 In this lecture we will learn to appreciate the Galilean transformation and its built-in assumptions. A decent understanding of the past will help us to set the stage for the present. We'll learn a way to derive the form of the correct transformation between frames of reference, respecting the postulates of special relativity, and we'll learn how to begin applying this transformation and see that it is in fact consistent with the postulates of special relativity. It does end up being entirely self-consistent, and it gives us a basis for making predictions about the natural world, predictions that can be tested. The Galilean transformation was predicated on two assumptions, and these assumptions may not have been made very clear when you originally learned about this transformation. For observers in inertial frames of reference, that is frames of reference in which all observers agree that objects in motion are moving at constant velocities, time is assumed to pass in the same way for all observers regardless of the state of motion, and all observers agree that objects in each other's frames are in states of constant motion. I've drawn down here an example graphic of a representation of an object in motion with some velocity vector illustrated here that I'll use in a lot of the images going forward that will help us to think about these transformations. Now let's define two frames of reference that we can use as the archetypes for thinking about transformations of space and time information from one frame to another. Let's denote one of these as frame S, and we will always take frame S to be the thing that we call the rest frame. Now this is an arbitrary assignment. You can choose one thing to be at rest and not another or vice versa, but once you make that choice, you need to stick with your choice. You need to see that through to the bitter end. So for the purposes of illustrating the process of thinking about transformations of space and time information from frame to frame, we'll always take S to be the frame that is not in motion. Now in this frame of reference S, we will imagine that they carry along with them a coordinate system, like a framework of three lines that are at right angles to each other, that they'll use as reference markers for all spatial measurements, and the coordinates from their Cartesian coordinate system will be denoted X, Y, and Z. When they describe object velocities, they'll notate them as using the letter U. The letter V for velocity, as you'll see in a moment, has a special place in relativity calculations, and so to avoid confusing us as to what velocities we're talking about, we will use U to denote object velocities. Now we will define a second frame, S with a little prime symbol next to it, or S prime, that is moving relative to frame S at a velocity V, so everything in that frame is moving all at once in the same direction at the same speed, relative to S, and in that frame they too have a little framework, a little Cartesian coordinate system framework of X, Y, and Z, but they label their coordinates X prime, Y prime, and Z prime, and when they measure object velocities they denote them as U prime, to be consistent with the notation of their coordinate system. Now we will do the following to simplify our thought process going forward. It doesn't have to be this way, but we can set the problem up this way to make it easier for ourselves. We will assume that they have arranged their coordinate axes so that they are always parallel to each other. X is always parallel to X prime, even if X prime is moving relative to X, Y is parallel to Y prime, Z is parallel to Z prime. This makes it easier for us mathematically to relate things between the frames. We can't allow chaos to reign in all of this. We are going to further simplify for the purposes of our discussion here that frame S prime has a velocity V that is entirely and only along either X or X prime. It's entirely parallel to X and X prime and has no component along Y, Y prime, Z, or Z prime. Now in the Galilean picture of things, in all frames time is absolute, so that T is equal to T prime. The time measured in one frame, frame S, is equal to the time measured in another frame, frame S prime, always. That is the definition of absolute time. Now, I've recovered the picture of our two little frames of reference here and our blue object in motion, viewed from the perspective perhaps of one or the other frame with its velocity U or U prime. Now this picture built from the postulates of the Galilean or Newtonian approach to space and time, then allows us to define the equations transforming observations in one frame to observations in another frame. For instance, if we measure X prime and T prime and U prime and frame S prime, these equations will allow us to figure out what the people in frame S would see. And here are the equations. You've probably seen them in this form or a similar form in introductory physics. Because the motion of frame S prime is entirely and only parallel to X and X prime, there's a transformation between the X and X prime coordinate system that is given by the first equation. Measurements in Y are equal to measurements in Y prime and measurements in Z are equal to measurements in Z prime. And of course, because of absolute time, T is equal to T prime. Now object velocities are related between the frames using the velocity transformation equation, which if you use calculus, and I would invite you to do this as a simple exercise, you can prove from the first of the equations up here, X equals X prime plus VT. You can prove that this equation is the addition of velocities derivable from the coordinate transforms using a little bit of calculus in a few minutes of work on paper. But basically, the velocity observed for an object in frame S is equal to its velocity in frame S prime plus V, the velocity of frame S prime with respect to S. Now, there's a problem here. We know that the postulates of special relativity are more compatible with reality than the assumptions that were made to define the Galilean transformation in the first place. And if you play around with a Galilean transformation, you can pretty quickly find out that it violates the postulates of special relativity. So for example, and this can be