In our last lecture, we had applied the postulates of special theory of relativity and eventually arrived at Lorentz transformation. We do not say that we derive Lorentz transformation, we just say that we arrived at it. We presented a series of arguments on how Lorentz transformation could be arrived at. The real test that Lorentz transformation is believed on is on experiment. If we find an experimental support, then only we believe on Lorentz transformation. After that, we took an example, which was an old example of a light being emitted from the origin and we had earlier found out that under Galilean transformation, the speed of the light does not remain constant if we change the frame of reference, which was obvious because the classical mechanics was not designed for that. Then we applied Lorentz transformation on the same example and said that if we apply Lorentz transformation, then the velocity of light turns out to be same in the other frame also, which is what Lorentz transformation is based on or the second postulate of special theory of relativity is based on. So, that is the way we proceeded in our last lecture. Just to recapitulate, let us write or let us discuss the Lorentz transformation again. These are the equations governing Lorentz transformation. In fact, these are the equations. We had introduced two symbols. One is beta, which is v divided by c. Just to remind, v is the relative velocity between the frames and c, of course, is the speed of light. Then we defined another parameter called gamma, which was 1 upon under root 1 minus beta square. If we use these two parameters, the Lorentz transformation equations become like this. For x coordinate, its x is equal to gamma multiplied by x minus vt. y prime remains equal to y, z prime remains equal to z and t prime becomes gamma multiplied by t minus vx divided by c square. We had earlier said that in the non-relativistic limit, meaning when v is much smaller than c, this beta is negligible in comparison to 1 and gamma is essentially close to 1. Then x prime just becomes x minus vt, which is the standard Galilean transformation. These two equations are common even in the Galilean transformation. Here, if v is very small in comparison to c, this particular term is neglected or can be neglected. This gamma will tend to 1, then t prime will be tending equal to t, which is again the standard classical Galilean transformation. So, these equations effectively become important only when v is close to c, where this factor beta square cannot be neglected in comparison to 1. Today, we will try to discuss some of the examples, some other examples of Lorentz transformation. Specifically, two very, very important and popular results arrived out of Lorentz transformation. One of them is called length contraction. The length contraction as the name suggests means that the length of an object may not be constant if an observer in a different frame observes the length of the same object. In fact, as we will be seeing there is a contraction or length becoming smaller is associated with this phenomenon, which is called length contraction. But before we really derive it, let us understand a few terms because it is important to derive or to work with length contraction formula correctly. It is important that we are familiar with these things. The first term is the proper length. Proper length as the name suggests or as we have defined of a particular rod is the length measured in a frame in which it is at rest. So let us imagine that we have to measure the length of this particular rod. This is a rod, the length of which we have to measure. Now this particular rod is stationary in my own frame of reference. So I do not see any motion of this particular rod because as I said this is stationary in my frame of reference. So whatever is the length that I would measure for this particular rod would be called proper length. If this particular rod was moving in my frame of reference, then the length that I will measure of this particular rod would not be called proper length. So proper length has a specific meaning that it must be measured in a frame of reference in which this particular rod is at rest. And that frame of reference in which it is at rest, that particular frame of reference is called proper frame of reference. That is what we have written in this particular transparency. Proper length of a rod is length measured in a frame in which it is at rest and proper frame is a frame in which this rod is at rest. Now let us assume that this particular rod is along the x direction. Remember this particular x direction is the direction of the relative velocity between the frames. So let us assume that this particular rod is inclined or is put along the x direction or it is put exactly on the x axis to be more precise. So this is what I have written. Let us first assume that the rod is placed along the x axis. Then we will show just now that its proper length is the largest and if we go to any other frame in which there is a relative motion along the x direction, then its length will turn out to be shorter. That is why it is called length contraction. Now let us suppose that we have to measure the length of this particular rod. How do I measure? The question is very simple that I measure the coordinates of the two ends of the rod. If I measure the coordinates of the two ends of the rod, actually by subtracting these coordinates I can always find out what is the length of this particular rod. So if this rod's length is measured, I can measure what is the coordinate here. I can measure what is the coordinate here. Then take the difference of these two. This will give me the length of the rod. But the question which I am posing is that is time important. Let me explain what I mean to say that suppose I measure this particular length, this particular end or coordinate of this particular end of the rod, let us say at t is equal to 0. Now this end, this particular coordinate can I measure at any other time? Is time important for its measurement? See the answer to this question is that, no it is not important, so long this particular rod is at rest. So I can measure this coordinate now and go and have a cup of tea, come back and then measure the other coordinate. Still the difference of these two coordinates will give me the length of the rod because this rod is fixed in my frame of reference, it is not moving. So this rod is not going to change its coordinate as a function of time. So it is immaterial when I decide to measure the coordinate of this particular end and when I decide to measure the coordinate of this particular end. But suppose this particular rod was moving in my frame of reference, then the same statement is not correct because if I make measurement of this particular end now and make the measurement of the coordinate of this end little later during this particular time, this rod has been moving. So this rod is moving, I make a measurement now and by the time I go to the other end make the measurement, by that time the rod has been displaced from its position. So its coordinate has changed. Therefore if you still I decide to take the difference of the two coordinates that will not give me the correct length of the rod. So this is what I have said, measure the coordinates of the two ends of the rod, it is time important for the measurement, yes only if the rod is moving, then it is important that the two ends of the rod must be measured at exactly at the same time. The situation I have picturized here in the next transparency, I have depicted a moving rod. This red is the position of the rod at a given time, let us call time t is equal to 0. Now let us suppose at time t is equal to 0, I made the measurement of coordinate of this particular end and find out, find it out that it is equal to x1. Now I make some delay, I do not immediately make the measurement of this particular end but after some time I decide to make the measurement of the other end of the rod. Then by this particular time this rod has moved ahead. So at a later time what I would be measuring is probably this coordinate which is x2. So you can clearly see that x2 minus x1 is not equal to the length of the rod. It is a different question that from these two measurements also I can find out the length of the rod provided I take care of this displacement that the rod has taken or the displacement that rod has undergone during the difference of the time during which I have measured that is a different question. That also I will give an example little later but at this moment what I want to do I want to measure the length of the rod by just measuring the coordinate differences. And for that if the rod is moving in my frame of reference it is important that the difference that the two coordinates must be measured in the same time at the same time only under that condition the difference of the coordinates will give me the correct length of the rod. So this is what I have written if the coordinates are measured at two different time their difference does not give the correct length this is what I have said. There may be other methods of measuring the length of the moving rod some would be discussed later as I have mentioned that if I can calculate how much is the rod how much the rod has moved during the time difference that I have taken or the time I have taken in making a measurement of one end and the other end then of course I can find out the length of the rod that I will do little later. But if the coordinate difference of a moving rod has to give me length then for a moving rod the two ends must be measured at the same time that is what I have written. But the coordinate difference of a moving rod would give correct length only if the two ends are measured at the same time this is an important point to note because this is where I find a very large number of students make a mistake they do not bother about time they just decide and they just say that x2 minus x1 equal to the length which may not be correct because if the object is moving in the frame of reference then x2 minus x1 need not give the correct length until these two measurements were carried out exactly at the same time. Now, let us come back to the problem let us assume that this particular rod is at rest in S frame of reference we have already defined what are our S frame and S frame of reference Lorentz transformation is applicable when we impose certain conditions on S and S frame of references which we have discussed earlier that the axis should be parallel and the relative motion should be along x direction so that is anyway have to be valid. But what I assume in this particular case that it is the S frame of reference in which this rod is stationary it means it is S frame which is the proper frame. So, this rod AB which I am calling here is AB is at rest in S frame of reference obviously it means that in the S frame of reference this particular rod is moving I can always imagine that I have put a rod let us say in a train and the train is moving with respect to an observer on ground. So, a person sitting on the train feels that this rod is not moving, but a person sitting on the ground feels that this particular rod is moving remember this x direction is the direction of the relative motion this is in this direction that S frame is moving relative to S. Now, we have earlier agreed that relativity things are simple if we define events