Special Relativity | Lecture 3

[Music] Stanford University all right uh this evening I want to get started on talking about particle mechanics in the special theory of relativity how particles move Notions of momentum energy we will get to eal MC^2 and uh that sort of thing the next time or maybe even tonight a little bit we might even get uh move past that if we get finished with it and start to talk about classical field Theory um let me just do one thing on the Blackboard that we will need recurrently and over and over uh for various purposes mostly for purposes of showing that the relativistic formulas that we write down have the limit of the non-relativistic formulas that we may have written down that we did write down two quarters ago in classical mechanics we are going to want to show that when objects move slowly compared to the speed of light that the new fangled Rel relativistic theory of particle motion and particle mechanics reduces to the old fangled Newtonian okay for that we're going to need to make approximations which become better and better for slower and slower velocities and in particular the thing that we're going to have to approximate now I know you all know this but I'm going to do it just for completeness on the Blackboard we have this quantity which recurs over and over over 1 - V ^ 2 square otk of 1 - v^2 it really means Square < TK of 1 - v^2 over c^2 if we measure the velocity and units of the speed of light then it's just 1 minus v^ 2 but we remember that V represents the fraction of the speed of light of the object which is moving with velocity V so typically it's very small and we just want to write down how we approximate this for small velocities in particular how we approximate it to order velocity squared uh this is just an example of the binomial theorem I'll write it down once and for all we'll use it a couple of times tonight and I assume everybody knows it the binomial theorem says that 1 plus a small number Epsilon doesn't matter if it's small but uh but one plus a small number Epsilon to a power P where P can be anything as a matter of fact it can be positive negative imaginary complex doesn't matter that this can be expanded in powers of Epsilon a tailor series if you like if you don't know tailor series we just expand it in powers of Epsilon for Epsilon equals z it's just given by one and then the next term is p * Epsilon now I'm not going to work I'm not going to prove this this is something this is the standard binomial theorem and the next term everybody remember the next term P * h p choose P choose 2 which is the same as P * p minus1 over 2 factorial 2 factorial is incidentally just two times Epsilon squar and then it keeps going like that I'll write the next one and you can write all the ones after it it's a homework assignment to write out all the infinite infinite number of them what uh yeah pi over 3 factorial Epsilon cubed and so forth uh if Epsilon is not too big then this series will converge and the meaning of saying it converges is is that each term is significantly smaller than the previous in such a way that uh that it's a good approximation to uh to uh to terminate the series after an appropriate uh number of terms where're appropriate depends on just how much prec Precision you want but in particular uh the if Epsilon is really very small It's usually the case that the first term is the most important one of course it's the zeroth term which of course is the most important one but the zeroth term is awfully boring here uh we're going to find for the various applications that the zeroth order term the one is just too boring to be relevant to us and it's the first order term here which is important so let's terminate the series at that point of course it does break down if Epsilon is not too small but that's that's the uh all right so supposing we want to expand 1 - vun of 1 - v^ 2 where Epsilon is now we're now using Epsilon equals minus V ^2 in other words 1us Epsilon 1 plus Epsilon is the same as 1 - V ^2 and the power p in this case is equal to 12 square root of a quantity is that quantity to the 1/2 power and so this becomes 1 + Epsilon which is 1 sorry 1 plus P Epsilon which is 1 + 12 that's the p and then Epsilon is minus v^ 2 - v^ 2 which is the same as 1 - v^2 / 2 so that's a pretty damn good approximation for small velocities in particular for non-relativistic velocities to the < TK of 1 - v^2 is just 1 - v^2 / 2 that's where a lot of tools in classical physics come from if thought about as the limit of relativistic physics most of the tools and I'll show you some examples come from here I'll show you some examples as we go along uh the other example that we will need will be one divided by the same thing this case is the same as 1 - V ^ 2 to the power minus a half the only difference relative to what we just did above is that P is minus a half that has the effect of changing the sign here so that this becomes 1 + V ^2 / 2 two signs have conspired to cancel each other the sign from the half up in the numer up in the exponent and the minus v^2 here give us 1 + v^2 over 2 all right so that that that is just by way of uh of uh setting up little piece of calculational mathematics that we use in various situations all right now we're going to be interested in particles now a particle can be anything that holds itself together it doesn't have to be an elementary particle it could be the sun it could be a box of chocolate it could be um a cookie no cookie it could be anything and when you think about the location of uh the uh whatever it happens to be could be a donut let's take the case of donut um what we mean by the position of the object is generally the center of mass position when we talk about the mechanics of the motion of particles we're speaking really about the mechanics of the center of mass of an object in the case of a donut the center of mass is where there is nothing but that's okay we know where the center of mass is and motion according to the center mass is governed for example by by Newtonian mechanics FAL ma uh momentum is mass time velocity does it matter whether it's an elementary particle or a donut or cookie no its momentum is its total mass times its velocity its kinetic energy is 1 12 mv^ s in the toning in physics that's the energy of motion does it have any other energy if it's a donut does is does a donut have any other energy besides 1 12 mv^ s calories yeah calories exactly it has all the internal chemical energy and that sort of stuff that's inside the donut that part of the energy does not depend on its um overall motion on its state of motion it's a constant it's a constant with respect to how fast the object is moving so you can say that the uh energy of a moving donut is 1 12 mv^ 2 plus the internal energy and the internal energy doesn't depend on the state of motion okay so those are those are just some reminders about classical mechanics another reminder about classical mechanics is to go back and read the classical mechanics book where you will find things like the principle of least action which we will use tonight L grangian canonical momenta and all that stuff and we're going to use them so very quickly review in your head now the previous quarter of classical mechanics the important idea about the principle of Le least action was the idea of a trajectory in SpaceTime not just a trajectory in space but a trajectory in SpaceTime that if you plot position of a system it could stand X could stand for a onedimensional coordinate it could stand for a three-dimensional coordinate in fact it could even stand for the coordinates of a large number of particles and we can just call it space or coordinate space vertically we plot time and the trajectory of a system between two instance is a curve that goes from one point to another curve and we can call it a world line if it's a particle if it's a single particle we call it a world line we call it the world line of the particle now this concept is just as good a concept in nonrelativistic physics as it is in relativistic physics but it takes a certain Primacy in uh in relativity because of the of the connection between space and time that space and time can morph into each other by a Lorent transformation we tend to focus more on the idea of SpaceTime the trajectory of a particle is Tim like now what do I mean by Tim like let's define the notion of Tim like begin with just two points begin with any pair of points in space time and think of the separation of the coordinates the difference between the coordinates of the two points the difference of the coordinates now I'm speaking about space and time coordinates we can call them delta T and Delta X I discussed for a couple of times already the notion of the invariant proper time between the two points and we wrote that as Delta t^ squar minus Delta x squared is equal to Delta to squar where to where Delta to itself the square root of this quantity is the proper time connecting those two points now this quantity can be positive this quantity which I've written delta T squ it can clearly be positive or negative it is positive if the separation between these points is such that the T interval is larger than the spatial distance between them in other words if the interval is like so if the spatial distance is less than the time distance in that case Delta to^ squ is positive and the interval between these two points is called Tim like it's got more time in it than it's got space in it when an interval is Tim likee you can always find a Lorent frame of reference you can always transform to a to a frame of reference where the two points are at the same position where it hasn't moved in fact you simply go to the to the frame of reference where this line is at rest there always exists such a such a frame of reference and so if two points in space time are Tim like separation relative to each other you can always move in such a way that they wind up being at the same point in space that they haven't moved relative to each other okay that's one case the other case is where Delta X is bigger than delta T in that case the separation has more space interval than time interval look like this let's say it's at less than 45 degrees relative to the horiz to the horizontal incidentally I should I should say right now when I'm thinking about four dimensional space time three coordinates of space one coordinate of time what