Four-Vectors in special relativity

hi guys and welcome to this virtual lecture course on classical mechanics and relativity i'm dr andrew mitchell and in this lecture we'll be discussing the four vector formalism in special relativity four vectors are objects which live in four dimensional space time that's three plus one dimensions three of space and one of time and they transform under the full symmetry operations of the poincare group this means they transform under translations and rotations of our coordinate system as well as the lavender transformations compare this to regular three vectors which live in three-dimensional euclidean space vectors are things which transform under rotations and translations of our coordinate system when we go to four dimensional space time we can generalize this we have to include then the lorentz transformations which mix up space and time we'll be discussing different kinds of four vectors in this lecture we know of one already the space-time event this is specified by four coordinates three of space and one of time we see that this of course transforms according to the levent's transformation but there are a whole family of these and we'll be discussing how to derive different four vectors in this lecture including for example the full velocity the four momentum the four acceleration and so on this will set the theme for relativistic mechanics which we'll be discussing in the future lectures we'll see that there are certain objects which are relativistically invariant and are the same in all reference frames and this is very naturally and nicely explained within this four vector formalism in particular you'll see that there are two different flavors of four vectors the covariant and the contravariant four vectors we'll be discussing both of these and their interpretation as well as how they're connected through the minkowski metric tensor which encodes the hyperbolic geometry of the underlying four-dimensional space-time so in this lecture we'll be discussing a lot of the formalism we'll be discussing a lot of the notation for example einstein's summation notation and these contravariance and covariant four vectors and contractions of those objects so this will allow you to then go to the books and other lectures and so on and understand how to read these equations how do you read a relativistic lagrangian that's manifestly covariant for example we'll touch upon an example of electromagnetism at the end of this lecture as a concrete example so we'll be discussing this in the lecture we'll be making an outlook towards relativistic mechanics so let's get down to work in this lecture i'll be giving an introduction to four vectors as they appear in special relativity of course this can be generalized even further when we consider general relativity but that won't be within the scope of this course an event in space time is specified by three spatial coordinates and one time coordinate we can assemble these together into what's called a contravariant for vector this is denoted by a capital x and a superscript mu the greek letter mu there this superscript is referred to as the index and what we'll be discussing here is called the index notation by convention when we use a greek letter such as mu for the index this means it runs over the four space-time coordinates three for space and one of time the index mu can take values 0 1 2 or 3 so there's four coordinates but by convention we start labeling from zero so 0 1 2 and 3. the zeroth component is time and the one two and three components are the spatial coordinates x y and z so we can specify a specific component of our four vector simply by specifying the value of this index x superscript 0 for example refers to the time-like component ct whereas x1 refers to x x2 refers to y x3 refers to z all of these things lumped together is referred to as our x mu our 4 vector and a 4 vector corresponding to a space time event is called the 4 position just remember that when you see the superscripts 0 1 2 and 3 here these are not powers so for example x 2 here does not mean x squared it means the y component of the four vector since the x y and z components labeled 1 2 and 3 represent the spatial coordinates sometimes we refer to the four vectors in this form where we single out the time component in the zero slot and then we have a regular three vector corresponding to the spatial coordinates i'll refer to these as time like and space-like because as we'll see when we generalize to other kinds of four vectors we will still have four components but they won't simply be t xyz there'll be other things however we can still identify these as being time-like or space-like components by convention if we have an index which is a greek letter such as mu as i've given here this runs over the four components of the four vector if however i use a latin index like i or j then by convention this only refers to the space like components x y z in this case from the contravariant four vector we can obtain the corresponding covariant for vector by acting on it with the metric tensor of the space the covariant 4 vector follows from this equation on the right hand side we have the corresponding contravariant for vector x nu so i'm specifying this by the greek letter nu here and we're multiplying it by the metric tensor g mu nu this is a tensor because it has two indices and you see they're both subscripts they're lower indices then we simply sum over the values of nu because nu is a greek letter as i mentioned that means it runs over the four dimensions of space time meaning we sum over the four components nu equals zero one two and three new therefore is contracted in this equation we get rid of the new because we're summing over it what we're left with is something involving mu and that's what we see on the left-hand side when you've seen these equations a few times you'll get used to this the idea is we sum over the value of nu here and we're left with a mu the mu is downstairs and so it appears downstairs on the left hand side of the equation and x with this lower index mu is the covariant for vector we refer to the contravariant four vectors as having an upper index and the covariant four vectors as having a lower index is the greek letter upstairs or downstairs that's the thing which symbolically distinguishes between the controverian and covariant four vectors g mu nu is the metric tensor which defines the geometry of the underlying space in our coordinate system we discussed the concepts of the metric tensor and what they mean in one of the first lectures of this course and i'll say some more about what that means in special relativity very shortly but first let me introduce the so-called einstein summation convention according to einstein's summation convention we can write this sum here in this rather simplified form we just leave out the summation sign we just write g mu nu and x superscript nu and the summation over nu is actually implied by the notation specifically there's an implicit sum over repeated greek indices where one of them is lower and one of them is upper what do i mean well in this expression we see that nu the greek letter nu is repeated in these two terms and furthermore one of the indices is in the lower position and one of them is in the upper position if we ever see something like that then einstein's summation convention tells us that we should just implicitly sum over the values of nu which means in practice nu equals 0 1 2 and 3. this is referred to as contracting the four vector so these two expressions are identical this is literally just a notation it's not there's no physics in here there's no mathematics it's literally just a matter of notation whenever you see this thing it means this sum however we'll see that using this notation cleans up and tides up our equations hugely and makes things look very nice and simple einstein once joked that introducing this summation convention was his greatest contribution to physics in the end it just makes the equations look nicer but you should always bear in mind that when we see things like this there's an implicit sum this operation multiplying a contravariant for vector by the metric tensor and contracting this index which means summing over this index one upper and one lower is referred to as lowering the index and you can see why that is you start off with a contravariant four vector with the index in the upper position and you end up with a covariant four vector with the index in the lower position so we can talk about lowering the index and as i'll show you in a moment we can do the inverse and raise the index we can take a covariant for vector and transform it into a contravariant for vector that would be referred to as raising the index as a mnemonic rule of thumb you can imagine these indices kind of cancelling imagine that we have one upstairs new and one downstairs new we can sort of draw a line through these things cancel them out and then what you're left with is a mu downstairs which is what appears on the left hand side of the equation however i want to emphasize that this is only a mnemonic it's only a way of remembering and figuring out what goes on in these equations it's not really that we're cancelling these things as you can see there's actually a sum over mu here but this is certainly a good way of remembering the result the full theory of the way this works is actually part of the mathematical field of differential geometry and this involves tangent vectors and fiber bundles and manifolds and so on and all this stuff we won't actually get that into that in this course um maybe i'll do a lecture