Amar G. Bose: 6.312 Lecture 02
DR. AMAR G. BOSE: After the lecture on Tuesday, I was reminded by the teaching assistants that the class runs till 12:30. I was trying to rush and hold you for a couple of extra moments after 12, and I didn't realize, I'd forgotten, it does run until 1:30. So that's how-- I mean, 12:30. That's how it'll be.
I was also reminded after the class by the teaching assistants that I forgot to mention that on Thursday 28th at 8:00 in 26100, there will be film shown by LSC, which is at my request. And it'll be admission by tickets, and this class will be given tickets to make sure that you get a seat. And it's a very important film, as far as I'm concerned. Some of you will have seen it already. I'm going to ask how many. The title of the film is Stand and Deliver. How many have seen it? Wow. OK. If you don't want to see it again, don't take tickets.
But I have seen it more than once myself. I'm going to give an introduction to it that evening because when I saw it, I was so impressed, I was led to believe it was true, and I called up the real teacher-- I think the vice president of research of our company, who was a former student here, brought it to my home one Saturday evening. And Monday, I called up the real teacher and asked if I could come out and visit him, and that started a relationship in which I saw better teaching than I've ever experienced in my life, and I learned something about the potential of students that I guess I never had believed in before. So if you haven't seen this, it is terrifically important that you do, because it will tell you something about your own potential that should last your lifetime.
Let's see. There were other announcements. Oh, yes. The use of Bibles. You can use them if you want, but it's you that gets the short end of that stick. As I've told you, the most valuable part of the subject is going to be the homework, and it's going to make you think. And the minute you have a crutch in front of you-- and many of the problems will be the same as have been given before, simply because we've tried to optimize those. And the minute you take advantage of the crutch, your muscles get weak, your mind gets weak. You won't benefit.
I don't mind, as I mentioned, talking together, discussing problems. But don't make it a leader and a follower situation. And when you come to do the homework, to write it up, do it yourself. If you do anything else, you will absolutely just lose what you could get otherwise.
Now we're going to start off today. You'll see a lot of equations on the board. Don't get scared. We're going to start with one-dimensional wave equations, or we're going to lead up to the wave equation, all in one dimension. About the simplest way you can look at waves. And from that we're going to go to something whose name will scare you, but the concept shouldn't. Vector wave equation.
Now ask all the questions you can about this when you do your homework. And in the tutorials and the recitations, we're going to go further than we actually are going to use, even in the subject. We're not really going to take much advantage of vector wave equations.
But I think I may have told you, or if I didn't, I will now, that I don't really care whether you ever need to solve an equation in acoustics in your entire life. I hope the subject is useful to you. So I'm going to show you some things that I want you to go through the reasoning process, and I want you to see how they were developed, because one day, you will be developing things that we don't know about. And having looked at the development process of other established results will help you. And that's really why we're going to go as far as something like this. We otherwise wouldn't need it at all for this subject.
And a further note before we get started. Most of what you have studied to this point has been material that's been developed by someone else. And really, what you've done, is you've followed the steps and justified to yourself that step B follows step A, and C follows B. But that's a totally different process then if you were developing it yourself.
And I can well remember sitting in classrooms here and thinking, oh my God. I'll never get to the level of the professor, who can just present this material, let alone the much higher level of the person who developed it. How would he ever have known to go to step two after step one? It's just-- I mean, there are a million things you could have done, and he chose this one, and it's just right.
Well it turns out, that's not the way things happen in life. What happens, even in mathematics, you want to go from here to here and you stumble around, you go all over the map , and eventually you come to this result. Now, then you know that from this hypothesis, you have this thing, and you've proven it. Once you know it exists, then you find eventually a very straight way of going from here to here. And that straight way is not at all obvious from step to step to step.
And that's what you're presented when you go to school. And you wonder, my god, how could anybody have thought of this after this point? It shouldn't be obvious to you. And if it's not, don't think that you're not as good as the person who did it, because he did it this way.
Poisson and other French mathematicians gave birth to statistics sitting in the gambling rooms in Nice and Cannes, on the French Riviera, literally counting on their hands, drawing figures, doing all sorts of stuff. When you try to prove something-- and I don't care whether it's in physics or mathematics, even-- your mind is going all different ways. I wonder, I wonder, I wonder. You're visualizing things, you're doing all sorts of funny things. And when you get finished, when you get over here and know that it's true, then you try to go this way. Part of the reason is, in your own ego, you'd love to make the most elegant presentation of the result. And you eventually find a way to do that. But don't think that the steps are obvious.
So what I'm going to try to do when we go through different steps is tell you what shouldn't be obvious to you at this step. What is the creative part, what shouldn't be obvious if you were doing it for the first time, only to build your confidence that it's OK to visualize things, to dream about things, to get wild ideas, to try them, to get to the end proof.
