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GEOL7333-Seismic Wave and Ray Theory - Day2 02-18-24 PM
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00:00 Yeah, some, yeah, good folks. Uh uh uh we're missing
00:08 , but we need to start up . And so, uh I will
00:15 you that um you owe me a tomorrow morning uh concerning what we talked
00:24 in this lecture this morning. uh maybe you thought about a,
00:29 good question um uh over lunch. uh uh I'm gonna ask you to
00:37 that, write it down somewhere and send it to me by email um
00:43 and we'll uh address uh those questions in the morning along with the questions
00:51 you are going to come into your from this afternoon's lecture. So
00:56 I'm gonna expect from each of you questions um tomorrow morning. Sure.
01:04 us resume uh this afternoon where we off this morning. And I will
01:09 you that I am always uncomfortable with concept of potentials, scalar potentials and
01:17 potentials. It's just that I don't a good picture in my mind.
01:23 uh What those mean I can I can understand displacements, I can
01:30 um uh uh pressure fields and so , but I don't really don't understand
01:38 uh the scalar potential what that And so there uh uh uh
01:45 this approach that we just did, the classical approach. That's what you
01:51 see in every textbook in uh separating out the uh PNS waves from
02:00 complete solution using hell Holt's theorem. there's another way which I'm gonna show
02:07 now and this way is more powerful more elegant. And furthermore, it
02:14 for anisotropic materials. So for anisotropic , we will learn on uh the
02:21 lecture that uh Mr Helmholz doesn't work us. We can still do the
02:28 decomposition into two parts, but those are both mixed PNS. Uh And
02:36 uh uh the, the only way get um um wave equations which apply
02:46 to uh uh specific with types in an isotype case is to use the
02:59 machinery, which I'm gonna show you . Uh which as I say,
03:04 uh uh I like better than the approach to showed you earlier the classic
03:11 . So let's uh let's do And I'm gonna need a, a
03:28 . OK. So let's start off the wave equation as we previously devi
03:34 uh uh devised it. So it's here uh lots of derivatives, uh
03:42 lots of different components M and N K and J. And uh this
03:49 a little bit uh different from the I showed you earlier because I don't
03:54 use symbol I for an index, gonna use symbol I for the imaginary
04:00 square root to minus one. um um we're going to solve this
04:11 intuition rather than uh the healt machinery we said before. OK. So
04:20 are going to uh uh uh employ no, the idea of plane
04:32 And so um I'm gonna show you that any result, any solution for
04:40 wave equation can be expressed as a of plane waves. It's just a
04:49 of the uh result by uh fourier any function can be expressed uh in
04:57 of its fourier spectrum. So, this context of fourier spectrum is a
05:04 of plane waves. And so the , any solution uh uh can be
05:10 as a constant times the exponential factor the uh oiler exponent, the Oiler
05:18 E with the phase factor uh uh phase, uh a term I times
05:25 phase factor. And this is uh way of writing what we did before
05:32 if we substitute. And so we make a guess that this is gonna
05:36 . And then we substitute that into equation of motion which is right here
05:41 then the derivatives disappear. And instead get uh um uh omega square and
05:49 get KL times KN and let's just up, up a minute. Uh
05:55 when we put this equation in into , let's look at the left hand
06:01 uh on the left hand side of wave equation. So uh here is
06:06 left hand side. And when we two derivatives of the um of the
06:12 , two derivatives of this, we minus omega squared and the derivatives of
06:17 the time go away because we've made assumption. And in a similar
06:22 we get a product of KL times in here and again, a minus
06:27 because we're doing it twice. And of this, uh I here,
06:34 these equations are, are, are uh called the Christophel equations.
06:41 afraid there's another German uh uh physicist . And it's basically the same equation
06:46 it's the wave equation. But with assumption that the solutions are gonna turn
06:51 to be plane waves. And this the solution for uh just a single
06:57 , those plant waves. And uh which one is it, it's the
07:00 which has the uh frequency omega. now uh we learned before that um
07:14 um this is gonna work, we're have uh um plainly solutions if we
07:20 omega equals V over K, uh , we should say omega squared equals
07:25 squared. And so let's just uh divide both sides by K squared.
07:31 we end up with V squared here K squared showing explicitly here. So
07:38 this is three equations uh uh uh equals 123 for the UN. And
07:45 , what's, what's not known The, the velocity is not known
07:49 that's gonna be the velocity of uh frequency component. And uh uh the
07:59 is given by these uh uh uppercase uh uh I'm gonna go back here
08:11 um you is a vector, this here is a scalar. And of
08:17 , we're assuming here that all of vector components propagate with the same scr
08:23 the same um ee exponential factor. , so the unknowns are the use
08:35 and uh uh and the velocity uh is one member of the class of
08:43 which have um which have frequency omega the omega is hidden inside the velocity
08:54 . So uh this is a AAA of, of, of this is
09:04 the eigenvalue equation. And uh uh mathematicians have spent their entire careers studying
09:16 like this. And so you can immediately one solution for this if uh
09:24 the polarization factor U is zero, that uh uh uh this solution is
09:32 valid. And this solution is, mean this is true uh no matter
09:38 is V but we, we don't no uh uh uh trivial solutions.
09:43 want n nontrivial solutions. And it's turn out these are only possible for
09:49 discrete values of V three, not . OK. So let us um
10:03 uh take the isotopic case with waves in the 13 plant. So uh
10:11 a, a vertical plant where uh like the screen where the three direction
10:17 vertical and the one direction is horizontal those uh waves are propagating in this
10:23 in any angle in this, in plane of the screen. And then
10:26 Christophel equations uh uh uh when uh spelled out or like this, let
10:32 back up. But here they So we're, we're just going to
10:36 that uh uh uh uh the N and M and only um uh Jimmy
10:46 that the, the NCCL and N can only be one or three.
10:52 what we mean by propagating in the play. OK. So when you
10:57 put that restriction in there and, spell out all three equations um
11:04 you get this. And so from , you can see that in the
11:12 equation, it only involves you Whereas the first equation has U one
11:17 U three, no U two and third equation has U one and U
11:23 , no U two. So what say is the 1st and 3rd equations
11:27 coupled together while the second one is . So we can just divide out
11:34 uh the U two here and uh uh uh find that um um let's
11:41 , let's back up here. All , if, if we uh rearrange
11:47 this expression here, we find this , and uh in this, in
11:53 case where the propagation is in the plan, the length of the uh
12:00 the wave vector K is uh K squared plus a three square square root
12:06 that. So, uh this ratio is one. And so uh uh
12:11 canceling that out, canceling out the the twos. We immediately find that
12:16 this case, uh we found that uh huh a velocity uh for uh
12:27 uh waves uh for, for this uh equation is very simple. It's
12:34 uh the, the square of the is view of a row. And
12:38 gonna call that the sh velocity Why is it sh uh because uh
12:45 uh uh polarization factor is out of plane, it's in the two
12:51 whereas the plane is the 13 So that uh no matter which uh
12:57 these ways are traveling uh in the , it's uh uh the polarization is
13:03 out of the planet. So we that the sh mode uh think of
13:08 the, the three, the three is vertical and the one direction is
13:14 and the two direction is also So these waves are polarized in the
13:19 direction out of the plane. That's it says. OK. So here
13:28 uh uh so that's easy. Uh here are the other two couple of
13:32 . And um so first, what gonna do is we're gonna look for
13:37 waves, we're gonna look for waves have uh uh uh once newly polarized
13:46 oh And that, what that means uh yeah. OK. The wave
13:56 has this form and we insert that either one of these two equations,
14:03 one of these two equations. And find that uh everything simplifies and the
14:09 uh uh comes down to, I'm gonna back up here, here's
14:14 MS and our M and uh M M and everything simplifies with this
14:24 And we find that the uh uh , the, the square of the
14:28 is mo row. So we previously that uh VP square and we also
14:34 it K plus four thirds M squared uh T plus four thirds new over
14:41 . And so uh presto change we have found that the P wave
14:47 , our assumption of functional variation is that is that we did find a
14:52 solution provided that the solution is given . And so we call back the
14:57 wave, we seek a transversely polarized . That is one that has components
15:05 this. I wanna um I want to contrast this vector with uh the
15:14 that the longitudinally vector that we uh before I'm gonna back up one slide
15:22 up two slides. So here it that our longitudinally polarized solution looks like
15:27 with A K one and A K . Now going forward, transversely polarized
15:34 is gonna have the K three here A minus K one here.
15:42 if you notice with this choice, now this vector dotted with this vector
15:48 a zero. So that means that polarization is perpendicular to the propagation.
15:55 we s we insert this into either those Christo equations, we get
16:00 the velocity becomes the SV velocity square new of a row. But now
16:07 polarization is in the plane of the . So we, we got another
16:14 . Uh And uh this SV wave the same velocity as the S this
16:21 wave has the same velocity as the wave only the polarization is different.
16:29 it worked. That's what it says . The SV wave velocity has the
16:35 velocity as um uh the sh So what that, what we just
16:42 is sheer waves travel the inside of isotropic body with a velocity which is
16:49 of polarization. It always has to motorization has to be perpendicular to propagation
16:58 it doesn't uh matter. O it's not restricted. So because of
17:12 invariance, we often just call them waves now. So we have found
17:18 a P wave solution and a transverse uh without using non observable potential
17:25 They're like that. And we're gonna it later for anisotropic rocks that uh
17:31 two sure polarization have different velocities. in other words, we can say
17:37 degenerate the degeneracy is broken when we that this is degenerate. That means
17:42 found SVVSV equals vsh we won't find for anisotropic neurons strong, I kind
17:53 like that solution, that approach better the uh uh solution with the
18:01 All we did was we made a and it was an educated guess and
18:07 , we guessed that the solution was be playing away. And we put
18:12 in and we solved the equation for plane waves directly without ever bringing in
18:23 functions, which I think none of have a good uh feeling. For
18:28 reason I showed you the potential functions is because that's the standard way of
18:33 it. And if you look at textbook, you'll see that. But
18:37 prefer this way I just showed So the next topic is what we
18:43 the inhomogeneous wave equation. So that that is um in my case,
18:54 is I think different than you um thinking of when you hear hear this
19:02 this word in homogenic. So right , this word in homogeneous is a
19:12 mathematicians word. But let me tell what mathematicians uh uh mean by
19:19 So let's start with the P This is as we derived it.
19:24 I just uh put all the terms the left side and put a zero
19:27 the right side. That doesn't bother . Now, because the unknown function
19:35 appears in all of these terms mathematicians it a homogeneous equation, a homogeneous
19:46 . Now, there is no indication in this equation where uh the source
19:54 . So now let's put in a , let's put in a point source
19:58 the origin of coordinates radiating equally in directions. Then the wave equation becomes
20:09 uh uh becomes this. And on right hand side, we have a
20:16 tr so this is the direct delta . This uh box is pointed in
20:25 wrong place. It should be pointing delta and uh uh this one should
20:32 pointing at um um at, at all this uh this a with a
20:39 uh is the acceleration at the How do I know it's, it's
20:44 acceleration. It's because it's uh uh I gotta have the same physical dimensions
20:51 this which is acceleration. And this delta function says that it's only happening
21:01 the source at R equals zero. this is uh this quantity is uh
21:07 similar to the chronicler delta function which a matrix. But this is a
21:15 and invented by the uh uh the physicist Dirac in the 20th century.
21:21 this is a quantity which is zero except that at um the origin at
21:28 equals zero, the value of this infinity. It's a spike at R
21:35 . It's not a spike that goes one, it's right that goes to
21:41 and away from the origin. This a zero. So it's really a
21:49 . It's not a, it's not narrow, um, um, uh
21:53 function, it's a spike and it's , has a particular property which is
22:02 , um, uh the area under spike is defined to be one.
22:12 imagine, uh uh uh in the , uh uh a spike goes up
22:18 infinity and, and uh uh everywhere from article zero, it's ab it's
22:25 zero. What's the area under that ? Uh That's uh infinity times
22:31 isn't it? And uh so we that to be uh the area under
22:36 spike is one. Now, let's at this equation. We got the
22:45 the unknown here. Yeah. unknown here. But here, it's
22:50 no unknown here since the unknown does appear on the right side, this
22:55 called the inhomogeneous vector wave equation. it's for P waves because this a
23:01 pointing in the same direction as the . And we're gonna find a solution
23:09 this in the next lecture. Uh now we're only gonna note that because
23:17 the origin articles here, that's a place. It's obviously better to use
23:22 laplacian operator and spherical spherical coordinates rather um uh partition cos I'm gonna back
23:33 , see right here, we have Laplace operator. We defined it in
23:36 of uh of uh Cartesian coordinates. because now we have a source uh
23:45 waves radiating waves from the source, obviously gonna be clever for us to
23:51 change from uh Lalas in terms of coordinates into the Laplace, in terms
23:58 radio components. And so this is uh the case where there's no angular
24:03 , it's gonna be radiating equally in directions. OK. So we're gonna
24:09 the solution to this in the next . But for now, I'm just
24:14 out that what we have found is definition of an inhomogeneous wave equation.
24:21 let me start uh see. um me, are you, are
24:27 there? Can you hear me? , I can hear you.
24:32 So this goes to you. Uh this true or false? The inhomogeneous
24:37 equation is the same as we derived , except it applies to a non
24:43 . That is a layered subsurface formation we have in the real earth.
24:48 that true or not? It's No. Well, it,
24:57 we didn't, as we were discussing , see, uh uh as we
25:00 discussing it, there's no layers in , there's no end homogeneity. Uh
25:07 um So we are not yet ready apply these things to the real
25:13 We are uh at, at, the moment, remember that when we
25:18 this vector wave equation at a certain , we assumed homogeneity, we assume
25:23 the medium was homogeneous and we haven't that. So this statement is false
25:32 . So um um uh leaving uh this goes to you um is this
25:39 or false? The inhomogeneous wave equation the same as we derived earlier,
25:44 that it has an extra term which the source of the weight. That's
25:50 . Of course. OK. all of this was for the inhomogeneous
25:57 equation. Now, let's look at , the non uniform wave equation.