left as an exercise for the student especially because Maxwell's equations are not something you get rigorous training on in introductory physics, you can show that Maxwell's equations, their forms are not invariant under a Galilean transformation. That would violate the first postulate of special relativity. Because if the equations, if Maxwell's equations have different forms that can be determined by experiment in different frames of reference, that implies that it's then possible to know whether or not you're in the absolute rest frame, for instance, in the mechanical view of the universe in the frame of the ether. It's a bit easier to see how the second postulate is violated by using a simple example. You can imagine that the object in motion on the previous slides is a beam of light and it's been emitted in the ether frame, which will take to be frame S, the absolute rest frame. And what you'll find then is that that beam of light will have a very different speed in all other frames moving with respect to the ether frame, the absolute rest frame, or frame S in our notation here. That violates the second postulate, that the speed of light must be observed to be the same by all observers regardless of their relative states of motion. So already the Galilean transformation is immediately shown to be at odds with the postulates of special relativity, which again are based on observational evidence. So here's our picture again of these two inertial reference frames with the blue object being studied by both of the observers in each of the reference frames, S and S prime. And we want to find a transformation of X and X prime, Y and Y prime, V, U and U prime and T and T prime between these frames that gets something that's compatible with the postulates of special relativity. And we don't have to change this picture to build up the correct transformation. We just have to apply the postulates of special relativity in constructing the transformation from one frame to the other. And these new postulates, the postulates of special relativity, will enable us to arrive at a mathematics that's consistent with observation. So our goal is to figure out what is the correct transformation. And we will continue to work with frames of reference wearing object velocities are observed to be constant. That is inertial frames of reference. That puts the special and special relativity. Now, it must be true that in two inertial reference frames, S and S prime, as depicted in the cartoon above, that because the object in motion will be observed in either frame to have a constant velocity, maybe a different magnitude, but both frames of reference will agree, yes, we have each observed a constant velocity for the object that we're studying. It must therefore be true that X in the frame S is equal to the object velocity times time if it's moving entirely along the X direction. And in the moving frame, frame S prime, it must be true that X prime is related to T prime by the observed velocity of the object in that frame, U prime. In order to further satisfy the first postulate of special relativity, it must also be true that the transformation equations represent a linear transformation between the frames. Otherwise, it can't be true that all frames observe object velocities to be constant. Let me demonstrate this. It's important, I think, to start exercising your calculus a little bit at this stage in the class so that you get a bit more comfortable with using calculus as a means to make predictions about the natural world. So let's begin by assuming an extremely generic form for the transformation between spatial coordinates in frame S and space and time coordinates in frame S prime. I've assumed that X is given by some unknown transformation with a spatial term and a temporal term. Each of these has coefficients. I'll talk about those more in a moment. And each of the coordinates in the moving frame is raised to some unknown power for space, it's N, and for time, it's M. And similarly, the time coordinate in frame S is related to the space and time coordinates in frame S prime. In a same way, there's some new coefficients, C and D, that enter in here. But again, I've raised X prime and T prime to various powers. They could be 2, could be 10, could be 20. We don't know. Now, let me comment on these coefficients. A, B, C, and D are constants here. We could always absorb some non-constant behavior in the coefficients into the function of X prime and T prime that we're using here. Now, I've used a simple function just raising the space and time coordinates to a single power. But you can draw the conclusion more generally that an arbitrary polynomial of X prime and T prime also won't work to satisfy the requirement that all observers agree that the objects moving at a constant velocity regardless of their frame of reference in inertial reference frames. So here's my generic pair of transformation equations. I don't know what A, B, C, and D are, and I don't know what N and M are. But what I can do is I can recall that any velocity, any object velocity, like the object velocity along the X direction in frame S, U, X, or the object velocity along the X prime axis in frame S prime, so U prime X, is defined by a derivative with respect to time. That is, U X is dx dt, or U prime X is dx prime dt prime. That's the definition of velocity from its most foundational aspects. So what I would like to do to motivate that this has to be a linear transformation between the frames is to simply take the above equations and turn them into statements about differentials of X and X prime and T prime and T, rather than just statements about the coordinates themselves. This is where you can dust off your calculus and see if you can arrive at the same answer that I get here. But the bottom line is that the differential of X is related to differentials in X prime and T prime by this equation. So the coefficients A and B remain unscathed, but you wind up with this new power of X prime and T prime due to transforming this into a statement about differentials in space and time, rather than just about space and time themselves. And then similarly, you get an equation that looks very much like that for dt, differentials in time in the frame S as well. Okay, so let's hold