generally it is always a good idea to define events. So, let us suppose we define events relating to the measurement of the length of this particular rod and these events I am defining here calling event number 1 and event number 2. So, this is my event number 1 let me just write show the rod here this was my S frame of reference and this rod was at rest here this was my end A this was my end B and this rod is stationary in S frame of reference being also observed by another observer which is an S frame in this S frame this rod is actually moving to the right as the frame of reference S is also moving to the right with respect to S. So, my event number 1 is that I make the measurement of the coordinate of this particular end of the rod and let us assume that this particular measurement is being done by an observer in S frame of reference. So, he makes a measurement of this particular the coordinate of this particular end of the rod this is my event number 1 then he goes here let me not say goes here but he also makes a measurement of the end B of the rod this I call as event 2. Now, as this particular rod A, B is not stationary in S frame of reference. So, in principle he cannot go from A to B and make the measurement the measurement of the coordinate here and coordinate here have to be done at the same time if the coordinate difference of these two events has to give me a correct length of the rod. So, this is what I have written here I have defined my events my event number 1 is observer in S measures the coordinate of A end of the rod and let us suppose this coordinate B X 1 and T 1 normally we have a tendency of now including time also as a coordinate because as we will be seeing going ahead in the course we realize that for many purposes time X as another coordinate. So, because this event number 1. So, I have put a subscript here 1 to this event number 1 occurs at a coordinate of X 1 at a position coordinate of X 1 and at a time coordinate of T 1. Similarly, the observer in S measures the coordinate of the B end of the rod and let us suppose he does it and finds out the coordinate to be X 2 and the measurement is done at time T 2. So, the coordinate of this particular event is X 2 T 2 while the coordinate of the first event is X 1 T 1 as we have earlier realized and discussed this X 2 minus X 1 will give me correct length of the rod only if T 2 is equal to T 1 because it is in S frame that the rod is actually moving. This is what I have written here X 2 minus X 1 is equal to L in S only if T 2 is equal to T 1 which I have said let us assume that this is equal to T rather than writing same symbol. So, I said let us assume that T 2 1 T 2 is equal to T 1 and is equal to T. Then these two coordinates I use Lorentz transformation to find out the information in S frame of reference. Let me write the Lorentz transformation once more here just to remind you it was X prime is equal to gamma X minus V T Y prime is equal to Y Z prime is equal to Z T prime is equal to gamma T minus V X divided by C square. This is what was my Lorentz transformation. Now, let us apply this Lorentz transformation on to the two events. If I apply let us say first event number 2 it means the value of X was X 2 and time T was T 2 I will get the coordinate of that particular event in S frame of reference. So, I will get X 2 prime the coordinate of the second event in S frame of reference. Similarly, if I substitute here X 2 substitute here T 2 I will get T 2 prime which is the time at which the second event occurs according to an observer in S frame of reference. So, this is precisely what I have written here. If you go back to the transparency it is X 2 prime is equal to gamma X 2 minus V T because T 2 I have taken to be equal to T this is what I have written earlier. So, this I have just put it equal to T then T 2 prime will be given by gamma T which is just T 2 which is equal to T minus V X 2 divided by C square exactly the same thing I put the coordinate position and time here for the first event put position and time for the first event I get these two equations. You can very clearly see that as far as S prime observer is concerned T 2 prime is not equal to T 1 prime. See this T is same as T but this X 2 and X 1 there has to be a difference if the length has to have some I mean if the rod has to have some length then X 2 cannot obviously be equal to X 1 therefore this T 2 prime is not equal to T 1 prime. So, what it means that according to the observer in S frame of reference these two events occurred at different time it should not be surprising to us because remember earlier we have discussed that simultaneity is relative I am I shall be discussing I shall be harping on this issue number of times saying that simultaneity is relative. It is in S frame that the two events occurred at the same time hence they were simultaneous according to S frame of reference these two events did not occur at the same time therefore they are not simultaneous. So, according to him the two events occurred at different time would it mean that S 2 prime will not agree that X 2 prime minus X 1 prime is equal to length no that is not correct because in S frame of reference this rod is stationary. So, it is not important whether the two ends should be measured at the same time what is important is that we just make measurement at any given time. So, even if the measurement of one end of the coordinate in S frame of reference was done in different time from the other end the rod being at rest in that particular frame of reference the time the position difference will still give me the current length. So, what I want to argue out that X 2 prime minus X 1 prime is equal to the correct length of the rod in S frame of reference irrespective the fact that T 2 prime is not same as T 1 prime. While in S frame we could not have done a similar argument I have to force the time to be same in S frame of reference only in that case X 2 minus X 1 would have given me the correct length. Now if I want to find out a relationship between the lengths measured in S and S all I have to do is to take the differences. So, I know that X 2 minus X 1 X 2 minus X 1 was the correct length in S frame of reference because these two times are same X 2 prime minus X 1 prime would anyway be correct length in S frame of reference even though the times were different. So, I will say that X 2 prime minus X 1 prime is L prime the length of the rod in S frame of reference and call X 2 minus X 1 as the length of the rod in S frame of reference. So, if I subtract these two you will just realize that this particular term V T here would cancel time being identical. So, you will just get X 2 prime minus X 1 prime is equal to gamma multiplied by X 2 minus X 1 this is what I have written in the next transparency. We note that T 2 prime is not equal to T 1 prime still X 2 prime minus X 1 prime is equal to L prime this is because the rod is at rest in S frame. Measurement time has to be adjusted to be same in S and not necessarily in S frame. I have taken the differences I get X 2 prime minus X 1 prime is equal to gamma X 2 minus X 1 as we discuss this is the length in S frame of reference this is the length in S frame of reference I call this is L prime I call this is L. So, this L prime is equal to this L is equal to L prime divided by gamma which is what I have written in this particular equation as we have seen earlier that gamma is to be always greater than 1 because you remember gamma was equal to 1 upon under root 1 minus V square by C square. So, you are always subtracting some quantity from 1. So, therefore, this particular factor is always going to be less than 1 and 1 divided by something less than 1 would always make a quantity greater than 1. So, gamma would always be greater than 1 therefore, this L would always be smaller than L prime. So, we say length is contracted it is called length contraction. Let us take an example slightly different example in which I do not make the rod along the X direction, but let it have a slightly different orientation. So, in this particular example, I have assumed that this particular rod is lying actually in X prime Y prime plane. I still assume that it is the S prime frame of reference which is the proper frame of reference for this particular rod. So, this rod is actually stationary in S prime frame of reference, but the only difference I say that this rod is no longer along this particular direction, but making some angle you know somewhere here like this. So, my rod is like that AB and this is S prime frame of reference. So, I assume that the end A of this particular rod is at the origin here. I assume that length of the rod is 1 meter. So, effectively in S prime frame of reference, I know the X coordinate of A which is 0 0, S coordinate of B which I can find out very easily. If I know the length and if I know the angle and what I have to do now is to find out the length of this particular rod in S frame that is what is the question. So, let me read the statement of the question or statement of the example. There is a 1 meter rod AB the one which I have said the length has been given to be 1 meter rod and it is kept stationary in S prime frame of reference which is the proper frame of reference. With end A at origin, the way I just now shown in the picture in X prime, Y prime plane making an angle of 60 degrees. So, I know the theta value which is 60 degrees with X prime plane with X prime axis I am sorry. What would be its length in S? If the relative speed between S and S prime is 0.6 C. So, I have given some numbers. So, we can calculate the answer a numeric answer. Also, find the angle that the rod makes with X axis in S. So, according to S observer at what angle this particular rod is inclined with respect to X axis. So, this is essentially the question. The same picture which I had drawn on this particular piece of paper you have S frame of reference this particular rod is at rest with A being at the origin, this end B being here, this angle being given is 60 degrees and this length is being given to be 1 meter. Now, like before I will define the events and we will define the events to ensure that these two events occur at the same time in S frame of reference because it is an S frame that this particular rod is moving. Also, because here Y coordinate is also involved see in earlier example we did not bother about the Y coordinate because Y was 0 in S frame it had to be 0 in S frame of reference. But here because this particular other end of the rod will have a different value of Y than the first end of the rod B end will have a different Y than the A end. Therefore, it is better to talk about the Y coordinate also. So, we will define our events. So, I said like before we have to define events relating to the measurement of the ends of the rod and these events have to occur at the same time in S. Let the coordinates of these events be again like before even is X 1, Y 1, T. So, this coordinate occurs this particular sorry event occurs at coordinate X 1, Y 1 and T according to an observer in S frame. Similarly, event number 2 I have put subscript 2 happens at X 2, Y 2 and at time T these two times I have put same to ensure that actually the measurement of these coordinates will give me the correct length of this particular rod. So, in principle as we will be discussing if I take the difference of this and take the difference of this and square and add and take the