 In our last lecture, we had discussed some examples, which involve momentum and energy and their relationships, how we change from one frame to another frame. And specifically we discussed the center of mass frame, which we call C frame. There is in contrast to C frame, we have a L frame, which we call as a laboratory frame, which need not be the frame in which the initial momentum is 0, initial momentum of the set of particles is 0. While in C frame it is 0, we also said that some problems become easier to solve if you go to center of mass frame. We also discussed that some of the experiments, high energy experiments are performed in the center of mass frame in order to save energy. So, this is what we have said about the last lecture. We discussed some examples involving energy and momentum relationship. We also discussed the concept of C frame and worked out some problems related to it. Today we would like to go little ahead and introduce a new concept, which of course in classical mechanics is well known, which is the concept of force. In Newtonian mechanics, force is very important thing. We all know what is force. So, let us discuss what happens to the definition of force once we come to the special theory of relativity. If you remember, there are two ways in which we can talk of force in classical mechanics. One, of course, force is a vector quantity. So, we always write with this vector sign F. We can write that F is equal to the rate of change of momentum. When we say rate of change, it always means the time rate as far as this definition is concerned. So, force can be written as time or rate of change of momentum, because in classical mechanics P is given by mass into velocity where mass is supposed to be constant. Therefore, equivalently we can write this relationship in the form of mass into acceleration, because the rate of change of velocity with respect to time is given as an acceleration. So, in classical mechanics, we can use either of the two ways to express the force, either as a rate of change of momentum or mass into acceleration. We shall just now see that in relativity, we have to choose one or the two and in principle, in fact, in relativity, we choose the first one. In relativity, we define force as the rate of change of the momentum. So, we take the first definition, not the second definition, because as we shall see that the two definitions are not same. Specifically, as you know that in momentum, it relates to something, in fact, it relates to the mass in which there is a gamma u factor. We will just now discuss that. So, this is what we said. The force in special theory of relativity is equated to time rate of change of momentum. So, the definition that we have taken in relativity for the force is force is d P d t, where P is the momentum of the particle. Let us see what does it mean. Let us try to take this example forward, this definition forward and try to get little bit more insight into how our concepts will change with this particular definition of momentum, I am sorry, this definition of force and also the way relativity defines momentum and other quantities. So, as we have said that force in special theory of relativity, we have said earlier, is defined as gamma u m naught u. So, u, if a particle is moving with a speed or rather with a velocity u in a given frame of reference and m naught is the rest mass of the particle then gamma u m naught u is defined as the momentum of the particle. This is what we have discussed earlier in our earlier lectures. So, I must take the rate of change of this particular quantity if I accept this definition of force, which is force is equal to a rate of change of momentum, which essentially means that force has to be written as d d t, which is the time, which is the rate of change of the time derivative m naught, which is here u, which is here and gamma u I have expressed as 1 upon under root 1 minus u square by c square. So, force must be written in this particular fashion, which essentially means that I have to take a time derivative of this particular quantity, where now u appears at two different places, one here, another here. In classical mechanics, in non-relativistic mechanics, this was the only place where u was appearing and once we took a time derivative, it just straightway gave us m into acceleration m into d u d t, which I can write as an acceleration. But now, if I take this definition, I have to take this particular quantity, take the derivative of this quantity also. So, it is a function in which there u appears twice. Therefore, I must differentiate by parts in order to get the actual expression for force applicable for special theory of relativity, which I have done in the next transparency. So, we will now differentiate by parts. As we have seen, the force that we have written was force is equal to d d t of m naught u under root 1 minus u square by c square. So, I am trying to differentiate it by part. So, first I take u, take the derivative of this quantity plus I take this quantity multiplied by derivative of u. This is what I have written in this particular transparency. So, u multiplied by d d t of m naught under root 1 minus u square by c square plus m naught divided by under root 1 minus u square by c square multiplied by d u d t. It is just simple