 So I've created a Slack channel where you can post questions. And also, I can copy questions. So I get questions from emails. And I will anonymously post those questions onto Slack with the answers. And so on that Slack channel, I'll be constantly trying to answer those questions in real time. And also, I'll be updating the lecture notes, fixing typos as we go along, and also perhaps elucidating points which were unclear that either in the lectures and the notes to help verify some issues. OK, so today the goal will be to essentially set up the correct effective field theory for electrodynamics in order to calculate the radiation from a binary in-spiral, electromagnetic binary in-spiral. And then tomorrow, we'll start off with a real problem going into GR. But once we've established all this machinery, that will be relatively simple. So let's remind ourselves about where we started off. So we're considering some charge distributions that are bound. And we want to calculate radiation at infinity from such a system. And we said one way to do that is to calculate the vacuum persistence amplitude. So here, I'm just writing it down for, say, a general scalar field like that. We would actually call that usually the partition function in a quantum field theory class. So in our case, for the electromagnetic field, can you hear me? Yes. Oh, OK. I'm lucky things went out. OK. Then that would be the vacuum persistence amplitude. And we said that we could write this as e to the minus e of j t, at least for static sources. And here, we're replacing j will be our, will actually be j will be basically our world lines. So j will be replaced by axonist derivatives, where we said we have some world lines, x i of lambda i, where each particle is parameterized by some affine parameter. OK, I see. Sorry. So people are joining the Slack channel in real time, which is fantastic. But I can hear it each time someone joins. I have to figure out how to turn it off, because it keeps beeping on my computer. Must be some way to turn off the notifications. All right, well, maybe, yeah, if you could hold off joining the Slack channel, at least till the end of this lecture, since if you have questions, now you can just put it in the chat. And then you can join the Slack channel afterwards, because every time somebody joins, it beeps on my computer. So it's hard for me to hear if anything's going on. So if you could just hold off on that just for a little bit. I'm sure there's some way to turn it off, but I don't want to take the time to figure it out at the moment. OK, so we're coupling our electromagnetic field to these world lines, which we're going to treat as sources. So the logic will be, well, first we'll treat them as sources, and eventually we'll elevate them to dynamical degrees of freedom by doing the path integral over them. So the action for the electromagnetic field, and here I'm working in the Feynman gauge, because it's convenient. That's our gauge fixing term. And now because the electrodynamics problem is Gaussian, the action is quadratic in the fields by completing the square in the exponent, we can actually do the integral exactly. So z of j just becomes e to the minus i of 2, where interestingly enough, so gf is the Feynman propagator. So gf in momentum space is going to be minus, sorry, g mu nu over k squared plus i epsilon. So at this point, you should be puzzled already, because this is a complex valued function. When k squared goes to 0, it picks up a complex contribution. And clearly, we're doing a classical problem. So normally, the propagator, the two-point function, should be real. If we're really interested in real propagation. But let's follow our noses. As particle physicists, this is what we do. And certainly, if we include quantum corrections, that's not a problem. We should be able to take the classical limit and still get the answer. So we're going to just plow ahead and not worry about that. And then we will worry about that. We'll come back to this issue of the i epsilon later. Now it's important to note that I didn't have a choice in my i epsilon prescription. Because to make the path integral convergent, I needed to add an i epsilon a mu squared, where the choice of this sign is fixed by the convergence of the integral. So I had no choice. I had not done anything by hand here. It's not ad hoc, OK? Good. OK, so what are our sources? Well, j mu of x, I'm going to write as v mu of lambda d lambda d4 x minus x of lambda, OK? So let me label since we have more than one particle. Now, notice that I've already worked in the point particle approximation, right? So I've taken my blobs. I've reduced them to point particles. Now we know, as we discussed last time, that we can include finite size effects by adding more terms to the action. And we'll come back to that in a little bit. OK, so now let's plug this in. So this z of j, we define as e to the i omega of j, OK? So omega of j will be equal to minus i over 2. Let me cut and paste to save some time. Several terms. I'll have d4k over k squared plus i epsilon. And I should change this sign here. So this sign actually was point in the notes that sign was wrong. That was pointed out to me by a diligent student, who I appreciate very much. And