 Hello. In our last lecture, we had discussed some of the issues of classical mechanics. We started with the concept of a frame of reference and we discussed what we mean by inertial frame. We said that when we talk of Newton's laws of motion, Newton's laws of motions are equally valid in all inertial frames. Therefore, one does not have to be a specific to a given frame while describing our motion. So long, the frame is inertial. Unfortunately, when we came to the electromagnetic theory, our conclusions were somewhat different. We realized that electromagnetic theory, it appeared as if all the frames of reference or all the inertial frames of reference were not equivalent. There are some frames in which the velocity could be given by C is equal to 1 upon under root epsilon naught mu naught. This is what we had discussed because epsilon naught mu naught appeared to be fundamental constants. Therefore, we thought it is probably only one frame of reference in which this speed can be given by this particular expression. Otherwise, the fundamental constants themselves will become frame dependent, which we did not like as an idea. So, we thought that probably there is one preferred frame of reference in which the expression of light is given by the standard formula involving the fundamental constants C is equal to 1 upon under root epsilon naught mu naught. We are not very happy with this situation because it appeared that in mechanics, all the inertial frames seem to be equivalent, but in electromagnetic theory they are not. Is it so that different branches of physics really play a different type of game? Normally, we do not expect. We expect that nature should be uniform. Different parts of physics should tell us the same thing. So, if in mechanics, all the inertial frames turned out to be equivalent, we expected that same thing should have happened. But here are lot of problems with electromagnetic theory. So, this is what I am trying to recapitulate. We had discussed that from mechanical point of view, all inertial frames appear to be equivalent. However, electromagnetic theory seems to suggest that there is a preferred inertial frame of reference in which the speed of light is given by the fundamental constants. We said that we must go back to an experiment to find out whatever we are saying is true. So, the issue which has to be really tested is that, is there any special frame of reference? In fact, this inertial, this special frame of reference we decided to call by ether or it was decided at that time to call it by ether. And we thought that it is only in this particular frame of reference that the expression of velocity of light is given by c is equal to 1 upon under root epsilon of u. So, all we have to do is to test now to find out an experimental mean by which we could test whether this particular high cost is correct. It means that if you go to any other frame which is different from the ether frame, then the speed of light should be different and normally should be given by the standard velocity addition formula which is very well known in classical mechanics. So, essentially what I am trying to say is that if I change my frame of reference from ether, even though the frame of reference is inertial, the speed of light would be different and this will follow the standard additional velocity additional formula which is well known in the classical mechanics. What is the implication of the hypothesis? Let us try to understand what the ether hypothesis implies. Let us take a very simple example and assume that we have an object and a beam of light in ether medium. Let us just try to understand whatever we are trying to say as far as this hypothesis of ether is concerned. So, in this particular figure I have shown one particular object which is here and let us imagine that I am talking of my ether medium. Everything is given in this particular ether medium. This object I am calling as an object O. This V is the speed as measured by in person in the ether medium. The C of course is the speed of light which is also given in the ether medium and because I am talking in terms of ether medium, therefore the magnitude of this velocity must be equal to 1 upon under root epsilon or E1. Remember I repeat that this particular picture is assumed to be in ether medium. This velocity V is also in the ether medium. The C is also in ether medium and because this is in ether medium, therefore the magnitude of the velocity must be given by the fundamental constants. Now, suppose there is an observer sitting on this particular object O. What is the speed that it will measure for this particular light? As we have just now said that this must be given by the standard velocity addition formula which means that if an observer is sitting on this particular object O, he or she would measure the speed of light to be different and that would be given by C minus V. This is a standard velocity addition formula which I will write as Vm, the velocity as measured, velocity of light as measured in the object frame O. This M I have written just to imply that this is the velocity I would measure while sitting on this particular object if my hypothesis of ether is correct. I want to make a claim under this particular hypothesis is that even if V is same, of course the magnitude of C is also same. If I