in future when we talk about general relativity in which we'll go into the mathematical details more fully however in special relativity we'll see that this is actually rather simple and we don't need to get into all of that detail so we've just seen that if we have the contravariance for position x new then we can lower the index and obtain a covariance for vector x mu by doing this contraction where the einstein summation convention over the uh greek letter nu here is implied g with the two lower greek indices mu and nu is the covariant metric tensor it's the metric tensor that when contracted with a contravariant four vector gives the covariant for vector on the other hand when we're raising the index as in the second equation we take a covariant for vector and contract it with the contravariant metric tensor to give a contravariant for vector so what is the relationship between these two different metric tenses well we can actually find the relationship by first lowering the index and then raising it again that returns our contravariant full vector back to a contravariant four vector in this first part here we are lowering the index of the contravariant for vector x mu to obtain x nu then we raise the index again to obtain x sigma this time so by comparing the left-hand side of the equation and the right-hand side of the equation we see that the product of the covariance metric tensor and the contravariant metric tensor is actually the identity matrix and therefore the contravariant and covariant metric tensors are actually the inverse of each other we see that they must be the inverse of each other because we can lower and raise the index and get back to where we started in terms of the index notation we can take this product with the einstein summation convention implied here where we're summing over the four values that the index nu can take here which is repeated one upper one lower and we obtain the identity matrix this might look a little bit funny we're specifying here two indices again one is upper and one is lower that's what you get after you contract the new in this expression so just as if you were to have two matrices if you multiply a matrix by its inverse you get the identity matrix this is something very similar here in index notation so far all of this is very general and here we have a totally general metric tensor for an arbitrary geometry of our space time in an arbitrary coordinate system let's now be a little more specific and use the actual metric tensor that applies in einstein's theory of special relativity in cartesian coordinates in special relativity we use flat minkowski spacetime this is the geometry of four-dimensional space-time that we've been discussing so far in this lecture course if we were to go to general relativity we'd see that space-time could be curved but that's beyond the scope of this course so here we're just going to concentrate on the special case arising in flat minkowski space time in special relativity in that case the generic metric tensor g mu nu becomes a specific metric tensor the so-called minkowski metric tensor given by this eta mu nu traditionally this is given by the letter eta another greek letter i'm afraid for our purpose here we can consider the minkowski metric tensor simply to be a four by four matrix this is a bit of an over simplification of the full theory but it suits our purposes here so we can regard the metric tensor as a matrix let's call it eta matrix i'll denote matrices in this lecture with this double line underneath and an element of that matrix eta subscripts mu nu is simply an element of this matrix in particular a to mu nu corresponds to the muth row and the newth column in special relativity this takes the special form where the eta matrix is this diagonal symmetric matrix with ones or minus ones on the diagonal since we specify an element using mu and nu which both run over the values 0 1 2 and 3 we can label the rows and columns of this matrix 0 1 2 and 3 0 1 2 and 3 so that we can identify that the 0 0th element is plus 1 whereas the 1 1 2 2 or 3 3 elements are all -1 all the off-diagonal elements are just 0. we saw this same object appear when we were talking about the matrix formulation of special relativity in the previous lecture as you know the metric tensor encodes the geometry of the underlying space in this case this minkowski metric tensor encodes a hyperbolic geometry of the dimensional space time as we discussed previously we say that the signature of the metric is plus minus minus minus we're basically just reading off the values on the diagonal of this matrix so we have plus one minus one minus one minus one this is referred to as the signature of the metric i should mention here that there are different conventions about this is also possible to define the minkowski metric in terms of the minus of this object that would have a signature minus plus plus plus this is also perfectly valid you get exactly the same results when you compute any physical properties it's simply a convention however we have to pick one convention and then stick with it and in these lectures i'm choosing the signature plus minus minus minus which corresponds to this specific form of the minkowski metric tensor but just bear in mind that in other lectures or other books you might see the opposite convention now it turns out that the fact that this metric tensor is a diagonal matrix with just plus or minus ones on the diagonal makes things a lot more simple as we'll see now firstly because we have a diagonal matrix it actually doesn't matter whether we specify mu or new as the row or column so we can be a little bit careless about this we don't have to worry too much about whether the first index corresponds to the rows or the second index corresponds to the columns it's a symmetric matrix furthermore because of the special properties of this matrix the contravariance and covariance metric tensors are actually identical i mentioned earlier that they are the inverse of each other and you can see because this matrix just is a diagonal matrix with plus or minus ones clearly the matrix is its own inverse so again this simplifies matters in special relativity finally because this is a diagonal matrix it makes matters rather more easy when we're doing index contractions as an important example let's work out the covariant four position from the contravariant four position using this minkowski metric tensor in special relativity so we start out with the definition of lowering the index of our contravariant for vector remember that because we have this repeated greek index nu one upper and one lower this by the einstein summation convention implies this sum over nu so if we write out the sum we get eta mu zero times x zero plus eta mu one times x one plus eta mu2 times x2 plus eta mu3 times x3 however because the metric tensor is a diagonal matrix it means that only one of these four terms will survive if we're considering x 0 on the left hand side then only this first term will contribute that's because eta 0 0 is plus 1 whereas eta 0 1 8 0 2 8 0 3 are all zeros on the other hand if i have x 1 on the left hand side then only this term will contribute because only eta 1 1 is non-zero a to 1 0 a to 8012 a2 1 3 as you can see from the second row here are of course all zeros and so on so you can immediately see what happens here when i'm writing down the covariance four-vector for the position on the left-hand side here we're going to get an object that is almost exactly the same as the original contravariant for vector but just with some pluses and minus signs in particular the time-like component is going to get a plus sign and the space like components are going to get minus signs so when we put all this together we can write our contravariant for position x mu as ct xyz this is the four vector whereas the corresponding covariant for position because we do this contraction with the uh the covariant metric tensor eta we get ct minus x minus y and minus z so finally we see that after all of this first the contravariant four vector has uh time-like and space-like components in this form ct and a three-vector are whereas the corresponding covariant for vector simply has a minus sign on the space like components and that's really the only difference between the contravariant and covariant four vectors within the theory of special relativity in flat minkowski spacetime as i mentioned things get more complicated in the curved space time of general relativity but we won't be getting into that in this course so the only thing we need to remember here is that the space like components get an extra minus sign when we're talking about the covariant for vector so now we're in a position to talk about the generalized dot products of four vectors as we talked about in a previous lecture this is basically the inner product of these four dimensional space-time vectors these also called lorentz scalars which are a generalization of the usual scalars in our three-dimensional euclidean space when we generalize to four dimensions we also have the uh the notion of a scalar when we talk about a dot product being a scalar product for regular three-dimensional vectors we have the generalized dot products of our four vectors being a lorentz scalar and those are obtained simply by taking the inner product of our four vectors so what does this mean in practice well consider the following object this is the contraction of two four vectors one of them is a contravariant four vector and the other is a covariant for vector i say it's a contraction because we have a repeated greek index mu here one of them lower and one