WEBVTT
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Hello, in our last lecture, we had described
time like and space like events, we had discussed
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that, what are the basis on which it was postulated
that space greater than speed of light, would
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not be allowed in special theory of relativity.
So, let us quickly recapitulate, we had given
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an example of pool events and we had said
that if delta x is the coordinate difference
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between these two events, and delta t is the
time difference between these two events.
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Assuming delta x and delta t to be positive
then if delta x turns out to be smaller than
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c delta t, we call these events as time like
separated events. Just to remember, when we
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say, time like, it is time factor delta t,
which is large. Similarly, if delta x turns
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out to be larger than c delta t then we call
these events as space like separated events,
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when we say space like, delta x is suppose
to be larger, just to remember.
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And we had mentioned that, in case we want
the causality not to be violated it means,
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if in a frame of reference two events occur,
the second event is an out come of the first
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event. First event is the cause of the second
event, then you would like that in any other
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frame of reference, we should always have
the same order of the events. We cannot see
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in any other frame that whatever has whatever
is the result or whatever is the out come
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has come first and the reason for that is
coming later.
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They have given an example that suppose, I
shoot somebody then in my frame of reference,
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shooting of the gun, if it is event number
1 and that person being hit by the bullet
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is event number 2. Obviously, event number
2 has to occur after event number 1 then we
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should not be able to find out a frame of
reference, in which it so happens that, that
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persons observes that, the person is hit by
the bullet first and the bullet is shot later.
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So, the same order of events must be maintained,
if first event is the cause of the second
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event so this is what, we called causality.
So, we do not expect from the physics point
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of view that, causality should be violated
and that is possible only, if we do not allow
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space greater than speed of light, so that
is what, is the physics reason for this particular
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conclusion.
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Just to mention again, I am writing here,
for time like separated events, time interval
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is pure and time order cannot be reversed
of course, if it means that, time order cannot
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be reversed it means that, we cannot find
out a frame of reference, in which delta t
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0. It means, we cannot find a frame of reference,
in which the two events occur at the same
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time.
Similarly, for space like separated events,
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space interval is pure and position order
cannot be reversed by again the same logic,
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if it has to be reversed it means that, we
cannot also not find out the frame of reference,
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in which these events occur at the same place.
So, that is what, I have written, for space
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like separated events, space interval is pure
and position order cannot be reversed, one
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cannot thus find a frame, in which the two
events occur at the same place.
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So, this is what, we had discussed essentially
in our last lecture, the most general behavior
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of these space like and time like, they will
be discussing little later. So, let us postpone
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this particular discussion, until we evolve
certain other concepts, at the moment what
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I would like to give, is a few examples of
space like and time like events and the cause
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and outcome related events, just to make our
ideas clear. So, let us go to one or two very,
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very simple examples.
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Let me take the example number 1, let us assume
that an event occurs in a frame of reference
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S, at origin at time zero and event it could
be for example, lightning, striking or anything,
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whatever we can imagine of. So, one event
occurs in a frame S at origin, origin means
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x is equal to 0, y is equal to 0, z is equal
to 0 and I also assume that, this occurred
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at time t is equal to 0.
Again the same frame S, a second event is
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formed to be observed, which is at a point
A and the time difference, it means it occurs
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at 0.01 micro second later, micro is 10 to
power minus 6 second. So, it is 1 into 10
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to power minus 8 seconds later, another event
is found to be observed at point A, not at
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the origin and the x coordinate of A has been
given as 5 meters, while y and z coordinates
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are same.
So, we have two events, one occurring at x
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is equal to 0, another occurring at x is equal
to 5, the first event occurring at time t
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is equal to 0 and the second event occurring
at time t is equal to 0.01 micro second.
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The question is, will it be possible to find
a frame, in which these two events would occur
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at the same place slash same time, find the
speed of such a frames. So, first we have
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to resolve, whether it should be possible
to find a frame of reference, in which these
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two events occur at the same place or at same
time or whatever it is. And once, we find
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out then we have to find out the speed of
that particular frame of reference, in which
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these events either occur at the same place
or at same time, depending upon the situation.
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The problem is essentially simple, all we
have to do is, to take delta x
and compare it with c delta t. If delta x
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happens to be larger than c delta t, then
we know that, it is a space like event, if
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it happens to be smaller than c delta t then
we know that, it is a time like separated
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events. Then accordingly, we can conclude,
whether it is possible to find a frame of
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reference, in which these two events occur
at the same position or at same time so we
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have given everything so let us look at this
particular thing.
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In S, the x and t co-ordinates of the events,
I have just listed, I have removed y and z
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reference of y and z coordinates because that
is not really material as far as this particular
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problem is concerned. Because, in both the
events, y and z’s are 0 so with a event
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number 1, which occurs at x is equal to 0
and t is equal to 0, S has been given in the
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question where, event number 2, which occurs
at point A, which is at a distance of 5 meters
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along the x direction from the origin.
So, x co-ordinate is 5 meters and as we have
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mentioned that, it occurs at a time 0.01 micro
second it means, the time was 1 into 10 to
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power minus 8 second. So, we have written
by event table, my event number 1 is occurring
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at 0 0 and event number 2 is occurring at
5 meters and 1 into 10 to power minus 8 second.
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Now, all I have to do is, to find out this
delta x then we have to find out this delta
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t then multiply by c and compare the two,
that is all I have to do. Anyway, it is very
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easy to see that, for the first event x was
equal to 0, for the second event x was 5 meters
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so; obviously, delta x is 5 meters, x 2 minus
x 1 is 5 meters. Similarly, t 2 minus t 1,
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t 1 occurred at time t is equal to 0 I mean,
event 1 occurred at time t is equal to 0,
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event number 2 occurred at time 0.01 micro
second so t 2 minus t 1 there will be 1 into
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10 to power minus 8 second.
