 In this lecture we will learn the following things. We will learn how to define kinetic energy and momentum while incorporating special relativity. We will learn about the nature of mass and the concept of intrinsic mass. And we will learn about the relationship between energy, momentum and mass. Now let's take a look back at Newton's second law from the perspective of classical physics and in particular have a look at momentum or classical momentum in the context of this discussion. So in introductory physics you are introduced to the concept of momentum roughly as follows. Historically it was observed that there appeared to be a conserved directional quantity associated with motion. This quantity which we call momentum is well defined in the classical domain of physics that is low velocities and large scales by the product of the mass of an object m and the velocity of the object u vector. So we arrive at the definition, the so-called classical definition of momentum, by taking the product of these two things m times u and that gives us p, the momentum or linear momentum of that object. Now in a closed and isolated system perhaps with a whole bunch of different objects i equals 1 to n, it is observed that this quantity overall is conserved, that is the sum of all momenta of all objects in a closed and isolated system can be written as a singular number, the total momentum and that total momentum remains constant no matter what happens inside that closed and isolated system. Now when a system is not closed and isolated for instance subject to some net external force f, then the full beauty of Newton's second law is observed to be obeyed by the system. That is that the net force acting on the constituents of the system is just given by the change in momentum of that system divided by the change in time or dp dt. So Newton's second law f equals ma can actually be rewritten in terms of momentum concepts as just f equals dp dt. Now of course we need to bridge from classical physics to modern physics and to do that I want you to start thinking a little bit about the laws of physics and their invariance under transformations from one inertial frame of reference to another. Recall that one of the postulates of special relativity is that the laws of physics should not depend on what frame of reference you are measuring them in. They should be the same for all frames of reference and the consequence of that of course is that you can't tell if you're in an absolute state of motion. But the benefit of that is it preserves the forms of the laws of physics for all observers regardless of whether or not they're moving. So if one subjects the classical momentum concept consideration moving from one frame of reference to another, imagine a second frame of reference observing an object moving at speed u prime and that second frame of reference s prime is moving at relative velocity v to the original frame s. Now imagine that this is all closed and isolated and in the rest frame the velocity of the object is u and in the moving frame it's u prime and the conservation of momentum will hold and so for instance if we take the momentum observed in the rest frame for this object so p equals m times u and we use the Galilean transformation from classical physics to move to what we observe in the moving frame. We find that of course the moving frame will observe p prime equals m times u prime and we can relate the momentum in the moving frame and the momentum in the rest frame using a Galilean velocity transformation changing u prime to u minus v and then distributing that inside the definition here. So when we do that we find out that the momentum observed in the moving frame is related to the difference between the momentum observed in the rest frame and sort of the frame momentum itself m times the velocity of the moving frame. Now if we then consider changes in momentum in the moving frame with respect to universal and absolute time so dt prime or dt it doesn't matter which in the classical view of physics we just wind up taking the time derivative of momentum in the moving frame and if we distribute that time derivative to the two terms on the right hand side above we find out we have du dt and dv dt. Now since the moving frame is moving at a constant velocity relative to the rest frame dv dt is zero that is the moving frame is not accelerating with respect to the rest frame it's moving at a constant velocity with respect to the rest frame so this second term is zero and we see that we recover exactly dp dt in the rest frame. In other words dp dt in the moving frame is the same as dp dt in the rest frame this is Newton's second law and so we find that this transformation in classical physics leaves the form of Newton's second law invariant at least under Galilean transformations assuming that's the correct term for transformation of space and time and velocity. Now this should all work in domains where the speeds are low compared to that of light but we know that the original definition of momentum was predicated on experiments and observations that were all done in that low velocity large-scale regime of investigation that is sort of the human scale of speeds and sizes we also know that that wasn't quite correct the Lorenz transformation not the Galilean transformation gives the correct way to define relationships between frames great well let's just take the classical definition of momentum and apply the Lorenz transformation the correct transformation between frames so when we do this of course we find that the momentum is equal to mass times velocity and we want to view this in the moving frame where the momentum in the moving frame should be mass times the