change the direction of speed of light, then both the magnitude and direction of the speed measured in object frame will turn out to be different. This is what I am trying to show by very, very simple vector addition formula. Let us look at this particular figure. This is the same V which I have drawn in the last transparency and this is C which is the speed of light. This is also C which is the speed of light. This is also in ether frame. This C is also ether frame. All that has become different is that in this particular first figure, this C is being shown to be directed in this particular direction while in the second figure, C is being shown to be directed in this particular direction. The magnitude of this particular C is same. This V, both magnitude and in direction are same. Again, I mentioned that this particular object is moving with a velocity V as seen in the ether frame and the C that I am talking is also being measured in ether frame. My question is, what will be the velocity of the speed as it would be measured by an observer sitting on O frame or the object frame? As we have just now said, the formula that we have to use is Vm is equal to C minus V which is this particular formula which I have written here also. It is this particular formula which will give us what is the speed of light as measured in these two cases. In the first case as you can see that this will be Vm. Remember, when I am saying Vm is equal to C minus V, it means that Vm plus V is equal to C and if you look at this particular figure, standard velocity addition formula Vm plus V is equal to C. If I walk in this particular direction, then walk in this particular direction, my net displacement is here. This is the way we normally remember our velocity addition formula. Therefore, I expect here Vm plus V is equal to C. Similarly, here Vm plus V is equal to C. So, this Vm is given by C minus V. As I can see that this Vm is pointed out in this particular direction, this Vm is pointed out in this particular direction and as is also very clear, the magnitude of this Vm is much smaller in comparison to the magnitude of this Vm. So, this is what I was trying to explain saying that if the velocity of light is measured in different direction, I would find that not only the direction which is of course the direction of measurement, its magnitude will also become different. So, one conclusion which I want to draw that if my ether hypothesis is correct, then if I am in any other inertial frame which is also in which I am taking the measurement, I will find that in that frame, the magnitude of the speed of light would be direction dependent. Depending on which direction I measure the speed of light, I will find magnitude to be different. This is what I was trying to say even if I am talking of the same V, even if my object or inertial frame is the same inertial frame. So, this is what I have reiterated in this particular transparency that if ether hypothesis is correct, then in O frame, one should measure different values of speed of light when the direction of light makes different angle with the direction of motion of the object in ether and the measured speeds should be consistent with the velocity addition formula. Now, let us imagine that this object about which I was just now talking is actually Earth. And for simplicity, let us not go into the too much of complications as of now. Assume that this V, this particular Earth is moving with a constant velocity in this ether medium which I am calling as V. We know that this is not strictly spring correct because the motion of Earth around Sun is fairly complicated. It also has a spin moment. But let us ignore all those things. Let us imagine that Earth is really an inertial frame. Let us imagine that at this moment it has only velocity, an orbital velocity V which I assume at moment to be constant because then only I can assume that the Earth is an inertial frame. So, let us imagine that this object is Earth about which I have just now talked. Now, on this Earth, let us try to shine light in two different directions. As we have just now said, I will find that the magnitude of these light, magnitude of the speed of these light which are travelling in two directions would be different. Let us make some specific cases and try to understand this particular thing a little bit. This is what I have said. On Earth, if we send two light signals, one along the velocity direction and another perpendicular to that frame, then they should travel the different speeds when measured on Earth. So, what I am trying to say is that let us choose two directions, one along the direction of the motion of Earth in the ether medium. So, this is my Earth which is moving in this particular direction with a velocity assumed to be constant. Then I sent one light signal along this particular direction which is the direction of V. I sent another signal which is perpendicular to this direction of V. So, one light signal goes like this, one goes like this. Remember, these signals are being emitted with respect to an observer sitting on Earth. So, the speeds of these two will not be seen, but what I have mentioned earlier would be Vm. And as we have discussed earlier, the magnitude of this Vm would be different because they are making different angles relative to the velocity V which