differentiation by parts. Now, let us look little more closely at this particular equation and try to see whether I can make meaning out of it. So, next transparency that is what I am trying to do. I am trying to look at this particular first term, which is d d t of under root 1 minus u square by c square. This m naught which was appearing outside, I have brought it inside. I have multiplied by c square and also divided by c square. The idea is that I can express this in terms of the new definition of energy, because I know this particular factor multiplied by m naught multiplied by c square can be written as the energy of the particle, total relativistic energy of the particle, which is defined as gamma u m naught c square. So, this is my gamma u m naught c square. So, this whole quantity can be written as the total energy of the particle divided by c square. So, the first term in that particular integration by part can be written as u multiplied by d d t of e by c square, where e is the total relativistic energy of the particle. So, just expressing in a different form. Now, d u d t, we know that can be written as an acceleration. If you are in a given frame, if you see a particle moving, the rate of change of velocity of that particular particle in this frame is obviously defined as an acceleration. So, a has to be expressed as d u d t, which means that the force now can be expressed as u multiplied by d d t of e by c square, which we have just now discussed, plus d u d t I have written as the acceleration of the particle. So, m naught a divided by under root 1 minus u square by c square. So, it is very strange. We find that with this new definition of force, force is given in terms of summation of two quantities, in which only one of the quantity here depends on the acceleration. Here, the first quantity depends actually on the time rate of the total energy of the particle, which is in total contrast with the classical mechanics. So, let us look at some of these things. Let us first discuss this at this particular equation and try to see what modifications I have to make in terms of our understanding in relativity about the force, which is quite different from the classical mechanics. So, this is what I have summarized in the next three points. First thing that we realize that f naught is not given, f is not given by gamma u m a. Had I taken other definition of force is equal to mass into acceleration, then probably I would have had a tendency of writing force is equal to gamma u times m naught a. But as we have said that in relativity, we do not take that particular definition, but take the definition of rate of change of momentum, which has caused us an additional term in the force. Therefore, it is very clear that in this case, force cannot be written as gamma u m m naught a in general, unless of course, the first term is 0. Remember, there are two terms here. This is the gamma u m naught a. This term is an additional. If this term happens to be 0, then it is a very different case. Otherwise, it is always sum of two terms and therefore, f is not written in terms of just gamma u m naught a. Remember, as I have mentioned earlier, just in the beginning of this lecture, in classical mechanics, these two definitions are equivalent. I could have written force is equal to mass into acceleration or I could have written force as rate of change of momentum. Here, I take only force as rate of change of momentum and not in general f is equal to m a, which I might have a tendency see that because m depends on the speed of the particle relativity. So, just replace m by gamma u m naught. It does not work that way. The second thing which is also very interesting is that the acceleration may not be in the same direction as force. Let us look here. This force is a vector quantity. Acceleration is a vector quantity. So, is u, which is the velocity of the particle? Now, we can always have a situation in which the particle is in a moving in a particular direction and acceleration of the particle is in a different direction or we apply a force which is in a different direction. Therefore, which is not the same direction as the direction of the velocity of the particle. For example, a particle could be moving this way, moving this way and I may decide to apply a force in this particular direction. Now, classical mechanics would not have made any difference. If the force was in this direction, the acceleration would also be in the same direction because f was equal to m a. But in this particular case because the force also depends on this direction of u. Therefore, the force will no longer be exactly in the direction or rather acceleration will not be exactly in the direction of f. It will also depend on u, which is in total contrast with the classical mechanics, where force was just in the same direction as the acceleration. Here, force will produce an acceleration, which also depends on the direction of the speed of the particle in that frame. This is what I have written. We also note that with this definition of force, the acceleration may not be in the same direction as the force. Of course, there is another third thing which I would like to mention about in this particular equation is that if force is constant, it need not mean that acceleration is constant