In our last lecture, we had discussed the velocity transformation formulas. We eventually started from Lorentz transformation, putting the transformation into a differential form, then eventually evolved and worked out the velocity transformation. Then we also discussed an example of relative velocity to emphasize what we mean by relative velocity. In classical mechanics, sometimes we are not, we are a bit casual in defining relative velocity, but relativity one has to be very careful that when we say a relative velocity, it means we have to make a transformation to another frame and find out the velocity of the particle and this relative velocity would never exceed the speed of light. Then we also discussed a particular example in which there are three different frames of references were involved. We have been earlier doing generally two frames of references. So, to make problem little more difficult, we had involved three different frames of references and saw that how one has to take care of looking into transformations from one particular frame to another particular frame of reference, but so long we do things in a proper way, we should not have any difficulty. So, this is what I have written that let we recapitulate. We worked out the velocity transformation. We discussed examples of a relative velocity and those involving transformations between three frames of references. This is what is the velocity transformation equation. Just to recall, again I mentioned that v is the relative velocity between the frames of references. u x is the velocity of a particular particle which need not be constant. It is the x component of the velocity of the particle. We are supposed to know u x, u y, u y is the y component of the velocity of the particle as measured in S frame and u z is the z component of the velocity as measured in S frame. Then the velocity of the same particle measured in S frame of reference, the components will be given by the x component will be given by this equation, the y component will be given by this equation, the z component will be given by this equation. What is interesting here is that if we look at u y prime, it also depends not only on u y, but also depends on u x.   We also discussed inverse velocity transformation. It means if the velocity components are given in S frame of reference, how to find out the velocity components in S frame of reference? As we have always been telling the prescription is very simple that instead of v use minus v change prime quantities to unprime quantities and unprime quantities to prime quantities. So, this is what basically we had discussed in our last lecture. In today's lecture, we will discuss one example which is little more complex, little more complicated involving multiple events and multiple frames of references. The idea of discussing this particular example is to see what type of care we must do or what is the way in which we should attack these problems when there are multiple events and multiple frames. As I have been telling that first thing that one has to look is that be consistent in your frame of reference. We should not mix up the frames of reference which many times in classical mechanics we are not very careful. So, we should picturize the scene looking purely from the point of view of one frame of reference. Then the chances are less that we will make mistake. The second thing that we do is to identify the events. What are the important events? Try to write the coordinates of those events including time. Time also we are treating as a coordinate and these coordinates once we have written in a particular frame of reference, let us try to find out whether we have got all the coordinates. If we have not got the coordinates, whether we have got an equivalent information in some other frame of reference, then we keep on making transformation and eventually we fill what we call as an event table. It means all the events should be known in all the frames of references. Their coordinates should be known in all the frames of references. So, once we have defined these events, the coordinates of these events must be known in all the frames of references that are of interest to us. So, let us look at this particular example. The example statement is little longer. Let us go slowly and try to understand it. A person gets sick while moving in a train. Of course, and because we are dealing with special theory of relativity, everything moves with a constant velocity here. There is nothing like and nobody stops, you know, everyone is moving with constant velocity because that is the way we solve the problem. Because once we, if a person stops or anything happens like that, you know, then it is no longer in per view of special theory of relativity because it does not involve a inertial frame. So, all of our velocities are constant and they never change as far as the examples are concerned. So, there is a person who has got sick while he is sitting in a train and of course, train is moving with a constant velocity and it is a relativistic velocity. And this speed is 0.6 c in plus x direction as seen by an observer in a given frame. Let us assume that this is a ground frame or earth's frame which we also assume to be inertial.  So, let us put in a specific symbols. So, because we are talking of person, so I am calling this as a p-frame. Now, this person once he gets sick, he sends a distress signal by emitting light. So, he emits light asking for help and at that particular instant of time when he gets sick and he sends the signal, he is 30 kilometers away from the origin in the plus x direction. That is what we have written. A person gets sick while moving in a train, the speed of which is 0.6 c, which is obviously relative to the ground. In the positive x direction, call this frame p. He shines light asking for help when he is 30 kilometer away from the origin in plus x direction. Now, there is a hospital which is located at the origin. Let us assume it is again the ground frame and we call this particular frame of reference as h-frame because this is the hospital frame which is actually same as earth frame, if you want to call it. So, hospital is located at the origin and as soon as this particular person receives the light signal, he tries to look for an ambulance and he finds by that time fortunately that there is just an ambulance which is just passing by him and going in the direction of the sick person, but with a little larger speed 0.8 c. So, he instructs the person that or he instructs the ambulance that you go and help the person. That is what is the problem. A hospital is located at the origin called the h-frame. As soon as the light reaches the hospital, an ambulance called a-frame, a person sitting in ambulance I am calling that as an a-frame. So, there are three frames. There is a p-frame which is a person's frame who sends a signal. There is h-frame which is hospital frame, which I am calling as earth frame and there is an ambulance frame, a person sitting in the ambulance which we are calling as a-frame. So, an ambulance called a-frame is found in front, which is moving in plus x direction with a speed of 0.8 c. So, earlier speed of the person was 0.6 c. Now, the person in the hospital instructs the ambulance to go and help the sick man. So, they go and help that person. Assume no time delay between receiving the signal and giving the instruction and no slowing down of car and train. So, let us assume that as soon as the signal is received instantaneously without any time delay, he could instruct the driver of the ambulance that go and help the person. And the