So we saw last time that if you had, on a frictionless surface, a spring and then a mass-- spring, mass, spring, mass-- that if you pushed one end of it, that would compress the spring, which would give a force to the mass, which would give it an acceleration, and then it would start to move, compress the next spring, et cetera, and that, if you want to call it, a wave, the thing you moved here, would eventually go down, and you'd see one mass moving, then the other mass moving-- each one moving, and then coming back to where it was. Each one moving and coming back. And it would look like a wave travelling down.
And we gave a representation of that wave in terms of springs and masses, but we said, look. Air has spring to it. If you enclose a volume and squeeze it, it's certainly a spring. And it has mass. Like a kilogram per cubic meter. So they're all in one here.
Now we want to derive a wave equation. Wave equations are all similar in many ways. They involve two kinds of energy storage. In the case of this spring and the mass, it was energy stored in the spring, potential energy and kinetic energy in the mass. And what actually happens is energy goes in the spring, then goes out of the spring when the spring expands, goes into the mass and form a kinetic energy, goes into the next spring in terms of potential energy.
So we're going to have all of this happening in the air, in this particle. And the particle, remember, we're calling is an imaginary surface following a constant mass. Point number one in what's not obvious. If you were deriving a wave equation for the first time, had never seen a wave equation in any other discipline, just the thought of tracking a constant mass is a real creative step. You could struggle in many ways before you realized that step and did not wind up with a wave equation.
By the way, Raleigh, Lord Raleigh, 1877, 1878, I think, wrote two volumes. The Theory of Sound. And this is in a time, of course, when you only could use a flame to measure the pressure, because as the pressure wave goes by, the length of the flame changes, or you could, at most, have a moving diaphragm that would scratch on a piece of paper. No instruments to speak of. And he was excited about finding out what the velocity of sound was, and out of this came the most beautiful theory, and some of the highest theoretical treatises on acoustics that exist today. It's probably one of the best ever written, in 1877. Imagine this.
So you would never have realized that the thing to do, probably, first, was to get a little mass and track a constant mass. And that's what our particle is going to be. A very, very small, imaginary surface around the constant mass. As the sound wave comes along, the mass gets squeezed, but the pressure starts at this side, so it also gets translated. So squeeze, translate. Squeeze, translate. All it happens to it.
Now, we're going to do here with one dimension. So that means all the motion, the velocity, the displacement, and the change in density are all along one axis, we'll call the x-axis. And before we get started, let's talk a little bit about the notation.
There is a so-called reference text. The only reason that I've assigned that thing is it still is one of the best books for tables. For example, what materials on the walls absorb how much energy at what frequency, et cetera. And it's a nice thing to have later on. That's about the only reason. If I want you to look at any of it at any time specifically, I'll tell you. But the notation used in the book, they don't too well distinguish between things like instantaneous and complex amplitudes, et cetera. And so that's not going to be similar.
Now, we basically are going to be dealing with two principal variables in acoustics-- pressure and velocity. Velocity will be the particle velocity of this little tiny imaginary surface. Instantaneous quantities we'll represented by a small letter, complex amplitude is by the cap letter.
Now, the other way is velocity and complex amplitude of that. These, in general, for the one-dimensional case, these are functions of x and time, as are these. Functions of x. Complex amplitudes are never, never, never functions of time until much later on in the term, when we tell you. Special case.
In general, more dimensions, this is a function of xyz. The instantaneous functions are a function of x and time. Now we can look at the sound that's due to the sound wave, if you wish, as a perturbation on the barometric pressure, which we'll call p0. p0 is the ambient pressure plus little p. So the total pressure is this, ambient plus the part due to the sound wave.
Just to get a little idea of the relative proportions of these for sound waves that we will be dealing with-- by the way, in this subject, we'll probably restrict ourselves almost always to sound waves in the frequency range. That is typically associated with hearing 20 to 20 kilohertz. The p0 up here is about 10 to the fifth newtons per square meter.
The threshold of hearing, measured at about 1,000 hertz, threshold of hearing for the human being, is about 2 times 10 to the minus fifth threshold of hearing. 2 times 10 to the minus fifth newtons per meter squared. And the threshold of pain for the human is roughly 120 dB higher than the threshold of hearing. 120 dB is what? 20 log to the base 10 times the pressure ratio. So that means 10 to the 120 dB is 10 to the sixth greater. So the threshold of pain is about 20 newtons per meter squared.
So the loudest sound that I'm sure you have all heard from concerts, not acoustical instruments, is only 20 newtons per meter squared, compared to 10 to the fifth. So if I were to draw the perturbations of the sound wave on the ambient pressure, you wouldn't see the line. 10 to the minus fifth, roughly, the difference between even the loudest sound and the ambient pressure.
So you can think of the ear as an extremely sensitive barometer. That's basically the way you can look at it.
Now the density of this little particle, total density, this is total pressure, is ambient plus the other one. Total density would be the density in the absence of a sound wave plus the part that's due to the sound wave. And another thing that changes, of course, is the volume. If the density changes, remember, you're tracking a volume of constant mass. So as a sound wave goes by, pressure increases. Guess what? The volume decreases. So the volume of this particle-- volume total-- is the volume in the absence of the sound wave, plus tau. Unbelievable.