so this is what our result's going to look like, OK? And I still have, sorry, and I have e to the ik dot x minus y. And then I still have my two delta functions, d4 x minus x1, number 1, d4 y minus, sorry, yeah, y minus x2, number 2, OK? Good, all right. So let's explore out what this means, what this object looks like in the non-relativistic approximation, OK? So let's consider the non-relativistic approximation, because that's what we're going to be interested in, as that's the, so in electrodynamics, we could solve this problem relativistically. It's linear, it's simple, it's not a problem. But non-relativistically, it won't be so. So to warm up for GR, let's do the non-relativistic expansion in the electrodynamics case, OK? So the first thing we're going to do is we're going to choose lambda to be the time. And we're certainly free to do that. And let's look at the leading order result. Actually, you know what? Let me take it back. Before we do the non-relativistic approximation, let's explore this a little bit, what the meaning of everything is here, OK? So first of all, we see that there's going to be a real part and then there's going to be an imaginary part. And remember, we said this thing's going to go like e to the minus i ej of t. And this is, by definition, e to the i omega of j. So we're going to try to extract the energy. So the real part of the energy, so the real part of w, will be the energy or the potentials. And the imaginary part of w will be related to the decay rate or the power loss, OK? So how can we see that? Well, if we write k squared plus i epsilon, we can write this as minus i times the imaginary part, sorry, times delta k squared plus the principal part of 1 over k squared, OK? Thought of as a distribution. So the first thing we can see is that the imaginary part puts the photon on shell, right? It imposes k squared is equal to 0, OK? That means it's a solution to the equation's emotion. It's a propagating photon degree of freedom. So this corresponds to radiation, OK? Now, if we're interested in the piece, independent of i epsilon, this is going to be the real part. And this is going to be the potential, OK? So in fact, so I'm not going to go through this in the notes. But if we look at the imaginary part, so the imaginary part, I'm sorry, I'm not going to go through some detail because it's in the notes. And it would take a little bit too much time. So this we can see this is trivially true, OK? And this follows from just coming all the way back here. And taking the imaginary part, we get a, if I go to momentum space, this gives me a delta function of k. And then the fact, so this, remember that the famous identity, sorry, that the delta function on shell condition, OK? Which is just a phase space integral. So this on shell condition just converts it into a phase space integral, OK? So I don't know. Yep, there we go. Sorry. Good. So what I show in the notes, which I encourage you to look at, is so how are we going to relate it? We can relate the imaginary part to the power because we can see that the probability to emit nothing, so this is the vacuum persistence amplitude, is just the square of z. This is the probability for no emission. It's just going to be e, sorry, e to the minus d3k over 2k j squared, OK? And that just comes from squaring this vacuum to vacuum in the presence of the source. That's nothing like this, OK? So what I show in the notes is that if you want to calculate the power loss, then all you have to do to get the power loss is to weight the probability. So we know that from quantum mechanics that the real part or the imaginary part of the energy or the real part of w of j is just minus gamma t over 2, right? Because any probability, any decay probability just goes, sorry, any decay amplitude. And so the square that gives you the decay rate, so gamma is the width. Remember, this is the imaginary part of the energy, right? So what I show in the notes is that the power loss is just given by 1 over t e to the k over 2k times j of k squared. So I haven't shown you that here. Here, all I've told, the only thing we actually calculated was a vacuum persistence amplitude. What we really want, if we want to calculate the power loss, is the probability to emit n-quanta weighted by the energy of each quanta. And as I show in the notes, to get that, it's exactly given by this result, OK? All right, good. So now let's consider some examples. So let's work out an example. So we're going to start with, this is the exact result. Haven't made any approximations yet. That's our exact result. And we've claimed that once we calculate the real part, that will be the energy or the potential of the two sources. And once we calculate the imaginary part, that will give us the power loss, OK? So that's almost what exactly, so what we're really interested in as an observer infinity is the power loss, because we're measuring the power. We would also like to be able to measure the phase. And at this point, we don't have the mechanisms, the power to calculate that. And we'll have to do that. But for the moment, actually, even at LIGO, the first thing they really calculate is, or the first thing that they really are interested in is the power loss. From the power