If I am multiply this by c, to compare with
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this and if I take c is equal to 3 into 10
to power 8, will be 3, 10 to power 8 will
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cancel out 10 power minus 8 so this will become
3 meters. So, as I see very clearly that,
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delta x is 5 meters while c delta t is 3 meters
that is what, I have written in this particular
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transparency.
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Delta x is equal to 5 meters, delta t is equal
to 1 into 10 to power minus 8 second, c delta
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t is therefore 3 meters therefore, delta x
is greater than c delta t it means, these
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events are separated space like. Because,
delta x is turning out to be larger than c
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delta t and if the events are separated space
like, it is not possible reverse the position
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order of the events it means, it is not possible
to find a frame of reference, in which these
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two events will occur at the same position.
Though it is possible to find out the frame
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of reference, in which these two events will
occur at the same time that is, you see and
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verify whatever we are telling.
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This what I have mentioned just now, the events
are space like separated hence, sign of space
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interval cannot be reversed but that of time
can reversed. How do we find this particular
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thing, just use Lorentz transformation.
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This what I have mention again hence, it is
possible to find a frame, in which the two
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events occur at the same time but we cannot
find one, in which the events occur at the
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same place.
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Now, we go to Lorentz transformation assume
that, there is a frame reference, frame of
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reference S prime, which is moving a relative
to S with a velocity v and I know that, it
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is possible to have delta t prime is equal
to 0. All I have to is, to find the speed
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of that particular frame of reference, for
which delta t prime will turn out to be 0.
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So, I write delta t prime is equal to gamma
delta t minus v delta x by c square, this
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is standard Lorentz transformation. Now, what
I have to do, I know my delta t, I know my
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delta x, how to find out a v, for which delta
t prime is equal to 0. I substitute delta
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t prime is equal to 0, substitute therefore,
delta t 1 into 10 to power minus 8 second,
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I substitute for delta x 5 meters, solve for
v, this is what, I have done in this particular
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transparency.
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I have written delta t prime is equal to gamma
delta t minus v delta x by c square, for delta
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t I have substituted 1 into 10 to power minus
8 second, for delta x I have substituted 5,
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for c square I have substituted 9 into 10
to power 16 assuming of course, that c is
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equal to 3 into 10 to power 8 meters per second.
Just solve this equation, we get v is equal
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to 1.8 into 10 to power 8 meters per second
obviously, this is speed, this is smaller
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than speed of light.
It means, it is physically possible to find
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out the frame of reference, in which these
two events would occur at the same time it
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means, the time difference between these two
events will turn out to be 0 in this frame
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of reference. And that frame of reference
would be moving relative to S, with a speed
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of 1.8 into 10 to power 8 meters per second.
Obviously, it is clearer that, if v would
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have been larger than this particular value,
if you look at this particular equation, if
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v would have been larger than this particular
value, delta t prime would turn out to be
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negative.
And time order of the events would get reversed,
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which is possible in this particular phase
because the events are space like separated
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and not time like separated. Now, some of
you can ask question, why cannot I do the
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same exercise and try to see, whether I can
really make delta x prime equal to 0 and if
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whatever we have said is correct, I should
be able to get a speed greater than speed
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of light.
Then only, it will be possible for me to make
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delta x prime is equal to 0. So, let us just
look into that to satisfy ourselves that,
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whatever we are saying is correct.
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So, I do exactly the same thing instead of,
delta t prime making equal to 0, I try to
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make delta x prime equal to 0. So, let us
try to find v, a speed of frame of reference,
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in which I assume if at all possible that,
these two events occur at the same position.
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So, now, instead of writing delta, the transformation
Lorentz transformation corresponding to time,
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I will write a Lorentz transformation corresponding
to space.
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So, that is what it gives here, delta x prime
is equal to gamma delta x minus v delta t,
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which is the Lorentz transformation corresponding
to the x coordinate. I substitute my numbers,
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delta x y is equal to 5 meters, I substituted
here 5 meters, delta t was equal to 1 into
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10 to power minus 8 second, I substituted
the value of 1 into 10 to power minus 8 second
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here multiplied by v.
I solve for this particular thing, we can
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see that gamma will turn out to be is equal
to 0, this one when divided by 5 make will
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will make this 5 and this 10 to power minus
8 then it goes to numerator, it will become
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10 to power 8.
And v will turn out to be 5 into 10 to power
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8 meters per second, which is obviously greater
than speed of light. So, it means, if at all
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it was possible for us to have a frame of
reference with v greater than 5 c, which in
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this particular case is 5 into 10 to power
8 meters per second then it was possible for
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us to find these two events at the same position.
But, as we have already seen that, on the
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ground of physics and also some extent on
the mathematical ground, we have rejected
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speeds greater than speed of light. Therefore,
it I would generally conclude that, it is
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not possible to find out a frame of reference,
in which these two events occur at the same
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position.
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This what, I have written here hence, we see
that it is not possible to find a frame with
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v less than c, in which the events occur at
the same place. Now, we have just now said
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that, it is possible in the particular case
to revert the time order, to reverse the time
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order, it is possible to find out a frame
of reference with v less than c, in which
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the time order delta t turns out to be negative.
It means, events number 2 occurs before event
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number 1, we have said also that, this is
not possible if these two events are cause
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and outcome related. It means event number
1 and event number 2 have to be totally independent
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events, event number 2 cannot be a cause of
event number 1, if this particular condition
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has to happen, let us just look and try to
clarify our ideas little further.
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That is what, I have written here, can the
two events described in the event be cause
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and effect time for example, let us imagine
a particular case.
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That, this was my origin O and this was my
point A, and this distance is 5 meters, everything
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being described in S frame. Now, let us assume
that, event number 1 was one car, which is
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moving to the right was been found at origin,
we have solved many such problems. So, let
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us assume that, events number 1 was that,
this particular car, was found at x is equal
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to 0 and this car is moving with a constant
velocity v along the plus x direction.
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And let us assume that, event number 2 is
that this same car being found here at x,
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is equal to A. Now; obviously, these two events
are cause and outcome related, if the car
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was not starting from this particular point,
car would have never this particular point
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A. Therefore, for event number 2 to occur,
event number 1 was essential, the car must
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have passed through this particular origin
and must have come to point A.