I'm using that only because it's used in a lot of textbooks, and it's a symbol, and believe it or not, after Thursday, it's going to disappear and never reappear, I believe. So it'll be eliminated very quickly. But I use it only because you see funny things like that-- I mean, you would expect it to be dv.
One of the problems, if you want to call it a problem, that we'll have in this subject, is we get involved in mechanical disciplines, electrical disciplines, acoustical, magnetic, and all symbols of the alphabet are used. And so for velocity, thank god, most of the acoustical people use a u anyway, and it doesn't confuse with voltage, which we will use for a V.
OK. So these are the symbols. Now let's see what we can do with our one-dimensional model. Imagine that you have here our little particle, dx long, da area, cross section area. This point here could be x plus dx, and this, we'll say, is x. This has a density rho-- well, before a sound wave comes along, it's rho 0. But rho 0 times the density times the volume is the mass. And that's a constant. So that's actually equal to rho times v-- the rho total times the v total. Because as you squeeze it, the density, of course goes up, the volume goes down, but the mass, which is the density times the volume, has to remain a constant for this thing. That's the definition of what we're looking at.
Now, no motion this way at all. So I don't have to worry about anything here. There's no differential pressure here and here. This is a plane wave created by a huge wall, and it's just moving this way. No differential pressure, no motion in the other direction but the x direction.
So now we know that this thing has a mass. We know that it has a compliance. What do you think you would do if you were trying, for the very first time, to write a wave equation? You know-- let me tell you, let me say that you would know that these wave equations are some sort of differential equation. You know that the body has mass. What kind of physical law or constraint would you bring into it?
There's mass inside there. There's going to be some sort of motion. Now, don't make me as who's buried in Grant's tomb. Yeah?
SPEAKER 1: [INAUDIBLE]
DR. AMAR G. BOSE: That's a darn good one. You're involving motion and a force. Let's write Newton's Law for this little thing. p of x plus dx. The pressure at this end minus p of x, the pressure at that end. If I took all of that and I multiplied by da, guess what? That's the net force on this. The pressure on this side minus the pressure on that side. A net force should be equal to ddt by Newton, ddt of the mass of that particle and the velocity. But the mass is a constant. It's rho 0 v0. So that's equal to ddt of-- well, I can bring the rho 0 v0 out. Rho 0 v0 ddt of u, velocity. That's what Newton's Law has to say.
Now, I said one more thing. That we know we're hunting, ultimately, for differential equations. This thing reminds you of something pretty close to a differential equation, if you just made a minor modification of it. What does this expression look like, almost? Something like dp dx. Except remember, pressure is a function of x and t, and you've only incremented an x here, so it looks like a partial. And the only thing we have to do to do that is put down here a dx. And of course, not to change the side, put up there one.
And so when you know you're hunting for a differential equation, and you see something this close, grab it, for god's sake, and recognize that I've got something here. All right?
Let's take that. And that is partial of p, and the limit partial of p with respect to x da dx is the total volume. Times vt. And that's equal to ddt-- oops, rho 0 v0 ddt of u.
Now, total derivative-- u, again, is a function of x and t. So this, I hope you know or have seen and can review from calculus, that ddt, the total derivative with respect to u, when u is a function of not only t, but another variable, can be expressed as partial of u with respect to t plus partial of u with respect to x dx dt.
Now, if you got just this far, and you'd had any experience at all before, you would say, oh my god, I see something happening here. Namely, dx dt is you. And this is a product of two different terms in u. What kind of an equation does that make this whole thing? The world, in math, an equation is divided into two types-- non-linear. You better believe it. When you have products of the variables, or squares, or powers of the variables-- so you go, oh my god. If I keep this stuff, I've got a non-linear equation. And of you who have had any experience in that know that what you can do with it is very, very limited, and only in special cases. And we'll see why it's so easy to work in the world of linear as we go through the subject.
So you would love to get rid of this darn thing. If this were the first thing, if you were deriving this so-called wave equation which we're going to get to for the first time, you would take one look at that and you'd say, gee. I hope I can neglect it. Now you don't know. You have no idea. But you do know that if you keep it, you're going to get a pile of equations that if you can do anything with, you're lucky. You can, it turns out. But boy, that's a different field.
So what you would do-- the vast, the world of things we work with is linear, and you pray that it's linear. You look at the curves that we had up here, and you say, hey. Very little things are happening with that sound wave. Maybe I'm going to be lucky. Maybe these quantities of velocity and whatnot-- these things might be small. Maybe I'll be dealing with low enough frequencies.