loss, you can get all the information. I'm not all the information, but mostly information that you want. OK, so let's do a simple example. And we're going to do it in the non-nulltavistic limit, again, to prepare ourselves for the real deal, GR, OK? And we're going to do the simple example in two ways. First, we're going to do it by taking the full theory result, the exact result, and expanding it. So in QED, we have that ability. In GR, we will not. Then we'll use that intuition to build an effective theory and re-derive that same result directly from the effective field theory. And the reason we want to do that is because we need to understand how to build the effective theory. Once we understand it in QED, doing it for GR will be trivial. So for our sources, again, we're going to take, so j mu of k is going to be the Fourier transform. So we're going to use the Fourier. So we're going to use our delta function. So there is our source. And so now we're going to make a non-ralthavistic approximation. And we're going to choose lambda to be t, right? So now, excuse me, w is 1 half dt prime. Those are the two times for the two different world lines. There is our propagator. And then we have e1 squared, 1 minus v1 squared, e to the ik, x1 of t minus x1 of t prime. So we have to make sure that we use different names for our affine parameters. I'm calling 1t and 1t prime. Then we have the same thing with 1 going to 2. And then we have e1, e2, 1 minus v1. Let me put this in explicitly. Plus, again, 1 goes to 2, which in this case gives an identical contribution. So I've used the d4x integrals. And that's why we end up with the positions in the exponents. OK? OK. So I should put it, we're making the non-ralthavistic approximation, and we're going to keep up the order v squared. So the first nontrivial correction. Now, how do we know that there's no linear piece in v? And the answer is that the symmetries forbid such a term, because we know the system is time reversal invariant. So the first correction has to go like v squared. If we were looking at some other effects, which did not preserve time reversal invariance, like a dissipative effect, friction, something like that, then that will no longer be true. OK. So the first thing we do is we know that in the static limit, let's consider the static limit. So in the static limit, we're going to take the b to 0. We're going to add two static charges. And then if we look at the exponents and we look at the k0 piece, we're going to have t minus t prime in the exponents. And so this is going to lead to delta k0 once we integrate over t. So we can see that the static limit is associated with vanishing energy. And I should say to gain just a little bit of intuition, what we're essentially doing here is we're considering the exchange of one photon. So electrodynamics is linear. So that's all you get is just one photon exchange, and you get nothing else. When we get to gr, this will get dressed by nonlinearities and so forth. But for the moment, that's all we have. So what this is telling us is that in the static limit, there's no energy transfer between the photon carries no energy so that we know that in general, away from the static limit, the energy transfer is going to scale like some power of the velocity. And we further borne know that k, just by dimensional analysis, the only scale around is r. So we can see that in the non-relativistic limit, k0 over k is over order v to the n, where we haven't fixed n, but we will in a moment. So that means that we can treat k as being a small quantity that with which we can expand. Now let's consider the terms we have in our expansion. We have these terms that go like e1 squared and e2 squared, and then we have the cross terms. Let's get some physical insight into what the meaning of those terms are. So let's consider the e1 squared term. So we have something like dt dt prime 1 over k0 squared minus k squared. And then we have e to the i k0 t minus t prime e to the i. Now we're going to take the small velocity limit. So if we drop that term at leading order, then we can see that this thing is going to be proportional to, sorry, and then I have a dk0 d3k. I do the time integrals. So now I'm going to have d3k over k squared, delta of k0, actually, OK. Actually, I'm sorry, let me do the k0 integral first just to clarify things. So then we have 1 over k squared, and then we have delta of t minus t prime dt dt prime e to the i k dot x1 of t minus x2 of t prime, OK. So we can see t is equal to t prime. This goes to 0. And then we end up with this very sick looking integral. That's infinite, OK. That shouldn't surprise us. So this is just the self-energy of the charge, OK. So that's the usual self-energy in classical electrodynamics. And from the point of view of quantum field theory, this is a divergence, which is a power law divergence. And we can just absorb that divergence into the charge, OK, or into the mass, in this case. So from the point of view of quantum field theory, of course, we could do this integral, say we have to regulate it. We could do it in dim reg, dimensional regularization. But in dimensional regularization, that gives 0, right? So in dimensional regularization, every scaleless integral vanishes. So