of them upper so of course this object really means that we sum over mu and of course now we can just write out component by component what this product corresponds to because we know term by term the elements of our contravariant and covariant for position term by term we just get these four objects and if we just plug in the actual values of this these things we see that this is equivalent to ct squared minus x squared minus y squared minus z squared the fact that we have one covariance and one contravariant for vector means that we get these minus signs here we get a relative minus sign between the space like components and the time-like components when we do this contraction notice that we can also write this product on the left-hand side here in terms of two contravariant four vectors if we also include the metric tensor in our minkowski spacetime which is this eta that's just following from the definition of our covariant four vector in terms of the contravariant one so we can kind of see that this is basically just like a regular dot product of the four vectors but because we've got this extra metric tensor in here we get this extra minus sign on the space-like components whenever you see a contraction of two four vectors like this imagine that we're calculating the square modulus length of the vector but in our four dimensional space time because we have the uh minkowski metric tensor encoding the underlying hyperbolic geometry of space time we have this extra minus sign coming in this is like the pythagoras theorem but generalized to fourth dimension the fourth dimension being time and entering here with this relative minus sign now just as x mu is a given for position of a particular event we can also take the difference between two four vectors by the property of linearity of our four vectors we can then write delta x mu simply as the difference between our four vectors for two events a and b in component form we then just have the difference term by term for each component for event b and event a we can therefore express this as c delta t delta x delta y and delta z these are the differences in the time coordinates in the x y and z coordinate between our two events importantly because of the property of linearity when we take these differences we obtain an object that is again a 4 vector delta x mu is a 4 vector and of course we can do the same thing for the covariant 4 position therefore if we write down the lorentz scalar for these four vectors then we see it is simply c delta t squared minus delta x squared minus delta y squared minus delta z squared and we recognize from the usual three-dimensional euclidean pythagoras theorem that delta x squared plus delta y squared plus delta z-squared is itself simply the distance between two points in space delta r squared this object is not a four-vector it's a scalar furthermore we see that here we've obtained the space time interval as we saw in a previous lecture delta s squared which we now know to be this lorentz scalar object is something that's the same in all inertial reference frames it doesn't depend on the reference frame we're in and it's invariant to the lorentz transformations we will see that emerging very naturally within this index notation involving four vectors when we consider the lorentz transformations of four vectors which we're going to do now how does a four vector x mu in a reference frame s look in different reference frame s primed we'll assume as usual that s and s primes are reference frames in the standard configuration meaning that we have aligned axes and the origins can coincide at t equals zero and that there's a relative velocity between s and s primes in the x direction if s is an inertial reference frame then s primed is the initial reference frame as well provided that the velocity v along the x direction is constant in that case we know what the lorentz transformations are and indeed in a previous lecture we assembled these into a single matrix transformation now we're going to look at the four vector version of the lorentz transformations using the index notation generically we have a lorentz transformation that takes a contravariant for position x new in reference frame s and converts it into a contravariant for vector x mu in reference frame s primed which i denote using this primed on the left hand side here the object that performs this lorentz transformation is this thing here lambda and it has one upper index mu and one lower index nu as before we can think of a kind of rule of thumb mnemonically at least where we imagine cancelling these two indices new here and then that leaves us with a single upper index mu which we then see on the right hand side so this is a little bit different from the contractions we've seen previously which for example raise or lower the indices or eliminate the indices altogether as in the lorentz scalar contraction here we see a transformation we turn one contravariant four vector into another contravariant for vector and we do that by contracting out this index nu here and we're left with an index mu of course this is just a mnemonic we don't really cancel out these news we use the einstein summation convention we have a greek index one upper and one lower and so really it is implied here that we sum over the four values that mu can take 0 1 2 and 3 of this product following from the original lens transformations it's easy to work out what this object here lambda should be in special relativity it helps to regard this object's lambda as an element of the matrix lambda here with a double line underneath again to indicate that it's a matrix the fact that we have one upper index and one low index has a subtle meaning we won't get into that here we'll actually just regard this basically as an element of a matrix with a row mu and a column new and we can actually do this because the matrix lambda is again a symmetric matrix meaning that it doesn't really matter which one we regard as mu and which one we regard as new it can either be the rows or columns or columns and rows in practice when we're calculating these objects it doesn't really matter the lorentz transformation matrix lambda mu nu is given by this four by four matrix here this is specifically in the case of a lorentz boost along the x direction that means our two reference frames are connected by relative velocity v along the x direction beta is therefore the relative velocity in units of the speed of light v over c and gamma is the lorentz factor one over the square root of 1 minus beta squared all of this was covered in a previous lecture where we derived all of these equations the purpose of this lecture is to reformulate this physics in terms of four vectors in the index notation again when referring to a row mu and a column nu of this matrix because mu and nu are greek index indices which run over 0 1 2 and 3 we can label the rows and columns 0 1 2 and 3. 0 1 2 and 3 and when we pick a specific element of this matrix we can therefore refer to it by its rows and columns mu and nu so we have the lorentz transformation defined in this way where we have an implicit sum over nu which contracts that index it converts one contravariant four vector in a reference frame s to another contravariant four vector in reference frame s primed the four vector in question is the four position here ct xyz and our laurence transformation matrix is this four by four object from the previous slide so let's just show by brute force calculation that this expression here really is equivalent to our familiar lorentz transformations let's try to calculate c t primes in terms of quantities in the unprimed reference frame ct primed is of course x naught in the primed reference frame x naught gives me ct and in the primed reference frame i would get ct primed so this object is given by lambda 0 nu times x nu and of course here we have implicitly the einstein summation convention over the values of nu and when we write out that sum we get these four terms that's really all this notation means this is just a compact way of writing out all this stuff now we know x naught x one x two and x three are just the four components of the four position in the unprimed reference frame c t x y and z respectively likewise we have components of the lorentz transformation matrix lambda where the first index corresponds to the row the second index corresponds to the column again we can just look up the elements of this matrix very easily we need the 0 0 the 0 1 the 0 2 and 0 0 3 element of this matrix so just substituting those values in we have x naught is ct x1 is x x2 is y x3 is z we have lambda 0 0 is this component gamma lambda 0 1 is this component minus beta gamma lambda 0 2 and lambda 0 3 are both zero so actually these last two terms don't contribute overall therefore we simply have that ct primed is equal to gamma c t minus beta gamma x or i could just write that by factorizing out this factor of gamma as gamma into c t minus beta x which is indeed the equation that we're looking for that's correct one can similarly repeat this procedure for x1 x2 and x3 in the primed reference frame thereby obtaining the full lorentz transformations and indeed we'd find that we would recover the equations that we previously derived this is capturing the same physics it's completely equivalent it's just a different formulation it's a different notation four vectors are in fact defined as being objects that transform under the full symmetry operations of the so-called poincare group this involves three different types of symmetry operation the first is translations or displacements if you like in space and or time