So what I would do if I were doing it the first time, I would say-- I'd put a little note here. Neglect and verify. So I'm going to neglect it, and you're going to verify it in the homework. Namely, we're going to take away this term, and we're going to go and get a model, which no longer represents the real world. Because that is what the math told us from the simple model. We're going to get a solution to this model, and then we're going to go back and say, if I apply that solution to this equation, could I have neglected this? And lo and behold, it'll turn out you could have, for all the things that we will be dealing with. If you go out of the bounds of this and you get into shock waves and whatnot, you're going to have to get to nonlinear things. But that's how you would do it the first time through, is hope to god that the linear theory's going to work. Neglect it, but make a note to verify.
So finally we got out of this expression. Partial of p with respect to x is equal to rho 0 v0 over v total times partial of u with respect to t. Now let's see-- oh, ho, ho. I have done something incorrectly.
Now, take a look at this. Which way is this force? Assuming that the partial is positive, which means this term is greater than this term, which way is the net force on this? Yeah?
SPEAKER 2: [INAUDIBLE] to the left.
DR. AMAR G. BOSE: Exactly. If this pressure is higher than this, which makes this gradient, as indicated here, positive, the net force is that way. The coordinate system is positive to the right. So it's minus that gives you the mass times acceleration, which is defined in the positive direction of the coordinate system. So there's a minus sign sticking out here. Let's put it over here, then.
OK. v total. Now v total is v0, remember, the volume in the absence of a sound wave, plus the little tiny change that comes with the sound wave. But this is a multiplier on the whole right hand side of the equation. So if I just replace that and say that vt is approximately equal to v0, I make an error in this side, which is only the error between the change in volume due to the sound wave and the total volume.
If there were more terms-- I mean, suppose I had other terms down here. vt, some other things. Well, let's take a worst case. vt minus v0 down here. And I said, well, I'll let vtv equal v0. Suppose I had this in the numerator. Then I can't do that. Obviously, this becomes infinite then, and the little tiny term is very valuable.
So when it's a multiplier on the whole thing, you only introduce a percentage error equal to the percentage error the approximation between vt and v0. As long as there aren't any subtractive terms or other terms in there. So I can say, this is approximately equal to v0.
And finally, we can then write, partial of p with respect to x is minus rho 0 times partial of u with respect to t. That is what we got out of Newton's Law. Call that 1. And when I make a box, it's more or less so that I can see it to get and find it on the board up here, being this close to it.
Don't think that any boxes are to be remembered. There are certain things in the subject if you remember just because you use them. Your parents never told you to memorize your name, but you all more or less can remember it. That will happen with certain things. Not this one, but it will happen with many things. But don't think boxes are important because they will be useful in solving particular things.
That takes care of everything we can get out of Newton's Law applied to that. Now, the thing is also that that particle that we're following, of constant mass, is also compressible. What do you think, we ought to bring in some physics on that? Any ideas?
SPEAKER 3: PV equals nRT?
DR. AMAR G. BOSE: PV equals nRT. It's a relationship between pressure and volume, and it all starts from that. That's right. But the relationship between something that I squeeze in terms of the volume change and the pressure isn't always the same. Depends on how I squeeze it.
You remember this from--? You had it. Guaranteed. Yeah?
SPEAKER 4: Is it a Hooke's law, compressible spring?
DR. AMAR G. BOSE: Hooke's law is something that, when we're all done, we're going to have an equivalent of Hooke's Law in f equals kx, if you want, for this thing. That's right. The real problem is, what's k? k is different if you compress it in two different ways. Does anybody remember this from physics at all?
SPEAKER 5: Adiabatic versus isothermal?
DR. AMAR G. BOSE: Adiabatic versus isothermal, exactly. Isothermal is the one that means constant temperature. If you had a jug with a disc on top, and you pushed it down, and it was a nice metal can, so conductivity to the outside was very good, and you pushed it very slowly down, it took a day for you to move it an inch, you could say that the temperature in that gas was always constant. If you took that plunger and went like that, turns out that the temperature would rise. You've compressed the gas, you've increased the pressure in it.
And turns out-- again, you'll see this on the homework. The diffusion, the time for the temperature to equalize-- in other words, it's become hot. Now how long does it take for that to flow out? Is slow, compared to the speed with which you compress it, then it's adiabatic. And that's exactly, as you will see from your homework, what happens in a sound wave. Sound waves going much faster than you could compress any piston in a jug.
And so as the sound wave goes by, remember, we said that an area becomes compressed. Well, the area just adjacent to it is now expanded, just like that spring model with the springs in the masses. When this spring is compressed, the previous one is already expanded. So what would happen, is you'd have a hot temperature here and a cool temperature here. If that temperature could equalize very fast, if the thermal diffusion was fast enough, then you would have just one temperature for the whole thing, because this would be down, this would be up in temperature, where it's compressed. And then a second later, of course, it all changes again.
If the thermal diffusion was fast enough, compressions and sound wouldn't be adiabatic. They wouldn't be adiabatic. They'd be isothermal. Well, it isn't fast enough, as you'll see, and the compressions that go on with sound waves are, in fact, adiabatic. And again, from your physics, you may remember that for adiabatic compressions, p, v, total pressure-- I'm going to use the Ts for a moment. These are all going to disappear. The sub Ts are going to go away within a day or so. pv to the gamma is a constant, where gamma-- I don't expect you to remember this, but I think you should remember where to look it up. Gamma is the ratio, specific heat at constant pressure to specific heat at constant volume. And for air, which we're going to be dealing with, air, gamma is approximately 1.4.