Delta X stands for is the three components of the vector Delta X and Delta X squ means the sums of the squares of those three components Delta X S Plus Delta y^ S Plus Delta z^2 okay I suppose we could if we lik if we want it to be uh fancy we would put a little arrow over here to indicate that it's a three Vector the Delta X is a three Vector all right now if the three Vector Delta X is bigger than delta T then the interval becomes space likee space-like intervals are as I said are more space in they are time and in this case you cannot find a reference frame in which the two points are at the same position in space what you can do is you can find a reference frame in which they are at the same time you can always find so that means that there's no invariant significance to one end being later than the other you can always find the frame of reference in which the two points are at exactly the same time and even worse I've drawn this so that point B let's call it b and a point B appears to be later than point a but there exists a reference frame this you can work out it's very easy to work out you can find a reference frame in which point B is earlier than point a so there's no invariant significance for points which are separated by space- likee interval there is no invariant significance to which one occurred first uh one being later than the other or being simultaneous this is the relativity of simultaneity okay particles move on Tim likee trajectories everywhere along the motion of a particle if we break the trajectory up into little intervals every little piece of that trajectory is timel that's just a statement that particles can't nothing can move faster than the speed of light we've already seen how frustrating it can be to try to get something to move faster than the speed of light you try to compile up a whole bunch of velocities Each of which is 910 the speed of light you know uh stationary Observer shoots out something with 9/10 the speed of light the moving thing shoots out something else with velocity 9/10 of speed of light something else shoots out a third thing 9/10 of speed of light you never get anywhere it's like Zeno's Paradox you just get closer and closer to the speed of light without ever getting there and so it seems like a reasonable hypothesis that nothing ever goes faster than the speed of light and that's a good thing because it would get very confusing if we said a particle goes from A to B but we can't make sense out of which one is earlier the starting point of the particle or the end point of the particle would get us uh very confused we get into all sorts of questions of uh of time travel and so forth but the physical fact is nothing including a photon can move faster than the speed of light and I would generalize that and say that information information is carried by physical systems uh signals signals it's signal information from one place to another can also not propagate faster than the speed of light we can take that as a postulate but um if it were otherwise we'd get into some very paradoxical and inconsistent uh uh situations all right so that's the notion of Tim like and space like the notion of the world line of a system and the world line of the system is made up out of small Tim like intervals now let's focus on one of those timik intervals let's redraw it over here is's one of the Tim like intervals I imagine in your head that it's a small interval because eventually we're going to let that interval get smaller and smaller I don't think we talked about did we talk about relativistic four velocity we we didn't do it we we didn't do it in any detail all right all right so along the trajectory the particle moves from one point to another point with the separation Delta X but Delta X is a four-dimensional Delta X it consists of a delta T and A Delta of the three other coordinates which of course call Delta with an arrow above them that's Delta X mu sometimes I will call these spatial Delta X's sometimes we'll call them Delta x i when you see an i it just runs over space when you see a mu it runs over all four coordinates of space this is a this is an a um an interval along the trajectory here has four components to it and it defines a Delta TOA between the two end points of the interval Delta to is equal to the square root of of Delta Delta t^2 minus Delta x^2 incidentally just uh once or twice every so often let's put in the speed of light just to remember where it goes T has units of time x has units of space to make this thing and and um since this is called proper time let's assume that it has units of time this one's okay it has the same unit as the left- hand side but this one is not okay it has units of length this one has units of time where do we put the speed of light divide by c^ S right I'll do it now and then every so often do it once more over here uh and then stop doing it because it gets annoying now we have the concept of velocity now basic ordinary velocity is the derivative of x with respect to T but there's a notion of four-dimensional velocity it is a four Vector it transforms as a four vector Vector Delta X mu transforms as a four Vector Delta to is an invariant it doesn't transform every Observer no matter how they're moving ascribes the same Delta TOA but different Delta x m to this interval the velocity ordinary velocity ordinary velocity uh let's call it V with a vector here the ordinary velocity is it's components are simply dxi by DT or the limits in the limit that the Delta X's get small it just becomes the limit of Delta XI / delta T that's a three-dimensional notion of velocity there is and of course this also has has uh this this should be written this way or VI I is equal to dxi by DT all right that's one notion of velocity a more invariant relativistic notion of velocity is instead of using T down here refer the separation of coordinates to the proper time instead of to the coordinate in instead of the the um uh the frame dependent ordinary time that gives rise to something called four velocity it doesn't have three components it has four components so it has a mu up here the MU stands for the four components incidentally mu runs from 0o to three zero is the time component this is equal to DX mu by by not the to but not the T but the to so there are four of them and they constitute what is called the four velocity the four-dimensional analog of the Velocity first question is what is the connection between the four velocity and the ordinary velocity so let's work that out nonrelativistically of course there are only three components of the Velocity so that must mean something funny happens to one of the components here and we'll see what happens all right let's start with DX sub I by DT let's start with V subi that's equal to DX dxi by DT but let's write it as DX I by D to D ided by DT that's a triviality I've just divided and multiplied by DT but it's useful sorry uh yes this is correct all right now DX I by DT what is that that's usei usei where I now I'm in particular concentrating on the space components of U times DT by DT by DT by DT we'll come back to DT by DT in a moment in fact we're going to come back to it right now but let's go let's consider the um uh yeah before we go ahead let's consider exactly this quantity you not what is UN not un not is the fourth component of this relativistic velocity and what it is is it's DT by DT that's exactly what it is DT by D to the fourth component or the zeroth component I should say of X is just T where is it T and U is just DT by DT DT by DT is just the inverse of this so we're interested in figuring out what d by DT is let's figure out what d by DT is D is the square root of one no it's the square root of DT ^2 minus DX squared that's just rewriting this thing here but with differential notation d t^2 by time d x s and this means all three components the sums of the squares of all three components let's take out of the square root DT Let's Take It Outside the square root of T and let's Factor it off this stuff and then we get in here one minus dx^ 2 by DT ^ 2 or just DT * the < TK of 1 - V ^ 2 right so what do we know we know that DT by DT is equal to the square < TK of 1us V ^2 that's what this 1 minus v^2 is or the square root of 1 - v^2 it's the rate of change of proper time relative to ordinary time if V is close if V is small by comparison with the speed of light if it's close to zero then this is close to one in other words in a non-relativistic limit DT by D to is just one there's no difference between to and T if the object is moving with close to the speed of light there can be a large difference all right let's see where were we I uh ah all right so what we want from here is this equation over here I've crowded too many equations together I was working on this here's where I was working all right so first I'm I'm making a little bit of a mess out of this a mess out of the Blackboard in any case all right first of all this is equal to DT by DT what is DT by DT DT by DT is square of 1 minus V ^2 what is DT by DT one over that right so you not whatever it is is one divided the sare < TK of 1us V ^2 it's a thing which none relativistically just goes to one in the non-relativistic limit where V is much smaller than uh than the speed of light this is just one that's why we never think about it we never think about DT by DT because ta is T and DT by DT is just one good all right on the other hand relativistically we worry about it what about here V subi is u subi d by DT but d by DT is again just which way is it d by DT isun 1us v^2 divided by let's see yeah V subi is D XD to D to DT looks like it's multiplied why does that surprise me no it doesn't surprise me that's correct correct that's correct so now we can assemble together what is the meaning of the four velocity in terms of the normal velocity first of all you not the time component of the four velocity is just 1 over the < TK of 1 - V ^2 what about the space component use of I that we read off from here that's equal to V subi ided by that same squ otk of 1 - V squar all right so this is the basic computation you would do to find the components of four velocity again if we're talking about the non-relativistic limit where V is very close to to zero then the square root is Trivial and u subi and v subi the two Notions of velocity are the same as you get up closer and closer to the speed of light it looks like USI is