 In this session I will be hosting and there will be hosting. Can you hear me? Okay, it seems that you are not co-host yet, or are you are? Yeah, I am. Maybe we can wait just one minute because I see that the number of participants is increasing. Okay. All right, are you ready? Yeah. Okay, perfect. Maybe we can start the recording then. The assistants. Hi, CTS assistants. Maybe you could start the recording now. Yes, I'm here. Sorry. No problem. I couldn't hear you. Yeah, we're going to start now. So the recording. Perfect. You can start. Okay. All right. So we are very happy to have Ida, who will be delivering the third lecture on additional waves from the reunions. Sorry, on F-234 additional waves. Okay, thank you. Okay, so let's pick up where we left off last time. We were discussing power or the power radiated in a QED binary. And just to remind you of the logical flow. We started by calculating in the full theory for the exact theory, because QED is Gaussian, we could solve it exactly. And if you remember, we were able to calculate J exactly, sorry, W of J, where this was the Feynman propagator. And then we interpreted the real part of E as a set of potentials in the non-relativistic approximation. And the imaginary part was proportional to the width. And this allowed us to calculate the power. Okay. So we finished off by showing that the power loss was proportional to this over some time period. And this was proportional to the second derivative of the set of the multiple of the dipole moment where P was the dipole moment of the binary system itself. Right. So, essentially, we have orbit. And then it formed some effective particles with dipole moment. This has some dipole moment and the rate of change that dipole moment this gives you the power loss. And that's all done in the full theory. And in the full theory, we were able to discern that the potentials were due to a photon which had a propagator look like this. And we call this, this was a two point function of the potential propagator. AP has momenta K mu that scale as V over R and R. And we said that the radiation, I think we use dotted lines for the potential. And the radiation was real in the sense that it was on shell. And for the radiation field. We said K goes like you are and all the components are small compared to our due to the V suppression. Okay. And then we so we use this to build the EFT where now L was a function of X, X dot a potential and a radiation. Okay. And then we integrated out the potential to yield some action that only depended on the radiation field and and the positions where this included the dependence on the leading order cooling potential and then we had the sub leading order order V squared potential, which dependent on X and V. Okay. And now what we'd like to do is to figure out how to calculate the power loss in the effective theory so in the full theory, we just did it because we could solve it. But in general, as we said before, when we get to GR that won't be possible anymore. So what we'd like to do is be able to calculate the, the power loss in the effective theory. Okay. So, if you remember, we said the first, the prime directive of the effective theory is is that every term scale homogeneously V, which we accomplished. And the other, the secondary directive is that all terms must obey the symmetry. It must be invariant under the symmetries. Okay, now in our case we've broken Lorentz invariance by taking non relativistic limit, but we should still have gauge invariance. Okay. So our effective theory should still be gauge invariant. So let's remind ourselves what our effective theory looks like. So we have L s, we still have our kinetic term, which we haven't expanded out in the yet, but it's relatively easy to do that. She said my notation is always a to mu nu is one minus one. Okay, so this is the kinetic term. To expand that out, we're going to get. So there's a sum over I here for each world line. We're going to get some over I one half and the I squared plus relativistic corrections to that. And then we have minus the potentials. And then we calculate the last time, and then we still have J mu a new bar. Remember J mu, there's a sum over I for all the particles. Okay, and there's a charge for each particle. Good. Now, we said that we have to every term must scale homogeneously. And at this point, these terms we know scale homogeneously, and we can certainly expand this so each term scales homogeneously but what about this guy. Okay. So, we said the wavelength of the radiation to K radiation goes like V over R, which is much, much greater than one over R. Okay, so our dipole expansion tells us that this has to be much less than one. Right. So we have to multiple expand. So this is of order V. So instead of just inequality, let me say this thing's over V. So we must multiple expand the radiation field that is forced upon us by the the prime directive. Okay, so we're going to write J mu the radiation. So this becomes some over I, I, the new I'm going to suppress that the time argument now because just to save space. Okay, so this is notice now this field because of the delta function here. So this is an integral over DT, which I've taken to be my, my, my affine parameter to be T. Okay, so now this field here is evaluated on the world line. Okay, but now we're forced to multiple expand. So let's let's write this out. So we have DT, some over I, the I, let's write it first. V zero, I, a zero minus V, I, I. Okay, and our multiple expansion is going to lead to, oops, sorry. Okay, so notice I've only kept up to, I've only kept up to order V squared. So let's, so let's see if we can simplify this. Now, the final result is supposed to look a gauge invariant. Right. So we need some coupling that when we're done is going to look a gauge invariant. So we better be able to we better be able to write this in a gauge invariant form. Okay, so the first first term remember this guy is one with our affine parameterization. So we have the first term is DT, a zero q total. That's a coupling of a zero to total charge. And then we have another term. So in this term, let me integrate this term by parts. So then I have plus that AI dot zero T minus V. Sorry, I'm missing a, I don't know that I'm missing an index here. This should not be here. Sorry. Right that comes from the expansion of this term. This X dot grad a zero zero T plus dot dot dot dot. Okay, but these two terms remember that he I is D zero, I minus D I a zero. So the second term just gives me I X I E I that into E zero T and this is nothing but P that. Okay. So we we end up E of course is gauge invariant. So our effective action as effective Q a zero of zero and T plus P dot E zero and T and so why is this gauge invariant the first term does not look gauge invariant. However, a zero goes to a zero plus D zero or gauge transformation. And since Q is a conserved quantity, therefore it's a total derivative and we can drop it. Okay, so we end up with an action which is explicitly gauge invariant. And if you wish you could carry out this expansion further, and you would get the quadrupole dotted into the first derivative electric field and you'd get the magnetic dipole and so forth and so on. And you're sure you'll get a gauge invariant result. Because we've worked in a background field gauge. Even if you calculated quantum corrections you would still get a gauge invariant result but of course everything we're doing here. Sorry about that. Okay, so good. So now we have our effective theory. So what is our effective theory. So we have sum over I one half m i v i squared minus