those equations here for a moment and consider them. So what I would like to do now is use these equations to relate the observed velocities in each frame of reference. And so to do that, I'm just going to take the ratio of the above two equations. Why? Because when I do that, I get dx divided by dt, which on the left side is just the definition of the velocity of the object in frame S, ux. And on the right side, I get something that's a fair bit nastier than that, but we will simplify it into something that looks a bit more familiar in a moment. So notice that I have dx prime and dt prime both in the numerator and in the denominator of this ratio. So what I can do next is I can divide the top and the bottom of the right-hand side by dt prime. Doing so eliminates dt prime from the right-hand terms in each part of the ratio and creates a dx prime dt prime in the left term of each part of the ratio. And that should look very familiar because dx prime dt prime is by definition the velocity of the object in the prime frame, in the S prime frame. And so we arrive at this final relationship that relates the observed velocity of the object in frame S prime to the observed velocity in frame S. But there's a problem here. Unless n equals m equals 1, the above equation will always leave a lingering space and time functional dependence on the right-hand side, which violates the first partial of special relativity. The speed observed in frame S, even if the speed in frame S prime is constant, will not be observed to be constant because it will depend on where in frame S prime the object is at any given moment. It will have a space and time dependence that is rather nonlinear. And so we're forced to conclude that in order to be compatible with the basic idea that we're looking at inertial reference frames and relating object velocities in inertial frames of reference, where there are no net forces that can cause changes in the state of motion of the object, we're really forced to choose a linear transformation from frame to frame. That's what we had in the Galilean transformation, but we are still stuck with it even here in special relativity. That's a good thing because it vastly simplifies the mathematics. We're going to now begin to build up the mathematics of the transformation. Now that we've accepted that we need a linear transformation from frame S prime to frame S, for example, we get a very simplified pair of equations. A equals ax prime plus bt prime, and t equals cx prime plus dt prime. But we don't know what these coefficients are. They may be trivial. They may be zeros or ones, but we need to figure it out. So now that we've established the linearity of the transformation, we need to nail down a, b, c, and d. So we need to think of some special limiting cases of this picture where we can isolate these constants, maybe one at a time or in small batches, and in doing so, figure out what they are in order to be compatible with the postulates of special relativity. Now this is a standard trick in algebra. We have four unknowns in two equations. We're going to need four special cases to solve for all the unknowns, and the postulates give us the framework to define those special cases. So let's pick special case number one, where we take the moving object, the blue ball, and we pin it to the origin of the coordinate system of frame S prime. So in frame S prime, the object will always be located at 0, 0, 0 in the X prime, Y prime, Z prime coordinate system. It's moving along at the same speed as the frame itself, and so it's observed to be at rest in frame S prime. In frame S, however, what we see is we see the blue ball pinned to the coordinate system of this frame, and they're all moving together in a velocity V to the right along the X axis. So in frame S, it's observed that the object is moving at V, so U equals V in frame S. Now when we do this, we have a simplifying situation for X prime. X prime will always be 0, because this thing is pinned to the origin of frame S prime. And so we can simplify the above equations to the following. We have now that X is just equal to BT prime in this special case. T is just equal to DT prime in this special case. And then the velocities X equal UT and 0 equal U prime T. Prime are the resulting equations from this special case. So let's take the first two equations, the one for X and the one for T, and divide them. And then let's use the third equation, X equals UT as a substitution to eliminate one of the unknowns. So when we do this, we wind up with X divided by T is equal to B over D. Go ahead and check this yourself. And then from the velocity equation we get that X over T is equal to U. But in this special case the object speed is also the frame speed V. So we wind up with X over T equals V. And as a consequence of that we get the first constraint on our coefficients. Whatever B and D are, their ratio is equal to V, the velocity of frame S prime. Now let's choose a second special case. And you might have guessed that this would be the next thing that we would do. We pin the blue object to the origin of the rest frame coordinate system. We put it at 0, 0, 0 in frame S. R, 0, 0 in 0. And now we observe that blue object from frame S prime. Now from the perspective of frame S prime, which is moving to the right at speed V, the blue object appears to be falling further and further behind their frame. Its velocity appears to be negative V from the perspective of an observer in frame S prime, the moving frame. So with that in mind and fixing X to 0 in the general equations up here on the right, we can simplify the equation set to the following. 