much uh bigger than vabi is that right yeah usbi is much bigger than vabi or visi is much smaller than USI okay this is the notion of four velocity this is the notion of for velocity and everywhere along the trajectory of a particle the particle is characterized by a position you can add in the time at which you characterizing it so that's a four Vector of position and also a four Vector of velocity now the four Vector of velocity doesn't really have four independent components it only has three independent components let's go through why that's true let's take the yeah okay I'll explain very quickly why it doesn't have four independent components there is a relationship between the three components of spatial velocity of um U sub I and U subn and it's just a relationship that U subi let's call it U squared which happens to be v^ 2 over square otk of v^2 over 1us v^2 that's the space component squared is V ^2 / 1 minus v^2 if [Music] I seems like you can just see it from the definition of u z yes we can yes we can but I'm I'm halfway through it so let's finish it let's take un squar minus this thing here all right so that put a minus sign here + 1/ 1us V ^2 I'm making this more complicated than it is the answer is that the right not square root just one minus v^2 oh boy what's the right hand side one 1 - v^2 1 - V ^ 2 1 so all four velocities no matter how fast the particle is moving or not moving the square of the uh the square or the difference of the squares of components time component and space component is always equal to one that's always equal to one and that's why there are not four independent components of the four velocity only three independent components of the four velocity they're related by a single relationship like this otherwise they're independent could you also say that conceptualize it by saying that differentia respect to one of the components say it again can you also can you alternatively conceptualize the dependen by saying that it's differentia with respect to one of the comp yeah yeah sure there's an easier way to I I said this badly Let's uh let's uh let's go through a very simple version of it all right Delta to^ 2 is equal to Delta t^2 minus Delta x^2 okay boy did I make a mess over there let's just divide both sides of this by Delta to squ we get one is equal to Delta t^ 2 over Delta to^ 2 minus Delta x^ 2ar over Delta to^ 2 Delta to delta T by Del Delta to that's un not so the first term here is just un not squared and the other one is U Vector squared okay so it's that's that's all that went on here a complicated way to say this right but let's go back now to the uh to the world line here's a world line it's characterized by a four velocity and a four Vector only one redundancy and that's that the four components of youu are not completely independent we want to know what law governs the motion let's take a free particle one with no forces on it what law governs its motion and the law that we're going to abstract from a quarter's worth of classical mechanics is the principle of least action all right the principle of least action the only thing new is that the action is going to have we want our laws of physics to be the same in every reference frame that means we should try to cast them in terms of quantities which are the same in every reference frame if we have a principle that says that in going from here to [Music] here the particle goes along a trajectory which min minimizes or maximizes minimizes some quantity called the action we are doing good business if we make sure that that action doesn't depend on which frame of reference it's evaluated in we would like the action to be something which is the same in every reference frame we know of only one and that's one fact the action should be invariant and the other thing that we've learned from experience is the action should be built up incrementally as a sum over the trajectory of little infinitesimal pieces for each uh segment of the trajectory that we can abstract from classical mechanics action being a sum or or an integral better yet an integral over the trajectory and the thing of which it is an integral should be a quantity which is the same in every reference frame there's really only one thing that is invariant in going from a small from a position to a displaced position and it is the proper time along the trajectory the proper time from here to here is an invariant meaning to say that all observers no matter what the velocity they're moving with will not agree on the Delta X's but they will agree on the Delta tow so a good guess and it's the right guess right guess for the action is the sum we'll convert that to an integral in a moment of all the little Delta tows from one end of the trajectory to another end of the trajectory all the Delta tows that is the action not quite not quite and that action has to be minimized holding the end points fixed remember your classical mechanics the action principle you hold the initial and the final configuration fixed and you search for the trajectory that minimizes or makes stationary the action between the two end points of the trajectory all right so you wiggle around the path until you find the path of least action since the action is composed out of invariance everybody will agree uh that a given path minimizes the action all right so that's our uh that's our principle add them up okay so what does that mean that means we have a freedom incidentally we have a freedom we could multiply the action by a number any number it won't make any difference uh why not because if if uh if a thing is stationary with respect to changes of the trajectory If I multiply it by 10 or seven or minus 15 it will still be stationary of course if I multiply it by a negative number I will have I will turn a POS a maximum to a minimum but I won't change the fact that that trajectory makes the action stationary yeah what happened to energy why why are you getting ahead of the game we start with action energy we derive we will do that we will do that we will before the evening is over we will discuss why we didn't start with energy but the answer is it's not invariant uh it's not even invariant in ordinary uh Newtonian physics the energy of a particle is not independent of its state of motion so if you see a particle standing still and I see the particle moving we ascribe a different kinetic energy to it so it's not an invariant concept it's a good idea to start our discussion of particle motion by talking about invariant that everybody agrees upon and in that way make sure that our laws of physics are the same in every every reference frame okay so yeah uh could you state again why a a relativistic invariant is a good choice for action or a good guess for it's a good guess it's a good guess simply because it's the same for every Observer no matter how they're moving that's what proper time was proper time was the particular combination of Delta x's and Delta T's between two points that everybody will agree on the value of it okay it's the lur it's a thing which is invariant under the lorence Transformations if we want our laws of physics to be invariant it would be a good idea to base them on quantities which all observers say are the give have the same value for so for example if I just say you know you know it's analogous to the following this is entirely analogous to the following idea supposing I am interested in the shortest distance between two points on the Blackboard how what do I what do I do if I'm interested in the shortest distance between two points obviously I search for the shortest distance I I search for the curve which has the shortest distance between the two points now might we find that we get different answers in different coordinates inate systems that coordinate system or that coordinate system or whatever no of course we won't and the reason is distance in the ukian plane is invariant with respect to rotations of coordinates in particular if we think of the distance from one point to another along the curve to be a sum of a lot of little incremental distances and everybody agrees on the incremental distances then the Curve that we get will not depend on the coordinate system that we use to evaluate it why it's an invariant concept the shortest distance between two points what we do then is we add up all the little ukian distances here and we're doing exactly the same kind of thing adding up all the little proper times along the trajectory in fact the formulas are almost identical apart from a sign all right now we have a freedom we could multiply this action by a number number excuse me yeah um you may have said this but I don't remember it um we arrived at proper time delus Del square root of that square root of that yeah was there a reason is there a reason why you inherently choose a positive square root the negative s oh we will always choose the positive square root yeah um uh yeah proper time is by its very definition a positive quantity it's the time rid along the clock from the past to the Future and uh oh we could use the negative root we could if we use the negative root everywhere it would simply change the sign of the action that's okay what would happen if you change the sign of the action it would not change the equations of motion at all instead of saying we were looking for the minimum of the action we might say we were looking for the maximum of the action but remember to minimize a function or to maximize it exactly the same equation you just say it's stationary so it doesn't matter whether you take the action and multiply it by a number including the possibility of a negative number it won't change the uh the trajectories if the rule is the trajectories should make stationary the action stationary uh means it could be a minimum or a maximum I guess also if you had a coordinate system in which the x coordinate didn't change then if you chose the positive score then the proper time is the same as the time yes yes that's true that is true all right and I'm going to put in one more little sign one more little change I'm going to put a minus sign in here now as I said multiplying by minus M changes nothing it's a convention it's a complete convention but the convention was established not by Einstein when he was doing relativistic mechanics but long before Einstein by people who had no idea what the special theory of relativity was but who had some