ambulance person immediately rushes towards the sick man. Anyway, he was anywhere going and there is no slowing down anywhere. So, whatever are the speeds, they are always constant because that is what is necessary to maintain that all the frames of reference that I am talking are inertial frames of reference. So, no slowing down and no time delay in conveying the information from the hospital frame of reference to the driver or the ambulance frame of reference. Now, these are various questions which have been asked. Question number one, what are the positions and times of the person sending the signal in h and a-frame that is the hospital and the ambulance frame of reference, what is the position? Some of them will be simple questions, some of them will be little more difficult questions. Let us see. The second question which has been asked, find the position of the person in h-frame which is hospital frame at t is equal to 0 and in a-frame which is the ambulance frame at t prime equal to 0. Number three, find the time in h-frame and a-frame when the ambulance reaches the person. So, eventually the ambulance reaches the person and we have to find out what is the time in h-frame and a-frame. This is basically the problem that we are going to work it out. So, I have sort of picturized this particular thing in this particular transparency. So, this particular person, this is sort of a generic figure. So, just to sort of understand, we will draw a specific figures relevant to a particular frame of reference little later. So, this is a hospital which is supposed to be situated on ground and this we call as h-frame. This is my car which is moving with a relativistic speed of 0.6 c going towards the right which I am calling as p-frame because this person is sitting in this particular train and that is where he gets sick. Let me call this is my ambulance which I am calling as a-frame which is also moving to the right with a speed of 0.8 c. Of course, this information of 0.8 c, this speed of 0.6 c both are given in h-frame which is a ground frame that is the way we have described this particular problem. Now, we have to ask why we have to answer various questions as we have discussed just now. So, this is just a picture to give you an idea and the distance was 30 kilometers at the time when the light signal was sent from this particular person towards h which we call as a distress signal just to give the information that there is something wrong with me please come and help me. As I said our first job in all these problems is to identify events. So, which are the events which are important in this particular case and then eventually start looking and filling the event table by putting the coordinates of all these events. So, if you look at the statement of the problem and think a little bit, you will realize that there are three events which are important in these events. In this particular problem, there are three events which are important in this particular problem. First which I call as event number one is the sick person sending the distress signal. So, person is sick. So, first thing that he does is to send the distress signal. He sends the signal that is the event number one. Then the signal is received by the hospital that is event number two. So, hospital receiving the signal is event number two. Of course, as we have said that it is the same time he instructs the ambulance. So, instructing the ambulance etcetera everything occurs exactly the same time. So, they also they are all included in that event number two. So, even number two you also find that the ambulance is just in front of hospital that is also an event number two. So, event number two comprises of various things in that sense. Now, event number three is that actually the person who is sick, he is actually able to the ambulance is able to reach that particular person.  So, these are the three important events as far as this particular problem is concerned. Sick person sending the distress signal, event number one, hospital receiving the signal, event number two, then ambulance reaching the sick person, this event number three. Now, let us choose one particular frame of reference and try to write the coordinates of all these three events in that particular frame of reference. If you realize that in this particular problem, most in fact all the information has been given in H frame that is the hospital frame or the ground frame. So, it is much easier to write the events in this particular frame itself. So, what I will do to make it systematic, it is much better to go systematically while solving relativity problems. What we will do is to write the coordinates of these three events in H frame, the hospital frame that is the ground frame. So, that is the way first we will do, then we will look into other frames and then we will go ahead with the particular problem. So, let us first try to write the coordinate of event number one in H frame, the hospital frame. It has been given in this particular problem that the light signal was sent by the person when he was at a distance of 30 kilometers. So, obviously and of course, we have said that hospital is situated at the origin. So, the coordinate of this particular event, the x coordinate of this particular event is obviously 30 kilometers. That is very, very clear. We should also look at about the time aspect because as far as the question is concerned, though this is nothing specifically has been mentioned about in the problem of time, but we have to choose an origin for time and looking at the Lorentz transformation, we have to choose the time when the origins coincide as we have always said. So, it is quite simple to realize that most of the problem is given between H and A frame, the hospital and the ambulance frame and it so happens that when the light signal is received in H frame that is also the time when the ambulance is passing by its side. So, let us call that time as t is equal to 0 in H frame and that will also automatically turn out to be t prime equal to 0 in A frame that is the ambulance frame. So, time is chosen at that particular instant. But we realize that though the hospital received the signal at time t is equal to 0, the event must have occurred before because light took finite time to reach this particular person. So, as we have said in our earlier lecture that it is not important when I get the information, what is the of importance at what time the signal or what time the event took place. For example, if somebody is departing from railway station or one particular by a flight and somebody comes and tells me that some of your relatives is going away and somebody has gone to see him off, then he comes back, he says, oh flight was in time, really departed at 9 o'clock. It does not mean that you got the information. Obviously, you did not get the information at 9 o'clock, you might have got up at 10 o'clock or half past 9. But it does not mean that the event occurred at 9, half past 9 or event occurred at 10 o'clock. You know that event occurred actually at 9 o'clock. So, similarly, once I have received the light signal in my watch, time t is equal to 0. But I also realize that this light would have taken finite amount of time to come from 30 kilometers. So, this event must have occurred earlier. So, when I have to find out the time of this particular event, I have to take this particular factor also into consideration. So, this is what I have written in this particular transparency. That in H frame, event number 1 occurred at because the event number 1, so I am calling it this as 1, at 30 kilometers