velocity in the moving frame well if we insert into this the relativistic transformation of velocities in special relativity we wind up with this nasty thing over here the mass times u minus v over the quantity one minus u v over c squared that's the thing we have to insert that contains the velocity of the object as observed in the rest frame and of course the relative velocity of the two frames all right well fine so let's then transform this into a statement about differentials so if I try to write the differential of p prime in terms of the differential of p the momentum in the rest frame if I do the calculus on this I wind up with this horrible looking thing here and then of course if I do dp prime dt prime which would be the change in momentum with respect to time in the moving frame that's related to the change in momentum with respect to time in the rest frame by this horribly velocity dependent thing here this is bad why is this bad it's bad because it totally violates the first postulate of special relativity the forms of the laws of physics must be invariant across all inertial reference frames but here we see that one frame has that forces just equal to dp dt but in the other frame that very same law is horribly velocity dependent this is not good and rather than throwing the whole concept of momentum out the window what we should do is stop and ask ourselves did we really define momentum the conserved quantity associated with degree of motion did we do that assignment correctly in the classical regime of physics did we just get the wrong definition is m times u to naive a definition of momentum given now what we know about space and time and invariance in special relativity now in order to come up with a more appropriate and physically correct definition of momentum that is relativistic momentum there are many many alternative approaches to finding the correct definition of momentum textbooks gloss over this because in many cases the framework for coming up with the exact form of this is not really approachable to students at the level of a student taking this course so I had to cherry pick a methodology to motivate where the definition of relativistic momentum might come from and I prefer the method that comes from my colleague Darren Acosta so let's assume that the problem in the original definition of momentum was that of the definition of time used in the time derivative of space momentum was defined as mass times velocity of an object u velocity is the derivative of space with respect to time so perhaps it's that time definition that's the flaw in the original definition of momentum after all that definition of time did not regard changes from frame to frame as having any appreciable effect on time dt was not necessarily invariant from frame to frame and in fact could have been the root cause of the problem we saw on the previous slide however there is in fact a time unit that all observers regardless of their states of relative motion can agree on they can agree it exists and it can be measured the same way in a specific frame every time and that is proper time denoted with the letter tau so if two events occur and those events are observed by all observers in all frames of reference all observers agree that proper time will be observed in a frame where the two events happen at the same spatial location that is the definition of the proper time it is the shortest time duration measured in any frame by any method of measuring time durations using two events now it's always possible to find such a frame if you're not in the frame where proper time is defined you could always accelerate yourself in such a way until you enter the frame where the regularly occurring events that will be used to define passage of time occur at the same place the time in any other frame is going to be given by the relationship between time in that frame and the proper time so in any other frame the time t for a frame moving at velocity v with respect to the proper time frame is simply given by gamma the gamma factor associated with the motion of that frame relative to the proper time frame times the proper time duration tau now we're talking about inertial frames of reference moving in relative constant velocities with respect to one another and so as a result of that the gamma factors involved here will not be time dependent they are defined using constant velocities of objects or constant velocities of frames relative to one another or both so consider an object moving at velocity u with respect to the proper time frame that in and of itself that object would be a frame of reference that's in relative motion to the frame in which proper time can be observed so let's trade the old time derivative in the definition of momentum that is momentum equals mass times the first derivative of space with respect to time for the derivative with respect to proper time that is momentum will now be defined as mass times the first derivative of space with respect to proper time now we want to convert that into any other frame specifically into the frame where the momentum is being measured which may not be the proper time frame and to do that we just substitute for detail with the relationship between it and dT and if you do that you'll find that you now have mass times the first derivative of space with respect to time times a factor of gamma so if this is a better definition of momentum one that preserves the second law from Isaac Newton under transformations from frame to frame then we should be able to show that and the definition that we get from this exercise using proper time derivative instead of just the plain old time derivative is that the momentum of an object