is the velocity of Earth in the ether medium. Let us try to make a calculation and try to find out what are the velocities in the perpendicular, in the parallel and the perpendicular direction. My next figure shows the same thing. This is assumed to be Earth which is moving with the velocity V in this ether medium. There is one light signal which is being sent perpendicular to the direction V which I am writing as V perpendicular. There is a light signal which is being sent parallel to V which I am writing as V parallel. Now, remember the velocity of light C as per my hypothesis is C only in ether frame. So, let us suppose this is my C. Remember what I am measuring is this. Let us look at perpendicular direction first. What I am measuring is this particular thing, but this is a result of two velocities, components. One is velocity of light C which is in the ether medium and one V which is because of the fact that this particular Earth is moving with the velocity V in the ether medium. Now, if I go back to my old equation which I had written earlier which is Vm plus V is equal to C. Therefore, this Vm in this particular case would be my V perpendicular. So, I must get V perpendicular plus V is equal to C. So, what I am measuring is this component V perpendicular which is an outcome of a vector subtraction between C and V. Now, let us come to the parallel case. If I come to the parallel case, situation is exactly similar except that this V parallel and V both are in the same direction. But this V parallel must have been obtained as a result of subtraction of C and V by the same formula Vm is equal to C minus V which I have just now written here. Now, in this particular case, my Vm now becomes V parallel and this V parallel must be given by C minus V or C plus V must be given by V parallel. So, if we look here, V parallel plus V must be equal to C. So, this is the velocity component that I am measuring which I have shown here in this particular figure by this V parallel vector. This V parallel plus V vector must lead to C which of course is given by the fundamental constants 1 upon under root epsilon or mu. Let us try to find out what is V perpendicular and what is V parallel. If one knows vector, it is very easy to find it out. All I have done is reproduced these two figures from my previous transparencies. Now, we realize that V perpendicular must be equal to C minus V, same expression which I have written earlier. If I look V perpendicular, this is VC and if I look at the magnitude of this, you will realize that this makes a right angle triangle. This is a hypotenuse. So, C square of magnitude must be given by V perpendicular square plus V square. Therefore, V perpendicular square must be equal to C square minus V square using the standard theorem. Therefore, this is what I have written as the magnitude of V perpendicular must be given by under root C square minus V square where I have used the notation that when I do not write any vector sign on the top of it, I mean only the magnitude of it. So, magnitude of C is being represented by just C without a vector sign. Similarly, the magnitude of V is being represented by V without a vector sign. So, this is just the magnitude. C is also just the magnitude. The magnitude of V perpendicular is given by under root C square minus V square by just looking at this particular triangle which is a right angle triangle. Now, let us evaluate V parallel. V parallel is still simpler because V parallel and V both are in the same direction. So, we realize that V parallel is equal to C minus V by the same formula that we had discussed earlier. So, V parallel magnitude is just simple magnitude C minus V. So, this is what I have written that V parallel magnitude is C minus V. As we realize and as you have been mentioning that this V perpendicular and V parallel would be of course different in magnitude. Now, let us try to do one more thing. We have just now seen what will be the velocities or just now evaluated what will be the velocities if a light signal is sent on earth perpendicular to the V direction and parallel to the V direction. Let us just reverse these directions. Let us imagine that light signal is perpendicular to V but not in the original direction but in the opposite direction. Similarly, for V parallel we will assume that it is not really parallel but anti-parallel to the velocity direction. Let us evaluate these velocities. Solution is exactly similar as before other than that I have reverted this particular vector. So, this V perpendicular which was earlier pointing towards up is not pointing towards bottom. This V parallel which was pointing towards right hand side is not pointing towards the left hand side. That is why I have written here V perpendicular with a negative sign here. So, this is V minus perpendicular. This I have written is V minus parallel. Now, my idea is to calculate this V perpendicular minus and V parallel minus. I will use exactly the same terminology or same method the one which I have used to calculate V perpendicular and V parallel. All that has happened is the directions of the vectors have changed. Again like before these two figures have been copied from the earlier transparencies. Now, you realize that in this case hardly anything has changed because this C is still the happiness of this right