which an SI system turns out to be 3 into 10 to the power 4 meters. So, the x coordinate of the event number 1 is 3 into 10 to the power 4 meters. The time for this particular event has to be found out by finding out how much time light took to travel to the hospital. And as I know the distance is 30 kilometers, the light must have taken 30 kilometers divided by its speed that much time to reach there. And obviously, this event would have occurred in negative time because time was 0 when the light signal was received at the hospital. So, therefore, this time t 1 will be equal to 3 into 10 to the power 4 which is the distance at which the event occurred divided by the speed of light. And if you calculate this number assuming c is equal to 3 into 10 to the power 8 meters per second, you get this time t 1 as minus 1 into 10 to the power minus 4 second. It means this event must have occurred at time 1 into 10 to the power minus 4 second before the light reached. So, this is a point which has to be realized that what we have to look is at the time when the event actually took place. And that time is a negative time even though the person at the hospital received the signal at time t 0 to 0. So, I have found out what is t 1. Now, let us look at e 2, the second event for which to find out the x coordinate and time coordinate is very simple. Because when he received the signal, obviously, he was at its origin. He was in the hospital. The person who was the person who received the signal at the hospital, he was obviously sitting in the hospital and therefore, was at the origin. And we have already said that when he received the signal, time was equal to 0. So, the x coordinate of the second event is 0 and the time coordinate of the second event is also 0. It is the most simple. That is what I have written here. e 2 is equal to x 2 is equal to 0. The second event, so I put 2 here. Second event, so I put 2 here. So, x 2 is equal to 0. t 2 is equal to 0 is the coordinate for the event number 2. Of course, you have said given that the hospital is the origin which has already been given in the problem. Now, let us come to event number 3. This is a little more tricky. But remember, everything has been given in the H frame. So, it is not that all that tricky. In fact, you do not require Lorentz transformation here. So, I still do not know what at what value of x the person actually caught.  Also, we do not know the time we have to work it out. I realize that during the time light was approaching the hospital, this person must be moving to the right. Let me just draw a figure to explain this particular thing. So, if we say that this is my hospital and at t is equal to 0, this is where was my train. This distance was 30 kilometers, but light took certain amount of time to reach from here to here.  So, at time t is equal to 0, this particular train would be somewhere on the right hand side. So, this is the situation when this particular person sends a light signal. It took certain amount of time for light to reach here. During this particular time, this particular train compartment move to the right and be somewhere here. Then there was an ambulance which went and tried to catch this particular person. By the time this particular person in the ambulance goes and catch this person in the train, this event obviously would have occurred at a further further of value of x, because this person continuously keeps on moving with a speed of 0.6 c. So, let us first find out the position, this particular position at time t is equal to 0, where the person was situated at time t is equal to 0. Remember this distance was 30 kilometers. So, if distance was 30 kilometers, all I have to find out that during the time light travel from here to here, how much this person went from this side to this side. I have already known the time that light took to reach here. This train is moving with a speed of 0.6 c in this particular direction. So, 0.6 c multiplied by the time will be this distance. That is what I have calculated, added this to 30 to get the distance at t is equal to 0. So, that is what I have written, the position of the person at t is equal to 0 is this original 30 kilometers plus the time that light took to reach the hospital, multiply by the speed of the person which is 0.6 c. If you just add these numbers, it turns out to be 4.8 into 10 power 4 meters. It means at that instant of time when the light reached the hospital, this particular person was at a distance of 48 kilometers, his coordinate was 48 kilometers. But remember this is still not x3, because x3 involves catching the person. It will take still further more time before the ambulance will be able to reach, because this is the time when ambulance was still at the origin of H. So, it will still take further a certain amount of time for the ambulance to reach this person. Therefore, obviously, what I have calculated here as 48 kilometers is not x3. I have to still find out x3. So, this is the picture which I had drawn on the paper. Same thing has been drawn here. At time t is equal to t1, he was at a distance of 30 kilometers. At time t is equal to 0, it had gone 18 kilometers further off. So, this distance was 18 kilometers. And at really time t is equal to t3, when this ambulance will come here and catch him, the distance has to be further off. So, I have to find out this time t3, and I have to find out this value of the coordinate, not this value of the coordinate. So, this is, I have just found as an intermediary step. But eventually, the event is this particular person being caught by the ambulance, and that will occur at a different value of x. So, let us try to find out that particular value of x.  Of course, this time is also in H frame. Remember, at this particular moment, everything is being talked in H frame, no other frame. I am not confused with any. I have written in the first transparency H frame, and all those things are being talked only in H frame. I am not looking at any other frame. I am not even thinking about any other frame, the A frame or the P frame. I am only thinking about H frame, trying to do all calculations in H frame. Once I try to describe into different frame, then I will think only in that particular frame. I will not think in terms of H frame. This is something which is very, very important not to confuse between the frames when we are trying to work out the problems of special theory of relativity. So, let us assume that time in the watch of the person sitting at the hospital was t3. When the third event took place, it means when ambulance reached the sick person. Already, we had agreed that this distance was for t8 kilometers. And at this particular time, the watch of this particular person showed time t equal to 0. And if time t is equal to t3, it means during this particular t3 time, this person was also going in front with a speed of 0.6 c. While this particular ambulance, which is here, was travelling to the right with a speed of 0.8 c. And in the same time t3, the distance covered by this particular person, this particular person in the sitting in the ambulance will be same, will be this to this. And in the same time, this particular ambulance will travel a distance from here to here. Original distance was 48 kilometers. So, 48 kilometers plus in t3 time whatever distance this particular train travels, that will be this distance. And that distance must be same as the distance travelled by this particular car in time t3. Of course, this car travels with a speed of 0.8 c. Hence, this distance travelled by car in time t3, this distance travelled by the train in time