Amar G. Bose: 6.312 Lecture 07
AMAR BOSE: I just looked at some of the homework, want to congratulate you. Just flipped through it, there's a big improvement in the neatness, and that really helps the people who are grading, and really helps you, because then they can follow through and see where you made just a numerical error or a thought error, and help you on it. Thanks a lot.
What I think I'm going to do here this start with this, because I really don't know how long it's going to take. And then if we get time, we'll put some more of the networks in, this is a little bit better way of looking at it, on the board. Simple sources. In network theory up until now, you've seen that the impulse has a big role to play. If you put an impulse, something that is very narrow and very high, into a network-- very narrow compared to the time in which the response of the network is varying-- you find out that that's a very useful tool.
And the reason it's useful is really that if you had any x of t to put into the network, and get out some-- and this has some impulse response, h of t. If you had any x of t, you could always break that into small pulses, narrow pulses. And if you knew the response to one by super position because it's a linear system, you get the response to all. So the output of a system can be regarded as a sum of impulse responses to all of these impulses that make up x of t.
Now in space, there is an equivalent. In acoustics, we talk about it as a pulsating source, a little point source, the limit of a sphere as it goes down, a sphere who is pulsating with time. And if you know the response somewhere out here to this little pulsating sphere, then you would know the response to a sum of pulsating spheres, which might be, as you're going to see, an array which you can cause to broadcast in a certain direction much more than it broadcasts in another direction.
Also if you know just the response to that simple little pulsating sphere at some distance, if you had an irregular object here that was reasonably small compared to the wavelength, or compared to lambda over 2 pi, you're going to be able to use that by superposition, and make up this object as a sum of all those little pulsating spheres. So it's very similar to what you do in time domain in networks, and this pulsating sphere has a big role to play.
So let's look at this. We'll say it has a radius of r0, and we're later going to shrink that radius down to be appropriately small, which we'll talk about. We would like to get way out here, some very far away from this, the pressure resulting from the pulsation of this. So we know, you know, you've derived on your homework, that p times r, pressure time r for the spherically symmetric case, the p times r satisfies the same wave equation that x satisfies for a one-dimensional wave.
So as a result, p of r and t can be expressed as p plus-- of course, that's a function of frequency, but I don't think I'm going to bother writing it all the time, you know that now. I mean, can be a function of frequency. It doesn't have to be, it can wind up as a constant. p plus over r e to the j omega t minus r over c. p plus because it's the positive going wave.
So this looks exactly like what you would see if r were replaced by x and you had a plane wave. You have a spherical wave, it goes down. And by the way, though we haven't talked about energy yet, whenever you see something radiating like this, you realize that out here, the same energy is going through a sphere that's this large that came out of that one. And so an energy, I've told you, is going to be related to pressure squared.
Well, the area of a sphere as it goes out is related to the radius squared. And so the pressure has to drop off. You know that the pressure out here is going to be less than the pressure there, and it has to drop off as 1 over r squared. So that p squared drops off as-- pressure has to drop off as 1 over r so that p squared drops off as 1 over r squared. So whenever you see something spherical, watch out.
OK, now we know that velocity satisfies-- we can go back to the wave equation for it-- but you know that velocity satisfies the same kind of partial linear equation, and so it has to have the form for the spherically symmetric case, u of r and t is equal to u of r and omega. e to the j omega-- well, e to the j omega t. This, by the way, could be written, if I take the r out here, e to the j omega minus j omega rc, that of course is p of r and omega this times e to the minus j omega r over c. And this is similarly, we know, partial differential equation, uj omega t general solution, and some complex function of space, in this case just r, and omega.
OK, so we would like to find-- if we can find this, we have the pressure for all r. We'd like to find this, ultimately, in terms of the velocity or something very close to that, that will define of this surface.
OK, so Newton's Law relates it to radiant of p is minus rho 0 partial of u with respect to t. So let's just plug that in and see what we get. Radiant of p. Well, in spherical coordinates, in this spherically symmetric case, radiant of p is equal to a partial of p with respect to r. So let's take the partial of p here, of this, with respect to r. Well, let me, so that I don't have to fool around inside here, let me pull the r all out. p or r and t is equal to p plus over r e to the minus j omega r over c, e to the j omega t.
Now, partial of this thing with respect to r is a product of two functions of r. So first one would give me minus p plus over r squared e to the minus j omega r, e to the j omega t. And that's one of the terms. I have a product, so the second one is this thing times-- whoops, it's not a magnitude-- is this times the partial of that with respect to r, which is minus j omega over c. And then the whole function over r e to the minus j omega r over c, e to the j omega t.
All of that, by Newton's Law, must be minus rho 0 partial of u-- I'm sorry, yes-- partial of u with respect to t. Here's u, partial of u with respect to t, this is a constant out here, drops out of j omega. j omega u of r and omega e to the j omega t. u to the j omega t's are gone. And I guess can now factor out the p plus, try to get an expression for that. See, there are minus signs everywhere, so my result will be all plus signs in that case. So p plus-- yes?