18:46.450 --> 18:53.450
So, these two events are related, event number
1 is sort of a cause for event number 2 because
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car must pass through the origin then only
it has to reach at A. Now, if this happens,
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can these two events lead to situation, which
is given in this particular problem, involving
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delta x is equal to 5 meters and delta t involved
in equal to 1 into 10 to power minus 8 second.
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Is it possible that, these two events are
separated of course, delta x is given as 5
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meters and the time interval between these
two events was 1 into 10 to power minus 8
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second. That is what, I have written here
in this transparency, can the two events described
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in the example be cause effect time. For example,
can the event 1 be a car passing by origin
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and event 2 be the same car passing at A.
If this happens it means, and if the time
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separation between these two events is 1 into
10 to power minus 8 second it means, the car
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must have taken 1 into 10 to power minus 8
second in of course, S frame, we are talking
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of only one particular frame, which is S frame.
So, it means, the car must have taken 1 into
20:10.360 --> 20:17.360
10 to power minus 8 second to reach from O
to A and the distance between O and A is 5
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meters.
Therefore, the speed of the car must be 5
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meters divided by the time, which is 1 into
10 to power minus 8 second, which will give
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you 5 into 10 to power 8 meter per second,
which is obviously, larger than speed of light.
20:44.600 --> 20:51.600
Therefore, if these two events for example,
were related like this then this car must
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travel with speed greater than speed of light
and if this is not happening then it is not
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possible in a situation like this, for this
particular situation to arise where, delta
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x turns out to be larger than c delta t.
And remember, only in a situation where, delta
21:10.499 --> 21:16.989
x is larger than c delta t, it is possible
to revert time interval and because these
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two situation cannot lead to this particular
delta x be greater than c delta t. Therefore,
21:22.720 --> 21:29.720
we always say that, time order cannot be reversed
as far as, these two events are concerned.
21:29.769 --> 21:35.529
Because, these two events relating to the
motion of the car cannot lead to this situation
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and only if they though have let to situation
of delta x being greater than c delta t, it
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would have been possible for me to find a
frame of reference, in which the car would
21:45.539 --> 21:52.539
have reached A first before starting from
O. Hence, you can see that, by limiting the
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speed to a value less than or equal to c,
we have avoided a situation, in which causality
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would be violated.
22:05.230 --> 22:10.309
Same thing I have written here, in this particular
transparency, if that happens, the car has
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to travel a distance of 5 meter in in 1 into
10 to power minus 8 second, as seen by an
22:16.129 --> 22:22.850
observer in S, which would mean it is speed
would be 5 into 10 to power 8 meters per second,
22:22.850 --> 22:29.220
which is greater than c.
Now, let us take a situation, make it little
22:29.220 --> 22:33.549
more realistic, let us assume that car is
not really traveling with speed greater than
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speed of light. So, I change my, my delta
x is fix as 5 meters, I change my delta t,
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let me still call my event number 1 as the
car passing through O and event number 2 as
22:46.830 --> 22:53.830
car passing through point A. But, I assign
a speed to the car, which is less than speed
22:54.419 --> 22:58.059
of light, this what I have written here.
22:58.059 --> 23:05.059
If car has to have speed less than c then
delta x has to be less than c delta t and
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the events would become time like separated.
If the events are becoming time like separated,
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as we have discussed, it is not possible to
revert the time order in any other frame of
23:15.600 --> 23:21.840
reference. And therefore, it is not possible
to find the violation of causality it means,
23:21.840 --> 23:27.669
we will not be able to find a frame, in which
the car reaches A first before staring from
23:27.669 --> 23:31.419
O.
So, let us this, let us assume that the speed
23:31.419 --> 23:37.519
of car, as seen in A I am sorry as seen in
S is 2 into 10 to power 8 meter per second,
23:37.519 --> 23:41.769
this is some value, which I have taken some
arbitrary value, which is of course, less
23:41.769 --> 23:48.769
than speed of light. Of course, delta x is
still 5 meters and if the car’s speed happens
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to be 2 meters per 2 into 10 to power 8 meters
per second. I can find out, how much time
23:54.850 --> 23:59.090
this particular car would take, to reach point
A.
23:59.090 --> 24:06.090
which is of course, given by the distance,
which is 5 meters divided by the speed, which
24:07.889 --> 24:14.889
is 2 into 10 to power 8 meters per second,
as I have described earlier. So, this time
24:18.409 --> 24:25.409
delta t would be given by this, which I can
simplify 5 divided 2 is 2.5 into 10 power
24:29.019 --> 24:36.019
minus 8 second. So, in this case, we will
find that, delta t turns out to be equal to
24:38.669 --> 24:45.669
2.5 into 10 to power minus 8 second now, let
us compare delta x with c delta t.
24:47.139 --> 24:53.129
That is what, I have written here, time to
reach A in S would then be given by 5 into
24:53.129 --> 25:00.129
2 into 5 divided by 2 into 10 to power minus
8, which is 2.5 into 10 to power minus 8 second.
25:00.470 --> 25:07.229
I have to just multiply it by c, to find out
what is my c delta t, if I take c is equal
25:07.229 --> 25:11.729
to 3 into 10 to power 8 meters per second,
this 10 to power 8 and this 10 to power minus
25:11.729 --> 25:18.729
8, their product will lead v 1.
So, 2.5 multiplied by 3, which gives me 7.5
25:19.450 --> 25:26.440
meters obviously, c delta t is larger than
delta x and therefore, these are time like
25:26.440 --> 25:32.359
separated events. Now, it is possible to find
out a frame of reference, in which these two
25:32.359 --> 25:37.460
events occur at the same position but now
I will not be able to find a frame of reference,
25:37.460 --> 25:44.379
in which these two events occur at the same
time, let us just see.
25:44.379 --> 25:49.549
The speed of the frame, in which the two events
occur at the same place now, I do exactly
25:49.549 --> 25:55.889
the same thing but slightly different numbers.