V of x i and x i dot which we calculated plus Q a zero and these all these guys all get integrated so notice it here. So we should put a DT and these fields all depend on T. Okay. And now, and remember this is composed of the radiation field, and we can power count each one of these terms as we did before. And this would be leading order this would be order V and so forth and so on sorry V squared, no V, sorry V squared, V squared. And now we can calculate anything we want. So remember the the the algorithm is So remember these peas contain our functions of the Xi's So in order to calculate the power loss we first have to calculate the equations of motion for the Xi's solve them and plug them into these peas. Okay, so let's calculate the power loss. Well, how do we calculate the power loss we know how to do that we calculate the probability to emit one photon. We square it, and that will give us the probability to emit one photon. So what we're going to do with we're going to integrate this thing, weighted by the energy, and this is going to be the power loss. Okay. So, another easy way another way of saying this is this thing is proportional to the imaginary part of this Feynman diagram. And the reason I want to write it like this is because it's going to look exactly like the calculation that we did before. Right. So. So here, these, these are each insertions of P dot E. Right. So look at this. This amplitude that we're going to take the imaginary part of. So we're going to have two insertions of P dot E. So we're going to have a DT DT prime P of E T P dot E prime. Okay. And just using standard Feynman technology, you know, we get one with contraction between these two electric fields. So if we open up these, if we open this up, this P depends on T, and then this P depends on T prime. Okay. And if we use fine engage remember the fine engage propagator is proportional to G mu new. So, which is to say that the contraction the Witt contraction to a's like this. So there'll be a Witt contraction between these two terms, and they'll be a Witt contract between these two terms but no cross terms. And then if the propagator of course will go like K squared plus I epsilon. And then if we take the imaginary part. This is going to give us our delta function. Okay. So this is the 4k, which leads to the 3k. Once we use the delta function. And then we could calculate these with contractions. And just do the algebra and you're going to get P of T dot P of T prime. Okay. So this is the magic part of M. And remember P, we need to wait, wait, the integral by the energy so I'm going to get another factor of the energy here with this integral that happened sometimes. Okay. And then of course, you can do these integrals and you just end up with DT one time left integral left over and P. Okay. That's exactly what we got. Classically. All right. So, let me pause now for questions so this is a logical stopping point. What we've just done is we've re-derived the exact, the expansion of the exact result by first taking the effective theory. And then calculating it within the effective theory for the purposes of the next step. We're talking about one other small thing before we move on to GR. So let me let me pause here for questions. So it's a question by Max. Hi. Could you please comment a little bit on and so if we will now go to quantum theory where exactly what this previous story changed. So that's right. So, so notice that I've, we've taken the, we're ignoring, we're essentially ignoring finite mass effects, right? So our sources are like taking the electron mass to infinity, right? So, in fact, so maybe this is a good point to discuss this, we've been drawing Feynman diagrams with these lines here, right? And so for instance, our potential we drew was like that. But that's a little bit misleading because this implies that the, the sources are propagating that they have propagators. But we're treating the sources as classical objects. So a more correct diagram would just be this where these are just in sources of classical sources, right? So the difference between a quantum source and a classical source is a quantum source of course has some propagator that looks like this. But a classical source actually has a propagator that looks like this where V is the velocity. So what we're doing is we're assuming basically when we look at these exchanges that there's no quantum fluctuations in the lines, okay? Now, where would the first quantum correction come from? And that's because we're considering, you know, very massive objects. Because in the end that's what we're interested in, right? We're not going to try to do small black holes, we're going to try to do astrophysical black holes. We can't do small black holes because no one knows how to do that. Where by small, I mean on the order of the Planck mass, the short-shield radius. Of course, you would have quantum corrections coming from electrons that would correct the Coulomb law. So for instance, you would have an electron loop here. And the net effect of those electron loops will be to take every E and replace it by E of R, right? So this is, these are going to give logarithms of R. Well, not, yeah, in X space. We'll give you logarithms of R. And so E of R will just be the running where E of R is the running. And in QED, that's really the only correction. Of course, I mean, you'd have to calculate to all orders in quantum corrections, you need to include all the corrections to this. But that's the only, if you knew the self-energy of the photon, you would be done, right? Now of course, at distances long compared to the electron mass, these effects are all suppressed, right? So from the point of view of effective field theory for R, much greater than one over ME, the coupling just freezes, right? The coupling stops to run because at, you know, at these long distances, you really, you know, the, any effect of a massive particle is going to be suppressed by the mass, right? So the corrections will be down by RME. So in any classical problem, all those quantum corrections just get absorbed into defining alpha at infinity at R equal infinity, essentially, which is one over 137, right? If you start to prove distances on the order of the electron mass, then these will be corrected by logs of R, right? But these guys are going to be down by RME. Did that answer the question? Yes, yes, yes. Thank you very much. Okay, so before we go to GR, I can't resist, especially since this is a particle physics summer school and not necessarily a GR summer school. I wanted to show you how you could use these results in QED to learn something interesting. Okay, so let's talk about finite size effects. So what are the first finite size effects in electrodynamics? Well, there's terms that go, we said, like F squared. And I'm not going to worry about the RPI for the moment and putting in the square root of these. So this is what the finite size action looks like. Okay. So what are these finite size effects? Well, first let me rewrite them as CE times E squared plus CB times B squared, where you can work out that CE is minus 2 C1 plus C2. And CB is minus 2 C2. So remember F squared is just some combination of E squared and B squared and B dot F also can be written in that way. It can be decomposed in terms of EMB. Okay, so how would we fix these coefficients? These are some unknowns. Well, we'd have to do some matching procedure. So we fix CE and CB by what's known as a matching procedure. Okay, so how do we do that? So let's take an example. Suppose we had a spherical conducting shell, and we wanted to