4. Kronecker Delta

Added: 2026-05-11

[00:00:12]A very good morning and welcome to this course on the foundations of continum mechanics. Today we'll be talking about the chronicer delta operator and its application in the vector dotproducts.

[00:00:24]So let's get started.

[00:00:30]So let us start with the first vector operation that is vector addition.

[00:00:43]So we'll write this in the symbolic notation because we are used to that um till now.

[00:00:51]So the vector w is a sum of the vectors v and u in uh the symbolic notation.

[00:01:06]So if I expand this we get this expression.

[00:01:36]So as we know that the basis vectors E1, E2 and E3 are orthonormal.

[00:01:45]So they're orthogonal that is they're independent. So what that means is that the components of E1 on both sides have to be equal. And uh same is the case for the components of E2 and E3. So we get W1 is equal to V1 + U1.

[00:02:04]Similarly, W2 and W3 are v_sub_2 + u2 and v_ub3 + u3 respectively.

[00:02:14]So this uh gives us the general expression in the initial notation for vector addition which is wi is equal to vi + ui.

[00:02:34]So this I'll say is the initial notation representation.

[00:02:52]So in this expression uh I is a free index and this expression is general in I in the sense that I can take any value. So this expression is true for i = 1 2 and 3 all values. So this is the way we write vector addition in the initial notation. So in this course we'll be using this way for writing that sorry for writing that w is equal to vectors v plus vectors u. So let's move on to the next vector operation that is scalar multiplication.

[00:03:47]So scalar multiplication simply is that the vector v can be written as some scalar alpha * the vector u in the symbolic notation.

[00:04:05]Now if I write this in the initial notation this simply becomes vi is equal to alpha ui and this is true for all values of i. So remember i is a i uh this expression is general in i in the sense that i can take any values from 1 to 3. So this is true for all values of i.

[00:04:34]So next uh we'll move on to vector multiplication.

[00:04:57]And uh in vector multiplication we will start with the famous vector dotproduct.

[00:05:14]So before we actually start uh discussing the vector dotproduct we first need to understand an important symbol that is chronicer delta. So we will park ourselves here and from this point onwards we'll talk about the chronicer delta symbol and then we'll come back to the vector dotproduct.

[00:05:40]So let's uh discuss this in a fresh page.

[00:05:53]So the chronica delta symbol is written as the Greek lowercase alphabet delta with subscripts I and J and this is defined as follows. So delta I J is equal to 1 if the indices I and J are equal and this is equal to zero if the indices I and J are not equal. So remember this is a definition.

[00:06:28]So it's a very important and a very useful symbol. And uh the motivation behind delta J as you may already have guessed by now is the vector dot product. So uh let's see how this helps us in the vector dot product.

[00:06:47]So if I write EI EJ where E is um a unit vector. So EI is a unit vector in the I direction. So this is equal to and this is uh by and not by definition but uh by the nature of the unit vectors. So if I is equal to J then uh this becomes uh I mean if the value of I is equal to value of J that is uh if I and J are both equal to 1. So this is E1 E1. So this becomes equal to 1. So if I equal to J by values then this is equal to 1. But if I and J are not equal by value then it's equal to zero.

[00:07:44]Okay. And uh we can easily see this because uh they are orthonormal.

[00:08:03]So we can see that E1 E1, E2 do E2 and E3 do E3 are all one and um any other combinations they're all zeros. So we can see that ei actually is equal to the chronica delta delta j.

[00:08:23]So this is where the chronica delta symbol helps us um use it in the vector dotproduct.

[00:08:31]We'll use that later. But before that uh there's an important property of the chronica delta that's called the substitution property of delta j and that is something we'll discuss No.

[00:09:28]So let us uh try to understand this property.

[00:09:33]So it says if delta J is multiplied with another tensorial quantity and one of the indices of delta is repeated. So let's underline this. So one of the indices of delta is repeated.

[00:09:49]Then we can simplify the expression by getting rid of the delta and replacing the repeated index. So replacing the repeated index by the other index of delta. So let's understand this uh through examples.