Delta x still 5 meters but delta t has now,
25:55.889 --> 26:02.729
become 2.5 into 10 to power minus 8 second,
I use exactly the same expression delta x
26:02.729 --> 26:09.729
prime is equal to gamma delta x minus v delta
t. Delta x being 5 meters, delta t being 2.5
26:10.909 --> 26:17.789
into 10 to power minus 8 second multiplied
by v, I can solve for v, this gamma would
26:17.789 --> 26:23.429
of course, cancel because there is a 0 here.
This 5 divided by 2.5 will give me 2, which
26:23.429 --> 26:29.350
is 2 into 10 to power 8 meters per second,
which is of course, smaller than c, this result
26:29.350 --> 26:34.639
was also obvious, this turns out to be same
as the speed of the car, which we know. Because
26:34.639 --> 26:40.320
in the speed of the car, when if a person
sitting on in the car will observe both the
26:40.320 --> 26:44.389
events to be occurring at the same position,
we have noted many such problems earlier.
26:44.389 --> 26:50.729
That, if a person is sitting in the car, he
will find that O is approaching him and the
26:50.729 --> 26:56.539
point A is approaching him and as far as,
his concern, he or she’s concern, both the
26:56.539 --> 27:01.019
events are occurring at the same position.
Therefore, delta x is 0, which I am able to
27:01.019 --> 27:05.389
find out of course, in this particular case,
with a realistic speed, which is less than
27:05.389 --> 27:09.570
c, which is equal to 2 into 10 to power 8
meters per second.
27:09.570 --> 27:14.710
But now, if I want to make delta t prime equal
to 0, I will not be able to do it, unless
27:14.710 --> 27:18.799
I allow speed is greater than speed of light,
let us try and see.
27:18.799 --> 27:25.129
Of course, just the same narration, which
I have said earlier, this is the same as the
27:25.129 --> 27:30.239
speed of the car, as expected. Now, the speed
of the frame, in which the events could occur
27:30.239 --> 27:34.749
at the same time would be given as follows.
27:34.749 --> 27:40.749
Which is given by, delta t prime is equal
to gamma delta t minus v delta x by c square,
27:40.749 --> 27:46.979
I substitute the new numbers. For delta t,
I have 2.5 into 10 to power minus 8 seconds,
27:46.979 --> 27:53.979
delta x is same 5 meters, I workout this particular
equation, I will get 4.5 into 10 to power
27:55.649 --> 27:58.919
8 meters per second, which is obviously larger
than speed of light.
27:58.919 --> 28:05.090
So, only if they would having the frame travelling
with this much speed then only, it would have
28:05.090 --> 28:12.090
been possible for us to find out that, the
two events occur at the same time. Because
28:12.320 --> 28:19.210
of physical reasons, I have avoided, I have
said, the speeds like this are not allowed
28:19.210 --> 28:24.249
therefore, I would conclude that, it is not
possible to find a frame of reference, in
28:24.249 --> 28:29.580
which the two events occur at the same time.
And of course, it would mean that, it will
28:29.580 --> 28:35.169
not be possible to find a frame of reference,
in which the time orders are reverse. Therefore,
28:35.169 --> 28:42.169
causality is is excepted, it is not rejected,
causality has to be has to maintain, has to
28:43.090 --> 28:49.059
be maintained and we have to maintain this
causality, as we have seen that, we have to
28:49.059 --> 28:53.710
restrict speeds to speeds less than equal
to speed of light.
28:53.710 --> 29:00.710
What I have written hence, when the events
are related as cause and effect, the time
29:01.090 --> 29:06.070
order cannot be reversed, unless the speed
greater than c is allowed. Instead of this
29:06.070 --> 29:09.729
particular train, you could have taken any
other example and still you could have seen
29:09.729 --> 29:16.729
that, the causality will not be violated.
And in the case, they has their their events,
29:18.099 --> 29:22.960
which are related as cause and effect then
they will necessarily turn out to be time
29:22.960 --> 29:28.529
like events and in that case, time order cannot
be reversed.
29:28.529 --> 29:33.749
Let us take one more example, this is still
a simpler example and example like this, we
29:33.749 --> 29:37.940
have been discussing earlier.
29:37.940 --> 29:42.960
There is an old example of a ball being thrown
in a train compartment of proper length L
29:42.960 --> 29:49.009
plane it means, there is a particular compartment,
which is moving, it is a relative to earth
29:49.009 --> 29:55.489
assuming, earth to be a inertial frame of
reference. And the length, as measured by
29:55.489 --> 30:00.519
an observer in the compartment is L prime
because that is a proper length because the
30:00.519 --> 30:06.609
compartment is at rest and in it is own frame.
One is ball is thrown towards the front wall
30:06.609 --> 30:09.399
and another is thrown towards the back wall.
30:09.399 --> 30:13.899
This is what, I have shown in the picture,
similar picture we have drawn earlier that,
30:13.899 --> 30:19.059
this is a person, which throws a ball to the
first, towards the front of a wall, the front
30:19.059 --> 30:24.179
being described like this. If this is the
direction of the speed then this wall is the
30:24.179 --> 30:29.629
front wall, this wall is the back wall so
person throws simultaneously one ball in this
30:29.629 --> 30:36.080
side, another ball on this particular side.
And my event number 1 and event number 2 are
30:36.080 --> 30:40.849
corresponding to the events of the ball being
received at the two walls.
30:40.849 --> 30:47.849
So, this is what, I have described here, event
1, ball reaching the back wall of the train,
30:47.970 --> 30:54.970
event number 2, ball reaching the front wall
of the train. Now, if we remember this type
30:56.979 --> 31:01.159
of problem, we have done with ball then we
have done with speed of light, light being
31:01.159 --> 31:08.159
flashed at the two ends, we have done many
times. Because, the ball is being thrown from
31:08.529 --> 31:15.529
the center of the compartment therefore, each
ball has to travel a distance, a horizontal
31:16.149 --> 31:22.200
distance along the x direction, as L prime
by 2.