26:02 is what uh Rosado was uh uh was uh uh thinking about for applying
26:09 to a non uniform um uh OK. So uh oh mm keep
26:21 in your mind that when we say uniform media that's different from uh inhomogeneous
26:27 equation. OK. So where did assume that the, the medium was
26:33 uh uniform on page 63 we assumed uniform when we took, when we
26:41 this quantity here does not variable with . So we took it outside of
26:47 uh derivative. That's where we assumed , right? And that, that's
26:55 if and only if the medium is , where does the medium uh
27:00 These are the properties of the medium . This is the property of the
27:04 here. But let's not uh um not worried about that here. If
27:12 going to um we wanna take this outside of this derivative, we can
27:18 do that if the uniform is if, if the medium is
27:23 Now, let's think what happens in real world where the medium is non
27:31 , maybe the medium has layers, it has salt bodies. Who knows
27:35 it's got in there. And so uh we're just gonna take this same
27:39 where we started with before and we're uh I make uh apply chain rule
27:45 . So we get um um this what we got before, but now
27:50 a new term depending on the non of the medium and see how we
27:57 chain calculus. So uh this second is still here but now the,
28:03 j uh uh derivative operating only on , that's here. If you're not
28:09 about this, you need to uh review uh chain little calculus. And
28:15 there's some help for you in the . Now, in the earth,
28:23 can be all, there can be of in homogeneity. So let me
28:28 you this um um uh piece of and describe to you uh uh uh
28:38 it is. So this is the from a particular type of a uh
28:45 instrument and it shows 23 m of bore hole and it's called war hole
28:55 two. And this one was uh uh done by Baker Atlas. I
29:01 know whether uh uh they have the version of this tool or not.
29:07 slimmers, they did, who But the tool is lowered into the
29:13 hole and then it's pulled up steadily as it pulls, as, as
29:18 pulled up, um there is uh uh uh part of it is rotating
29:24 like a, a one revolution per at every 10 seconds that, that
29:30 uh uh rapid revolution uh about the of the world. And as it's
29:38 , it's putting ultrasonic waves into the um uh well into the mud,
29:46 waves are propagating through the mud, the side of the borehole wall,
29:51 back to the tool and being recorded the uh right close to the
29:57 So it's uh it's recording uh normal reflectivity from that ultrasonic wave as it's
30:06 out uh uh from the rotating So then the data is presented in
30:12 way, this is called the unwrapped . So this goes from 0 to
30:17 degrees around the tool. And then tool is being pulled up like salt
30:23 that you can see there's bands here white where there's no data. That
30:26 because there's a structural element in the which present prevents any data from being
30:34 at this asthma. But it's not problem. You can see that uh
30:38 the, the different colors here um the intensity of, of, of
30:47 uh represents the amplitude of the reflected . So it represents the reflectivity of
30:54 borehole wall. And uh uh you uh uh uh here, it's high
31:02 , it's low here, it's intermediate . So you can see the layering
31:06 clearly as this thing is pulled up 23 m of. Oh,
31:14 what I wanna do is zoom So when we zoom in on
31:19 we see lots of small scale laying small scale. And now I'm gonna
31:27 in further and further. And you see that uh uh no matter how
31:36 you zoom in, there's always there's in homogeneity on all scales.
31:43 maybe not so clever for us to assuming uh homogeneous media when we're deriving
31:54 wave equation. Now, Mr Hook this in the 17th century. Uh
32:06 uh we might do this when we're thinking about um measuring uh properties of
32:12 in the laboratory. But when we're uh properties of rocks in the
32:18 we need to um we need to that uh uh the possibility that the
32:25 really is homogeneous. So that means the wave equation which we um spent
32:35 much time arriving, it's not gonna very good. Uh Because we made
32:44 fatal assumption right there at the end the medium was homogeneous was uniform.
32:53 there might be lots of indication lots um of um context inside the real
33:00 , especially the sedimentary crust where that's true. So let us consider a
33:08 that has horizontal layers on. So derivative with respect to X one or
33:13 two are gonna be zero. let's consider only vertically traveling P
33:20 So the, the displacement in the direction and the two direction also
33:24 So in that case, the uh the, the non homogeneous wave equation
33:32 like this. This is the term we looked at before. That's the
33:36 term. But now we have a which involves only the variation of the
33:43 m with respect to the vertical And look here we have a single
33:49 of displacement. Whereas here we have two derivatives of displacement. So we're
34:03 talk about a solution to this later the afternoon. But you can see
34:08 uh right here, thi this is very typical situation where we have horizontal
34:15 and kind of vertically traveling P And you can see there's this additional
34:20 which was not included in our discussion the uh uh plane wave earlier.
34:26 Normally uh uh we think about these only. But in the real
34:32 you can see that there's likely to more terms here depending on uh this
34:41 . OK. So um uh are you there? Yes provision.
34:49 . So um this goes to is this true or false? The
34:56 homogeneous wave equation is the same as derived earlier, except that it applies
35:02 non homogeneous uh subset formation. Is statement true or false? Mm I
35:10 say it's, it's, it's false it's not the same occupation. Oh
35:16 don't know. Well, not the question but uh yeah. Yeah.
35:20 . It implies that is that is layer media. So it would be
35:27 then that uh Well, no, , I like your uh your first
35:32 . Uh uh um Yeah. Let's see here. Um mhm.
35:46 question, if I under I am the question is that if the question
35:50 the same and for me, I saying it's not the same.
35:54 you are correct. It's not the . Mhm. OK. So now
36:00 think about solutions to these equations. I am going to show you an
36:06 simple situation. But even though it's it's very simple, it's as,
36:14 as simple as it can be, it's very complicated. The the the
36:20 earth is not gonna be simple, it's gonna be complicated like this.
36:26 . So what do we have We have uh uh a arrival times
36:30 a wave as a function of offset uh the source is gonna be uh
36:35 and it's, it's gonna have move like so and look at all these
36:41 of arrivals that are like to let happen. Um This is taken from
36:46 book uh by Sheriff and Gel Dart uh uh um 30 years ago.
36:53 Now you might know the name He was a professor in this department
36:59 for many years and he just died few years ago. I had the
37:03 to know him. I suppose that else here in the room uh knew
37:08 . Um But um Utah did you Bob Shar. Oh Yeah. So
37:13 were lucky. But, uh, , he died shortly after you arrived
37:17 he had a long illustrious career, , uh, as a professor at
37:23 of Houston. And furthermore, before , he had a long illustrious
37:29 uh, as an oil finder, , for, um, uh,
37:35 Chevron, I think. Yeah, think it was for Chevron. And
37:40 , uh, after he quit he joined the University of Houston and
37:45 did an amazing thing. I don't any other person uh who's ever done
37:52 . Um but he uh gave money the university to establish a faculty,
38:00 a special faculty chair or highly a faculty. And he did that three
38:07 . So he gave millions of dollars the university as a faculty member and
38:14 didn't get any of this mo uh money, he gave it to the
38:18 and the university uses it to augment salaries of uh distinguished professors.
38:26 for example, Professor Stewart uh sits a chair which is uh uh no
38:34 after Bob Sha and Professor Castano Uh uh so uh and then
38:42 I think there's one more chair. uh so uh I, I'm a
38:46 but I didn't give millions of dollars the university. I don't know any
38:51 anywhere in any university who ever gave of dollars back to his uh his
38:58 university and get this. So Sheriff this and then he uh uh he
39:09 do it after he retired, he gave it while he is a
39:13 . And so they were sheriff chairs this and that, and he,
39:17 was not holding any of these sheriff . They were, they were going
39:21 his friends and colleagues and, after he gave the money,
39:28 we had a, um, a term started. So the university
39:35 uh, uh, um, uh, se send us all,
39:39 , at the beginning of the term said, I, if you want
39:42 , a parking permit, uh, , uh, you have to pay
39:45 it and the price is, uh, uh, a couple of
39:49 a term, um, for each . And so he got one of
39:57 letters and I was outraged that the would charge him for parking a couple
40:03 $100 per, uh, uh, , uh, semester when he had
40:08 given the, the, uh, university millions of dollars. I thought
40:12 was outrageous but he was, uh, he was a humble man
40:18 he said, don't worry about Uh, uh, I'll, I'll
40:21 for my own parking. So, only did he pay for the salaries
40:25 his colleagues, he paid for his parking and it was way out at
40:30 end of the, uh, at , at the edge of the
40:33 I thought it was outrageous. he should have been given,
40:37 uh, uh, an honorary parking . Right. Next to the university
40:42 his name on it. And uh uh it should have been a uh
40:47 uh the, the, the university have shown better grace than to uh
40:54 his money and charge him for Anyway, that was Bobs. And
40:59 uh you can buy this book. If you don't have this book,
41:02 should do. It's a good book exploration geophysics. It's about um this
41:07 inch and a half, thick and of good stuff. Now, uh
41:12 look at all these uh branches So here is the direct wave,
41:16 starts at the origin and it goes here. Uh uh There's a
41:21 it's a straight line here. So can deduce that uh um the uh
41:28 un uh uh in this model Um uh The velocity is uniform.
41:39 , this graph doesn't show anything about . This is not wiggles, this
41:44 just the time of arrival of these phases. And so this one is
41:49 horizontally at a very slow speed And uh it's, it's hitting uh
41:56 know, it gets to a um receiver at 1000 m, it gets
42:00 at this time. So the uh velocity is uh 650 m per
42:07 OK. Then there are refraction. I, I said the, the
42:13 was um uh uniform but I think wrong here. Um uh These are
42:20 and uh we will talk more later the um in the course about what
42:26 refraction is. But I can tell that a refracted way it goes down
42:30 then it goes horizontally. When when it uh when it uh encounters
42:37 uh layers, it turns and goes and then to get back to the
42:46 , it turns back vertical. So a refracted way. And at this
42:51 , you should have uh uh many questions in your mind. Uh uh
42:56 what makes it act like that? all that will be clear within the
43:01 few lectures. And so those are two events here and here, B
43:10 C and you see there are also straight lines. And so, Denfd
43:22 and F are ref reflections, not , but they're reflections. And so
43:29 have hyperbolic move out like so uh look at ee is hyperbolic move
43:35 So G is a dipping reflector uh where uh the, the reflection is
43:44 a horizontal layer uh but it's And so you can see that uh
43:50 has a different hyperbola than you see . And furthermore, uh look at
43:55 reflection here, the, the uh minimum time is happening at vertical incidents
44:03 offset. But down here for the reflector, the minimum time is happening
44:08 over here. So because it's that causes this behavior, we'll talk
44:14 about that later in the course. there are multiples H and I,
44:21 multiple is something that goes uh uh down and comes back and then goes
44:26 down again and comes back up. many different uh possibilities for these uh
44:34 echoes inside the layered medium. And just showed two of them here then
44:43 J and K, these are ground and airways. So the ground roll
44:49 this one. So this is what we used to call uh railways.
44:57 are surface waves, they travel very . See it's slower than the direct
45:02 . So this direct wave uh uh the one we did it, that's
45:07 A that travels with a B velocity 650 m per second. But uh
45:14 here, this ground wall is, I said VP, here, I
45:19 show that is V uh R for and then uh for K that's even
45:27 . So that is the sound uh of the wave going through the
45:32 So that's 330 m per second. when you have say a dynamite uh
45:41 land, um uh normally the dynamite uh is um uh exploded uh several
45:54 below the surface of the, of ground, maybe 10 m below,
45:57 20 m below. And uh so uh um it, it explodes and
46:03 through the near nearby rock, but the parts of it is going up
46:08 when it hits the uh uh the surface it moves the surface up
46:13 down and you, if you're standing , you can feel it, but
46:18 better not be standing there because normally bunch of, uh, uh,
46:22 , uh, water that comes out the shot hole. When you do
46:25 , you don't wanna be standing But if you were standing close,
46:28 could feel it in the bottom of shoes as it comes up. But
46:32 you're standing, um uh 100 m or 1000 m away, you can
46:38 feel it uh uh as it comes and makes a wave through the air
46:45 at the sound velocity of, of the velocity of sound and air.
46:50 a um a slow arrival. And shown here. And then finally,
46:58 have diffraction which come uh uh uh uh I will tell you more about
47:06 later in the course s different from , different from reflections. These are
47:14 . Now, each of these lines the arrival times. It's the time
47:19 the peak of the wavelength for each . So the wavelets are spread
47:27 for example, oops go here. uh le let's think about this
47:31 Imagine here here is a reflection and uh this is the peak, this
47:37 the arrival time at the peak of wavel. Well, you can imagine
47:41 if we draw the wavelet in the wavel, it is gonna extend
47:44 here to here maybe and it might with the WAV from here to
47:48 So all of these um these um are shown much more simply than they
47:56 gonna appear uh on our instruments because simply the peaks of the arrival
48:02 They don't show the wavelets at Yeah. A also this graph doesn't
48:08 anything about amplitude. So we are to be interested in amplitudes in this
48:17 . Now, here is a question um philosophy, not physics. This
48:26 philosophy and the question is what is difference between signal and noise?
48:33 so a lot of people have thought a lot about that. And uh
48:39 uh these are the definitions I like um uh that uh noise is what
48:46 call signal that we don't care but maybe we should care about it
48:51 noise is actually a signal that we understand. Maybe we should understand
48:56 For example, U Utah is looking uh uh railways. So for most
49:03 uh the last 100 years, railways been considered noise and nobody was interested
49:10 those, but he is thinking of a signal because he wants to understand
49:15 information they carry. So uh this here that one person's noise is another
49:22 signal. And as we get to a more mature profession or learning how
49:28 understand and make use of parts of signal that we previously threw away as
49:42 , I'll give you some more examples that later. So uh I think
49:52 is um um uh Mercedes uh turn . So uh Mesa, I post
50:00 to you primary reflections, reflections. of primaries, not of multiples,
50:06 primaries from flatline reflectors, we have move out which varies with offset is
50:12 linear are hyperbolic. And so we here two possibilities of linear and two
50:20 of hyperbolic. So uh um I think it's linear. Uh
50:29 now uh these are reflections. So uh uh uh uh right now you're
50:37 in your Schlumberger mode. Now, want you to think your western Chico
50:42 where you doing surface seism uh surface look at reflections. And are those
50:51 arriving linearly or not? Let's go here. OK. So here are
50:59 reflections here and there, there's one see that one is not linear.
51:05 . Yes. So the refraction are , the refraction are linear but the
51:14 are hyperbolic. So it's got to either C or D. So
51:18 which do you vote for the Yeah. So I go with C
51:27 . OK. Next question is uh one goes to li li these are
51:32 reflections, not primary reflection but multiple . And so um um uh which
51:44 we um uh so the reflections, they're gonna be Haker by, they're
51:53 gonna be linear. So, which one of these? Uh uh
51:58 you think that the uh the soonest is gonna be at zero offset.
