.260
of velocity, of course once we apply the force
the velocity will change, magnitude in general,
23:05.260 --> 23:12.260
both in magnitude and direction or at least
one of them that has to change. But if it
23:12.390 --> 23:18.310
always happens that force is always perpendicular
to the direction of velocity, that F dot u
23:18.310 --> 23:22.040
will always be equal to 0, that has we have
said the first term will become 0 and then
23:22.040 --> 23:25.760
will have a simple case of F is equal to m
a.
23:25.760 --> 23:32.030
The force will be given by F is equal to m
a, of course m is we have said, is a velocity
23:32.030 --> 23:38.240
dependent mass. So, it must be written as
gamma u times m naught a. So, this is would
23:38.240 --> 23:43.700
be the case or which we thought that was could
have been one of the definition of force.
23:43.700 --> 23:49.510
In principle with this definition of force,
is applicable only when force is perpendicular
23:49.510 --> 23:54.790
to the direction of u, in that particular
case F is actually in deep given by gamma
23:54.790 --> 24:01.490
u m naught a.
So, some times because this gamma u m naught
24:01.490 --> 24:06.440
is called transverse mass because this will
be the mass of the particle. Now we can write
24:06.440 --> 24:12.960
the force equal to mass into acceleration
with this mass gamma u m naught, only when
24:12.960 --> 24:17.380
the force is applied perpendicular to the
direction of velocity of the particle. Therefore,
24:17.380 --> 24:24.380
many times its termed as, a transverse mass.
Now, we must we wondering, whether we have
24:25.380 --> 24:29.240
a situation if you remember electronic magnetic
theory, we know that this situation is called
24:29.240 --> 24:35.790
quite common. If there is a charge particle
in the magnetic field then the force on that
24:35.790 --> 24:39.760
particular particle is always given by u cross
B e u cross B.
24:39.760 --> 24:46.760
Let us say q u cross B is a force on the charge
q, if it is moving with a velocity u and because
24:53.080 --> 24:58.540
this is the cross product u cross B, it means
this force will always be perpendicular to
24:58.540 --> 25:02.410
u. So, this magnetic force will always be
perpendicular to the direction of u.
25:02.410 --> 25:08.040
So, let us just take a simple example of this
particular case, when a charge particle is
25:08.040 --> 25:12.900
under the influence of the magnetic field,
just give as an example of the case in which
25:12.900 --> 25:16.610
the force is actually in the same direction
as acceleration.
25:16.610 --> 25:23.610
So, this is the example which have taken,
an electron is moving with a speed of 0.8
25:25.890 --> 25:31.980
c. In a circular motion under the influence
of a magnetic field of 1.5 tesla, we have
25:31.980 --> 25:38.980
to find the radius of the orbit. This case
as I said, simple because force and the acceleration
25:40.790 --> 25:42.530
under the same direction.
25:42.530 --> 25:49.530
So, let us first find the force. Force will
be given by q u cross B as we have just now
25:49.900 --> 25:55.400
said, because this is an electron. So, q is
equal to the just charge of an electron which
25:55.400 --> 26:02.400
is e and because u and B are perpendicular
each other and this is cross product. So,
26:04.820 --> 26:09.500
u cross B will just give you u multiplied
by B. Because of u magnitude of u multiplied
26:09.500 --> 26:13.800
by magnitude of B, multiplied by sign of the
angle between the two, sign of the angle being
26:13.800 --> 26:19.810
90 degrees, this give you one.
Therefore, the magnitude of the force will
26:19.810 --> 26:24.570
be euB I am putting the value of e after taking
the approximate value 1.6 into 10 power minus
26:24.570 --> 26:30.750
19 coulomb. u has been given, it was as 0.8
c. Multiplied by the magnetic field which
26:30.750 --> 26:36.640
has been given as 1.5 tesla. This gives me
that the net force on the charge will be 5.76
26:36.640 --> 26:43.640
into 10 to the power minus 11 newton. Of course,
this will always will directed towards the
26:44.680 --> 26:48.100
in a direction perpendicular to the direction
of the velocity, which actually will provide
26:48.100 --> 26:51.540
the centripetal force for this particular
electron to move in a circle.