t3 plus original distance 48 kilometers must give me, must be equal and therefore, I can evaluate t3. This is what I have written in this particular transparency. If t3 is the time taken when the ambulance reaches the sick person, then the original distance 48 kilometers plus the distance moved by the train in time t3 will be 0.6 c multiplied by t3 because the speed is 0.6 c must be equal to, because as far as the ambulance is concerned, it also travels with the same time t3, but with now a speed of 0.8 c must be equal to 0.8 c multiplied by t3. So, from this you can calculate the time t3. If you take this to the right hand side, 0.8 c minus 0.6 c becomes 0.2 c. So, which comes here in denominator and then I calculate time t3, which turns out to be equal to 8 into 10 to the power minus 4 second. So, this is the time for the even number 3 because this is the time when the person is actually caught by the ambulance. So, t3 that is why I have written this as t3. This is the time for third event, which is taking place at time 8 into 10 to the power minus 4 second. Now, if I have to find out the x coordinate, if I have to find out this particular distance from the origin, this will be just t3 multiplied by the speed or 48 plus 0.6 c multiplied by t3. So, any of the method will give me the coordinate x3. That is what I have done in the next transparency. The coordinate of the person at the time of event 3, that is the ambulance catching this particular person, sick person occurred at a time 0.8 c is this speed of the ambulance multiplied by the time, which we have just now calculated, which was here 8 into 10 power minus 4 second. This is what I have calculated. If I multiply this by this, I find out the distance at which this event or the coordinate at which this particular event would have occurred, which is 19.2 into 10 to the power 4 meters, which happens at 192 kilometers is quite far off. So, at the distance of 192 kilometers or at the coordinate of x3 is equal to 19.2 into 10 to the power 4 meters, event number 3 occurred. Now, I have filled this particular event table in H frame, the hospital frame. So, which I have written here in the my next transparency, this is event stable in H frame of reference. Event number 1 occurring at a distance of 30 kilometers and at a time of minus 1 into 10 to the power minus 4 second e2 simple 00 e3 occurs at a distance of 19.2 or at a coordinate. It is better to say coordinate. It occurs at a coordinate of 19.2 into 10 to the power 4 meters and at a time of 8 into 10 to the power minus 4 second. So, I have filled the event table in H frame of reference. Once I have got all these information in one frame of reference, then it is a very simple job to transform these information to any other frame. All you have to do is to use Laurence transformation. So, now let us try to fill this particular table in A frame, the ambulance frame because H and A obviously satisfy all the conditions or all the criterion which we have imposed for using the Laurence transformation. Just simply use Laurence transformation and I get all these events into A frame. So, that is what I have been doing here. I go to A frame. So, now stop thinking about H frame. Start thinking in terms of ambulance frame. Make the frame H and A satisfy the criteria to use Laurence transformation directly, which essentially mean that I have to put t prime equal to 0 exactly at the same time when the origins were concerned. And also I assume that that was the origin of the ambulance because I have to transform from these two frames. So, I have to use the relative velocity between these frames, which is a relative velocity between the hospital and the ambulance frame because the velocity has already been given in the H frame. So, same is the relative velocity because that is how we have defined relative velocity. So, v is 0.8 c. Substitute in this expression for gamma. c square will cancel with this c square. This gamma turns out to be equal to 5 by 3. So, the gamma that I am going to use in this particular equation will be 5 by 3 and v will be equal to 0.8 c because I am going from H frame to A frame and the relative speed between these two frames is 0.8 c. Even the first event, it occurred a distance of 30 kilometers. So, this is x minus vt because t is negative minus 1 into 10 to the power minus 4. So, this becomes plus. So, 3 into 10 to the power 4 plus 0.8 c into 10 to the power minus 4 multiplied by gamma. So, gamma x minus vt. That is what is going to give me x 1 prime which turns out to be equal to 9 into 10 to the power 4 meters. It means according to an ambulance frame, this event occurred at a coordinate of 90 kilometers, 9 into 10 to the power 4 meters. Same to time transformation, t 1 prime will be equal to 5 by 3. The time which is minus 1, it is gamma t prime minus vx upon c square. So, this is t which is minus 1 into 10 to the power minus 4. v is 0.8 c. x was 30 kilometers 3 into 10 to the power 4 meters divided by c square. Substitute the value of c. You will find out that this time is minus 3 into 10 to the power minus 4 second. So, according to the observer in ambulance also, this event occurred before this particular person had reached the hospital. So, he was trying to drive towards the hospital and this particular event of this sending signal being sent occurred at 3 into 10 to the power minus 4 seconds before this particular person in the ambulance was to reach the hospital. Or a better word, the hospital to reach him because as far as he is concerned, he will feel that hospital is approaching towards him, for the hospital to reach him. So, we sort of again mix up the word, we say this person will reach hospital because that is the way we visualize if I am sitting in a car. I always visualize that, I am going to reach this particular destination. But actually what I visualize when I am sitting in the car that the hospital is coming to me. Though we are somewhat casual in speaking and saying that I am going to reach the hospital. As far as even 2 is concerned, if you want to do a Lorentz transformation, you can do it, but it is not really necessary because I know that when x is equal to 0, t is equal to 0, x prime has to be 0, t prime also has to be equal to 0. So, I just put e2 as x2 prime is equal to 0, t2 prime equal to 0. Of course, I must remind you that this prime side thing I have used for the A frame, for the H frame, I have not used prime. Actually, we will be discussing later about the P frame, then I will be using double prime. So, just to make it clear. Now, let us look at even 3. Let us try to find out the coordinates of even number 3 in the ambulance frame. Strictly speaking, if one is smart and one has understood things quite well, to find out the coordinates of even number 3 in ambulance frame, one need not have to use Lorentz transformation. One can find it by a shorter route. But let us not do that at the moment. At the moment, let us just try to write Lorentz transformation and work find out or work out the coordinates of even number 3 in ambulance frame. Later, we will discuss which is the shorter way of solving this particular, this particular part of the problem or just finding out the coordinates of e3 in ambulance frame. So, let us look at this transparency for even number 3. I have to find out x3 prime and p3 prime. x3 prime gamma x minus vt. Gamma, we had already calculated 5 by 3 for x3. That is the coordinate of the event in hospital frame of reference. It occurred at 19.2 into 10 power 4 meters. v is 0.8 c. Time was 8 into 10 power minus 4 second. So, it is gamma x minus vt. It so happens. Just