it basically means that we can shift the origin of our coordinate system through space and also we can consider a different reference time to be our t equals zero together this forms an abelian league group of translations on space-time secondly we have rotations in space and actually these form a non-abelian lead group finally we have lorentz boosts these are transformations that connect two uniformly moving bodies the lorentz transformations allow us to go from one inertial reference frame to another moving at a constant relative velocity the full group of these lorentz boosts corresponds third symmetry category here taken together the second two here rotations and lorentz boosts make the lorentz group however the full symmetry of special relativity is the poincare symmetry which involves all of these symmetry operations lorentz scalars are objects which respect this poincare asymmetry that means that they do not change when we perform translations or rotations or we perform the lorentz transformations this in turn means that lorentz scalars are the same in all reference frames we refer to them as lorentz invariants or relativistic invariants so let's look in a bit more detail to these lorentz scalars these are objects which do not depend on the reference frame so let's now consider two four vectors let's call them a and b if these objects are genuine four vectors then it must mean that they transform according to the lorenz transformation in that case we can write down these two equations one for a and one for b we start off with contravariant four vectors in the on prime reference frame a new and b new and then by contraction with the lorentz tensor here we're able to obtain the corresponding contravariant four vectors a mu and b mu in the primed reference frame provided that we have two such four vectors then we can form the product a mu b mu one with a lower index one with an upper index this is a contraction because we see here the repeated greek index one lower one upper so we have implicitly the einstein summation convention to consider here with summing over the different values that mu can take so this is basically like the generalization of the regular dot product of vectors in 3d space to the four vectors in space time this is an inner product and it will see that it's actually a lorentz scalar meaning that it indeed is something that doesn't depend on the reference frame now if that's to be so then it means that this object this product should be the same in one reference frame as another reference frame if i write these four vectors down in a given reference frame as i have done here a and b then i can transform to a primed reference frame a different reference frame moving relative to the first one with a constant velocity v and if this really is a lorentz scalar then these two objects should be the same so let's prove this relation and notice here that this is completely general it holds for any two four vectors and for any two inertial reference frames so how do we go about proving this well we'll just take the left hand side of the equation here and we'll substitute in our two expressions for these lorentz transformations and thereby obtain something in the unprimed reference frame and we can compare that with the right hand side of the equation but before we can do that we actually need to convert our covariant for vector in the primed reference frame to a contravariant for vector in the primed reference frame that's because the lorentz transformation equation that i've written down here is for the contravariant four vectors whereas in this equation we see that a appears actually as a covariant for vector so i need to raise the index of this object first before we can proceed that gives me this following form where i've now converted the covariant a mu to a contravariant a nu and again we assume that we're contracting over this greek variable nu here on the right hand side i'm writing the same product of our four vectors in the unprimed reference frame but here i actually switched the dummy variable mu to the dummy variable alpha so what do i mean by the dummy variable here well mu and alpha don't actually feature in the end result that's because it's a repeated index and i'm implicitly summing over it so in this first expression i implicitly have a sum over mu taking the values 0 1 2 or 3. in the second expression i implicitly have a sum over alpha taking values 0 1 2 or 3. so whether i call that label mu or alpha doesn't matter i'm still summing over specific values and in the end i get the same result no matter what i called variable so on the left hand side i've now simply inserted the lorentz transformations for my a and b contravariant four vectors in this first part written in blue i have a contraction over this dummy label sigma here which leaves me finally with a new uh in the primed reference frame and for the red piece we have the transformation of the b4 vector i see that from the contraction of this dummy index alpha here i'm going to end up with b mu in the prime reference frame which is exactly what i want so on the right hand side to make the comparison i have to raise the index of this covariant for vector a alpha and i'm going to turn it into a contravariant for vector a sigma and that will leave me with this expression now i can tidy the expression up a bit on the left hand side notice that these objects are not themselves matrices they can be regarded as elements of matrices but each one of them is in fact just a simple single scalar number therefore i can rearrange these objects at will and that leaves me with this expression and so on the left hand side and the right hand side of the equation i see the same factors a sigma and b alpha a sigma and b alpha both of them are contravariant four vectors with the same indices so if this equation holds true and therefore if the product a mu b mu is actually a lorentz scalar then that implies that this factor here must be equal to this factor so i have this expression and in fact it's very easy to check that this equation is indeed satisfied let's just see that this equation actually makes sense we have a lot of indices in here and there's actually several summations implied on the left hand side we have an implicit sum over the variable new and we also have a sum over the variable mu these are both repeated indices mu mu is repeated one upper and one lower and new is repeated one upper and one lower so really we can write this thing as the double sum over mu and nu after all of that on the left hand side we're therefore left with one sigma and one alpha and that's exactly what we see on the right hand side of the equation here furthermore with the explicit expression for the minkowski metric tensor eta and the lorentz transformation matrix lambda we can simply verify that this equation holds an easy way to do this is to recognize that on the right hand side of the equation here we basically have an element of a matrix and on the left hand side we have an element of a matrix product mathematically the equation in the box here is identically equivalent to saying lambda eta lambda is equal to eta where lambda and eta appearing in this equation are now treated as matrices so to prove this we simply write out our matrix identity here and substitute in for the lorentz transformation matrix here written in black and the minkowski metric tensor eta written here in blue if you just multiply out these three matrices you'll indeed find that you recover this eta on the right hand side where we have to remember that here gamma is 1 over the square root of 1 minus beta squared with that identity and these matrices at hand we can actually prove that this identity is correct so the punchline of all this is that if a mu and b mu are genuine four vectors then this object the contraction a mu b mu is the same in all reference frames it's a lorentz scalar and this holds for any pair of four vectors a and b could be any four vectors they could also be the same four vector in particular the space time interval delta s squared which is defined as this object must therefore be a lorentz scalar and is the same in all inertial reference frames we saw in the previous lectures that this plays a very important role within the special theory of relativity so we've now discussed at length the properties of one such four vector the four position which describes an event in space-time we have four coordinates uh t x y and z but what about other four vectors we've just spent some time to prove some general relations about relativistic invariance involving pairs of four vectors but so far we've only identified one four vector in fact as we'll see there's a whole family of four vectors there's the four velocity the four acceleration the four momentum the four gradient the four potential the four force there's many different objects which can be assembled into four vectors and which transform according to lorentz transformations one way of generating these four vectors is to take an existing four vector and differentiate it so let's start off with the four velocity which is obtained from the four position by taking the time derivative but here we have to be a little bit careful what are we differentiating with respect to we differentiate with respect to time but time in which reference frame do we do d by dt or d by dt primed in different reference frame what time do we differentiate with respect to if time itself is a relative quantity if the fourth velocity u mu is going to be really a four vector then it means it must transform according to the lorentz transformations therefore if we start out with the 