under root I will get the length of the rod in S frame. Again, we have to carry out the Lorentz transformation to find the corresponding coordinates of these two events in S frame of reference. The only difference in this particular case is that I already know the coordinates. So, I can write this equation makes not such a great difference, but just to ensure that we understand the situation well. Now, we applied for the second event the Lorentz transformation which is X 2 prime is equal to gamma X 2 minus V T just like in the earlier example of length contraction. Only thing I say that if this particular length was 1 meter this angle is theta then the X coordinate of this particular rod will be 1 cos theta because 1 is the length of the rod the Y coordinate of this particular rod will be equal to 1 sin theta. So, this is the X coordinate this is the Y coordinate this will be 1 cos theta this will be 1 sin theta because theta has been given to be equal to 60 degrees. So, 1 into cos 60 which is half this will be half meter and 1 sin theta will be equal to 2 3 by 2 meters. So, this particular coordinate the Y coordinate would be equal to 2 3 by 2 meters while this X coordinate will be equal to half meter. So, this is what I have written in this particular transparency that X 2 prime is equal to gamma X 2 minus V T which I know is equal to half meter Y 2 prime is equal to Y 2 which I know is under root 3 by 2 meters. Similarly, I can take the Lorentz transformation of the X coordinate which gives me X 1 prime is equal to gamma X 1 minus V T this coordinate is equal to 0 Y 1 prime is equal to Y 1 is equal to 0 because I know in S frame of reference and A is at origin. So, X 1 prime and X 2 prime I am sorry X 1 prime and Y 1 prime both must be equal to 0. Like before we take these two events to be different in S prime their time will be different in S prime, but I have not specifically written in this particular case the equations involving time difference where they the time transformation because as far as the problem is concerned it is asking only for the length it is not asking for time. So, therefore, I have not written the equation relating to time transformation I have just written the X transformation because for the problem that is not important. Only thing I have to ensure that the time of the two events is same in S that I have ensured by putting the two times to be same that is all. Like before I take the difference of these two just like what we did in the earlier case I will find gamma X 2 minus X 1 to be equal to 0.5 because remember one was 0.5 another was 0. So, the difference is just 0.5 similarly Y 2 minus Y 1 was equal to root 3 by 2 because one was root 3 by 2 another was 0. So, this is this and of course for velocity equal to relative velocity equal to 0.6 C I get gamma is equal to 1 divided by under root 1 minus V square by C square which is 0.6 square this becomes 0.36 1 minus 0.36 under root it becomes 0.8. So, this is equal to 1.25. So, gamma is equal to 1.25 I mean this 0.6 gives you very clean number for gamma. So, in many of the problems to make the mathematics simple making the algebra simple now we choose the gamma value which is something 0.6 0.8 that this gives clean value of gammas. So, I substitute these values of gamma here and find the length as I have said earlier the length is given by X 2 minus X 1 plus Y 2 minus Y 1 square look back at this particular rod if I know the coordinate of this particular rod even if they are not 0. So, what I have to take the difference of the X coordinate and the difference of the Y coordinate this length plus this length this length square plus this length square will give me this length square if I take the under root I will get the length of the rod. So, this is what I have written here the length of the rod in S is thus given by X 2 minus X 1 whole square plus Y 2 minus Y 1 whole square we just now seen from the previous transparency here that X 2 minus X 1 is equal to 0.5 divided by gamma and gamma is equal to 1.25. So, I have written here is 0.5 divided by 1.25 square plus here there was no gamma there is just root 3 by 2 because Y prime is equal to always Y if I calculate this number we get under root 0.91 which is approximately equal to 0.95 meter. Hence the length of the rod in S frame of reference would turn out to be approximately equal to 0.95 meter and not 1 meter. I have to calculate the angle and angle again if I have to calculate this angle I have to calculate this distance this distance divide by the 2 this will give me tan theta. So, this is Y 2 minus Y 1 this X 2 minus X 1 divide the 2 and that will be equal to tan theta.  So, 1.25 comes into the numerator if you calculate this number approximately turns out to be 2.165 which gives theta approximately equal to 65.2 degrees. So, according to an observer in S this rod is inclined not at 60 degrees but 65.2 degrees. I hope you would have appreciated one particular aspect from this particular derivation that the gamma factor arrived only in the X component and not along the Y component. Therefore, actually it is only the X component which get got contacted and not the Y component. So, this is a point which I have mentioned here specifically that we note that only X component of the rods length is contracted and not the Y component. Similarly, not even the Z component because Y prime turns out to be equal to Y Z prime turns out to be equal to Z. Therefore, there is no gamma factor involved gamma factor involved is only in the X component and that is the factor which is the relative velocity direction along which the contraction would really occur. Now, let us take another example. There is an observer in S and he finds that a lightning strikes at a distance of 20 kilometers from origin at t is equal to 0. So, for example, I am sitting here in this particular place and 20 kilometers away from me lightning strikes it could be any other event but I am sort of taking as a lightning. So, at 20 kilometers away some lightning strikes and that was time t is equal to 0 as far as my I am concerned. So, I am sitting at my origin and I find out that at time t is equal to 0 lightning had struck at distance of 20 kilometers away. Remember at this time t is equal to 0 according to me the origin of S was also passing by my side. So, the observer in S would feel that the origin of S prime passing by his side and this event occurring at t is equal to 20 kilometers happened exactly at the same time which is t is equal to 0. Now, the question is at what distance from origin did this event of lightning occur as far as the observer in S prime frame of reference is concerned. This S prime observer is moving relative to me with a speed of 0.6 c. So, let us read the statement again according to an observer in S it lightning strikes at distance of 20 kilometers from origin at t is equal to 0. At what distance from origin did this event occur in S prime which moves with a speed of 0.6 c in frame S. Let me first try to give a wrong answer for this particular example and my experience tells that very large number of student have a tendency of giving this wrong answer. Then we will see what is the correct answer and why the wrong answer is not really correct. So, standard wrong approach to this particular problem is essentially applying blindly the formula of length contraction. So, I may start thinking that at t is equal to 0 and t prime is equal to 0 the origins of S and S prime were coincident. Let us imagine a stationary rod in S frame that in S frame there is a stationary rod which is 20 kilometers long very long rod which is extending from the origin to the place of lightning. I can always imagine that there is a huge long rod which is going right from origin 20 kilometers away and this is stationary in S frame. Now, I expect that the length of this particular rod would be contracted in S because this length of the rod is proper in S frame. So, probably I can apply length contraction formula. Let me just read before I go to the next transparency. At t is equal to 0 and t prime is equal to 0 the origins of S and S prime were coincident. Imagine a stationary rod of 20 kilometers extending from origin to the place of lightning in S. The length of this rod would be contracted in S frame. So, we just now said that gamma is equal to 1.25 for v is equal to 0.60. So, gamma is 1.25. So, this 20 should be divided by 1.25 because this is a contracted length and that answer is 16 kilometers. Therefore, this particular lightning must have struck in S frame at a distance of 16 kilometers from origin. Let me read. So, the coordinate of the lightning in S frame would be given by the contracted length 20 divided by 1.25 which is 16 kilometers. I am again repeating this is the wrong answer. I have written boldly wrong answer. Let us try to find out first correct answer. Correct answer. Whenever there is a confusion apply Lorentz transformation directly. We are not going to make a mistake. Apply Lorentz transformation. I know the coordinate of this particular event which is at x is equal to 20 kilometers, t is equal to 0. I can always find out what is x prime and what is t prime by applying Lorentz transformation. So, this is the correct answer. The coordinate of the event in S is x is equal to 20 kilometers and t is equal to 0. Hence, using Lorentz transformation, the x coordinate in S will be given by x is equal to gamma, x minus vt. x is 20 kilometers, t is 0. So, this particular quantity becomes 0, 1.25 multiplied by 20, 25 kilometers. In fact, this length appears not to be contracted but extended. So, according to the correct answer or according to the answer which you have obtained from Lorentz transformation, this particular event occurred at a distance of 25 kilometers from the origin, not at 16 kilometers, not at 20 kilometers but at distance of 25 kilometers. The length does not, has not contracted. Is it, is it not, does not it look funny? Let us try to find out what is, what is the reason why we are getting different answer. Let us try to understand and let us try to really think that what, why this earlier result was wrong and why this particular result is correct. Why this discrepancy? Look at time. Let us do first a time transformation. t prime is equal to gamma, t which is 0, minus v which is 0.6 here, rate of velocity between the frame, x coordinate of the event which is 20 kilometers, 20 into 10 to the power 3 meters divided by c square. If I take c is equal to 3 into 10 to the power 8 meter per second, I find t prime is equal to minus 5 into 10 to the power minus 5 seconds. What does this negative, see there is a minus sign here. So, this thing becomes minus. What does that minus sign means? It means this event occurred before 0 time. Remember, according to s prime observer also at t prime is equal to 0, his origin was coincident with the origin of s. And this event t prime, this particular event of lightning struck, according to s prime at t prime is equal to minus 5 into 10 to the power minus 5 seconds. It means it appeared before. It means his, the origin of s and s prime were not coincident by till that particular time when the event occurred. The two origins became coincident at a later time, at 5 into 10 to the power minus 5 seconds after this particular event occurred in s prime frame of reference. This is somewhat shocking, but this is what will be the perception s prime frame of reference, observer setting on s