determine what are the values of CE and CB for that conducting shell. So the idea is the following. So we solve the full theory, and then we match onto effective theory to determine the Cs. Right, so what do I mean by that? So we're free to match anything we wish, any observable that we want, which is to say whatever observable you take in the effective theory should match the full theory. So, and once you match those coefficients, you're done. You can use them for any process. So you can take one physical process, match for the Cs, and then use the effective theory in any physical process you want, because once you know C, you're done. The game is over. So it's easiest, you should always choose to match using the simplest possible observable. So what we'll do is we'll place the sphere in some dipole field, in some background field, and then we'll ask for the response. Okay, so we're going to write AMU as AMU background plus delta AMU. Okay. And we're going to calculate this delta AMU. So what we do is we take our finite size action. So we take S finite size. And now we're going to write it as E or delta E, E background plus delta E squared. Let me just do the CCE piece in this way. And we're going to calculate the linear order in delta E. So this is going to give E background squared plus E background delta E background plus the piece that we're interested in. Okay. So in the effective theory, we can calculate delta E. So this E background is acting like a source is equivalent to some J. So this is just like J dot E. So I know how to calculate delta E. It's just going to be the convolution of the Green's function, where now this is E. Right. That's how you that's how you solve the boundary value problem and electrodynamics. So in the notes, I'm not going to go through the details, but this, this, sorry, that's not such, there's no delta E back to delta E. So this field is going to take the form at long distances of, there's going to be some dipole field and electric and magnetic dipole field at long distances. Okay. And then we can just read off the induced. So this is the induced dipole moment. This is the induced magnetic moment. Okay, so I can do this calculation as a function of E background. Right. So that the induced dipole moments will be functions of the background. Right. So now we know this is the background. So this comes out of the calculation I haven't shown this. And so now all I have to do is solve the full theory so I can go to Jackson and solve the problem, the, the, the solve the boundary value problem for a sphere in a background field and it goes like this. So this allows me to this tells me that C E must be minus R cubed over two. Okay, and I could do the same thing with CB. So this procedure of matching is, is, is as a is a bread and butter calculation of any effective field theory. Okay, assuming that you can calculate in the full theory so in a counter example that would be the chiral Lagrangian in in in QCD. And they are the coefficients we can't match because we don't know the full theory. So the only way we can fix those coefficients. Well, we know the full thing but we don't know how to calculate analytically at least. So in that case the coefficient have to be fixed from experiment. So you take some observable use these these operators to calculate the prediction for that observable in terms of the coefficients. You measure it in an experiment and then you extract the value of C in this case, which is such a useful pedagogical example. You can calculate the full theory it's just a Jackson problem, put a charge sphere in a background electric field, calculate how it deforms. Right do the same thing in the effective theory which is what we've done here. So you take the answer to and that allows you to just extract the value of C E. And if you did the if you in the notes I show actually that CB can also be extracted and only differs by factor. Okay, so what use with these finite size effects of what use are they and now I'm going to show you before we go back to gravity waves. So now you can do some interesting molecular physics using these this result. So let me consider two neutral molecules. Okay, and their distance are is much greater than this are. So our effective theory should apply. Well, what's the force on these two molecules. Okay, so our action. Now each one of these molecules could be different. So I'm going to give each one of them a different polarizability so these C's are, you can see from this definition the C's are proportional to alpha. Okay, so in principle these could be two different types of molecules it doesn't matter. And we could ask well what is the force between these two guys well we know how to calculate potentials in our effective theory. So we at our disposal we only have these terms because these are so q is zero. So the the there's no P and P is zero. So there's no P that either know nothing there's the only interactions are due to the polarizabilities. So what can happen is is that what's the first contribution to the to the potential. Okay, well from Feynman diagramology, we know we have two lines coming off right because it's quadratic and e squared and now I have to contract the lines in some way. And that's all we can do. Okay, so we can calculate this diagram. But before we do that, let's just, you know, let the power of this formalism is manifest by the fact that even without doing a calculation we can do. We can determine what how the force depends on our for two neutral molecules. Well, because we know it goes like CE CE, so V of R is going to go like CE squared, and then there's going to be R to some power here. Okay, it's going to be some power law force. And we know that CE goes like our cube, just on dimensional ground so even if we didn't do the matching, we would still know that CE goes like our cube, why is that well from reasoning and effective field theory right, we know that the small expansion parameter is is R over R. Right that's our first that's our starting point. So remember, we had two expansion parameters in the first place we took the finite size object and we shrunk it to a point. And that gave us an R over our expansion. And then we did the static, the small V expansion, and we. If we drop the finite size, then that would mean we'd only be working to leaving order and are in little R over big R, keeping the first finite size effects are the first corrections and little R over big R. So these guys have to scale in some with some power of R because that's our expansion parameter small big R. So we have two dimensional grounds. So this has a DT here and in our units where h bar equals C equals one everything has energy scaling F scales like energy squared. Sorry, energy, let me think, let me math, use math so we don't get confused electric field. Right and then we have a DT that scales like one over mass. So that must mean oops got too many duties. So this is master the force. So this has to scale like master the minus three, which scales is which is our cube since our scales of one over mass. So even if we had not done the matching, we know exactly how the seas scale with our. Okay. So now, even if we never, if we didn't know anything about how to do these calculations, all we need to know is that two insertions of C is all we need to get a force, because now we know this thing goes like our cubed squared. And then to make have units of energy, it has to go like R to the seven. Okay, so we've just determined that V of R goes like C E squared over R to the seven. And that's pretty cool because there's been no physical input