that's a little bit weird. So is it infinite or is it 0? And the answer is from the point of view of effective field theory, it doesn't matter. Because any time you have a power law divergence, it's telling you that it can be absorbed into a local counterterm. And there is no logarithms involved. So I don't really have time to go into this. But basically, whenever you have a power divergence, you can just toss it. It has no physical effects. It can always be absorbed into a counterterm. And physically, or I should say technically, of course, it had to be true. Because if I used a cutoff, I would get one answer. And if I used dimensional regularization, I get 0. And it shouldn't matter how I regulate my theory, I need to get the same physics, right? So it's clear any time you get a result, which depends upon the regulator, assuming the regulator is not breaking any symmetries so you don't have to worry about anomalies, then you should get the same answer with any regulator. So if you're getting 0 and 1 and thinning the other, that means it just can't be physical, which means it has to fall out of any prediction for any measurable quantity, which is another way of saying that it can be absorbed into a counterterm. So that means we can come back here and we can toss these guys. So we can toss this guy and we can toss that guy. And let's just look at what we're left over with. So we're left with the cross terms. So what do the cross terms look like? Well, let's look at the leading order piece. So we're going to set v to 0. And we're just going to look, so this is a purely static limit. And in that case, we have minus 1 half dt dt prime d 3 k k squared. So this is w. And then we have a factor of 2 because we have two cross terms, e1, e2, e to the i k t minus t prime, e to the minus ik, sorry. And then we have, yeah, so this is k0. So I still haven't done the k0 integral. So sorry, I realized I just introduced some notation. So in my lectures and in my notes, ddk with a bracket means ddk with a 2 pi to the d. All right, so now I can again do the dk0 integral. That gives me a delta function and kills one of the 2 pi's. And I have to be careful because this is a plus sign. Where did this plus sign came from? That comes from this minus sign, e1, e2, delta t minus t prime, e to the i k dot x1 minus x2, OK? Hopefully I've tracked the signs right. There's a possibility that I have not. Now I could just do this. So this is do this integral over time. And then I just get the Fourier transform. There's one time integral left over that's just going to give me a t. And this is equal to minus e times t. So e is d3k, or it's just a Fourier transform. OK, so that's a sanity check. We've just re-derived the Coulomb force law. So you might want to do, if you're not familiar with this integral, and the easiest way to do this integral is to first put in a mass and then take the limit when the mass goes to 0. So you might want to do that. Just take you a few minutes in your spare time. Now let's go to order v squared, or next to leading order. And there are two sources of v squared corrections. So let's come back and to see where those two sources of v squared corrections are. OK, so we're ignoring now the self-energies. So there's a v squared correction here. And there's also a v squared correction hidden in here. OK, so there are two sources of v squared. The first one is just the explicit v1 dot v2. And the second one is from the fact that k squared. I know you have a question. Uh-huh, OK. Matt? Yeah, hi. So my question is about this order v squared corrections. So you put lambda equals g, right? But it seems to me that from there, you also should get a v v squared correction because it is proportional to tau times this gamma factor. So you should get v squared from that as well. Yeah, so in this, so in general, you write there are v square corrections from the affine parameter, writing it in terms of t. But in this case, there isn't. So in the case of, so certainly the m, this, from the kinetic energy, you will get that. But in this case, you don't, right? Because I'm taking lambda just to be the time, not the proper time here, right? But you're right, in the kinetic piece, you will get that. OK, I see. Thank you. Yeah, OK, good. So here, so this guy is a v squared correction as well. If we take v, k, sorry, to be of order v times k, OK? So remember before we said that k0 is of order v to the n. And we didn't exactly specify what n is. But now I claim that n is equal to 1. And you will see that quite clearly because, remember, k in x space is just d by dt, right? k0 is a d by dt, right? In other words, I can replace in my any integral I have here, starting at this point, the very beginning, any k0 I would have say here, I can just replace by a d by dt. And then by an integration by parts, that's going to give me a v when it hits the x's, OK? So k0 will scale like v. So this term will be order v squared, which is of the same order of the term that we're keeping, OK? So the first term will just be proportional to the Coulomb potential, right? We don't have to redo all that work because if we come back here, we could see that it's just the whole thing goes like 1 minus v1 dot v2, right? So there's no effect on the