angle triangle. This V minus perpendicular is still C minus V. Changing the sign has not changed its magnitude. It is still given by magnitude of C square or the magnitude of this square is still given by the magnitude of C square minus magnitude of V square by the standard theta. Therefore, we write this V minus perpendicular also as under root C square minus V square which has not changed from my previous value. However, there has been lot of change in V minus perpendicular parallel because as you can see from the figure itself earlier minus V my V parallel was very small but now my V parallel has become very very large because to this I must add V to get my C. If I go this much then I must revert back to get this C and because this magnitude of C is same and I have to actually subtract this V from the magnitude wise then only I will be able to get this C. Therefore, it is very clear that using this particular equation my V minus parallel now will be C plus V. So, what we will find out is that this particular velocity in minus parallel direction has increased. It has become C plus V. I repeat in V minus perpendicular and V plus perpendicular there was no change as far as the velocity direction is concerned. It means if I am sending a beam upwards like this and or in the river direction towards backwards it makes no difference it will still give me the same speed of light. However, if the light travels in this particular direction this will travel the different speed then when light is travelling in opposite direction. It makes a difference when light speed of light is being measured in the direction of V. Now let us discuss the famous Michelson-Morley experiment. Michelson-Morley was one of the great experiments which was designed to test the ether hypothesis and this is a very very remarkable experiment because as we shall be seeing it gave a negative result and this negative result itself was a pock making and changed the way we look at physics today. It is a very interesting experiment using essentially the same ideas that we have just now discussed. Let us look at the experimental setup. In this transparency I am showing a rough sketch of this particular Michelson-Morley experiment. By S I mean we have a source of light here and let us assume that this is a monochromatic source of light. We know how to create a monochromatic source of light. Even at that time of Michelson it was known how to create a monochromatic source of light. We have a monochromatic source of light as light is coming being emitted by this. For the moment let us assume that this particular direction is the same direction as the direction V that I have been talking which is the direction of velocity of earth in ether medium. For simplicity we can discuss the other cases later. Now here we have what I have written M is a half-silvered mirror. By half-silvered mirror we mean that it does not reflect all the light. It reflects some of the light or approximately 50% of the light and 50% of the light it allows to be transmitted. And if I put this particular mirror at an angle of approximately 45 degrees from the direction of the velocity of light then we will find that approximately half of its light will travel in this direction which was the original direction of the light while half of the light or half the intensity of the light will travel in a direction perpendicular to the original direction of light because the angle of incidence has to be angle of reflection therefore and I have taken this particular angle as 45 degrees or I have designed this particular angle as 45 degrees therefore this light beam now would travel at an angle of 90 degrees from my original direction. So this beam is divided into two beams one traveling upwards one traveling towards right and as I have said I have assumed that this right direction is actually the direction of V. This direction therefore is a direction perpendicular to V. Now here we have two mirrors one is M1 another is M2. These two mirrors are the normal mirrors they are not half silver mirrors so they will reflect most of the light which is being incident on them. And let us assume that these two mirrors are being put in such a way that this light is normal to them one is incident normally to them. If the light is incident normally to them the light will be reflected back and therefore this light will go up and will come back. Similarly this particular light or towards the right hand side will also go and will get reflected from this particular mirror M1 which is also a fully silvered mirror and this mirror will reflect the light the light will go retrace its path backwards and will keep on coming backwards and again these two sources of light would meet this half silver light. Because this mirror is half silver mirror because this mirror is half silver again some of the light from this particular beam will be reflected in this particular direction and some may be transmitted let us ignore that particular beam. Similarly the light which is coming back from this particular direction part of the light will be transmitted part will be reflected let us ignore the reflected beam in this particular case. Let us look only at this particular beam which is now resulting as a superimposition of two beams one which is reflected from M1 and again reflected from M another