This lecture introduces tensor operations in continuum mechanics, focusing on the outer product (tensor product) of two vectors u and v, written as u*v, which produces a second-order tensor called a diad with nine basis components (e1e1, e1e2, ..., e3e3). The outer product is not commutative (u*v ≠ v*u) because the basis components are distinct. The lecture establishes a key pattern: inner products (dot products) reduce tensor order by 2, while outer products add orders. Various products between vectors and diads are demonstrated, including u·(v*w) producing a vector, u×(v*w) producing a second-order tensor, and diad-diad products like (u*v)·(w*p) producing a second-order tensor, (u*v):(w*p) producing a scalar, and (u*v):(w*p) (vertical double dot) also producing a scalar. These operations are fundamental in continuum mechanics for deriving equations in fluid mechanics and turbulence analysis.
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L8. Tensor OperationsAdded:
Good morning and welcome to this course on the foundations of continuum mechanics. Today we'll be talking about tensor operations. So in this lecture today we'll talk about the operations involving vectors and tensors and uh this will give us lots of practice on initial notation. So let's get started.
So the first product we'll talk about today is called the tensor product.
The tensor product is also called the outer product.
Right? So if you have two vectors u and v their outer product or tensor product is simply written as u * v. So please note that notation notation wise there is no dot or no cross here we just write a u and followed by a v.
So if I was to use the initial notation, this would be UI EI* VJ EJ.
And if I move the scalar components to the left, so this becomes UI VJ followed by EI EJ.
So uh now the question is what exactly is this? We are used to having just one ehat uh like e i hat which uh would expand typically to a vector if you write all the components. But what exactly this quantity where you have e i ej without any dot or a cross in it. In fact, this is the tensor product or the outer product of the unit vectors ei.
And this is called the diode of ei and ej. So the quantity e i ej is called the diad.
And uh when I say e i ej that is the outer product or the tensor product of ei and ej. So let's expand the expression because as you can see in this expression we have repeated indices you have ui ei and then you have a vj ej. So if I was to expand this by doing a sum over the indices I and the indices J, this would expand to a total of nine terms added together. So let's write them down.
So we can see uh there's a sum of nine independent terms here and uh this looks very much like a second order tensor because in a second order tensor you have a total of 3 to the^ of two that is nine terms. So this has nine terms, nine mathematical terms in the expression which is which gives us a hint that this is a second order tensor which has 3 to the^ of two terms in it.
And if you compare this with uh typically a vector, a vector would have three terms. Something * e1 plus something * e2 plus something * e3. Here we have something * e1 e1 plus something * e1 e2 plus something* e1 e3 and so on all the way to something* e3 e3.
So uh this definitely tells us that these uh combination of unit vectors E1 E1 E1 E2 E1 E3 and so on they are the basis of a second order tensor like we had in a vector. So I can say E1 E1, E1 E2, E1 E3, E2 E1 and so on all the way to E3 E3 are the bases in a second order tensor.
So with that logic we can say that any second order tensor can be written as a linear combination of these nine basis components.
We also note here that the bases E1, E2 and the bases is E2 E1 are different so they are not equal so they're different basis so what that means is that the outer product of E1 and E2 or the outer product in general is not commutative Y.
This is an important point because uh sometimes we confuse this like in dot product which is commutative but the outer product is not commutative because the unit uh because the basis E1 E2 is not equal to E2 E1 and so on. So they are not commutative they are different bases.
Now sometimes just to avoid notational confusion we write the outer product or the tensor product of uv explicitly as u * v. So rather than not using any symbol we use a cross with a circle around it which is a symbol also used for outer product. Now in this course while writing the outer product we will not use this cross with a circle symbol but uh some books do that because sometimes it is confused with a slight dot that may appear in the print.
So this is called the outer or tensor product of U and V. And the quantity UV is called the diad of U and V. So this quantity UV is called the it's called a diad.
Why do we call this uh outer product a tensor product? Well, that's because you started with two vectors u and v and that gave you as a result a tensor which was a second order tensor. So this is the reason that we call it a tensor product because you end up getting a tensor of higher order by combining uh tensors of lower order. In this case, uh we are showing a vector. But as we'll see later that this tensor product can also exist between tensors of higher orders. But for now, let's stay with the diode of two vectors.
So I can also write that u v is not equal to v u for the same reason that the outer product is not commutative. So I'll be using the word outer product, tensor product and the diad of two vectors interchangeably and I would really want to do it so that you get used to these terms.