4 position which is a 4 vector then we must differentiate with respect to something that is the same in all reference frames in fact the 4 velocity is obtained by taking the proper time derivative d by d tau remember the proper time is the wrist watch time to a an observer at rest at the origin of his own reference frame the proper time is a relativistic invariant this can itself be seen from the relativistic invariance of the space time interval delta s squared that's because delta s squared is equal to delta s primed squared in any reference frame following from the fact that delta s squared is itself this contraction over these four vectors and we've just proved that that object is relativistically invariant since delta s squared generically is defined as being c delta t squared minus delta r squared then we could simply consider a different reference frame in which the observer was at the origin at rest in his own reference frame the time coordinates in such a situation is the proper time and since we're at the origin of our coordinate system then delta r squared is equal to zero and therefore delta s prime squared in this particular reference frame is simply c delta tau squared this is when we're at the origin of our own rest frame but since we've just established that delta s squared is the same as delta s prime squared we know that in any reference frame the space-time interval is basically simply equal to c delta tau squared so we've just proved that apart from this factor of c the space-time interval is actually equal to the proper time and therefore the proper time is indeed also a relativistic invariant what this means is that when i take the lorentz transformation of delta s or equivalently now delta tau i get the same result this implies that if i take the proper time derivative of the 4 position this will again be something that transforms like a 4 vector why because the 4 position is a 4 vector and we're taking the time derivative the proper time derivative which is invariant to the lorentz transformation so we defined the full velocity u mu as d by d tau of x mu this is then guaranteed to transform like a four vector using this relation that delta s is equal to c delta tau i can actually also write this as simply c d by d s of our four position x mu let's evaluate this for velocity using this definition first of all we use the chain rule to write the proper time derivative d by d tau in terms of the coordinate time derivative d by dt then we incur this factor of dt by d tau and by the time dilation formula we know that this object is simply the lorentz factor gamma which of course is a function of the relative velocity between our reference frames here we're talking about two reference frames one measuring a time t and the other measuring a time tau we have an observer in a given reference frame with a given time t and then we have an observer at rest at the origin of their own coordinate system and of course they're measuring the proper time tau the velocity v is simply the relative velocity between these two reference frames so by the chain rule we can write that before velocity u superscript mu which is the contravariant for velocity is given by gamma dx mu by dt where t here is really the coordinate time now since the 4 position x mu is itself given by c t x y and z we can actually now perform explicitly this differentiation with respect to the coordinate time t so i will obtain that the contravariant for velocity u mu is the factor gamma multiplied by c times dt by dt dx by dt dy by dt and dz by dt as the four components this obviously gives gamma into c and then simply the velocity components the regular three velocity components v x v y v said along the x y and z directions respectively i could also express this as gamma multiplied by the four vector which has components c in the timeline component and then the velocity vector v in the space like components this v vector here is the regular non-relativistic 3 velocity this 4 velocity is a genuine 4 vector and therefore it does transform according to the lorentz transformations meaning that i can write u mu in a primed reference frame and related in the usual way to you knew in the unprimed reference frame by performing this contraction with the lorentz tensor i can also consider the lorentz scalar product given by this contraction again of course the einstein summation convention over the repeated index mu here is implied following the usual rules we can multiply this out we would find that it is equal to gamma squared into c squared minus v dot v otherwise known as c squared gamma squared into 1 minus v squared over c squared interestingly this object can be seen to be exactly simply c squared and that's because the definition of the lorentz factor gamma is 1 over the square root of one minus v squared upon c squared so this factor gamma squared out the front here cancels with this factor in the bracket leaving us overall with a factor of c squared now as emphasized on the previous slides these lorentz scalar products things that look like this involving the contraction of a contravariance and a covariant four vector are the same in all inertial reference frames so this object this lorentz scalar product the value c squared here is something that is the same in all inertial reference frames in fact we could have got this result in a far simpler way and this is actually a kind of teachable moment because this is a trick we often play in special relativity when dealing with these four vectors and contractions what we can do is consider a specific reference frame in particular let's consider the co-moving or rest frame of a particle we switch to a reference frame which is just moving along at the same velocity as our particle in this rest frame it looks like the particle has zero velocity in this prime reference frame we would say that v primed vector is simply zero i can now evaluate this object this lorentz scalar object here in the primed reference frame and i know that that object because v primed is equal to 0 in the primed reference frame must simply be equal to gamma squared c squared that's because by our definition of the contravariant for velocity here we have gamma c and v if v is equal to zero in the reference frame s primed then this component is just zero when we perform the contraction we're going to get a factor of gamma twice a factor of c twice we multiply those two things together and we get gamma squared c squared however if we're in the co-moving rest frame that means v is equal to zero and so the factor gamma is actually also equal to one in this rest frame so overall we find that this contraction is equal to c squared but since these objects are the same in all reference frames we know that c squared must also be the result when we consider it in any reference frame and of course this is the explicit resolve result that we recovered um by doing it for a general velocity v so you see this all works out we will actually use this trick of evaluating these scalar products in specific convenient reference frames many times and this is the first example that i wanted to provide of that technique let's now move on to consideration of the so-called for momentum this is denoted by the four vector p mu it has four components and it's a genuine four vector that transforms under lorentz transformations it's simply given by the four velocity that we've just calculated and that we know is a four vector multiplied by the mass of the particle and i'm giving specifically this mass here a subscript zero to indicate that this is the so-called rest mass of a particle i mean here it's the mass of a particle in its own rest frame we'll see what that means and why that's important later the rest mass is basically object we would normally call the mass of a particle this is a scalar quantity all observers would agree what the mass would be in the particle's own rest frame because we take a 4 vector and multiply it by a scalar quantity we obtain another 4 vector so the 4 momentum so defined is indeed a 4 vector the definition of the 4 momentum here makes the analogy clear between the regular non-relativistic momentum which is just the mass times the velocity here we have the rest mass multiplied by the 4 velocity often you will see this take the form mc in the timeline component and a vector p in the space like components where we define here m as the relativistic mass meaning the rest mass m naught multiplied by the lorentz factor gamma likewise we have a regular three-dimensional vector p here which is defined as being the relativistic three momentum not the four momentum but the three momentum and it's relativistic because again we see this lorentz factor gamma appear it's gamma times the rest mass times the velocity so here m naught v would be the regular non-relativistic three momentum we multiply it by the lorentz factor gamma and we obtain the relativistic version so there's some kind of consistency in the notation here we have m naught for the rest mass we have m which is gamma times m naught is the relativistic mass we have m naught v would be the non-relativistic three momentum and then we multiply it by a factor of gamma and we get the relativistic three momentum we can assemble these things together into a single four vector we multiply the relativistic mass by the speed of light m c that goes in the time-like component of the four momentum and then we have the relativistic three momentum in the space-like components of the relativistic four momentum so what is the physical meaning of the time-like component of the four momentum p zero is equal to m c well written out in