31:22.200 --> 31:27.519
And both of them travel with the same speed,
as seen in the S prime observer, as seen in
31:27.519 --> 31:34.460
the S prime frame of reference. Therefore,
both the walls, both the balls would reach
31:34.460 --> 31:41.460
the wall together it means, delta t between
these two events will be equal to 0. As per
31:45.159 --> 31:52.159
the first event is concerned, I assume that
the center of the compartment is at the origin
31:53.049 --> 31:59.129
then the first event occurs at a distance
of minus L by 2, second event occurs at a
31:59.129 --> 32:01.710
distance of plus L by 2.
32:01.710 --> 32:08.710
Let me draw my compartment here, this is the
center point, a ball is thrown through the
32:10.889 --> 32:17.889
right and a ball is thrown through the left,
this distance is L prime by 2, this distance
32:20.950 --> 32:27.950
is L prime by 2. Therefore, if I calculate
this speed as I said u prime, this speed I
32:29.590 --> 32:35.149
have said as u prime, if I have to calculate
the time that it takes then all I have to
32:35.149 --> 32:40.349
do should divide this L prime by 2, by this
speed and because this distance is same as
32:40.349 --> 32:43.729
this distance, this speed is same as this
speed.
32:43.729 --> 32:49.109
So, the time when this ball hits here and
the time when this ball hits here, these two
32:49.109 --> 32:56.109
times must be same. So, if this I call as
t 1, if this, I call it, this time I call
32:56.450 --> 33:02.840
as t 2, t 2 minus t 1 must be equal to 0.
But, this event as far the first event is
33:02.840 --> 33:09.119
concerned, it occurs before the origin, at
a distance of minus L prime by 2 and at a
33:09.119 --> 33:14.440
distance L prime by 2 but it is in the negative
sign. This events occur in the plus x value
33:14.440 --> 33:20.019
at a distance of L prime by 2 so if I take
this as the difference, the difference will
33:20.019 --> 33:27.019
be turning out to be L prime so therefore,
delta x prime will be equal to L prime.
33:27.429 --> 33:33.529
So, this is what, I have written here, the
delta x prime is equal to L prime and delta
33:33.529 --> 33:38.659
t prime is equal to 0 of course, delta t prime
is obviously equal to 0. Therefore, delta
33:38.659 --> 33:45.659
x prime is greater than c delta t prime because
L prime is positive and non zero. Therefore,
33:45.919 --> 33:52.729
because delta x prime is greater than c delta
t prime, these events are space like separated
33:52.729 --> 33:58.940
events and therefore, it possible that, the
time order gets reversed as far as, these
33:58.940 --> 34:03.979
events are concerned but it is not possible
that, the position order gets reversed.
34:03.979 --> 34:10.979
Let us assume as frame, which is the gone
frame an observer is observing while sitting
34:13.050 --> 34:18.450
in the ground, this particular train. I applied
the Lorentz transformation of course, I have
34:18.450 --> 34:22.679
to apply inverse transformation here because
my S prime frame of reference is moving with
34:22.679 --> 34:29.679
speed v. If I say, delta x equal to gamma
delta x plus v delta t, plus sign we got this
34:31.619 --> 34:38.619
inverse transformation, delta x was L prime,
delta t prime was equal to 0 therefore, delta
34:40.309 --> 34:45.480
x will turn out to be equal to gamma L prime.
I would not discuss this particular issue
34:45.480 --> 34:48.919
because we have discussed this particular
issue earlier, as we have said that, it is
34:48.919 --> 34:53.169
not possible to find, when this particular
x, delta x is 0. But, I have escaped, this
34:53.169 --> 34:59.460
case I have taken a specific example shown
so if I take delta t, delta t is equal to
34:59.460 --> 35:06.460
gamma delta t prime plus v delta x prime divided
by c square, delta x prime is L prime, this
35:08.190 --> 35:12.220
is 0.
So, delta t turns out to be equal to gamma
35:12.220 --> 35:19.220
v L prime by c square so we can see that,
the time difference S, in a S frame of reference
35:20.559 --> 35:27.559
will turn out to be v gamma v L prime divided
by c square. Of course, depending upon v,
35:28.430 --> 35:33.630
v happens to be positive, this delta t will
turn out to be positive, if v happens to be
35:33.630 --> 35:38.440
negative because I can always assume that,
this particular train was moving other way
35:38.440 --> 35:43.430
that, this delta t prime will turn delta t
will turn not to be negative.
35:43.430 --> 35:48.779
Therefore, depending upon the sign of v, delta
t can be positive or negative, as we have
35:48.779 --> 35:53.599
seen this particular example. Now, I would,
all I would like to emphasis that, these two
35:53.599 --> 35:58.109
events, although they appear to be thrown
by the same person, from the center of the
35:58.109 --> 36:05.109
box and same time, these are not cause and
effect related events. Because, the motion
36:06.250 --> 36:09.950
of the first wall is independent of the motion
of the second wall, it is not because that
36:09.950 --> 36:14.029
the ball one must thrown, that ball two has
to be thrown.
36:14.029 --> 36:19.670
It just matter that, we decided that person
sitting there, decided to throw the two balls
36:19.670 --> 36:26.170
at the same time otherwise, he could have
thrown one ball earlier than the second ball.
36:26.170 --> 36:31.710
In no way, the motion of throwing of the first
ball is related to the throwing of the second
36:31.710 --> 36:38.529
ball A for, as far as, these two events are
concerned, they are not cause and effect related.
36:38.529 --> 36:45.049
Though it appeared as with the same man is
throwing the two balls, may be one is outcome
36:45.049 --> 36:52.049
of the second, of the cause of the second
event, this is not correct because these two
36:53.079 --> 36:58.349
big balls being thrown are sort of independent
each other.