Many of you may be seeing this outer product of two vectors for the first time. Till now we were only used to two kinds of products between vectors either the dot or a cross productd but uh this is something new for us. But that doesn't mean that this is not used. It is very much used in the field of continuum mechanics and that is the reason we are studying that as a part of this mathematical background in the course. So you will see that it is useful and it is definitely very popular as well in continuum mechanics. Okay. Uh now that we've been using the word outer, let's also clarify the word inner which we have used in the past and compare the two.
So the inner product between two vectors u and v is written as in the symbolic notation as u dov. And we know that when we do that uh what we get is a scalar if we write in the initial notation this is ui vi. So this expands to u1 v_sub_1 plus u2 v2 sorry plus u3 v3 which is just a number. So this is a scalar and scalar is a zero order tensor.
So we will see some sort of a trend here or uh a pattern here and then we'll try to generalize. So u is a first order tensor or a vector. So I'll say um order one first order v is also a first order and when we did an inner product between the two we got a zero order.
So I can write some sort of an equation which says 1 + 1 and then a minus 2 equal to 0. So what we are doing here is that we are taking the input tensor orders. So which is 1 + 1 because u was order one v was order one and then you add them. But when you do an inner product, the inner product results in a minus2.
So these are input tensors.
This minus2 is because of inner product and this is the output tensor.
Of course, we're talking about orders here.
Um I will talk about this in more detail in later this week but for now let's remember that we have a pattern here when we talk about inner products. Now the inner product is also represented as matrix uh as a in matrix form. So let's write the matrix representation.
So in the matrix representation we write u1 u2 u3 as a matrix which is one row and three columns and then we write uh v as v_sub_1 v2 v3 as a column vector which is 3 + 1 and when you multiply this you know that you get just one element which is u1 v1 plus plus u2 v2 plus u3 v3 which agrees with this 1x 1 that's uh from matrix multiplication. So the inner product between vectors u and v which is basically the dotproduct of u and v that is u dotv is represented as um u written as a row matrix v written as a column matrix and we get a single scalar term as a result.
So let's now move on to the outer product.
So in case of outer product as we have just seen in this lecture that is written symbolically as U * V.
This is a first order. This is the first order and this results in a tensor of second order.
Okay. So this is second order. So let's write our um the pattern that we notice here. So 1 + 1 this is because of the input tensors is equal to two that is the output tensor. So we see that there's no minus2 here because the minus2 happens because of an inner product in the orders. Since there's no inner product, it's an outer product. So the input tensor orders simply get added and we get a second order tensor as a result.
So let's uh go back and compare with the inner product. So inner product we had minus2 here but that is not there in the outer product.
Let's also look at the matrix representation.
So in this case we will write U as column vector which is three rows and one column. And we'll write V as a row vector which is one row and three column. So when you multiply a 3x 1 matrix with a 1x3 matrix you will get a 3x3 matrix.
So you have u1 vub1 u1 v2 u1 v3 and so on.
Right? And this uh agrees with the nine terms that we had. So let's go back and have a look at the sum of the nine terms that we had here as you can see. So these nine terms are exactly what we have except that we are simply writing the components. We are not not writing we not multiplying them by the basis vectors and adding like you do in vectors where rather than writing u1 e1 + u2 e2 plus u3 e3 you simply write the vector as u1 comma u2 comma u3 or simply as a row vector u1 u2 u3.
So let's move on to the next kind of product. So this product is called the vector diad product. So what we're doing is we are multiplying a vector with a diad and a diad is something that we already have just defined. So but in fact vectors and diodes can be multiplied in different ways. So we'll have sub sections here. So the first way we'll multiply a vector u with a diode which is a diode of two vectors v and w is using a dot product that is an inner product. So again uh these things will be very new to many of you seeing uh a product between vector and a higher order tensor that is a diad and then the products can be of different kinds.
All right. So this is a first order tensor and this is a second order tensor because it is a diode of two vectors as as we have just seen. So let's write in the initial notation.
So the way this product is defined is that you take the dotproduct between adjacent terms or the adjacent vectors.
So in this if you look at the left of the dot you have a nei and to the right of the dot we have nej. So the dot happens between ei and ej.
So I'll move all the scalar components that is ui vj and w k to the left because they are simply scalar components. they can be moved to the left and then we are left with is ei dot ej and then we have an ek remaining remember don't move around the vectors because the order in which the vectors appear is also important here so just to be safe if e k is to the right of ej it stays as it is so this from uh the definition of chronica delta is so ei delta ej and then we apply the substitution property which results in getting rid of the j in the u term. So you have a ui vi w k e k and we can see that this whole thing is nothing but a vector that is it is a tensor of first order.