52:03 So, what I'm gonna do is gonna defer this question uh because you
52:07 see that uh uh it depends upon which multiples we're talking about. So
52:13 sort of a trick question, talk about multiples in the weeks to
52:23 OK. So I'm gonna uh uh uh Utah is a, a specialist
52:29 uh railway, I'm gonna put this Utah. Um And, and for
52:38 model, remember it's, it's, a laterally homogeneous model. And so
52:43 of these answers is uh that is . Yeah. So it's a,
52:54 uh um if for uh so immediately , he knows he's looking at linear
53:01 move out, not hyperbolic, move and for his choices between A and
53:06 uh but uh it, it knows knows that uh at the shortest
53:11 it's uh uh zero times and you get a different time if it's uh
53:16 refraction. OK. So, so is the topic which um Professor
53:27 I got a question about the, no, the no noise and the
53:32 , you said that probably noise is that we don't understand. But they
53:36 about the multiples, multiples are, mean, should be considered noise always
53:42 not. Well, no, uh used to when I was your
53:45 everybody thought that the multiples were But in the past, um let's
53:51 20 years, we realize that there are signal and uh uh uh uh
54:00 can learn more about the earth if uh image the multiples. So,
54:06 uh it's always a good idea to multiples from primaries. But uh the
54:12 idea was uh throw away the multiples analyze only the primaries. That's a
54:17 idea. But here's another good Throw, throw away the primaries and
54:22 the multiples. So, uh there , you learn uh certain things about
54:27 uh which uh uh you learn certain from that, which um we'll not
54:36 into now, but we'll touch on later. So uh there's a perfect
54:42 of how um we converted a certain of arrivals from noise to a signal
54:52 uh getting smarter and also by uh better data and so on, but
54:57 by getting smarter and by having better . Mm So let me bring you
55:05 this uh um very interesting topic of . And I have a um a
55:15 interesting episode in my own career about which uh we're not talking about
55:21 Uh I don't think we, we uh talk about my own experience with
55:29 um uh later in the course. right now, um I,
55:34 I'm gonna give you some uh very ideas. So let's consider a s
55:41 over a place like this. Here's surface of the earth and we got
55:45 place A and a place B and got a vector source at each
55:54 And at both places, we have , this is uh uh uh the
55:59 at a measuring data that came from and this is the receiver at the
56:07 at B receiving uh data that came A. So this is what the
56:19 theorem of elasticity says. It says form this vector product between the force
56:26 a dotted with the displacement of A from B and that the product has
56:35 be equal to the force at B with the displacement at B coming from
56:44 . So uh fairly simple statement of . So um um the proof of
56:53 is given in this uh old book a eh love which you might want
56:58 look up now, that is the theory of uh theorem of elasticity.
57:08 We are more familiar with a special called the scalar reciprocated the and so
57:15 applies to P waves only. And the P waves are uh polarized in
57:20 same direction as the uh propagation, can uh we can uh dispense with
57:26 vector uh dot product here. And uh so if we have only uh
57:32 vertical sources uh uh uh uh see I said it wrong. We
57:41 up. Um uh We're not dispensing the um that product, but we're
57:49 that the force has only a three . And uh so uh the,
57:55 the data is arriving um uh mostly . So we have only the three
58:02 of displacement. And so these scalar now the same. And we're gonna
58:07 o obviously, we're gonna arrange it the forces are equal. So what
58:12 says is that the displacement of a from B equals the displacement of B
58:19 from a, another way to say is if you interchange the source and
58:25 position, the data is the So I think that most of us
58:33 sort of familiar with this concept in . Now, it wasn't always like
58:40 when I was uh uh uh before my time before we were
58:44 uh uh we used to do split surveys with every common midpoint eliminated illuminated
58:51 both directions. Can you see all uh uh uh um our heads down
59:02 ? So we, we would have uh uh uh common midpoints illuminated from
59:09 directions. But now we know because the uh of the reciprocity theorem,
59:16 we can obtain the same information with half of the receiver effort using offend
59:22 . So we only have to shoot the left here and we're uh uh
59:28 receiving uh all these uh things and is uh all, all the data
59:34 received at these various receivers and it so that they all have the same
59:41 midpoint. So we uh uh once we realized that uh uh uh
59:50 the rest there, we only had do this. We didn't have to
59:54 the sources over here receiving over here it shows here. So that's uh
60:01 an important thing. We could not marine acquisition without this theorem because the
60:09 in marine acquisition, we always have source on one end and all the
60:14 going the other way, we never go from a source here in,
60:20 these receivers. We only have sources . If we did put another source
60:40 and radiated into these, then uh theorem says that the, the traces
60:45 be the same. So uh we do it, we save a lot
60:50 money and, and, and we do it. Yeah. Uh There's
60:59 of implications of this. Oh And the way, I should tell you
61:01 this theorem is a very deep It applies to all sorts of
61:08 It applies to the situation where the is laterally in homogeneous. Maybe there's
61:17 rocks below A and slow rocks below B even. So the theorem hole
61:23 holds in the case that uh walks an isotropic. It holds uh
61:30 in almost all cases that you can of. So it's a very deep
61:37 with hardly any objectionable assumptions in very powerful theory now because it's so
61:49 it makes for more efficient imaging because uh of this argument, uh if
61:54 have a straightforward imaging algorithm, uh you have to uh uh that has
62:00 cost uh in the computer, it a cost which depends upon the number
62:05 source positions. However, some acquisition have many more sources than they do
62:13 . So if in the computer, interchange the roles of sources and
62:17 these algorithms are more efficient. So we save money not only in the
62:26 but in the computer by realizing But uh uh the scalar reciprocity theorem
62:31 applies to py or sheer waves or waves. We need the full vector
62:38 theorem instead. And uh we will about that later in this course.
62:44 let me show you a AAA very consequence of all this. This is
62:51 the Eisner reciprocity paradox named after uh a good uh geologist of the previous
63:00 . His name was Elmer Eisner, for Texaco and he proposed this
63:07 And I'll first, I'll explain the to you consider uh two D propagation
63:13 the plane of the screen and consider you have an elliptical reflector and consider
63:20 a, a fluid here. So no she only P waves and electrical
63:25 with the source and the receiver at two F side. So what are
63:30 two FF side? These FCI are places inside the ellipse? Such that
63:36 you draw a straight line from this , from this focus to here and
63:42 to here, that's exactly the same of string as if you go from
63:47 to uh to Y and back to receiver, no matter where you put
63:52 and Y, it's always the same of, of a string between uh
63:59 uh two faux side, very special . OK? And furthermore, when
64:05 have a string like this, you equal angles between the sources of your
64:09 . So that um uh um these lines are like ray paths and because
64:17 the same length of ray pa pa it's the same travel time. So
64:21 the rays emitted from the source arrive the receiver at exactly the same
64:27 That's the, that's the a property the ellipse and the two full
64:33 it's just simple geometry. Uh This proved uh thousands of years ago.
64:40 now this is what uh uh uh said, suppose you have this sort
64:44 a situation and now you interchange source receiver. So the sources here and
64:49 receiver here, it's very clear that recording will be same, the same
64:55 symmetry should be obvious to everybody that that's gonna be the same.
65:01 , let's consider that you remove the of this reflector here, so that
65:06 of the energy from this source gets to outer space, the other half
65:12 inside the ellipse and it all collects the receiver all at the same
65:17 but half of it gets lost. . Follow me up. Now let's
65:24 source and receiver. So now we the source here and the receiver here
65:31 now 95% of the energy is inside uh app the ellipse and only a
65:40 bit sneaks out past the uh receiver gets lost to outer space. But
65:49 the reciprocity theorem says is this recording the same as in this previous case
65:56 half the energy is lost. Isn't amazing here? Half the energy gets
66:01 here. Only a little bit of gets lost. But this, what
66:06 theorem says is this recording is the as the previous case, even though
66:15 the energy got lost here and hardly , any of it got lost
66:20 Isn't that amazing? I think that you all should find this
66:28 And so Eisner posed this as a and uh the pages of uh uh
66:37 uh the journal Geophysics and he said is the, is the reciprocity theorem
66:45 or valid or not. And if not valid, where did it go
66:51 ? So there were famous geophysicists lining on both sides of this argument arguing
66:58 each other in the, in the of Geophysics. Some said yes,
67:04 said no. And so um on uh uh the argument and then uh
67:15 said yes. And here, here's reason why and they gave a very
67:19 mathematical arguments to show that the reciprocity must be true. And others gave
67:28 arguments saying that it's obviously wrong. , I think uh uh most of
67:35 would probably look at these two this one and the previous one.
67:40 we think, well, couldn't possibly true. The debate was finally resolved
67:49 uh Professor Clair Bug at Stanford which is in Central California and his
67:58 student uh Dellinger who later became my at Amaco and BP still is working
68:07 and they did the following calculation they here that here's the picture of the
68:13 . You see, here's the ellipse uh hardly any of the uh energy
68:18 lost. Here's the ellipse with half the energy getting lost. And so
68:23 uh calculated these two I know these and you can see here the uh
68:34 energy is coming in underneath this curve there's no energy from the gap
68:39 That's uh these angles here and there's uh energy arriving from the gap.
68:46 this, that's this gap here. that's the energy. And they found
68:51 that in fact, uh uh the of, of energy under this curve
68:57 approximately twice the amount of energy under curve just like you would think
69:05 But nonetheless, the pressure from these is about the same for the three
69:13 figures. It is about the So why is this uh uh you
69:18 , what's the difference between energy and ? You will uh uh be uh
69:25 that the uh the pressure comes from derivative of the energy. And what
69:32 measure in our receivers is the not the energy. So even though
69:37 energy here is twice the amount of here, the pressure pulse is the
69:45 in both cases. So um uh encourage you to read this paper by
69:51 B and Dellinger in 1987 plastic So here's a quiz. I think
70:07 uh uh it's the turn of uh . Carlos. Is this true or
70:14 ? The reciprocity theorem is just a result with no practical application in our
70:22 . It's false. Yeah, that's . I just uh uh uh I
70:26 gave some uh good examples. we only have to do uh in
70:30 surveys. We only have to do acquisition. Uh Well, we don't
70:36 to do split spread acquisition. Those the reciprocity there. Um uh Roda
70:44 few uh is this true or false general? The rest positive theorem can
70:50 paraphrased as if you interchange source and positions. The data recorded is the
71:00 . It's true. That is uh common mistake, mistake that this is
71:09 good uh a good statement of the case, the scale of reciprocity
71:14 But uh uh this is uh but asked you about the general reciprocity
71:21 So now uh le le this is same question for you except that this
71:27 here. This is about the scale respiration here. Uh So, um
71:35 uh you know, already the answer this is um oh no, this
71:42 a trick question uh uh relief. What, what's the answer for this
71:49 ? Uh So you made uh you the classic student mistake, you were
71:54 too quickly, didn't read it And so uh uh uh this is
72:00 because it's about the energy and this is about the data. So this
72:07 a good statement, the scale of theorem about the data. That's
72:12 But the question says the general it does the scalar theorem, but
72:17 got the energy down there. So one is F. So, so
72:21 a trick question. You have to about uh every, every single
72:26 So I'll, I'll tell you um a little bit about my own experience
72:31 this. Uh I spent uh half career with Amaco and then the other
72:37 with BP after BPF Amaco. So uh in the Amaco half of my
72:45 , I spent half of the uh in the research center in Tulsa.
72:50 in those days, we were the research center in the industry. And
72:55 uh at a certain point, I from uh research to uh the exploration
73:01 in Houston. And who, so had been there just a few weeks
73:08 we got a request from our colleagues Amaco Norway. And they said Conaco
73:15 done a new kind of survey. We, we want to do a
73:22 thing on Amico Feet and the new of survey was a converted wave survey
73:29 a marine environment. So what you is you put out um uh uh
73:37 on the C four, brand new in those days, that was about
73:45 . I would say brand new idea , to invest the money to put
73:50 on the sea floor and then you a normal source um behind the boat
73:56 the service. And it's gonna be down key waves, converting somewhere in
74:01 subsurface to sheer waves. And since have the receiver on the sea floor
74:09 three vector components of motion on the floor, you can detect this converted
74:15 wave coming up. If you have in the at the sea surface,
74:20 can't detect those sheer waves because they make it through the uh water
74:29 So our friends in Norway said, do that uh at Amao. So
74:36 was part of the team that uh the uh acquisition and guided the
74:43 And one of the first things we when we looked at the data was
74:49 when the interchange source and receiver, did not get the same traces.
75:00 when we formed up a common midpoint , it was not symmetric, we
75:06 slow arrivals on one side and fast on the other side. And so
75:14 was puzzled about this. And I not uh uh noted inside Amao as
75:26 an expert imager using size of And I uh that was true.
75:33 , I was never an expert but had friends, there were experts.
75:38 so I went with them and I them, I went to them and
75:40 showed them this data and very clearly symmetric common mid black gathered. And
75:50 said, well, you know, reciprocity theorem says that it's got to
75:56 reciprocal, it's gotta be symmetric. must have messed up the geometry uh
76:02 . Somehow you go back and fix your geometry and you'll be OK.