viewed from a reference frame is given by the gamma factor of that object relative to that frame times its mass times its velocity as observed in that frame now again i want to be careful here because the gamma factor that appears here is very specific it has to do with the gamma factor associated with the velocity of that object viewed in the frame of reference the object itself could be viewed as a reference frame of course but because we're going to start talking about transforming object velocities into other frames moving at speed v relative to the one where we measured it it's extremely important to realize that there are suddenly going to become multiple gamma factors in your equations some of those gamma factors will relate to the observation of the object and the passage of time relative to its frame of reference and some of the gamma factors will be related to the relative motion of other frames of reference relative to the one in which you're defining momentum and if that all seems confusing it is and the only way to get better at this is to practice practice practice so the gamma factor here i've denoted especially with a subscript u to indicate that it is not the velocity of another reference frame v that appears in here but rather the velocity of the object itself u and so this gamma factor is defined as one over the square root of one minus u squared over c squared that's what gamma with a subscript u is going to refer to now this redefinition of momentum can be demonstrated with a lot of algebraic pain to leave newton's second law invariant and in fact this is accepted to be now the correct definition of momentum i leave it to the viewer to go through the exercise sketched out on a previous slide to transform the momentum of an object observed in one frame into another frame moving at velocity v with respect to that first observing frame and show that the form of newton's second law dp dt remains invariant from frame to frame now any good definition of momentum will hopefully respect the observations of the past that at low velocities the classical definition of momentum seemed to be good enough if special relativity is the more correct general framework for describing space and time then in some appropriate limit in this case low velocity of the object we should be able to recover the classical definition of momentum so let's give this a try and i'm going to begin by writing the gamma factor for the moving object gamma with a subscript u as a binomial expansion and i've used this before in an earlier lecture so hopefully the rhythm of this will begin to look familiar the binomial expansion is very useful for carefully step by step exploring what happens when you send a parameter of the theory in this case the velocity of an object relative to that of light closer and closer to one of its limits so we'll start by writing down gamma subscript u with its traditional definition of one over the square root of one minus u squared over c squared and then we can use the binomial expansion approach to write it instead as a series of terms of increasing powers of the velocity over c so the first term is just one the second term is one half u squared over c squared etc after that you have terms of order u to the fourth over c to the fourth u to the sixth over c to the sixth and so forth those terms matter when u over c is very close to one but when u over c is very close to zero those higher order terms really don't matter so much compared to the lower order or leading terms in the expansion so now let's write relativistic momentum using this series expansion of the gamma factor so i have momentum is equal to gamma subscript u times m u which is now this series expansion times m times u and you'll notice now that i have an extra u to multiply into the series expansion if i take m times u and distribute it to every term in the series expansion i wound up with something that looks like this the leading order term now has a dependence on velocity but the sub leading term has a dependence on velocity cubed over c squared and then the terms after that are velocity to the fifth over c to the fourth or velocity to the seventh over c to the sixth etc and as u approaches zero that is as the velocity of the object gets much much much much much lower than the speed of light essentially as its velocity is sent towards zero any terms that depend on u cubed over c squared or higher in this expansion are going to vanish they're going to approach zero much faster than that leading term of m u the leading term will dominate the series expansion as u over c gets very small so i can start from this expanded version of momentum using the binomial expansion and in the limit that the velocity is much much less than the speed of light only the first term in the series will survive the one that's largest compared to the others as u over c goes to zero and that's just m times u we have recovered the classical definition of momentum in the limit of velocities that are small compared to the velocity of light so we can proceed similarly now having had some measure of success with looking at momentum as the quantity the directional quantity of motion thinking about kinetic energy which is the scalar or directionless quantity associated with motion that can also be conserved so let's begin to think about kinetic energy in special relativity did we really have the right definition in the old days one half mv squared is that the relativistically correct definition of kinetic energy well we can start by looking at the relationship between external forces changes in states of motion work and kinetic energy when an external force acts on an object and displaces it over some for