prejudices about the way classical Newtonian mechanics works so we're going to come back why this minus m is there we'll come back it's for the purpose of matching onto non-relativistic formulas it would not change anything if we didn't put it there but in order to match with non-relativistic formulas we'll put a minus n there for the time being okay so let's let's uh take that to be the action and now what we do is we the the interval is the trajectories have already been imagined to be broken up into intervals and now we move the the points around in space and time and look for the trajectory which minimizes or makes State I'll say minimize which minimizes the action and that will be the true trajectory that's the principle of least action but before we do that let's convert this to an integral let's convert this by taking the limit in which the differential in which these finite differences here are replaced by differentials so this will become then minus M times the integral of D along the trajectory the integral and I'll explain exactly what this means in a moment the integral of sum of little infinitesimal detals from one place to another there's a reason why the integral is written by this curly symbol here you know what the curly symbol means sum right and Delta to just becomes dtau from one end of the trajectory to the other from the initial end to the other end and now we can write down a formula D is equal to the square [Music] root of 1 minus no DT ^2 minus dx2 and D x^2 means the X2 plus the y^2 plus d z^2 now this is a funny thing an integral with a d to and underneath a a thing in a square root there I bet you've never seen an integral like that maybe you have where the differential things are underneath the square root but it's easy to fix it's easy to make it look more standard factor out a DT factor out of DT and put the DT on the outside that means under the square root we have to divide by dt^ 2 and this just becomes one and what about DX by DT that's the good old fashioned velocity squared so this becomes 1 - velocity squar or 1 - XI do squared XI do 2ar means x do2 + y do 2 plus Z do s all right I won't bother writing the sum and I won't bother writing X Y and Z it's just x suby dot and Dot now means derivative with respect to ordinary time derivative with respect to ordinary time so by factoring out a DT I converted this to something a more recognizable than it might have been before it's an integral of something that depends on the velocity remember about the ordinary principle of least action and the ordinary principle of least action for an element for a particle for example the action is an integral over a quantity called what the lunian and the lran the depends on what positions and velocities what about for the case where there are no forces then it only depends on the velocity not the positions and here's what we got we have something like that here we have that the action is min - M * the integral of 1us the square of the Velocity from the initial starting point it's called A to B so so we have a l grangian now and now we can just take off and do all the things that we would do with a lran let's write what the lran is Lan is minus M which I say is a convention minus him over here times the square Ro T of 1 minus now I'll write it X do s+ y do 2 plus Z do s it's not as simple as the usual um the lunian that lrange first wrote down the lunian that L what lrange write down he just wrote down for the Luni in the kinetic energy he didn't even know he may I don't even know if he knew it was energy he wrote down for the lunian just 1 12 mx. squared okay that's nice and simple lran Lan of lrange elabel that was just m x do^ 2 over two 12 M v^2 where X do s does mean X do s + y do s+ Z do s this is the lran lrange would have written down this is the L grangian that a relativistic version of lrange would have written down what's the connection between them well the first connection between them and the most important for our purposes right now oh well before I say that what is the value of this lunian the value of this lunian is that it encapsulates the motion of a free particle in a way that is independent or that is the same for every reference frame is guaranteed because this quantity the action is independent of reference frame making it stationary you'll get the same answer in every frame so this is the translation of that to L grangian language okay first let's ask what uh what this is approximately like not approximately but what it's like in the limit where the velocities are much smaller than one one being the speed of light so this is one a minus M Square t of 1us v^ 2 velocity squared that we go back to the good old binomial theorem the binomial theorem tells us that theun of 1 - v^2 is approximately equal this is approximately equal to minus M times 1 + V ^2 / 2 so first of all there is a contribution which is just minus m it doesn't depend on anything it doesn't depend on position it doesn't depend on velocity it's just a number what does a adding a number to the lran do does nothing adding a number to the to a lran does nothing because the trajectory the action of the trajectory uh you know comparing different trajectories the constant piece here doesn't depend on the trajectory doesn't depend on the X doesn't dependent on velocity plays no role at all so it's not important to us right the important piece oh do I have I'm sorry minus right 1 minus v^2 over 2 all right so apart from the constant piece the piece with the juice in it the piece with juice in it is plus mv^ 2 over 2 - M + m v^ 2/ 2 in other words it's just the L grangian of lrange in this way we can be certain that the laws of motion that we derive in this way will in the limit of small velocity be the same as uh good old llas lran Newton and so forth so far we haven't put any in anything that resembles a potential energy or or a force law so far we're simply working with a free particle okay so that's a good thing now next concept I'm not going to work out the equations of motion I will tell you right now the equations of motion have the brilliant uh ability to Simply make this particle move on a straight line with constant velocity we'll see that but the equations of motion are not the interesting things uh for this evening's lecture the interesting thing is to identify the Notions of momentum and energy that's where we want to go we want to understand where the relativistic formulas for momentum and energy come from from they're not quite the same as the non-relativistic formulas so let's go through them momentum and energy are important because they're conserved if we account for all the particles in the universe or at least all the particles in a closed system they may be interacting with forces and this and that but the net momentum is conserved what was that a consequence of you remember what that's a consequence of translation invariance all right so translation invariance and translation invariance is true here um we're we're actually in a special case of L grangian mechanics everything we said about lran mechanics lonians hamiltonians conservation laws it all applies here it's just that the lran is a little bit strange it has a square root of 1 - x do squ plus y do sare plus Z do square but other than that the rules of the game are the rules of L grangian hamiltonian whatever we whatever we derived previously and we can use it all right so the first question then is what is the meaning of momentum it will be conserved it will be conserved because this s Gran is translation invariant it only depends on the velocities it does not depend on position it's translation and variant the momentum will be conserved as a consequence of everything we learned the first quarter okay so I would like to identify for example P subx the X component of momentum so we go back we open up our book on classical mechanics and we go back to the definition of momentum not the definition which began by saying momentum is equal to MV but the definition which came from the lran point of view and that was do anybody remember the definition of momentum in terms of lran everybody have it at their fingertips was parti L partial Q dot part the derivative of lran with respect to the component of velocity derivative of Lan with respect to x dot I won't bother redoing classical mechanics in front of your eyes here it's a little bit too late for that at this point thing to do between now and next week is to uh is to review this I should have told you last or last week p subx is partial of Val with respect to x dot same thing for p sub y p subz and so forth so let's apply that lran is minus M times the square root of this thing here all right so this is going to be first of all is a minus M then there's a derivative of this square root with respect to x dot all right what do you get when you take the derivative of a square root you get one over a square root one over a square root of everything inside the square root which happens to be 1us V ^2 yes there's a half thank you there's a half there now we have to take the derivative of the thing inside the square root with respect to x dot all right that's going to give us a minus from here so there's another minus sign plus and then the derivative of this thing with respect to x dot which is twice x dot so that will eat up the um the the two in the denominator and I will put an x dot here or m v subx ided the square Ro TK of 1 minus the total velocity squared well we've seen this before have we not right here V subi in this case v subx / < TK of 1 - v^2 is just U subx so what do we get we get p subx here it is p subx is nothing but the mass not times the ordinary velocity but by the ordinary velocity / > 1 - V ^ 2 which is u subx the relativistic four velocity times the mass is equal to uh to the X component of momentum likewise for the Y component and so forth and so on the Y component piece of Y is equal to m u m u suby and likewise for z m is rest mass m will m is the rest Mass m is mass okay let's uh let's talk about a uh convention uh the use of the term Mass rest mass and all that stuff um there's an old convention and a new convention the new convention is older than I am um more or less nobody that I know who does physics used is the term rest Mass anymore rest mass is an anachronism that's the word anacronismo uh undergradu textbooks cannot seem to get their head around the