36:58.349 --> 37:05.349
Hence, it is possible to reverse the time
order but I could have gone into this slightly
37:07.849 --> 37:12.049
different situation. When I could have made
these two events later, let us see, let me
37:12.049 --> 37:13.630
first read, what ever I have said.
37:13.630 --> 37:18.490
Note, these events are not cause and effect
type, if there was single ball thrown and
37:18.490 --> 37:23.170
the events would have been related to the
motion of that single ball, then they would
37:23.170 --> 37:27.839
have been cause and effect related and then
only they will be time like separated.
37:27.839 --> 37:33.260
Let take an example here, suppose, this person
would have been sitting at the end of the
37:33.260 --> 37:39.680
compartment and was throwing a ball from this
particular end towards the front and event
37:39.680 --> 37:44.940
number 1 would have been throwing the ball
and event number 2 would have been receive,
37:44.940 --> 37:50.930
ball being received at the front end. Then
these two events are cause and effect related
37:50.930 --> 37:55.700
because unless this ball would have been thrown,
the ball would not have reached here and in
37:55.700 --> 38:02.260
this case, we will find that the events are
time like that is what, I wanted to emphasis.
38:02.260 --> 38:07.450
So, let look at this situation and assumes
that, event number 1 is ball thrown from the
38:07.450 --> 38:14.450
back wall and event number 2 is ball reaching
the front wall of the train. I can find out
38:14.900 --> 38:21.819
exactly the same way, what are delta x and
delta t corresponding to these two events
38:21.819 --> 38:26.270
of course, delta x prime is L prime.
38:26.270 --> 38:33.270
Because the first event occurred at the back
of the compartment.
38:36.380 --> 38:40.569
The ball was being thrown here, so first event
occurred here let assume that my origins now
38:40.569 --> 38:44.799
here. So, the first event occurred at this
particular point, second event occurred at
38:44.799 --> 38:50.849
this particular point, these two are separated
by distance of L prime. Therefore, delta x,
38:50.849 --> 38:57.720
S seen in observer S prime, the compartment
frame is obviously L prime so delta x prime
38:57.720 --> 39:03.289
has not change, which is L prime.
But, but the ball takes a finite amount of
39:03.289 --> 39:07.950
time to go from this particular end to this
particular ends now, delta t prime is not
39:07.950 --> 39:14.369
equal to 0, delta t prime is equal to the,
will be equal the time taken by the ball to
39:14.369 --> 39:20.990
travel a distance from here to here in the
x direction and that distance given by L prime
39:20.990 --> 39:22.420
divided by this speed.
39:22.420 --> 39:27.150
This is what, I have written here in this
particular transparency that, delta x prime
39:27.150 --> 39:33.980
though is equal to L prime but delta t prime
is now L prime divided by u prime because
39:33.980 --> 39:39.579
this is the time that, this ball will take
to go from the back end to the front end,
39:39.579 --> 39:46.180
as seen in observer, as prime. And if I calculate
delta x prime minus c delta prime, I have
39:46.180 --> 39:53.180
to multiply this by c and if I multiply by
this c and take L prime common so I can write
39:53.819 --> 39:59.220
this, I take this L prime out.
So, in the bracket, I will left with this
39:59.220 --> 40:03.599
L prime has to be taken out so this will evolve
here, L prime has to been taken out so it
40:03.599 --> 40:10.150
will be 1 divided by u prime, I have multiplied
by c so to be c divided by u prime. And because
40:10.150 --> 40:17.150
c is expected to be larger than u prime therefore,
this quantity is negative therefore, c delta
40:17.480 --> 40:22.950
t prime is larger than delta x prime. And
these two events become time like separated
40:22.950 --> 40:27.039
events then it is not possible to find a frame
of reference, in which the time order of this
40:27.039 --> 40:32.260
two events will gets reversed.
Now, let us try to calculate delta t, just
40:32.260 --> 40:39.260
to see that, this cannot become negative so
let us, let me comeback to the frame S, apply
40:39.329 --> 40:42.160
inverse Lorentz transformation.
40:42.160 --> 40:48.880
I write delta t is equal to gamma delta t
prime, which is L prime divided by u prime,
40:48.880 --> 40:55.329
as we just now, discussed plus because this
was an inverse transformation then v delta
40:55.329 --> 41:02.230
x is any way here, delta x was L prime divided
by c square. So, this should this is what,
41:02.230 --> 41:08.779
will be the value of delta t, as seen in S
frame of reference, what I am doing, gamma
41:08.779 --> 41:13.289
is any way out of this particular bracket.
I take L prime also, out of this particular
41:13.289 --> 41:17.730
bracket and I also force u prime outside this
particular bracket.
41:17.730 --> 41:24.250
I take L prime divided by u prime, common
from this particular bracket, if I take here
41:24.250 --> 41:30.089
then this factor will become 1, if I am taking
L prime anyway, L prime here but there is
41:30.089 --> 41:35.549
no u prime here so I have to multiply by u
prime. So, this factor becomes equal to gamma
41:35.549 --> 41:42.549
L prime divided by u prime multiplied by 1
plus v u prime by c square. As you can see
41:44.049 --> 41:50.750
that, this delta t cannot be negative of course,
it cannot be negative, if v is positive but
41:50.750 --> 41:57.750
even if v is negative, this particular factor
because v and u prime, both are going to be
41:58.130 --> 42:03.720
smaller than c individually.
Therefore, this factor cannot exceed c square
42:03.720 --> 42:10.289
therefore, this factor has to be always less
than 1 and even if it is of negative sign,
42:10.289 --> 42:17.289
the whole quantity delta t will always lead
me in positive value. Hence, what I have written,
42:17.400 --> 42:22.529
delta t cannot be negative, irrespective of
the sign of v, so long this speeds are lower
42:22.529 --> 42:29.160
than c and therefore, it is not possible to
find a frame, in which the time order get
42:29.160 --> 42:34.420
reversed.