SPEAKER 1: [INAUDIBLE] minus p plus over r squared [UNINTELLIGIBLE] e to the minus j omega r over c?
AMAR BOSE: Over c, thank you. OK, thanks. Now p plus out. I have left here 1 over r squared. I'll factor e to the minus j omega r over c out. I get plus j omega over cr. So far, let's see. e to the minus j omega r over c equals j omega rho 0 u of r and omega.
OK, let's make it one step simpler yet. Well, let's not. It can work here. I'm going to define something that we're going to see again and again in acoustics called volume velocity, which is nothing more than the particle velocity times the area through which the particles are moving, definition. Volume velocity is particle velocity u times the area through which the particles are moving.
So volume velocity for Vv of, in this case-- I'll use this symbol for the complex amplitude of volume velocity, just a sub-v to remember it's volume velocity. Volume velocity of r and omega is equal to, in this case for the sphere, 4 pi r squared. That's the area of the sphere at any point r times the particle velocity. Any questions? OK. So we can use this in here, and I will get on the right hand side j omega rho 0, substitute for this, volume velocity or r and omega over 4 pi r squared. OK.
Now expression for p plus-- remember, p plus is really what I want, because when I get p plus I'm done. I can express the whole of the pressure anywhere out here in space as a function of the volume velocity in space, and then I'm going to shrink down and say, I want to get p plus as a function of the volume velocity at the surface of this tiny sphere. Why I want to do this is because it's going to turn out that that's the whole thing that dictates-- that's all I have to know about this sphere, what's its volume velocity, to get the pressure anywhere else.
Now that's p plus then is-- if we can get all this stuff straight-- j omega rho 0 volume velocity of r and omega, OK, over 4 pi r squared. That takes care of that side. Now I have to bring this stuff down underneath. So I'll take out a 1 over r squared here, 1 over r squared I'll factor out. 1 plus j omega r over c. Oh, there's a fellow here that goes over as e to the j omega r over c.
Now remember that omega over c is-- and you should always, not remember this, but know how to derive it-- 2 pi over lambda. So I can express omega over c as 2 pi over lambda. And why do I want to do that? Because I want to get a feeling for r relative to j omega-- I mean relative to omega over c. OK, so this goes out, this goes out, and we finally have p plus is equal to j omega rho zero volume velocity of r and omega over 4 pi times, 1 plus omega over c is 2 pi over lambda, j times 2 pi over lambda times r. Whoops. All times e to the j omega over c times r. Yes?
SPEAKER 2: [UNINTELLIGIBLE PHRASE] lowercase r's look like capital v's.
AMAR BOSE: Capital v's? Yes, I will try to do that. Good. And when I make the capital v, what I will try to do is put tails on it. OK, let's see, have I done it? Yeah, I think that's it. Now that's OK for solving for p plus in terms of the volume velocity at any point r. Now what I want to do, and I have to go very slowly over this, because if I don't get it across correctly, it'll be very confusing why I'm shrinking down to r0, and after I get that, I'm going to an expression for p at any point r.
You see, this expression is for p plus. When I have p plus from the top equation here, I have pressure everywhere. But I want p plus in terms of the volume velocity at the surface of the sphere of radius r0. And I'm going to now put on the additional condition that r0, the radius of the sphere, is much, much less than, guess what, lambda over 2 pi. The sphere is much, much less than the wavelength of what your radius divided by 2 pi.
And let's, then-- when I put that assumption in, this will be able to be simplified, and low and behold, I will have p plus in terms of the volume velocity of, not some point out at r, but r0, the surface of the sphere. So that volume velocity belongs to the sphere.
OK, so we're going to let r go to r0. r's going to go to r0, but r0 is much, much less than lambda over 2 pi. So downstairs here, this imaginary part, this is a vector-- always draw pictures if you have a problem-- this is a vector, and the imaginary part is going to be very, very small compared to the vector, so the vector is approximately unity.
Got rid of this complex thing down here. So this fellow is going to go out, neglect it because of this condition. Oh, I should have put that in, 2 pi over lambda for omega over c. Let me do it. j 2 pi over lambda times r. Now this r becomes r0. I'm looking at the surface of the sphere.
And so this whole exponent is very, very small. e to the j something is a vector that is 1 unit long and at a small angle. If in fact the angle is very, very much less than 1, then the whole thing can be substituted by unity. So that becomes unity, downstairs becomes unity, and we finally have p plus is equal to j omega rho 0 over 4 pi times volume velocity of the source. Oh, but I'll write it this way, volume velocity of r0 and omega, but this is Vv of the source s.
So go back to the expression for pressure that we had at the beginning. We now have p or r and t anywhere. Now watch out. I went to r0 to get this relationship. I'm taking this p plus, the complex amplitude, and I'm sticking it in here to get the pressure at any point r. p of r and t is equal to p plus, which is j omega rho 0 over 4 pi, and there's an upstairs there, e to the j omega t minus r over c times the volume velocity, god bless it. OK. Vv or r0 and omega from here-- that was all part of p plus-- e to the j omega t minus our r over c. So there is the expression for pressure.