So it turns out that an adiabatic compression is a little stiffer than one that's isothermal. When it's adiabatic for a given volume change, you've got to apply more pressure, as you can see from the gamma being positive. So that's the law that's going to govern the compressibility, if you wish, of this gas, and let's apply it to the little volume of gas. Or to, in fact, you don't even need to do any geometry to go after this one, because it's independent of geometry. This thing has no geometry in it. This is a scalar here.
Pressure. Let's take just the differential on both sides. I'm looking for a differential equation. It's a good idea to try this. dPT times the product, vT to the gamma, plus pT, gamma-- taking the differential of that second term-- gamma vT to the gamma minus 1 dvT is equal to 0.
OK. Now let's get all the things separated over here. I won't even bother separating them. Let's see. dPt is equal-- I'll take all the rest over to the other side. Minus pT gamma over vT, because you have downstairs, vT to the gamma, upstairs, vT to the gamma minus 1, and dvT.
Now remember, vT was a constant plus this tau. p was a constant plus that. So dpT I can write as dp equals minus-- pT, this is a multiplier on the whole side, I can write that as p0 with negligible error in this. I mean, I could put approximate signs here, but what I'm actually doing is I'm creating a model, and in the model, they won't be approximations. The model, it's real. They're approximations of the real thing. But since I'm going to be dealing with a model after now, I'm not going to bother with the approximations. I'm just going to say it's equal to that. Gamma v-- total, again, multiplier on the whole term with no subtraction terms on it, the whole side. So I can write that as v0. And dvT up here is just top.
dvT is d tau, sorry. I'm going to use more space than I wanted to here.
Now, if I wanted to take the derivative on both sides with respect to T, I could write here, dp dt is equal to minus rho 0 gamma over v0 d tau dt. And that's what I get out of the Gas Law. I'll just put that down as number 2.
Now there's only two things that happened here. We have potential energy getting squeezed. We have kinetic energy motion. But look what we've done. We've introduced more than two variables with two equations. We have pressure, we have velocity, and now, darn it, we have a tau in here. Which is relating volume.
But there's another equation, which is, in acoustics, called continuity equation. It's really a bookkeeping type of thing. And we'll call it what they call it-- continuity equation. If you look at that little element of link dx and here at this end, x plus dx, you have a velocity of-- the velocity is u equals u of x plus dx, and the velocity at this end is u of x.
Well, think about it. Velocity-- and this is an area da, same little particle-- if this thing is going faster than this-- let's say they're both going in this direction. If this side is going faster than this, guess what? The volume is getting bigger. So you can relate the variable velocity to the tau, the change in volume. And now you'll have three equations, three unknowns, and bang, they're all set up to produce the wave equation.
So u of x plus dx minus u of x times da, the area, times dt. Velocity times this little bit of time is the distance. The distance times the area is the increase in the volume, dvT, which is equal to d tau.
So that, again, you know you're hunting for a differential equation. You have something that's looking like the very definition of it, so you grab it, and right away, realize you've incremented a variable that's a function of time and pressure, but you've incremented it in time and space. You've incremented in space, so put down a dx here, a dx over here. And you've now identified a partial derivative, partial of u with respect to x is equal to-- well, let's see.
dt I can take down over here, and I'll have d tau dt. Now I have left da dx. Well, da dx, now I'm going to go a little faster, because we went through it before. da dx is the total volume, but the total volume, when it's sticking in a term, multiplying a whole side, could be approximated by the volume in the absence of a sound wave. So 1/v0. If that's too fast-- we went through it before, but if you want to slow down on that one, I'll do it.
OK. So this fellow came from the Gas Law. Three equations. If you solve these three equations together, out comes the wave equation. Don't want to solve the three equations together yet. What I want to do is go and derive the-- well, let me just think. Maybe I ought to--
Six of one, half dozen of the other.
I think I'm going to take a detour, and before I get the one-dimensional wave equation, I'm going to derive these equations in vector form. Now, hang on. I know a lot of you are thinking, oh my god. I told you we're not going to use vector form. Maybe once or twice in the homework or something.
But why would anyone, whether we're going to-- what's the fundamental reason that anyone would bother with a vector equation? Let's say a vector wave equation? In any discipline? Any idea? Yeah?
SPEAKER 6: Because the real world is three-dimensional?
DR. AMAR G. BOSE: Yeah, I mean, that's true. But problem is, I could give you a three-dimensional equation here just as well, An xyz kind of thing. So it couldn't be for that reason, because I could do that very well. Yeah?
SPEAKER 7: Can you use linear algebra to solve equations?
DR. AMAR G. BOSE: Linear algebra, I can also use linear algebra to solve [UNINTELLIGIBLE] differential equations here. So the answer is yes, but that's not the real reason.