This is what was sort of discussion of giving
42:34.420 --> 42:41.420
some example about time like and space like
events. Now, let us go little ahead, in Lorentz
42:42.690 --> 42:49.690
transformation, enough ,as it given me all
the information about relativity or do I require
42:50.410 --> 42:57.410
something more. What I would like to mention
here is that, this was only the first step,
42:58.700 --> 43:05.700
just doing Lorentz transformation is not enough
in order that, the postulate special theory
43:07.269 --> 43:12.210
relativity that, all the laws of physics remains
same in all the frames.
43:12.210 --> 43:19.059
It is necessary to do something more and what
I would like to say is that, atleast I would
43:19.059 --> 43:26.059
have to redefine my momentum. So, now, I will
give you a very simple example, it is a classical
43:27.880 --> 43:33.500
mechanics, this particular example solved
in a very, very simple fashion, which we call
43:33.500 --> 43:40.500
as, completely inelastic collision. In a completely
inelastic collision in a classical mechanics,
43:41.829 --> 43:47.380
we always say that, the momentum has to be
conserved but the mechanical energy is not
43:47.380 --> 43:51.079
conserved.
Actually, if you take the total energy that,
43:51.079 --> 43:55.819
should be conserved there also but it is converted
into some sort of energy, which is not really
43:55.819 --> 44:00.180
mechanical. And traditionally, in the classical
mechanics, whenever we are solving collision
44:00.180 --> 44:06.150
problems, we apply conservation momentum of
course. And conservation of energy, which
44:06.150 --> 44:12.809
is happens to be the conservation of mechanical
energy, only those cases where, the collisions
44:12.809 --> 44:18.559
are elastic.
It is possible to have non elastic collisions
44:18.559 --> 44:23.849
or the example, which I have gave you as completely
inelastic collisions, in which the two bonds
44:23.849 --> 44:29.880
come together and gets stuck to each other,
in which we do not conserve the mechanical
44:29.880 --> 44:36.880
energy, we conserve only the momentum. Now,
we take this particular example of completely
44:37.569 --> 44:44.569
inelastic collision, as given in a particular
frame and of course, in this particular frame,
44:45.010 --> 44:51.140
we will say that, momentum is conserved.
Now, what I will like to show that, if I change
44:51.140 --> 44:56.029
my frame of reference with different frame
of difference then in that particular frame
44:56.029 --> 45:02.710
of reference, momentum need not be conserved,
if we just take only the Lorentz transformation
45:02.710 --> 45:08.900
into consideration. Therefore, something else
is need to be done, to make conservation of
45:08.900 --> 45:14.539
momentum, which we believe to be one of the
fundamental principles of physics to be valid
45:14.539 --> 45:21.539
in all initial frames. So, I take one specific
example, the example is essentially very simple
45:23.569 --> 45:28.940
but even in this particular example, we will
show that, this violets the universality of
45:28.940 --> 45:31.710
conservation of momentum
45:31.710 --> 45:38.710
So, this what the example I have, title I
have to need to redefined in momentum, consider
45:39.700 --> 45:45.930
a completely inelastic collision in S frame.
So, let us assume that, there is a frame S
45:45.930 --> 45:51.450
in with one particular particle of mass m,
is found to trouble to the right with a speed
45:51.450 --> 45:58.450
of 0.6 c, in the same frame another mass travels
to the left with the same speed, which is
46:01.859 --> 46:08.859
0.6 c. Now, the observer S finds that, these
two masses collide and then get stuck to each
46:12.390 --> 46:18.109
other so what we call, completely inelastic
collision. So, this was a situation before
46:18.109 --> 46:24.960
collusion where, one mass was travel to the
right with the speed of 0.6 c, another mass
46:24.960 --> 46:31.960
exactly identical, was travel into the left
with speed 0.6 c. They have a head on collision
46:32.430 --> 46:39.430
and then two masses get stuck together so
what is remaining is, just a single mass of
46:41.140 --> 46:47.460
total mass 2 m.
The two masses stuck together and this particular
46:47.460 --> 46:54.460
mass comes to rest, and as you can very easily
see that, this will conserve momentum.
46:55.349 --> 47:02.349
Because, if I take the momentum of the first
particle, that will be given by m into 0.6
47:04.089 --> 47:11.089
c, if I take the momentum of the second particle,
the magnitude of that will also be m into
47:11.460 --> 47:18.460
0.6 c. But because this speeds are in different
direction, opposite direction therefore, the
47:19.130 --> 47:24.510
momentum direction, momentum being that will
be opposite so this will be the net momentum
47:24.510 --> 47:29.450
because this magnitude is same as this magnitude
so this should be equal to 0.
47:29.450 --> 47:36.450
So, this will be what, I called as a initial
momentum it means, according to conservation
47:39.049 --> 47:45.630
momentum, the final momentum must also be
equal to 0 because mass is 1 0 obviously.
47:45.630 --> 47:49.569
Therefore, it means, the mass must come to
rest, because then only, it is possible for
47:49.569 --> 47:54.869
momentum to be conserved or to become 0. So,
this speed has to become 0. And that is what,
47:54.869 --> 48:00.000
I am saying that, after the collision, the
two masses are come together and got stuck,
48:00.000 --> 48:05.490
and have come to rest.
Therefore, the final momentum is also 0, so
48:05.490 --> 48:10.509
we have a situation where, initial momentum
is also equal to 0, final momentum is also
48:10.509 --> 48:14.420
equal to 0 therefore, momentum is conserved.
48:14.420 --> 48:21.420
So, as you have seen just now that, initial
momentum is 0, final momentum also 0 therefore,
48:23.890 --> 48:29.890
in S frame, the momentum is conserved. Now,
let us try to look this particular collision,
48:29.890 --> 48:34.599
from one particular frame on specific frame,
S frame, which happens to be moving with the
48:34.599 --> 48:41.599
same speed S, the initial speed of the first
particle, which is 0.6 c. And we show that,
48:41.859 --> 48:47.730
in this particular frame, this conservation
is valid, even if you can show violation in
48:47.730 --> 48:51.240
one particular frame of reference, the law
gets valid.