This is volume velocity of the source. At any point r and p in terms of the volume velocity of the source with one approximation, namely, that for r0 much, much less than lambda over 2 pi. Which is the case, because we're shrinking at any frequency, I don't care what it is. The pulsating sphere is defined to be the limit of a shrinking sphere, just like impulse is defined in time to be the limit of something that got narrower and narrower and higher and higher.
Any questions to here? This is a very important result. OK. Now what I'd like to do is take that result and apply it to an irregular shape, but a special irregular shape, an irregular shape that itself is small compared to the wavelength. In other words, let's see, at say 100 hertz, if this irregular shape were radiating 100 hertz, 100 hertz has a wavelength of about 11 feet, so if the irregular shape were this big, that's quite small compared to the wavelength, or a sixth of it.
Now there's something that we get into when we do this that we have to really think about this surface. Here's a surface, we'll define the point r somewhere way out here-- this is r-- where we're measuring we've taken some point, doesn't have to be the center of gravity but some reference point on the surface. And we're much further out than my drawing has room to indicate. And this is p of r theta phi and t, normally.
This is a now irregular shape, so it's a three dimensional. It's not spherically symmetric anymore. Now what I'm going to do is to say well, we're going to look at this little element ds and we're going to integrate all over it. This is ds. The surface here is area s. We're going to get the contribution from each point on this as if it were a point source. And it had a certain volume velocity, Vv of that source.
Now the volume velocity of this thing may not be the volume velocity of this. In general, if you take a musical instrument for example, you take a violin and you look at what's happening, people think that everything comes out the f holes, the sound, but that's not exactly the case. If it were, you could make the whole thing out of steel it would sound the same. Not quite, because the resonances on the inside may be different, but the whole surface is vibrating. People talk about the lost art of the Stradivarius, had a lot to do with the varnish on it.
And if you look at how it's vibrating, it doesn't go like that. All different parts of it are moving in different directions at different frequencies. So this part might be coming out, the part over here might be going in, et cetera. And now I'm going to take the super position of a lot of these-- all of these ds's summed over or integrated over the whole surface. So that's easy it looks like. p of r theta phi and t is equal to j omega rho 0 4 pi-- now I can't take the r anymore, because the r is h. The r is h that goes to that ds.
And I can compute h by establishing some coordinate system relative to this point. And then I go over here to the ds, and the knowledge of the angle of this and r gives me h, however you want to set it up. So h is something that depends upon where that ds is, so I've got to take that inside the summation. And the summation in the limit is an integral Vv of, I'll call it h, and omega. Whatever h is, that depends on the volume velocity of that.
ds, that little element ds, and I'll sum over all the elements. And let's see, e to the j-- well, I can get the j omega t out, that doesn't depend on anything. I can get e to the minus j omega r over c. And all of this thing, then, times-- the whole integral times e to the j omega t. And let's see, that should do it. I just summed the volume velocities of all the elements on this thing over s.
Now if we look at this thing carefully but with insight, it falls apart into something very, very simple. Again, omega over c is equal to 2 pi over lambda. Oh, this isn't r, this is h, right? That's h because the distance of this element ds from here is h. Now let's look at this h. If the whole surface is small compared to lambda over 2 pi. Let's take, suppose this was the biggest dimension of it-- it doesn't have to be in line with anything, but let's suppose this was the biggest dimension, and I'll call that dimension d. No to be confused with the d of ds.
If d is much, much less than lambda over 2 pi, then what it says is the contribution out here, of this point over here, this little ds over here, relative to the contribution of this, the delay or the phase shift that you've gone through, from this element to get to here relative to that, from this element to get to here, isn't very important.
You can consider the phase shift that you get-- look, you could make h up, for example, just to see this on scratch paper on the side, let h equal, let's say r plus some delta h. If I did that, this would be a delta h. r plus the vector delta h would give me s-- would give me h. But if the delta h has a maximum of the distance from wherever r stops to any element in here, then in magnitude delta h, much must be much, much less than-- sorry, delta h is certainly less than d. The worst it could be is d if you put r in one boundary of it, and you got all the way across to the other boundary.
So if d is less than lambda over 2 pi, the magnitude of delta h is certainly much, much less than lambda over 2 pi. So the phase shift that is due to the relative difference in distance from this point to this and from this point to that you can not worry about. In which case, this expression here can go down to e to the minus j omega r over c. Which is very nice, because I'm integrating this r, as far as I'm concerned, for the integral is a constant. Any questions? Yes?
SPEAKER 3: So this is the case if all of them are pulsating in phase, so it's not like a violin?
AMAR BOSE: No. This might be pulsating this way and this one-- wait a minute-- this one going like that at the other end. I'm not talking about that phase yet, because that phase is in here. I'm talking about from whatever this fellow is doing, the phase difference between what this fellow is doing and what this is doing, caused by the difference in distances, is small.