whatsoever. Right. This was just purely based on dimensional analysis and power counting. Okay. And, and the, the, the basic assumptions of effective field theory. This is known as the chasm your pulled our force. And in the notes I show you how to calculate the coefficient here by calculating this diagram. Now, this may puzzle you a little bit, because normally when you learn atomic physics quantum mechanics, if you take a class in quantum mechanics, you find that the first force between neutral atoms is the Van der Waals force. And that potentially goes like one over R to the six. So what's going on here. And the answer is as so the question is to how much can we trust this result. Well, these guys are atoms and they have some excitations. And so we're ignoring all those excitations. So in the language of effective field theory we've integrated out all the excited states. So it better be that we can't excite those excited states and our problem. Right. So if we bring these guys close enough together, then the typical wavelength of these potential modes we know goes like K will be one over R. Right. And if the energy carried by the photon is of order the first excited state, then this theory will break down and then include the excitations of the atom. And when you do that you'll get the art of the six and this is actually I actually derive that in the notes for those of you who are interested. And sort of the cool thing about this whole program which has never really been explored is you can start to look at nonlinear effects. So for instance, are the prime directive tells us that we could also have terms like ease either the fourth. Right, which will be higher order in, let's call it before, which would be higher order, and these will lead to some interesting nonlinear effects, because you could draw a diagram like this. Right. And that and I don't know, you know what's been calculated and what hasn't in atomic physics but it would be fun to explore what the effective theory can tell you about say nonlinear optics and things like that. But that, as far as I know no one, no one has looked at that. Okay, so this is a good point to stop in this lecture. Let's take a break and then when we come back, we will start on relativity on GR. Are there any questions at this point. What about we take the question in five minutes after the break. We leave five minutes to think about it and maybe there will be more questions. Sounds good. Okay, see you in five minutes. Right. Could you remind me, how should I pronounce your, your first name, ida or ira? Ira. Okay, sorry, because before I think I. It's okay. No problem. Okay, so that's a question by please go. Yeah, this probably relates to question I asked on the first day. But so we calculated the coefficients for spherical objects. Do they change if we drop the spherical assumption, because then, no, we. Yeah, I mean, just, okay. But by, by how much. Yeah, that's also. Let me show you what happens if it's not spherical anymore. Can you see my screen? Can you see my screen? Yes. Okay. Oh, I saw in the chat that there seems to be a problem with lag. Is that still happening? Yeah, a little bit. Was it like that yesterday too? Or is it just today? I think only today. Then I don't, yeah, then I don't know what's going on. I'm sorry about that. So, so if you don't have spherical symmetry, then. Then then there's a preferred direction and we have to change the action. So it's not just that the coefficients change. But now we have new terms in the action, right? So in particular. If, if the, if the object has some spherical, has some breaking of spherical symmetry, then, then the object will have a permanent dipole moment, right? So now I can add term to the action that looked like this. Now remember, we don't get confused between the point particle action and the point particle, the binary. Remember what the, the logic was, it went like this and then like this. So we started off with two finite sites objects. We reduce it to an action for two point particles. Then when we multiple expanded, we left with one over P dot E where this P is the P of the total system. Okay, now here we assume that there was no P dot E. In other words, the individual particles didn't have any multiple moments, right? So, so if the, if the individual particles were oblong, then a member this action was just j mu a mu, right? But if this guy, if this guy had spherical, something which broke spherical symmetry, I'd have to add more terms, which knew about that preferred direction, which we haven't included, right? So, and in fact, you could show that by gauging variance, you would have to generate a P dot E term where P was proportional to the permanent moment, right? And then if you had other moments, you know, if you had some really complicated object, no with what it's doing that weird. If you had some complicated object with some weird shape, then this would have all sorts of multiple moments and you'd have to include all of them. So I might, you'd have to include all of them to, to include those effects, right? Did that make sense? I just don't see, like, looking at the final thing we want to achieve. Is this a problem? Like, will this be a problem that we have to kind of know this, the object in advance? Absolutely. Yes. So in general, if you don't know what the object is, you have to include all those terms, right? And in fact, we'll talk about what happens, for instance, what if the object is spinning, right? Then we're going to have to include those terms as well. So you can imagine how complicated it gets, right? Because in principle, in fact, you know, the dipole moment could start to change with time and it gets more complicated. So we've just taken the simplest example of a spherical object. And for LIGO, for instance, that's an excellent approximation because black holes and neutron stars are, you know, are are spherically symmetric. When they start to spin, of course, then they're no longer spherically symmetric. So we'll have to, if we have time tomorrow, we'll talk about spin and how to include it. That may partially answer your question. Yeah, yeah, that makes a lot of sense. So I'll just wait and see if tomorrow is clear also. Okay. Thanks. All right. Next question is my amendment. Hello. I was curious about one statement that was in my head. It is to confirm, I'm asking now. Can I say that what we are doing right now is a kind of an Euler-Heisenberg Lagrangian supplemented with the information of how a classical source inserted in this Lagrangian. So the Euler-Heisenberg Lagrangian is purely quantum mechanical. The ideas behind it, of course, are very similar in the sense that, so in Euler-Heisenberg, you, at distances long compared to the electron mass, you integrate out the electron. So in the full theory, you would have this and this would match on to a theory that went like F to the fourth, right? So it is certainly similar, but this is purely a quantum mechanical effect, whereas so far we have not included anything quantum mechanical. But in spirit, it's similar, yeah. So Euler-Heisenberg was one of the first sort of effective field theories that was ever written down. For them, it was just sort of a model. They didn't really have control of anything, but the ideas were all there. I see. But if I see the diagrams we draw up to now, these closed loops of photons, can I say that they are quantum effects as well? Yeah, so good. So we'll talk about that within the context of GR. So certainly in QED, this would certainly be a quantum effect, right? And we'll see, for instance, in GR, that