instance straight line distance s vector the action of accelerating this object under the influence of an external force represents itself a unit of energy being imparted to the object and that energy is known as work work done by an external force changes the kinetic energy of the object it was in a state of some kinetic energy maybe zero and then a force acted on it and accelerated it and now it's in a different state of kinetic energy because its velocity has changed that means that the work done by the force has had some action in changing the kinetic energy of the object and according to the work kinetic energy theorem the change in the kinetic energy of an object is directly proportional to the work done by the external force now the work done by the force on the object displacing it over for instance a linear distance s vector can be written as the dot product of that external force and that displacement now i'm taking some shortcuts here with the form of the work equation this is for a constant magnitude force displacing an object over a straight line distance that's not the general form of the work equation and i will use the general form of the work equation in a moment so let's assume a constant force acts on an object from the perspective of an observer in frame s and the of course the the form of that force and its relationship to the momentum of that object and the changes in momentum of that object will be given by newton's second law the force is equal to the change in relativistic momentum with respect to time this is now the correct definition of momentum in that frame and used in any other frame preserves the form of newton's second law which is f equals ma or f equals dp dt now let's say the force acts over a small displacement a differential of a path ds vector and at any moment it's related to the velocity of the object and the time over which the displacement occurs via the fact that the object velocity is the change in the path position divided by the change in time in that frame in other words u vector is ds vector dt we can write the differential of work the little bit of work done by that constant force over that little bit of displacement by thinking about the definition of work itself in a more general form that is the little bit of work done in displacing the particle over a little bit of path ds vector by a constant force f is given by the dot product of f and ds vector now by newton's second law this has to be equal to the first derivative of the relativistic momentum with respect to time that is what the force should be equal to and again that thing is dotted into ds vector the little bit of displacement but we can replace ds vector with its relationship to the instantaneous velocity of a particle under the action of this external force ds is just going to be equal to u dt now to simplify this dot product i'd like to assume that the change in momentum is in the same direction as the force that's applied on the object so the force is entirely directed in the direction of the displacement or the change in momentum or the change in velocity and as a result of that the dot product trivially becomes the product of the magnitudes of the two vectors to find the total work done by the force which is to be related to the total change in kinetic energy if i can find the total work being done by this force i can absolutely relate that to delta k the change in kinetic energy and perhaps arrive at the form for the kinetic energy what we're going to do is we're going to integrate both sides so by the work kinetic energy theorem the change in the kinetic energy of the object whatever equation that is is given by the work done by the force on the object and that is going to be the integral of this equation here the sum of all the little bits of work should add up to the total work and so that equates to taking the sum of all these little bits here and if i pull out all the constants and all of this i'm going to wind up with the mass times the integral of u times the quantity d gamma u u the dts have cancelled out here in this dot product leaving us with just a differential of the gamma u times u well that doesn't look like a very pleasant integral but there is a way that we can get this into a more pleasing form one that's more easily solved i'm going to start by rewriting this relationship delta k equals w equals m times the integral of the speed times the differential of gamma u times speed to get this into an easier to solve form we're going to integrate by parts to get a final form for the integral this is using the trick that the for instance integral of u dv is equal to u v minus the integral of v d u so let's make some identities between this more general form of the equation and the specific stuff that appears in the integral up here i'm going to identify u as being equal to u that's straightforward i'm going to identify v as being equal to gamma times u when i do that i can then write u times v which i need here as gamma u times u squared and then i need v times du well v times du is just going to be equal to gamma sub u times u times du that's pretty straightforward try this on your own this will help you dust off your integration by parts but you'll find that the integral becomes the following the change in kinetic energy is given now by substituting in using the integration by parts trick as m times gamma u times u squared evaluated at the end points of velocity the initial velocity ui and the final velocity uf minus the mass times the integral of gamma u du again evaluated between the initial and final velocities and if you work through all this you'll get an equation that looks something like this you have this first