So let's write our pattern here in terms of the orders. So what we started with is this equation which had a U dot VW.
So let's write it again. U dot VW. So this was first order.
This was second order.
And what we got as a result was a vector which is a first order.
So our equation is 1 + 2 - 2 is equal to 1. So this 1 + 2 is the input tensors.
This final one is an output order of the output tensor and this minus2 is because of the inner product.
Right? In fact, if I had to complete this equation here, um, if I go back from the initial notation to the symbolic notation, I have a UI vi, which is basically a u do v. So remember, we are learning to go back u and forth. So u dot v and then a w k is simply the vector w just uh to write the complete equation in the symbolic notation.
Now many of you may be thinking that uh where in the world would we get things like diads or things like a dot product between a vector and a diode. Now again as I said these are quite uh common in in continuum mechanics. So I'll give you a quick example on uh what kind of work in terms of dealing with equations in continuum mechanics would result in these quantities.
So let's uh deviate a bit and take an example.
So let's say you have an equation u like dw by dt where w is a quantity which depends on the independent variable time a vector quantity which depends on independent variable time alpha w is equal to zero. It's a very common equation you would see in mechanics. So what we do is well the right hand side is a zero vector by the way but I'm just writing a zero. So what we'll do is we'll multiply both sides by the vector v. So this becomes v * dw by dt plus alphaw is equal to still a zero.
And when I move that v inside what we'll get is d. Let's say v is not a function of time. So this becomes d of v * w upon dt. Now this v * w is nothing but the diad of the vectors v and w plus alpha vw because you multiply it from the left. So it becomes a diode like this.
And now it may happen that we may take a dot product on the left hand side with a vector u.
So this will become a d by dt of well let's say u is a function of time so we don't move it inside so it's u dot d by dt of vw plus alpha is simply a scalar then u dot vw and this is equal to a zero. So you can see that uh by playing around with the equation by multiplying it with different terms we get these term where we have a product of a vector and a diode that is an inner product of a vector and a diode. And again if you look at equation of fluid mechanics for example when you write the momentum equation in a general vetorial form you do get these uh terms or when you work in turbulence where we deal with the derivation of uh the turbulence kinetic energy or uh turbulence dissipation rate or some other higher order turbulence parameters then we get these kind of mult multiplication between vectors and diodes and so on. So this as I said it's it's quite common. You will see that in advanced courses in continuum mechanics.
So don't assume that this is just for theoretical purposes. It is very much used.
Right? So let's go back to our um work that we were doing. So we just talked about one kind of product between a vector and a diode that is a dot product. So let's talk about another kind of product which is the cross productduct. So U cross VW.
So let's go into the initial notation.
So this is UI EI crossed with VJ EJ time W K E K. Again the cross happens between adjacent vectors EI and EJ. So first I will move all the scalar components to the left. So UI, VJ, WK and then we have an EI cross EJ and then we have an EK.
Okay. Again the ordering matters. Make sure EK stays to the right.
Okay. So this is what we get. And now as the trend is I'll move the epsilon to the extreme left.
Then we have a UI, VJ, WK. And then we have the vectors EQ and EK. Now what is EQ?
This is nothing but the outer product of EQ and EK that is a diode of EQ and EK.
So this looks to be a second order tensor. So whatever the output of this cross product was a second order tensor this whole thing. So let's write the pattern. So the input was a first order.
This was a diad of vector. So this was a second order and the output is a second order. So in this case uh our equation does not work because we neither have an outer product and neither we have an inner product. So this is a different kind of product. So the input is 1 + 2 and the output is some two. So we have a minus one here.
That's what the vector um in this case the cross productduct results in. Right?
So this was a product between vector and diodes. So if we have if we can have a product between vector and diodes, why don't we have product between diodes and diodes? So let's do that.
So three is a diode diode product.
So again the first kind is UV dot WP. So we have two diodes UV and WP and we take a dotproduct between them.
So let's write in the in addition notation ui ei vj ej dotted with a w k e k and eq eq. So in this case the dot happens between adjacent vectors. So this becomes ui. So let's move all the scalar components to the left. So, UI, VJ, W K and P Q. And then we have let's follow the order. So, we have an EI first and then we have a delta J K followed by and E K.
So, let's check again. We have ui, vj, w k pq which are the scalar terms.