prime frame of reference that this particular event occurred before the origins of the two frames were coincident. This is what I have written. So, according to s prime, lightning had struck 5 into 10 to the power minus 5 seconds before the origins of the two frames coincided. In s prime, the events of the two origins coinciding, if we call this as another event, that the two origins being coincident, this event of the two origins coincident and the lightning striking are not simultaneous though in s they are. According to an s observer, lightning striking and the origin of s prime passing both occurred at time t is equal to 0 same time. Therefore, the events were simultaneous. According to t prime, the first the event of lightning struck before the second event that is the coincidence of the origin occurred later. They were not simultaneous. Simultaneous is relative. This is what we have been discussing by giving many examples that this is something which is very different from the classical ideas to understand that simultaneity is not relative, is relative in special theory of relativity. According to s prime, the origin of s is approaching him with a speed of 0.6 c. If we are looking with respect to an observer in s prime frame of reference, this particular origin s was actually moving towards him and only after 5 into 10 to the power minus 5 seconds will the origin s reach his origin. So, this origin is always coming towards him and it will take 5 into 10 to the power minus 5 seconds before this origin will really come and meet him while lightning has struck earlier, while lightning has struck here when the origins were not coincident. This is what I have written. According to s prime, the origin of s is approaching him with a speed of 0.6 c and only 5 into 10 to the power minus 5 seconds after the lightning is striking will the origin of s reach his origin. I wanted to make one particular point which sometimes again I find that students are somewhat confused. See, it is not important when the observer in s prime gets the information of lightning because the information will take certain time to reach. What is important is the time of the event. Suppose one of our friends is departing from railway station at a particular given time and I come to know that his train departed exactly at 10 o'clock. This information may not reach me exactly at 10 o'clock. The person who has gone to see him off may come here, may take 1 hour to reach my place and may give me information that the train had departed actually at 10 o'clock. It does not mean that the event occurred at the time when I got the information, event had occurred earlier. See, what is important is the time of the event. I know that the train departed at 10 o'clock even though I might have got the information late. So similarly, an observer in s prime may have gotten this information later because it always takes finite amount of time for the information to reach him. But all this time if he tries to calculate and take into account, he can find out at what time the event actually takes place. So for example, from the railway station if somebody sends me light signal saying that train has departed, I know how much time the light would have taken to reach me here and I can calculate at what time this particular event would have occurred. So when I am saying the time of the event, it is important to realize that this is the time at which event actually occurred. So this is what I have written. Note when the observer in s prime gets the information of lightning is not important. What is important is the time when the lightning took place as per his calculation. Now as we have said here that after 5 to 10 power minus 5 seconds, the origin of the two frames will reach and at that time according to s prime will be time t prime equal to 0. I can find out what is the distance here. How do I find the distance? I know that after 5 to 10 power minus 5 seconds, this origin will reach him, will reach here and I know what is the speed which is 0.60. So I can calculate what will be this distance at the time when lightning struck. So what can find out what is the distance of the origin of s from the origin of s prime at the instant lightning struck according to s prime observer. Remember all these things I am talking now with respect to s prime observer. We should not never confuse the frame of reference. When I am talking one frame of reference, let us assume that we are sitting in that frame of reference and try to get all the information in that frame only. So I calculate this particular time 500 to the power minus 5 seconds multiplied by the speed with which the origin is approaching me. This distance is 9 kilometers. So let us now recall our ideas, readjust our things, assume that lightning struck at point P and a stationary rod OP of 20 kilometer length is at rest in s as we have said earlier. This length is proper. This length would indeed appear to be contracted in s prime and would indeed be given by 16 kilometers. About that particular aspect there is no problem. The problem is only that this particular length contraction whether my coordinate gives me the correct length or not. This is what I have written here. But according to s prime the lightning occurred before his origin coincided with s. Hence the coordinate of the event does not measure the length of their ordnance frame. While what I had calculated by length contraction is the length, but the question that has been asked is what is the coordinate of that particular event and that coordinate is not equal to the length of the rod. So I have given the situation in a picture. When the lightning strikes, assume that lightning strikes at that instant the picture according to s, the picture that s observer will draw. He will draw that the origin of O prime and O were coincident at that time and at a distance of 20 kilometer this lightning struck. This is the way s is going to picturize the event. But according to s prime the picture will be somewhat different. According to s prime he will feel that he is 9 kilometers away from the origin. This event actually occurred at 25 kilometers but then this rod OP according to him will be 25 minus 9 kilometers which is 16 kilometers which is indeed the contracted length that we had observed. So that is what is the difference of perception. According to this observer s prime he was 9 kilometers away from the origin when the event occurred. So the coordinate of this particular event is actually 25 kilometers what we obtained from Lorentz transformation. Being 9 kilometers away from here 16 kilometers is actually the length of the rod OP which is indeed the contracted length. Before I end let us discuss one more important aspect of this particular Lorentz transformation which is also very popular which is called time dilation. For discussing time dilation we have to define a proper time interval just like we have defined a proper length. Proper time interval we can define between any two events. There could be any two events one event two we have defined many types of events. We will be defining also in the course of lecture many other events. If these two events occur at the same position in a frame of reference then that time interval which is measured in that frame of reference is called a proper time interval. So let us assume that two events appear to be occurring exactly at the same position. So I am sitting here something happens here then again exactly at the same point something else happens here exactly at the same point. So if I take the time difference between these two events that is what is called proper time interval and I measure the time interval between these two events. What I am trying to say is that if anybody else who is moving relative to me measures the time interval between these two events he will find that this time interval is larger that is why it is called time dilation like the eye specialist dilate the eye of the ball becomes bigger. So similarly this particular time interval becomes dilated it has become larger that is why it is called time dilation. In any other frame the time interval between these two events would appear to be larger than the proper time interval that is why it is called time dilation. Now let us assume that it is the S frame we have already discussed what is S frame and S frame that is the S frame in which the time interval is proper. It means these two events must have occurred at the same position in S frame. If they have occurred at the same position it means even their coordinates X coordinates would be same. So it means these two events even one and even two occurred at the same value of X and let us suppose the time as measured by an observer in S is T1 and T2 for these two events. So T1 minus T2 or T2 minus T1 whatever you want to call the difference of the time interval is the proper time interval as measured in S because in S the two events occurred at the same value of X or same position. Now I apply Lorentz transformation if I apply Lorentz transformation exactly like before I get T2 prime is equal to gamma T2 minus V X upon C square this X I have taken to be same. So this X I have not written X2 for the first event I have written T1 prime is equal to gamma T1 minus V X by C square this X being same. I take the difference of these two again these two will cancel out I will get T2 prime minus T1 prime is equal to gamma T2 minus T1. This is what I have written here the time difference between these two events in S is therefore given as T2 prime minus T1 prime which is equal to gamma T2 minus T1. This T2 minus T1 was the proper time interval because this was measured in a frame of reference in which the two events occurred at the same position. So this is proper time interval gamma being greater than 1 T2 prime minus T1 prime will be larger than T2 minus T1 and therefore time interval will be dilated. Many times this proper time interval between the two events is written as tau this is one of the symbol which is very very commonly used. There are one or two small comments which I would like to make first that you will note that to apply time dilation only the X coordinate of the two events had been used. Strictly speaking if the Y coordinates of the two events were different one event occurred here and another event occurred at some other height. Here another event occurring at some other height but at the same position is still time dilation formula is applicable. So strictly speaking that time interval is not proper. We will define exactly correct definition of proper time interval little later but as far as application of time dilation formula is concerned only the X coordinates of the two events need be same. So that is what I have written. We note that to apply time dilation only the X coordinates of the two events have to be same. Though the general definition of proper time interval all the coordinates have to be same. As we said we shall discuss this aspect in detail later. The second point which I want to mention is that if I apply Lorentz transformation for X coordinate according to S prime observer these two events would indeed appear to be occurring at different position. Like in the length contraction according to an observer in S prime the time difference the two events was not same. It was different. Here the X will be different but this is not so shocking because even classically that is what we see. See if any event for example occurs here even classically we do not have to go to relativity. If an event occurs let us suppose somewhere here and one frame was here another frame was let us say also supposed to be here they observe these two events exactly at the same value of X. But the second event when it occurs the second observer has moved to the right therefore the event the X coordinate has changed. Therefore the X coordinate being different is not all that shocking. This is also expected classically. It is a time difference being different in different frame that appears to be much more shocking. So that is where I will end my lecture just by giving small summary. We discussed two important consequences of Lorentz transformation namely the length contraction and time dilation. Then we gave some examples of length contraction and one where one can make an error in the direct use of the formula. Thank you.