term m gamma u times u squared plus the second term which looks a bit nastier mc squared times the square root of 1 minus u squared over c squared and we are to evaluate this at the end points of the motion so let's do ourselves some favors here and assume that the initial speed of the object is zero that means that the initial kinetic energy must also be zero whatever the equation for kinetic energy is that's got to be true the final speed will just set to be u some final speed u that we achieve and at that point the kinetic energy is k so substituting all this in we find out that the kinetic energy is equal to m times gamma u times u squared plus mc squared times the inverse of gamma u minus mc squared and rewriting this doing some algebraic gymnastics with the gamma factors and mc squared you'll find that this can be simplified to this lovely little equation here the kinetic energy of a particle is simply given by the quantity of its gamma factor minus one times mc squared now i'm going to let you show that last step on your own it's good practice for the gamma factor gymnastics that you'll often have to do in these problems we find that the relativistic kinetic energy is just gamma u minus one all times mc squared m is just the mass of the object c is just the speed of light and gamma u is its gamma factor relative to the frame in which the object is being observed you can use the binomial expansion trick once again and i encourage you to try this on your own in the limit that the velocity is much much less than the speed of light and you'll find that the expression reduces to one half m u squared the classical definition of kinetic energy these quantities for momentum and kinetic energy have all the right behaviors they don't look like what they looked like in their assumed classical forms they reduce to their classical forms in the appropriate limit and they leave laws of physics invariant where they can be applied now we've looked at momentum and we've looked at kinetic energy but what about the total energy of an object in special relativity in classical physics the total energy of an object was just its kinetic energy and if it wasn't moving it was said to have no energy now that's not entirely true if that object was being acted on by an external conservative force it's possible that that object could have some potential energy associated with it for instance if you raise a ball up in a gravitational field it has some now stored potential energy if you let the ball go it will be released and turned into kinetic energy but for a force-free situation an object at rest really had no defined energy in classical physics is that still true well we can start by just simply noting that as before the total energy of a body in any system is composed of at least two parts a kinetic part describing the energy associated with its motion and a potential part describing any energy that is stored internally in the system and that could be released by some means now the total energy then is the sum of these two pieces so I will use capital E to denote total energy k to denote kinetic energy and u to denote potential energy or stored energy we see that kinetic energy in special relativity is the difference of two pieces k is equal to gamma mc squared minus mc squared so if we rearrange the above total energy equation and then plug in this expression for kinetic energy we arrive at an interesting preliminary conclusion so if I take k and solve for that using the above equation I find that k is equal to the total energy minus the stored energy and if I substitute in with this equation I find that k is also equal to gamma u mc squared minus mc squared and by identifying and relating terms in these two equations for k I can draw the conclusion that the total energy of an object is given by gamma u mc squared and the stored energy of an object even one that's at rest is mc squared it's mass times the speed of light squared so by this identity the total energy of an object in special relativity is given by gamma u mc squared and in the limit that the object is at rest we see that the total energy becomes not zero but mc squared mass times the speed of light squared and we note that the same quantity mc squared has been identified in the above exercise as a kind of energy stored somewhere in the object what's particularly remarkable about this exercise is that by our own means we arrive at a conclusion that albert einstein too arrived at in his miracle year in 1905 it's one of the most profound conclusions drawn from special relativity that mass is itself a form of stored energy and even when a body is not moving its total energy is not zero but rather decreases to a minimum given by e equals mc squared and this latter equation is one of the most famous in the history of science it is an equation that would lead to the development of nuclear weapons nuclear power plants the pet scan a non-invasive medical invention the particle collider and many other technologies taken for granted feared or loved in the modern world for an indivisible fundamental particle for instance the electron is a pretty good example of this we've never seen that the electron is made of anything else one has to conclude then when when it's at rest its energy is the result of some kind of intrinsic mass a fundamental property of matter just like electric charge appears to be a fundamental property associated with matter now it's interesting it's interesting to ask yourself well how much energy if i could find a way to convert it into some other form is contained in the mass of an object well consider the fact that a typical ish human being has a mass somewhere in the realm of 60 kilograms