SPEAKER 8: Reverse chalk dust?
DR. AMAR G. BOSE: Great, you people do it just for that? Only professors when they're [INAUDIBLE]. Any ideas? Vector theory? You studied-- you had some vectors in math, some of these like Dell notations and things like that. How many have been exposed?
Ah, OK. That's interesting. If you're unexposed to it, and most of you were exposed to it, and you don't know-- I'm not saying it's your fault, but don't know why. There is an extremely fundamental resource. I'll start the sentence. Vector equations are independent--
SPEAKER 9: [INAUDIBLE]?
SPEAKER 10: Of coordinate systems?
DR. AMAR G. BOSE: Of coordinate systems! That is the reason. In other words, if I have a vector equation for any physical process, and I'm one of the oddball 60 or whatever there are coordinate systems, all I have to do is go to a tabulation of this, and get it in this elliptical spheroidal, or whatever I want. And there it is. I have all the equations.
Now, you've all have the experience in calculus at some point of trying to take some expression in rectangular coordinates and put it in polar coordinates. And you were probably extremely lucky if you got the right answer the first time, with signs and all this and that. Just going from two-dimensional rectangular to polar is a situation which you can easily, if you're like me, come out with the wrong answer. Now you can go to any coordinate system you want, once you have a vector equation.
And so the real reason we're going to do vector equations right now is that you see two things. One is a very interesting thing in the physical process, that if you have done a one-dimensional, you can very easily go to the vector, and from the vector, you have all dimensions and all coordinate systems. So looking exactly at what you've done here, the vector equation almost hits you in the face. And then you have this incredible power that you can go to any other one. And in disciplines that may come up in the future, in your own careers, think about that. Don't try to jump to a different discipline, and say, oh my god, vectors. How do I-- no. Do it in one dimension. Look at the one dimension as we're going to do, and you'll see it'll come out in the vector.
Now let's do it. Let's see. I want to save most of this. Vector equation. Let's look at this one. Carefully look. What we have done here for this one dimension is really multiply a pressure times this area. The pressure that's in here, the pressure that is on the inner surface of this with the dA.
Now, the reason that this is a different sign from that, is that dA would be going the other way if you assigned this to be a vector that's normal to the surface. So what this expression is, is really a one dimensional expression for the integral of pdA. If I had took an expression like this, the integral of pdA-- pressure is a scalar in here-- and I took it over a-- now, you know, it's no longer a one-dimensional situation. It's just some blob, a tiny blob that's our particle, if you wish. We do this integral, integrate it over this surface of that particle, reduces exactly to this expression in one dimension. p times dA plus p times dA. I didn't bother to integrate it over all of this surface, because it's all the same, and it will all cancel out. Whatever was going this direction and that direction is going to be the same, because there's no pressure variation this way. So the most important thing is to realize, oh my God, generalize on what you have in one dimension, what is it here, and go back and check that in one dimension, this exactly reduces to that. pdA.
So pressure times area. That's the force. Net force. God knows which way it's going to be. I mean, it might wind up over here like that. And that should be then, by Newton's Law, ddt of mv. Mass is rho 0 v0, which is a constant. It's really exactly the same as I did over here, before it's the rho t vt. But that can be approximated, rho 0 v0. Times du dt. But the only trouble is now that du is a vector.
That is the statement of Newton's Law in an integral form. Now, I'm not looking for integral equations. Don't like them. Much harder to play around with than the differential equations. Want to get a differential equation out of this. This side looks like it's not going to be a problem. This side, big problem, it looks like. Anybody know how to get that integral out of there and get it into something that involves derivatives?
SPEAKER 10: [INAUDIBLE]
SPEAKER 11: Green's Theorem?
DR. AMAR G. BOSE: Green's Theorem? There is a more elementary way of thinking about it. I'll give you a last--
SPEAKER 12: Differentiating the whole equation?
DR. AMAR G. BOSE: Differentiating the whole equation? Yeah, now, the trouble is-- I have a hard time, because I don't know what this pressure is all over that area. Let me give you one hint. This time, I won't give your first word. I'll give you the last word. Law. Not the kind that there are too many practitioners of it. But come on. Law, law, law. Physics. Sophomore year. [UNINTELLIGIBLE]
Gauss' Law. Now, I don't give a darn if you remember what Gauss' Law is. But the interesting thing is that it's a way of getting this integral into its derivative right away. What it says is that the integral of a scalar over a surface is the integral of the gradient of the scalar over a volume. The integral of a scalar over the surface of a volume is equal to the integral of the gradient of the scalar over the volume. So this becomes the integral over the volume, which in our case is the total volume of that little thing here, this little particle that we're looking at. A gradient of p dot dA. Sorry. dv.
In other words, I take the gradient of this scalar-- this whole integral-- this integral, by the way, is a vector, right? It's a vector times a scalar. This is also a vector. This is a scalar, this is a vector. I take the gradient, and I integrate it over the volume, and I get this by Gauss' Law.