0 equals AX prime plus BT prime. The T equation doesn't get affected by any of these choices. We wind up with 0 equals UT and X prime equals negative VT prime substituting in for U prime with negative V. Now let's employ the first and third equations, namely this simplified first equation for the coordinate X and this X prime equals negative VT equation to further get the constraint on coefficients. So if we do that we wind up with the first equation telling us that negative AX prime equals BT prime, that's the consequence of the first equation. From the third equation we get that X prime is negative VT prime and if we combine these two things together we find out that AVT prime equals VT prime. T prime entirely drops out of both sides using the substitution and we find out that V equals B over A. Go ahead and try this yourself. I'm going through this a little bit fast but of course you can pause this at any time and work through the algebra on your own. And so we arrive at our next batch of constraints. We already knew from the first special case that V equals B over D. From the second special case we get that V over A and if these two constraints are simultaneously true then it must be true that A equals D. So now we have really constrained ourselves down. So here's the third case we'll look at. What if the object is a beam of light? Now this is the first time that we will definitively deploy one of the postulates of special relativity, specifically the second postulate. Because if the object in both frames of reference is a beam of light then by the second postulate of special relativity observers in both frames must observe the velocity of the object to be exactly C 2.998 times 10 to the 8th meters per second regardless of their relative motion. So what happens to our equations as a result of this fact that according to observations and encoded in the second postulate all observers observe that the beam of light moves at the same speed regardless of their state of relative motion. The equations simplify as follows. The first two don't really change at all but because the last two are statements about velocity and their relationship to space and time measurements it must be true that X equals CT but also X prime equals CT prime when the object in motion is a beam of light. So let's combine the first two equations, the X and T equations and substitute using the information from the third equation at the bottom. When we do that this one gets a little nasty at first. We wind up with X divided by T equals this horrible ratio over here not looking very promising so far but X over T is just C by the third equation and if we divide the top and bottom of the ratio on the right hand side by T prime we wind up with X prime over T prime in both the numerator and the denominator and X prime divided by T prime is just C, the speed of light. So we wind up with the speed of light here the speed of light here and the speed of light here and now what we can further do is take the previous constraints relating B and D and A we can substitute those in and go through a simplification process and when we do that we find out that C is equal to B and we finally arrive at C equals A times the velocity of frame S prime relative to S over the speed of light squared so all of our constraints allow us to eliminate B eliminate C and eliminate D from the equations up here in the top right of the screen B is equal to A, V C is equal to A times V over the speed of light squared and D is just equal to A itself so so what we have done now with special case one, two and three is we've eliminated three of the unknown coefficients in favor of the fourth A and all we have to do is come up with one more constraint that allows us to figure out what A is well here's our last case and the last case is a basic assumption about the transformations first of all we chose that the transformation of X and T to X prime and T prime from the perspective of observations in frame S prime being mapped onto observations in frame S have a certain form but because it shouldn't matter which frame we pick to be the one that's at rest and the one that's in motion we should get the same transformation equations if we had started with frame S prime being at rest and having frame S be the one that was in motion the only thing that should change between observations in frame S and observations in frame S prime is that the relative velocity of the two frames changes sign that's the only thing that should change when you alter the perspective of which one is at rest and which one is in motion and so as a consequence of that we should be able to eliminate the unknown A and figure out what it actually is so let's start by writing down X and T in terms of the coefficient A and all the other things we've already sorted out so I've effectively just copied these two equations down here next let's rearrange and rewrite these equations not as solutions for X and T but solutions for X prime and T prime as if we were trying to figure out what the person in frame S prime would have seen if frame S was chosen to be the frame that was in motion now a lot of algebra is involved in this and I will leave it as an exercise to the viewer to try this out to practice their chops at algebraic manipulation in order to get what we want but the bottom line is that if you work this through you will find out that X prime is given by this nasty function of X and T and T prime is given by this equally nasty function of X and T now these equations tell us what should have been observed in frame S prime given observations in frame S treating frame S as if it had been the frame in motion and frame S prime as if it had been the frame at rest but all that should change when switching perspectives on the problem like this is you should get the same equation differing only by a minus sign on any term with V in it then we're forced to say A is equal to one over A times the quantity one minus V squared over C squared to the minus one if you rearrange now and solve for A you find out that A must be equal to one over the square root of one minus V squared over C squared this quantity shown here this may not look pretty but this strange thing one over the square root of one minus V squared over C squared shows up all over the place of special relativity calculations and so it's given a special name we don't leave this as coefficient A it gets the symbol gamma the lower case Greek letter gamma shown here in the lower right gamma is defined to be this strange beast here one over the square root of one minus V squared over C squared so let's take a look at the final form of the Lorenz transformation the mathematical transformation that obeys all the postulates of special relativity if we are making observations in frame S prime and relating them to frame S which is taken to be at rest then these equations here tell us what we want to know we measure things in