Hello everybody. Welcome back to another After Effects tutorial. In this one, I'm going to show you how to create a glass morphism effect, but for video. I've shown you how to do it in Photoshop for still images. This one's for video. And if I move this little iPhone-type glass morphism effect around, you'll see here that it moves with it and the blur stays underneath it. So how did I do this? It's pretty simple. Let me show you. It'll take about two or three minutes maximum. So let's just start from scratch. So there we go. We've got this video here of this lady. She's just dancing around doing her thing. Note right away that the video is very, very bright and it's very, very sharp. That's deliberate because blurry videos don't look too good with glass morphism effect. Okay. The next step is making sure you are not selected on the layer, but you clicked outside of it. You now want to go up here and grab your rectangle tool. You want to select a fill color of, well, it doesn't matter what color it is. I'm going to go with a nice purpley blue color like this. And I'm going to turn the stroke off just like this. This little white rectangle with the red bar through it means no stroke. So I've got fill, no stroke. Now I'm going to go ahead and just draw your rectangle and I'm going to make it something like this.  So there we go. We've got this rectangle. I've drawn it. Beauty. Now the next step is we want to actually blur this out. But before I do that, there's a little button here that's an adjustment layer and it applies to just this layer. So I'm going to click that and then, well, nothing's happened yet. We want to go to effects and presets. I got a sneeze. Hold on one sec. Oh, I'm okay. Excuse me. Ah, I'm all right now. All right. Walk it off, Curtis. You want to go to the effects and presets and then go to fast box blur. Select that drag and drop it on top. When you do that, nothing happens right away. But you got to go ahead and change the blur radius.   Let's go to 40. This is really, really glassmorphistic, if so to speak. So there we go. We've got the, we've got the general look here, but it's not quite ready. What we're going to do now is we're going to duplicate this layer, go to edit, and then go to duplicate. And when we do that, we're going to get an extra set of blur on there. And that's actually not what we want. So I'm going to go into the layer and I'm going to go into the effects. I'm going to click on the fast block box blur. And that's a mouthful and I'm just going to delete it. But what I am going to do is I'm going to go up to the top here and I'm going to switch the fill in the stroke. So instead of having a fill of purple, I'm going to actually create a fill of nothing. I'm going to remove the fill, but I'm going to add the stroke. So I'm going to add a stroke in. And for this one, I'm going to go with a stroke of let's say something like 15. Let's go. Yeah, it's 14 pixels. Now you don't see it yet, but here's what you got to do next.  Now it is very rectangular. So it doesn't maybe quite look right unless you've got like an iPhone five or something. But if you want to get to that modern look, you just want to drag drop down contents, go to let's see where the heck it is. It's under transform rectangle. And we're just going to go ahead and add and actually not we're going to go to rectangle path. And we're going to just add in surround this I'm just going to ask and scoop this up to something like that. Let's go to 60. And then if I click out, look at that we've got a nice little frosted glass glass morphism effect applied in just a few minutes. Thanks for watching.


So let's talk a little bit about cyanide.  Cyanide's an exception. Is there cyanide in this room right now? Yes. There's some cyanide inside each of us. What we're going to talk about with cyanide is a little bit about where it's found, a little bit about its history. And using the general paradigm that we've talked about before, we'll talk a little bit about its physical chemical properties, about its toxicokinetics, its absorption, distribution, biotransformation and elimination, about its mechanism of action, that is the toxicodynamics of cyanide. And then, finally, of course, the most important thing is the clinical presentation because we need to know certain things about the agent in order to treat the patient. But the bottom line is we have to recognize what the patient has and treat the patient. Now, as I said, cyanide is found everywhere. Cyanide's an interstellar space. It's been on the earth for a long time. Organisms have had to develop strategies for dealing with small quantities of cyanide in the environment, in the external environment and in the internal environment.  If you're a smoker, you have twice the quantities that you see on the slide. Now, cyanide's one of those things where, in small concentrations, not only can the body handle it, but in fact, the body actually requires it in larger concentrations. Once you overwhelm the body and pass a certain threshold, it becomes dramatically toxic. It's been a favorite of assassins for long periods of time. All sorts of great stories that we have about cyanide. And the other thing I wanted to remind you about is that with the nato codes, what's the nato code for cyanide? For hydrogen cyanide, it's AC. And for cyanogen chloride, it's CK. Sometimes, in referring to cyanide, the cyanide ion itself, we use CN with a little negative cyanide ion. Don't confuse that with the nato code CN, which stands for chloracetofinone. It's one of the riot control agents, and you'll hear about that from Colonel Newmark. Not meant to be confusing. I'm trying to outlay fears about this. Okay, now, in ancient Egypt, cyanide wasn't known as an isolated compound, but it was known that certain extracts, they had in Egypt what they called the penalty of the peach. And it was an extractive from peach pits, and that was administered to condemned criminals. And that was the method of execution. The Romans had something very similar. They used cherry laurel water. But, again, it was an extract from cherry pits, and we'll find out that plants, in just a few moments, we'll find out that plants contain cyanogenic compounds, compounds that release cyanide under certain conditions. But it wasn't really until 1782 when Fonshella actually isolated cyanide as a distinct compound. He purified it, noted it, didn't realize that it was poisonous until four years later when he accidentally dropped a flask of the stuff and became the first known fatality from purified cyanide. Sir, you may remember from the history lecture, from Dr. Seidel's history lecture, that Sir Lyon Playfair wanted to put cyanide in shells to be lobbed onto Sebastopol during the Crimean War. Napoleon III wanted to dip his... In fact, he did dip his soldiers' bayonets in cyanide. They're by antedating by, I guess, about 100 years the use of cyanide in bullets by the Symbionnes Liberation Army. Never mind that that would have vaporized, wouldn't have done any good, but they dipped their bullets in cyanide. Cyanide, obviously, has a very colorful history. In warfare, it was used in World War I by the French in a compound called Vincenite. It wasn't very effective. Not because cyanide is not very effective, but because the way in which it was put into the munitions was substandard. Munition wasn't filled as much as it should have been and because of the bursting charge. Cyanide, hydrogen cyanide, is extremely volatile, but more than that, it's explosive under certain conditions. And when the munition burst, most of the cyanide went as well, and so it was hard to achieve good concentrations on the battlefield simply by... for logistic considerations. It is possible to achieve higher concentrations if you know what you're doing. And of course, you're in World War II. We don't have documented evidence of its use on a battlefield, although there are certain reports that the Japanese may have used this. We're not really sure about that, but of course everybody knows that in World War II the Germans in gas chambers used Zuclone Bay, which was hydrogen cyanide absorbed onto a physical medium and then that. Now, that's slightly different from what happens in gas chambers today where a cyanide salt is dropped into an acid and releases hydrogen cyanide. Not quite the same mechanism, but the same result. And in the Middle East, we know... well, we know that the Iraqis used nerve gas. We know that the Iraqis used mustard. And part of the reason that we know that is because the compounds were persistent enough to stay around. We found munitions with them. Not so with cyanide, so we don't have any physical evidence for that. It is highly probable that people like these died in the 1980s from cyanide. Now, off the battlefield, do you all know what happened in 1978 in Guyana? Yeah, the Jonestown mass suicide where people added cyanide to great flavored Kool-Aid. 1982, Tylenol. What happened in 1982 with Tylenol? Right, somebody took apart the two halves of the gelatin capsule and put cyanide in there and caused several deaths, and that's why we have Tylenol caplets now instead of Tylenol capsules. But we still use cyanide for judicial executions. Cyanide is a very important component of smoke from fires. Theoretically, anything that has carbon and nitrogen in it together has the potential to produce cyanide on combustion. Wool, silk, lots of plastics, especially the polyurethane, polyacrylate kinds of plastics generate significant quantities of cyanide. There's enough cyanide that comes from a typical sofa that was manufactured maybe 10 years ago to cause fatality if you're in a room where that sofa is burning. Industrial processes, cyanide is used all over, just like phosgene. Cyanide is used for gold electroplating, for photo processing, for a variety of industrial processes and it's used by the tons. And you find it at home, you can, silver polish can generate this. PCP labs, cyanide is found in many, many places. Now, remember I said that some plants, remember I talked about cherry laurel water and the penalty of the peach? Well, it turns out that plants, especially some plant pits, contain compounds which, although non-poisonous in the form in which we find them in the pits, can be converted into cyanides, that is they release cyanide upon digestion. We have enzymes, there's an enzyme called emulsin, which we either produce ourselves, some animals produce that, or it's released by bacteria that are in the gut if the pH is right. Now, where do we find these