5. Tutorial - Week 1

Added: 2026-05-06

[00:00:12]A very good morning and welcome to foundations of continue mechanics. Today we'll go through the first tutorial in this course that is tutorial for the first week. So let's get started.

[00:00:24]So the topic for this week's tutorial are tensors, initial notation, summation convention and chronica delta, particularly the substitution property of chronica delta.

[00:00:40]So we'll be solving 10 problems in the tutorial today and these problems will help you grasp the course and the concepts better as well as help you solve the problems that will be provided in the assignment as well as to prepare for the exam.

[00:00:56]So let us start with the first problem.

[00:00:59]The problem says that we have to expand CI JI.

[00:01:04]Now the first thing in solving problems like this is to check whether there's a repeated index and whether we can apply the summation convention. So in this case we can see that the index i appears twice.

[00:01:22]So we can apply the summation convention.

[00:01:25]So c i j i can be written as sum i goes from 1 to 3 because uh we live in a three-dimensional world c i ji and this then we expand so we write c1 j1 plus c 2 j2 + c3 j3 and that's it. This is the expanded form. There's nothing more we can do here because uh J is u a free index. So this term is general in J and uh we have no knowledge of what the quantity C is. So we have to leave it at this stage.

[00:02:18]So let's uh try more problems now.

[00:02:28]So the second problem is that we have to expand and simplify or should we we should say just simplify the expression delta j time uj vi.

[00:02:50]So again before uh we do anything uh we will have to expand but in this case what we see is that we have a chronica delta and uh we don't need to expand here because we can apply the substitution property in delta. So because if we expand it then we'll be wasting um our energies as we learned that we can apply the substitution property here. So in this case what we see is that the index i and index j they're both repeated indices in the overall expression.

[00:03:30]So if uh you would want to go back to the lecture notes on the substitution property and have a look at its definition please do that. I'll help you recall. So if an index u in um the delta appears as a repeated index in the whole expression uh then we can get rid of the delta and we can replace the index with the other index of delta. Now remember the repeated index should be outside delta in the sense that there's an i here and there's an i here. The two indices of delta if they're repeated then it's simply equal to three as we saw in the last lecture.

[00:04:09]So uh let's not worry about that. So we see here sorry that uh the index i is the repeated index. Let's uh do it that way. So the index i is repeated.

[00:04:26]So in that case we can get rid of the delta. So in the simplified expression there's no delta. There's only u and v.

[00:04:34]And we replace the index i in the remaining expression with the other index of delta which is j. So this becomes j. This already was a j. So we get a u jj v j which you know is simply u1 v1 plus u2 v2 plus u3 v3.

[00:04:56]So when we see a delta u with uh the possibility of substitution rule that can be applied we apply that we don't expand it because ultimately it's going to result in the same expression. So why waste the energy again with practice we will learn how to identify these situations. So in this case if you leave the problem I leave the answer at this stage then it's all right you don't have to further expand because the problem said simplify not expand but if you expand it that is also fine right now uh many students will ask that why did we take I as repeated why not take G as repeated well uh we can do it so there's no preference as such so let's try that let's try solving that way.

[00:05:52]So I'll say u another way.

[00:05:58]So delta i j uj vi. So in this case we say that okay the index j is repeated.

[00:06:06]So we apply the substitution property which says you get rid of the delta. So there's no delta. There's only U and V.

[00:06:13]And you replace the index the repeated index that is J with the other index of delta that is I. So the J that was under U gets replaced with an I. And we already had an I. So UI VI. And you already know that uh this also expands to U1 V1 plus U2 V2 plus U3 V3. So ui vi and u j vj they are equivalent because i and j in these expressions are simply dummy or repeated indices. They don't have a meaning um by themselves as such.

[00:06:53]Right? So let's move on to the next problem.

[00:07:04]So the third problem says that we have to simplify delta i I delta k.

[00:07:20]Right? So in this case we see that uh the indices i and k are both repeated but they appear with the deltas. So we cannot apply the substitution property.

[00:07:32]substitution property requires that the indices two indices of delta are different. So let's write that that will help us identify the approach.

[00:08:02]So the substitution property can be applied when the two indices of the chronica delta are different. So in this case uh the two indices of both deltas are the same. So we cannot apply the substitution property but we can expand.

[00:08:18]So delta II individually is simply uh delta 1 1 plus delta 22 plus delta 33 which is 1 because by the definition of delta if the two indices are the same it's equal to 1 + 1 + 1 this is three so delta i i is a three and similarly delta k is also three because uh that's the same thing just dummy I mean indices I and K. So delta I I * delta K K is 3 * 3 which is equal to 9.

[00:09:03]So let's uh try another problem.

[00:09:20]So the problem says that we have to simplify delta J delta J I.

[00:09:26]So let's see what this simplifies to.

[00:09:29]Now in this case we can apply the substitution property.

[00:09:34]Let's write it down.