frame S prime and we get X prime and T prime and then we combine them using these equations to get X and T the observations that should be made in frame S if on the other hand we're making observations in frame S we invert them to observations in frame S prime we flip the sign of V and change all the coordinates around and we get basically equations that look the same up to a minus sign on terms that just have V in front of them this number gamma always lurks out in front of everything and we see here another interesting thing that's going on we see that in transforming space and time from one frame to another space and time measurements in one frame get tangled up to compose the space and time measurements in the other frame in a sense space and time in special relativity are not separate entities we treat them as separate in introductory physics but the lesson of special relativity is that we should have been thinking of this all as one framework space time not separate frameworks of space and time this entire time now this multiplicative factor gamma depends on the relative velocity of the two frames and as you'll see it's effectively a measure of the degree of the relativistic effects how much you need to take into account special relativity to solve a problem correctly between two frames of reference and again it's given by this funny combination of velocity of the frames relative to each other and the speed of light one over the square root of the quantity one minus v squared over c squared so let's build a little bit of intuition about the meaning of the gamma factor it appears everywhere in relativity calculations at least in special relativity it gets absorbed into other things in what is known as general relativity the general theory of space and time which we will only cover in the most shallow way in this course it's largely unavoidable for all of the physics calculations you're going to do going forward so let's try to understand it a little bit better and to do that I think we can build some intuition by playing around with the quantity gamma at various limits of its observable nature so for instance what is gamma for a frame s prime that's at rest with respect to frame s well we would expect to find that the two frames are the same since they are then in the same state of motion well we already know that frame s prime has a velocity of zero with respect to frame s that the terms with vt in front of them will vanish but what happens to gamma well if you plug in zero for v in the function gamma you find out that indeed for v equals zero we observe that gamma is one in other words the multiplicative factor in front of either the space or the time coordinates when relating those to space and time coordinates in frame s all become coefficients of one in other words you're in the same frame so you should get the same space and time measurements that's good that's what we would hope would happen now on the other hand what is gamma for a frame s prime that achieves a velocity of exactly c the speed of light relative to s so this would be like imagining a frame of reference that's pinned to a beam of light moving at the speed of light and it's another very weird special case and it's weird because what happens is that the gamma function takes on its biggest possible value infinity you wind up with one over the square root of the quantity one minus c squared over c squared c squared over c squared is just one so you wind up with one over the square root of one minus one which is one over zero which is infinity zero goes into one in infinite number of times so that's the upper limit for gamma so far as we know it's impossible to travel faster than the speed of light there's no observational evidence that anything does travel faster than the speed of light and so we are led to believe and special relativity encodes this that the fastest speed in the universe is that of light and so gamma takes this special value of infinity at that speed so as we can see gamma is a frame velocity dependent number and it has a well defined range at the low end its smallest value it can take is one and at the high end the largest value it can take is infinity and it can take all numbers in between that depending on what v is I think it's useful to graph this albeit perhaps in a way that's not terribly familiar to you this is a plot of the value of gamma so the so-called gamma factor on the y-axis as a function of the frame relative velocity v on the x-axis and so you can see here that I have chopped off the low end of the x-axis at about 10% the speed of light why? because gamma has a value that's so close to one that generally speaking you don't have to worry about it being different from one now that's not true in all cases and we'll look at some of those cases going forward in the class but generally speaking if you are at about 10% or less the speed of light you cannot expect to really observe what are called relativistic effects that is effects that distinguish an observation from what you expected from Galilean or Newtonian relativity above 10% of the speed of light however gamma can begin to take values that are appreciably distinct from one and you can see here that when you get to values that are about half the speed of light which occurs roughly here on this graph this is one times ten to the eighth this is two times ten to the eighth right here and as a result of that you can see we've now appreciably started getting gamma factors that are above one by about 20% or so when we get to about two thirds the speed of light two times ten to the eighth meters per second we've achieved gamma factors that are about 50% bigger than one so 1.5 and as we begin to approach closer and closer and closer to the speed of light we see that the gamma function takes on increasingly larger and larger and larger in larger values spiking up to infinity at exactly the speed of light this plot will help you to understand why it is that we just didn't notice these deviations from the Galilean or Newtonian view of the universe for most of the history of science and humanity and that is because the laboratory experiments that we were effectively conducting as a species were all done at speeds that were far less than that of light and so we really