calculation. It's no this different than it was for the Coulomb potential. But for the second term, we're now going to get something like dk0, d3k, k to the fourth, k0 squared, e to the i, k0t minus t prime, minus ik dot, OK? So we can rewrite this as plus d by dt, d by dt prime. Let me write now this is delta x just to save some time. And now there's still a dt prime integral here. So now this becomes, and now these guys are evaluated at the same time, OK? So I can then just perform this integral and take the derivatives. And the result looks something like, so you've got something like v Coulomb times 1 minus v1 dot v2 minus e1 e2 over 8 pi r v1 dot delta x over r v2 dot delta x. And this is the first, so these are the first, these are the order v square corrections to the Coulomb potential. So just in general, so this is to do the k to the, 1 over k to the fourth integral, I'm going to give you a magic formula that you can prove for yourself as a fun exercise for a general power alpha in d dimensions. I'm doing it here in d dimensions. The reason I'm doing in d dimensions is because eventually, when we get to more complicated integrals or complicated situations, these integrals will diverge logarithmically. And as opposed to power divergences, logarithmic divergences are physical. So in fact, one of the reasons I actually originally got involved in this whole business was because when people were using standard calculation methods of GR to calculate these post-nuttonian corrections, they kept running into divergences. And this was a bit of a mystery. And they introduced ambiguity parameters. And it got rather complicated. And Walter Goldberger and I, my collaborator, not realize that, wait a second, from the point of view of quantum field theory, we see divergences all the time. It's not a problem. We don't have to introduce the ambiguity parameters. You just have to introduce counter terms. And you can introduce counter terms once you've built up the correct effect of field theory. So it's kind of the motivating factor behind trying to come up with this formalism. And you get logarithms in a classical calculation. And you get 1 over epsilon. And you get a renormalization group. And you can sum logarithms just like you do in quantum field theory, except these logarithms are not due to quantum fluctuations. So it allows you to use all the tricks of renormalization group in classical field theory to make predictions for logs at higher orders in all the power that comes from doing the RG. OK, so maybe this is a good time to take a break and to see if there are any questions that we want to talk about before we go on. No? OK, so I think this is a good. So when we come back, we'll talk about how to do the radiation. And then we'll talk about how to build up the effect of field theory. Perfect, so I'm back in five minutes, OK? Yeah, all right. But there is a question if you want to take. Yes, I have a question. So since we are in classical field theory, I'm wondering where does this correction to the collop to come from? So is this due to some reaction for us coming from the photon, or what is it? Yeah, I think it's just a purely relativistic correction. Ah, OK. In fact, you can think of, so the k0 squared correction is a correction to instantanity. It's the first correction due to the finite propagation speed of the light that's being exchanged. OK. It's the best way to think about it. OK, I went whenever you want to. OK. OK, so sorry. So let's talk now about the radiation. So the radiation comes from the imaginary part of the energy, or the imaginary part of w. So let's go back to our master formula for w. But now let me, instead, let me not multiply it out. Let me just write it like this. And then I, sorry, and I still have my exponentials. So why do I want to write it like this? Because it'll clarify the physics a little bit. So now we're interested in the imaginary part. So the imaginary part will be proportional to delta of k squared. Excuse me for one second. Today is the last day of school, and my children are home. So I have to tell them to try to be quiet. I'll be right back. OK, sorry about that. OK, good. So we said that k0 is of order v times k, which is order v over r. OK? So k0, which is now of order k in this, sorry, so hold on. Sorry, I need to come back. I need to take a step back for a second. So in the case of potential modes, or in the case of the potential, we found that there was a hierarchy that k0 was of order vk. OK? But now we can see that k0 is exactly equal to k because it's on-shell radiation, right? But the wavelength of the radiation, we know, is going to be k0, is going to be of order omega, right? Which is of order v over r. OK? So we can see that for radiation, we have k mu is of order v over r v over r. And for potential, we have k mu is of order v over r and 1 over r. OK? So at the very beginning of the first lecture, we talked about the fact that we're going to break up the field into long distance and short distances. And one thing we can see immediately is that the radiation is a long-distance piece and the potential is a short-distance piece, OK? So in a few minutes, when we do the EFT, that will be very important. So part of the reason I wanted to go