[00:10:11]So we have an expression deltaig j u.j and we want to use the substit substitution property here to simplify this expression.

[00:10:25]So let's uh read the property again. So if delta i j is multiplied with another tensorial quantity. So that is true. So delta j is multiplied to another tensorial quantity which is a first order tensor J and one of the indices of delta is repeated. So delta has two indices I and J and we can see that the index J is repeated.

[00:10:50]So that's uh that condition is true as well.

[00:10:54]Then we can simplify the expression by getting rid of the delta. So the simplified expression will not have any delta. it'll only have a u and replacing the repeated index of um repeated index by the other index of delta. So the repeated index here is j.

[00:11:18]So let's write that down.

[00:11:24]So we'll replace this by the other index which is J uh which is I right. So this is the simplified expression that we have here and remember that in the simplified expression you cannot have a J um as an as a free index because J is a dummy or a repeated index. So if uh we were to simplify this then j will be gone right.

[00:11:58]In fact uh let's expand it. It's not very difficult to understand this property by expansion. So let's do that and get a feeling of it. So if I look at let's expand the left hand side.

[00:12:16]So u you have delta j u.j J and this is basically J going from 1 to 3.

[00:12:28]This is from the Einstein notation or the summation convention.

[00:12:52]So which expands to delta i1 u1 plus delta i2 u2 plus delta i3 u3. Now we can see that in each mathematical term there's only one free index i. So I can take any value. So let's say the i value was uh equal to 1.

[00:13:14]So let's take case one that I is equal to 1. So in that case the expression becomes delta 11 1 u1 plus delta12 u2.

[00:13:33]Now delta 1 1 is a 1 but delta 1 2 and delta 1 3 zeros. So what we are left with is simply U1.

[00:13:45]Similarly, if I was equal to 2, then the expression would have been U2.

[00:13:54]So we can see the pattern here that the in the final expression will ultimately become ui which is equal to the right hand side.

[00:14:13]So let's uh look at one more example.

[00:14:33]So in this example we have delta j ui vk.

[00:14:38]So in this case the index i of the delta is a repeated index.

[00:14:47]From the substitution property, we can simplify this expression by getting rid of the delta. So we only have u and v now and replacing the repeated index of delta um with the other index. So uh the repeated ind sorry the repeated index um in the delta with the other one. So I was a repeated index. So in the simplified expression we'll have I replaced by the other index of delta which is J. So the final simplified expression is U.j VK.

[00:15:26]Right. So now that uh we understand this uh chronicer delta symbol and its substitution property let us go back to our vector dotproduct.

[00:15:52]So if I was to use the symbolic notation then I would have written u dov and I would have expanded this. So this would have become u1 e1 plus u2 e2 plus u3 e3 dotted with v_sub_1 e1 plus v2 e2 plus v3 e3 and when I expand this I get u1 e1 dotted with a v1 e1 so that basically be becomes u1 v1 * e1 e1 which is a one.

[00:16:40]Then I take this u1 e1 dotted with a v2 e2 that becomes u1 v2 * e1 dot e2 which is a zero. So what we can see is basically we'll end up with u1 v1 plus u2 v2 plus u3 v3 all the terms where we have an e1 e1 and an e2 doe e2 and an e3. E3. So this is something we already know. We have been doing that for a long time. But let's uh do it using the initial notation. That is what is important today.

[00:17:28]So if we use the initial notation. So remember we are starting with vectors u dov that's the expression we want to solve for.

[00:17:43]So u can be written as ui ei remember we are using the summation convention so I don't need to put a sum symbol here and v vector can be written as vj ej again this is exactly the same as what we had above where I had written this expression because it ultimately expands to this above expression.

[00:18:19]Right? U so this is where u the beauty of the initial notation with all its uh rules such as summation convention they can be seen. So you can see that we don't have the sum symbols here but we will still open this these brackets.