never would have noticed these effects to begin with it was only when we began playing around with light and things that can move very close to the speed of light subatomic particles that we began to get ourselves into trouble with the intuition we had built up on human experience prior to that but we see now that the postulates of special relativity predict that we should have expected deviations from that Newtonian or Galilean view of time where time is the constant between all observers it's not it's the speed of light and we can see this effect encoded in the gamma function now as has been teased in the previous lecture on the basics of special relativity that is the postulates and their consequences this theory of space and time has some consequences that can feel surprising to the average human being for instance objects in motion relative to what we consider to be a rest frame that is the frame we denote S will appear contracted along the direction of travel we can actually show this as a prediction of the Lorenz transformation now that said to appreciate this particular effect even from the Lorenz transformation really requires you to think extremely carefully about what it has ever meant to measure the length of something I feel that discussion is best saved for class time as in class time we can get very hands on with the concept before we start plunging into calculations where the language you would use to describe the recipe for attacking this particular question may not feel very natural because you haven't really thought about what it means to measure the length of something especially the length of something that's moving relative to you so instead in this lecture let me concentrate on frames in motion relative to the rest frame S which will also observe a passage of time that relative to S seems slowed this effect I labeled time dilation and we're going to formally calculate it now and finally I'll also look at events that are simultaneous in one frame and show that that is not guaranteed at all in another frame that's moving relative to the first let's explore time and simultaneity in this lecture so how does one consider the passage of time in different frames of reference well to measure time we have to define a clock of some kind a regular pattern of events that all happen for instance at the same reference point in a frame a clock that's at rest in frame S prime and it provides regular information so for instance pulses of light at different times T1 prime T2 prime T3 prime and so forth always with regular intervals between them but that clock is always pulsing at the same position X prime so X1 prime equals X2 prime equals X3 prime whenever the time measurements are established and what we want to know is what's the time between the pulses observed in the rest frame so that clock is in a frame that's moving back to a frame that we agree is at rest we call it to be the rest frame what does a observer in the rest frame observe the time to do in the moving frame well again you want to relate time observations between the two frames but to do that you need to use the Lorenz transformation which takes space and time information from the moving frame and translates it into time information in the rest frame so we want to take the pulse at T2 and transform its time into what the person in the rest frame measures their so-called T2 and we want to take the pulse at T1 and transform that into the rest frames T1 and to do that we have to use this equation this comes from Lorenz transformation now we have a simplifying fact here and that is that the clock in frame S prime is always pulsing away at the same location X1 prime equals X2 prime so if we were to combine these two equations to calculate the duration of time between T2 and T1 we might do the following we might take T2 minus T1 and try to figure out what that is in terms of T2 prime minus T1 prime well because the clock pulses at the same location in frame S prime X2 prime and X1 prime terms cancel out and we're left with this equation which relates the durations in time observed in the two frames by a gamma factor and so I can write the two time durations delta T in frame S and delta T prime the time duration in the moving frame and I can relate them and they're related again by a gamma factor and I find that if I take the ratio of the time duration observed in the rest frame and the time duration observed in the moving frame that they will differ they will not be a ratio of one and the ratio however will be given by the gamma factor which takes a value of one but only in the special case with respect to each other at any relative speed greater than that gamma takes a value that's greater than one now until you get to very high speeds it's not appreciably greater than one but nonetheless it's not exactly one unless you're at rest with respect to each other and so that we see now that in a frame that's moving relative to another durations of time will always observed to be shorter than in the rest frame the duration of time observed in the rest frame is greater than the time that's observed in the moving frame for the same pair of events and the degree of dilation of time depends again on the ratio of v over c specifically through the gamma factor time in the moving frame will appear to the rest frame to pass more slowly now another expected effect due to special relativity is that events that are simultaneous in one frame of reference may not necessarily be simultaneous in another now we already explored that a little bit even under classical velocity situations but we can revisit that idea here under the Lorenz transformation so for instance consider two events like pulses of light which are observed to be simultaneous in frame s prime the moving frame the events have coordinates x1 prime t prime and x2 prime t prime so what is the time between the events observed in the rest frame does the rest frame observe that they are also simultaneous that is also at exactly the same time well we can start by relating the times in the rest frame to the space and time measurements in the moving frame for t2 and t1 again we're just writing down the Lorenz transformation here between observations in frame s prime and the observations we want to establish in frame