there'll be potentials that look like this because GR is nonlinear. And that looks like a loop, but it's really not a loop. But we'll talk about that. We should really think about this as a diagram that looks like this with no loops. Okay, so we will talk about what's quantum and what's not quantum in the context of GR in just a few minutes. Okay, but up to now, I can still think that we are in the classical realm. Totally, totally classical, yeah. And we'll remain classical. The only thing I ever drew which wasn't classical was this. The Casimir-Poldar force. That's quantum mechanical. Okay, okay. And it's because of how you resolve that effective vertex. Yeah, so this loop right here, any closed loop that doesn't involve an intermediate classical line will be quantum. So for instance, that's quantum. This is classical because this line isn't really there, right? This should better off be drawn like this, right? But notice that here, there's still a loop even if I draw those lines as classical sources like that. So this is quantum and this is not quantum. Okay, thank you very much. All right, next question is very much. And so since you brought up the Casimir-Poldar force and I was curious. So could you in this formalism also include like a finite temperature effects? Absolutely, sure, yep. The only difference is you would need to include finite temperature propagators for the photons. And then the calculation would go through exactly the same. You could get the finite temperature force as well. Okay, nice. Okay, thank you. Okay. So let's press on to GR. So now let's do GR. But before we do that, we need to do one little piece of housekeeping that is important in building any effective field theory. Because the fact that you can use the equations of motion, use of the equations of motion at the level of the action is allowed. And that's important because remember the algorithm is you write down every term in the action which is allowed by the symmetries. And then you can just eliminate. So any term in the action which vanishes using the equations of motion can be thrown away. Which as I mentioned in the first lecture is a little bit counterintuitive quantum mechanically because the quantum mechanically we know we have virtual particles which don't obey the equation of the motion right off shell modes. So we're allowed to use the equations of motion at the level of the action, because the action is some off shell quantity. It induces virtual effects. And the answer is, there's, there's something called the equivalence theorem, which is not the equivalent serum of GR, but the equivalent serum of quantum field theory, which states that any physical observable then in particular the S matrix is invariant under canonical field redefinitions. And that's going to allow us to use the equations of motion to eliminate any term in the action that vanishes by the equations of motion. So let me give you an example. So suppose I have a scalar field theory that looks like thing like this. Okay, so I'm making it as simple as possible. I've made a massless scalar field and the equations of motion are the leading order equations of motion or box flies equal to zero. Now I have some operator here and this operator will be say suppressed by some mass scale. So this is a higher dimensional operator. So C is going to go like some power of one over M, where this N is fixed by whatever the units of FR so F is some polynomial in the field if I knew the power. Then I could choose M to give me the right answer. And if we're looking at energies much much less than M, then the operator will be suppressed. So this would be an analog to one of our operators that, you know, like, like our, like our E squared operator something like that although E squared doesn't vanish by the equations of motion but I can imagine. I can imagine say here is a good example. I can write down a term that went like D mu F mu new squared. Okay, that would vanish by the equations of motion. So here is not F mu new. This is just some functional field. So, let me make a redefinition Phi goes to Phi minus C times F of Phi. Let's plug it into the action. So now L becomes Phi minus C F of Phi box by minus C F of Phi plus C box by F of Phi plus order C squared. So we're going to we're going to so this is going to give us back Phi box Phi minus C F of Phi box Phi plus C box Phi F plus order C squared. Okay, I've eliminated it from the action up to corrections of order C squared. But if remember if I'm calculating in a systematic way and I'm only calculating up to order C I don't care about the C squared terms. Right. So any operator which vanishes by the equations of motion can be eliminated up to sub leading pieces and if you're calculating the high orders you'd have to include those pieces. Okay, and so why why are physical observables independent of these field redefinitions. And the answer is because remember that when we do a path integral. This is just a dummy variable. And it's not as if we have to worry about the boundary conditions on these fields we take them off to infinity. So there's no boundary conditions so I can make any field redefinition I want as long as the Jacobian is trivial. And so as long as I have a redefinition of this form, then everything just goes through, and it allows me to eliminate all terms in the action which managed by the equations of motion. So this is a very important result in in in field theory in general, but especially an effective field theory because it allows you to make your life a lot easier by not having to write down terms in the action which managed by the equations of motion. So that that reduces the calculation burden in it in any effective field theory that you're working in. Okay. Are there any questions about that. There's one by Max. Yeah, so could you please explain about why we can make make use of this in the case of GR button button but not in the, in the electric magnetic case. I don't know you can use it in the electric magnetic case just that I did so for instance in the electric magnetic case. We might have thought that we could write down in a term like this. Right. But that's zero by the equations of motion so we don't have to include it. Okay. Okay, thank you. Yeah. All right. Hi, so if Jacobian is not trivial, probably we will have something like costs and can we still use this, the definition in this case. Yeah, so if you have a non trivial if it's a non-clinical transformation you can get ghosts. And then you have to be very careful. So there's, there's, there's been work on this. So if you make the ghostly couple, it gets, it gets a little bit tricky. Is it not heavy, like go son, not heavy in this case. No. Not necessarily. But if you want on the slack channel, I'll put a reference to a paper which discusses this issue. Okay, it would be very good. Okay, thank you. Okay, so let's, let's talk about GR now. And let's apply our algorithm to GR. So the first thing we need to do is we need to understand our power counting parameters. So that's the first order of business, whenever you build an effective theory. Okay, now, GR is more complicated because we have more dimensional quantities so we have. We have R, we have R, we have M, and we have G Newton. Right. And from our first lecture, we talked about the fact that that as V goes to zero we approach flat space. So it makes sense for us to expand around flat space. So we're going to use V as our expansion parameter. And we need to figure out how everyone else scales with V. So one way of doing that is just to remember that if we have a bound state. Then we