idea that the idea that the mass of a particle is a tag that goes with a particle which characterizes the particle and not the motion of the particle the mass of an electron if you look up mass of an electron you won't get something that's different if the electron is moving a stationary you will simply get whatever it happens to be I don't remember the mass of the electron in kilograms but it's some small number uh so the way that mass is used in modern language is it's a number that goes with the particle but it is exactly what used to be called the rest Mass what used to be called the rest mass is now just called Mass what used to be called mass is what energy energy divided by the speed of light squared or something we use the term energy for the thing which characterizes uh the moving object and the energy at rest we just call the mass now there's the speed of light Factor that's just a conversion factor and we'll come to it all right so when I when when we talk about the mass of a particle we mean what used to be called the rest Mass from now on I will never use the term rest Mass I will always use the term Mass meaning the energy at rest okay good right that was a it's an important distinction I I I normally don't even think in a language of rest Mass so I didn't mention it but it was good that uh Warren Warren yeah Warren brought it up uh so that we don't get confused okay now let's talk for a moment about four vectors four vectors are the things like Delta X and delta T supposing I have a four Vector in this this case it would simply represent a little interval that's the four Vector of displacement I guess you would call it the four Vector of displacement that is a delta T and A Delta X now supposing I told you that Delta X as opposed to delta T that Delta X was equal to zero could that be an invariant characterization of this interval what happens to Delta X when you Len TR trans form it changes right in particular it'll pick up a little bit of delta T so it cannot be an invariant description of this Vector a vector with Delta x equals z would be a vertical Vector but in some other frame of reference that we're moving past the past the stationary axis the same Vector would be tilted in other words if you were moving past it the beginning and the end point would be separated by some Delta X so the statement that Delta X is equal to zero that might have some significance to a particular frame of reference but it can't possibly be an invariant uh distinction what about the statement that Delta X and delta T are both equal to zero or all four of them not just uh not just Delta X and delta T but Delta Y and Delta Z can that be is that an invariant uh statement all components of this Vector are equal to zero of this four-dimensional Vector are equal to zero well if you think about the way things transform a Delta X when a transforms will pick up a little bit of delta T A little delta T will pick up a little bit of Delta X but if delta T and Delta X is zero in one reference frame then it will be zero in every reference frame so saying that a vector or four Vector in particular is zero the whole thing all components of it that's an invariant statement to say some specific component or even a specific subset of components less than all of them is equal to zero that is not an invariant statement the same is true of the four velocity the four velocity here are its components is it an invariant statement to say that U subi is equal to zero I meaning for just the space components no it isn't because if you what it means to say that U subi is equal to zero is just that you're in a frame of reference where the particle happens to be stationary change frame of reference UI will not be zero okay um so in general if you want to if you want to express invariance statements you want those statements to be about all components of a four Vector a four Vector means a thing like Delta X or U you want it to be a statement which is true for all the components you want it to be a full-fledged Vector statement full-fledged four Vector statement okay let's come to the concept of momentum conservation momentum conservation we've so far identified three components of momentum only three components those three components of momentum happen to be proportional to the components of four velocity they happen to be proportional to the components of four velocity but that's incidental for a moment the important thing is that momentum is conserved if I have a collection of particles and we add them all up the initial momentum must be equal to the final momentum now that's a vector statement that's a vector statement that the three components of momentum and let's label them this way initial must equal the total the total initial momentum must equal the total final momentum or we can subtract and just write the law of momentum conservation says that this is equal to zero initial momentum minus final momentum is equal to zero this in itself is a vector equation is it an invari iant vector equation if it's true in one frame is it necessarily true in another frame not unless we can make an equivalent assertion about some fourth component if there is a fourth component that turns the momentum into a four vector and then we say all four components are conserved then it becomes an invariance statement then it becomes an invariance it's only an invariance st statement if we can identify a fourth component and say that the fourth component is also conserved a fourth component such that the three components together with the fourth component add up to a four Vector well it's pretty darn clear what the fourth component of this thing has to be huh what must the fourth component be it must be M time U we have four three components if we didn't have the M there we would have three components of the four velocity and the fourth component of the fourth velocity would just be the thing which fills out the full four Vector so question is there a significance to Mu subn whatever it is we can give it a name we can call it p subn uh we can give it a name is there a significance to it and do and is it something that we already recognize is something which ought to be conserved and of course the answer is yes there is a fourth thing which is conserved for particle mechanics it's energy so it's natural to ask the question is this the energy is this fourth component of four momentum nothing but the energy of the particle well how do we decide that how do we decide what the energy is anybody got a clue about how to decide what the energy is ah look at the hamiltonian right and that's the only way that's the only systematic systematic procedure look at the hamiltonian so the next step is to figure out the hamiltonian we're going to erase some of the Blackboard here and go to the hamiltonian and a hence from the point of view of Relativity all the work that we did in the same quarter was just to set ourselves up for this kind of problem right so let's go to the hamiltonian I know that not all of us remember what the hamiltonian is so I will remind you the hamiltonian is given in terms of L grangian again I don't expect you to remember it but I do expect you to be able to look it up and uh and uh remind yourself where it came from why this was an important quantity it's a conserved quantity it's an important quantity uh I will write it down for you it's the sum of all the coordinates sum of all the coordinates of a system of the Velocity associated with that coordinate I sometimes put the indices upstairs and sometimes I put them downstairs pay no attention to where I put them at the moment and uh what does it multiply remember the momentum associated with that same direction I'll put that one upstairs just for variety and then but that's not finished what else is there minus minus the Lan that's the hamiltonian as I said if you want a systematic way to think about mechanics where we don't just make up stuff as we go along we fall back on the basic principles of mechanics allrange Hamilton and so forth okay so now we can write it down we now know what it is let me write the lran over here and the P's what is p subi p sub I is equal to m x subi dot / < TK of 1 - V ^ 2 that is the same as M USI but I'm not going to use it that way I'm just going to use it the way it is here p ofi is that and the lran is - M sare < TK of 1 - V ^ 2 where that means vx2 + V y^ 2 + V c^ 2 that's L Gran we have everything we need now to calculate the hamiltonian or what is essentially the energ energy okay so we have then summation of I XI dot times P subi that's m x subi do sared over I think squ 1us v^2 is that right that's the first term here the XY do s picks up one XY Dot from here 1 x by do from here the mass over here and now minus the lran so minus the lran is plus M Square < TK of 1us v^2 looks like a bit of a mess but it is not it is simple first of all XI do squared that's just the total velocity squared x do S Plus y do S Plus Z do s That's What I Call v^2 so the first term here is just not even a sum anymore it's just m v^ 2 over < TK 1us v^ 2 then we have to add plus M times a square root in the numerator how we going to add a thing with a square root in the numerator to a thing with a square root in the denominator well we'll put everything in the denominator so we have to multiply and divide by square otk of 1 minus v^2 that gives us in the numerator 1 - v^2 and then squ < TK of 1 - V squ and the denominator there we are okay now it doesn't look so fearsome anymore because we notice immediately that MV s over here cancels mv^ 2 over here and the whole thing is M upon Square < TK of 1 minus v^2 all right that's the hamiltonian so let's add to the momentum a fourth component here which sorry P not which is M and do you recognize what uh what this thing is this is UN not so indeed we find that Poss apart from a possible yeah when you wrote down the no no no no no no this formula for the hamiltonian is extremely General all systems that we've ever studied this is the formula that was the General general formula for any lran any action principle we derived this the conservation of H in in very very general grounds right depending on what you put the the depending on what you put in for the lran it might be non relativistic physics it could be any number of things good all right so now we're in business the conservation of momentum becomes the conservation of four momentum