In other words, just imagine that they're both going in in phase, then they're going like so. Then I'm saying that the phases, when added up out here, I could consider them all to be at a distance r. The shift due to the difference in the distances of the two points to here is immaterial. And I'm saying that, then, independent of what they're doing, like that. That phase shift, what they're doing, differences is already in here. OK? You sure?
SPEAKER 3: Yeah.
AMAR BOSE: OK.
SPEAKER 3: Except, doesn't that mean that it should be-- why isn't v, volume, dependent on h, omega, and t?
AMAR BOSE: This fellow? This is the complex amplitude. So he has in it-- OK, yeah. Now-- oh, I told you I couldn't put the r under this because that was h, so I better put h in here, inside the integral. Now I'm going to tell you a lie, but I'm still going to tell it to you, that I can substitute r for this h. And that's a lie some time, and I'll tell you when it's a lie.
But let's look at the physics of making that substitution. It says that when I'm way out here, and now I have to make an additional assumption that r is much, much greater than d, the distance of the biggest dimension of the physical device. I'm way out here. I have a device that's this size, and I'm out there in the other room measuring the pressure.
What it really is saying that I'm going to count that the strength, in other words, this has to do with the strength that each differential element is-- oh, at this point I ought to either or start over. This quantity here should be u times ds. OK? Because what I want to do is take the velocity at each point, integrate it with the area, which gives me the volume velocity of that point.
In other words, to get volume velocity, I'm taking the volume velocity of each point and summing it over the whole sphere. Well, what is the volume velocity of this point? It is the velocity, u, of that point times the area of that point. That is the definition of volume velocity. So how I stuck volume velocity in there by copying that, we won't speak about.
OK, now what I'm really saying now by trying to say that I'm going to let h be equal to r is, gee, the strength, which here's the contribution, this is the volume velocity here of the point here. By the time it gets out here, it goes as 1 over the distance, which is 1/h. But I'm so far out that I'm going to say 1/h for this fellow and for that fellow really can be represented as 1/r.
Anybody not like that? I've already told you it's false, but it sounds pretty good. So where is it a problem? It's a problem if I just had a dipole source. For example, let's suppose I had-- this thing was actually two sources, just two point sources. And suppose that this one had a volume velocity of plus 1, and this one had a volume velocity of minus 1. And now I'm way out here, and I say, well, I'm going to give, say, that the distance from here to here and from here to here is essentially the same, it's r.
Well, then what do I get out there? I get 0. Because the volume velocity here gives me the same signal out there as the volume velocity there in magnitude, but they're equal and opposite in phase, the sources are going like that, and so I get a 0 out there. But if I admitted that, wait a minute, this distance is a little larger than this distance, then I don't quite get a 0.
And so when you make an assumption like this inside the integral, what you're doing is you're saying, I will get some funny 0's that aren't really 0's. In other words, if I find out what my pressure is out here at some point it'll come down to-- if I've made a polar pattern, which we haven't talked about, it might come down to 0. But the real thing will have a little bit of-- it won't go all the way to 0. So I get perfect polar patterns, perfect in the sense I see them going to 0.
But in reality they won't, because this volume velocity of minus 1, let's say, contributes out here more strongly than this one, and so they don't cancel. That's the error I make. But look, my god, I'm going to be able to simplify things enormously. So when you're looking at expressions like this, and you say, well, I'm a mile out here, h is really the same, a mile plus 1 foot isn't any more, so I'm weighting this thing accordingly, be careful.
Understand what you're doing, because it's a sum of terms as long as you're inside this integral. There's no problem if you had an expression, some rational function, for example, and you had an h down here and you wanted to change it to r, just one term, no problem. The error would be 1 foot in 1 mile, no problem. But if you have a sum of terms, each of which has a different h, and you pull out the h, you are making an approximation that you better be aware of.
And what can it cause? Yes, it can cause nulls to be. Only when the terms up above are adding to 0 do you get in trouble.
Now let's see what simplification comes from this, though. We get equals j omega rho 0 4 pi r-- r isn't anything to do, r's fixed in this integration, I'll bring him out. I get integral of u, of h and omega ds and then times e to the minus j omega. I already told you why this can go to j omega r.
Well, I can write it this way. e to the j omega t minus r over c. And this integrated over the whole thing is the volume velocity of the source, of that source. Some of it's going in, some of it's coming out, going all different directions, different phases. But the integral over that is the net volume velocity of that source. So this becomes something like we saw for the pulsating sphere-- 4 pi r volume velocity of the source e to the j omega t minus r over c.
Now that is a very fundamental-- and so I can see this thing no longer depends on anything, I have these other angles, p of r and t is equal to that. Looks just like the expression for a pulsating sphere, the only thing is this is now the volume velocity of the whole thing. What are the approximations? Two, in this case. That you're looking far away from the sphere, and that the distance, the largest dimension of this irregular surface, was much, much less than lambda over 2 pi. This fellow plus-- somewhere I wrote down d-- yeah, there it is.
Now there's something that I've done here that is easy to do, and it's as wrong as wrong can be. I've made an assumption that none of us have thought about. And the first time I saw this, I never thought about it either. I've applied some basic result incorrectly. This wasn't a screw up. This was on purpose. I've said that each of these little elements can be considered like a pulsating sphere as far as their contribution over here goes.