glycosides? What plants? Well, for one thing, leitreal, the anti-neoplastic compound, which isn't why it's highly touted in Mexico against cancer, well, it was supposed to act because in fact it produces cyanide. If you were just to inject leitreal into a vein, you don't get any effect. But if you ingest leitreal, which is the way it's taken, and it gets to the duodenum and the pH rises, then you get the activation by emulsin, the release of cyanide, and then you get toxicity. Who else? Well, cassava. Cassava is the same name in Africa given to manioc or tapioca in Brazil. It's the major source of carbohydrates for 300,000 people in the world. And if that's improperly processed, you can get cyanide toxicity. And what that cyanide toxicity leads to is a series of chronic symptoms and signs and diseases. There's something that used to be called Nigerian ataxic neuropathy, now we generally call it tropical ataxic neuropathy, where there's a variety of bulbar and other neurological signs, probably because of cyanide release from improsper processing of cassava. There's tobacco amblyopia. The reason that many people who smoke have problems, subtle problems with night vision is probably because of the elevated cyanide in the blood and other tissues of smokers. Now, I remember studying cyanide many years ago and being confused by all of the biochemistry and how does it work and how does the therapy work, and I really couldn't understand it until I sat down and simplified things. Remember how I said all real teaching and learning involves some simplification of data. Cyanide enters into a variety of reactions with proteins and metals and all sorts of compounds in the body, but for our purposes there are two reactions that are really important, two general reactions. They are the following. Cyanide loves to react with metals, especially transition element metals. And where do we find those in the body? Well, we find molybdenum, we find zinc, we find copper in small amounts in various enzyme systems in the body, but especially we find iron in the body. Where do we find iron in the body? RBCs. Okay, RBCs. And in what form? It's hemoglobin. Excellent. But the iron in hemoglobin is in a ferrous or plus 2 form. Cyanide likes to react with ferrous ion, but it would prefer to react with ferric ion. It has an even greater affinity for ferric ion. Do we have ferric ion in red blood cells? Not normally. Normally, most of the hemoglobin is in fact in the form of straight hemoglobin, but very small quantities in the body are converted into what we call methemoglobin, which is where the iron is in the plus 3 form. We have vanishingly small quantities of that in the body normally because it doesn't carry oxygen. But when hemoglobin is present in the body, cyanide is really attracted to that. Keep that in mind. Now where else in the body do we have iron compounds? Well, in various enzymes, and notably in the cytochromes inside cells. What are cytochromes good for? Oxidation. In fact, the so-called oxidative phosphorylation. The cellular respiration where we... that's really where we use oxygen and generate ATP. That's important. And that compound actually cycles back and forth between a ferric and a ferrous form. And so cyanide will combine with both forms, but in our case, we're really dealing with the combination of the ferrous form. Now the other thing, sulfur-containing compounds, cyanide will react with almost anything that has sulfur in it that has two sulfurs connected by a partially covalent bond with polarity. Don't worry about the details of that. They're called sulfanes. The point is this is catalyzed by an enzyme called rodentase. One of the few enzymes that doesn't end in an A is it ends in an E. And it's an irreversible reaction. So when cyanide goes through that process, it has gone forever. Whereas this reaction with metals is a back-and-forth kind of thing. Now here's an example. Here's a compound called hydroxycobalamin, or vitamin B12A. It exists in the body. It contains cobalt. Cyanide reacts with that cobalt, and the resulting compound is cyanocobalamin, which is just another name for vitamin B12. So here's a case where cyanide participates in the normal of the metabolism in the body. And we need it. Cyanide, again, with sulfanes, as you'll notice, that's a one-way reaction. And the rate-limiting step for that reaction is not usually the amount of enzyme. It's the amount of sulfur in the body. Keep that in mind, too. If we supply more sulfur to the body, more cyanide reacts with the sulfur that we administer. Just keep that in the back of your mind. Now, the gases that we're worried about on the battlefield are AC and CK. AC is hydrogen cyanide. That's the needle code for hydrogen cyanide. It's an acidic compound in the sense that it has the capacity to dissociate into hydrogen ions and cyanide ions. In point of fact, that dissociation is so weak that the resulting solution, hydrocyanic acid, doesn't even turn litmus paper pink. And so we talk about the cyanide ion as if that's the bad actor. And in fact, it by itself does do bad things to the body. But hydrogen cyanide with the hydrogen attached can also act in the same way. Very volatile. And the fact that the acid is called hydrocyanic or prusic acid reminds me to tell you that the reason that cyanide is called cyanide. What does cyanide mean? What does the root mean? Blue. The same root. There's the Greek Quran from which we get cyanosis. So cyanide, cyanosis, hey, that's an easy association, right? It's an association I want you to break because there's no association. Cyanide is not classically associated with cyanosis. The reason we call cyanide cyanide is because the starting material that von Scheller used for his synthesis was ferrocyanide, which is prussian blue. And so the resulting compound, he added a blue name to it. Cyanide is not blue. It does not cause cyanosis except in the terminal stages. Now, it's very volatile. The gas rises, it's lighter than air. And its boiling point is about 78 degrees Fahrenheit. So in this room, I think some of you are feeling a little cold. It's probably not 78 degrees in here. In what form would we find cyanide? In what physical state would we find cyanide? Okay, it'd be a liquid. We haven't yet reached its boiling point. But we're pretty close to the boiling point and so it would evaporate fairly quickly. So we would encounter cyanide as a liquid and a gas. And it's possible to come in contact with liquid cyanide. It's been used in assassinations that way. It's thrown in the face, immediate death. But on the battlefield, more likely it would be contact with the vapor from that liquid. Now people describe classically this. Oh, cyanide, yeah, that's the one that smells like almonds. Smell is a very subjective sensation. In fact, the first people that smelled this thought it smelled like burning rope, you know, all sorts of things. But more than that, there are a couple of things that are wrong with smell. Quite apart from the fact that it's difficult to come up with an objective description of the smell. First of all, about half of us have a genetic defect that prevents us from detecting that odor at all. So it may be well that some people can say, oh yes, I can recognize cyanide unless you've smelled it. You don't know whether you can smell it or not. The other thing is, what's the sense of our five senses that accommodates the most quickly? It's olfaction.   And so smell is, except for some things like mustard, is not a very good warning sign. Onset time dramatically rapid. And look at the LCT50. 2,500 to 5,000 milligram minutes per cubic meter. By comparison, phosgene was 6,000, so this is much lower than that. Oh, half as much as that. Look at the variation. One of the things that we'd said before is that CT is a useful estimate of the effects of a gas or a vapor. And it's fairly constant. The reason that this has so much variability in, and actually the variability extends on both ends of that 2,500 to 5,000 number, is because when small quantities, the body can deal with cyanide. It's been dealing with cyanide for years. It's only when you overwhelm that. So if you were to present the same amount of cyanide to somebody over a long period of time, no problem. The body can metabolize that. If you present the same amount, the same CT product to somebody, over a period of seconds, it overwhelms the defenses and kills the patient. So that's really the chief exception to our idea that CTs, CT products, are good estimates by themselves of exposure. Cyanogen chloride, CK. CK is just like cyanide with chlorine added. It dissociates to form chlorine and cyanide. So what do you think the additional effects are going to be now that you've had the pulmonary agents lecture? What are the effects of chlorine? They're irritation, aren't they? Of the upper airway, and you'll know this. In fact, cyanogen chloride has sometimes been used as a warning or irritant gas, marker gas for other things that are even more toxic than it is. And why? Well, it's pretty toxic, too, isn't it? Well, the LCT50 is 11,000 milligram minutes per cubic meter. So it's twice as potent as cyanide, right? Half as potent as cyanide. Remember, as the LCT50 goes up, it takes more of the substance to kill you. Now, you'll notice from this graph, here again, this is a graph that you've seen before, where I've put agents in their order of introduction into warfare. So we have chlorine in it, about 6,000 milligram minutes per cubic meter. And we have phosgene, which is about 3,000, and then cyanide with its variability, but let's say that it's 2,500, which is more potent cyanide or mustard? Mustard's the next one, which is H. Mustard. Mustard is more potent than cyanide. And you've heard that point made by Colonel Hearst, that, yes, we think the cyanide's bad, well, mustard is more potent. So what makes cyanide such a bad actor? Cyanide's far worse than mustard on one account. Well, because of its speed of action. Rapid onset. If somebody dies from mustard, that death is unlikely to be immediate. It's going to be down the road. The first 24 hours, if you have the massive exposure with this formation of pseudomembranes in the trachea and the bronchi, but more likely it's going to be a few days or maybe even weeks down the road. But cyanide, if you get enough of it, death is almost instantaneous. Well, within minutes, anyway. And that's why it has such a bad reputation. Now, we talk about cyanide and we say it's a blood agent. And remember, in the introduction I said, yeah, one of the classifications is blood agent, but later why that's a bad name? The reason that it's a bad name is that it implies that cyanide does its dirty work in the blood. And it does do some things there, but the chief action is elsewhere. Now, why were these agents called blood agents to begin with? Well, it's the same distinction we made at the beginning of the lecture. The first agents used in World War I besides the minor use of riot control agents were what? Chlorine. Phosgene. The pulmonary agents. And the pulmonary agents act locally. Now, here we then have cyanide introduced. Cyanide is distributed systemically throughout the blood. So therefore it's a blood agent. So it's correct to say it's a blood agent. But is mustard distributed through the blood? Mustard came along a little later. Sure, it was. It is. Are nerve agents systemically distributed?  