76:07 I went back and checked everything and was all correct. And uh uh
76:13 finally, I read the uh out desperation, I read up on the
76:20 theorem and I realized that this statement we have right dear, that's a
76:29 statement of the scale of reciprocity theorem only to P waves. But I
76:37 not looking at P waves, I looking at converter waves. So I
76:40 I needed the general reciprocity theorem, was that vector relationship which I showed
76:47 with the dot products. And so uh uh so that vector relationship uh
76:54 that for a converted wave survey, do not expect um symmetrical spirit thread
77:06 spread gathers. No, you should um the differences in a split spread
77:12 , especially if the subsurface is non . And it turns out that when
77:22 analyze the various components of what we uh uh measuring and so on
77:27 the, the general reciprocity theorem, it was true, it did not
77:32 our data. It, it uh uh referred to other components, other
77:38 components. So uh uh not uh episode remains very uh very strong in
77:48 mind. And then I'll tell you more thing. Um We had the
77:57 done by a product by uh one the major acquisition companies and they did
78:04 good job. And they said, you mind if we uh learn how
78:10 uh image such converted wave data in and then we'll develop our own techniques
78:16 we'll sell services to other clients. But we need to learn how using
78:22 uh your Amaco data, of they acquired it for us, but
78:26 was our data and they would have looked at it at all without our
78:33 . So we gave him permission. so the then uh uh they did
78:39 thing and we did our thing and there came uh an opportunity to present
78:47 maybe a year later at the annual of the European Geophysical Society, which
78:57 held that year in Geneva Switzerland. so I went over there and,
79:01 gave my um uh results and it uh extremely well perceived uh well
79:09 Um Yeah. Um In fact, won an award, I think we
79:13 the best paper award at the But meanwhile, on the exhibition four
79:22 um uh acquisition company at a booth at the booth, the young man
79:29 was um had done the processing of data using his own method was showing
79:39 his results different from my results to who passed by. And they,
79:45 I said, they were very different my results. And so what he
79:49 was to his uh people who are was uh uh those people in Amao
79:55 know what they're doing. He was recent phd um recipient. Uh and
80:02 phd had concerned converted ways. So thought himself to be a real expert
80:09 he knew that I was not, had never um thought about converted waves
80:16 all before, before this data came way. And so I, he
80:24 AO doesn't know what they're doing. at a certain point standing in his
80:29 around the booth was my boss and boss was unhappy to hear um these
80:39 for Amoco, especially coming from our contractor. So after the convention that
80:46 man was con was summoned to come Tulsa and explain himself and I was
80:52 to come to Tulsa from Houston to myself. And so during those sessions
81:00 uh direct discussion, we found out he had not realized that um the
81:09 midpoint gathers were not symmetrical. So had assumed that because of reciprocity,
81:15 should be the same. And so um he processed them the same where
81:21 should have processed and different. So uh his results were wrong. Ours
81:28 right. And shortly after that, was no longer employed by that
81:33 It's a good um oh good story shows that when you have science,
81:42 results like this, which uh uh and you don't understand uh the
81:48 Um You should get together as colleagues behind closed doors where the bosses are
81:53 watching and roll up your sleeves and out the science together and then uh
82:00 resolve the differences that way in So like he didn't do that and
82:05 paid the price by losing his So uh let us then uh summarize
82:15 lesson. Uh Today we have learned the previous lessons uh led to the
82:22 wave equation for fluids. And uh make more realistic assumptions inside solids.
82:29 get the vector wave equation. And particularly simple for uniform isotropic solids.
82:37 equations don't have anything in there about source. But when you put the
82:41 in there that uh uh makes uh uh uh uh an additional term.
82:48 furthermore, if the uh subsurface happens be in ho non homogeneous, non
82:55 , then uh uh the equations are and how we're gonna get different uh
83:02 types of solutions, direct waves reflected retracted ways, multiple ways diffraction
83:09 all of these things. And typically of those are uh you know,
83:15 on our uh instruments and we need learn how to separate out the ones
83:20 interested in from the ones we're not in. And then finally, a
83:26 interesting topic of elastic reciprocity. So brings us to the end of um
83:35 about that lecture? So, I this is maybe a good spot
83:43 It's 330. It's a little let's break for 15 minutes and come
83:48 at uh three at 345. Houston . I think it's the same in
83:53 . Yes. Uh Carlos, is the same in Colombia? No,
83:56 is 440. Here would be OK. Uh uh So not a
84:03 difference. So we'll see you back in 15 minutes and we'll take up
84:06 next lecture. OK. So, welcome back folks. Uh uh let
84:15 begin with um uh lecture four and remind you that uh you are going
84:24 be sending me this afternoon this Uh A question, one from the
84:31 uh lecture and one from the afternoon . So, uh uh uh
84:39 I'm only a little bit behind schedule . Uh We're doing fine on the
84:44 . So let us now consider are subject of elastic body waves.
84:59 So here are our um lesson We're gonna learn about the p how
85:08 wave equation has P wave solutions of types. Now, what we've talked
85:14 before was we derive the wave equation now we're going to find solutions to
85:21 . Ok. Uh And it's not be as easy as you think.
85:25 , but we, uh, well, well, we'll see as
85:30 go along. Uh, we're also see how, uh we can,
85:38 uh uh the, the concept of waves comes in naturally and how these
85:45 waves can be summed together in uh different ways. And uh we're gonna
85:52 able to make up any solution to problem in terms of sums of plane
85:59 . Now, those are playing How about so, uh uh when
86:03 have sources in the real world, always have sources and that we have
86:08 radiating from those sources and there's no waves, right? If you have
86:15 , a localized source, you're gonna waves which radiate a wave from that
86:22 , uh waves that are radiated and from that source curved wavefront, you're
86:29 gonna have plane waves anywhere. So you should be thinking, why are
86:34 dealing with plant ways? Well, reason I said is that we can
86:40 instruct any solution to any problem out plane waves. So it, if
86:46 we understand uh the plane wave well, that's the main thing we
86:52 to know, we can always postpone later putting them together to get way
86:57 solutions to more complicated problems involving, know, curved wavefront. Now,
87:05 the title of this course, we it's called seismic waves and ray.
87:11 so here, finally, we, begin to talk about wave fronts and
87:16 . Finally, they finally ready to about that. And of course,
87:21 have uh you are familiar with the of move out. This is an
87:25 term, move out. Uh uh back to the thirties, I
87:29 Uh And uh uh that's an essential of our modern acquisition design is move
87:41 . Now, when we have waves in the subsurface with different paths,
87:49 gonna cross each other and interfere with other. So what happens then do
87:55 ricochet off of each other, you , like build your balls on
87:59 on, on a pool table or they pass through each other like
88:05 So that's an important topic. All is P wave stuff with most of
88:11 data is P waves. But we're gonna have to think about sheer waves
88:16 converted webs. And then uh uh would say that most of us think
88:24 uh P wave about about psychic wave using a simplified mental model that
88:31 that we refer to as a convolutional . So that's we're gonna take this
88:38 uh towards the end of this And I would be surprised if we
88:43 to that today. I suppose that gonna get to that uh probably tomorrow
88:51 then all of that is wave To be honest, we're not really
88:56 in wave propagation as geophysicists. We're in the images that we can make
89:03 of these rays. Imagine we have uh an acquisition um survey out there
89:13 we got lots and lots of receivers lots and lots of sources and we
89:18 all the sources and record all the , zillions and zillions of ones and
89:24 this mind boggling amount of a number ones and zeros coming from all of
89:29 shots and all of these receivers. what we wanna do is uh make
89:36 image of the surp surface out of the best we can make and let
89:41 guide our uh bosses in deciding where drill the wells which are gonna find
89:48 oil, which are gonna find our for our salaries. That's really what
89:52 interested in, not the wave propagation , but to do there to get
89:58 , we have to understand the wave . So that's what this course is
90:02 . So starting off with that Wayne Pits, OK? Consider the
90:09 wave equation. This is in the in the ocean with one unknown,
90:16 unknown is the pressure and it's gonna uh varying uh in three dimensions and
90:22 . So here is the wave equation here now. Well, let's find
90:27 to that. Uh uh First, find only uh uh uh uh waves
90:33 are traveling in the X three So that means that this term is
90:37 be zero and this one is gonna zero. And so, uh the
90:42 is gonna simplify it down to So that looks pretty easy. Uh
90:47 Here's what we're gonna do. Instead doing a formal solution in the way
90:52 mathematician might do it, we're gonna things the way a, a physicist
90:57 do it. We're gonna guess the . And then we're gonna see if
91:01 , our guess is a good But you know, by plugging our
91:06 for the solution back into the original and see if it works.
91:11 So let us then guess that the pressure as a function of depth and
91:19 is given by E to the T Z over V where the E is
91:25 number we talked about. And you this question mark that means we don't
91:29 whether this is valid or not. go, we're gonna see if that
91:33 . So what we do is we this expression into the left side of
91:37 previous expression, the two derivatives in to time. And uh uh uh
91:45 here is a special property of, uh Oiler number that when you do
91:53 , when you take two derivatives with to time of this, it's the
91:57 as you start off before it doesn't a thing. Now, the right
92:05 of it is uh uh right side the wave equation is V squared times
92:10 derivative. You take two of the of this function with respect to Z
92:17 it brings down a minus one over squared because we did it twice
92:23 And then uh uh uh this V cancels out this uh V squared and
92:29 minus one squared makes uh uh minus . And so uh we end up
92:35 only this term here, which is same as we had here. So
92:41 the, the right hand side of . Uh uh Excuse me, thi
92:46 is the left side, this is right side, the two are
92:49 And so it works. So um is there a problem? So
92:56 me ask you um uh uh uh three of you, do you see
93:03 problem at all with this cos We found out that it um it
93:12 satisfy the wave equation. But is a problem? Do I hear any
93:21 uh proposing uh problem? Well, got one, look here on the
93:30 side of this equation, we have pressure and on the right side,
93:34 dimensions. So that can't be because that's not physically correct. We've
93:40 to have the same physical dimensions on sides of the equation. So we
93:45 put in a multiplicative constant right here guess what? It still works.
93:51 , is there still a problem? at this thing here. Does anybody
94:00 a problem with this equation. I see a problem whenever you have
94:08 exponent, the exponent has to be . And so this exponent is not
94:16 , this exponent has the dimensions of . So at me, it makes
94:20 difference whether the time is measured in or whether it's measured in uh uh
94:26 or whatever, you, you cannot an exponent with uh uh with physical
94:34 to it. So what we're gonna is we're gonna include a factor up
94:39 with dimensions of one over the OK. So uh that um uh
94:46 exponent is now dimensionless. And we're call that the angular frequency. And
94:55 uh uh it, uh uh we're allow this thing to depend the,
94:59 , the constant here uh should be to depend on frequency. So we
95:05 go through the same thing. we put this proposed expression into the
95:10 of motion. We work through the uh the uh the calculus. And
95:16 the left side is the same as had before, but with an omega
95:21 , um uh you know, in and um on the right side is
95:27 we had before, but now it's an omega squared over V squared.
95:31 here we're multiplying by V squared. again, that cancels out and
95:36 the right side equals the left here's the left side right there.
95:41 it still works. Now, the is, is there still a
95:48 Look at this expression here and tell , do you see any problems with
95:57 solution? It does solve the wave . But is there a problem?
96:05 , I see a problem. It's the kind of solution we want because
96:09 unstable. Look what happens at what , as time goes to a large
96:17 . Uh uh So as time goes infinity, this thing goes to infinity
96:22 as the distance goes to infinity because this minus sign here, it goes
96:27 zero. So that's not the kind a uh of a solution we
96:32 which, which becomes infinitely large at times. So what we want is
96:39 wave, we want something oscillator. the way we do that is we
96:43 in a minus uh uh put in , uh uh the imaginary number I
96:48 in here. So if you're not with the uh this eye, uh
96:58 it's a strange kind of number. call it an imaginary number, but
97:03 that's not really a good way to it, but that's what everybody
97:08 Uh And I would rather call it , but it's uh the official name
97:14 this is imaginary. And so, uh you should look up um um
97:23 eye in the glossary, what you do if you uh uh uh the
97:29 people can do that by clicking So there is a formula. Oh
97:37 there, there's still, there is formula which uh which expresses this oiler
97:46 to the I omega T minus uh zero V that uh uh can be
97:52 uh in any textbook on complex algebra be given by this expression.
98:00 it's the cosine of the same um without the I. So there's
98:06 I here, it's the cosine plus times the sign of the same
98:12 But now you can see very clearly this is a complex number with a
98:16 part and an imaginary part. And of them are uh functions of E
98:23 Z uh in this form. And so a again, it's possible for
98:30 to verify that this works. So what we're gonna, we're, we're
98:35 putting in an eye right here. the left side of the equation um
98:40 when we do our derivatives with respect time, we bring down I squared
98:47 omega squared. So the I squared minus one. So there, it
98:52 the right side brings down uh Linus over V times omega minus I omega
99:04 V squared, V squared cancel. again, we get the uh uh
99:11 uh the, the right side equals the left side, it still
99:19 Now, is there a problem, anybody see any problems with this?
99:37 . So here's a pro I see problem on the left is an observable
99:42 on the right is a complex So you can't have and observable with
99:50 complex numbers of uh our instruments are instruments. They don't observe any complex
100:02 . So don't worry about this. , I promise you that when we
100:07 the initial conditions and boundary conditions to any particular problem, the resulting pressure
100:14 be real. So that's a promise can hold me to that.
100:21 Now, this solution is a one plane wave traveling vertically and it is
100:31 in the direction of plus Z. let's do the geophysical thing and say
100:36 plus Z is pointing downwards. So solution is never gonna give us a
100:42 way of coming back up, it's gonna go down. And so how
100:48 we know that the wave is going ? It's because the phase remains constant
100:53 larger time if the Z is also . So uh the, the peak
100:59 the wave uh is arriving at a uh time at greater depths. So
101:07 can have waves waves that come back by putting it in here plus or
101:14 . And you can verify that this works. So you see, uh
101:22 have uh fixed up the solution more more and it still looks pretty much
101:29 we had before. But now it uh uh reasonable. Um But
101:35 is it the only solution? we could use different values of
101:42 So this is actually representative of an family of solutions. Each member of
101:47 family has a different frequency and maybe different coefficient in front. Now,
101:58 suppose the velocity V depends upon Well, uh uh it's gonna turn
102:05 that, that, that still And so when we have the
102:09 depending on the frequency, that's um uh uh that's called velocity dispersion.
102:17 means that uh waves of one frequency ahead of waves of the other
102:22 That means that uh the shape of wavelet changes as it moves through the
102:29 . So uh we're gonna take this , not now, but in chapter
102:38 , OK. Now, since the is linear, then the sum of
102:43 or more of these solutions from the of different uh solutions with different
102:48 Here's an example, uh Here's uh an example with the frequency equals omega
102:55 , here's a fr uh sample. Here's a solution with omega equals
103:01 We already decided that each of these uh is a solution. Now,
103:07 learned that, that some of these also a solution and you can verify
103:11 still works. Now, why do call these plane waves, we call
103:20 plane waves because uh uh there's no perpendicular to the direction of propagation.
103:25 these are propagating in the vertical they're the same. And for all
103:30 and all wives, she, she see any X's or any Ys here
103:35 all. No, that in the of the 19th century, in the
103:46 of the 18th century, there was very uh important result found by this
103:55 mathematician Fourier. So that's pronounced in French way, Jean Baptiste Joseph
104:05 So, but he proved that any function of time and space may,
104:10 be represented as a summer of plane , that's really remarkable. Any solution
104:19 be represented as an appropriate sum of waves with different uh coefficients for each
104:26 wave. And that what this means that physically by finding the family of
104:33 wave solutions to the one D scalar equation, we've already found all the
104:39 to that uh equation. All we to do is find the coefficients and
104:43 all the waves together. These coffi are called the spectrum. And we
104:49 these coefficient, we find them by uh uh uh uh fitting the coefficients
104:56 the initial conditions. And the boundary . Obviously, if you have a
105:01 source, that's gonna have different um spectrum than a weak source. And
105:08 a source can be have a variety different frequencies in the source.