and if by some means all of that could be converted to another form of energy like kinetic energy or chemical energy or radiation then the above equation tells us the energy in joules that this represents e equals m times c squared which is 60 kilograms times something that's about nine times 10 to the 16 meters squared per second squared this yields a total energy in joules stored in your body in the form of mass energy is 5.4 times 10 to the 18 joules now for comparison the energy the little sliver of energy that reaches the earth from the sun every second a tiny bit of the total energy that the sun can emit and yet the same energy that keeps our planet warm and hospitable to life as we know it that energy is 10 to the 17 joules the stored energy in the form of mass energy in your body is a factor of 10 more than that and if it even a fraction of it could be converted into some other form of energy it represents a terrifying amount of potential so let's do an example of this sort of hidden energy of matter by considering the mass that's lost by a uranium nucleus during fission the process of breaking that nucleus into pieces nuclear fission was itself first discovered by physicists and chemists Otto Hahn and Fritz Strassmann and this was done in Germany in December 1938 if you know anything about your history this was the period of Nazi rule of Germany now the observation of nuclear fission especially the fact that the uranium nucleus was observed to split into nearly two equally massed parts was a bit of a mystery and it was explained very quickly thereafter by physicist Lisa Meitner and her nephew Otto Frisch the physics community came to understand that what was going on here just in breaking up the nucleus of a uranium atom was the potential of a vast power that lays in the hearts of all unstable atoms to be unleashed on humankind so consider the process shown at the left this little blue ball is supposed to represent a neutron one of the components of a nucleus they can be freed from the nucleus and fired at other nuclei a neutron striking a u235 nucleus will set off a chain of events that results sometimes and it breaking up into roughly equal mass pieces a nucleus of the element krypton krypton 92 and a nucleus of the element barium barium 141 now the mass of the unsplit u235 nucleus is given in atomic mass units using this number and i'm keeping the precision on purpose because small differences when it comes to mass energy matter a lot now the masses of the daughter nuclei krypton 92 and barium 141 are 83.798 atomic units and 137.327 atomic units respectively now i should note that for purposes of conversion one atomic mass unit is given roughly as the mass of a proton 1.6605402 times 10 to the minus 27 kilograms now if you check the daughter masses do not add up to the parent mass mass is not conserved in this process it's lost in the fission process and the amount of mass that is lost is roughly 14 atomic units even accounting for the fact that three neutrons get produced in the fission process that only adds up to roughly three atomic mass units that's still 10 atomic mass units or so of energy left over that could be converted into forms like kinetic energy or heat now since we've checked that the daughter masses don't add up to the parent mass we recognize that there's missing mass energy here and that mass energy that's missing is about 2.1 times 10 to the minus 9 joules about a billionth of a jewel now that doesn't sound like much but consider what's going on in this cartoon at the left three neutrons have also been produced in this process three neutrons that are bullets that can be fired at other u235 nuclei that might be lingering nearby for instance if you highly enrich uranium to greater than 90 pure u235 it's possible to set off a reaction of events that cannot be stopped and has catastrophic consequences this process can initiate what is known as a chain reaction as you multiply the fission process over and over and over again using these neutron bullets that get produced from the first fission process so for instance the first split makes three to the one neutrons the second generation of splits makes three to the two neutrons because each of these neutrons goes on to split a uranium nucleus that produces three neutrons so that gives you nine the third generation gives you three to the three or 27 a typical chain reaction in purified u235 can go something like at least 40 to 50 generations before this device will blow itself to pieces that's a multiplicative factor of about three to the 45 or three times 10 to the 21 so you're taking the energy left over from one split and you're multiplying it by about 10 to the 21 now those neutrons won't all go on to split uranium nuclei some of them will be thermalized and will result in dumping thermal energy into the body of the material or into the surrounding air around it if the energy of those neutrons is converted to heat from collisions you'll find that this level of multiplication is sufficient to explain the explosive yield of the very first uranium atomic weapon codenamed little boy which was equivalent to about 13 to 18 000 tons of trinitrotoluene or tnt being dropped on a single city that's 54 to 75 trillion joules of energy that weapon devastated the japanese city of Hiroshima at the end of world war two so we can see that a little bit of mass energy goes a long way and it can have positive applications in society it can have negative applications in society but all of this stems from the revelation that energy and mass are not distinct from each other now in classical or Newtonian Galilean physics there is a relationship between momentum and kinetic energy we know that k