So to say that cyanide is a blood agent is correct, but it's no longer very specific, because that term could also equally well apply to mustard and to nerve agents. And as I said again, the other reason I don't like it is that it implies that it somehow binds with something in the blood and that its mechanism of action is there. Now, we know that carbon monoxide, for instance, binds with oxyhemoglobin to form carboxyhemoglobin. Now, cyanide will react a little bit that way, but its main site of action, we think, is in the cells. And what it does is to interrupt that process of cellular respiration to which we alluded a few minutes earlier. Here is the mitochondrion. Here we have the various elements of the cell and we have the nucleus up here and here are some ribosomes and here's the endoplasmic reticulum and the Golgi apparatus and all of this. What we're really interested in is this little organelle here with the outer membrane and the inner membrane and the crusty. This is a mitochondrion. And the mitochondrion is the powerhouse of the cell. That's where oxygen is finally taken, converted into hydroxyl ions and hydrogen ions and where ATP is taken and generated and ATP is generated. So this is where energy is produced in the cell and where oxygen is used. And how does that happen? Now chemically it happens through a process that we'll talk about in just a few slides, but we say that cyanide binds to an enzyme in the mitochondrion. The enzyme complex is called cytochrome A3 and this particular enzyme in that complex is called cytochrome A3. You don't have to remember this, just the general idea. But what's important to know is that it's a stable binding, but it's not irreversible. Cyanide can bind to that and it likes to bind with metals. And here's a metal ion in the ferrous form most of the time. What are the two reactions that cyanide likes to participate in? It likes to react with metals and sulfur. Ooh, very good. Okay, that's what I wanted you to remember. And cyanide actually has a higher affinity for ferric ion if it could find it. Okay, what do we do? We interrupt oxidative phosphorylation. This is the terminal end of the electron chain process that you may remember from biochemistry and it's not my intention to go into this in a lot of detail. Just to remind you that this is where ADP is taken and changed into ATP. This is where oxygen is taken and changed into essentially the components of water. This is where energy is generated. And this is also where cyanide acts. So now you can tell me what cyanide does. What would be the effects on a cellular level of cyanide? Here's shutdown energy production in the cell. So no more ATP. What about the oxygen presented to the cell? Can't be used. So the cell says high to it and buy to it and the oxygen goes into the venous blood. So we're not going to have any ATP generated. There will be no extraction of oxygen from the blood. So venous blood can be just as healthy pink looking as material blood. Now using oxidative phosphorylation that's what happens when we get compounds from the Krebs cycle from the Cystric Acid cycle from oxidative metabolism. Is that the only way the cell can generate energy? No. We can generate energy via glycolysis. We have fermentation cells can ferment. The only problem is when they do that they generate lactic acid. Because if that gets into the blood we have an lactic acidosis and a high anion gap. Now here are normal mitochondria here are mitochondria that are unable to use oxygen. They swell, the inner membranes get bloated. There are real problems with that. The mitochondria just don't work anymore. And that's what happens with cyanide. Now at an organ level, cyanide affects various things. I'm going to ask you, what is the organ or tissue in the body that receives more blood and oxygen per gram of tissue than any other tissue in the body? The brain! Wrong. The heart! No. The liver! Ooh, good guess, not right. The lungs! That's another good one. The tissues in the body that receive more oxygen and blood per gram of tissue than any other are two small nubbins of tissue at the bifurcation of the common carotid into the internal and external carotids. They're the carotid bodies. They're there to monitor the partial pressure of oxygen in the blood. And when they sense that there's not enough oxygen they send signals to the brain to say, spit out epinephrine! Send messages to the adrenal medulla to spit out epinephrine to increase the heart rate and to increase the blood pressure. And they say to the brain, send messages to the respiratory center to increase the rate of breathing. And so one of the first effects of cyanide poisoning will be intense stimulation from these carotid and their similar receptors on the aortic arch, these chemoreceptors. Now cyanide damages, cyanide is systemically distributed to every cell in the body. And it affects almost every cell in the body, but it affects some more than others. It affects the nervous system. And eventually the neurons in the respiratory center of the brain die. What happens when the neurons in the respiratory center of the brain die? Right, central apnea. And if untreated, that is the mechanism of death for people with cyanide intoxication, even though the cause of death is cellular hypoxia. But the mechanism of death is, cellular hypoxia in the respiratory center that the diaphragm doesn't work. Now if you can ventilate these people, you can actually keep them alive long enough for cyanide to kill the heart, which it does, given enough time. But what happens first in a patient that receives a really high concentrate now, if you're splashed in the face with liquid cyanide, some people collapse immediately from that. We have case reports. And probably from massive release of epinephrine generating a terminal cardiac dysrhythmia. But in the usual case, inhalation of high concentrations causes, first of all, stimulation of the carotid bodies and a brief period of hyperventilation. Now if I were to say, gas, what's the first thing you do? Hold your breath, stop breathing. Now this is such an overwhelming urge that people exposed to cyanide may have difficulty obeying this first and very important part of the mask donning drill. The onset is very fast. We have this transient increase in rate and depth of respiration. The blood pressure goes up. Why? Because of the epinephrine that's released and the heart rate goes up. And eventually, we have from hypoxia, we have convulsions, we have rigidity, we have apostatnes, which is the head back like this. We have trismas, which is the tepne of the balakja. And all this happens within a minute, within a half a minute or so. Patient loses consciousness, goes into convulsions, and then because of hypoxia, the heart rate slows, the blood pressure drops, the heart stops, and the patient dies. But after this respiratory depression. Now, what about, yes? Do you have a question? Yes. Did I get some of these clinical trials or have they done autopsies for prisoners that get executed via the gas chamber? I mean, is this honest? Yes, the answer to that is yes. And I don't have time to discuss this in detail, but there is quite a literature on people, on gas chamber victims who have been, who have been observed medically during executions. We have an even greater wealth of data from animal experiments, but this has been confirmed in humans, too. This is what happens. Not everybody's the same, but classically, this is what happens. Now, in the skin, what about the skin? The skin can look nice and pink, even flushed. Why? Because blood in the veins and the capillaries still has plenty of oxygen in it. It's not for the same reason in carbon monoxide poisoning, because there, the compound itself, carboxyhemoglobin, is kind of a cherry red color. But here, it's simply because oxygen is not being extracted from the blood. You may have this bitter almond smell on your breath, but remember, cyanide poisoning is not characterized by cyanosis early in the course. Now, terminally, yeah, all sorts of things can happen, and the patient can turn blue. And not everybody has a classic presentation. So some people may be cyanotic at the beginning, but classically, no cyanosis. Now, it's often said that cyanide is an all or nothing thing.  And so, why do we even worry about therapy? Well, there is a continuum with cyanide. We know there is a continuum because we know that there are people with tropical ataxic neuropathy from eating just a little bit more cyanide from baseline.  But even if you inhale quantities that are not immediately lethal, you can develop ataxia, headache, confusion, panic, a variety of symptoms that are really nonspecific and some that you might expect to find on the battlefield anyway. So, the classic signs.  Well, that's one sign, you know, the skin, the profound metabolic acidosis, the odor if you can detect that. Now, the patient himself or herself is going to have this transient hypertension, this transient tachypnea as well. You know, it's a patient that goes ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah, ah. And within a few minutes, it's all over. Now, what does that remind you?  Okay, whatever. What difference, does that remind you of anything else that we've studied? What else can look like that? Nerve agent. We hear Dr. Seidel's stories of people who just collapse like that. Do nerve agent casualties convulse? Sure they can't. Do nerve agent casualties, you know, so how do you tell the difference? What about the differential diagnosis? That is, what about the A on our asbestos list? What happens if it's cyanide rather than nerve agent? If you see a patient acting like that? Now, for very long, it can happen that nerve agent casualties can stop breathing within a little while, within a few minutes. So can cyanide casualties? No cyanosis. Okay, that's one key. No cyanosis. Now, I say that, that's classically true. Now, there will be some cyanide patients who will be cyanotic. But yeah, if you see somebody who has that presentation but is not cyanotic but is not breathing, cyanide moves up the ladder a little bit in the differential. What else? Myosis. Nerve agents cause myosis. Does cyanide cause myosis? Not at all. In fact, the patient, pretermily, will have the fixed and dilated pupils characteristic of hypoxia. There will be some medriosis. Not that cyanide specifically causes that, but they'll get that from hypoxia. Okay, and we could go through this whole mnemonic about, okay, what state would we, if we do come to the conclusion that cyanide, what state are we likely to be thinking about? Vapor, right? Or gas, if it's above 78 degrees. Okay, body sites, we're mostly talking about inhalation, although there are plenty of other ways, other routes of getting cyanide. The effects, are they local or systemic? Well, they're systemic, they're dilated throughout the body. That's what makes it different from the inhalational agents that Colonel Baxter told you about. And of course, we can run the gamut from mild to severe cyanide intoxication. Time course, these patients will not get better if they're already convulsing, unless you can provide some good treatment. And let's talk about the treatment. You already know the first step is to protect yourself. And with any agent, there's general supportive therapy. Some agents have specific antidotes. And some agents are so important that you need to know about the specific antidotes without consulting a book. There are two classes of agents like that. Because there are only two classes of agents that kill you so dramatically quickly that you don't have time to consult a book. What are those two kinds of agents? Nervin cyanide, they're the two people that can cause somebody to go down like that really quickly. And can kill them very quickly. Let's talk about what we would do generally. Of course, for any agent, terminate the exposure, separate the patient and the agent, put a mask on the patient, drag the patient out of the area, whatever. Airway breathing and circulation, extremely important.  