Can anybody figure out what's wrong with that? It looked so good that nobody questioned it as I went. In other words, I got pressure by adding up the sum of the volume velocities here. Yeah?
SPEAKER 4: Maybe each ds isn't independent from each other?
AMAR BOSE: Each ds isn't independent from each other-- the ds's really are, because they are different spots on a surface. And the volume velocities could be anything. The volume velocities may be dependent, but whatever they are, I grabbed them in the integral. So that one-- yeah?
SPEAKER 5: Half the stuff is going inside?
AMAR BOSE: Half the stuff is going inside-- we're getting hot. Namely, you're talking about the surface. That's-- Yeah?
SPEAKER 3: Surface radian won't give you a spherically symmetric--
AMAR BOSE: Aha. You see, I applied superposition to this thing. Imagine this thing is some sort of [UNINTELLIGIBLE], applied superposition to it, and I said, each one of these is a pulsating sphere. But the pulsating sphere was derived as a sphere that radiated out into free space, no obstacles in the room. But when you apply superposition to this, this surface here, by superposition, this one has 0 volume velocity. All these have 0 volume velocity. This is the active one.
Get its contribution, then you set it to 0, get the next contribution, just like when you did in the time domain over here somewhere, we drew-- I erased it. We had an x of t, and got the response to this pulse here with all the rest being 0, then we put everything else to 0, got the response to the next pulse, et cetera. Well, this is, if you want, a pulsating sphere sitting on a rigid object. And that rigid object is anything but the space that we're talking about. A pulsating sphere sitting on that waste basket there is not a pulsating sphere in space.
So this brings us to a very interesting point, and I'm going to just tell you about this, and tell you that you've all witnessed this before, every one of you. But I'm not going to work it out. It's a lot of manipulations and the reason that you'll see. It's worked out in the classical-- these used to be the classical problems that were given to students to work out. And I'll mention that in a minute, in Morse you'll find it's worked out, for example.
Forget acoustics for the moment and go to something that you've all had an opportunity to witness. It's gone through, you've not thought about it, but-- think about a pond, a shallow pond, in which you drive a stake into the pond, a stick, maybe, so big in diameter you drive it in and let it stick up. And so the thing is sticking up here. I'm looking down on the pond now, so this is the stick, but it's sticking up above the water.
Now you throw a pebble in here. And what happens is you see the ripples in the water going along like this, and going out in a circle from that. I'm sure as kids you all threw stones into a pond. Now this stake that's here, if you do this, you find out that the waves go right on by. They look the same over here as they did over here. Now if you instead built something that was, let's say, a platform in that water that was 30 feet by 30 feet or something, guess what you would expect to see out here? Something very different than a wave going on out here.
So the way you look at this in math, or physics, whenever you want to call it, is that when a wave hits something, it causes another wave to be radiated. We already saw that in a plane wave hitting a wall. p plus goes down, and if it hits a rigid wall, it generates a wave coming back which happens to be exactly equal to p plus-- the p minus was equal to p plus.
So when a wave of any type hits a surface that is rigid in its discipline-- in acoustics, we're talking about a rigid wall, in electromagnetics we're talking about a short circuit or the other extreme. Whenever it meets something that isn't looking like the space it was radiating through, that something generates another wave. And this is the problem of scattering, for example, in physics, where a wave comes in here, could be anything, could be a light wave, hits an object which is big compared to the wavelength, which we see every day in physics in our room with respect to light. This, the pressures that hit here result in a wave going back called a scattered wave.
And the field out here is the superposition of what came in and what came back from the scattering. In our tube, we saw that, we've gone through it completely. That what exists in here is p plus plus p minus. But the p minus came about-- you only put a p plus going down here, the p minus then came about because it hit this. And the resultant field in here was p plus plus p minus.
That's exactly the same thing as it is here, except the geometry gets much more involved and complicated. If the object that it sees is very small compared to the lambda over 2 pi, there's not much of a wave, first of all, that goes out from it. It's so, so little that it doesn't perturb what's going on around this surface. And so what happens is, this is a classical problem in acoustics where you, for example, put a plane wave incident on a sphere.
You can imagine, even though it's a sphere, the geometry's bad enough because you're coming with a plane wave in x and bang, it hits this thing, and it radiates, it generates a scattered wave, and then you have to superpose those. That's a lot of manipulations, and in fact, it isn't even-- you won't even find all of the manipulations in Morse, you'll probably have to go to some papers to see how it's actually solved. But what you discover is exactly what you discovered in the pond. That if the dimensions of this are small compared to the lambda over 2 pi, the wave just goes right on by.
This isn't small compared to lambda over 2 pi, I'd better draw this field a little smaller, because this is lambda over 2-- if you're looking at crests the way I would depict a plane wave. There's the sphere. So it just goes on by. So what happens is, in the other disciplines too, waves ignore objects which are small compared to lambda over 2 pi. That's basically it.