[00:09:49]So which says that if one of the indices of a delta is repeated um then we can get rid of the delta. Now we have two deltas here and you can see that in the first delta I is repeated in the first delta J is repeated in the second delta J is repeated and the second delta I is repeated. So we have four options to apply the substitution property. The first uh I mean first choice we have to do which delta should we take and get rid of and the second is which index of the delta we have chosen is to be taken as repeated. Now the beauty is that you can do anything. You can take you can choose any one of the four choices and you will still get the same answer.

[00:10:35]I'll show you uh by taking a couple of choices. So let's uh take the first delta as uh the delta for applying the substitution property.

[00:10:51]take first delta and take i as repeated.

[00:11:01]So I'll write the expression. So delta i j delta ji. So what we have done is we have taken this as the delta and we have taken the index i as the repeated index. So as per the property what we do is we get rid of the delta.

[00:11:23]So we get rid of this one and what we are left with is an expression which is uh the remaining part that is this delta and the repeated index in the remaining expression is uh where in in that expression I is replaced by J. So this becomes delta JJ.

[00:11:45]So here there was an I that's replaced by a J. And this we already know is a three from the last problem.

[00:11:55]And as I said whatever choice you take is going to result in the same answer.

[00:12:01]So let's take the second delta and take G as repeated.

[00:12:21]So in this case we'll take this as a delta and we'll take J as repeated.

[00:12:31]So what will happen is that when we apply the substitution property the uh delta will vanish. So the delta we have is this one. So this will go and what we are left with is the remaining part which is in this case also a delta and the repeated index that is j is replaced with an with an i. So this becomes delta i i i which is also a three.

[00:13:03]So I uh request you to please uh pause this video and try the other two options where you take the first delta with j as repeated and the second delta with i is repeated and appreciate the fact that all these choices result in the same expression.

[00:13:24]So let's try the next problem.

[00:13:29]Problem five.

[00:13:50]So we have to simplify this expression.

[00:13:54]Now this has three deltas this time and we can see that uh whatever delta we choose both its indices are repeated. So the first delta has indices I and J and we can see both I and J are repeated in the whole expression in the whole mathematical term. In the second delta we have J and K. Both J and K are repeated. and the third delta has k and i and both k and i are repeated in the overall mathematical term that involves the three deltas. So again as I said that the choice is yours you will get the same answer. So we go from left to right. So we'll take the first delta on the left as the delta and the remaining expression as the expression which uh is considered to be some uh expression. So we will take the first delta and let's take I as repeated. So we'll say take first delta and the index i as repeated.

[00:15:01]So the expression delta i j delta jk delta ki will simplify to. So we have this as the delta and uh we have I as repeated.

[00:15:18]So we'll get rid of this delta in the simplified expression.

[00:15:23]So what we are left with is delta j k and delta ki. But we have to replace the repeated index that's k with the other index of the uh delta that we have taken that is the other index is j here. So this k gets replaced with a j.

[00:15:45]This is not the final answer yet because we still have two deltas and we can still apply the substitution property here. So what we'll do is we'll take delta jk as the delta for applying the substitution property.

[00:16:22]And we will take the index j as repeated.

[00:16:38]So the expression that we had delta j k * delta kj j this uh will reduce because we have this is taken as a delta and the index j is taken as repeated.

[00:16:54]So we have delta kj left but uh we are replacing j with the other index that's k. So this becomes delta k right. Uh now this can simplify further because delta k is simply equal to a 3 because when you expand it it becomes delta 1 1 + delta 22 plus delta 33 which is equal to 1 + 1 + 1 which is a 3.

[00:17:26]So this is the final answer.

[00:17:29]So this whole expression delta i j delta j k and delta k i multiplied together is equal to three. And see these problems that I've given are usually expressions you experience when you're trying to solve problems in tensor algebra and tensor calculus. So we will see these expressions that will appear uh in front of us. So this uh these results that we are getting will be useful. But of course you don't have to remember this. You have to learn how to simplify using the rules such as summation convention and substitution property of the in the initial notation in tensors.

[00:18:14]So let's move on to the next problem which is problem six.

[00:18:37]Now this is a problem that has uh been provided to confuse you a bit because we have two deltas multiplied together.

[00:18:47]But this time there's only one index that is repeated.

[00:18:51]So if you choose the first delta as the delta for applying the substitution property, you can only take J as a repeated index because I is not repeated. Similarly, if you chose the second delta as the delta for applying the substitution property, then only the index j is repeated. K is not. You can still choose any of the deltas, but the repeated index has to be the one that is actually repeated. So, let's choose the second delta as the delta. In fact, uh let's do it left to right. That's uh what typically I do. So this is taken as the delta and we take um well there's only one option here that J is repeated.

[00:19:37]So I'll suggest you uh when you try it uh you try to take the second delta as repeated but I'm taking the first one.

[00:19:45]So this uh simplifies to so we get rid of the first delta. So we only have the second delta left that is delta jk. But we replace the repeated index that's j with the other index that is i. So this becomes delta i k.

[00:20:03]And there's nothing we can do further with this because both indices of uh this delta are free indices. So there's no further simplification that is possible at this stage. We leave it at this.

[00:20:21]So let's uh go to the next problem.

[00:20:35]So this is a true or false uh kind of a problem uh where it's asked is it true that EI. EJ is equal to EJ.