OK. Now why do I want to trade one integral for another? Because when I shrink the particle, this integral is easy to evaluate. When I shrink the particle, the integral is the integrand, gradient of p. And that's the reason. Otherwise you wouldn't want to trade one integral for another one.
So gradient of p is equal to rho 0 v0 ddt of u. Uh oh. ddt of u. That was the one that gave us this problem before. Where was it? This thing. When we tried to get the total derivative of u, we wound up with something like this.
And now we have a worse problem. It's a ddt of a vector. I do not expect anyone either to have remembered it, or seen it, maybe. So I'm going to tell you what it is. It's partial of u, the vector, with respect to t, plus u dot a funny-looking thing-- u dot del operating on u.
This, of course, is u and a u, and it's a non-linear term in which we're going to say, neglect the darn thing and verify. And it's going to be the same-- I'll say a few more words about it first.
Let's let shrink this fellow here now. Shrink the volume down. I have gradient of p times the volume. Gradient of p times vt. I should write not as a capital. vt is equal to-- aha! I did the same thing twice now. Remember, I forgot the minus sign? This, I pointed, reduced exactly to this in one dimension. So the minus sign had to be there. So minus, minus, minus, I'll stick it in here, vT, rho 0 v0. Is it too much in one step to say, you know this is multiplying all one side, and so this is going to be v0?
I better not do it in one step. Here we go. Rho 0 v0 times-- now, neglect and verify. Can't verify until you get to the bitter end, and then plug in the kinds of things you're dealing with, and see if that's a good approximation. So this becomes partial of u with respect to T.
OK. So finally, we get gradient of p is equal to minus-- there's a v0 on both sides. Minus rho 0, partial of the vector u with respect to t. And that is the vector Newton's Equation. Call it 1a or something. If we called the other one 1, that's the equivalent.
And of course, again, always check back. Look, what's the gradient in one-dimensional rectangular coordinates? It's a partial of the variable with respect to x, yes.
SPEAKER 13: What happens with the ddt in the first line?
DR. AMAR G. BOSE: First line. Here.
SPEAKER 13: [INAUDIBLE] the ddt in the front?
DR. AMAR G. BOSE: Oh, oh, oh! Duh. [UNINTELLIGIBLE]. That's what happened to it. I did it twice. So to go from here down to here, you take this derivative. Let's write it correctly. It's the derivative of mass times velocity of the product. Thank you.
So very similar gradient, and then at the final check is that gradient of p is just partial of p with respect to x in rectangular coordinates. In one dimension.
OK. Next. What did we do? We did the Gas Law second. OK. Gas Law-- let me just see how that looks. Adiabatic compression. Now, having seen this once, I know that everyone in this room is bright enough that you can give me the vector equation straight off. Not even any mental computations for this one. Where is it? Here it is. This is what it was in one dimension. Anybody want to give it to me? No? I mean, I won't give you any time, because you can think in time, and you don't need to think.
SPEAKER 14: It shouldn't change.
DR. AMAR G. BOSE: Shouldn't change. What's the fundamental reason?
SPEAKER 14: Differentiating the time that it takes.
DR. AMAR G. BOSE: Sure. There's no spacial derivatives in this thing at all. It's not a function [INAUDIBLE]. Done. Continuity equation, we have it. 2 and, we'll call it 2a.
We have only one left, and that's the bookkeeping thing. Continuity equation. Let's just do it parallel, right here. Look at this one-dimensional case. Again, just go back to your one-dimensional case to get the vector. Look at this expression here. What is it? It's a special case of the integral of u, which is now going to be a vector, dot dA. That's exactly what this much of it is. So if I had over the surface s, u dot dA both being a vector, I would get exactly what's in the circle here in one dimension. Because I've multiplied this u, which was in this direction. I've dot producted it with the vector out here, and the vector going back here [INAUDIBLE] in the negative direction. So that over the surface is exactly what we have over there.
And that, then, must be equal to d. Let's see. Hold on a second. u dot dA. dt times dt-- well, I'll take that over to the other side-- is equal to d tau dt.
Now, u dot dA. A vector integrated over the surface of a body-- there is a theorem, a law-- it's called [INAUDIBLE]-- that gets that into a volume integral. And then the reason we want to go again to the volume integral is because, aha, you shrink the particle, as we're happy to do. And then it's only the integrand that's left, and that's in terms of derivatives.
So does anybody remember the theorem that gets you from here to a volume integral? Divergence Theorem. Yep. I mean, I'm happy that somebody remembered it, but. Divergence Theorem says, if you have a vector integrated over the surface, it's equal to the divergence of the vector integrated over the area.
Now, let's integrate over the volume. Divergence of a vector is a scalar. The dot product of the vector with this is a scalar, and it's d tau dt. So that goes into this, and finally, when we shrink that volume, then, this becomes divergence of u times v total is equal to d tau dt. Or divergence of u is equal to-- how did we express that thing? We just took the v over to the other side. I'm going to use v0 for vt. It's multiplying the whole side. 1/v0 times d tau dt.