So our assumption here, that d was very small compared to lambda over 2 pi, is what adds the validity to this. And without that statement explicitly, I could never say that I will treat this whole surface as a sum of little point sources, because point sources, little pulsating spheres, are radiating into free space. This is not, it's radiating into an object-- part of it, part of it's going out. But the object is small compared to the wavelength, and so all the fields are the same as if this object weren't there, and I can consider this thing by just looking at its surface.
Now if this thing was a solid-- issue closed on that one, but new issue. If it's a solid that's radiating like that, somehow, expanding and contracting, don't ask me how, it's not an issue, but somebody brought up the case, but what about inside? So you can think of a shell that's doing that. Well, as long as the surface of the shell is moving with the velocity here, and velocity times ds is a volume velocity, everything we've done is all right. I don't care what's going on inside.
The poor shell cares what's going on inside, though, because if that's a sealed volume, for example, whatever's driving that shell has to drive the shell a lot harder, because he's compressing what's inside, in general. But as far as the fields outside are concerned, no problem with this expression. That's the same thing, for example, in a pulsating sphere, not a point, but pulsating sphere, finite radius. If you were going to devise some mechanism to make that thing expand, you'd have to drive it a lot harder because there are spherical waves, by the way, which go inward, just as there are ones that travel out. But as far as finding the field outside goes, that has nothing to do with it. The only thing that counts is what's the movement of the ball that's radiated.
OK. So somewhere we came to the bottom line of this, that a body that is small compared to lambda over 2 pi, if you look far enough away from it, you find out that it behaves just like a pulsating sphere, and the only thing that's important about it is the volume velocity. What is the volume velocity? It's velocity times the area. It's the volume of air that you would move if you were sinusoidal and you integrated over 1/2 of a sinusoid of this volume.
So now this has an application to things you have experienced also. You've all heard sound from a loudspeaker, for example. Well, turns out-- here goes part of the folklore-- loudspeaker like this, put it in a room for low frequencies where the wavelength is large compared to the box, 50 hertz, for example, there aren't too many boxes that are big enough to be large compared to a 50 hertz, 30 hertz wavelength. At 50 hertz you have a wavelength of about-- well, let's go to even 30 hertz, 30 hertz, it's about 33 feet, or something like that. And 1/6 of 33 feet, most boxes are small compared to that.
So it turns out that for the low frequencies, if you took the box that's sitting here pointing at you and turned it around, it wouldn't make any difference. If you do this experiment outside, for example, you take a loudspeaker box and stand outside so that you don't have reflections, and let a person stand and listen to a 30 hertz or 40 hertz somewhere, low frequency, and then turn the box all the way around, he won't hear any difference.
So people think that the thing is pointing at them, and they get the idea that that's important, but no, it isn't. At low frequencies, the sound radiates as if it was a pulsating sphere. Because the sound radiating out of this thing totally ignores this box, because it's small compared to the dimensions, and so it doesn't matter which way you point the thing. All comes out the same.
At high frequencies where the wavelength of the radiated signal is small compared to the box, it's a totally different thing. When you go high enough frequency, then this thing looks like an infinite wall to the radiator and so he radiates straight out. So that's an interesting problem, that at high frequencies something like that would radiate quite directionally, like a flashlight.
And in fact, you can think of acoustical sources this way. At low frequencies, think of them as a light bulb. So if you make the analogy to light, think of them as a light bulb in which they're radiating omnidirectionally. And at high frequencies, think of it as a flashlight. They all tend to beam when you go to high frequencies. We'll see more of that, and we'll get a better insight to why that happens later on.
OK. Now this result, let me write down so I can use it again. Oh, let me also write down in case I get time at the end, I'd like to talk to you about near fields, which I promised you last time. Versus far field.
Let's look at a simple array of these point sources, namely, two point sources like so. This kind of thing in physics I think you have seen before, in light probably, in diffraction. It's the basis of a diffraction grading, for example. Let's call this r to the center of it. And if you worked any of these problems, by experience, you will set up your coordinate system such that it has symmetry about it. This being d, would you call this distance here d/2. And the distance down here is d/2.
So d/2, d/2, we'll let this be volume velocity 1. Volume velocity 1, volume velocity 2. And we'll take an easy case, but it's not just for the heck of it. This is a case that's used all the time in practice. I'm doing is for two pulsating spheres, but how many times have you landed at airports and looked up, and on the ILS system, you'd see dipoles in an array, bunch of them like this, usually somewhat of a curved array, and the whole array is sitting there going like that.
You won't see that much longer, because as you're going to see here, you can have a fixed array. The mechanical stuff-- excuse me to the [UNINTELLIGIBLE] people, but the mechanical stuff isn't so reliable. It eventually wears out, and the electrical, you can just change the phases of a fixed array, and cause the beam to go up and down, which is the kind of thing you need on an ILS system.
So basically, you'll see an array-- I don't know what Logan still has, whether they have the old fashioned thing or not-- dipoles in this direction, just center-fed rods going into the board, and the thing'll be sitting there doing that. Well, from what you're going to see here, you don't have to do that. Now that it's getting easy to build amplifiers and whatnot for each dipole, you can steer the whole array electrically. So what we're going to do here has many applications outside of acoustics.