full we have that m c is equal to gamma m naught c which in turn is m naught c multiplied by one over the square root of one minus v squared over c squared this is the definition of the lorentz factor gamma this is the exact definition of the time-like component p0 of the four momentum let's now consider specifically the non-relativistic limit where the velocity of the particle v is much much less than the speed of light c in that case we know that v squared over c squared will be extremely small and we can perform a taylor series expansion of this gamma factor we've done this a few times before the result should be familiar by now we see that we get this leading factor here m naught c multiplied by 1 plus a half of v squared over c squared and then higher order corrections we're at which are actually of order v upon c to the power of four so if v is much much less than c then v over c to the power of four is extremely small and those are the terms we're going to neglect in the non-relativistic limit if i now multiply both sides of the equation by an additional factor of the speed of light c on the left hand side i would get m c squared on the right hand side i get m naught c squared plus a half m naught v squared and then there would be some higher order corrections here which are beyond the non-relativistic limit which i'm neglecting this is an extremely interesting equation on the right hand side here we see a factor half m naught v squared this is precisely the non-relativistic expression for the kinetic energy of a particle in the non-relativistic world m naught is literally just the mass we don't have any concept of a relativistic mass there so a half m naught v squared is of course the non-relativistic kinetic energy the dots here representing the terms that we neglected are therefore obviously the relativistic corrections to this kinetic energy all of the neglected terms here have at least a factor of v squared with v the velocity and they have units of energy for this equation to be dimensionally correct that implies that these other two terms are also energies they have units of energy now if we consider a particle at rest where the velocity is equal to zero then obviously the kinetic energy a half mv squared is equal to zero therefore m c squared here is equal to m naught c squared when we're at rest and this can be regarded as a rest energy this object has units of energy and since the particle is at rest there is no kinetic energy m north c squared can therefore be regarded as an intrinsic rest energy of the particle it only depends on its rest mass which is the same for all observers and the universal constant c squared which involves the speed of light again this is something that is the same to all observers so the quantity m naught c squared is a relativistic invariant it's the same to all observers it is also a significant contribution to the overall energy that's because the speed of light c is so large if we imagine one kilogram of mass the equivalent rest energy would be 25 million kilowatt hours that's a lot of energy and of course this is at the heart of all nuclear power generation so the total energy of a free particle and what i mean by free particle here is that it's one that's not subject to any potential it's simply moving around in free space the energy of a free particle is clearly this intrinsic rest energy when the velocity is equal to zero plus the kinetic energy of the particle which accounts for the contribution to the energy when v is not equal to zero in the non-relativistic world this is simply a half m naught v squared that's the kinetic energy for non-relativistic physics the relativistic corrections for the kinetic energy i will again represent by these dots here so what we've identified here is that this whole object is equal to the total energy of a free particle it consists of an intrinsic rest energy m naught c squared plus a non-relativistic kinetic energy plus the relativistic corrections to the kinetic energy overall this is the total energy of the relativistic particle and all of that stuff is actually equal to m c squared where on the left-hand side of this equation m is the relativistic mass overall therefore we've derived that the total energy e is equal to m c squared this is einstein's famous mass energy relation this equation is true in all reference frames we have to just bear in mind however that the m here is the relativistic mass so if we are in a given reference frame and we see a particle moving by with a given velocity v then the total energy can be written as m c squared where m is m naught gamma c squared and the gamma here is as usual one over the square root of one minus v squared over c squared and v is the velocity of that particle so the energy here depends on the speed of the particle it also depends on our reference frame the v appearing in this equation that enters in the gamma factor here is the relative velocity between us and the object so what we see here is that the energy of a particle depends on the reference frame it is not itself a relativistic invariant m naught c squared which is the rest energy the intrinsic rest energy is a relativistic invariant m c squared where m is m naught gamma the relativistic mass is something that depends on the reference frame this object is the energy of the free particle in a given reference frame this equation e equals m c squared is of course a very famous equation and is key in determining the physics of relativistic kinematics we can now use these definitions of the four momentum and einstein's mass energy equivalents to derive an even more powerful and general formulation called the energy mass momentum relation this will be the fundamental equation of relativistic kinematics and we'll see that it's extremely useful when doing actual calculations in relativistic mechanics the first step is to note that we can identify the four momentum either as m c in its timeline component or using einstein's mass energy equivalence we can write this as e over c and again in the space like components we have the relativistic three momentum as before since we've identified the form momentum as a genuine four vector this of course implies that we can perform a lorentz transformation on it using this usual formulation we can therefore transform the four vector from one reference frame to another in particular we can transform to the rest frame of a particle in which the velocity v primed is equal to zero this implies that the relativistic mass m is actually the rest mass because we're in the rest frame and also the relativistic three momentum p vector here is equal to zero because our velocity is equal to zero in the rest frame therefore we can write the relativistic four momentum p mu primed as m naught c in the timeline component and zero in the space like components and equivalently we can write that as the energy e over c in the timeline component but in the rest frame we'll call this the rest energy e rest again we have zero in the space line components this is just following from these two equations in particular note that the energy e and the relativistic three momentum p vector both depend on the reference frame they are not relativistically invariant they are components of a four vector they're not a four vector themselves on the other hand the lorentz scalar formed by making the contraction of the contravariant and covariant for momentum is a relativistically invariant object it's the same to all observers in different reference frames according to the usual rules for making these lorentz scalars if i have a 4 vector which is of this type here drawn in red then i obtain e upon c squared for the time like component and then minus p squared for the space like components the minus sign coming in here because we have one covariance and one contravariant for vector on the left hand side of the equation i'm computing the same object but in the rest frame of the particle then i will use this expression for the relativistic for momentum in the rest frame and in this case the lorentz scalar is a very simple object it's just m naught c squared that's because the space like components are equal to zero so here we've utilized the relativistic invariance of these lorentz scalars to derive this interesting equation after some small rearrangements we then end up with this expression that e squared is equal to p squared c squared plus m naught squared c to the four this is just rearranging this expression a little bit and this is referred to as the fundamental equation of relativistic kinematics it relates the total energy in a given reference frame e to the momentum in that reference frame p and the rest mass m naught in the next lecture we'll be utilizing this expression extensively when we're studying relativistic mechanics let's consider now a quick example of a photon before momentum as usual the contravariant for momentum we'll define as e over c in the timeline component and the relativistic three momentum p in the space like components interestingly in the case of photons we have that the mass is equal to zero photons are massless particles however they still have a momentum from de bruy's equation we know that the wavelength of the photon nu is given by the speed of light c times the momentum p divided by planck's constant h and of course by the planck einstein relation the energy of the photon is related to h times that frequency nu which is therefore equivalent to the speed of light times the momentum this implies that for our massless photons our four momentum p mu can actually be written simply as p in the timeline component and p vector in the