have a balance between the kinetic energy and the potential energy. This is also called the virial theorem. This is actually true for bound states. So we're only, we're building and we're crafting our effective theory to describe bound states. And that's crucial in the sense that we're using this relation, which is only true in certain kinematic configurations. So our theory will be limited to those configurations of course those are the configurations that we're interested in. We see that V squared scales as GM over R. Okay. And we're going to take that to be much less than one. Now, we have another expansion parameter. That's going to allow us to differentiate between quantum and classical, right. And in particular, we're going to have an expansion parameter that goes like h bar over L. That's pretty small, right for a for a classical orbit. That's that's an excellent approximation to drop that. So, we know that L goes like R and V. Right. And we just said that R goes like GM over V squared. So L goes like M. R goes like GM over V squared. So GM squared over V, which I'm going to write as M squared over M plank squared one over V. So M squared over M plank squared is going to go like V times L. I'm going to remind you that G Newton goes like one over 32 pi M plank squared. Okay, so usually in GR, you'll never see M plank because you need an H bar to go from M plank to G Newton. That is particle physicists were usually trained to think about M planks and not G Newton's. Okay, so So our two expansion parameters are going to be V and one over L. So we're going to do a double expansion. So how should things so the leading order action as well as has units of of angular momentum so the leading order action should scale as L. And then V to the zero. So every term in our action should be a sum of terms. Each one that scales is L to the alpha V to the beta. Okay, so it's a sum of terms. It's called operators. Okay. And we're going to end in the end, we're going to drop any term that goes like any terms in the action that don't scale like L. There'll be some terms that scale like one over L. And these will be quantum. So now we do exactly what we did in the case of of electrodynamics. And we start off with our so the first thing we're going to do is start off with a point particle action so step one is to write down the point particle actions. Okay. So we know how to do that. That's minus minus MI. And notice here. Now we're doing GR so we can't just take that to be flat space. And then we have the bulk action. And let me put in the canonical normalization. Okay, and then plus finite size terms which we'll get to in a second. Okay. All right now since we're expanding around flat space the V equals your limit we're going to write G mu new. As equal to a to mu new. Plus h mu new over M plank. Plus h bar mu new over M plank. This will be the potential mode. And this will be the radiation mode. Okay, just like we did in electrodynamics. Okay, so now let's expand let's expand it out. So let's take our square root and let we're going to take. Sorry this at this point this is still the lambda. And this should be the lambda I here for each guy. And now we're going to take our affine parameter to be T. So now the kinetic term. So I'm going to contract it with this is the contraction with this guy. And then we have plus. And now we can expand the out the action in, in the fields. And so this thing is going to look like just to give you a sample of what it looks like. Rather ugly ugly mess. Excuse me. I pass your question by max. Okay, yeah. Yeah, so I would like to ask you about this, this expansion of Jim Jim and you. In the case of Maxwell, I was, I was thinking of this potential mode as just having one degree of freedom. But now here in the gravity case, I would think that so, at least in the relative risk the case that the potential mode should have more than one degree of freedom right. I guess it in the non relativistic case really on the end up with this thing which basically Newton potential or is this around thinking. I'm not exactly sure what you mean. But so the potential mode doesn't really isn't really a dynamical degree of freedom. Right in the sense that at leading order it's propagator doesn't have any doesn't know about any time dependence right. So if you think of it more of as an auxiliary field that we're going to solve for and remove from the theory, the radiation mode will have two degrees of freedom the usual to transverse degrees of freedom of the graviton. I don't know if that address the question I wasn't quite sure I understood the question. Yes, yes, I mean, I don't see what you mean but I mean in the next case yes I saw I thought is this this long to the model is really a non propagating freedom. So in the sense that you really have a function but this function that doesn't obey equations of motions. But in the GDR case I thought that they are more than just one of those functions which, which, which is not propagating. The, yeah, so the, the member that potential mode is just not the zero component. It's also transverse components of the potential mode as well. Right. So it's not, it's not, it's not a gauge. It's not a statement about gauge is a statement about the dynamics of a bound state. Right. In other words, you, you certainly like in QED you can choose a gauge where a zero is equals zero if you wanted to. So let's imagine we did we could have gone through our whole analysis with QED in the 80 equals zero gauge. It's a little bit tougher to do technically I mean, not hard you could do it but and then you would still have AI potentials right. So it's important not to conflate the, the gauge degrees of freedom as unpropagating degrees of freedom from the potential modes, which are non physical, right. In the sense that you, in the sense that they don't they don't exist as asymptotic states if you're doing it in QED. They're off shell modes. Having problems mirroring. Okay, so I could expand this guy out and to save some time I'm not going to write it all out but it you know there's there's terms that look like there's linear terms in H with some function of V and then there's quadratic terms in H, some other function of V, and then there's the kinetic terms to save time I'm not going to write them out but you know this is the mb squared plus m squared plus mv to the fourth with the right coefficients and so forth and so on. Okay. So, now, this H is the full H. So we're going to write this H as H plus H bar. Right, and then we're going to do what we did before we're going to write H as D3K, H of K of T we're going to do the partial Fourier transform and then we're going to carry out the exact same program, we carried out previously. So if we look at the potential mode. So L of the potential, it's going to look much like it did in the case of QED, just with different contractions, because we have more possibilities. So as before the time derivatives will be sub leading so this is in order V square correction V square correction, the propagator the two point function will come completely from here, and we'll scale is one over K square just as it did in the past. So in fact the propagator in momentum in momentum space times a tensor structure P mu nu alpha beta, which is just written in terms of. So this thing is just is just. Okay. So, so now let's, we can go through all our scaling our scaling arguments as we did before to read off how H of K scales as V to the one half. And this is in momentum space in coordinate space, which we'll also use later at scales. Also be the one half but over our, and