calculate this number. You will get it equal to 0. Let us look at the time t3 prime. t3 prime is equal to gamma 5 by 3 t is 8 into 10 power minus 4 second minus v which is 0.8 c, relative velocity between the frames. x 19.2 into 10 power 4 meters divided by c square. Gamma t minus v x divided by c square. If you calculate this number, this turns out to be 4.8 into 10 power minus 4 seconds. If you look at this particular value of e3, probably you would have realized that the shortcut that I was trying to mention. If you are really sitting with respect to an observer in the ambulance, what you would feel that the car was approaching towards him. The car means the train compartment was approaching towards you. Like hospital was initially approaching, the hospital passed by. Then you have the train compartment which is coming towards him and then you catch them. That is the event number 3. So, according to the observer in the ambulance, hospital was approaching him. The compartment, train compartment in which the sick person is sitting is also approaching So, event number 3 occurred when he was sitting right there and what is the event number 3?  They are combining at the same point. So, obviously, this particular event occurred exactly when he was sitting here. Similarly, event number 2 is the event when the light reached this particular person sitting on the hospital and exactly at the same time the ambulance was also passing by. So, this event number 2 also coincided with this particular ambulance coinciding ambulance coordinate coinciding with the hospital coordinate. As you can see here also if you go to the transparencies, x2 was equal to 0, x3 is equal to 0. Both the coordinates turned out to be same because both the events are occurring exactly in the same position. So, for person sitting in the ambulance, the hospital reaching him occurred when he was at his origin. Obviously, the train reaching him will also occur when he was at his origin. So, without working out Lorentz transformation, I could have immediately guessed the answer that x3 should be equal to 0. Now, you would have realized that I could have calculated t3 also very easily because just now we said that event number 2 occurred when x2 was equal to 0, event number 3 occurred when x3 was equal to 0. It means both the events occurred at the same position. In both the events occurred at the same position, then the time interval between these two events must be the proper time interval. Remember, it is in the A frame that both the events occurred at the same position. So, time interval is proper in A's frame of reference, not in the hospital frame of reference. Because in the hospital frame of reference, this particular catching is done at a distance quite far away while the second event occurred at x is equal to 0. But in the ambulance frame, both these events occurred exactly at the same position. Therefore, the time interval t2 and for event number 2 and even number 3, the time difference between these two is a proper time interval in ambulance frame of reference. Therefore, in hospital frame of reference, it must have been dilated. So, if I know the gamma values, knowing the time interval which I have already calculated, I have already calculated t1, t3. I can always, I am sorry, t2 and t3. I can always find out t2 minus t3 or t3 minus t2. And from that particular time interval, I can always find out t2 prime minus t3 prime or t3 prime minus t2 prime by just using the gamma. That is what I have written in the next transcript. It is just a picture showing whatever I have said that this is somewhat different from the scene which was being looked by a person in the hospital frame of reference. As far as the ambulance frame of reference is concerned, the first event occurred farthest off. So, remember it is just contrary to the hospital frame of in hospital frame of reference, the chronology of the events were different. The first event occurred when he was closest to hospital. Second event occurred when he was further away from the hospital. Third event occurred when he was still further away from the hospital. But in the case of ambulance, it is different. In the case of ambulance, first event occurred, this particular train compartment was farthest off from there. When the second event occurred, that particular thing has come nearer to him. And the third event occurred at that time, it was just close to him. It was just at the same place where he was situated. So, the distance as the events keep on coming keeps on decreasing in the frame of reference of the ambulance. Also I would like to realize, I would like to mention that as far as the event number 2 is concerned, at that time the origin of ambulance was coinciding with the origin of the hospital and not with the train. Train was still further off. So, this is this picture which is showing this behavior that at first event, this particular train was further off him and supposed to be moving towards him. And the second event when it occurred, it was still not at the origin, it was further off from him. And it was the third event when this particular train compartment has reached this particular ambulance. Again I repeat, it is important that we should be clear about our frames of reference. We should be able translate our thinking with respect to an observer who is sitting in that particular frame of reference. Then we are not likely to make an error. So, this is what the shortcut I was describing. We could have found out the coordinates of E3 without using transformation also. The x has to be 0 because it occurs at the origin of A frame. This is what we have just now described that it has to occur at x is equal to 0. So, nothing surprising, I need not have worked out the Lorentz transformation equation. The second thing, we use time dilation. We can use time dilation for the time part. We realize that T3 minus T2, it means the time difference between third event and second event as seen in the ambulance frame of reference being proper time interval between E2 and E3. It is actually in the ambulance frame that this particular time interval is proper. Therefore, in this particular frame of reference, which is the hospital frame of reference, T3 minus T2 must be gamma multiplied by this particular time interval because this time interval is proper. So, T3 minus T2 must be a dilated time. So, if I calculate from this, substitute these particular values. I have just now calculated T3 minus T2 prime, which was 8 into 10 power minus 4 for T2 and T2 prime times were 0 anyway. So, this is equal to 0. This is also equal to 0. I substitute in this particular expression. I get T3 prime is equal to 4.8 into 10 power minus 4 second, which is exactly the same result, which we had obtained here. So, we can very easily see that we get the same result without actually using Lorentz transformation. We can just use the time dilation formula, still get the same result. So, now I know all the events and I have filled the event table in the ambulance frame of reference. So, this transpose issues the event table in a frame. Event number 1 occurred at 9 into 10 power 4 meters and at a time of minus 3 into 10 power minus 4 second. Event 2 occurred at x is equal to 0, t is equal to 0. Event number 3 occurred at x is equal to 0 and at time 4.8 into 10 power minus 4 second. So, my event table is filled both in h frame and a frame. Now, let us look back at the question, see that whether we have answered all the questions, if there is something still remaining. Are all the questions answered? This is the first question, which we had asked. What