because it depends, the first term depends on u. In fact, in the second term also there is a gamma u. So, in principle, as speed of the particle changes, the same force may cause different acceleration while in the classical mechanics, because mass was velocity independent. So, if your force remains same, acceleration will always remain same. But here, acceleration will keep on changing because once we apply force, the velocity will keep on changing and depending on the velocity, the acceleration will also keep on changing. This is put in contrast with the classical mechanics. Of course, we can understand if we have to say that our speed of light is our ultimate limit and the particle will never be able to cross the speed of light. It means once we are applying the force and as the particle is trying to approach the speed of light, you will obviously expect that it cannot accelerate. This velocity cannot change that much. So, the acceleration will not be that high once we are approaching to the velocity of light. So, therefore, in that sense, from realistic concept, we can understand it, but from classical mechanics, it is in total contrast. So, this is what I have written. Even if the force is constant, the acceleration is not as instantaneous velocity is also involved both in the first term and in the mass term. So, both the terms involve velocity. Therefore, you will not have, if the force remains constant as a function of time, it does not mean that acceleration also remains constant as a function of time. Now, let us look little more closely at the second term. I am sorry, the first term, the term which involved u or which involved the rate of change of energy, whether we can give a little more physical meaning to the rate of change of energy or the term dE dt. Now, in relativity, we have said that this total relativistic energy E can be written as sum of kinetic energy plus the rest mass energy M naught c square. The rest mass energy we have discussed was a total relativistic concept. So, M naught c square is the rest mass energy of the particle plus the kinetic energy of the particle will give me the total relativistic energy of the particle. So, if I just look at this time derivative of E upon c square, I can write this as time derivative of M naught c square plus k, just substituting for E k plus M naught c square. So, this is what I have just substituted here divided by c square. Of course, the first term would give me dDt of M naught and M naught is the rest mass of the particle, which is of course not expected to change as a function of time. Therefore, this particular derivative will give me 0. Therefore, this term can be simply written as dDt of k upon c square, which is what I have written here. So, dDt of E by c square can also be written as dDt of k upon c square because E and k differ only by the rest mass energy and rest mass energy does not change as a function of time. Now, let us look, go back to our classical mechanics and try to see how this particular k or rate of change of k is interpreted in classical mechanics. If one remembers the work energy theorem, one would realize that kinetic energy is results because of a work done by a given force. Let us look at the other transverse. So, as we have said, let us suppose there is a particle which is moving in a given direction and we apply a force, then if in a given time delta t, this particular particle moves by distance delta r and we take a dot product of f and delta r, this is what is called work done by the force and this by work energy theorem should be equal to the change in the kinetic energy. So, if the particle was moving without any force, of course, it is constant in the kinetic energy has to remain constant, it will not change. Once we apply a force, once the force is applied, whatever is the displacement in a given time delta t, if this displacement is delta r, then f dot delta r, f dot delta r is called the work done by the force and this is equal to the change in the kinetic energy because now under the application of the force, the particles kinetic energy would change and the change in the kinetic energy in the same time delta t will be given by f dot delta r. This is a well known classical result which is called work energy theorem. So, this is what I have written. If the displacement of the particle in time delta t under the influence of a force f is delta r, of course the particle could be moving even before we have applied the force, it does not change thing. As we have said, f need not in general be in the same direction as u. Then by work energy theorem, the change in kinetic energy delta k is given as delta k is equal to f delta r. So, I can divide this by delta t both the sides and assume that delta t tends to 0. So, these deltas can be changed to derivatives differentials. So, that is what I will be doing next. The rate of change of kinetic energy therefore would be given by delta k divided by delta t which I can write as f dot delta r delta t. Delta r by delta t is the total displacement of the particle in time delta t divided by delta t. Therefore, this will be given the velocity. So, this is the velocity of the particle. So, delta r delta t can be written as velocity of the particle and if I make delta t tend to 0, I can take this as instantaneous