So this distance here is different from this distance here. If this is far enough away, this point here, r is much, much greater than the total distance between the two. Then I would have for these two, these lines are all parallel, in essence, if that's far enough away. So if they're parallel, I would have here d over 2 sine theta, where theta is the angle from the array to the point up here. In other words, the angle between here and r is theta. That's d over 2 sine theta there, and another one there.
Now I have two point sources. I know now, relative to here, that this distance is r minus d over 2 sine theta, to a very good approximation. This distance here is r plus d over 2 sine theta. So two pulsating spheres-- yeah, a pulsating sphere is supposed to radiate into free space, but who cares, this little fellow here is so small he can't even see him. So his radiation is not disturbed. The scattering from this one is negligible.
So I can superpose the radiation from this point source with the radiation from that point source and I've got the result. The only problem is it's a bit messy. But the smoke clears on these things, and it comes down to something simple. Whoop. Well, doesn't matter. OK, so let's write. Now there's no dependents this way, about the vertical axis there's symmetry of course. So it's p of r, theta, and t. And I can write that as a sum of the contributions from each.
And now volume velocities. The actual practical use of these, generally, when you make antennas in E&M that steer, like the ones we were talking about, you have the same amplitude, you want to get the same power into each one of them, and you just vary the phase. So I'm going to let v1 equal to Vv, volume velocity, e to the minus j. I'm going to keep my symmetry in this thing. e to the minus j theta. So it'll have an angle of-- theta, what's this. phi over 2. I'll take it phi over 2. I'll have a difference of phi between the phases.
Vv2 is the same magnitude of volume velocity e to the plus j phi over 2. So I have volume velocities of each point source. I have to superpose them to get the whole thing. So it's j omega, using the expression for the radiation from a point source, j omega rho 0 over 4 pi r. But r now-- let's take this one first, OK, the Vv1. r is this distance from the first point source down.
So r is r minus d over 2 sine theta. j omega rho 0 over 4 pi e-- OK, volume velocity is Vv magnitude e to the minus j phi over 2, phi over 2. That's the volume velocity. Now I have e to-- where the heck is that expression, I should-- yeah, e to the-- oh boy-- j omega-- r over c, but r is all of this, unfortunately. e to the j omega r minus d over 2 sine theta omega r over c, plus another term like that.
Plus j omega rho 0 from the second source, 4 pi, the r downstairs is r plus d over sine theta. And the top is volume velocity, which is magnitude of volume velocity e to the plus j phi over 2, e to the j omega over c times r plus d over 2 sine theta. Boy, if I haven't made a mistake in this. I remember-- yeah?
SPEAKER 6: [UNINTELLIGIBLE PHRASE] exponential negative? Like, e to the minus--
AMAR BOSE: This fellow?
SPEAKER 6: No, on the top one.
AMAR BOSE: e to the minus, yes. Why isn't it negative? Because i goofed.
I remember my sophomore year when I was a student here, they used to have a course in mechanical engineering in which you had to do compression waves and whatnot, in bars, and you wrote differential equations until you were blue in the face, pages. So I went to the instructor and I said, look, you know, this is unbelievable, the amount of-- seemed like dog work-- and if I could get a good grade on the quiz, could I be excused from the homework?
So he told me, he said, look, I'll tell you yes, but I'll also tell you, you'll pay a price for it. Because the more times you go through all this stuff, the more orderly you become, and the less careless you become. So just listen to me, and one day, you'll remember. And I've remembered it so many times, because when I get involved in stuff like this, probability I'm going to come out with the right answer the first time is much smaller than it would have been if I'd done all that work.
OK, let's see. All this stuff is times an e to the j omega t, I believe. Let me just see. Let's go back and recoup here, and see what we did. This is the pressure, I want to make it the sum of the two expressions for each of the pulsating spheres. Since I have a t up here, I've got to have at the end of all of this thing an e to the j omega t. OK. Multiplying all of that.
I think that is OK. Now let's go. OK, I can get all this business out here, put all the exponents together somehow, and then pray. Let's see. p of r and theta and t is equal to j omega rho 0 4 pi volume velocity. OK, I got all of that.
I'll even put d to the j omega t in there to get rid of. So he's out. OK, here we go. Two terms. Let's see if I can squeeze the first one under here. r plus d over 2 sine theta. Now the top. Oh, it's a minus, sorry. Top term, I'm going to try to put all these exponents up there. I get e to the minus j omega-- you know what, I have an e to the minus j omega r here, over c, and one up here.
So why don't I get that stuff out also? e to the minus j omega r over c, out. Now I'm going to a little bit simpler expression. First term in it is r minus d over 2 sine theta. Top is, oh boy, e to the-- I'm going to put it this way-- e to the plus j omega over c times d over 2 sine theta. Please watch carefully, because you can correct what I'm doing. Minus phi over 2. With a little bit of luck, that is the first term. It's a plus d over 2 sine theta, yeah, I got that one. It's a minus phi over 2, got that. What? Oh, the j should be outside. Thank you. Plus j, that's very important. OK.