I have a story that I often tell, and I don't have a lot of time about a medical student that found his dog collapsed, administered mouth to snout respiration. He himself collapsed, was taken to the hospital, somebody smelled bitter almonds on his breath. He got there, he was transported, he was revived, he had cyanide. And so elimination through the breath is actually important in cyanide. You'd think that oxygen wouldn't help these people. Because they can't utilize oxygen. But in point of fact, using 100% oxygen does seem to help them. Maybe it's a mass effect of having so much oxygen that pushes the cyanide off the enzyme. We're not sure. But don't neglect to give oxygen, even though you'd think it wouldn't do any good. Correct the metabolic acidosis and observe, because these patients can sometimes crump after they have been apparently resuscitated. What are the two reactions that cyanide likes to participate in? Likes to react with? Metals. Metals and sulfur. That's what we're going to do. We're going to give it a metal, and all we're going to give it is met hemoglobin. And we're going to form met hemoglobin from the hemoglobin that's already in the body by administering a compound, a nitrite, that will generate met hemoglobin and the cyanide can react with that. And the second thing we're going to do is to use this enzyme rodinase. We're going to administer a sulfur donor. We're going to let the enzyme already in the body convert some of that cyanide to a less toxic compound by reacting it with a sulfur. Let's talk about the first process. You know what happens here, even though this looks complicated. Oxygen is taken in by the lungs. It reacts with hemoglobin to form oxihemoglobin. And in the tissues, oxihemoglobin gives up oxygen and becomes the oxihemoglobin again. Both of these are in the ferrous form. Cyanide doesn't like them very much. Cyanide, given to the body, if it doesn't see, if all it sees is this and this, it'll go to this because it likes to combine with the cytochrome A3. So we want to give cyanide something else, something that's definitely in the ferric form. And we can do that by administering a nitrite. The nitrite generates, changes some of this oxihemoglobin to met hemoglobin. The cyanide then has a choice and it says, can I react with this or this? I'd rather react with this, and it does. It forms a temporary depot of a compound called cyanamin hemoglobin. How many of you have Windows 95? Use that. Okay, so you all have a trash can. I used to have to restrict this to Macintosh users. You have a recycling bin. Okay? When you put your e-mail or a document you don't want in the recycle bin and then you say, ooh, I really wanted that. Can you get it back? Sure, you open the recycling bin and out it comes if you want it to. That's exactly what I want you to think of this as. This is a recycling bin. Because this is a reversible reaction. If all you do is give the nitrite, this will stay here for a while and then it will come off and you'll have more nitrite in the blood again. And that may be the reason for delayed toxicity in some patients that appear to do well and then crump later on. How do we use it? We use it in a kit that used to be called the Lily antidote kit, cyanide antidote kit. Now somebody else who prepares it now and it's called the Pasadena cyanide antidote kit, but it's a standard kit that's used in emergency rooms and it has some sodium nitrate. It actually has in it and I'll let you see this afterwards. It has some aspirol or purls in here. They're just, they're little vials of amyl nitrite. You know all about amyl nitrite. And that's given by inhalation by crushing the vials and putting it under the patient's nose. What's wrong with that? In a military setting. Well first thing they're going to have a mask on and the other comment is to the point too. They may not be breathing and there are other problems with amyl nitrite so it's not a component of the military antidote kits. Instead we use another nitrite. We use sodium nitrite. Converts some of the oxy hemoglobin to methemoglobin. We're not really sure how it works because it has a therapeutic effect before we can detect really elevated levels of methemoglobin. So maybe there's another mechanism of action for this. But for purposes of instruction let's say that it acts by this route by generating methemoglobin. This is how much you give. Now I can ask you to memorize this and remember I said this is one of the, this cyanide is one of the components where you need to know what to do immediately. But you don't necessarily have to remember the dosage if you have a kit and what that means is you administer one of these vials and then you administer the second antidote that we'll discuss. If the symptoms recur, symptoms and signs recur, you administer half of the second little vial and that's the easiest way to remember this. What you need to do is to find an antidote kit and practice with it. So there's the antidote kit and this is what we've done. We've created a temporary depot, we've put something in the recycling bin. Now what we want to do is empty the recycle bin. That is, we want to get rid of the cyanide once and for all and we can do that by adding a sulfur donor. And the sulfur donor we use is the second component of the military version of the antidote kit. It's the sodium thiosulfate. This reacts with cyanide via the enzyme rodentase to form thiosyanates and sulfites which are less toxic and which are excreted in the urine. And here's the, this is what we do as fairly safe medicine, used to be, here's the administration. You can either memorize that or you can just remember that in the old kits you give one of these great big vials, 50 mils in the vial, or in the new kit you give the entire contents of one of the bottles, IV. And then if symptoms and signs recur you give half of the little vial and half of the second big bottle. Okay. Now, studies have been shown, that show that this combination can detoxify up to 20 lethal doses of sodium cyanide even after breathing is stopped as long as the heart is still beating. The chances of recovery by utilizing this method are very good. So just as nerve agent casualties can be immediate, they can appear almost to be dead and they can be dramatically salvaged by appropriate administration of nerve agent antidotes. So cyanide casualties can be dramatically resuscitated even almost at the point of death if you use the antidotes and use them correctly and have enough experience with the kit to know, okay, this is what's where, this is where I start the IV. Now, in other countries, especially other NATO countries, other antidotes are used. But you know, you can understand how they work because you know how our antidotes work. There are other hemoglobin formers and there are cobalt compounds and a variety of miscellaneous things. The other hemoglobin compounds, the other hemoglobin formers work the same way that nitrates work. They generate metahemoglobin. The one used in Germany is four-dimethylaminol phenol or DMAP. It generates metahemoglobin really quickly. So we generate this depot really quickly. The only problem is the depot degenerates really quickly too. So we can have more cyanide back in the blood. But this is what's used in Germany. Now, there's another compound, the second one on the list, peraminopropiofenone, PAPP. This forms metahemoglobin too, but it forms it so slowly that it's of very little use after cyanide has been presented to the body. So why do I even talk about it? How could we use a compound like that? For pretreatment. In fact, some of the studies in the institute that we're doing in the institute are studies leading to pretreatment with PAPP and other similar agents against cyanide. Not yet fielded, but studies are in progress. Now, in Britain, France, and Holland, cobalt compound is used, chelocyanar, which is cobalt edta or dicobalteditate. And this is what's cobalt? Cobalt's a transition metal, isn't it? Cyanide will react with that. And that's used as an antidote, but the adverse effects are sometimes pretty serious. Laryngeal edema, that can be a serious problem. And what is that? That's simply cobalt toxicity. Now, in France, France uses the precursor to vitamin B12. You've already heard about this. Hydroxylcobalamin can react with cyanide to form vitamin B12, but it's really difficult to add to administer adequate amounts because the solution's fairly diluted. This is a big molecule, and cyanide's a small molecule and this is stoichiometric reaction. Okay, so what do we know about cyanide? I've presented you a lot of biochemistry, well, as little as I could, but as much as I could to show you that it's interesting and to show you the mechanism of action and why the antidotes work. What do you need to remember about cyanide? It's a gas, okay? Usually, in fact, it could even be a liquid and a vapor. And above 78 degrees Fahrenheit, it will be a gas. Does it act locally or systemically? Systemically, dramatically. And what makes it such a bad actor? Why does it have such a bad reputation? Speed of onset. Speed of action. Short onset. Speed of death. Okay? Not really a blood agent, even though we call it a blood agent. It's a cellular poison. And what does a patient look like? He looks almost like a nerve agent, casually. But probably won't have a cyanosis. Will not have myosis. Few other discriminating signs unless you have time to draw blood for labs. And you won't, in this situation, you don't have time to draw blood for labs. You have to diagnose quickly and treat. And how do we treat? We treat with nitrites. Don't say nitrates. Get it right. Say nitrite. And with phyosulfates. We'll talk to you about how they work. Do you have any questions? Sir, how many successful resuscitations have they had with the kit? Oh, this kit has been used in... Okay, on the battlefield, we don't have... I'm not aware of any situation on the battlefield because we don't know about battlefield use of cyanide before the 1930s when this kit was developed. And in fact, before that, they used to use methylene blue, which is actually what you'd administer after giving too much met hemoglobin form. But we have quite a body of information from emergency rooms where people who have ingested toxic quantities of cyanide have been treated with this regimen with success. And it seems really to work. Now, some people do well if they're coming out of this. Some people do well just with observation. But we have an enormous quantity of literature of people who have been treated well with the antidotes and survived in this country and in many other countries. Yes. Can you use any of the antidotes, I.M., or via YouTube? No. No. The antidotes are not well-absorbed I.M. and in particular for DMAP, used in Germany. If it's given I.M., it causes necrosis of the muscle. So you never want to give that I.M. But these antidotes are I.V. antidotes.