26:51.540 --> 26:55.920
I mean in a case, a particular particle is
moving in the circle, there has to be a centripetal
26:55.920 --> 27:01.170
force and if I assume that the speed of the
particle is constant then there has to be
27:01.170 --> 27:04.390
constant centripetal force acting towards
the centre, always towards the centre of the
27:04.390 --> 27:11.390
circle. It is a well known classical results.
On this particular case, I can find out the
27:11.820 --> 27:16.780
acceleration because my expression of force
into force relation between force and acceleration
27:16.780 --> 27:22.480
is comparatively simple, and acceleration
will be just given by force divided by the
27:22.480 --> 27:29.190
transverse mass, which is gamma u m naught.
We know for the case and we have done many
27:29.190 --> 27:34.960
problems earlier, there for the case when
u is equal to 0.8 c, then gamma u turns out
27:34.960 --> 27:35.720
to be 5 by 3.
27:35.720 --> 27:41.000
So, this 5 by 3 have used here. This is the
value of the force which we have just now
27:41.000 --> 27:46.350
found out in previous transparency is 5.76
into 10 power minus 11 Newton. So, this is
27:46.350 --> 27:52.900
what I have written here, divided by m naught,
rest mass of the particle. I am taking approximately
27:52.900 --> 27:59.780
the mass of the electron as 9.1 into 10 power
minus 31. Of course as I say this is the gamma
27:59.780 --> 28:04.450
u, 5 by 3 which is here.
If I take this, this acceleration turns out
28:04.450 --> 28:11.450
to be 3.80 into 10 power 19 in cgi in si units.
Now we know from classical mechanics that
28:14.390 --> 28:19.010
if a particle has to move in a circle, there
has to be centripetal, there has to be real
28:19.010 --> 28:24.960
force because this acceleration amount of
this particular particle is equal to u square
28:24.960 --> 28:29.280
by R assuming of course u to be constant,
and the magnitude of u to be constant, direction
28:29.280 --> 28:34.850
is not constant is always changing. So, velocity
is always changing, but this speed is constant.
28:34.850 --> 28:39.420
So, this acceleration, in this case particular
particle this particular electron has to move
28:39.420 --> 28:42.850
in a circle. The total acceleration of the
particle must be equal to u square by R, where
28:42.850 --> 28:48.200
R is the radius of the circle. I have just
now found out, what is the acceleration of
28:48.200 --> 28:54.500
the particle. So, I can substitute it to equal
to u square by R. I know u, I can always find
28:54.500 --> 29:01.500
out what will be the radius if electron has
to be move in a circle.
29:01.520 --> 29:08.520
The radius turns out to be u square which
is 0.8 c square divided by this 3.8 into 10
29:09.100 --> 29:15.960
power 19, which we have obtain earlier as
an acceleration. So, this must be the radius
29:15.960 --> 29:21.020
as we have seen this comparatively simple
case, where we found out that acceleration
29:21.020 --> 29:26.780
is in the same direction as the force and
only difference which we have to make to taken
29:26.780 --> 29:33.780
into account in relativity is, instead of
using m naught have use gamma u. Now let us
29:33.860 --> 29:40.860
consider the second case that is the case
in which there is a straight line motion,
29:41.790 --> 29:47.740
this also very interesting case. So, let us
discuss this particular case little bit more
29:47.740 --> 29:49.440
in detail.
29:49.440 --> 29:56.220
This is my case two, a one-dimensional motion
in which force is applied in the direction
29:56.220 --> 30:00.400
of the velocity of the particle.
30:00.400 --> 30:04.600
What happens in this particular case, this
was my expression of course to rearrange the
30:04.600 --> 30:10.090
first term which was there, I have put it
second term just to make it looks very simple.
30:10.090 --> 30:17.090
So, I have put F is equal to m a plus u into
F dot u divided by c square, this was my original
30:17.410 --> 30:24.410
relationship between the force and acceleration.
everything is in the same direction. So, I
30:24.440 --> 30:30.240
need not write vectors F dot u, because F
is the same direction as u and F dot u will
30:30.240 --> 30:34.150
be F, Multiplied by magnitude of F multiplied
by magnitude of u multiplied by cosine of
30:34.150 --> 30:38.950
the angle between the two, and because angle
between them, between the two 0 between the
30:38.950 --> 30:41.600
two vectors is 0. Therefore, that is equal
to 1.