conservation of X momentum y momentum Z momentum and the conservation of energy before we go on we should figure out about what this new concept of energy has to do with the old concept of energy how different is it than the energy that we've already established okay so let's the one uh thing that we're going to have to be careful about is the factors of speed of light um they are conventions largely they're conventions if the energy defined in one way is conserved it won't make any difference to its conservation if we in order to make it match with units if we multiply it by the speed of light or something conservation is conservation and we will probably have to do that okay let's look at mu that is M over the square < TK of 1us V ^2 that's what we're calling the energy of the particle now of course if we want to restore some well before we rest restore yeah before we restore units let's use the um uh binomial theorem let's use the binomial theorem to approximate this here it is here's the binomial theorem for 1 / 1 - v^2 and it gives us 1 + v^2 2 all right so this becomes M plus M v^2 over 2 this of course is completely recognizable again it is just the ordinary Newtonian kinetic energy had Newton known about energy this would have been what he would have written down for kinetic energy it is what kinetic wrote down I have no idea who first uh thought of this as energy but uh somebody did what about this term over here well first of of all this equation is not dimensionally consistent not dimensionally consistent here you have an m and here you have an M * a velocity squared to make it dimensionally consistent all we have to do we can do it in a number of ways right to make it dimensionally consistent but to give it units of energy we're thinking that it's the energy to give it units of energy this already has units of energy it's just a good old kinetic energy in order to give this term units of energy we have to multiply it by some power of the speed of light what power of the speed of light well we already know this thing is missing two powers of a velocity compared to this one so what's it going to be it's going to be MC s all right I won't I will um I will just write that this is the energy so here we see coming out of the basic principles of mechanics the conservation of a quantity which when it's evaluated for a system at rest is just MC s for a system at rest its energy is related to its inertial mass to its mass by mc^ S how do we know that this is the inertial mass well it's the inertial mass because it enters into the energy the same old way that inertial mass entered into the old energy it enters into momentum at least for slow velocities in the same way that the inertial mass entered in here and so this must be the inertial mass time c^ 2 that is that's the origin of um Einstein had another uh another way to think about it he he asked himself in a collision of objects of various kinds given he he worked a little bit differently he didn't know about nus theorem he uh had to work from the ground up so to speak he talked about collisions of objects objects come together break up and Scatter and go flying off and he said look I know that the momentum is conserved what do I have to think of as the energy in order that the energy and momentum together would form a complex that would be conserved uh in any frame of reference we've short circuited that and uh found the the formula for the energy by standard methods okay so that's the origin of eal mc² it doesn't apply only to a particle it applies to any object uh that can be identified as a closed object that its energy when it's the object is at rest now what does the object being at rest mean inside here there are lots of molecules moving around like crazy so in that sense this object is not at rest on the other hand its Center of Mass is at rest if I hold it at rest what we mean what we mean by um by the velocity here is the velocity of the center of mass and what's left over when you set the center of mass velocity to zero is the rest energy the rest energy is related to the mass by a factor of speed of light another way to say that is that the speed of light is just a conversion factor from one way in of defining energy to another it's just a conversion factor um silly but want to make it dimensionally consistent so you multip C if I would have yeah you could multip by C over 10 million a lot no no no no no no it has to reduce to this when C is one okay right good point good point that's right on Purely dimensional grounds we never could have deduced that this would be one here but on the grounds that it has to reduce to this when C is equal to one we say there's only one thing we could do with it so that one in the bomal expansion is important oh yes absolutely absolutely there are higher order terms missing right yes there are higher order terms and they are simply very very small when the velocity is well I mean for some purposes we might care about them but the leading piece for slow you know we could evaluate them and we could see how small they are for a baseball moving it 90 mil an hour and believe me they would be really really small so right physical way to think of those corrections to the energy yeah the con well the connect they corrections to everything everything gets Corrections those Corrections are quite significant when the object is moving close to the speed of light in fact I mean they become huge let's just think about it for a minute um if I were to plug in the speed of light here well this would get pretty darn big but it wouldn't get any bigger than this but it's perfectly finite on the other hand if I plug it into this formula V goes to one the kinetic energy of an object moving with close to the speed of light becomes infinite that's another way to see that you can never get up because it would take an infinite energy to get up to the speed of light also an infinite momentum these U's also are V's divided squ > 1 minus V ^2 so as you push the particle faster and faster its momentum and its energy increase and explode diverge become infinite when the velocity becomes a speed of light that means it would take an infinite Force to accelerate a particle to a speed of light and infinite forces are simply not part of nature I I was one um the conservation law right four momentum conservation of four momentum that's exactly right okay yeah yes we're going to come back to that in a minute right right that's just a statement that momentum that energy is determined in terms of momentum if you have two particles with opposite velocities does the mass of the combined system is it greater than the Su of the two typically yes yes of course um if you wanted the mass of the combined system by that I mean the energy of the combined system when the combined system is at rest now what does it mean the combined system is at rest it means exactly what you said that the two objects might be moving in opposite directions then the energy is definitely larger than the sum of their masses so I in this convention the rest mass iser than the rest Mass all right now we have to be careful about definitions some things are simply p purely definitions some people might react at this point well what I mean by the rest mass is simply the sums of the rest mass of two particles that that is not a very good uh invariant concept at all another person might mean no I don't mean that what I mean is think of the thing as a unit and discuss its energy in the frame of reference where its total momentum is equal to zero that would be what we call the center of Mass frame or the rest frame of the combined system now since all ordinary systems are made up out of particles that are moving hither and thither all over the place uh a invariant concept is the energy of a system in a reference frame where its momentum is zero if we call that the mass then the mass of the composite system in your case in your particular case would be greater than the sums of the uh of the two masses if there were forces particularly attractive forces it might be less than uh but that's another case yeah what what does the mass term do when you're included in the phase of a way function gives an e to the I m^2 c m c s times time right so I'm got well I mean I have trouble forming this into a question but I mean the way we got this expression was just a lot of abstract mathematics bres H let's not call it abstract let's call it systematic Systema what I mean the uh This MC s term is very significant right I it's I mean it describes things like an atomic bomb for example I mean it's yeah the thing that's conserved is the total energy not the individual masses of particles so let's let's talk about an example oh you know before I talk about the example uh let me first take a little digression into massless particles massless particles are a little bit strange here's the energy of here's the energy over here m / 1us v^2 now what's the velocity of a massless particle one okay we're in trouble well maybe the trouble is not too bad the numerator is zero and the denominator is zero it doesn't tell us what the answer is but at least it tells us at worst it's Z over zero that's better than one/ zero okay right maybe not a lot but it's a little better than one over zero all right it doesn't seem like such a good idea to try to think of massless particles in terms or the energy of massless particles in terms of their velocity since all massless particles move with exactly the same velocity can they have different energy if they all move with the same velocity well yes they can and the reason is because 0 over zero is not determined thinking about massless particles is not good to think about them or to distinguish them by their velocity all their velocities are the same okay we have to think about it differently the way to think about massless particles and in fact we can think about any particle this way but uh but to think about massless particles in particular is to think of the energy as a function of the momentum instead of the Velocity okay we could even do we even do that quite often in ordinary mechanics when we write the energy of a particle kinetic energy it is 1 12 mv^ squ okay that's one way of writing it but another way to write it is momentum squared over twice the mass this is I'm talking now not about relativistic mechanics just um oldfashioned mechanics we write the energy