are the positions and times of the person sending the signal in h and a frame? It means basically when event number 1 occurred. Event number 1 occurred, I have already known the coordinates. I know the value of x, I know the value of time, both in h frame, both in a frame. I have to just pick up those values and plug it here. That is the answer. Second part, find the position of the person in h frame at t is equal to 0 and in a frame at t prime is equal to 0. See, at h frame at time t is equal to 0, I had calculated the intermediate position. Remember, we go back to this picture here. I had already calculated at time t is equal to 0, this distance is 48 kilometers. See, originally it was 30 kilometers. During the time light took to come to here, this has moved further away by 18 kilometers. So, this distance is 48 kilometers. So, this I have already calculated. As an intermediate step, though it was not really related to an event, but we had already calculated. So, this distance is 48 kilometers. But what is the position of the person in a frame at t prime is equal to 0? I have not calculated yet. Because this particular event, whatever we are talking, is occurring at different time and different values of x. So, this need not be correct. I have to still calculate this particular value. Third thing, find the time in h frame and a frames when the car reaches the person. When car reaches the person or the ambulance reaches the person, this car means ambulance. When this ambulance reaches the person, obviously this is time e3. This is an event e3. So, whatever is the value of x that we have calculated, we have calculated both in h frame and a frame. So, we have to just pick up those numbers, plug it here. Event number 3 tells me the solution of the last part of the problem. We have to still find out what is the position of the person at t prime is equal to 0. Let us go back here. According to A, that is an ambulance person. Even occurred, the event number 1 occurred when the person was at 90 kilometers away at time t prime is equal to minus 3 into 10 to power minus 4 seconds. Remember, these were the coordinates of event number 1 in ambulance frame. So, it means the distance of that particular person who had sent the light signal was 90 kilometers, but the time in the watch of that particular observer sitting in the ambulance was minus 3 into 10 to power minus 4 seconds. The problem is simple. I am looking only at the f frame. The question is that if there is a particular train which is at a distance of 90 kilometers away from me at a time minus 3 into 10 to power minus 4 seconds, where it would be at time t prime equal to 0? I can find out if I know what is the speed of that particular train relative to me. So, I have to do a velocity transformation. If I do a velocity transformation, I can find out what is the speed of that particular train which is coming towards me. Then I can find out how much distance it would have traveled in time 3 into 10 to power minus 4 seconds. Subtract that time that that distance sorry from 90 kilometers, I will find out where this particular person would be situated at time t prime equal to 0. So, during this time as I have said the person was traveling with a relative speed of ux prime. Let us calculate ux prime. Apply a velocity transformation formula. I have to go from the frame of hospital to the frame of ambulance relative speed v is 0.8c. ux, this is the speed of train which I want to transform. I want to find out what is the speed of train in ambulance frame of reference. So, ux is the speed of train as seen in the hospital frame of reference ux. So, this is ux 0.6c minus v which is 0.8c 1 minus ux into v divided by c square. So, there was a c here, there was a c here that cancels with that c square. So, you get this value of ux prime which is minus 0.2c divided by 0.52. Obviously, there is a negative sign here because as far as the observer in ambulance is concerned, the train is approaching one. It is not going away. So, direction is minus x. It is towards minus x direction. That is why there is a negative sign. Now, the distance traveled by the person in a frame as we have just now discussed during this particular time until the time what shows t prime is equal to 0 in his watch will be this particular speed with which this train is moving or coming towards him multiplied by the time that I am talking which is 3 to 10 power minus 4 second which comes up approximately equal to 3.46 into 10 to the power 4 meter. So, this is the time, this is the distance that the compartment would have moved in ambulance frame of reference towards him. The original distance at time 3 is equal to 10 to the power minus 4 seconds was 90 kilometers. I have to subtract this distance from 90 kilometers to find out what will be the coordinate or what will be the position of the person in the frame of reference of the ambulance at t prime equal to 0. That is what I have done in the next transparency. The distance of the person in a frame at t prime is equal to 0 was the original distance 9 into 10 to the power 4 meters minus the distance that has been traveling that this train has traveled in the frame of reference of a in the time 3 into 10 to the power minus 4 seconds I get this distance approximately equal to 5.54 into 10 to the power 4 meters. So, this is the situation as far as ambulance frame is concerned. At the first event the distance of this particular train was 90 kilometers. At the time of the second event t prime equal to 0 this distance was 54.55.4 kilometers and at the third event this particular distance was 0. So, I have calculated all the distances sometimes it becomes easy when we picturize the thing. Remember if I would have just I would just like to make two points. If you would have just transformed the first part I will not get the same result as 55.4 kilometers because that particular times are very different. So, that is one thing. The second part is the second point that I wanted to emphasize is that when I calculated the time intervals or when I calculated the coordinates of all the events in hospital frame I never use Lorentz transformation. Lorentz transformation was not required because all the information was given in the frame of reference. So, in principle this problem could have been a standard classical mechanics problem other than the fact that I have used that speed of light is same other than that it does not require anything special. If all the information has been given in my own frame of reference I do not require a transformation. Only when I go to different frame transformation is needed. Now, let us though it has not been asked in this particular problem let us spend a little bit of time in finding out things about P frame the person's frame the person who was sick. If I have to find out the absolute times I will be somewhat in trouble. It is not really a trouble but if one wants one can do it. But the way the situation has been described this person never matches either the origin of ambulance or the hospital at any time. And if I have to use Lorentz transformation directly then I have to resynchronize the watches so that the times becomes equal to 0. For example, if I want to transform from hospital frame to this particular person's frame then I have to make their time 0 when their origins were coincident or if I have to transform from P frame to the ambulance frame again I have to make time t is equal to 0 when their origins coincided this can become somewhat cumbersome. But if I use in the differential form if I take the differences then I do not have to bother because that particular position of the origin they does not matter if I am taking the differences because that cancels out. So, if I have to ask or if I have to get some information in the person's frame of reference the sick person's frame of reference if it is given in the differential form I did not re-synchronize my watches I can still get the answer. So, just one quick example about this thing. Suppose in the problem we had also asked that find out the time in the that the ambulance took to reach P in P's frame after sending that is the signal.  