105:19 So just think when you have a blast, that's an impulsive source and
105:26 is composed of a wide spectrum of uh all in sync with each
105:35 And so um that's uh one type source, here's another type of
105:44 Um uh a Viber size source in Viber size source. We have uh
105:51 uh suitable for uh land acquisition of truck drives up to the source
105:58 stops, lowers the pad lifts the of the truck off the wheels onto
106:05 pad and then vibrates the whole truck the pad sending vibratory signal down into
106:13 earth. It doesn't do just one . It makes a frequency sweep,
106:22 starts with low frequencies and sweeps up high frequencies and then it stops,
106:28 . The wheels drives up to the shock point and does it again.
106:32 here we have our laboratory signal lasting the source lasts for about 10 seconds
106:40 that goes down into the earth and , that gets recorded when it eventually
106:47 back to the receivers. That's what recorded 10 seconds of vibrations. But
106:53 have clever techniques using the facts that just developed here about um uh superposition
107:00 waves. We have clever techniques for computing and effective impulsive source from this
107:11 laboratory source. So that's um a outside of, of uh um this
107:21 , but you will encounter that concept in your course on image because
107:28 if you're having AAA source that lasts 10 seconds, when the uh that
107:35 those waves get to the receiver, gonna be lasting a lot more than
107:38 seconds. Why is it? Because be director waves uh arriving through a
107:46 of of uh 10 seconds, but other ways arriving by different ray
107:52 overlapping each other. All this complicated uh the noise of all these waves
108:01 uh uh at different frequencies, different , everything else. And we know
108:05 to disentangle all of this so that can construct good images out of that
108:11 of data. No notice that this oscillates forever, no matter how um
108:27 me long times are and how far distances are it? Right? It
108:33 it oscillates forever. That's not the of a solution we want, we
108:39 a solution which is localized in for example, from an impulsive
108:45 And this one goes on forever. Monsieur fourier guarantees that any localized signal
108:56 be decomposed into a song of plane which combine instructively into a localized wavelength
109:05 they combine destructively at long time. even though this thing goes on
109:12 we're still gonna be using that and not like a wavelet, we want
109:16 wavelet which uh uh uh uh which localized in time. And so even
109:24 we know from before yet that we construct such localized wavelengths out of these
109:32 which gone forever just by making the song isn't that remarkable. So all
109:40 these facts, we found out about vertically traveling plane waves. So now
109:46 think about the 3D scalar case where have waves traveling in all directions.
109:51 all we have to do is uh uh take is replace the second derivative
109:57 respect to Z with the laplacian operator again on P. And then the
110:05 of plenary solutions has these members. uh uh uh uh The infinite number
110:12 different frequencies are possible and different uh wave vectors. And uh all that
110:20 insist is that the length of the vector given by this sum uh has
110:31 OK. It is related to the the uh frequency in the F with
110:38 form that we said before. So verification requires that this uh um identity
110:47 . So you uh uh point out that the, this is a condition
110:52 the magnitude of the wave vector not its direction. So that now we
110:59 a proposed solution that works for any in any direction of travel. And
111:04 fact works for weighted sums of all in all directions of travel.
111:11 So that was all for propagation inside ocean, right? This is the
111:22 wave equation. OK. So that's but it's, it's not what we're
111:27 it's not all we want, we need to know about vector wave
111:32 uh inside solids. So uh uh we have this, this is for
111:39 waves all night. So the way looks like this where we have
111:46 the VP squared and since we're going all directions uh the PO operator and
111:53 unknown is the uh the particle displacement that he wave as a vector.
112:02 we're gonna do a similar thing we're assume that this uh uh yeah,
112:10 that, this can be solved by a so by uh by a plane
112:17 solution like this. And then, it's linear, that is because we
112:22 the unknown only appears to the first in all these terms, then we're
112:28 be able to uh uh instruct any solution by appropriate sums of plane waves
112:37 this. So does this work in more complicated situation? Oh yes.
112:43 we, we put this expression into for the left side and for the
112:47 side. And so uh uh as as we have a omega squared equals
112:54 squared times the square of the length the wave vector. It's gonna
113:03 OK. No, think that's, gonna think about some properties of wave
113:11 of uh solutions like this. And to uh do this, let's uh
113:19 back to the simple case of um wave propagation in the vertical direction
113:26 So we have only only Z And furthermore uh this is now a
113:32 and the direction of propagation. The of, of polarization of the uh
113:39 motion inside the wave is also in vertical direction. And uh uh so
113:46 expression here in the exponent is gonna be given by the sage function for
113:53 B waves which is given by this up clock. I jumped ahead a
114:00 bit. So let's think about a given place at two different
114:05 So uh uh here is the difference phase at those two different times.
114:11 uh uh time, one and time , the same place Z in both
114:17 . And of course, the same in both cases. So that difference
114:21 simplifies down to uh the frequency times difference in times because this other term
114:28 out. Now, let's uh uh that the two times correspond to two
114:35 peaks. In that case, this uh change in uh this change in
114:42 is equal to two pi that's the the uh uh number of radiance uh
114:48 two successive peaks, that's what's equal this. And so we conclude then
114:54 the uh uh uh the difference in is given by two pi omega just
115:02 here by omega. That's the cyclic one over the frequency, which is
115:08 period. And so uh uh there's discussion of that um a word and
115:19 glossary. Now let's do the other . We're gonna look at two different
115:26 at the same time. So at given time, the difference in the
115:30 in two different places is given by expression here. So we have the
115:35 time here and here and we have places here. And here simplifying that
115:42 uh we get this. And so the two places correspond to two successive
115:48 , then this difference is again two and so the uh um um
115:59 solving this equation, we find that uh this difference in uh positions is
116:10 to the VP over the frequency that's wavelength. And so now we can
116:18 why it is that we call VP velocity of the wave, we just
116:23 out that the wavelength is given by and the period is given by
116:27 And combining these, we find that velocity is the wavelength divided by the
116:34 previously this, we used this symbol just as AAA shorthand for this square
116:40 here. But now it's see, see that this quantity is the,
116:45 velocity of the wave it carries the forward by one wavelength in one period
116:52 it says here. So finally, understand we uh we understand what,
117:01 is the implications of physical meaning of um notation that we put in
117:09 if VP before up until this it was just notation. And now
117:16 see it as the interpretation, it's uh velocity of the wave. It's
117:23 the velocity of the particles within the , it's the velocity of the wave
117:30 . So let's think about this. uh qui so let's see. Uh
117:37 Yeah, le le I think it's turn says uh is this true or
117:41 plane wave may be defined in this ? Mhm Well, so let's
117:54 is this a way let's think about uh uh at, at, at
117:58 times, at uh uh very long , this is large. So the
118:03 thing gets to be infinitely large at times. So that's not what you
118:08 call away, right? So, order to get a wave out of
118:14 , we have to have an eye here and has to have an eye
118:18 it that makes it uh o oscillate with OK. So uh uh Carlos
118:27 is this true or false that this one D scalar wave equation has plane
118:34 solutions of this form here. Um can see that um but things
118:55 Well, so again, you made same mistake that Lily made uh uh
119:01 closely here, there's no eye here there's no I here. So these
119:10 uh uh uh qualities which are not , right? Uh uh uh uh
119:18 t gets to be large, this thing gets to be infinitely large.
119:22 uh the only way that uh you're get waves out of this is to
119:26 the eye in here. OK. , uh uh say that um uh
119:35 uh is this one true or Uh This is for the 3d scalar
119:41 equation. Does it have plane wave like this? Which are sums of
119:46 like this goes on forever here. uh um uh each term has um
119:54 coefficient, a leading coefficient which depends the frequency and it's the same frequency
120:01 the exponent and the uh wave vector a function of that frequency. And
120:07 over here we have another term with different frequency and this goes on,
120:12 sum goes on forever. It is statement true or false. Look at
120:16 carefully and see if there's any tricks there. I, I think
120:25 it, it's true. Yeah. . So you, you are quick
120:30 . Uh, uh uh, maybe quick. Uh uh uh when you
120:34 to the final exam, uh uh be so quick. You got plenty
120:39 time in the final exam. I'll read it carefully and see if
120:43 any tricks in there. OK. , but this time you're correct.
120:49 . Um um uh uh back to um le le um Is this true
120:58 false in this 3d scalar wave The wave vector is related to the
121:06 by this expression. I didn't hear false. Yeah, that's, that's
121:12 dead uh dead false. Uh it's uh wrong in lots of
121:17 Uh uh So if uh either of rest of you uh are confused by
121:22 , uh go back and look at relationship between the wave vector and the
121:27 vector and you'll see that it's not . OK. Um Carlos. Um
121:36 about this? True or false? I repeat the web equation, the
121:41 is related to the cyclic frequency uh this way. Is that right?
121:47 . OK. Do that is But uh um uh if you uh
121:51 back through the slides. You'll see we never did say this explicitly.
121:55 said it implicitly, but you are uh uh you're correct. Uh uh
122:03 , you're showing uh a good understanding what is explicitly taught. OK.
122:12 , all that's good for plane but in uh our experiments, we
122:17 not have any, any plane waves all. All of our waves have
122:23 wavefront. And why is that they because they spread away from a localized
122:31 . And uh so it's uh uh enough to call our sources or point
122:38 . And so those point sources are shed waves uh in all directions and
122:43 all gonna have curved wavefront. The way you can get um uh plane
122:49 is by having, for example, um uh uh uh vibrators uh s
122:59 every uh foot along the, the in both the X and the Y
123:05 and it goes on forever and trigger all together at the same time that
123:10 send AAA plane wave going down vertically a plane wave. But that's obviously
123:17 . Nobody's gonna do that. we do not have plane waves,
123:22 have spreading wave. So as we uh in the previous lesson, when
123:29 have a uh a uh a source there, uh we need to augment
123:35 uh uh the homogeneous wave equation with inhomogeneous term, we call it in
123:41 because it does not have the unknown it. They're known as P.
123:46 we're gonna a augment it with this here. Uh So this is gonna
123:51 a scalar in, in the scaler . So uh all of these are
123:58 and uh uh this uh time dimension we have here uh uh tells you
124:05 the source varies with time. Maybe impulsive or maybe not. Uh uh
124:10 all uh described by this time dependence this quantity here, but it's all
124:16 at the source. So at the has uh uh uh this uh a
124:22 function. Remember it's the Dira it doesn't have any subscripts here.
124:28 the direct delta, it's equal to at the source at, at A
124:33 equal zero and it's equal to uh , excuse me, it's equal to
124:40 at the uh R equal zero. for our not equal zero, it's
124:47 to zero. So it's a it goes up to infinity and it
124:52 the area under that spike is That's the definition of the, the
124:59 Dalton. It's got uh the height infinity, a breadth of zero and
125:06 times zero is the uh area under spike and it is defined to be
125:12 one. OK. So now what gonna do is redefine this uh uh
125:22 uh decompose this a into the uh into things like this. And um
125:29 so, uh the question is why we rename the source strength in this
125:34 . And uh it's to um uh see uh these things have um physical
125:42 , which we can understand when you at this, I would say it's
125:47 to say what is the physical dimension A. But when you look at
125:51 this way, you can see it's uh the velocity or separating out a
125:55 of VP squared, that's velocity here's pressure, here's distance somehow.
126:02 is RZ or that's gonna be a distance. We'll just define that
126:07 Now recall that delta of R has physical dimensions of one over RQ,
126:19 did that come from? So this a 3d problem. So uh uh
126:26 uh when I uh when I'm thinking the, the area under that
126:32 what I do is I uh make integral over uh uh from zero to
126:40 at all angles. And uh so uh integral is gonna be an integral
126:46 a volume. And um um so , I want the interval to uh
126:56 give me up, I want the of delta over a volume to be
127:03 . So the, the integral has dimensions of wi and so delta itself
127:10 have the uh the physical dimensions of over volume. So that, that
127:15 has dimensions of what? So the of this stuff inside the um bracket
127:28 pressure over times squared. So uh here is our n homogeneous uh wave
127:42 . And so we're gonna guess the . So here is the, guess
127:47 , it's a little bit more complicated we had before. It's gonna have
127:53 sum of plane waves. So every of these is a plane wave
127:57 uh, that it seemed to be bit a bit strange, uh,
128:02 , um, uh, uh, can see these are radiating waves,
128:07 gonna assume that it can be made a um a, a summation of
128:12 waves somehow by making the right sum . We're going to um yeah,
128:21 curved wavefront out of this. And gonna sum to uh uh a large
128:28 here. And uh then out in , we're gonna have um um we're
128:35 normalize this by the same number of . And then we're gonna multiply by
128:41 uh that characteristic distance to this R divided by the radius. So this
128:49 gonna yield geometric spreading as this wave to farther and further radii, it's
128:56 oscillate according to this and it's gonna decrease in amplitude by this. That's
129:04 make the geometric spreading. So that's guess. So it's just a guess
129:11 vector four a decomposition. And we in here explicitly the geometric spread.
129:24 how are we gonna check the validity this gas? Well, we're gonna
129:29 into the wave equation, this gas . So, on the right,
129:33 gonna put that into the wave equation the wave equation you remember has the
129:39 operator in there. So it's gonna um uh that's uh the LA plus
129:49 Dell Square is gonna operate on our for uh for A P. And
129:55 uh so it's, it's uh the operator is not gonna touch this because
130:01 depends only on uh uh omega and one is only on Omega in
130:06 The op the Lalas operator is uh to um uh operate on um uh
130:18 expression here. Let's back up, our guest. And so the whole
130:26 operator is gonna uh uh uh is have a term coming from this R
130:32 from this A. So here is uh the first one and here's the
130:39 one. Uh uh let's back up . Um So we operate with the
130:50 operator on this term here. We're this one over here there. It
130:55 , we ignored it and we operated the exponential with the laplacian operator.
131:00 got minus K squared times the original function here. Now, we're going
131:07 uh ignore this, the, this part using chain W calculus here
131:12 we're operating with a low plain operator this one over here right here.