equals p squared over 2m go ahead and try it yourself if you've never seen this before convince yourself that this is true in classical physics p equals mv k equals one half mv squared do the substitution there's a relationship between kinetic energy and momentum now in the more correct description of space and time given by the special theory of relativity we have kinetic energy mass energy and momentum what is the correct relationship between these things let's begin with the momentum equation that is momentum is equal to gamma u times m times u let's then insert a sort of clever multiplicative one multiply this equation by c over c which has the effect of just multiplying the equation by one but allows us to distribute the c in a useful way we can take the denominator one over c and move it to the left and associate it with the velocity of the object u so we wind up with a term of u over c in this equation now we know that the equation for total energy has c squared and gamma u minute and gamma u depends on u squared over c squared they're related to each other so i recommend you try squaring this above equation square p which then squares this thing on the right hand side gamma u times m times u over c times c and when you do that you get this equation here now if you then use the fact that u squared over c squared can be related to gamma by one minus one over gamma squared you can then insert that and you find that p squared is equal to m squared c squared times the quantity gamma u squared minus one now if you stare at this for a moment you'll notice that this equation has a piece in it that's awkwardly close to e squared e squared the total energy would be given by gamma squared m squared c to the fourth so multiply both sides of this equation by c squared we wind up with p squared c squared on the left this is going to be equal to m squared c to the fourth times the quantity gamma squared minus one if we then distribute the m squared c to the fourth into the parentheses we wind up with this equation and we can identify the first piece here as e squared and the second piece here as m squared c to the fourth or the square of mass energy so putting it all together we find that energy and mass and momentum have a relationship to each other and it's an elegant relationship between an object's total energy its momentum and its mass energy and special relativity and that relationship is given by this quadratic equation e squared equals p squared c squared plus m squared c to the fourth now this equation allows us to think about some cases of certain kinds of particles and one very interesting special case is to look at particles that have no intrinsic mass now the electron is a particle with intrinsic mass the muon is another example of a particle with intrinsic mass albeit 207 times that of the mass of the electron but we can ask ourselves what if there is a particle out there in nature that has no intrinsic mass can it exist and if it did exist what would its properties be well let's take a look at that we can use these relationships to study this very special case now it will turn out that photons which are the particles involved in light have never been observed to have an intrinsic mass they behave as if they have no mass at all so let's go ahead and take that exact limiting case of m equals zero and if we plug that into the energy momentum and mass energy relationship we find that we're left with e squared equals p squared c squared that is we can take the square root of this and say that the total energy of a massless particle is given by its momentum times the speed of light the total energy of a massless particle is entirely energy of motion in other words if such a particle could be stopped from moving you would have to interpret it as them ceasing to exist their total energy would suddenly become zero but of course that violates the conservation of energy you can't just make energy go away without consequence so this implies that such particles can actually only be stopped when they're removed from the natural world by being absorbed into another process now you might then feel emboldened by this and say aha well this is great i'm going to go ahead and figure out what e and p are for massless particles but then you very quickly run into a problem and that is that e depends directly on m and p depends directly on m as defined in special relativity and so you get no useful information from these equations from special relativity special relativity can't give you otherwise useful information about what the total energy actually comes from and what the momentum actually comes from for such particles so what is it that defines energy and momentum of a common particle like a photon which so far as we know has no intrinsic mass no mass energy well to answer that question we're going to have to wait a little bit longer and see as we enter the next phase of this course so to conclude this lecture let's look at what we have learned we have learned how to define kinetic energy and momentum while incorporating the principles of special relativity and in doing so we've learned something deep about the nature of mass and we've learned to appreciate that there is intrinsic mass in nature and that mass in general is associated with a kind of internal energy of all objects an object at rest does not have zero energy it has internal energy given by mc squared we've also learned about the relationship between energy momentum and mass we've looked at some applications of the relativistic concept of energy momentum and mass and we've left ourselves with some questions that we can hopefully resolve by delving deeper into nature in the next phase of the course