[00:18:39]So you can see that ui and vj they are simply scalar components of vectors. So they can be separated out.

[00:18:48]So this becomes ui vj and then we have the term ei dot ej.

[00:18:57]Put this in brackets.

[00:19:00]So there's no sum symbol but still we are comfortably doing the algebra.

[00:19:05]And now uh we already know that ei. EJ is the chronicer delta symbol. So this becomes delta J.

[00:19:17]And we can also see that uh one of the indices of delta is repeated. In fact both of them are in this case. uh but uh for um our application let's assume that I is the repeated index of delta.

[00:19:37]So then what we get is if uh one of the indices of delta in an expression that multiplies delta with some other tensor is repeated we can get rid of the delta.

[00:19:47]So in the final expression we'll only have u and v and we replace the index i with the index j. So we get a U.J VJ and you can easily see that uh this will expand to u1 v_sub_1 plus u2 v2 plus u3 v3.

[00:20:14]So this is uh getting the expression using the initial notation. So the whole aim here was to let you do an algebra on the initial notation.

[00:20:30]So here we applied the initial notation, the summation convention and the substitution property of delta all in one go.

[00:20:45]Right? So uh before we close this lecture I'll want to give you one more example in u the initial notation particularly with the use of delta.

[00:21:09]So the problem says simplify the expression delta I.

[00:21:14]Okay, so this is independent of the vector dotproduct. We are done with that. So how do we proceed? Now we can see it's an initial notation. It involves a delta. So the first thing we have to do is to apply the summation convention because as per the convention, this has to expand to something, right? So we apply the summation convention. So this is actually equal to sum i going from 1 to 3 delta i i which expands to delta 1 1 plus delta 22 plus delta 33. And now we apply the definition of the chronica delta symbol which says that delta 1 1 is equal to 1, delta 22 is equal to 1 and delta 33 is equal to 1. So this simplifies to a value of three. So delta I I is equal to three.

[00:22:14]Now the reason I discussed this case is that uh many many students do this mistake where they say that delta I I is equal to 1 and the mistake they make is that they say that since the indices of delta are both equal so by definition it is equal to 1.

[00:22:35]So the mistake here is uh that first you have to apply the summation convention and expand it only then you apply the definition. So we will write this rule because this will confuse us at many stages. It's uh it may be clear at this stage but when you have a complicated expression we end up sometimes making this mistake.

[00:22:58]So the rule says always expand the summation first then apply definition.

[00:23:23]So if you remember this rule or you use this rule when you solve problems practicing them you will be comfortable with expanding complex expressions involving multiple symbols such as delta and others as well that we'll discuss in the next lectures and uh the use of summation convention in the initial notation.

[00:23:46]So this uh brings us to the end of this lecture. Today we discussed the chronica delta symbol and its substitution property. Uh we did a couple of examples to reinforce our understanding and we then applied that to the vector dotproduct to understand how to use the uh the delta symbol and its property in uh with the use of initial notation. So we did some algebra with the initial notation today.

[00:24:18]By the way, uh this chronicer delta symbol is uh also known as the substitution operator sometimes.

[00:24:40]because of the reason that uh we almost always substitute the delta by removing it and uh replacing the index in the expression. So it's sometimes called the substitution operator.

[00:24:55]Right? So in the next lecture we'll go beyond vector dotproducts. Uh that is we'll be studying tensor products in the coming lecture in the next week. uh next uh we'll have a tutorial this week where we where we will practice problems from the topics that we have done or discussed in this week.

[00:25:17]So before we close this lecture, let's do a quick concept check.

[00:25:24]So the first problem asks us that is delta J the same as delta j.

[00:25:33]So go to the definition of delta and see if you can answer this question.

[00:25:39]And the next question says that we have to simplify this expression that is delta j * sigma j k. So again uh to simplify this expression you look at the definition of the substitution rule or the substitution property and go step by step and try to simplify this expression by getting rid of the delta if you can.

[00:26:05]Thank you very much and I'll see you in the next tutorial.