Now you have said signal let us say at a given time how much time after I really got the help it means the ambulance reached me. So, I ask for help after how much time I got the help. So, if I have to ask this particular question remember I am still talking about event number one that person I am now imagine that I am the sick person I am sending the signal I sent the signal that was event number one remember same event even number three ambulance reached me this is even number three. Now probably you would have realized that if I am the sick person I am the person who is sending the signal obviously I am always sitting at my origin when I receive the ambulance again I received at the origin these two events in the P's frame occurred at the same position therefore time interval between event number one and three number one and three this time difference is a proper time interval in P's frame. So, in the hospital frame and ambulance frame this must be dilated if I know this particular thing it is very quick I can find out the answer. I can choose because anyway I have found out the time differences between even an E3 both in H frame and A frame I can use any of these things use corresponding gamma values and find out time difference in P's frame. So, let us just take one quick thing let us assume the hospital frame let what I have written here we realize that even E1 and E3 took place at the same value of xp this now happen just to P in P's frame hence this time interval is dilated in H and A frames for P frame I have put a double prime. So, T3 minus T1 this is a proper time interval T3 minus T1 this is the hospital frame of reference must be equal to gamma pH let us be very careful that which gamma I am using because I am transforming in three different frames it is between the person and the hospital because T3 minus T1 is in the hospital frame of reference. So, I have to use gamma by using the relative velocity between P and H which was anyway given to be equal to 0.6C the person was moving with the speed of 0.6C in H frame and we know that gamma pH we have done in many problems is 1.25. So, this T3 double prime minus T1 double prime which is the time difference between event number 1 and even number 3 in the persons frame of reference in the sick persons frame of reference will be given by this time interval if you calculate the time difference between event 1 and even 2 one was in minus 1 another was in 8. So, it becomes 9 to 10 to the power minus 4 divided by gamma which is equal to 7.2 into 10 to the power minus 4 second. So, this is the proper time interval and this is the time interval which will be measured in the sick persons frame of reference which I am calling as P frame of reference. Just to check if I multiply this by the gamma between A and P person and the ambulance then I must get back the time interval which I have calculated between E1 and E3 in the ambulance frame of reference just to check if I have thought everything right. To use that particular gamma I have to use the relative velocity between the persons frame and and the ambulance frame of reference. So, I have just written just to confirm we can find the time interval in A frame using this proper time interval between E3 and E1 in P frame persons frame exactly the same thing this particular time interval is proper. I have to find out T3 prime minus T1 prime I have to multiply by gamma PA I had calculated the relative velocity as 0.2 divided by 0.52 c c square cancels out gamma PA turns out to be equal to 13 divided by 12. Therefore, T3 prime minus T1 prime will be equal to this proper time interval multiplied by 13 divided by 12 which is equal to 7.8 into 10 power minus 4 second. If I go back to my event table this is what you have to have seen 4.8 and 3 this is minus this time difference is actually 7.8 into 10 power minus 4 second this is what I had expected. This matches with whatever we have evaluated using Lorentz transformation. In the end I would just like to give my summary of course there is in this particular summary we have only solved one particular problem which is a long problem involving three different frames of reference and three different events. So, we discussed one example involving multiple events. Thank you.


So we go back to our IAR software where we left off and If we now scroll down to the app body task Which starts about line one seven five and just before the infinite loop We need to paste in our Commands for clock gating So if we go and copy Commands here, so if we highlight this and copy them and if we paste them into our code Like that Do a bit of tidying up and now we can build our project and We want to go project Download download active application again And that will put the latest version into our board We then reset our target board on my board now I am getting about six milliamps in stop mode and about eight milliamps in run mode so That gives us about another 15 percent Reduction slightly more on mine. I'm about 20 percent reduction in stop mode And I'm about 15 percent reduction in play mode there on mine So so you can see now the current consumption is slowly coming down steadily within our project So what else can we do inside the hardware to actually reduce the current consumption? so inside the GPIO cell of every pin on the device we have a a pair of protection diodes The to pull up and pull down resistors, and we have a Schmidt trigger So this Schmidt trigger is going to consume power So the best way to disable that is to enable the unused IO pins to go to analog So as you can see the analog line comes off before the smith trigger And this can be done inside the cuba mechs tool environment So there is an option here in to disable all unused gi pios and set them to be analog inputs On most devices that would save us some current consumption But as you can see from the snapshots here from the data sheets Analog mode is actually the reset state of all GPIO pins in the STM32 L4 device So we can still make the setting inside the cuba mechs tool which would normally make a current consumption saving for any other device So if you go into the project settings and code generator Inside the cube and x tool There is a tick box where you can set all three pins To be analog to optimize the power consumption of this device As all pins are already default. This is not actually going to make us any change on our Audio player application. So the current consumptions we have at the moment Are going to remain exactly the same on the STM32 L4 Any other device that would make a substantial difference as well Disabling all those extra components inside the GPIO peripherals So we can also let the peripherals do The job using the DMA and put the core into a sleep mode So if you look at a typical example there, the core would be asleep The ADC would be streaming data via the DMA into SRAM Then the core wakes up for a short period of time to process the data that you've just received And then the core goes back to sleep after starting the next process Which would to stream that process data out via