this guy scales to be the one half. And we're going to use that to determine the scaling of the terms in the action. Okay. So, I haven't, because I didn't write out the exact form of these f's, I'm going to tell you what they are now. But if we look at V to the zero, there's a term that goes like minus M over two M plank H zero zero at order V to the one minus M over M plank V I H zero I would be squared I'm not going to go any higher is minus M because this is where the first sort of non trivial things shows up. Okay, so let's let's let's calculate the Newtonian potential and check our scaling. Okay, so let's look at both of these terms and see how they scale with L. So if we look at L to the, let's look at this guy here. Let's let's power count that term. So it goes like DT M over M plank H zero zero. Okay. So, we said that M, let's collect our scaling rule for the angular momentum. Okay, so this tells us that M over M plank scales as the square root of VL. So an H zero zero if this is in coordinate space, we said scales as root V over R. Okay, so this guy scales as R over V. This guy scales as V the one half over R. And this guy scales as square root of V times L. So this, the whole thing scales as. So, this guy goes like R over V squared V L. And then square root of V over R, which goes like the square of L. V to the zero. So this, this leading order term goes like V to the zero square root of L. So let's calculate the leading order contribution to the potential. Well, that comes from two insertions. Let's call this of L V zero. Right. So this goes like, then we have minus M one over two M plank minus M two over two M plank. So let's insertion of this two of these guys. Sorry. One on the top line. This is M one, and this is M two. And then we have the contraction of H zero zero. And then we have E to the I K zero T minus T prime. So let me do this in momentum space. That's the propagator for this line. Okay, and then P zero zero zero, right, because we're contracting an H zero zero here, and an H zero zero here. Okay. So this just becomes you can just do this integral. And it becomes M one M two over four and plank squared. And then there's a not keeping track of the real factors. And then there's a one over R. So, as you might guess, this just gives back V is G. M one M two over R. Okay, just as we got before. Now. So let's check the scaling. So, we could try to work out the scaling here, but we don't need to because we're already done. So since each one of these insertions we said scales is the square root of V, square root of L. This V has to scale with L, right, because this scales is L this scales L. That's the point of making sure everything is homogeneous. Once we know everything is homogeneous, then we can just read off the power counting at our leisure. Okay, I think I've gone over seven minutes, and we wanted to leave time for more questions. This might be a good place. So I mean, I realized we're taking a giant hammer for a tiny nail. We've just, we've just re derived Newton's equations but the very next step will be the first one where it's non trivial. Perfect. I spent a lot of time for questions. So we're going to start the recording now as usual. If there are no questions I can keep going or if people are tired and they want to stop then we can just continue tomorrow. Whatever the organizers want to do. It's way one more minute to see if there are questions now, otherwise maybe you can do more minutes if you wish. Okay. Yeah, I'm happy to, you know, we'll get a little bit further if we can finish something today, so. But I don't want to, you know, people have other time commitments. I don't want to. Basically we have time booked for this lecture. Oh, we had time booked for the questions anyway, right? Okay, yeah. So it's funny. There are no questions for the moment. Take more than more minutes and then we do 10 more minutes of questions afterwards. All right, sounds good. Okay. Okay, so that's the leading order potential. Let's, let's calculate the order v squared corrections to Newton. Okay. Well, since we already have our action up to order v squared. We can calculate with impunity now up to order v squared. So what kind of diagrams can we get. And again, when I draw these lines what I really mean is this. This is the same thing as that. But the lines are sort of a bad habit that I've inherited from my particle physics upbringing. So we can have one insert insertion of LV one on each line. Right. Or I can have one insertion of LV two and one insertion of LV zero and of course symmetric. Right, those are all those are all there. But in addition, we have some interesting contributions from these quadratic terms. So, I think I. Oh, sorry. That's why. Sorry. There's also this term here. So look at, let's look at these contributions. And let's look at how these so that looks a little strange because it looks like they may not scale properly with L. But before we do that, we can see that what those contributions are going to are going to contribute. Okay, and plus symmetric. So let's let's power count it. So what we said was each one of these, each one of these guys. It scales like root L. And so that gives us potential that scale like L, which is good, which is what we want. Actually say the integral dt skills like L. But what about these guys so let's look at these terms here. So this guy we know these guys we know scale is root L. So it better be that this guy scales L to the zero otherwise we're going to get nonsense. Right. So let's look at how these guys scale. So we've got dt. H zero zero squared. And over and plank squared, for instance, that's one of the terms the other terms will scale in the same way. Right, and we said dt skills like our over V. We said H zero zero in coordinate space scales as root V over R. So we get a V over R squared. Then we get an M over M plank, and then a left residual one over M plank. Okay. Now we said M over M plank. We worked out up here. So let me insert an M here. I'm going to go down stairs to cancel it. So now this the V's cancel, and we have one over RM times M over M plank squared. So this goes like one over RM times V times L. But let me put in a V here and a V here. So this RM V is L. So this goes like V squared. That's exactly what we would expect. This guy goes like L to the L to the zero V squared. And then the potential that you generate will scale as L V squared, which is exactly what we want. Okay. So calculation these diagrams is just relatively straightforward to do. There's one other correction that we need to include that had we not included we would get the wrong answer. And that is the diagram that looks like this, which we call the Mercedes diagram. Okay, so this is an insertion of L V zero L V zero L V zero. And here we have a bulk three gravitan vertex. Okay. So let's work out the scaling of that bulk three gravitan vertex to see if it's consistent with what we have. So this vertex is going to go like one over M plank before X and squared. DH, DH, because remember R has two, two derivatives in it. And we're pulling out the cubic term so should have two derivatives and three fields. Okay. Actually, no, there's let's gotta get the units right. So I think it should be M plank right so H has units of mass so this is one, two, three. That's right. So one, one power of M plank outside. Okay, so let's figure out the scaling. This is an excellent excellent exercise. Now these guys were taken to be potential. So they're going to go like this. Right. So how does D for X scale well D three X for potential mode we know goes like one over our cube and