30:41.600 --> 30:46.590
So, these becomes magnitude of F multiplied
by magnitude of u. Everything is the same
30:46.590 --> 30:52.000
direction so this I can just write F multiplied
by u square of course divided by c square
30:52.000 --> 30:56.180
and this a can also write without vector sign
because I realize that everything is in same
30:56.180 --> 31:01.400
direction. There is only one particular motion,
the only one particular time direction.
31:01.400 --> 31:08.400
So, I can write this equation in a scalar
form by writing F is equal to m a plus F multiplied
31:09.310 --> 31:14.400
by u square by c square. Now we can find out
the relationship between the force and the
31:14.400 --> 31:20.260
acceleration because I can take this force
term on the left hand side and find out in
31:20.260 --> 31:24.680
this particular case, what will be the relationship
between the force and the acceleration. In
31:24.680 --> 31:31.500
case, there is a motion just in a single line
in case of one dimensional motion, it is very
31:31.500 --> 31:35.650
simple. Let us just put in the second transparency,
that is what I have put.
31:35.650 --> 31:42.650
So, this is the expression which had written
earlier F is equal to m a plus F divided by
31:43.610 --> 31:49.730
F multiplied by u square by c square. This
term is taken on the left hand side, I take
31:49.730 --> 31:56.730
F common. I get F multiplied by 1 minus u
square by c square, must be equal to m a.
31:59.780 --> 32:04.680
This particular quantity here I can divided
by here remember this m is also gamma u m
32:04.680 --> 32:10.330
naught.
So, therefore, this I can write as F is equal
32:10.330 --> 32:15.210
to m naught and this is a gamma u which was
one upon under root 1 minus u square by c
32:15.210 --> 32:19.470
square and there is another factor of 1 minus
u square by c square which is coming from
32:19.470 --> 32:25.470
this here. So, when this is divided remember
here this m naught here for m. So, m naught
32:25.470 --> 32:30.730
and the additional gamma u have incorporated
here and this becomes now 1 minus u square
32:30.730 --> 32:37.730
by c square to the power 3 by 2.
Very interesting result, because it says that
32:38.380 --> 32:44.400
in this particular case force is related to
acceleration. Force is in the same direction
32:44.400 --> 32:51.400
as acceleration, but the mass of the particle,
the way mass will behave will be m naught
32:51.430 --> 32:56.780
multiplied divided by 1 minus u square by
c square to the power 3 by 2. Its gamma q
32:56.780 --> 33:01.610
gamma u q while in the case of transverse
motion, I am in the force was perpendicular
33:01.610 --> 33:07.110
to u. It was just gamma u, but in the case
of one dimensional motion its gamma u q. So,
33:07.110 --> 33:12.420
this is generally called as a longitudinal
mass.
33:12.420 --> 33:19.420
So, we often define longitudinal relativistic
mass of the particle as m naught divided by
33:19.740 --> 33:25.630
1 minus u square by c square to the power
3 by 2, which is equal to gamma u cube m naught.
33:25.630 --> 33:30.640
So, with the motion the particle is an one
direction the mass turn out to be gamma u
33:30.640 --> 33:35.330
cube times m naught. As far as relationship
between the force and acceleration is concerned
33:35.330 --> 33:39.280
it is been very, very clear, when the relationship
between the force and acceleration is concerned
33:39.280 --> 33:44.350
this is way you should write immediately.
33:44.350 --> 33:51.350
Now, let us look little bit more carefully
we have lot of what we call as kinematic equations
33:52.790 --> 33:56.830
in the classical mechanics, we are if you
know the initial velocity you can find out
33:56.830 --> 34:01.640
the velocity final velocity if you know the
acceleration at a given time, you can find
34:01.640 --> 34:06.290
out the distance etcetera, etcetera. So, let
us look at some of these equations and try
34:06.290 --> 34:11.579
to get little bit more insight of, what is
happening when a force is applied in a one
34:11.579 --> 34:17.270
dimensional case on a particular particle.