velocity of the particle. This is why I am writing f dot u and this deltas I am replacing by d because I am assuming that delta t is tending to 0. So, I can write this as dk dt is equal to f dot u. Now therefore, the expression of force can be written more clearly. The way we wanted to write it containing two term, the first term has been reinterpreted in terms of f dot u. So, it will be written as f is equal to u multiplied by f dot u divided by c square plus m into a where of course, m is gamma u m naught. Let us remember it. So, this m is also speed dependent or the velocity dependent mass. It is a gamma u m naught. So, this is the way we write. We can thus write the expression of force in terms of acceleration as f is equal to u f dot u divided by c square plus m a. Of course, one can see that in non-relativistic limit, this expression will reduce to standard expression. In non-relativistic limit, this m can be assumed as m naught because gamma u will be very close to 1 because once u is very small in comparison to c, gamma u is very, very close to 1. Therefore, m will be equal to m naught and if u is very small in comparison to c, there is a c square in denominator. These values will be very small in comparison to c square. Therefore, even the first term will be neglected and therefore, f can be written as f is equal to m naught a. In a way, we are used to writing classical mechanics. Therefore, you can see that it really reduces to a standard expression in the non-relativistic limit. Now, let us look at some special cases in which the force happens to be in the same direction as acceleration. See, we have just now said because of this complicated equation that we have, I will call it complicated, in which there is a u also involved, there is a u involved here, there is u involved here. In fact, there is a work run by the force involved. Then only we get force is equal to mass into acceleration. So, you have two terms. Now, what are the cases in which force and acceleration could be in the same direction? That can happen in two cases. One of the cases will be where this term is 0. If this term happens to be 0, then f dot u, for example, when f dot u is 0, it means force is applied perpendicular to the direction of the speed, instantaneous velocity. Then this term will be 0, then f will be given equal to mass into acceleration, f and a will be in the same direction. And there is a second special case where we have only one dimensional motion. It means the particle is moving in particular direction. You are applying also the force in the same direction. So, particle speed changes also in the same direction. All the motion is confined in the x direction. So, instead of, for example, this particular case, which I have written here, I take u direction to be this and force direction is also in the same way. So, the motion of the particle is in the same direction. It is only a one dimensional motion. The particle lever bends and keeps on moving in a straight line. If it keeps on moving in a straight line, then in that case, this vector u and this vector a is in the same direction. Therefore, force will also be in the same direction as the acceleration. So, there are two special cases where I will see that force and the acceleration are having the same direction. Let us discuss these two cases because these two cases are interesting and are of course comparatively simple. This is what I have said. We can see that there are two special cases when force is parallel to the acceleration. Let us take case one. If the force is always perpendicular to the instantaneous direction of velocity, of course, once we apply the force, the velocity will change magnitude in general, both in magnitude and direction or at least in one of them it has to change. But if it always happens that force is always perpendicular to the direction of velocity, then f dot u will always be equal to 0 and as we have said, the first term will become 0 and then we will have a simple case of f is equal to m a. The force will be given by f is equal to m a. Of course, m as we have said is a velocity dependent mass. So, it must be written as gamma u times m naught a. So, this is would be the case or which we thought that was could have been one of the definition of force. In principle, with this definition of force is applicable only when force is perpendicular to the direction of u. In that particular case, f is actually indeed given by gamma u m naught. So, sometimes because this gamma u m naught is called transverse mass, because this will be the mass of the particle, you know, we can write force is equal to mass into acceleration. With this mass gamma u m naught only when the force is applied perpendicular to the direction of the velocity of the particle. Therefore, many times it is termed as a transverse mass. Now, we was wondering whether we have a situation. If you remember electromagnet theory, you know that this situation is quite common. If there is a charged particle in the magnetic field, then the force on that particular particle is always given by u cross b e u cross b. Now, let us say q u cross b is the force on the charge q, if it is moving with a velocity u. And because this is a cross product u cross b, it means this force will always be perpendicular to u. So, this