through the full theory calculations through the exact calculation was to teach us what are the short-distance modes and what are the long-distance modes. And so especially for those of you who are familiar with effective field theories, you need to integrate out all the short-distance modes to get the long-distance physics, and that's what we're going to do. But again, we haven't got there yet. Let's take one step at a time. So what we do notice, however, is that k, which is border 1 over r, is order v. And it's therefore much less than 1, OK? So this is just a statement that the wavelength of the radiation is parametrically longer than the distance scale set by the distance between the two objects, OK? So the radiation with some lambda is much greater than r. But this is nothing but the multiple expansion. This allows us to perform the multiple expansion. So we're going to take k dot x is much less than 1, OK? So now let's look at our w in the multiple expansion at leading order. What happens? Well, let's come back up here. So at leading order, we're going to take, so dx mu vt is just going to be 1, 0. And then we're going to end up with d4k k squared plus i epsilon. So at leading order, I'm going to take, remember, e to the ik dot x at leading order will be 1. So I can drop all those guys. And then I'm just going to end up with e to the ik0 t minus t prime. And then the products of the ease, again, all coming from our master formula here, OK? So now if I take the imaginary part at leading order, this is going to give me something like d4k delta of k squared, where I've dropped everything which is sub-leading. Now this thing, so this is going to give me d3k over 2k as usual. And then I still have the dt prime and the dt. There's no scales involved. So this is some ill-defined divergent quantity, which of course, we can adjust as well just set to 0. And this is just a statement, as we'll see in a second, that there is no monopole radiation. Well, at the moment, that may not be clear. But physically, it's due to charge conservation. Any time a field couples to a conserved quantity, it's not going to radiate, right? The charge can't change. So essentially, all we've done here is asked whether or not actually, yeah. I mean, you can actually show this as explicitly 0, even without regulating it. And that's just a statement that if my source has no time dependent configuration, if the charge is constant, the charge can't change. The charge can't change. It can't radiate, right? Radiation comes from time dependence of the charge distribution. So we need to get radiation. We need to go to next leading order, OK? So that means we need to keep the first piece in our expansion of the plane wave. So this should be familiar to you from quantum mechanics, where this is how you calculate radiation from an atom, right? And you do it in the multiple expansion by taking the fact that the radiation has wavelength, which is parametrically for non-relativistic systems, which is parametrically larger than the size of the system. So now if I keep this term, then the imaginary part of W will look something like, now there's a sum over the particles implied, OK? But so this is the decay rate. And if I take this integral and I weight it by the energy, that will give me the power loss. So weight by the energy. If I weight this thing by k, so I add to this a k, right? Then this thing is equal to, and you can do this as an exercise for yourself, as the inner product or the dot, the square of the second derivative of the dipole moment of the system where the dipole moment is. So this is called dipole radiation. And I have skipped steps in going from here to here because of time limitations. But this is in the notes. And the crucial point to understand is we've done a multiple expansion. The leading term in the power loss, we get from the imaginary part and weighting it by the energy of each emission. And then we just get back the standard result for dipole radiation of electromagnetic system, OK? So if we were incredibly naive and just said, OK, let's do the simplest thing we could possibly do, we've calculated the corrections to the potential, and we calculated the power loss. So then in order to calculate the signal that we'd be expected to see the detector, we would solve the equations of motion, take those solutions to the equations of motion, plug them in here, and read off the power loss. OK. So let me pause for a sec because now the next logical step is how do we derive from an effective field theory? So how do we build an EFT to reproduce this result, which seems pedantic, but will allow us, when it comes to GR, just to jump in immediately and solve the problem, OK? So are there any questions? OK, yeah, we have a half hour to go over the questions, so let's say that for now because I'm kind of running out of time. So how do we reproduce this result in the effective field theory? OK, so the basic rules are, one, every term in the action should scale homogeneously in your power counting parameter. This is the prime directive in effective field theory, right? So why is that? Because your goal is to calculate systematically. And in order to calculate systematically, you have to keep track of every source of a