[00:20:46]EI.

[00:20:48]Now of course uh you know that uh the vector dotproduct is commutative. So this is true. But we can also see this from the angle of the chronica delta because we know that ei is equal to delta j.

[00:21:13]Again, you can go back to the lecture notes and have a look in the previous uh in in this week's lectures, but um I can help you recall. So, EI. EJ is only equal to one when the two indices I and J are the same because uh EI's are unit vectors and they are linearly independent because they are uh they form a set of orthonormal bases. So E1, E2 and E3 are the three mutually perpendicular directions in the right hand cartian coordinate system.

[00:21:52]So this is E1, this is E2 and this is E3. So EIJ will be equal to one only if I and J are equal. And that is the same as the definition of the chronica delta.

[00:22:04]So EIJ is delta EJ. Similarly, EJ do ei delta ji. But we already know or we can see easily that delta j is the same as delta ji because by the definition of delta. Let's write it down.

[00:22:29]So this is equal to 1 if i is equal to j and this is equal to zero. if I is not equal to J. So we can see easily here that uh the definition of delta does not depend on the order of the indices I and J here. So delta EJ is equal to delta J. And by that logic using these two expressions that we have written we can see that EI EJ is the same as EJ dot EI. So the answer to this problem is yes.

[00:23:18]So let's move on to the next problem.

[00:23:37]It says a state true or false. The statement is scalers and vectors are tensors and uh this is true.

[00:23:55]Why? Because uh scalers are actually zeroth order tensors and vectors are first order tensors.

[00:24:12]And this is how we learned about tensors. We started with an intuitive understanding by defining scalers as zeroth order tensors because we need only a magnitude in this case. In vectors we need three components and in higher order tensors depending on the order of the tensor we need 3 to the power of n components. So if n is zero that is in scalers we need 3 to the^ of zero components which is 1.

[00:24:50]We need 3 to the^ of one component that is three and in higher order we need 3 to the^ of n components to describe the tensorial quantity.

[00:25:02]Right? So the statement that scalers and vectors are tensors is true.

[00:25:18]So the next problem says there's an expression given to us which is sigma j time ej.

[00:25:35]And there are two parts to the problem.

[00:25:38]The first part says that the index i is the dash index and we have two options to be chosen from pre/ dummy.

[00:25:56]The problem B says that the index j is the dash index. And again we have the same two options free slash dummy.

[00:26:14]So again think about it you already know the definition of free and dummy indices.

[00:26:20]So in uh the expression sigma ej ej the index j is repeated or dummy. So we answer the part B as J is the dummy index and uh the index I is free because it appears only once in the mathematical term. Now I keep replacing expression and mathematical term because we only have one term but we had if we had multiple terms in the expression then we'll have to deal with each and every term separately. So in this case since we only have one mathematical term in the whole expression we can we don't need to specify the term separately but we can see that I is a free index in this case because it appears only once.

[00:27:13]So now we move on to the last problem of this tutorial that's problem 10.

[00:27:24]So the number of components in an nth order tensor are and we have four options.

[00:27:44]The first option is 2 ^ of n.

[00:27:48]The second option is n squared.

[00:27:51]The third option is n cubed. And the fourth option is 3 to the^ of n.

[00:27:59]So in an nth order tensor the number of components well let's say sigma j is a second order tensor. So we take an example and we have two indices I and j. I can take values from 1 to 3. J can take values independently from I from 1 to 3.

[00:28:28]So we have a total of 3 * 3 9 combinations possible. And we've already written this as a matrix as sigma 1 1 sigma 1 2 sigma 1 3 sigma 2 1 sigma 22 sigma 2 3 and sigma 31 sigma 32 sigma 3.

[00:28:44]So we have nine components and that is basically 3 * 3 which is equal to 3^ 2 components in a in a third order tensor. CI J K we have three independent indices. So we have a total of 3 * 3 * 3 which is 3 cubed components.

[00:29:19]So we can see that the general expression will be 3 to the^ of n for an nth order tensor.

[00:29:37]So the answer is d 3 ^ of n.

[00:29:44]Now uh some people may get confused that uh the first option option a is 2 to the^ of n. But that will be true if you live in a two-dimensional world. But as I mentioned and I've been mentioning again and again that that we live in a three-dimensional world and uh therefore we will always uh assume that the indices of a tensor can go from 1 to three and we also draw our cartesian coordinate system as with the three basis vectors E1, E2 and E3. three.

[00:30:24]So unless it is specified otherwise, we'll always assume it's a 3D world. So the number of components is 3 to the^ of n, not 2 to the^ of n.

[00:30:38]So that brings us to the end of this tutorial. I hope uh after these 10 problems you have got some understanding and confidence in working with mathematical expressions that involve tensors with indicitial notation involving summation or repeated indices applying summation convention and the use of chronica delta's substitution property.

[00:31:07]I will repeat for the nth time that practice is what what will make you perfect. So please practice problems.

[00:31:14]Try to do these problems even after looking at the video yourself so that you are able to solve the assignment with confidence and solve the problem towards the end of the course correctly as well. So thank you and I'll see you next week in this course.