magnetic force will always be perpendicular to the direction of u. So, let us just take a simple example of this particular case when a charged particle is under the influence of a magnetic field just to give us an example of the case in which force is actually in the same direction as the acceleration. So, this is the example which I have taken. An electron is moving with a speed of 0.8 c in a circular motion under the influence of a magnetic field of 1.5 Tesla. We have to find the radius of the orbit. This case, as I said, is simple because force and the acceleration are in the same direction. So, let us first find the force. The force will be given by q u cross b as we have just now said because this is an electron. So, q is equal to the just charge on electron which is e and because u and b are perpendicular to each other and this is cross product. So, u cross b will just give you u multiplied by b because u magnitude of u multiplied by magnitude of b multiplied by sin of the angle between the two, sin of the angle being 90 degrees, this will give you 1. Therefore, the magnitude of the force will be e u b. I am putting the value of e, I think in approximate value 1.6 into 2 power minus 19 coulomb. u has been given to us as 0.8 c multiplied by the magnetic field which has been given as 1.5 Tesla. This gives me that the net force on the charge will be 5.76 into 10 to the power minus 11 Newton. Of course, this will always be directed towards the in a direction perpendicular to the direction of the velocity which actually will provide the centripetal force for this particular electron to move in a circle. I mean, in case a particular particle is moving in a circle, there has to be a centripetal force and if I assume that the speed of the particle is constant, then there has to be constant centripetal force acting towards the center, always towards the center of the circle. This is a well-known classical results. Now, in this particular case, I can find out the acceleration because my expression of force into acceleration force and relation between force and acceleration is comparatively simple and acceleration will be just given by force divided by the transverse mass which is gamma u m0. We know for the case and we have done many problems earlier that for the case when u is equal to 0.8 c, the gamma u turns out to be 5 by 3. So, this 5 by 3 I have used here. This is the value of the force which we have just now found out in the previous transparency 5.76 into 10 to the power minus 11 Newton. So, this is what I have written here divided by m0, the rest mass of the particle, I am taking approximately the mass of the electron as 9.1 into 10 to the power minus 31. Of course, as I say, there is a gamma u which is 5 by 3 which is here. If I take this, this acceleration turns out to be 3.80 into 10 to the power 19 in Cgi, in SI units. Now, we know from classical mechanics that if a particle has to move in a circle, there has to be a centripetal, there has to be a real force because this acceleration amount of this particular particle is equal to u square by r assuming of course u to be constant. I mean the magnitude of u to be constant. Direction is not constant. It is always changing. So, velocity is always changing but the speed is constant. So, this acceleration in case this particular particle, this particular electron has to move in a circle, the total acceleration of the particle must be equal to u square by r where r is the radius of the circular. I have just now found out what is the acceleration of the particle. So, I can substitute it to equal to u square by r. I know you, I can always find out what will be the radius if this electron has to move in a circle. The radius turns out to be u square which is 0.8 C square divided by this 3.8 into 10 to the power 19 which we have obtained earlier as the acceleration. So, this must be the radius as we have seen this comparatively simple case where we have found out that acceleration is in the same direction as the force and only difference which you have to make to take into a counter relativity is that instead of using m naught, I have used gamma u m naught. Now, let us consider the second case. That is the case in which there is a straight line motion. This is also very interesting case. So, let us just discuss this particular case a little bit more in detail. So, this is my case 2. A one-dimensional motion in which force is applied in the direction of the velocity of the particle. What happens in this particular case? This was my expression. Of course, I have slightly rearranged the first term which was there. I put it second term just to make it looks little simple. So, I have put f is equal to ma plus u into f dot u divided by c square. This was my original relationship between the force and the acceleration. Now, because everything is in the same direction, so I need not write vectors f dot u because f is in the same direction as u and f dot u will be f multiplied by magnitude of f multiplied by magnitude of u multiplied by cosine of the angle between the 2 and because angle between them between the 2 is 0, between the 2 vectors is 0. Therefore, that is equal to 1. So, this becomes magnitude of f multiplied by magnitude of u. Everything is in the same direction. So, this I can just write f multiplied