So, let us assume that a constant force is
34:17.270 --> 34:23.139
applied to a particle which is moving initially
with a speed of u naught, in the same direction
34:23.139 --> 34:28.000
as that of velocity. With this exactly the
case which we have just now said, that the
34:28.000 --> 34:34.779
particle is moving in this particular direction
with the velocity u naught and a time t equal
34:34.779 --> 34:38.659
to 0, I apply a force which is also in the
same direction F.
34:38.659 --> 34:43.349
So, by motion what were all the change in
the motion the same direction. So, everything
34:43.349 --> 34:49.119
within the same direction, so this is what
is the case? My question is that I applied
34:49.119 --> 34:54.539
the force for a given time what happens to
its velocity.
34:54.539 --> 34:58.980
So, let us assume that after time t, we applied
we leads to let us assume the force is constant.
34:58.980 --> 35:05.980
A constant force is being applied. I am interested
in finding out what happens after time t.
35:06.039 --> 35:09.680
So, let us assume of course the velocity of
the particle will change, direction will not
35:09.680 --> 35:16.680
change because force and the velocity are
in this same direction, but its magnitude
35:16.950 --> 35:23.950
will change. So, let us try to find out or
let you let after time t the speed of the
35:24.599 --> 35:28.170
particle or velocity, whatever you have to
call it, because it any one dimensional with
35:28.170 --> 35:35.170
the speed u t be the speed after time t. How
do I find u t? I have to integrate this equation.
35:38.769 --> 35:45.769
This was my force equation, acceleration have
written as du dt rate of change of speed or
35:47.339 --> 35:51.700
rate of change of velocity as acceleration.
This is just have written as gamma u cube
35:51.700 --> 35:58.700
m naught. This dt I will take on this side,
du I let it remain in this side and integrate.
36:00.660 --> 36:03.259
This is the way we derive those kinematic
equations, standard kinematic equation in
36:03.259 --> 36:08.499
classical mechanics. We just assume force
should be constant, live force equal to mass
36:08.499 --> 36:12.180
into acceleration, then integrate and then
obtain various equation what happens that
36:12.180 --> 36:16.190
the speed of the particle at a given time
t, what happens to this displacement of the
36:16.190 --> 36:20.220
particle at a given time t, we are doing exactly
in the same thing except that our expression
36:20.220 --> 36:25.279
have not become little more difficult.
So, this is what I have to integrate in order
36:25.279 --> 36:32.279
to find out, what will be u t at a given time
t. Therefore, I am just integrating it integral
36:34.749 --> 36:39.480
from 0 to t and this is this is the initial
velocity u naught, this is the final velocity
36:39.480 --> 36:46.480
after time t, u t m naught du 1 upon u square
1 minus u square by c square to power 3 by
36:46.789 --> 36:53.069
2, just integrating this expression is just
written here. Now in order to integrate let
36:53.069 --> 36:56.519
me not use this definite integral, let me
use still indefinite integral. It is comparatively
36:56.519 --> 37:01.660
easy. Then I will put the limits later. So,
this is what I am doing next, I am taking
37:01.660 --> 37:07.799
the indefinite integral of this equation of
the right hand side. Let me solve the integral
37:07.799 --> 37:11.690
in the indefinite form then I will put the
limits later.
37:11.690 --> 37:15.720
So, let us solve the integral without putting
the limits. So, I will just put integral F
37:15.720 --> 37:20.549
dt of integral m naught du 1 minus u square
by c square to power 3 by 2. There is the
37:20.549 --> 37:27.410
standard substitution to solve this equation,
we put u is equal to c sine theta then we
37:27.410 --> 37:34.410
take the differentiation we get du is equal
to c cos theta d theta. This du is now being
37:35.599 --> 37:42.599
replaced by c cos theta d theta, it is very
simple integral. For this u square, I am putting
37:44.640 --> 37:51.640
c sine theta. Let me just write it here.
37:52.549 --> 37:59.549
In the denominator we had one, we had m naught
du and here we had 1 minus u square by c square
38:03.700 --> 38:10.700
to the power 3 by 2. If I put u is equal to
c sine theta so this becomes, I am just looking
38:11.749 --> 38:18.749
at the denominator. 1 minus c square sine
square theta by c square to the power 3 by
38:26.200 --> 38:33.200
2, this c square will cancel. So, I will get
1 minus sin square theta to the power 3 by
38:38.140 --> 38:40.519
2. 1 minus sin square theta is cos square
theta.