in terms of the of sorry we write the energy in terms of the momentum that's another way an alternative way to express the energy okay what happens if we do the same thing in relativity can we find the expression for the energy of a particle in terms of its momentum well it's a very simple trick the simple trick is just to use the fact that you not [Music] squared minus u^2 let's let's add them all in minus u y^ 2 minus u z^ 2 that's right you said it Us z^ 2 is equal to one we worked out that before and now let's I'm going to set C for the moment I'm going to set c equal to one again cal 1 all right if I multiply let's see the um yeah the components of momentum all four of them are the same as the components of four velocity except for a factor of mass so what I want to do is multiply this equation by M2 M2 M2 M2 M2 and then this becomes this is the square of P squar but p is just the energy so this becomes the square of the energy minus and each one of these is the square of the corresponding component of momentum e^2 minus P S I forgot to multiply this side by m^2 e^2 minus p^2 = m^2 in terms of the four velocity it's u^ 2 minus U Vector 2 = 1 in terms of the corresponding components of the four momentum energy and momentum it's e^2 - P ^2 = m^2 or I can write that as e^2 or E = square root of P ^2 + m^2 I just I moved the p squared over to the right hand side and took the square root let's put back the speeds of light I'm going to leave this as an exercise for you I'll tell you what the answer is uh this is the square root of p^ 2 becomes p^ 2 c^ 2 and m^2 becomes m s c to the 4th check that out it's just an exercise and getting your units straight and there it is that's energy in terms of momentum square of momentum and the mass now from this formula we can immediately take the limit that the mass goes to zero we had trouble thinking about a photon in terms of velocity but we have no trouble in thinking about photons in terms of this formula what does it say it says yeah that's right energy becomes equal to the square root of the square of the momentum time the speed of light which is just the speed of light times the magnitude of the momentum let's call it magnitude of momentum square root of p^ s is just the length of the momentum Vector energy for a massless particle is simply the magnetude of the momentum Vector but in order to keep uh uh the units correct you have to multiply by the by the speed of light this is true for photons it's approximately true for neutrinos neutrinos have a tiny little bit of mass it's true for gravitons it is not true for particles which move significantly slower than the speed of light uh all right now that we know how to express the energy of a massless particle we could solve the following equation the following problem let's take a particle I'm going to give this particle a name it's called positronium happens to be uh an electron and a positron in orbit around each other but it's not important what we call it just it's not important what it is it's important what we call it we'll call it positronium it's an electrically neutral particle and it has a mass its mass is approximately the mass of two electrons but I don't care whether whether you call it two electrons or not it's a little bit is it more than two electrons or less than two electrons less than and why is it less than it's bound yeah it's it's right it's got a little bit of kinetic energy so that makes it a little bit more than two electrons but it has even more negative potential energy all right so whatever the whatever the positronium particle is it doesn't matter what it's made of it's simply an electrically neutral particle with a mass of I don't know I'm not going to look at I'm not going to ask you to look up the mass of the electron we know uh yes let me ask you to look up the mass of the electron that yeah it's five it's 0. 51 M but uh it's it's something in jewels I was hoping somebody would give it to me in Jewels what is it11 is that true I don't remember have no idea I mean I know it's 9.1 * 10 that that's the electron Mass all right so 9.1 that's 1.8 uh 9.1 some number of Jews that's the positronium so there's positronium and if you leave positronium around for a while positronium will Decay what will it decay into it will make two photons two photons will go flying off positronium will disappear the two photons will just go off in other words you've made um uh electromagnetic energy out of them you've made uh just plain old electromagnetic energy out of the patronum patronum disappears and becomes in its place two photons going off separating from each other can you calculate what the mass moment not the mass what the um energy and uh and momentum of those two photons are right so let's go through the exercise this is something which would not make any sense at all in non relativistic physics in non-relativistic physics the sums of the masses of particles are always unchanged chemical reactions happen some chemicals turn into other chemicals and so forth but if you weigh the system if you weigh the mass of the system the masses the sums of the ordinary masses never change okay that's non-relativistic physics here the sums of the ordinary masses is changing this has a mass of whatever it was in Jew uh this has a mass of whatever it is in kilograms and and each of these is massless the photons has no Mass uh so the sums the numerical sums of the masses of particles is not conserved not if this process happens and this process does happen but the right rule is not that the sums of masses is conserved it's that the energy and the momentum is conserved so the first question is what is the momentum let's do momentum conservation first let's assume that the positronium atom atonium particle is at rest in our frame of reference if it's not at rest in your frame of reference just move your butt and get into the frame in which it is at rest okay right get into the frame in which it is at rest so it's at rest how what's its momentum if it's at rest zero so the total momentum of the system to begin with is zero now at the case to two photons the first con conclusion is the photons must go off back to back in opposite directions if they don't go off in opposite directions it's quite clear the total momentum will not be zero so they have to go off in opposite directions uh by the Yeah they will go off in opposite directions and they must have equal exactly equal moment backto back equal and opposite momentum why the initial thing had uh zero momentum the final thing has zero momentum all right so that means that this particle goes off with a momentum p and this particle goes off with a momentum minus P good now let's use energy conservation the energy of the initial molecule not molecule positronium atom is its mass whatever its mass is times the speed of light squared in the this is not the mass of the electron not exactly very very close to it but not EX exactly a little bit less than twice the mass of an electron but it is the mass of the positronium atom so you get out your spring balance or whatever it is your your your scale you put a positronium atom on it and you weigh it and that's the mass of the positronium atom and when it's at rest in the rest frame it has an energy equal to mc^ s that has to equal the energy of the resulting two photons going off so this has to equal the energy of the two photons the energy of the two photons is the same because they have the same magnitude for the momentum and it's just equal to twice the speed of light times the momentum times the absolute value of the momentum P so that tells us what the absolute value of the momentum has to be it's m c over 2 that's the momentum of each one of the photons that's that's uh the simplest example of calculating a relativity Decay and calculating what happens this is the momentum of the photons to go out we could convert this using a little bit of quantum reasoning we could convert this to a wavelength for the photons going out if we use the connection between wavelength and momentum that we learned last quarter we could convert this to a statement about the uh the wavelength of the uh photons it's very small it's much much smaller than Optical so this is the mechanism by which a mass turns into energy of course the mass was was always energy but it was in this Frozen form of rest energy of the uh of the positronium molecule and when the patronum atom decays it results in two photons going out the two photons collide with the atmosphere spere with everything else and heat up they can be absorbed by electrons generate electrical currents and so forth and so on so it becomes of the energy gets converted from this Frozen form of rest Mass into a dynamical useful form of uh uh electromagnetic energy yeah um how is spin conserved well uh if the mo if the atom to begin with has zero angular momentum and these guys are going off back to back they better have opposite angular momentum so so if the uh if the uh electron in positron or lined up together it won't Decay no no it don't Decay um it'll always Decay always the cas it's always you have enough Freedom you um you have enough freedom in the photon uh angular momentum to make it to make it go yeah can you explain why particles have spe well let's put it this way in order to have a finite energy in order to have a finite well- defined energy if V goes to one the denominator goes to zero the mass in the numerator had better also be zero that's all you said it' be the same particles the resulting parles oh uh some fraction of the time the positronium atom for example will Decay into a pair of gravitons now that percentage of the time is 10 to the minus I don't know some large number but uh some percentage of the time the electrically neutral positronium atom will Decay to just a pair of photons H gravitons as I said don't look for it in the laboratory yeah that decay has to be instantaneous because ofation also instantaneous means what what does it mean that have's two two photons emium Decay into those because of time inst going well let's see so what experimental question would you ask you could tra you could count you could detect the photons in uh in some sort of chamber that leaves tracks and you could trace the you could trace the two photons back and you would discover that they both come together at exactly the place where the posr where the positronium was and so