by u square of course divided by c square and this a also I can write without vector sign because I realize that everything is in the same direction. There is only one particular motion only in one particular direction. So, I can write this equation in a scalar form by writing f is equal to ma plus f multiplied by u square by c square. Now, we can find out the relationship between the force and the acceleration because I can take this force term on the left hand side and find out in this particular case what will be the relationship between the force and the acceleration in case there is a motion just in a single line in case of one dimensional motion. It is very simple. Let us just put in the second transparency that is what I have put. This is the expression which I had written earlier f is equal to ma plus f divided by I am sorry f multiplied by u square by c square. This term is taken on the left hand side. I take f common. I get f multiplied by 1 minus u square by c square must be equal to ma. This particular quantity here I can divide by here. Remember this m is also gamma u m naught. So, therefore, this I can write as f is equal to m naught and there is a gamma u which was 1 upon under root 1 minus u square by c square and there is another factor of 1 minus u square by c square which is coming from this here. So, when this gets divided, remember here this is m naught. Here it was m. So, m naught and that additional gamma u I have incorporated here and this becomes now 1 minus u square by c square to the power 3 by 2. Very interesting reason because it says that in this particular case force is related to acceleration. Force is in the same direction as acceleration, but the mass of the particle, the wave mass will behave will be m naught divided by 1 minus u square by c square to the power 3 by 2. It is gamma q, gamma u q. While in the case of transverse motion, I mean when the force was perpendicular to u, it was just gamma u, but in the case of one-dimensional motion, it is gamma u q. So, this is generally called as a longitudinal mass. So, we often define longitudinal relativistic mass of the particle as m naught divided by 1 minus u square by c square to the power 3 by 2 which is equal to gamma u q m naught. So, if the motion, the particle is in one direction, the mass turns out to be gamma u q times m naught. As far as relationship between the force and acceleration is concerned, let us make it very, very clear. When the relationship between the force and acceleration is concerned, this is the way it should be written. Now, let us look a little bit more carefully. In fact, we have lot of what we call as kinematic equations in the classical mechanics where if you know the initial velocity, you can find out the velocity, final velocity. If you know the acceleration at a given time, you can find out the distance, etc., etc. So, let us look at some of these equations and try to get little bit more insight of what is happening when a force is applied in a one-dimensional case on a particular particle. So, let us assume that a constant force is applied to a particle which is moving initially with a speed of u naught in the same direction as that of velocity. This is exactly the case which we have just now said that the particle is moving in this particular direction with a velocity u naught. Then at time t is equal to 0, I apply a force which is also in the same direction f. So, my motion, whatever are the change in the motion occurs in the same direction. So, everything is in the same direction. So, this is what is the case. Now, my question is that once I have applied force for a given time, what happens to its velocity? So, let us assume that after time t, we have applied the, let us assume the force is constant. A constant force is being applied. I am interested in finding out what happens after time t. So, let us assume, of course the velocity of the particle will change. This direction will not change because force and the velocity are in the same direction, but its magnitude will change. So, let us try to find out, let u, let after time t, the speed of the particle or velocity, whatever you want to call it because any moment directly in the dimensional be the speed, u t be the speed after time t. Then how do I find u t? I have to integrate this equation. This was my force equation. Acceleration I have written as du dt, rate of change of speed or rate of change of velocity is the acceleration. This is just I have written as gamma u cube m naught. This dt I will take on this side, du I will let it remain in this side and integrate. This is the way we derive those kinematic equation, standard kinematic equation in classical mechanics. We just assume force to be constant, write force equal to mass into acceleration, then integrate and then obtain various type of equation. What happens that the speed of the particle at a given time t, what happens to the displacement of the particle at a given time t? We are doing exactly the same thing, except that our expressions have become little more difficult. So, this is what I have to integrate in order to find out what will be u t at a given time t. Therefore, I am just integrating it, integral from 0 to t and this is because this is the initial velocity u naught, this is the final velocity