So final check. What's divergence of u? This del operator, in rectangular coordinates, is i partial with respect to x plus j partial with respect to y, et cetera. So what is this in one dimension? It's partial of with respect-- this divergence comes out to be partial in one dimension of u with respect to x. So it's there.
You see that the difference in the vector equations-- it's really interesting. The difference in the vector equations and the one-dimensional are pretty small, in the end. Here, look at this. Gradient replaces that continuity-- I mean, the Gas Law, of course, is the same. And divergence replaces partial of u with respect to x, when u becomes a vector.
So now we have all the equations in no time. Let's see. I think what I'll do is the following thing. I will derive the one-dimensional wave equation, and then I'll try to get here a little bit early, and I'll put it on the board ahead of time. Because in deriving the vector one, it's exactly the same. You keep one eye on the one-dimensional, and you can go right down the line and get the vector. So I want you to see both on the board at the same time, which I had hoped to get to today. I won't, but I think it's worth going through just the one of them.
OK. One-dimensional. We have this fellow. We have this one and this to solve. We have three variables-- tau, u, and pressure. Now, I'm going to get a wave equation. I'm going to do the easy one. I'm going to get a wave equation in pressure. You can get the wave equation in velocity. One-dimensional.
So I want to get rid of tau from these equations, and I want to get rid of u. When that's done, I have an equation in terms of pressure that is going to be the wave equation. So let's see if I have them all on the board there. I can start right here.
Getting rid of tau. Let's see. I have it in this equation, too, and I have it in 3. Very easy. I have d tau dt. It's all set up. So let me take the first equation and substitute for d tau dt, this value I've gotten from here. So I get partial of p with respect to t. This is from 2 and 3. Partial of p with respect to t-- there's a minus sign here-- is equal to minus rho 0 gamma over v0. d tau dt is v0 times partial of u with respect to x.
OK. Now I have an equation in just p and u, and look, Newton's Law is an equation just in p and u. All I have to do is get rid of u. So we substitute again. Uh oh. u is partial with respect to x here, and it's partial with respect to t here. How do I get rid of u? Somebody?
Differentiate again. When in doubt, differentiate. Let's see. What would you like to differentiate this with respect to? x. So I get second partial of u with respect to x and t. If I differ it with this one? t. I get second partial of u with-- oops! But the order is different.
It turns out, as long as the mixed partial derivatives are continuous and in the real world they are, that you can interchange the order. And knowing that, you can do exactly what was said.
So to get rid of u, I will take this fellow with respect to t again. So I get second partial of p with respect to t is equal to minus rho 0 gamma, the second partial of u, but it's respect to t now, with respect to x and then t. Put t afterwards.
But that is equivalent to minus rho 0 gamma second partial of u with respect to-- the order reversed. If the mixed partials are continuous, and they are.
OK. That gets us into something which, if we do what you just told me on this-- oops. Where the heck is it? On this equation, we will be able to equate the second mixed partial with the one over there. So I will take the partial of this with respect to x. So the second partial of p with respect to x is equal to minus rho 0. Second partial of u with respect to x and respect to t. Rho 0.
Now let's just make sure I did everything OK. Looks all right.
Now I have this and this to equate, and bang, out goes you. So let's see what we are left with. Second partial of p with respect to t, which is here. Or I'll take the other one first, because I think the way you usually see a wave equation is this way. Take this, substitute in here this quantity that was derived from above. Let's see what happens. Minus rho 0, and there's a minus up here. Let's see. That's a plus sign. Rho 0-- well, I'll put it down. Rho 0, and I'm going to plug in for this expression over here, and I get rho 0. We have lost something. I don't know what it is yet, but-- second partial of p with respect to t squared. And there's a gamma down here.
Now, we have lost something-- this is very close to the wave equation, but this is wrong.
SPEAKER 15: [INAUDIBLE] Up there in equation 2, you substituted a rho for p.
DR. AMAR G. BOSE: Aha. That'll do it. Equation 2--
SPEAKER 15: It's supposed to be p.
DR. AMAR G. BOSE: This was supposed to be p0?
SPEAKER 15: Yeah.
DR. AMAR G. BOSE: That is a common mistake for me, all the time. In other words-- let's see now. Gamma-- that came from here-- dp dc-- yeah, p0, gamma over-- thank you. I would have been here for much longer than 12:30 to figure that one out.
OK. So let's get that straightened out here. Ta da, ta da, ta da-- which, yeah. This should be p0. Oh, and this is a rho. OK. so let's now see. This is p0, and that's it. That is the one-dimensional wave equation in pressure.
Now what I will try to do is put these steps on the board again at the beginning, and then we're going to do the vector solution for a wave equation. And once you have that, you've got the wave equation in any dimension. Now, please, it's easy, maybe, to follow the steps, but review them from your notes, and just make sure you really understand them. Good.