DT goes like Sorry, what am I saying. Okay. I'm just keeping the R here for units purposes. So if we're only interested in these Then we would just have this goes like B to the zero. And this goes like B to the minus one. Okay. Now, we're interested in the leading order piece here and remember that D zero is sub leading because it goes like V. So these are all going to be eyes at leading order. Okay. So in these H's we said a scale as root V over R. So, so let's keep track of the. So this guy, we're saying goes like one over V. This guy goes like root V over R. And we have 123 of them. And these derivatives each go like one over R because they're spatial. So if we're not interested in ours. Let's put put the ours back in when we're done just on dimensional grounds. Okay. So let me just write this as root V. So this whole thing has to be unit list. So this guy goes like R to the fourth. So if we're going to put the ours back in. And then we've got a one over M plank and roll R. Right to make a sorry and are downstairs to make it dimensionless. So this thing is going to scale like square root of V. And then let me put in an M over M plank. And then I've got an R and here. Okay. And let me add a V here. And the V here just so I have everything in terms of objects which I want. So now this goes like V to the three halves. M over M plank. One over L. So this is L. Right. And we said that M. Over M plank. So this is like root V root L. So we have V to the three halves. Square root of V. Square root of L. All over L. It's like V squared over root L. But that's exactly what we need. Because each one of these guys. Goes like root L. So this whole diagram is going to go like root L. The three halves V squared over root L. Goes like V squared L. Okay. So we definitely need to include this and importantly it is not quantum mechanical. Right. We can see it's not quantum mechanical because if it were, it would go like L to the zero and not L. So that's the point of all of all the power counting. So now we're almost done. So we have how many diagrams do we have? We have this one. This one, this one. The Mercedes diagram. And we still have one more diagram, which we should not forget, which comes from the corrections to instantanity that we've had in QED. So this is a K squared K zero squared. Over K to the fourth. And the calculation of these, this diagram is very tedious because of this vertex, the tensor structure, that vertex. The easiest way to do this calculation is, well, there, there, there are ways to avoid this, which, which I'll discuss. But if you want to work, if you wanted to do this calculation, it's best to just do all the tensor contractions in QED using X tensor or fine calc, whatever your preferred mode of calculation is. And the final answer looks something like this. When you add up all the terms. Okay, this guy comes, this is the Mercedes diagram. You can tell because it's got another factor of G. That comes from the bulk vertex. This guy looks a lot like the piece we had. The two guys in fact look like the piece we have in electrodynamics. And there's one piece that looks slightly different. And I should say these are all coordinate dependent quantities. The potential is a coordinate. So as opposed to QED, we need to be a little bit more careful in GR because we've chosen a coordinate system. We've chosen a gauge and harmonic gauge in this case. So I haven't, when I wrote down way back here. I neglected to mention that this is just one, I haven't chosen to tell you what my gauge choice is, but the gauge fixing condition will fix the tensor structure that propagator. So this result is a gauge dependent quantity. It shouldn't be surprising. It depends on your choice of coordinates. And in fact, there are many simpler coordinate system. You could choose coordinate systems, for instance, where that term is gone. The potential is not physical. It's not a gauge of great quantity. The power and infinity is gauge invariant. So no matter what coordinate system you choose for the potential, the, at infinity, whatever you calculate, the physical quantities you calculate won't depend on the gauge. And a typical physical quantity that you would measure would be the potential at infinity. So that this is just one convenient gauge to calculate. This method of calculation using Feynman diagrams is not simple. And so people have used this method to calculate up to V to the eighth. And once you get to the V the eighth, you get like 500 Feynman diagrams. And so clearly that's not the best way to calculate. We now know nowadays, for instance, that using amplitude methods, if you guys are familiar with those, there's methods to calculate all the quantities in one shot without calculating individual diagrams. And it makes things incredibly simpler because you don't have to carry around all, there's a lot of extra gauge dependent baggage. So if you use these S matrix techniques, you can calculate all these diagrams sort of simultaneously. And there are lots of other techniques that people develop subsequently where you can reduce the workload of having to calculate all the quantities in one shot. And so if you want to calculate, you know, that Feynman diagrams like this, there's tricks that you can use by choosing gauges that make your life a lot easier. But those are technical issues. And I was really here interested in, in conveying the important aspects of building the effective theory and the physics behind it. So that's a good place to stop. I should say this is this post Newtonian notation is V to the two N is called the NPN correction. So this correction order V squared is one PN. And this is, this is called the Einstein infel Hoffman potential. They were the first ones to calculate this. Actually, it's not true. It turns out Lorenz actually calculated it before then, before they did. And most people don't are unaware of that. In any case, that that's what that potential is. And that's the one PN potential. People have calculated now up to four PN. Five PN is the end goal. For reasons which why we'll discuss next time. Okay, so we have a good place to stop right there. All right. So now that we can take the questions. No. First one by our man. Yes, hello. Thank you for the lecture. So I was just curious about. So the last thing that you said, like about these methods to calculate easier the integrals. And then you mentioned the S matrix formalism. I mean, did you like thought about, I don't know, like cutting through the diagrams and stuff like this, that the usual things we do or is it, is there maybe some other technique not known to students, but. No, so the, yeah. So what's interesting is you can use all the usual unitarity sewing techniques, right? To calculate these diagrams, but it's actually a little bit easier than normally because you can throw out the quantum pieces. And so a lot of the tough integrals that you have to do, you can just ignore. Cause remember when you're calculating an S matrix element includes all the quantum pieces. And so, you know, when you reduced it to a set of master integrals, you could just pick out which master integrals are actually. Quantum mechanical and just throw them away. So it's actually a little bit easier than doing the full quantum computation. Although usually, you know, people who are experts in these in this field have code. That do these things and includes everything. And then they have to go, they go in by hand and pick out the pieces that they want. I don't know if that answered your question. Yeah, he says yes. Okay. Okay. Next one is Max. Yeah, thank you very much.