131:23 , um oh, you don't know . What is this, this is
131:30 Boston operator of one over R. so I'm here to tell you that
131:36 quantity here is a minus delta of and to um uh prove that to
131:43 , look up in the um look up Dr delta and you will
131:49 , find uh a discussion of how the uh the plus in operator one
131:57 R leads to the drag delta. , you know, it might be
132:04 the uh uh the math 101 I forgot which one look, look
132:09 both of those and we'll find some of this. So at this,
132:13 uh at, at this time, just want to um uh leave that
132:20 your uh outside research. So now put this into the wave equation.
132:29 um on the left side, uh gonna put our guess into the wave
132:35 right here. So on, on left side, we get uh the
132:40 sum and we get a, a omega N squared for each, for
132:45 uh term in this sum. When do the second route with respect to
132:51 , we get omega squared here. the rest of this is the
132:57 And, and uh uh in in the second term, uh uh
133:03 have the, the laplacian operator operating P that's just this, which we
133:10 in the previous slide. This is for the left, the left side
133:14 this equation. So let us collect on the left side. Um And
133:22 of this relationship here, this first vanishes a after we collect the terms
133:27 through this on your own later. so the left side is just this
133:34 . And I wanna compare that with right side. Uh here's the right
133:39 . And so you can see that , when they cancel out the common
133:43 , this is just a statement of three decomposition of this uh um source
133:53 of the, the time function of source is uh just given here by
133:58 fourier decomposition of the time series. so our uh the, the guess
134:04 we have works. So these are circular waves which geometrically spread out according
134:13 one over R and as they spread , they oscillate according to this function
134:21 , notice that the geometric spreading is same for all frequencies, whereas the
134:26 is different for different frequencies. So me see, I forgot who was
134:34 uh who went on? I think gonna pick on you since you're in
134:39 of me is this statement true or , false. Here's our uh expression
134:44 the solution for the wave equation with source which we call it um uh
134:51 homogeneous wave equation. And it says uh the statement is because of the
134:58 decay by one of raw geometric spreading just like attenuation. Uh Well,
135:08 he didn't say this but I you know that attenuation is different for
135:14 frequencies than for low frequencies. Think know that uh uh uh that should
135:22 familiar idea to you folks that low uh attenuate less than high frequencies.
135:35 that's not, that isn't true for uh uh uh the geometric uh um
135:41 spreading, geometric spreading goes as one all all frequencies. So uh uh
135:48 uh uh high frequencies and low frequencies agen generation is different for a higher
135:55 . And so this statement is OK. So uh Carlos question number
136:03 in the wave equation, here's the equation up here with the source turn
136:09 here. And the solution is given here in this wave equation. This
136:14 operator differentiates the one over R that's one over higher, which makes a
136:21 delta function. And that's what we to match the source term. Is
136:26 true or false? Well, uh go back to our discussion of the
136:36 few minutes and you'll see that when uh uh when we operate with this
136:44 operator on this one over R, what produces the direct delta function.
136:52 that's what uh that's what we need get. Spreading wa waves, spreading
136:58 instead of plane waves. So this is true, it's true.
137:05 a brace. Um question number if the source here, here's our
137:12 uh our guess for the plane wave or for the radiating waves uh uh
137:19 AAA scalar, a scalar situation like ocean. And this is going to
137:26 to a marine acquisition survey where we a source in the water, some
137:32 the question says if the source were directional source, like a vibrator instead
137:38 a radial source like an explosion or an air gun, the equation would
137:45 the same and this solution would still valid. Is that true or
137:56 I think it's, I think it's . Yes, you're right. And
138:01 false because uh uh uh uh if uh uh uh i in this
138:09 if we have a directional source, um um uh we'd have to have
138:15 source. Uh uh Let's see. me back up here is uh the
138:20 with the source, same in all . There's no angular dependence here.
138:27 uh if we had a directional this term would be different. And
138:31 this proposed solution would be different So you, so you're, you
138:36 correct. Now, actually, in the, in a real survey,
138:47 have say an air gun uh a survey. So we have an airgun
138:54 . So the air gun, you what it does is it, it
138:57 a pulse of compressed air out into water. And so that pulse is
139:03 of like um a dynamite plant. so that pulse sends soundwaves out,
139:10 normally we don't have just a single uh uh no, we don't have
139:18 single ergo, we have an array guns and we have um uh the
139:25 uh the, the, the, have the air guns deployed in a
139:29 an array um uh which is designed the operator to focus the energy
139:37 So we do have in that a directional source composite sending waves preferentially
139:55 . But each one of the separate um uh is uh isotropic like
140:02 So that directional source uh uh would uh a sum of turns which are
140:09 this. So again, the answer be false. OK. Now,
140:23 brings us to the topic of seismic . So let me ask, first
140:35 about the seismic waves. Has anybody a seismic wave? Well,
140:41 but you have seen waves uh on surface of uh uh of a lake
140:46 you toss AAA rock into the lake it makes ripples and the ripples expand
140:52 . Those are surface waves, not waves, but you can see them
140:57 your eyeball, you can see the of action. Yeah, in that
141:09 where you have a lake and a and you throw the stone into the
141:15 . Do you see any rays you wavefront? But do you see any
141:22 ? I don't think so. I think you see any rays. So
141:28 come we're talking about seismic rays? take this further. Has anybody ever
141:46 a ray of sound in the Like for example, right now I'm
141:50 , you don't see any rays of . You don't hear the rays of
141:55 , you hear the wavefront of sound it goes from my mouth or from
142:00 speaker to your ear, it jiggle wavefront, jiggle the, um,
142:06 receptors in your ear. And so responding to the waves of sound.
142:15 , uh, we don't have any of sound. How about rays of
142:21 ? Uh, has anybody ever seen ray of light? Well, you
142:29 have seen a flashlight but that doesn't a ray of light. How about
142:33 laser? A laser, does a put out a ray of light?
142:41 , you can see it, hold laser in your hand, shine it
142:44 the room and you see a spot the wall, but you also see
142:50 ray of the laser light. But gonna say that's not really what you
142:56 as you're looking at that ray of , what you're seeing is the
143:02 the way what you're seeing is the inside the laser beam scattering off of
143:10 particles in the air. And, you're not really seeing a ray at
143:15 . You're seeing scattering from dust particles the air. If you don't have
143:21 dust particles, very uh uh clean in the room, you don't see
143:26 ray, you see the spot on wall but you don't see the
143:31 So what I'm thinking is that raise imaginary, raise our, in our
143:36 only. And what's really real is waves. So let's develop that idea
143:49 . Here is the wave equation for N homogeneous scalar wave equation in one
143:54 of the solution. So it looks this here is uh the oscillator part
144:01 is the ge geometric part and this function is defined in this way,
144:11 surfaces of of constant phase are called . Now the gradient of the phase
144:19 back along the right. So if form the gradient of this, of
144:24 um uh uh phase function, it's , the gradient of omega T minus
144:30 dot R the gradient of that that's a minus K, the gradient
144:44 phase points back along the race, minus the thing back along the
144:51 Now, I would uh think that of our intuition about our data is
144:56 on ray theory. Not on way . Everybody has seen pictures like
145:01 And um all right, um nobody any difficulties with pictures like this but
145:07 they really exist? Have you ever one? I would say no,
145:11 would say that nobody has ever seen a ray of light and they have
145:18 heard a ray of sound. And instruments also don't hear, don't detect
145:27 or sound. Instead they detect Now, when we have a localized
145:36 , it generates a wavelet which is four a series of many terms like
145:42 . Here is the sum of many like this. That's the fourier
145:46 the sum. And so here is oscillator factor. Here is the geometric
145:52 factor. Now let us think about such a solution in this wavelength.
146:10 think of it as uh uh composed , of a product of uh of
146:15 amplitude function which very slowly and a which oscillates rapidly a as the wave
146:24 uh uh away from the source, arrival time varies slowly with distance according
146:31 the velocity of sound. So if uh um um take this song and
146:39 it into these two parts uh and and wavelength and the way it depends
146:48 the arrival time, tea this is um arrival time, not um
146:55 arrival time, not period. So ray theory is the behavior uh uh
147:05 the behavior of the wavelet, not amplitude. So we can say it's
147:12 high frequency limit of wave theory. um we don't see the rays but
147:23 like to think about the rays. uh um uh but the rays
147:29 are the rays obeying uh the wave or not. Let's um but this
147:38 , high frequency limit into the uh wave equation. No, that's what
147:47 doing here. Except at the source , the wave equation looks like
147:51 So at the source point, there's away from the source point, uh
147:57 zero. So let's think about only left side here. And we're gonna
148:02 in here uh separate the, the in terms of amplitude function times uh
148:11 . So the, the aptitude function only on distance. And the,
148:17 the wavel, it depends upon distance time. So we put this expression
148:22 here. And so that's what we've here. Here's a times W A
148:28 W same thing over here. And now let's use the uh chain will
148:33 to uh work that out. And so um uh the, this derivative
148:39 the respective time doesn't affect the A all. So the A is here
148:44 we got a second derivative with respect time of W only. Now this
148:53 the plan operator operates both on A on W because both depend upon
149:02 So um uh since it's AAA G an operator, uh uh uh chain
149:09 calculus tells you there's gonna be three which is a times the plus of
149:16 plus W times the plus of A another term of a gradient of a
149:23 with gradient of W. And we're assume that the amplitude very slowly and
149:34 wave book varies rapidly. So uh uh this greve is a small number
149:43 this Lalova is a, a very number. And so we're left with
149:48 this term. So this is a equation for W. So this is
150:02 WAV part owner. So here we're completely the amplitude and we have a
150:07 equation for W the same material property there. Now, remembering that W
150:15 a function of time minus arrival So you have uh uh um uh
150:22 wavelet uh leaving the source with a shape and that same shape more or
150:29 arrives at the receiver some arrival time , that's capital t later. And
150:37 , um uh uh let us uh form the second rive uh with the
150:45 the time of this wave function. so, uh this is the same
150:52 as we did before. This is wave equation in terms of W.
150:58 now we're gonna take the second And um um um I'm gonna leave
151:12 part, I'm gonna leave this part alone for a second. And I'm
151:16 operate with the local operator on W means it's gonna operate on this part
151:23 W. So that gives us uh uh two terms a second derivative with
151:31 time. Um uh Let's see Oh I'll just say that using chain
151:43 , this qu this low plus operator on W leads to um a,
151:49 gradient of this arrival time dotted with . And also um uh a term
151:57 which is uh uh the possum operator on me and with these uh coefficients
152:10 . And so I think uh what do is I'll, I'll leave you
152:14 work out uh that uh since uh by death, by assumption, the
152:23 function varies rapidly in time. Uh uh single uh derivative with respect to
152:30 is a smaller number than this 12 with respect to time. And so
152:37 we can uh ignore this term. we're left with this expression here,
152:43 is more conveniently uh written like Uh this equation is, is called
152:52 icon equation. That's a um a word. And it, I think
152:59 means proper, proper equation. And the ray theory approximation to the wave
153:06 . So here's the wave equation for wavelength. And with the assumption that
153:11 wavelength ra it varies rapidly in uh , we reduce this to uh the
153:19 equation which says that the gradient of arrival time dotted with itself is equal
153:27 one over the square of the Now, what we wanna do is
153:33 wanna uh uh uh eliminate the the time from this equation to find an
153:41 for the ray in space. And uh I have a series of algebraic
153:52 mathematical manipulation in the next few slides does this and it finds an equation
154:03 the ray in space. In my , these are complicated arguments and I
154:10 not succeeded in finding a simple way express that. So what I'm gonna
154:16 is I'm going to skip over those and let you go back together.
154:23 , I go back later and um uh uh follow this carefully in your
154:32 time. I think it's, it's self explanatory and ending up with
154:43 This is the equation for the ray . So this is the path uh
154:49 the uh uh ray is going to uh following through the, through the
154:55 uh uh uh uh at as a of this parameter. S so think
155:00 s as a time parameter and the path uh goes through in homogeneous rocks
155:07 this pathway. And this is what deduce that equation should be this,
155:14 is the, the, the ray equation. So uh I would say
155:21 uh you're probably gonna be happy that skipped over those intervening points. Uh
155:27 can go back and um follow along yourself. Although I don't think that
155:34 really crucial. Uh I can guarantee that there's not going to be any
155:41 on the final exam coming from that . And to think what this
155:49 here's our equation for the ray path , a complicated formation. So let's
155:57 of a special case where the velocities constant. Then this uh uh ray
156:02 equation uh uh since this velocity is , this gradient is zero. But
156:13 me, this is not a This is uh uh uh yeah,
156:18 , it, it, it is gradient. The velocity here is a
156:22 as a function of three dimensional And here we took the greeting of
156:27 . But because by assumption here, velocity is constant, this is a
156:32 . And so we're left with this second derivative of the great path with
156:40 to the uh parameter along the ray is zero. So that means that
156:44 doesn't curve, it means that uh a straight line. So that's
156:48 you know already, if the, the um medium is constant, then
156:54 are gonna be straight lines. That's we just grew in an earlier
157:01 So that's almost trivial. But I would say it's reassuring that we
157:06 um special uh that we find a case corresponding to what we no
157:16 So here's another special case, let's that the velocity varies only in the
157:22 direction. So now the ray equation this instead of the gradient here,
157:28 have only the variation with Z. now what we're gonna do is we're
157:34 form the cross product. We're gonna uh take the Z cross, the
157:39 side equals Z cross the right interchange the order of operation so that
157:46 Z cross comes inside this derivative and Z cross comes inside this derivative and
157:59 have Z cross Z. So uh you have uh a vector or making
158:04 cross product with itself, that's a . So we have this term along
158:16 zero. And uh uh in the in the slides that we um uh
158:26 over uh we define a slowness vector is um uh which uh the vector
158:35 lowercase P vector defined in this So that this result here says that
158:45 variation with S of the Z cross equals zero. It's CP right in
158:55 . So Z cross P is equal the uh gives the key component,
159:04 gives the X component in the X or X is the horizontal unit factor
159:09 the plane containing the ray, the slowness. OK. So let me
159:16 here. I think um let's go this again. If you have a
159:27 P and you cross with Z that's to them because of the definition of
159:32 cross product go back and look it . The definition of the cross product
159:37 that uh uh that gives you the X component of P and it's
159:42 vector there is the unit vector in X direction. So this quantity piece
159:47 X is called the horizontal slowness and often called the ray parameter. So
159:53 many uh textbooks, you'll see a of this ray parameter um uh uh
160:04 the X it's called P. But I'm gonna keep the X as a
160:10 . And now this expression says that variation of the horizontal component of the
160:18 uh uh parameter with the uh uh you go along the ray that is
160:26 . So what that means is that ray parameter is constant along the
160:32 So you know about this, this called, this is a Snell's law
160:37 I just derive Snell's law for the where um uh the velocity varies only
160:48 depth. And so when a ray going down through a, a medium
160:55 , which is varying with depth, turning like this. So that the
160:59 vector that the ray parameter defining this is a constant that is Snell's
161:09 And so we deduced that out of consideration of the property of array,
161:17 , in the sediment requests, the usually increases with that. So if
161:24 , in, in a case like , uh um um according to Snell's
161:29 , uh the angle here has to increase in order to keep uh um
161:37 uh the ray parameter con So since thing is increasing this one better increase
161:43 order to keep uh the ray parameter , that means that the ray bends
161:50 and then it reflects and the ray apart, getting uh back towards the
161:57 . And uh uh I I noticed the ray parameter is still constant,
162:02 remains unchanged despite the reflection. So is, you have, I know
162:08 seen pictures like this cartoons like this . And so, um uh if
162:16 rays, if the layers are are velocity as you go down, that's
162:24 the uh the seismic arrays look However, that's not always the
162:31 Here's a case where we have a layer. Why is it slow?