one which is reflected from M2 and then refracted from M. Now this particular beam is being observed by a telescope this is what is standard Michelson Morley experiment ok. We know that about Michelson there is Michelson has also devised an interferometer which is very well known as Michelson interferometer using essentially somewhat similar concepts in that sense the ideas are very similar. Of course here we are testing something which is very different. Now let me try to calculate what will happen to these two beams of light one which has gone upwards and got reflected from M2 and the one which has gone to the right and got reflected from M1. We realize that when the light was travelling upwards it will travel with the same speed as when it was travelling backwards and of course both these speeds will be different from the speed of light in ether as far as this particular light is concerned because this is travelling along the direction of V we realize that this velocity is going to be different from this particular velocity here which is travelling backwards because this is in a direction opposite to V. What I will do is to calculate the time that this particular light will take this beam will take to travel from this distance to this distance and come back and the light which travels from this distance from here this distance and come back and let us assume that these two lengths are same just for easiness let us assume that the two lengths are same so the lengths are same only thing which is different are the speeds so we calculate how much time the light would take once it is reflected from the original beam to go upwards and come back here and going this way and coming back here because these are the only two things which are different otherwise everything is same once they have come and joined this particular point the two beams are travelling together. Let us do that particular calculation and find out the time difference. This picture shows what are the velocity directions what are the velocity magnitudes which is just now what I have mentioned that when it travels upwards it travels with a speed of under root c square minus v square when it travels backwards it travels exactly with the same speed under root c square minus v square. On the other hand this particular beam when it travels in this particular direction it travels with a speed of c minus v when it comes back it travels with a speed of c plus v. Now we know the standard expression the time taken by a particular beam will be length divided by the speed. So let us first calculate the time taken for the light to travel in the parallel arm the arm which I have shown to be along the direction of v which is the relative speed between the earth and the ether medium. So as we have said the first beam it is travelling to the right it travels with a speed of c minus v. So the time that this should have taken if we assume that the length is l then it will take a time of l divided by c minus v to go from this particular original mirror let me write it here this was my mirror m this was my other mirror fully mirrored silvered mirror I am first calculating the time that it takes the light beam takes to travel this way this length I have assumed it to be l this is speed as I have said was c minus v. So the time taken for this will be l divided by c minus v then when the light beam travels this way the speed becomes different the length is same this is speed c plus v so this time will be l divided by c plus v. So the time taken for the light to travel from this point to this point is l divided by c minus v the time taken for the light to come from this point back from this point is l divided by c plus v. So the total time taken by the light to go from this point to here and come back will be the addition of these two times which is what I have mentioned here, L divided by C minus V plus L divided by C plus V. Now, let us simplify it a little bit. We just multiply both the sides by this side by C plus V numerator and denominator, this side also by C minus V both numerator and denominator. Then our denominator becomes C square minus V square, a standard algebraic simplification. There will be C plus V here, there is a C minus V here because this side I have multiplied and divided by C minus for V. This side I have multiplied and divided by C plus V, numerators and denominator. So, there is a C plus V remaining here and there is a C minus V remaining here. As we can see very easily, this V and this V both cancel. So, in numerator what will be remaining is only 2C, 2C divided by C square minus V square. This multiplied by L will be the time taken for the light to travel along the parallel arm. This I have simplified it further. This is what I have mentioned, L is equal to 2C divided by C square minus V square. I have taken C square common from this denominator. So, this C square I have written out here. This 2C I have taken out here. So, the numerator here becomes 2L multiplied by C because this 2C have been taken out of this particular bracket. So, what is remaining is only 1 here. C square has been taken out of this particular bracket. So, here what remains is 1 and this gets divided by C square. So, what is remaining inside this bracket is 1 divided by 1 minus V square by C square because there is a C here and there