correction. And by having each term in the action scale in a definite way in V, allows you to figure out what you need to do in order to include all corrections up to whatever order in V you're interested in calculating, OK? So in the full theory, you have nothing scaled homogeneously. Everything is just glob together. It's just everything you need. But we want to break it up into the pieces that we want, OK? So that is the prime directive. And how do we go about doing that? Well, we have to separate out all the pieces which contribute in different ways. So the first thing we need to do is we need to take our electromagnetic field, and we need to split it up into the two pieces which scale differently. Now, our intuition from going through this problem tells us exactly how to do that. There are two different types of modes in the theory, and they scale differently. So the full theory electromagnetic field, we can write as a mu potential plus a mu radiation, OK? So the momenta of this field scale as V over R, 1 over R, and of this field scale as V over R, V over R, OK? Now, mathematically, how do we do this in a sensible way mathematically? Well, we have to make sure, for instance, that we don't double count in our path integral. And we will discuss that later on. But more importantly, we have to figure out how to do this in a gauge invariant way. So the way we do that is we work in the background field gauge, which hopefully you're all familiar with from your quantum field theory classes, where we treat the radiation as our background field, OK? So we can choose a particular gauge fixing condition such that we can fix the gauge for the potential independently of the background field gauge, all right? So we'll fix this with a fine with a fine engage, as we did before. And this guy, by working with the background field method, we can choose any gauge we want in independent gauge. We could choose a fine engage again, OK? This is basically the Born-Oppenheimer type of approximation where you have fast moving degrees of freedom and slow moving degrees of freedom. The potential of the fast moving degrees of freedom and the radiation of the slow moving degrees of freedom. So we freeze the slow moving degrees of freedom and treat them as a background field. And then afterwards, after we integrate out the fast degrees of freedom, then the slow degrees of freedom are allowed to evolve dynamically. It's much like what we did when we took the point particles to be sources and then allow them eventually to be dynamical, OK? So now our partition functions, z of j, we can write as d potential deradiation. And then we're going to do the path integral over the fast degrees of freedom to get an effective action, OK? And the idea is we want to reproduce what is this theory. This will be our effective field theory. And we must choose as effective to reproduce all the long distance physics to the order of v of interest. Good, OK, so let's take, so now what we're going to do is we're going to take this decomposition and we're going to plug it into our action. Then we're going to derive what this action is. Then the next step will be to do the path integral by calculating Feynman diagrams to get rid of this guy and to generate that guy, OK? So the first thing we're going to do is we're going to take the potential and we're going to Fourier transform it, but only in space, not in time. So we're going to do a partial Fourier transform, OK? And then we're going to plug it into the action, OK? So that's our interaction between our source and this is j.a. And then we have the bulk action. First we have it for the potential field, OK? That is just f squared plus del dot a squared. And then we still have the action for the radiation field plus whatever gauge fixing term we're going to choose for the radiation, OK? So this is the first stage of the effective theory. And we want to make sure that in this stage of the effective theory, all our terms scale homogeniously. So we need to check the scaling of each term, OK? So how do we determine the scaling? So well, first of all, the field scales. So a scales as some power of v, a potential and a radiation. dt scales as some power of v and dx scales as some power of v, OK? Once we know how each one of these scales, then we'll be able to read off how each term in the action scales, OK? So first, some of them are easy by dimensional analysis. So we know that k for a potential, right? So these k's, so remember this k here, everything which is k, is acting on a potential field. So remember the potential is the guy that is, you could put p's here just to remind yourself, but then it gets a little confusing between p and k. So let me not do that. But at least here, the only thing we Fourier transformed are the potentials. And we already know how the potential momenta scale. They scale like 1 over r, so v to the 0. They don't scale at all. We know how d by dt acting on a potential field here scales. So d by dt acting on a k, we know goes like v over r, a k, OK? So therefore, we learned that d by dt is down by v, as we expected, over r. Now what about dt and dx? Well, dx basically is just conjugate to 1 over k. So we have to worry about, so here, we have to distinguish between how the