162:34 , maybe it's got in here a pathology maybe in here, it's got
162:40 , maybe in here, it's got , whatever is the reason this layer
162:45 slow. And so that means that to Snell's law, it's got uh
162:50 downwards and then this one is fast . So a again recovers and on
162:56 way up, it's gonna look the . So uh this uh um this
163:08 of smells law coming out of consideration the icon equation is really quite
163:15 It tells us how it raised a direction in the subsurface according to um
163:23 velocity that they're uh uh traveling they um uh they change with their
163:30 such that the horizontal component of the defined in this way is always a
163:36 . Now, you will re you recall from chapter one which I didn't
163:44 uh um directly, you will recall picture and this is a good time
163:51 uh uh review this picture. Uh I know that all of you have
163:56 courses in whole earth uh um uh which um uh as pictures in here
164:05 this. So you see uh the crust is here so thin on
164:10 scale, you can barely see And then there's the mantle and then
164:14 outer core and the inner core. so let's uh uh let's uh follow
164:22 cartoon in some detail. So this the outer core made out of liquid
164:29 . And this part here is the core. To me, this part
164:32 is the mantle. You can see wavefront going out here, can you
164:37 that? And you can, of , you can also see the rays
164:44 here is the inner core made out solid iron, solid iron and maybe
164:50 um uh elements as well. But uh uh for now it's let's
164:55 it's a solid. So now let's at the rays, the black rays
165:01 P waves and they curve upwards here they curve upwards here following Snell's
165:10 the red curves are sheer wave, red rays are sheer waves also curving
165:17 because of Snell's law. And of , uh these are slower, the
165:22 waves are slower than the P waves not evident on this cartoon. But
165:27 course, you know that and these curve up since for both P and
165:33 the deeper rocks have higher velocity. is that? Well, for
165:37 uh the higher pressures at those greater , I mean that the velocities are
165:43 and also the uh composition may uh if we're here. And so,
165:50 uh by the way, this slide um I got it from Professor Lee
165:55 this department. Now, let's look happens here at the outer core,
166:02 the outer core. Uh the sheer stop because these shear waves can't go
166:09 the liquid out of core. And , the, the P waves um
166:16 downwards. So that's following Snell's Law um um inside the core of uh
166:30 P wave velocity uh uh the of liquid iron is less than the P
166:36 velocity in the rock because the rock solid and the core is liquid.
166:41 for um uh I don't know for waves inside the core, this term
166:50 is zero. So uh so that the uh the velocity in the inner
166:55 is less than velocity in the So the rays refract down, but
167:03 as you go deeper inside the the pressure is higher and higher.
167:06 they, they curve back up following law and then you get the same
167:13 uh oh behavior on the way Now this is a cartoon uh not
167:26 data. So I'll just point out you that uh in the cartoon is
167:31 at this place here where we have an interface between the uh liquid outer
167:37 and the solid inner core, the should reflect up following snowball instead of
167:43 . But uh so that uh the is wrong here. No, I
167:58 you that rays are nobody's ever seen ray but race are, are mathematical
168:08 which describe the behavior of the uh the uh propagation in a way which
168:14 complementary to the wavefront. So uh wavefront are real but the rays are
168:24 constructs which help to describe the propagation the sound. Now, in our
168:34 , there is a need to be to calculate rate paths and we're gonna
168:41 that using the equation which II I derived a few minutes ago.
168:51 um I'll tell you a story about illustrating the need to um calculate
169:06 So, when I first joined this was in 1980 maybe before some
169:15 you were born. And in those , we were just beginning to explore
169:22 the Gulf of Mexico. And we that lots of places in the Gulf
169:28 Mexico, they have uh you within the sedimentary secrets, there are
169:34 bodies of salt and this is um a common occurrence, but uh um
169:44 it only happens in some places in earth and it, it happens because
169:49 geologic time, at certain periods of time, the Gulf of Mexico was
170:00 because of continental drift and plate The Gulf of Mexico was a closed
170:06 , sort of like the Mediterranean Mediterranean Sea is now. And uh
170:12 the, it was a warm water . And out of that warm water
170:18 was precipitated, eroded off the continents the basin and precipitated as salt.
170:28 so those salt bodies later got covered with mud. So the salt bodies
170:36 the sedimentary column and they have velocities are very different from the velocities of
170:45 surrounding sediment. So that means that rays go down through this salt
170:51 they refract, they change direction a . And so it turns out that
170:58 uh um more complicated story, but there are lots of oil reservoirs beneath
171:05 salt bodies. But in those early , we could not uh get good
171:13 of them because our um imaging algorithms not able to follow properly um are
171:27 as they went down through the The salt was like an irregular lens
171:34 the rays as uh the sound went . So that when they came back
171:39 the surface, we did not get images. So we had a lot
171:44 frustration in those early days drilling expensive in the wrong places because we could
171:54 see well beneath you couldn't get good below the salt because basically of Snell's
172:01 like we just showing hair. So a we thought OK, we need
172:09 have better imaging algorithms. Uh When first joined the company, the,
172:16 standard imaging algorithm was what we now Dix stacking. We would simply flatten
172:25 layers according to Dix's formula and we uh um stack the data common midpoint
172:32 . And that was our image. , that's a good enough algorithm if
172:36 have flat layers, but if you have flat layers, you need something
172:41 complicated. So the next idea we was something we call kerk off
172:48 So this course is not a course imaging. Mm You will learn more
172:56 imaging uh in your next course. co curve migration is uh an imaging
173:03 which is more sophisticated than uh N out corrections. And um it,
173:11 handles situations which are are are more than horizontal layers. And that made
173:18 improvement. But still, we had of um of uh poor images and
173:26 of expensive uh wells drilled in the place. And so uh we uh
173:38 try the next thing uh improve our imaging algorithms. So we would do
173:44 with a, a wave equation migration is now called reverse time migration.
173:53 that made an improvement but still not . So after making all these improvements
174:00 the imaging algorithms, finally, we to the conclusion, we need better
174:08 . Uh Actually, we thought to do, we need better data.
174:16 of our data is data coming from streamer surveys. And in those
174:23 all the um acquisition was what we call narrow ASU acquisition. So we
174:30 um streamers behind the um uh survey . And uh in the earliest
174:38 we had a single streamer maybe three long. And as uh time went
174:48 and we got better and better we had longer and longer streamers.
174:54 then we had the idea that we um put some special equipment in the
175:00 and we could have several streamers behind boat. And uh uh uh by
175:06 mid eighties, mid nineties, mid , we had a race of receivers
175:12 kilometers long and uh maybe one kilometer . And that was the data that
175:20 were using to try to image underneath salt bodies. And we were getting
175:27 results with our better acquisition and better where there better um algorithms but still
175:36 good enough. So at Amaco, asked ourselves what would happen if we
175:43 wide azimuth acquisition, suppose we were to illuminate the subsurface uh with uh
175:55 is going into our receivers from a distribution of as not just from behind
176:03 boat but from a wide distribution of . So we did a lot of
176:09 in the computer in the computer, made a model of subs server salt
176:18 and sent rays down through there uh with a narrow distribution of asthma and
176:24 that data and got lousy results. since we knew what was in the
176:29 , we knew what was down there find and we couldn't see it very
176:35 . And then we uh did in computer set rays down from the side
176:43 suddenly we got better images. And was that? But because it turned
176:50 that it turned out that we were bad images with narrow Asma acquisition because
177:00 salt and the overburden was bending the so that large parts of the reservoir
177:07 not being illuminated at all. All rays were going somewhere else. So
177:13 couldn't get good images of the reservoir of ray bending in the overberg.
177:20 we solve that problem by uh sending down um in the model from different
177:30 from the side. Wow, we good images, same algorithms, better
177:38 design. So we said, so what do we need to do
177:42 the real world to make white as ? And so we uh we found
177:48 uh several different ways to do One is by putting uh another source
177:52 off to the side. So you a regular source boat with the streamers
177:58 10 kilometers behind it and then off the side, you have another source
178:03 , no more receivers, but another and then that's firing also and you
178:08 get um rays coming in from the . And sure enough, that's more
178:20 , a lot more expensive. But meant getting better images and putting uh
178:27 well in the right place instead of wrong place. So that was
178:31 an a a an acquisition expense well doing. Then there's other ways to
178:39 wide, a aqui position. For , you can put ocean bottom receivers
178:44 the sea floor and then you can your boat, your source boat uh
178:49 you want and um get wide asthma that way. And so we spent
179:01 a small amount of money on the . All right, modeling this.
179:08 then we spent 100 and $50 million it out in the field of various
179:15 of wide as acquisition. And we out that they all work. Some
179:21 them are more appropriate in some circumstances than others depending on the size of
179:27 prospect and so on. But we out that everything worked why as an
179:34 works. So we uh during that , um BP bought AOL and uh
179:44 BP is a fairly open company, uh talked about it uh in the
179:51 and pretty soon every seismic acquisition contractor offering wide Asma services and they made
179:59 lot of money and their clients found lot of oil underneath complicated overburdens which
180:13 the rays of sound going down. um uh with a, a better
180:26 and with modern imaging algorithms, we're to see beneath a very complicated
180:35 And the only way that we could to that conclusion was by modeling in
180:41 computer tracing rays through a model in computer to see where the rays ended
180:49 and why it was that we needed wide azimuth acquisition. So that brings
180:57 back to this slide, which is front of you about calculating race.
181:03 let's um uh uh see about how would go about calculating ray pass in
181:09 overburden. So uh uh uh you to do that where the uh velocity
181:17 the overburden is complicated, for salt bodies and or when the reflectors
181:22 curved, so that um oh uh we're gonna do these calculations and in
181:29 cases, it's necessary to know or estimate the velocity field. And in
181:34 real, in the, in the , of course, you specify the
181:38 in the real world. You have estimate the velocity from the data.
181:43 an additional complication. Now, to these rate calculations, there are three
181:50 you might wanna use. One is shooting, one is called exploding reflector
181:55 the other one is called source to . So let's take those up
182:00 shooting, here's shooting. Uh So is the most intuitive way you simply
182:04 the ray pa equation which we just . Uh you have a sort of
182:09 conditions and those initial conditions specify the of the shot points and the takeoff
182:16 , for example, you might specify take off angle is zero degrees from
182:22 surface, one degree, two degree so on. And uh you might
182:25 this in polar angle or an ASU or both depending on whether you're doing
182:31 two D uh calculation or a 3D . By the way, do people
182:36 what is what we mean by 2.5 ? So tell us your time.
182:49 . So, so you have a D model, so you can show
182:53 model on the screen and uh uh behind the screen and in front of
182:58 screen, it's the same. So a two D model extended uh um
183:04 uh behind the screen and in front the screen, not very realistic,
183:09 ? But you can learn some things that. And then in that
183:13 you do 3d wave application. um, uh, that's called 2.5
183:20 and, uh, so, you'll, uh, you can learn
183:24 that from the glossary. Now, the problem that, um, when
183:34 specify the takeoff angles, that's like a rifle, right? You aim
183:39 rifle and you shoot at one two degrees, three degrees and so
183:43 . Uh But uh that ray, shoot off at two degrees that does
183:48 come back to the surface at any your receiver positions that come, that
183:53 comes back between the receiver positions just as you can see that uh um
184:02 you specify the uh initial angle, don't, you don't know where it's
184:06 uh come out until you do the and it always comes out uh between
184:13 uh uh positions where you have So here's another idea that uh it's
184:19 the reflecting exploding reflector. So here solve the very path equation with a
184:25 of initial conditions which specify the reflecting , the reflecting points, not the
184:33 points, reflecting points um on the horizon and the takeoff angles at the
184:41 . So usually what we do is take off angles of zero degrees.
184:45 you have sources and receivers are at same surface location and, and uh
184:52 uh um the, the rays reflect the reflector according to the curvature of
184:59 reflector. OK. There's a difficulty also that the emitted rays will not
185:06 at any of the receiver points that have. So here's an example shown
185:12 the next slide. So here we a curved reflector down here and we
185:17 uh uh uh uh it's call it uh an exploding reflector. So we
185:22 rays in e emitted normal incidence rays from this reflector going up to the
185:30 . And we have our receivers along . And you see that most of
185:34 rays miss the receiver, the rays um uh uh spaced at equal intervals
185:43 here, uh uh would take off normal to the reflector and most of
185:48 miss the reflector. So the way solve that problem is have more reflectors
185:53 the ground and more source points And so uh that's those are uh
186:02 affect solutions, but they, they be expensive, right? And then
186:06 more reflectors you have the more expensive now uh I want you to observe
186:15 , here's a place where the rays through this zone with high density,
186:22 lots of rays going through there. is that because of the curvature of
186:27 um reflector down here? So if had AAA receiver down here, you'd
186:33 a lot of energy. Well, course, you don't have a receiver
186:36 there. You have the receivers up to the surface and not at the
186:41 because this is a marine sit uh uh ex examples. So these receivers
186:47 towed slightly below the surface. But can see that these receivers in here
186:53 gonna be receiving a lot more energy these. Here here, the uh
186:59 energy gets spread out because of this down here. And here the the
187:04 gets concentrated because of this curvature down . And notice here these so the
187:13 are concentrated here. They uh they between the receivers and notice here these
187:23 lines of constant um arrival time here notice what's let me show you this
187:31 uh once around, notice what happens , it goes like this and it
187:36 and it doubles back and doubles back . So um this is called a
187:45 a trip application. But because of curvature down here, these rays are
187:53 are uh these uh see this ray is coming from here. This ray
188:01 is coming from here back of the rays coming in from the right
188:10 . It it comes to a cusp triples it back and triples back.