s now since the events are simultaneous in frame s prime t1 prime will be equal to t2 prime so if I now calculate the duration of time that passes in frame s t2 minus t1 I find a very interesting fact that it's not equal to zero it's equal to this thing on the right hand side here which depends on the velocity of the frames relative to one another v quite directly in this case not just through the gamma factor but gamma multiplied by v and what's particularly interesting about this is this question whether two events are simultaneous in one frame and simultaneous in another frame of reference that's in motion relative to it really picks at this interesting thing I mentioned earlier which is that space and time have to be treated as one framework space time and they can get tangled up in each other and what we see here is that because the events in frame s prime are simultaneous but not necessarily at the same location in space in frame s prime this creates a displacement in time between the two events in the rest frame in other words delta t in the rest frame is not necessarily equal to zero it's equal to gamma v over c squared times the spatial displacement of the two events in the moving frame delta x prime so we see that in the rest frame events cannot be observed to be simultaneous even when they are simultaneous in the moving frame unless those events happen at exactly the same position in space that is x2 prime equals x1 prime that's a special case or unless the two frames are not in relative motion to each other that is v equals zero in that case of course the two frames become indistinguishable and the whole discussion was moved to begin with but if the two frames are not the same frame of reference if the events occur at different locations in space in one frame but are otherwise simultaneous there they will not be viewed to be simultaneous in the other frame simultaneity of events could have been guaranteed in Galilean relativity even if they were at different locations in space because of the absolute passage of time time is not absolute and special relativity doesn't accept that as one of its postulates you find out that except under these very special conditions simultaneity cannot be guaranteed in another frame and again this is a really good example of how space and time are not inseparable from one another in special relativity in transforming observations in space in one frame you wind up with observations in time in another frame space and time get kind of up in each other and going from frame to frame a spatial separation in s prime becomes a temporal separation in frame s and I find this to be one of the more remarkable features of space and time as viewed through the lens of special relativity now one last question we can visit in all of this is whether or not it's possible to recover classical physics from this view of the universe in other words are the Galilean Newton view of space and time and relative motion totally gone were they totally wrong this whole time it turns out the answer is not really after all the Galilean transformation did work in real computation for centuries before special relativity was needed right I mean people were able to relate observations in different frames of reference at relatively modest speed compared to those of light so one of the things that should become evident from all of this and this is a general feature of a good theory of nature a good predictive description of nature that can be tested and even falsified a good theory of nature describes all new phenomena but also it accounts for the existing confirmed observations the old observations the old phenomena that we had all that experience with from which we built intuition what generally seems to happen is that when you find out that your current understanding of nature is wrong you find out eventually through enough observation and experiment and mathematical work the correct description of nature and you find out that your old observations were correct but in a more limited regime of nature in this case for instance low velocities so to recover the Galilean or Newtonian view we need only slow nature down from speeds close to that of light for example we can consider the special case of speeds between the frames of reference that are much much much much much much much less than the speed of light so that V over C for instance becomes a very tiny number approaching zero so let's look at what happens to the gamma factor using something called the binomial expansion now I'm going to illustrate the binomial expansion for this specific function here but in general there is a general form for the binomial expansion and you can use a math reference on the web paper book to explore the binomial expansion more on your own free time so let's begin by looking at the gamma function the gamma function was defined as one over the square root of one minus the quantity V squared over C squared so what we have here is we have a function of a number V squared over C squared that's bounded between zero and one V over C can either be zero for V equals zero or one for V equals C and as a result of that gamma takes values between one and infinity but this this parameter V over C that depends on the relative frame speed that thing is bounded between zero and one now one of the things that we can do is we can rewrite this gamma function as a more basic generic looking function by replacing V over C with a number alpha or in other words rewriting this in terms of alpha squared which is V squared over C squared and note that alpha squared is a number that is less than or equal to one and it's lowest value it can take is zero so we wind up with this more generic looking function one minus alpha squared all raised to the negative one half power in other words one over the square root of one minus alpha squared now if we apply the binomial expansion to this more generic looking function with this condition on alpha that it's bounded between zero and one then we find that we can write this function as a series expansion the first term in the series is just the number one the second term in the series depends on alpha and it's plus one half alpha squared now the binomial expansion allows you to keep adding terms that have higher