is a C square. So, I have cancelled one of the C and this expression is given by 2L divided by C multiplied by 1 divided by 1 minus V square by C square. So, this is the time light would take to travel along the parallel direction in going from half silvered mirror to the silvered mirror and coming back. I will simplify this equation a little further by assuming that V is smaller than C much smaller than C which is actually correct that the speed V whatever we are talking is much smaller than the speed of light. So, under that particular approximation this particular expression can be further simplified and because I have to compare this particular result with something else that is why I am looking for this particular simplification. So, what I have done this expression which I had written earlier 2L divided by C 1 divided by 1 minus V square by C square. I have written this particular expression as 2L divided by C which is still remaining there and 1 minus V square by C square to the power of minus 1. This is a very standard mathematical way of writing it. So, I write this as 2L divided by C 1 minus V square by C square to the power of minus 1. The advantage of writing this in this particular form is that I can use a binomial expansion and expand this into a series and because the series will contain higher orders of V square by C square I can neglect higher orders and retain only the first term and by binomial expansion this minus 1 will get multiplied by here. Therefore, what will be remaining is 2L by C 1 plus V square by C square. Of course, this is an approximation which we must realize because this is only correct and the limit that V is much smaller in comparison to C. This is what I have written in this particular transparency that this particular time turns out to be 2L divided by C multiplied by 1 plus V square by C square. Now, let us try to calculate the time in the perpendicular alarm. Now, we realized that the speed and the perpendicular alarm in both the directions, whether upwards or downwards is same which is given by under root C square minus V square. Therefore, time can also be obtained in a very simple fashion which is just 2L divided by under root C square minus V square which is what I have written here. Time in perpendicular direction this two factor is because the beam goes up and comes back and of course with the same speed therefore, this is 2L under root divided by under root C square minus V square. Again I am trying to simplify the same time by using the same approximation that V is much smaller than C. Therefore, I have decided to write this under root in this particular form. I have taken 2L by C out and written this particular thing in bracket as 1 divided by under root 1 minus V square by C square just taking this C square out of the under root. Once I take C square out of the root then this will become under root C square which is just C. So, what I am getting here is just 2L divided by C multiplied by this particular quantity. Remember here there is under root that is what is the difference from the earlier expression. Now, we know that this particular under root can be written as to the power of minus half. So, this is what I have written here. This is 2L divided by C multiplied by 1 minus V square by C square is everything to the power of minus half because under root is nothing but to the power of half and because this is in denominator. So, there is a negative sign and this becomes to the power minus half. Now, I do the same trick. I expand this into binomial series. Therefore, this particular factor and of course neglect higher order terms just retaining a term involving V square then what is happening? I must multiply this by minus half. When I multiply by minus half this minus becomes plus and this half becomes now V square by 2C square. If you remember in the earlier expression there was no 2C. There was only V square by C square. Here we are having V square by 2C square. So, we find that time perpendicular time taken by the beam in perpendicular direction is different from the time taken in the parallel direction. This is what I was trying to conclude. Let us calculate the time difference. The time difference which I have written here is T parallel minus T perpendicular. Of course, we will realize that time taken in the parallel direction is more than in perpendicular direction because here we had only V square by C square while here we had V square by 2C square because I have divided by 2. So, this expression has become smaller. Now, I am taking the difference of the 2. So, I have just taken 2L by C which is a common factor out and this T parallel we had 1 plus V square by C square. This T perpendicular we had 1 plus V square by 2C square because there is a negative sign everything becomes negative. So, this becomes minus 1 minus V square by 2C square. So, this sign 1 plus and minus 1 cancels out and you get 2V square sorry V square by 2C square because there is a V square by C square and minus V square by 2C square. So, this will give me V square divided by 2C square. This 2 factor will cancel with this 2 and this is what I will be getting. We will just write it here. Here 2L by C minus 1 minus V square by 2C square this 1 cancels out. So, what I will be having is 2L divided by C into V square by C square minus V square by 