the UART or an SPI we have a term inside the STM32 L4 which is called batch acquisition mode Which does exactly this type of process where the DMA and the SRAM are Powered and configured with the clock and the rest of the chip including the core are in sleep mode So you can optimize your power transfer of data that's been received from sensors as an Suggested application here on the screen and then processed and sent out to some sort of Low power UART or USART to the outside world If we look at our application we have here on our audio player We have the situation where the core is asleep The quad SPI is streaming information into a RAM block ready to be processed by the core in parallel a Second RAM block is being used with one of the other DMAs and streaming the audio signals out to our headphones Then our core wakes up Processes the batch of data that's just coming from the quad SPI ready to be sent out Then the core goes back to sleep As soon as DMA number two reaches the end of its area The buffers will switch around So that you get a seamless audio stream coming out of the SAI audio peripheral And then the quad SPI will then start filling the buffer be them with the next batch of data ready to be processed by the core And the cycle continues like that Going on inside the application


Hello students, I am Dr. Bhargesh Deshmukh from Walchand Institute of Technology, Swalapur Mechanical Engineering Department. This session is on design of spur gear. At the end of this session, the learner will be able to solve a problem on calculating rated power that a gear pair can transmit. Let us see the problem. It is asked to design a spur gear pair with 20 degree full depth Involved tooth consists of a 20-teeth pinion meshing with 41-teeth gear. The reduction is almost half. The module is 3 mm, m equals 3, while the phase width is 40 mm, 4-0. The material for pinion as well as gear is steel with an ultimate tensile strength of 600 Newton per mm square. The gears are heat treated to a surface hardness of 400 BHN. The pinion rotates at 1450 rpm and the service factor for the application is 1.75. Assume that velocity factor accounts for the dynamic load and the factor of safety is 1.5. Determine the rated power that the gear can transmit. See it is a very typical problem and it is asked to calculate the power, the rated power that the gear can transmit. You can recall what are the forces acting on the gear tooth. For a typical driven gear, Pn is the resultant force which is acting at an angle alpha, the pressure angle, at the pitch point. It has two components. One is Pt and another is Pr. The tangential force is given as Pt equals to mt upon d dash. The radial force is given as Pr equals Pt tan alpha, this radial force. But we know that radial force do not take part in the power transmission. Only the tangential force is contributing for power transmission. The tangential force is obtained from the equation of torque where torque mt equals 60 into 10 to the power 6 kW upon 2 pi n. The term kw is the power and n is the speed. If we know the speed and the power input, we can calculate what is the torque. The value of tangential component therefore depends upon the rated power and rated speed. Now in the question, it is asked to calculate the rated power. What do we need to do? Think upon it. The given data in the problem. The input speed is 1450 rpm. Zp number of teeth on pinion are equal to 20. Zg number of teeth on gear are 41. Module m equals 3 millimeter. Face width b equals 40 millimeter. Service factor cs equals 1.75. Factor of safety is given as 1.5. Surface hardness of the gear bhn equals 400. Sut of the material equals 600 Newton per mm square. With this available data, we need to calculate what is the rated power that the gear pair can transmit. Now to obtain the rated power, we need to obtain the rated torque. And to obtain the rated torque, we need to obtain the rated force. If I talk more precisely, to obtain the rated torque, we need to obtain the rated tangential force. That means our job is to find out pt somehow. Tangential force is given by beam strength or wear strength. Because we know that sb equals mb sigma by. But sb is nothing but maximum value of pt. Or sw equals bqdp dash k. But there also we know that sw equals pt max. Let us start solving the problem. Step one is beam strength. Can I calculate the beam strength? See the available data. Before beginning, I need to check this. Which element is weak? Either pinion or the gear. If the material for pinion and gear is same, then the pinion is weak element. Or it is weaker than the gear. The levies form factor is 0.32 for 20 teeth. You can refer it from the design data book. Permissible bending stress is one-third of sut. Use this formula. One-third of sut equals 200. And hence permissible bending stress is 200 newton per mm square. Beam strength is hence given as mb sigma by. y is 0.32. sut is 600. From that you have calculated sigma by as 200 is 240. 3 is the face width and 3 is the module value m. And hence we can get sb equals 7680 newton. Hence the beam strength is evaluated as 7680 newton. Next part is to calculate the wear strength. For the wear strength we need the ratio factor q, then k and the diameter. Diameter is the pitch circle diameter. q equals ratio factor that equals 2 zg upon zg plus zp. That equals 2 into 41 upon 41 plus 20. We can calculate q as 1.344. k 0.16 bhn upon 100 bracket square. That equals 0.16 400 by 100 bracket square. Then k equals to 0.56. The diameter is given as dp dash equals mzp. 3 into 20, 60. Hence we can calculate sw equals b q dp dash k. b is 40, q is 1.344, dp dash equals 60 and k equals to 0.56. sw equals 8257.54. Therefore the wear strength is around 8257.54 newton. Then the effective load. Effective load is governed by cs, cv and pt. v equals pi d dash np upon 60 into 10 to the power 3. We are calculating the pitch line velocity. Pitch line velocity is 4.5553 meter per second. Using this value we can calculate cv using Barth equation. cv equals 3 upon 3 plus v. That equals 3 upon 3 plus 4.5553. cv comes out to be 0.397. When we calculate cv, the p effective is given as cspt upon cv. cs is known, cv is known, pt we don't know. pt in the form of p effective we can get. We can find p effective and calculate tangential force to evaluate the rated power. This equation p effective is not known, pt is not known. cspt by some other equation we can calculate p effective. It is possible for us to calculate the tangential force pt to calculate the rated power. Beam strength equals 7680 newton which is less than the wear strength that equals 8257.54 newton. Hence beam strength is the criteria of design in the current problem. Remember students, you need to take the lower value of beam strength or wear strength. Compare these two, the lower value you need to select for the design. sb is also equal to p effective into fs, we know. Or sb is 7680, the value of p effective we have calculated as 4.41 times pt, fs is 1.5. From this equation we can calculate the tangential force. pt equals 1161. From that how to calculate the rated power? Rated power is given as first we need to calculate the torque mt equals pt tangential force multiplied by the radius dp dash by 2. We can get the torque. From the torque we can get the rated power. Rated power is 5.29. You need to write the answer as rated power that the gear can transmit is 5.29 kW. See how we have solved the problem. For the rated power you need to calculate the rated torque. And for the rated torque you have calculated the tangential force. This is what is the theme of the problem. Thank you.