38:40.519 --> 38:47.519
So, this becomes cos square theta to the power
3 by 2. This under root and this half goes
38:49.369 --> 38:55.440
away, this becomes cos cube theta. So, this
is what I have written here. Integral of F
38:55.440 --> 39:00.670
dt is equal to m naught for du as I have said,
we have written c cos theta d theta divided
39:00.670 --> 39:06.630
by cos cube theta, this c remains here, this
cos cube theta will cancel with one of these
39:06.630 --> 39:12.150
powers. So, this will come cos square theta
and eventually you will get 1 upon cos square
39:12.150 --> 39:16.210
theta which is sec square theta d theta.
39:16.210 --> 39:23.210
So, this equation becomes integral of F dt
is equal to integral of m naught c sec square
39:24.059 --> 39:30.609
theta d theta. We know that, sec square theta
when integrated it becomes tan theta. So,
39:30.609 --> 39:37.609
this equation just becomes m naught c tan
theta. This tan theta, I know sin theta divided
39:38.269 --> 39:43.059
by cos theta, I would like to express this
thing back in terms of u. So, what I will
39:43.059 --> 39:50.059
do I have written anyway c sin theta as u.
So, this I can put as c sin theta as u, this
39:51.759 --> 39:56.880
cos theta I will express as under root 1 minus
sine square theta and for sin theta again
39:56.880 --> 39:58.740
I will put u by c.
39:58.740 --> 40:05.740
This is what I have done in next transparency.
m naught c sin theta by cos theta, sin theta
40:07.089 --> 40:14.089
as we have said is this c sin theta is u,
which have written as u, this cos theta have
40:14.890 --> 40:20.849
written as under root 1 minus sin square theta,
because c sin theta is equal to u. Therefore,
40:20.849 --> 40:27.529
sin theta will be u by c. So, for sin square
theta itl becomes u square by c square.
40:27.529 --> 40:34.529
So, this integral F dot dt gives me m naught
u divided by under root 1 minus u square by
40:38.220 --> 40:45.220
c square. So, this is what I get as the result
of integration. Now we can put the limit,
40:48.489 --> 40:54.849
say what limits were from u naught to u t.
So, I take this same expression, I first put
40:54.849 --> 41:01.849
u t here I put u t here minus there I put
the limit u 0, which is u 0 here, u 0 here.
41:02.960 --> 41:05.630
This is what I am putting here.
41:05.630 --> 41:12.630
So, F multiplied by t is m naught u t divided
by under root 1 minus u t square by c square
41:12.960 --> 41:19.960
minus m naught u naught divided by 1 minus
u naught square by c square. So, this is what
41:20.650 --> 41:26.329
I will find that, for in this equation is
not in a simple form to express u t that of
41:26.329 --> 41:33.329
course that we can move later. But this definitely
gives, that if time if force is applied for
41:34.160 --> 41:39.539
time t this gives the relationship between
u t and u naught.
41:39.539 --> 41:45.200
Before, I try to derive this particular relationship
which probably will do in the later lecture.
41:45.200 --> 41:50.059
But let us try to look at this particular
equation and try to see the implication of
41:50.059 --> 41:57.059
this particular equation. So, for that we
have taken an example, consider a one dimensional
41:57.910 --> 42:01.910
motion which we have been considering in this
particular case.
42:01.910 --> 42:08.480
Let us assume that there is a particle of
rest mass m naught, which is subjected to
42:08.480 --> 42:15.480
a constant force F. First find the time its
speed takes to change from 0 to 0.2 c. I have
42:17.519 --> 42:23.359
put this particular problem in a fashion,
so that I can use this equation directly without
42:23.359 --> 42:30.359
actually working out for u t. So, initial
velocity has been given, which is 0. Particle
42:32.160 --> 42:38.660
is starts from rest and final velocity has
been given as 0.2 c and there is the particular
42:38.660 --> 42:44.609
force F, whatever might be the magnitude of
the force and its rest mass is m naught.