space like component notice that this is perfectly consistent with the energy mass momentum relationship that we've just been discussing which for a massless particle would imply that e squared is equal to p squared c squared or that e over c is equal simply to p so these equations are actually consistent here p is of course just the magnitude of p vector and p vector as before is the relativistic three momentum one interesting consequence of this is that when we form the lorentz scalar product involving the contraction of the contravariant for momentum and the covariate for momentum we actually simply get zero for this product and being a lorentz scalar this product must be equal to zero in all reference frames let's ask ourselves now how does the frequency of the photon new appear to a moving observer well let's transform to a moving reference frame which will be the primed reference frame we do this using the usual lorentz transformation as i've written down here let's consider specifically the time-like components of the four momentum in the prime reference frame that's p naught primed and therefore i use mu equals naught in this expression i have an implicit sum over sigma and then i obtain on the right hand side of the expression gamma p naught minus beta gamma p1 involving components of the full momentum in the unprimed reference frame what i've written down here is explicitly the expression for a lorentz boost along the x direction in a given reference frame we observe that the frequency of the photon is new this is the unprimed reference frame now we adopt a different reference frame for our observer this is the primed reference frame which is moving along the x-axis with a relative velocity to the unprimed frame and the relative velocity is a constant v p naught is equal to the magnitude of the total relativistic three momentum p whereas p1 is the x component of that vector let's say our photon is also moving along the x direction we're shining our lights along the x axis in that case the momentum only has an x component and so the magnitude of the three momentum of p is actually equal to p x then we can factorize this expression and we obtain this in this case we end up with our final result that p primed in the prime reference frame is equal to gamma into one minus beta times p in the unprimed reference frame we can now convert this into frequencies by using the debri relation up here in which case we obtain that's the frequency in the primed reference frame to the moving observer new primed is equal to the frequency in the unprimed reference frame nu divided by this factor d which is called the doppler factor d is just one over gamma times one minus beta and a little bit of rearrangement and also using the definition of gamma as one over the square root of one minus v upon z squared gives an explicit expression for the doppler factor which is the square root of one plus beta over one minus beta this is referred to as the relativistic doppler effect if i have some light shining along the x-axis and i am an observer also moving along the x-axis then the frequency of the light appears to change by a specific factor arising from special relativity which is this factor here is the relativistic doppler factor analysis of this equation tells you that if you are moving towards the light if the light is oncoming this corresponds to negative beta then we have a blue shift the frequency is increased towards the blue side of the spectrum whereas if we're moving in the same direction as the light then we have a positive beta and we have a redshift in the doppler effect this corresponds to the frequency decreasing the color of the light to the moving observer appears to be shifted towards the red end of the spectrum so let's now consider some other four vectors the four acceleration and the four fourths first of all we had the four position x mu and from it we obtained the four velocity u mu we did this by taking the proper time derivative we had to take the proper time derivative rather than just a general coordinate time derivative because we want this object to be a four vector and therefore we want it to be able to transform according to the lorentz transformations if we start off with the four position x mu as a four vector then we need to differentiate with respect to something that is the same in all inertial reference frames and the proper time is such an object in an exactly analogous fashion we can define a for acceleration by taking the proper time derivative of the for velocity we can also express that as gamma times d u mu by dt for a general coordinate time t we also considered the four momentum p mu which was obtained from the four velocity by multiplying by another lorentz scalar which is the rest mass m naught we can therefore define a four force as the rate of change of the full momentum but again we have to take the proper time derivative of the four momentum for this to be a valid four vector and therefore we have a relativistic version of newton's second law which relates the rate of change of the four momentum to the for acceleration this tells us that f mu is equal to m naught a mu and this equation allows us to relate forces on particles to accelerations of those particles within special relativity we'll be discussing more on this in the next lecture on relativistic mechanics what about relativistic calculus how can we take derivatives of four vectors well if we start with the four position which has components c t x y and z then of course we can take derivatives with respect to each of these components let's assemble those into a four vector this four vector will be called the four gradient the four gradient will look something like this one over c d by dt d by d x d by d y and d by d z in the timeline component we have one over c d by dt and in the space like components we have the usual three gradient which is denoted by this nabla vector symbol here it might be surprising at first glance to realize that this object is actually a covariant derivative i'll denote that with the differential operator d and the lower index mu here so as defined this is actually the covariance for gradient this might be surprising because usually we expect the covariant versions of the four vectors to have a minus sign on the space like components relative to the timeline component so why am i defining this one with all the positive signs as being the covariant version of the four gradient rather than the contravariant one well to understand that consider a function which is going to be a function of our space-time coordinates t x y and z according to the fundamental theorem of calculus i can write a small increment in the function d f in terms of a small increment along our independent coordinate directions dt dx dy and dz and the factors here are the partial derivatives of the function with respect to those coordinates d f by dt the f by the x the f by d y d f by d z if f is a scalar function then obviously d f must also be scalar that means that i should be able to write it in terms of a contraction of a covariant four vector and a contravariant for vector here we have a total differential for our four position d x mu which is of contravariant type and the covariance for gradient of our function f since we have a repeated index mu here one lower and one upper we have implicitly the einstein summation convention which means we contract this and obtain a scalar this lorentz scalar when you multiply it all out using these definitions here is precisely this expression here for the total differential df so this implies that because i have a relative plus sign between each of these components when i write down the fundamental theorem of calculus which gives me the increment in df this tells us that indeed this differential operator here should be of the covariant form once we have the covariant form of the four gradient set we can then easily obtain the contravariant for gradient this is the differential operator d with a superscript mu here that's the index in the upper position and this implies that we should have the relative minus sign here so it seems that for the four gradient the contravariance and covariance gradients are the wrong way round but all that's important is that we have a relative minus sign from going from contravariance to covariant and that's what we have in these expressions as a good mnemonic you can think of the um for grading as having the index downstairs because uh we're differentiating with respect to the coordinates and they are downstairs in this expression that's not really what's happening but it's a good way of remembering the result a particularly interesting quantity is the lorentz scalar formed by contracting the covariance and contravariant four gradients this is given the special symbol uh box squared and you can see it's this lorentz scalar object when we do the contraction of these two things that i've written down here we get one over c squared times d squared by dt squared minus the regular three-dimensional laplacian in space this uh nabla squared this whole thing goes by the name of the dalambashian operator and it's something that features very prominently in special relativity when doing a mechanics problems actually especially in the theory of electromagnetism this uh dallon version plays the role of the laplacian in regular mechanics it's an operator corresponding to the second derivatives of our space-time coordinates however because it's formed by this contraction here it's a lorentz scalar it means it's the same in all inertial reference frames so we here we have a differential operator that is the same