2C square. This will give me 2L divided by C into V square by 2C square. This 2 will cancel with this 2 and this is what I will get L divided by C multiplied by V square by C square. This is what I have written. So, the time difference between these 2 is L by 2 multiplied by V square by C square. Now, let us come to an actual experimentation. It is an actual experimentation. These angles that we are talking are not really especially the mirrors M1 and M2. They are never perfectly normal to the beam. They are making very small marginal angle difference. So, what happens the beam when it goes and comes back, the different parts of the beam are slightly deflected more or less and therefore, you find there is a continuous path difference between these 2. Now, this beam M1 which is reflected from M1 and the beam which is reflected from M2 because they take different time to travel. Therefore, when they come and superimpose on each other and are being viewed by the mirror, you will find just because of the small angle, you will find fringes. So, the fringes will be observed even if there is no time difference. You will always see the fringes. Now, because of this particular time difference, there will be a shift in the fringes because now the time difference will cause a small amount of shift in the original fringe pattern had there not been any time difference. So, what you would see is a bright fringe and a dark fringe. It means a bright portion of light where there is a light is bright and then there is a darkness. This is what is called standard fringes. So, these fringes would be seen, but because of this additional path difference, the fringe pattern would shift. This is what I have written here. If the mirrors M1 and M2 are slightly inclined, they would create a gradual phase shift and one would observe a fringe pattern in the telescope. The time difference in the arrival of the waves in 2 arms would cause an additional phase difference. So, there is already some phase difference and the time difference will be on the top of it. But the basic problem is that if I want to really check, if my hypothesis is correct, then I must be able to check and show that there is really this much of time difference between these two beams. If I am able to check it, if I am able to verify it, then I have proved my hypothesis. If I have not been able to do it, I have not been able to prove my hypothesis. So, the question is very simple, but there is a problem. The problem is that we have said whenever, even if there was no change in the time taken by the light, the two beams would have taken the same light, even that there would have been a fringe. Now, because of this time difference, there has been an additional shift. But how do I measure the shift? Because it is not possible for me to create a situation when these two beams take the same time, because this beam, this particular whole pattern has been devised with taking the motion of earthen to consideration. So, it is not really possible for me to what I call a find is zero of this particular feature, saying that this is where my original fringe should have been there. And now because of this, it has shifted by this much amount. Because it is not possible for me to create a situation or it is not possible for me to find a situation when there is no time difference between the two beams. The time difference has been created because of the motion of earth in ether medium. So, how do I find zero? So, how do I find whether there has been real shift? So, this particular problem was solved very nicely with a brilliant idea by Michael St. Morley. What they said has been shown in this particular transference. They said, let us try to rotate the whole apparatus. What will happen if I rotate? So, let us assume that we are rotating about this particular axis normal to this particular picture. So, this particular telescope has now come here. Sorry, this particular source has come here. This particular telescope has come here. This mirror has been rotated by 90 degrees. This particular mirror, which I was calling is M1 or M2 have now changed their position. There is one mirror here, there is one mirror here. Now remember, this was my original arm in which I was my life was travelling. Now, this has gone here after rotation of 90 degree. So, this arm which was originally parallel to the velocity direction has now become perpendicular to the light direction. This arm which was perpendicular because it has now been rotated by 90 degrees, this particular mirror has come here. So, this particular arm which was earlier perpendicular has now become parallel. So, because of this particular rotation, an arm which was originally perpendicular has now become parallel and the arm which was parallel has now become perpendicular. So, what has happened? A arm in which the light was taking earlier a larger time. Now, after 90 degrees of rotation, in that particular arm, the time taken will be smaller. Similarly, in an arm in which the light time taken was smaller, now will become larger. Now, if what I am trying to say is correct, then by this rotation, I would be able to see that fringe is now moving and if I am able to see its motion, then I have verified my hypothesis. So, this is what they did, is trying to create a rotation in this particular Michael