42:44.609 --> 42:48.930
I am interested in finding out how much time
it will take, which will depend on the force
42:48.930 --> 42:53.799
and m naught. But I am giving the I am interested
in finding out the answer, intervals of F
42:53.799 --> 43:00.230
and m naught because I want to really look
how the time value will vary, when the speed
43:00.230 --> 43:05.809
really approaching the speed of light.
43:05.809 --> 43:12.809
Now, we also like to make a similar calculation
if the speed changes from 0.2 c to 0.4 c and
43:13.230 --> 43:19.529
then 0.4 c to 0.6 c and then 0.6 c to 0.8
c means, interval speed interval is same,
43:19.529 --> 43:22.480
it is 0.2 c. So, you will go for 0 to 0.2
c.
43:22.480 --> 43:29.480
So, if we take plot speed, go from 0 to 0.2
c then 0.2 c to 0.4 c, 0.4 c to 0.6 c and
43:37.900 --> 43:44.900
then 0.6 c to 0.8 c. So, velocity intervals
are same. Force is same. I am interested in
43:48.779 --> 43:53.249
finding out the time, that particle will take
to go from here to here, go from here to here,
43:53.249 --> 43:58.759
go from here to here, go from here to here,
next what is the question.
43:58.759 --> 44:05.759
This is my question, force multiplied by time
is equal to m naught u t divided by 1 under
44:08.019 --> 44:12.940
root 1 minus u t square by c square minus
m naught u naught divided by under root 1
44:12.940 --> 44:19.940
minus u naught square by c square. This particular
quantity m naught divided by 1 minus one divided
44:21.249 --> 44:28.249
by under root 1 minus u t square by c square.
I am writing as gamma u t one at time t, one
44:28.920 --> 44:34.109
at time t is equal to 0.
So, gamma corresponding to this speed is gamma
44:34.109 --> 44:40.039
u t. gamma corresponding to the initial speed
is gamma u not. So, the same expression I
44:40.039 --> 44:46.450
have written as m naught multiplied by gamma
u t, u t minus gamma u naught. m naught has
44:46.450 --> 44:52.680
been taken common, u t is here. This particular
under root one upon this particular thing
44:52.680 --> 44:57.730
has been gamma u t, this m naught has been
taken out here, this u naught is present here.
44:57.730 --> 45:03.160
This 1 divided by 1 minus u naught square
by c square has been taken as gamma u naught.
45:03.160 --> 45:08.190
So, this is my expression.
So, for the first case I am starting with
45:08.190 --> 45:13.559
0 initial velocity. So, this term is 0. All
I have to do, should take gamma u t multiplied
45:13.559 --> 45:19.849
by u t. u t is 0.2 c for the first case, what
I have to find out? what is the gamma corresponding
45:19.849 --> 45:26.849
to this speed of 0.2 c? it means I will take
this as 1 upon under root 1 minus 0.2 square.
45:27.400 --> 45:33.559
I will substitute in this particular expression
and obtain what will be the time taken of
45:33.559 --> 45:38.049
course, this time will be this own quantity
divided by force. But as I said I want the
45:38.049 --> 45:42.170
answer in terms of both m naught and the force.
45:42.170 --> 45:49.170
So, as we said for the first case u t is equal
to 0.2 c 0 u naught is equal to 0. So, time
45:49.749 --> 45:56.569
which have set t 1 is equal to m naught c
divided by F, multiplied by 1.1 which is gamma
45:56.569 --> 46:03.019
gamma u t which is corresponding to 0.2, happens
to 1.021. It makes only 2 percentage change
46:03.019 --> 46:10.019
and even we are at the speed of 0.2 c multiplied
by 0.2 of c have been taken out here and the
46:10.430 --> 46:16.289
second term is 0. So, this gives me 0.204
m naught c bar F.
46:16.289 --> 46:23.289
So, the time taken for the particle to reach
its speed of 0 2 c will be given by 0.204
46:25.700 --> 46:31.150
m naught c upon a. This factor will be common
in all other my subsequence results. I am
46:31.150 --> 46:35.549
only will be comparing this particular factor.