48:51.240 --> 48:58.240
So, just one example contrary is enough to
say, that this particular law is valid so
48:59.390 --> 49:06.390
let us go to S prime frame of reference, in
which the relative speed v is 0.6 c.
49:09.609 --> 49:16.609
As we have seen, this tender formula for velocity
transformation is u x minus v divided by 1
49:19.410 --> 49:26.410
minus u x v by c square. In this particular
case, the relative speed within the frames
49:28.130 --> 49:35.130
has been chosen as 0.6 c, if we applied this
particular thing to the first particle, that
49:37.269 --> 49:44.269
first particle has a u x, which is equal to
same as 0.6 c. I substitute these two quantity
49:50.259 --> 49:56.069
in this particular formula, to find out this
speed of this particular particle in S prime
49:56.069 --> 49:59.789
frame of reference, which is what I have done
here.
49:59.789 --> 50:06.789
So, u x was equal to 0.6 c, v was equal to
0.6 c divide by 1 minus u x into v divided
50:08.579 --> 50:15.579
by c square, c square gets cancel, this becomes
1.36, what as you can see, numerator we have
50:15.819 --> 50:22.579
0.6 c minus 0.6 c, this gives me 0. So, irrespective
of the denominator of course, denominator
50:22.579 --> 50:28.029
is not equal to 0, I will get for the first
particle this S component of this speed is
50:28.029 --> 50:33.299
equal to 0, which is expected.
Because, if a frame of reference is moving
50:33.299 --> 50:37.680
along with the particle, with moving with
the same speed of the particle that particular
50:37.680 --> 50:41.910
in that particular frame of reference, the
particle would appear to be at rest. It would
50:41.910 --> 50:48.910
not be appear to be moving therefore, I am
getting this speed to be 0, which is obvious.
50:49.039 --> 50:53.059
Now, let us calculate the velocity of the
second particle before the collision. This
50:53.059 --> 50:58.789
particular particle was having the same speed
but in a different direction so my u x was
50:58.789 --> 51:05.789
equal to minus 0.6 c. So, therefore, I write
here as minus 0.6 c, v was any where 0.6 c
51:06.480 --> 51:11.910
and because this is negative so this particular
sign becomes plus. So, I get 1 plus 0.36,
51:11.910 --> 51:18.910
which when I calculate is turning out to be
minus 1.2 divided by 1.36 c.
51:21.759 --> 51:27.380
So, observer S prime would notice that, this
speed of the second particle of course, is
51:27.380 --> 51:34.380
in minus x direction and is equal to 1.2 divided
by 1.36 c. Now, let us try to calculate this
51:36.269 --> 51:42.869
speed of the two particle, which of course,
now turn into a single particle, after the
51:42.869 --> 51:49.150
collision. This particular particle was at
rest in S frame of reference so let us try
51:49.150 --> 51:56.150
to calculate this speed, here the speed in
S frame was 0, here v is 0.6 c.
52:00.490 --> 52:07.490
Therefore, which I am saying is the final
speed, u f x prime is equal to 0 minus 0.6
52:08.980 --> 52:15.980
c, which is v divided by 1 minus u, which
is equal to 0.6 c divided by c square. This
52:19.319 --> 52:26.319
quantity becomes 0 because of, 0 here, denominator
is 1 and you get minus 0.6 c therefore, the
52:27.890 --> 52:34.049
speed of the two particles, two combine particle
after the collision, as seen in S prime frame
52:34.049 --> 52:37.970
of reference will be turning out to be equal
to be minus 0.6 c.
52:37.970 --> 52:44.609
Now, let us try to calculate the momentum
of the particle, the initial and the final
52:44.609 --> 52:51.609
momentum, the first particle in S prime frame
of reference was having a 0 velocity, so obviously,
52:52.039 --> 52:59.039
it has a 0 momentum. The second particle was
having a speed of minus 1.2 divided by 1.36
52:59.509 --> 53:06.259
c, this multiplied by the mass, the momentum
of the second particle, which happens to be
53:06.259 --> 53:10.970
the total initial momentum as observed in
S prime frame of reference.
53:10.970 --> 53:15.309
This initial momentum as observed in S prime
frame of reference, which will be equal to
53:15.309 --> 53:22.309
minus 1.2 divided by 1.36 m c. Now, final
momentum, there is only one particular particle,
53:25.049 --> 53:30.529
which has a mass 2 m and that particular particle,
as we seen in the previous transparency, was
53:30.529 --> 53:35.400
moving with the speed of minus 0.6 c in a
S prime frame of a reference.
53:35.400 --> 53:42.400
So, this multiplied by the mass of the particle
which is of 2 m , the final momentum is turn
53:42.589 --> 53:49.589
out to be equal to minus 1.2 m c, as we can
clearly see that, this is not same as this.
53:50.509 --> 53:56.960
According to S frame observer, the momentum
of the particle has not conserved so according
53:56.960 --> 54:03.089
to an observer S frame of reference, this
collision really let to conservation of momentum.
54:03.089 --> 54:08.420
But, if everything what ever I am saying is
true then according to an observer S prime
54:08.420 --> 54:14.190
frame of reference, the momentum was not conserved
in this particular process. Hence, either
54:14.190 --> 54:18.259
we say that, conservation of momentum is not
a fundamental principle of physics, which
54:18.259 --> 54:25.259
I am not sure, anyone of us will agree, which
means that, we must change we must do something
54:25.559 --> 54:29.869
else so that, the momentum is also conserved
in S prime frame of reference.
54:29.869 --> 54:36.869
Therefore, we need to redefine our momentum
so in the event, we will summarize, what we
54:37.269 --> 54:43.890
have discussed today. We discussed two examples
related to space and time like separated event
54:43.890 --> 54:50.890
and specifically, we saw the cause and outcome
related events. Then finally, we discussed
54:51.559 --> 54:58.559
one simple example of completely inelastic
collision and we said that, we must redefined
54:58.740 --> 55:00.579
our momentum.
Thank you.