Now if we get the second term, we'll be all set. Second term is e to the minus j, I will write it, yeah, that's a minus d over 2. e to the minus j omega over c d over 2 sine theta minus phi over 2. And underneath, r minus d over 2 sine theta-- plus d over 2 sine theta. Wow, OK.
Now let's look and see. You get the mathematical answer, it doesn't mean beans. If you wanted to look at this thing and see what the polar pattern was-- I haven't defined polar pattern yet, but if you just look at that thing and say, OK, I want to go, I want to plot it as a function of theta, it doesn't give you anything, it doesn't talk to you. So approximations are in order, I hope.
Well, let's look at the denominator. Again, dangerous as heck. Whenever you have two terms, if you try to-- if it was only one term here, we could knock this off right away, because the d is much, much less than r. And we could just take that away and say, we're done. The only time you can't do that is when the numerators are nearly or equal and opposite. That's the old story that we just went through in the continuous body of point sources.
So if I do that, if I say that since this quantity is so small compared to this, that the strengths of the two pulsating spheres are going to be equal out here, I will have an answer that might give me a null, namely when this is equal to minus this. But if it's not equal to that, then I should be OK.
Now the way you would really do this, if you were doing it the first time, you probably wouldn't have that insight into it. What you would do is you would cross multiply the whole darn thing, get it over a common denominator. Then once you have it over a common denominator, you can neglect anything you want on the bottom that's small compared to something else, and you only make an error of the same percentage. So you'd multiply this thing up here, this one here, and you clear all the smoke. And basically, that's the way you do it if you didn't see, if you didn't have the insight to begin with, or if you hadn't done it before.
I was lucky to get to this place with no error, so I'm going to tell you that if you do all that, the only sacrifice you will make by neglecting this and that is that this will give you some nulls in this polar pattern that we're going to get to that aren't real. That there's a little bit of response there. If you go out and measure it, you'll find that the pressure isn't 0 at some angles when the polar plot tells you it is, and you can see why that approximation would give that error.
So I can bring an r downstairs. Now what I like about this is this is e to the j something plus e to the minus j something. Boy, that is the best of both worlds, namely, it's a cosine. So finally, we get p of r, theta, and t is equal to j omega rho 0 over 4 pi, volume velocity, which is common to both sources, 2 times this, because this is going to be-- well, I put the r downstairs, in other words, when I take away that, take away that, I have r, I can factor it out.
And what's left is 2 times the cosine. 2 times the cosine of omega over c d over 2-- I'm going to write omega over c here as 2 pi over lambda, just so you can see what the relationships are. 2 pi over lambda, so this becomes pi over lambda times d. Now cosine minus phi over 2, phi being the phase shift that's in the volume velocity. Times e to the j omega t-- hold it. Yeah?
SPEAKER 7: [INAUDIBLE] sine theta in that--
AMAR BOSE: Is there what?
SPEAKER 7: Sine theta in the argument of the cosine?
AMAR BOSE: Sine theta-- oh, yes. d sine theta. d sine theta minus phi over 2 e to the j omega t times e to the j omega t minus r over c. OK. Now what's of interest in these arrays is one thing. What is the polar pattern? Polar pattern means the following, that you go at a large distance from this, and you walk around on a circle and you measure the strength in this, let's say, pressure. You actually measure the logarithm of it, with an instrument measuring dB, but never mind. You measure pressure as you go around a circle.
Now the circle is not-- the circle is some fixed radius, OK, that I move around on. And the pressure as I move around here will change. Not due to r changing, because r isn't changing. And what I'll get is some sort of a plot like this. Here's where the source is, these two tiny little dipoles, and I'll get some funny looking thing, god knows, I mean, in general, I'll get something that looks like, maybe, like that as I go around there.
That tells me the pressure-- and the meaning of the polar plot is if I draw a radius from here to here, nothing to do with the radius over there, that the plot tells me that the strength of the signal, the pressure over here, is proportional to this length. If I go up here, it's very small. If I go along this axis in this particular plot, it's very large. If I go backward, it's pretty small.
So a polar plot, if you thought of a beam, like a flashlight or something, doing that, if you measured on axis, oh, that's a big strength. If you measured over here, the strength is proportional to the axis, what it was on axis is this over this. So the radius of the polar plot has nothing to do with the fixed radius of the expression for your pressure. Now the only thing you're interested in the polar part is the relative strengths on these axes, because if you get the strength on only one axis then from this polar plot, which is a highly normalized thing, you have it all over. If I know the strength here, then I know that this is 80% of it over here. So I've got.
OK, so what I'd like to do is just take-- I'd like to take two. How many of you have a commitment right after this? Six or seven. I won't like to take two. I'll just tell you what I'm going to do and won't do it. I would like to take two examples of this that are actually practical examples in which we pick distance and phase, and see what kind of polar plots we get out of them. We'll do that next time.