46:35.549 --> 46:42.549
Now, for the other case let us take u t is
equal to 0.4 c and u naught is equal to 0.2
46:47.180 --> 46:53.680
c. I put exactly in the same expression, force
now I have to take this factor as well as
46:53.680 --> 46:58.559
this factor into consideration, we just substitute
those numbers, it is very, very simple.
46:58.559 --> 47:05.089
State forward calculation, if I take that
number we get 0.232 m naught c by F. If my
47:05.089 --> 47:12.089
initial speed was 0.4 c and the final speed
for 0.6 c, then my this I have written as
47:12.940 --> 47:19.940
t 2, this is t 3 for third case, is 0.314
m naught c divided by F. If my initial speed
47:20.940 --> 47:27.940
was 0.6 c final speed is 0.8 c, after time
t 4 which is 0.583 m naught c by F.
47:28.739 --> 47:35.739
So, as we can see that with whatever the factor,
if we go between 0 to 0.2, the value which
47:37.079 --> 47:44.079
I get is 0.204. In this interval I get 0.232,
in this interval I get 0.314, in this interval
47:53.269 --> 48:00.269
I get 0.583. So, there is the speed interval
is same, this force is same. But this first
48:02.029 --> 48:07.430
half of the particle who travel the same speed
interval 0 to reach between the 0 to from
48:07.430 --> 48:14.430
0 to 0.2 c in 0.204 multiplied by whatever
is this factor. Between 0.2 c and 0.4 c it
48:20.019 --> 48:27.019
takes larger time between 0.4 c and 0.6 c
still takes larger time and between the 0.6
48:27.359 --> 48:31.259
c and 0.8 c it still takes a larger time.
So, yes the particle is approaching this speed
48:31.259 --> 48:38.039
of light. We are seeing that cover the same
speed that a to reach, to make the same changes
48:38.039 --> 48:43.180
in the speed to make the same changes in the
speed, you require larger and larger time.
48:43.180 --> 48:48.599
That is expected because now you are approaching
the speed of light and as we know that we
48:48.599 --> 48:54.900
cannot no really reach the speed of light.
So, as we are approaching here and here with
48:54.900 --> 49:00.589
the same force you will find that particle
is taking larger and larger time to make same
49:00.589 --> 49:07.349
change in the velocities, this is what the
interesting result that we see from this.
49:07.349 --> 49:14.349
So, this what we have said, as one can see
that for the same speed interval and a constant
49:14.400 --> 49:19.999
force, the time interval is different unlike
classical mechanics. In classical mechanics,
49:19.999 --> 49:26.130
if you remember, this was standard expression
in kinematic equation.
49:26.130 --> 49:33.130
We write v is equal to u plus a t, where v
is should be the final speed. This I would
49:33.579 --> 49:40.579
call as, u t minus u 0 plus a t using our
normally literature. So, u t minus u naught
49:41.519 --> 49:48.519
divided by a will be time and this a will
always be constant in the classical mechanics,
49:48.579 --> 49:54.239
if the force is constant because mass remains
constant in that case. Therefore, the same
49:54.239 --> 49:59.190
interval will be covered, same velocity interval
will be covered in the same time. While here
49:59.190 --> 50:05.099
is does not happen. That is what I have said,
that the time interval is different unlike
50:05.099 --> 50:10.869
classical mechanics. Further this interval
increases as the speed of the particle reaches
50:10.869 --> 50:13.509
close to speed of light.
50:13.509 --> 50:18.009
Now, I like to summarize what ever we have
discussed. We discussed the definition of
50:18.009 --> 50:23.609
relativistic force. We have said that this
force in general does not produce acceleration
50:23.609 --> 50:28.489
in the same direction. We discussed special
cases of longitudinal and transverse motion,
50:28.489 --> 50:33.549
I am calling longitudinal transverse motion
not sure with this as good way of saying,
50:33.549 --> 50:37.609
what I mean I think should be clear when the
force is applied in the direction perpendicular
50:37.609 --> 50:42.970
to you and when there is a one dimensional
motion we consider these two special cases
50:42.970 --> 50:46.309
in which the force happen to be in the same
direction as acceleration.
50:46.309 --> 50:47.329
Thank you.