that you could say look if they decayed if um if the Decay was peculiar and the positronium decayed first producing one Photon that went out to here and then later creating the other Photon which only went out to here they both moving with the speed of light and you trace them back they would appear to come from somewhere over here that doesn't happen if you trace back the trajectories of the photon and a uh uh as they leave tracks in a chamber or something like that they will come back and intersect that exactly where the uh where the initial positronium atom was I'm not sure if that's the question you're asking but I think that's the only thing I can make out of it hiding equation I to refer to what's that oh yeah I understand uh you concluded that e^2 minus P2 = M2 from that equation up above right and it looks like that equation is saying that 2 - u^2 - u y^ 2 - U z^2 is 1 yes why is that oh we went through that but I'll go through it again I'll go through it again for you we already had two versions of it but one of them was garbled and I guess an Ungar bled one just cancels out a garbled one and leaves no exp all right so do you remember this formula Delta to^ 2 is equal to Delta t^ 2 minus Delta x^2 okay oh here it is right over here right there it is good yes Kevin so you not yes is DTD that's correct Del Del right what does that mean about energy does that tell us anything about energy not not it doesn't it doesn't uh strike any deep chord in me uh you know I know it's true does it really ring something powerful only in the sense well let's see you want to D DET nothing nothing really uh visceral in that though um because at rest DT D is one motion it's not one and it's telling you how it's telling you how the energy changes as a function of velocity but nothing more than that yes zer [Music] yeah not if momentum is conserved um well in a laboratory situation in any laboratory situation of course there are contaminating effects um even if nothing more uh the presence of some tiny stray um field magnetic or electric field or something could corrupt the thing and uh and lead to a slightly different answer remember the rule is momentum is conserved for a truly isolated system a truly isolated system means uh that you've accounted for everything in the system and nothing else is interacting with it then the momentum of that system is conserved now if your laboratory is uh sufficiently isolated and the region of space is cold enough that there are no particles bombarding and so forth and so on the positronium together with the um and the photons constitute an almost closed system but you know the it's really not completely closed there grav if nothing else there's gravitational forces between the Earth and the particles they're completely negligible but they're not zero so yeah there are contaminating influences like like the inty principle that doesn't affect it momentum is conserved where it is the better you know the momentum and in particular the better you know the relationship between the momentum it may be that uh that you have less information about the relative positions but um that's a small effect and uh for our purposes tonight I want to separate quantum mechanics from classical but it's it's no it's a real question um right your um formula for energy was just an approximation well one there was an approx this was an well this is not an approximation that's right this well let's see where is it I lost it the yeah the equation E equals m that was an approximation so the non approximation is eal that's right this is the C in that equation oh okay let's do that okay ah over here all right so we have we begin with e equals where do we want to start m / > 1 minus V ^2 so we look at the units and in a minute well right now right right now a one and a v^ s have different units the only way to make v^2 have the same units as one is divided by c^2 there's no other way now the 1 minus v^2 over c^2 is dimensionally consistent however the whole formula is not the left hand side is energy the right hand side now has units of a mass mass and energy don't quite have the same units they're related by the square of a velocity we know that the units of energy the units of energy are equal to the units of mass times the units of a velocity let's put V squared an example would be kinetic energy is 1 12 mv^ s we know that's consistent all right so it tells us that we have to provide the square of a velocity here and that tells us that the C squ has to go over here right and then if youal Z get that's right then you get just plain MC squ right has to Decay into two photons and then what siman yeah yeah well that's that's a hard thing to check I mean if you if you if you had the positronium at rest now remember because of quantum mechanics there's going to be some uncertainty in its position okay and so there is a there's a region in which you may have known that the positronium was in then the positronium decays and um if you trace it back along the world line of the photon that comes out the two photons they will intersect now you could draw you could draw a space-time picture of this here's the space-time picture here's the positronium time goes up this is the world line of the positronium at rest then the positronium decays and a photon goes out this way whoops that's not very good Photon goes out this way and a photon goes out this way if you trace back along the world line of the photon you will find out that they will intersect first of all at the spatial location where the positronium was but they will intersect it at exactly the same time or at least at the same time if if we ignore Quantum uncertainties uh yeah what determines the axes along which the two photons come out of the Poss random sometimes that way sometimes that way sometimes that way that's Quant that's the randomness of quantum mechanics uncertain that's uh you know it's like a taking a spin orienting it this way and then measuring along the other axis sometimes this way sometimes that way so it comes out with some angular distribution but uh but randomly in different directions and at random times to some extent when you first L Lan right I think you picked it it was the one thing we had available the action principle being a statement about minimizing something or other about a trajectory will be invariant if you choose the action to be invariant right so we is there a more Bottoms Up way no no no so when we did the classical mechanics I mean I understood I hope this is right that we basically chose the action so that when you you get the oiler equations for that path integral that they provide the equations of motion Newton equations Newton equ yes indeed so but it didn't seem like there was anything analogist going on here maybe no the only reason we haven't put anything concept of force now all right what we could do is we could take this L grangian and work out the equations of motion and see what the equations of motion say but we already know what they say they say that the energy and momentum and therefore the four velocity is constant and that's all they say just as um Newton's equations when there's no forces around just say well I say that M you know mass times acceler ation is zero that's the same as saying the time rate of change of momentum is zero or that the momentum is constant all right so this will say that the momentum is constant nothing more or that the four velocity is constant well the four velocity being constant that means not only in magnitude but in Direction it says that the particle moves along a space a straight line in SpaceTime that the uh that the motion of the particle is a straight line through SpaceTime and that says it moves along a line in space with constant velocity that's the whole content of uh the oiler lrange equations for this particular case now we could try putting Force we could try asking how does this get modified if we try to put various kinds of forces in and we will do that one of the things we will do is allow the particle to have a uh charge and to couple it to an electric and magnetic field and see how it moves uh but so far that that's a little more complicated than anything we've done up till now up till now the whole content of the oil lrange equations is just momentum and energy conservation or straight line motion but not just straight line motion to two of Newton's Laws become the same law what are Newton's Laws a particle moves uh with along a straight line and the other one says the part and a particle moves with constant velocity here it becomes a particle moves in a straight line through SpaceTime which entails both straight line motion and constant velocity one more question now I got to go you talk about how in the limit velocity get very low that relativity um laws approach the laws right um or at least the laws of conservation as they would have been derived from Newton yeah and in I'm just trying to make I'm not really here but in a previous qu we were talking about Quant mechanics we're talking about in this other limit where you increase the mass you the the laws of Quant mechanics also approach yes but it's an entirely different limit one yeah right one is velocity is like number or something uh combination of various things but let's let's say the mass of the object and uh the important thing is the mass of the object and the smoothness of the uh forces that act on it it shouldn't be abrupt and sudden but that's another you know we got to separate things this this is got to do with the speed of light uh quantum mechanics has to do with plunk constant and you can have a situation which is highly quantum mechanical but nonrelativistic you can have a situation which is highly relativistic and not quantum mechanical and you can have both relativistic and quantum mechanical the two independent uh limits is this where they say that inity between I don't know no no there's no incompatibility I think you're thinking about gravity gravity there are puzzles about but uh we haven't gotten there yet so far we're not dealing with gravity at all we'll come to it okay I think it's uh time to go home for more please visit us at stanford.edu

Channel: Stanford

Views: 522,076

Keywords: mathematics, science, mechanics, field theory, quantum physics, classical mechanics, einstein, motion, momentum

Id: _R6J8Wb1Xek

Length: 119min 28sec (7168 seconds)

Published: Fri May 11 2012