in all inertial reference frames in particular you might notice this expression from the wave equation for light and that's of course no mistake this operator entering in the wave equation for light should be the same in all inertial reference frames because light propagates with the same speed in all inertial reference frames and we get the same wave equation in all reference frames so indeed this operator should be something that is independent of the reference frame we choose finally as an interesting example application of all of this i want to consider the case of the classical theory of electromagnetism here we have another kind of four vector um which is called the four potential a mu and it has uh the scalar potential in the time like component of this four vector and the vector potential in the space like components of this four vector the reason why this is a four vector and the fact that it transforms correctly according to the reference transformations is beyond the scope of the current lectures however i have a whole lecture series online um on classical electromagnetism where i go into the relativistic formulation so the link to that lecture will be in the description to this lecture check that out if you want to hear more for now we'll just take it as read that we have a four potential a mu now let's consider the mechanics of a single charged test particle in an electrostatic potential phi here it's important in this first part that we consider an electrostatic potential a potential that is not changing in time we can write down the lagrangian for this particle l which involves a free piece defined in the absence of any potential and then minus the potential of the system in the non-relativistic world the free lagrangian here would just be the kinetic energy however as we saw in the earlier lectures in the relativistic case this l3 is minus m naught c squared into one minus v upon c squared and if we have a charged particle in electrostatic potential then this potential u here is simply the charge q times the potential phi the action therefore is the integral of the lagrangian dt and we'd have obtained this expression importantly in the second term here we see minus q the integral of phi dt it's important to emphasize here that this holds in the electrostatic situation when we don't have time dependent uh electric and magnetic fields that's important because it basically pins us down to a specific reference frame it's the reference frame where the electric potential phi is at rest if we adopt a moving reference frame then the potential will appear to be changing and this then is not the electrostatic situation that i want to consider here so it's perhaps not too surprising that the action that i've written down here is not a relativistic invariant however if we think about the principle of least action and the principle of relativity together then this implies that the action really should be a relativistic invariant it should be something that is the same in all inertial reference frames why is that it's because when we use the principle of least action we minimize the action to find the correct classical path the principle of relativity tells us that observers in different reference frames must all have the same physics therefore different observers in different reference frames when they're minimizing the action should all pick the same minimum path if they're to get the same physics in turn this implies the action should be relativistically invariant so that they all choose the same minimum path if the action was uh frame dependent then different observers might choose different minimum paths and therefore they would get different physics so when we take together the principle of least action and the principle of relativity we're forced to conclude that the action has to be the same in all inertial reference frames this means that the action should be made from lorentz scalars we know that these are the objects that are relativistically invariant a valid action therefore would comprise terms that are lorentz scalars obtained by contracting four vectors the problem with the action that we've written down here in the electrostatic case is that this is not a contraction of any four vector this is not a lorenz scalar object and we have no right to expect that this action that i've written down here would indeed be the same in all inertial reference frames and in fact it's not so can we argue what the relativistic action should be for a charged particle subject to an electromagnetic potential we have to come up with something that is manifestly relativistically invariant and is made from lorentz scalars to do this we actually get a hint from the electrostatic case which i've written down here here we have phi times dt now phi is the time-like component of the full potential whereas dt here is the time-like component of the differential corresponding to the four position so let me write these down first we have d x mu the contravariant for vector and that has components c d t d x d y and d z now i'm going to write down the covariance for potential a mu and that has components which i can deduce from the contravariant for vector over here one over c times phi in the timeline component and then minus the x y and z components of the three vector potential minus ax minus a y minus a z i can now form a relativistically invariant lorentz scalar product here by taking the contraction of these two four vectors the covariant full potential and the contravariant differential dx mu here this would give me the following components phi dt minus ax dx minus ay dy minus az dz and we recognize that in the action we have just the first of these four components we have phi dt appearing but we don't have these other components to make the action a relativistically invariant quantity we cannot just have phi dt on its own we need to supplement it by these other three terms as well we only saw phi dt in the electrostatic case because it only involves the electrostatic scalar potential phi but in general of course we have also magnetic components corresponding to the vector potential a and therefore we need to include these contributions from the magnetic vector potential as well the argument that the action must be a relativistic invariant tells us the form in which the scalar potential and vector potential must enter into our action we conclude that the action should involve the following terms minus q integral phi dt plus q the integral of ax dx plus q integral a y d y plus in uh q times the integral of a z d z overall these four terms here can be lumped into this single term minus q times the integral of a mu d mu and this is manifestly a lorentz invariant object that is going to be the same in all inertial reference frames as we require notice also that when we go to the electrostatic limit we just recover this first term which is what we started with this implies that we have a relativistic electromagnetic lagrangian with this u here this generalized potential u being q times phi the scalar potential minus a the vector potential dotted into the velocity of the particle v how did i get this formula well the action is defined as the integral of the lagrangian dt so i need to pull out a common factor of dt from each of these expressions the first one phi dt already has that but the last three terms here are not in the form integral of something dt and so it's hard to identify straight away what the lagrangian actually is so what i'll do is simply multiply and divide by dt in each of these terms and recognize that dx by dt dy by dt dz by dt are the velocities of the particle along the x y and z directions then i have these terms which have a common factor of q times the integral of something dt if i analyze what's inside the integral there i have a piece which goes as phi i have an ax a y v y a z v z i can collect up these last three terms as the regular dot product of three vectors a dot v in here so overall um i have a relativistic electromagnetic lagrangian which must be of this form involving this generalized potential u as i emphasized in the beginning of the course this is interesting because it's a generalized potential which is actually velocity dependent so this is slightly unusual however if we plug this into the euler lagrange equations of motion then it generates for us automatically the lorentz force law which tells us how the force on a particle relates to the electric and magnetic fields so indeed this is the ultimate verification that this formulation works and so what we've learned here is that because the action has to be a relativistic invariant quantity the same to all observers it implies a certain structure to the kind of lagrangians that we can write down in particular the action can only be formed from pieces which are themselves lorentz scalars and as we saw these are basically contractions over four vectors in the particular case of electromagnetism the part of the theory which involves the coupling of charges and currents to the electromagnetic fields is basically given by this term here and the corresponding lagrangian involves this generalized potential so all of this was by way of a non-trivial example of the application of the the four-vector theory to electromagnetism so that's basically everything that i wanted to cover in this lecture on four vectors in the next lecture we'll be seeing how we can use some of this formalism to do actual mechanics problems we'll be doing relativistic mechanics in the next lecture

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Length: 89min 40sec (5380 seconds)

Published: Sun Nov 29 2020