Then we followed the order from left to right. So ei then ejek results in a delta j k. And then we have an e sorry this should be pq eq. So this is an E Q right. Uh so what we can do further is we can apply the delta substitution property. So this becomes equal to UI VJ W KQ and since we are applying the substitution property of delta so we'll replace the index K with an index J. So the index of W becomes WJ and then we have an EI and EQ in the same order as we had above.
So we can see that from the bases in this expression that this is a second order tensor.
So let's check the orders. So in the input we had two diodes UV and WP. So the orders were 2 + 2 and in the output we also have a second order tensor as we just saw. So the order is two but because of an inner product that is one dot we got a minus 2. So this 2 + 2 is input tensors or diads. The output order is here and this minus2 is because of an inner product.
Now let's talk about another kind of diode diode product which is called a double dot product.
Now this definitely is something very new. We'll be talking about dot product which is we just put one dot but we can also put two dots here now. So let's look at it. So we have vectors U and V.
I mean at the diode of U and V and then we have a diode of W and P and we put two dots like this horizontally. So this is called a double horizontal dot product.
So we have a second order 2 + 2. So that's the input order because we have two diodes in the input.
and we have two dots. So that means there's a minus2 minus2. So this minus2 and minus2 this is because of two inner products.
So the output should be zero right and this is indeed the case. So let's look at the definition of the double horizontal dot product. So if you have ui vj so you'll have a w and then this will be a j and you'll have an i here.
Okay. So this is the definition.
So again there's no need to fight this because this is a definition. This is how the double horizontal dot product is defined. Now if you want to compare it with the single horizontal dot product.
So in single horizontal what we did was we did a dot between these two. Okay. So this was a single dot product. So when we bring in a double dot product, we also do a dot between this and this.
So when you do that, I'll write it very quickly. So when you do a dot product between EJ and E K, all you end up with is a V J WJ. And when you do a double dot when you do a dot product between ei and eq you end up with a ui pi. So I just move this pi here just to maintain the same ordering. So I get a pi.
So this is the same as what we had here.
All right. So this is uh the definition of a horizontal double dot product.
Again this is not as arbitrary as it may appear at this stage. Uh but we'll see that later when we start using these products.
Now the third one in the category of diode diode product is the vertical double dot product.
So we'll call this the double dot product vertical double dot product.
So again this is a definition we don't have to fight this. So in the definition what we have is we have so in the horizontal one we did a dot between this and this. So here what we'll do is we'll do a dot between v and p and then a dot between u and w. So if I write the maintain the orders as it is in the initial notation although that's not required in initial notation. I have a UI VJ and then I have a W I PJ because V is dotted with a P and U is dotted with a W. So their indices have to be the same. Okay. So this is the vertical double dot product. In the symbolic notation, this is written as U dot W time V dot P. Now just one more thing before we close this lecture today.
The ordering of scalar components in the initial notation is not or does not matter. What I mean to say is that UI VJ in the scalar components times EI let's say EJ I can also write this as VJ UI EI EJ because these are simply scalar components here in this representation I cannot change the order of EI EJ because that represents an outer product between the bases and I know that outer product is not commutative but since ui and vj simply denote scalar components of vectors so they can be uh rearranged the answer is still the same. So be very careful that uh don't rearrange the basis components but you can rearrange the scalar components. So let us summarize today's lecture now. So today we learned about tensor operations particularly operations which involve uh vectors but result in tensors or which involve uh tensors of higher orders like diodes of vectors which is order two tensors resulting in tensors of various orders. Okay. So the input and output were tensors of different orders. So we started with the definition of the outer product also known as the tensor or the diode of two vectors. So we talked about the outer product.
This helped us construct a diode. And then using that we worked with some different kinds of vector diode products and we also dealt with diode diode products.
So I gave you some practice of initial notation and at the same time I defined some products of different kinds and also showed you a pattern equation where every dot or an every inner product results in a minus2 in terms of order and in the outer product there is no minus2. So the orders simply add as expected.
So that is the end of this lecture.
Let's have a look at some food for thought.
So the first question for you to think about is that is the dotproduct between the vector u and the diode vw is the same as uv the diode uv dotted with a w. Okay.
So the this is where we test the associivity. Is that true or not? So you will have to think about it. In fact, the way you can do is you can expand this using initial notation. Take the dot between adjacent terms and see what you get as an output and compare that with what we got today. The second FFT again is to check if the crossroduct between a a vector and diode is associative.
So thank you very much and I'll see you in the next lecture.
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