188:16 it's complicated in here because of this curvature in the surf that was down
188:28 . So these are feature uh uh features here happen um in the subsurface
188:39 , and they're gonna be there in some guise no matter how you
188:44 it because of the geometry here or you might have a similar situation caused
188:51 lateral variations in velocity in the over . You can tell from this uh
188:57 cartoon that the velocity here is a and all the complications you see here
189:03 from the curvature of the reflective. the the most in uh uh uh
189:12 way uh to calculate raise is called to receiver. And so you here
189:19 solve the ra at equation with a of boundary conditions which specify the initial
189:27 of the rays that is the shot and the final position of the rays
189:31 is the receiver points. And um is conceptually more difficult but straightforward to
189:37 computer. So uh that's all I say about um oh wow, about
189:45 repat calculations. But uh uh from example I gave uh you can see
189:52 watts uh uh large improvements in our happened in the last 20 years because
190:04 uh rep pa calculations like this uh realistic models with salt body in the
190:13 . And that led to major development acquisition technology, a acquisition technology which
190:22 call wide asthma acquisition. And that to a lot of discoveries of oil
190:29 gas beneath the salt bodies in the of Mexico and elsewhere in the world
190:35 show large economic consequences from these considerations , of repat distortions. So I
190:48 here a little quiz. Um uh uh let me start with BEA uh
190:56 this uh statement uh uh uh who a false or what it said,
191:05 of these answers is correct. It here that ray theory is the limiting
191:10 of wave theory in the limit of low frequency or high frequency or in
191:16 limit of isotropy, ignoring anisotropy or the limit of homogeneity, ignoring any
191:24 . Which of these special cases here ray theory. We say that I
191:37 thinking now we gave this answer explicitly half an hour ago, maybe,
191:53 , about a half an hour And where, what we said was
191:57 ray theory is the limiting case of theory corresponding to high frequency. And
192:05 you go back in the uh uh notes as you have them, you'll
192:09 where we stated that explicitly about half hour ago. And so um um
192:16 le you're up next here. It the icon equation is derived from the
192:23 equation assuming that same uh lemony Is that true or false? Remember
192:32 icon equation uh oh With, say again. He was very,
192:42 the icon. So the icon equation um uh we did derive explicitly and
192:48 it says is that the gradient of arrival time dotted with itself, the
192:55 is a vector to dot that vector itself and that uh dot product is
193:02 to one over the velocity squared. um so the question said that's the
193:11 equation And so uh we derive that from the wave equation using the assumption
193:18 high frequency we did. So, , uh what is this, the
193:29 equation? It's a scalar equation. uh w which is, it is
193:33 what we have here is a bunch equations. Uh ABC or D Carlos
193:39 me which one is the icon Not true professor. Yeah.
193:51 uh uh you know, I don't you folks for uh for stumbling over
193:56 uh ray theory um uh issues because pretty complicated um uh algebraically and I
194:08 over some. So I don't really you guys for stumbling here, but
194:12 answer was given explicitly uh uh 20 ago or so. It's this one
194:19 . So uh Carlos, I'm gonna you another chance here. Uh Number
194:23 , it says the Ray Paic equation the change of ray path along the
194:31 as the re as the parameter S according to the distribution of velocity.
194:40 that true? OK. I think is true. Yeah, that is
194:46 . That's a good word description of ray path equation. And so uh
194:50 for the rest of you go back , and see where we ended up
194:55 a lot of manipulations, we ended , we ended up with the ray
194:59 equation. And uh uh that's pretty like we describe here. So Li
195:07 um uh this one's for you. this true or false? The general
195:12 to the ray path equation is called Law. Is that true or
195:18 No. Uh uh uh um you're uh I think you're guessing.
195:23 And instead of thinking carefully, we St's Law as a solution to the
195:31 pa equation in this special case, the velocities uh change only with
195:38 only good that so when uh so you know, that's not a bad
195:46 in many cases in in in the . But not always. For
195:50 when you have uh salt bodies, a great example where the velocity
195:56 not just with depth but laterally as because of the presence of the
196:02 So uh this one is false. brace, this one is for
196:11 It says when the velocity does vary depth only. So this is gonna
196:14 us the Snell case, the rays up whether or not it reflects along
196:21 way. Now is is this statement or false? Do the rays bend
196:30 for this reason in parenthesis, whether not it reflects, I think it
196:40 true. Well, I showed you example where it's not true. I
196:44 you an example of the raise bend if it encounters a slow layer and
196:52 back in the uh slides and you'll a, a slide where there's a
196:56 a three layer model three layer cartoon the middle layer is colored yellow and
197:02 that yellow, it's uh uh So, in that case, the
197:07 been down, I guess I, maybe didn't read the question correctly because
197:16 , I don't see where it says this is slower, the velocity.
197:22 , I, uh uh it doesn't slower. Don't worry in here.
197:26 just says it varies with depth and says, uh, the rays been
197:31 , uh, it doesn't say it or faster. And so this is
197:35 good example of a question where you to think through uh uh exactly what
197:41 says and think through um exactly what learned and answer on that basis.
197:47 this one is false because if the goes with depth only, but some
197:53 the layers are slow. In that , the rays bend down.
197:59 this, this part about reflection, um what we call a red
198:04 Do you know, do you have um uh uh do you know that
198:11 phrase uh in Spanish or in Chinese we say that there's a red
198:18 That's a false clue. OK. uh uh uh uh I don't know
198:25 origin of that English phrase, but uh English speakers know that phrase that
198:31 there's a red herring, that's a clue. And so this here is
198:35 false clue. It, it makes think about reflections. And sure
198:40 uh Snell's law uh uh applies even a reflected raise only in the case
198:48 the velocity varies with depth only. uh uh um uh then snow's law
198:56 only to reflective ways. OK. we're getting short on time, but
199:01 think we have, have time to about uh move out. So uh
199:09 knows what move out is and in uniform isotropic layer with the horizontal
199:16 the move out follows the Pythagorean theorem is this uh you uh uh you
199:24 of this as uh uh uh uh move out equation. But when you
199:30 uh look at this more carefully with diagram, with these uh uh cartoon
199:36 mind, you'll see that this is the same as the Pythagorean zero.
199:43 , that's not a good representation of subsurface subsurface is not uniform like
199:50 So um if you have many layers gets more complicated, for example,
199:58 rays are gonna go down like. I'm coming here. So here I'm
200:02 a, a case where it's getting and faster. So the rays are
200:08 up and uh at every uh in layer, the angle is governed by
200:15 Law following this one right here. uh uh they uh refract at each
200:23 and then uh uh uh the arrival after all this travel is given by
200:31 the sum of the one way travel . OK. So the uh the
200:37 comes because you're going down and up one of these is a one way
200:43 time and you're summing over all the . I think that's pretty obvious.
200:47 the arrival uh uh uh the rival , the arrival distance is given by
200:54 formula. And you see it, involves the uh just like this.
200:59 involves the ray parameter, horizontal ray . This one also involves the horizontal
201:05 parameter. This formula is a bit . It sums up all of the
201:11 like this, the horizontal leg in , in each uh layer. And
201:17 have again twice of them because this the horizontal leg on the down growing
201:24 . Uh And uh uh it's the as on the upcoming layer. That's
201:29 same. And here you need uh gonna some overall layers. And so
201:37 uh we have these two are Actually, uh we put a receiver
201:43 X and we arrive, we measure arrival time uh T uh And what
201:49 don't know is the ray parameter which uh the ray is following uh down
201:56 . So let's eliminate ray parameter to the time arrival time as a function
202:04 offset. So first, uh uh is the arrival time as a function
202:11 the ray parameter um with the sum we talked about before. So we
202:19 convert this arrival time into a vertical uh arrival time divided by co
202:40 Uh Y You know, I'm a OK. Uh I need to talk
202:51 that more. So here's the vertical time and where does the coast state
203:00 from? It's the oblique uh um travel time in here, divided uh
203:07 uh uh into the vertical travel time gives you the cosine theta. So
203:14 comes from simple geometry. And then uh you know that the cosine is
203:21 to the square root of one minus squared. And you know that the
203:27 squared is given by the ray parameter uh uh uh the velocity using Snell's
203:37 . So that's all for the uh the arrival time. Now, for
203:41 offset, we do the same we start off with the uh this
203:47 uh for the summing up all the , the partial offsets in all of
203:53 layers. And uh we expressed a here as the, the product of
204:01 uh uh B times Z. Uh then uh let's see what do we
204:08 here. We take this uh um ray parameter out here and oh
204:17 Uh uh this time is equal to vertical time and divided by um um
204:25 cosine. And so that leaves us with the total offset is equal to
204:32 sum of all the partial offsets with expressions. Only involving key. See
204:43 no angle here, angles here, here, but we got rid of
204:48 angles and we're ended up with P . Now you noticed here that the
204:57 is an odd function of the ray . So if you had ray parameter
205:02 down from left to right, instead right to left, you had a
205:06 sign for the offset act. Now, here, the next thing
205:12 do is our old trick of the expansion. We simplify um this expression
205:21 a complicated expression involving uh the, ray parameter. Uh We're gonna assume
205:27 we have only small ray parameters. this is gonna be only a small
205:34 for vertical propagation to zero. And um uh and we're gonna consider only
205:47 offset acquisition. So uh that G means we're gonna use this tailor expansion
205:56 uh um um cool. But um do we use in it?
206:05 the initial offset is obviously a the initial um uh of the,
206:11 , the, of the, the of this derivative evaluated at uh uh
206:17 know PX equals zero, evaluated here the origin is uh while we take
206:23 through, we have an expression for as a function of B and we
206:27 take the derivative. So uh here that uh that expression I'm I'm gonna
206:34 up here is here is the expression T as the function of uh of
206:42 parameter to me. Uh Here is expression for offset as the function of
206:48 parameter. So now we just take derivative of that and um uh
206:55 we induce this and um yeah, . Let's see here. Oh
207:16 So I've given here this is directly the derivative. And I'm gonna assert
207:23 this is equal to uh the R average velocity times T zero. So
207:31 do, how do I come by ? Take this um uh T zero
207:36 divide it over here? And so you have a sum, a weighted
207:43 of the travel times the travel time going to be uh um um
207:51 excuse me, you have a weighted of the square of the interval
207:58 And what are the weights? The are the um uh the individual
208:04 And then at the end we divide T zero, which is the sum
208:09 the weights. So that is an MS average squared. Basically, we're
208:16 the uh uh uh uh uh we're doing the average over the stack
208:22 the square of the velocities. And gives us an R MS average
208:27 So that to a, a AAA approximation, we find that the offset
208:33 given by this expression here because that's zero. And this one is given
208:39 this. Now we're gonna do the thing with uh the time. So
208:46 a tailor expansion. And uh notice that um uh the, the small
208:53 now is is the square of the uh of the ray Pran taking the
209:02 of time with respect to P square work through the algebra. And you
209:09 uh that um that is also proportional uh the, the R MS velocity
209:17 . And this actually defines the R velocity squared. And that shows that
209:25 so that the time uh is given fir a first order tailor expansion in
209:31 small parameter P square. So then can substitute from the uh uh a
209:46 for uh the offset. We we that gonna back up here here
209:54 the expression for the offset. So just all for uh for the uh
210:00 parameter in terms of the offset, this stick that in here. And
210:08 find that in the end, we that T is equal to T zero
210:13 this uh correction term for the So that's our result. However,
210:24 I point out is we find out this is not such a good
210:30 And when you think about why it's , it's because it's never exact.
210:35 in a one layer case in one case, we have the hyperbolic move
210:42 equation. This is not the hyperbolic . This is the move out equation
210:48 time, not for the square So um uh a, a clever
210:57 to do to improve the approximation is we just square it and we uh
211:04 neglect the uh high order terms and uh uh now we find that the
211:10 time is approximately equal to the square the vertical time with this um corruption
211:17 involving only the R MS square. exactly in the one layer case,
211:23 uh um uh um uh that's the thera. And it comes directly from
211:32 assumptions uh one, the isotropic And uh uh uh because of these
211:40 , uh oh no, the ray is a constant. And then we
211:49 ourselves to small ray uh parameters. this equation right here was um uh
211:58 arrived by this gentleman here whose name Dix and he was one of my
212:05 when I was an undergraduate. So I know that you're familiar with that
212:11 Dick. And matter of fact, think I mentioned it to, I
212:15 it earlier in this uh um earlier in this lecture. And so
212:28 what we um learned here is that for the, for the, the
212:34 case, we have this approximation for arrival times as a function offset.
212:41 , now when you look at uh uh that's exactly the same as
212:46 uh a one layer case with the uh velocity, which is the VR
212:51 here. But I think this is um a confusing uh analogy because uh
212:59 the rays and our many layer case not straight. And uh so uh
213:06 to me, this is not a analogy to say that, that this
213:10 the same as a hypothetical problem. it was derived from uh a one
213:18 layer of coarse isotropic layers. So didn't see course here. But uh
213:27 we're gonna uh if the layers turn to be a, a fine fine
213:32 case that would be uh different. so uh let's leave that issue aside
213:40 now, we'll come back to that um in the 10th lecture of this
213:51 . Now this is an approximation of uh um turns out to be really
213:57 really good approximation. And it's good dipping layers and it's good for anisotropic
214:07 . If the move out velocity is as a processing parameter, not an
214:15 MS velocity. So if we just , if we just replace the arm
214:21 velocity with the move out velocity, a really good approximation. And we
214:28 it today every day every uh all the world. Uh uh by uh
214:34 uh OK. Finding for any real case, we find empirically we find
214:42 a velocity function which we call the out velocity function, which uh which
214:49 the uh the true arrival times uh obey this equation. And normally it's
214:58 that the move out velocity is not to the R MS velocity. But
215:02 I, I wanna return to that in the course. So uh this
215:17 begins a series of arguments which we have time to finish by the end
215:24 the day today, which is coming in three minutes. So I think
215:29 this is a good place to stop here where we introduced the concept of
215:36 move out velocity function. So let's here. I love, I,
215:44 know you have lots of questions from but uh we're gonna, you,
215:50 don't have time to um deal with today. So, um,
215:57 by email tonight, send me a from this morning and a question from
216:03 afternoon or maybe more than one and see you on Friday at one
216:12 And
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