after time t to u t, m naught du 1 upon u square, 1 minus u square by c square to power 3 by 2, just integrating this expression is just written here. Now, in order to integrate, let me not use these definite integral, let me use still indefinite integral, it is comparatively easier and then I will put the limits later. This is what I am doing next, I am taking the indefinite integral of this equation of the right hand side, let me solve the integral in the indefinite form, then I will put the limits later. So, let us solve the integral without putting the limits, so I will just put integral f dt of integral m naught du 1 minus u square by c square to power 3 by 2, there is a standard substitution to solve this equation, we put u is equal to c sin theta, then we take the differentiation, we get du is equal to c cos theta d theta. This du is now being replaced by c cos theta d theta, very simple integral. For this u square, I am putting c sin theta, let me just write it here to make it clear. In the denominator, we had 1, we had m naught du and here we had 1 minus u square by c square to the power 3 by 2. If I put u is equal to c sin theta, so this becomes, I am just looking at the denominator, 1 minus c square sin square theta by c square to power 3 by 2, this c square will cancel, so I will get 1 minus sin square theta to the power 3 by 2, 1 minus sin square theta is cos square theta, so this becomes cos square theta to the power 3 by 2, this under root and this half moves away, this becomes cos cube theta. So, this is what I have written here, integral of f dt is equal to m naught for du, as I have said we have written c cos theta d theta divided by cos cube theta, this c remains here, this cos theta will cancel with one of these powers, so this will become cos square theta and eventually you will get 1 upon cos square theta which is sec square theta d theta. So, this equation becomes integral of f dt is equal to integral of m naught c sec square theta d theta, we know that sec square theta when integrated, it becomes tan theta. So, this equation just becomes m naught c tan theta, this tan theta I know is sin theta divided by cos theta, I would like to express this thing back in terms of u. So, what I will do, I have written anyway c sin theta as u, so this I can put as c sin theta as u, this cos theta I will express as under root 1 minus sin square theta and for sin theta again I will put u by c, this is what I have done in the next class. m naught c sin theta by cos theta, sin theta as we have said is this c sin theta is u, so which I have written as u, this cos theta I have written as under root 1 minus sin square theta because c sin theta is equal to u, therefore sin theta will be u by c. So, for sin square theta it becomes u square by c square. So, this integral f dot dt gives me m naught u divided by under root 1 minus u square by c square. So, this is what I get as a result of integration. Now, we can put the limit, say our limits where from u naught to u t, so I take the same expression, I first put u t here, I put u t here minus then I put the limit u 0 which is u 0 here, u 0 here, so this is what I am putting here. So, f multiplied by t is m naught u t divided by under root 1 minus u t square by c square minus m naught u naught divided by 1 minus u naught square by c square. So, this is what I will find that of course, this equation is not in a simple form to express it in u t that of course, we can do later, but this definitely gives that if time, if force is applied for time t, this gives a relationship between u t and u naught. Before I try to derive this particular relationship which probably we will do in a later lecture, but let us try to look at this particular equation and try to see the implication of this particular equation. So, for that we have taken an example, consider a one-dimensional motion which we have in considering in this particular case. Let us assume that there is a particle of Rasmus m naught which is subject to a constant force f. First find the time its speed takes to change from 0 to 0.2 c. I have put this particular problem in a fashion so that I can use this equation directly without actually working out for u t. So, initial velocity has been given which is 0, particle starts from rest and final velocity has been given as 0.2 c and there is a particular force f whatever might be the magnitude of the force and its Rasmus is m naught. I am interested in finding out how much time it will take which will depend on the force and m naught, but I am giving the I am interested in finding out the answer in terms of f and m naught because I want to really look how the time will vary when the speed really is approaching the speed of light. Now, we would also like to make a similar calculation if the speed changes from 0.2 c to 0.4 c and then 0.4 c to 0.6 c and then 0.6 c to 0.8 c means the interval speed interval is same is 0.2 c. So, we go from 0 to 0.2 c. 