St. Morley experiment to see this particular fringe shift because they were not able to decide, they were not able to find out how to check the finite fringe shift. So, what I said in a rotated position, the time difference on the same arm will be same, but now with a negative sign because the arm in which it was positive, now it has become negative, it is taking longer time. So, the time difference magnitude wise will be same L by C multiplied by V square by C square, but this time difference will be negative. And if I rotate by 90 degree degree, then the total rotation will be just the difference of these two, which because of negative sign essentially implies an addition. Therefore, the total phase difference, the total, I am sorry, the total time taken would now be 2L V square, 2L multiplied by V square divided by C multiplied by C square. So, this will be the time difference between the two arms. This would cause, as I mentioned, a shift in the fringe pattern. So, as I rotate, if I keep on rotating, I will see my fringe is getting shifted and if I am able to observe this shift in the fringe pattern, essentially I have proven my hypothesis. Now, in order to find out how much is the fringe shift, I must divide this particular time difference in comparison to, I must divide by the time period what I say for a particular oscillation and this particular time period is inverse of the frequency. If you are aware, the frequency, the speed is given by frequency multiplied by lambda, which is the wavelength. This frequency I can write as 1 divided by T multiplied by lambda. Therefore, this T would be given by lambda by C. This is the time period of one of the oscillation in a harmonic oscillator, that is what we normally call it. So, this time difference must be compared with the time of one oscillation, must be divided by this. So, this is what I have done, delta T minus delta T prime must be divided by T and this T would be given by C by lambda. If I calculate this particular thing, this I get as 2L divided by lambda multiplied by V square by C square. So, this will be the amount of the fringe shift that we would be observing as a result of this particular experiment. Of course, this particular experiment has been performed number of times, but in one of the original experiments, these were the standard values which were taken. This L length was taken as 11 meters, not very large. The lambda was taken approximately as 5.5 into 10 to the power minus 7. And the speed of Earth in ethermuseum medium was taken as 10 to the power minus 4 C. If I give these numbers here, I will find that the fringe shift turns out to be equal to 0.4. I have just substituted the numbers in the earlier expression which I have written earlier. You find that there will be approximately half a fringe shift. Michael Sanmorely had done an error analysis and we are very sure that if such a fringe shift was really there, was existing, they will be able to observe this particular fringe. However, they never observed it. In fact, this particular experiment was repeated number of times in different seasons because you are never sure what is the direction of ether, whether earth is, how the earth is floating in ether, different seasons at different positions. It was never found to be true. We never found out a fringe shift. As I mentioned that this experiment had been repeated by, I mean, even in current years, not, I would say not very recent years, but even this particular, essentially this particular experiment has been repeated by better precision. And every time we found that the result is negative. There was no fringe shift which was observed. Appears that there is something really wrong with the ether hypothesis. This is what I have mentioned. That experiments were performed in various seasons but never gave positive results. Then there are a lot of attempts to explain why we are getting negative results of Michael Sanmorely and none of these explanations were really found to be very strong. We could never believe that Michael Sanmorely experiment could be understood, the failure of Michael Sanmorely experiment could be understood by these explanations. I would like to mention that other than Michael Sanmorely experiment, which is basically an experiment based on light, there have been other attempts also. One of the standard attempts is to look because if you remember in our last lecture we had mentioned that the basic starting problem was V cross V when we are talking about the Lorentz force and we said that what is this V that we are talking. At that time we said that probably this V that we mean is also the speed of that particular object, that particular charge in ether medium. So, if Earth is actually moving in ether medium, then if I put a particular charge, then it would have a V and therefore, in a given velocity it should experience a force of V cross V. So, there were some attempts which were made in order to assume a force on a given charge, all these attempts led to a failure. It appeared that there is no ether, it is nothing like a ether medium. So, this is what is the summary of whatever we have discussed today. Experimental attempts to look for motion of Earth in ether, they are not giving successful results. It did not appear that you could assign an absolute velocity to an object. Thank you.