and higher powers of alpha in them the next one will have a power of alpha to the fourth the one after that alpha to the sixth and so forth with different coefficients in front of them now because alpha is a number that cannot be greater than one as a result of that for the special case that alpha is much much much less than one in other words as alpha tends towards zero we see that indeed we recover the behavior that as these terms with alpha squared alpha to the fourth alpha to the sixth as they approach zero the only term left that really dominates in the sum is the leading term one and we see that gamma is approximately equal to one which is closer and closer and closer to zero and when alpha equals zero we get gamma equals one which we know is the limiting case of gamma for velocities of zero speed so that makes sense it's approaching the limiting case when v equals zero that's what it means for alpha to go to zero means v is going to zero too so what happens to the Lorenz transformation equations when we replace gamma in it with this binomially expanded version of gamma so let's start with x equals gamma quantity x prime plus vt prime let's substitute in the binomial expansion in terms of v over c being equal to alpha so now we replace gamma with this thing one plus one half alpha squared plus all terms with alpha to the fourth and higher in them and in the limiting case that v is much much less than c that is as alpha approaches zero these terms with alpha squared, alpha to the fourth, alpha to the sixth they contribute less and less and less to the sum until we're left with just one in front of the sum x prime plus vt prime so in the special case that the velocity we're considering for the relative motion of the frames is much less than that of the speed of light we find that we recover x equals x prime plus vt prime which is the old Galilean relationship between x and x prime and t prime now similarly I can take the equation relating x prime and t prime to time in the rest frame and I can substitute for the gamma factor using this binomial expansion I can also notice that there is a v over c lurking here in front of the x prime coordinate that's another alpha that sits in front of x prime so if I write that all out here's the binomial expansion of gamma here's that alpha that I've substituted in for the v over c that was lurking in front of x prime and what you'll notice is if I distribute this gamma to both the terms inside of this sum that the space term the x prime term always has an alpha somewhere in front of it multiplying it you can't escape it you don't just get a bare number like 1 multiplying the x prime coordinate whereas for the t prime coordinate there is a term in the expansion that just goes as 1 times t prime and so in the limit that the velocity is much much less than the speed of light all terms with alpha in front of them vanish to 0 and we're left just with t prime in other words in the low velocity limit t is equal to t prime and we see that we have completely recovered the Galilean transformation and we've reconciled with classical physics in the limit of low velocities this is why time appeared to be absolute in the original formulation of mechanics it's because when the velocities are much lower than the speed of light between two frames of reference you have a very hard time seeing these extremely subtle effects between clock measurements between the two frames but that is laid bare as a false perception of nature as you approach the speed of light in relative velocities between two frames of reference but we see that we can reconcile the old picture of space and time with this modern and correct picture at least correct as regards observation of the natural world simply by considering the limit of small velocities compared to that of the speed of light and we completely recover the old view the old view is nested inside the modern view as a limiting case so to review in this lecture we have learned the following things we've learned to appreciate the Galilean transformation and the assumptions upon which it's built we've learned a way to derive the form of the correct transformation between frames of reference the so-called Lorenz transformation that is the modern way of relating observations in one frame to observations in another frame of reference and we've begun to see how you start to apply this transformation by asking questions like what are the events and in what frame are they defined and is anything the same for those events are space measurements the same are time measurements the same in a given frame of reference to simplify the questions that we're trying to answer and then get the answers out of the Lorenz transformation and we see that we have arrived not only at a transformation that's consistent with the postulates of special relativity but which gives us a mathematical formulation for the intuition that we built off of the postulates that distance and time measurements are not going to be the same in different frames of reference even if all observers agree on the speed of light as a constant of nature so while we see that all observers must agree that light moves at the same speed regardless of their relative motion nonetheless observers in different frames of reference will disagree on lengths of objects the durations of time that pass and the simultaneity of events events simultaneous in one frame will not necessarily be simultaneous in another frame of reference except under very particular conditions we've also seen how to recover classical physics from special relativity by allowing the velocity of the relative motion of the two frames of reference to drop far below the speed of light so that these corrections from the original Galilean transformation all vanish and leave behind the Galilean transformation with its assumption of absolute time laid bare and we've seen how in that limit the Lorenz transformation exactly reproduces what were the original assumed relationships between space and time and velocities as encoded in the Galilean transformation we've recovered the past from the present and we can continue to use the present to build a foundation for making future predictions and that is precisely what we're going to do in the next section of the course