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 the 0.6 c to 0.8 c. So, velocity intervals are same force is same. I am interested in finding out the time the particle will take to go from here to here go from here to here go from here to here go from here to here that is what is the question. So, this is my equation force multiplied by time is equal to m naught u t divided by 1 under root 1 minus u t square by c square minus m naught u naught divided by under root 1 minus u naught square by c square this particular quantity m naught divided by 1 minus or rather 1 divided by under root 1 minus u t square by c square I am writing as gamma u t because there are two speeds one after time t one at time t is equal to 0. So, gamma corresponding to this speed is gamma u t gamma corresponding to the initial speed is gamma u naught. So, the same expression I have written as m naught multiplied by gamma u t u t minus gamma u u naught m naught has been taken common u t is here this particular under root 1 upon this particular thing has been written as gamma u t this m naught has been taken out here this u naught is present here this 1 divided by 1 minus u naught square by c square has been taken as gamma u naught. So, this is my expression. So, for the first case I am starting with 0 initial velocity. So, this term is 0 all I have to do is to take gamma u t multiplied by u t u t is 0.2 c for the first case. So, what I have to find out what is the gamma corresponding to a speed of 0.2 c it means I will take this as 1 upon under root 1 minus 0.2 square I will substitute in this particular expression and obtain what will be the time taken force this time will be this whole quantity divided by force. But as I said I want the answer in terms of both m naught and the force. 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 which I have said t 1 is equal to m naught c divided by f multiplied by 1.1 which is gamma gamma u t which is corresponding to 0.2 it happens to 1.021 it makes only 2 percent change you know we will know even we are at the speed of 0.2 c multiplied by 0.2 of course, c have it taken out here in the second term is 0. So, this gives me 0.204 m naught c over f. So, the time taken for the particle to reach a speed of 0.2 c will be given by 0.204 m naught c upon f of course, this factor will be common in all other my subsequent results. I am only will be comparing this particular factor. Now, for the other case let us take u t is equal to 0.4 c and u naught is equal to 0.2 c I put exactly in the same expression of course, now I have to take this factor as well as this factor into consideration. We just substitute those numbers this is very very simple state forward calculation. If I take that number we get 0.232 m naught c by f if my initial speed was 0.4 c and the final speed was 0.6 c then my this I have written as t 2 this is t 3 for the third case is 0.314 m naught c divided by f if my initial speed was 0.6 c final speed becomes 0.8 c after time t 4 which is 0.583 m naught c by f. So, as we can see that with whatever is the factor if you go between 0 to 0.2 the value which I get is 0.204 in this interval I get 0.232 in this interval I get 0.314 in this interval I get 0.583. So, the speed interval is same the force is same, but it is faster for the particle to travel the same speed interval 0 to reach between 0 to from 0 to 0.2 c in 0.204 multiplied by whatever is that factor between 0.2 c and 0.4 c it takes larger time between 0.4 c and 0.6 c it still takes larger time and between 0.6 c and 0.8 c it still takes larger time. So, as the particle is approaching this speed of light we are seeing that cover the same speed to reach to make the same changes in the speed to make the same changes in the speed you require larger and larger time and that is expected because now you are approaching the speed of light and as we know that we cannot really reach the speed of light according to the ledivity. So, as we are approaching here and here with the same force you will find that particle is taking larger and larger time to make same change in the velocities. So, this is what is interesting result that we see from this. So, this is what we have said as one can see that for the same speed interval and a constant force the time interval is different unlike classical mechanics. In classical mechanics if you remember this was standard expression in a kinematic equation of course the we write v is equal to u plus at where v used to be the final speed this I would call it as ut minus u 0 plus at using our nomenclature ut minus u 0 divided by a will be time and this a will always be constant in classical mechanics if the force is constant because mass remains constant in that case anyway. Therefore, the same interval will be covered same velocity interval will be covered in the same time while here it does not happen that is what I have said that the time interval is different unlike classical mechanics further this interval increases as the speed of the particle reaches close to the speed of light. Now, I would like to summarize whatever we have discussed. We discussed the definition of the relativistic force we have said that this force in general does not produce acceleration in the same direction. We discussed special cases of longitudinal and transverse I am calling longitudinal and transverse motion I am not sure whether this is a good way of saying what I mean I think should be clear when the force is applied in a direction perpendicular to you and when there is a one-dimensional motion we consider these two special cases in which the force happened to be in the same direction as acceleration. Thank you.