measure scales in radiation versus in potentials, OK? So notice that we don't have a d4x for potentials, because we Fourier transformed. The only place we have a d4x is for radiation, right? So the dx here will scale like 1 over the k of the radiation, which scales as v as r, this thing goes like 1 over k radiation, which goes like r. And then we know this is a scaling of v, because remember that k0 scales like k, this is for radiation, scales like v, right? So this guy, so dx is going to scale like 1 over v, OK? What about dt? Well, for the energy, dt scales like the inverse of the energy, because it's conjugate to the energy, or you can think about it. The only time scale is the orbital frequency. So dt is going to scale like 1 over omega, which goes like 1 over k0, right? So it's going to go like 1 over v, OK? I'm not worried about the units. The units are already always made up by r's. We're only interested in how things scale in v. So now we're all set to figure out what is the scaling of the field. To determine the field scaling, we make sure that the leading order piece scales as 1, OK? So this is how we read it off. So let's figure out first how ar scales, OK? So we have ar squared, and then we have a box squared, and then we have a d4x. And this thing should scale as v to the 0, OK? So we have ar squared, and the box scales as v squared, and the d4x scales as 1 over v to the fourth, right? Because the scales is 1 over k to the fourth, each k scales with a v. Box squared scales as k squared, each k scales with a v. So this has to be v to the 0. So therefore, ar has to scale as v, OK? And if we wanted to get the units right, we would know that it goes like v over r, OK? How does ap scale? Well, let's come back and look at the leading order piece for av. Now we know this guy is the leading order piece, because we know this dt is going to give us a v, and this guy is going to go like v to the 0. Remember, k, for potential, scales as v to the 0. It's a short distance guy. So now we have, now remember, we're looking at ap in momentum space. So we want to know how apk scales. So we have dt, d3k, then we have a k squared, and then k squared, OK? And this guy should scale as 1. So dt scales like what we just said, 1 over v, because it scales like 1 over omega, which scales like k, and k for potential, k0 for potential, remember, goes like v. Now the k, the 3 vector k for potential, scales like no powers of v, right? So we said that there are no powers of v for the 3 vector piece. So that gives us a v to the 0. This gives us a v to the 0. This gives us a v to the 0. So ak potential scales like v to the 1 half, OK? So that's the first step in the algorithm is to figure out how everything scales and make sure, and you figure out the field scalings by taking the leading piece and making it scale as 1. So now we could see, for instance, that this piece here in the potential is order v squared, OK? And sorry, I'm running two minutes late. So now that means this has to be treated as a perturbation to the leading order piece. So this tells us, let's see how we can just get back what we already know. If we look at the propagator for the potential mode, this leads to the propagator, which is 1 over k squared. And this leads to a correction, which goes like k squared, k0 squared, right? So if we look at the propagator, what does it look like? Well, it looks like k 1 over k squared. So I'm going to draw a line like that for my propagator. And then I'm going to correct the propagator by inserting one insertion of this operator and so forth and so on. That's rederived the geometric series for the propagator. But we've done so in a systematic way in the sense that each term in the action now scales uniformly in v. So we should really treat k0 as a perturbation, right? So now if we come back and we want to calculate systematically the potential, well, we have the leading order exchange of a Coulomb. Let me draw Coulomb propagators now to distinguish them from the paths in the world lines. Well, this diagram, this is a 1 over k squared propagator. That just gives us the Coulomb potential. But now we need to correct that propagator by inserting a k0 squared there. And this gives us the term that went like v1 dot delta x, v2 dot delta x over r squared. And then finally, the v1 dot v2 term is just going to come from back here, here, where we pick out the v piece that couples to this potential here. So that's just going to be we can draw that as this, one other diagram. So we have two Feynman diagrams, two non-trivial Feynman diagrams at next to leading order, v1, v2. So this is an insertion of v1, this is an insertion of v2, and this is going to give us v1 dot v2 times vCoulomb. So what we've done here is we've parsed the full theory into homogeneously scaling pieces and use that to calculate the v squared piece. So this seems like we're just, we are literally reinventing the wheel, but for a very express purpose. OK, so the next thing we'll do next time is we'll reproduce the power law from the effective theory, and then we will move on to GR. OK, so let's stop there for questions. I think I apologize for going a few minutes over. No problem. Thank you very much, Ida. Nice lecture, and now we can start the Q&A.