# Kronecker delta question

1. Sep 3, 2011

### quietrain

does anyone know why

djmdkl = dkm

when j = l

thanks!

can i interchange it to djmdkl = dkldjm ?

since l = j
do the "j" in dkjdjm signify summation?

so i get dk1d1m + dk2d2m + dk3d3m ?

2. Sep 3, 2011

### Pengwuino

Well, follow the logic the deltas imply.

If j=l, then the first delta function can be written $\delta_{lm}$ and you have $\delta_{lm} \delta_{kl}$. This combination is only non-zero when l = m and l = k which implies k=m, so you can rewrite the whole thing as $\delta_{km}$

3. Sep 3, 2011

### quietrain

oh i see thank you , so its by logic and inference?

4. Sep 3, 2011

### Pengwuino

Yes, although formally a summation (not in the way you put it, I don't know how you claim there's a summation in what you wrote) does occur but it's fairly simple logic as to why it's true.

5. Sep 4, 2011

### HallsofIvy

I woud say that "$\delta_{jm}\delta_{kl}$ with j= l" does not imply as sum but that $\delta_{jm}\delta_{kj}$ does, by the Einstein summation convention.

6. Sep 4, 2011

### de_brook

More to this is to think of the kronecker delta as a symmetric unit matrix and that allows you to write;
$\delta_{jm}=\delta_{mj}=\delta_{jl}\delta_{lm}=\delta_{lm}\delta_{jl}$

Last edited: Sep 4, 2011
7. Sep 4, 2011

### Fredrik

Staff Emeritus
It's not. For example, $\delta_{3m}\delta_{k3}$ is not equal to $\delta_{km}$ in general. It is when m=k=3, but not when m=k=2.

On the other hand $\delta_{jm}\delta_{kj}=\delta_{km}$, because this is a sum with three terms, and in every term, the first factor is 0 if j≠m and the second factor is 0 if j≠k. So if k≠m, all the terms are =0 because they all contain at least one factor that's zero, but if k=m, the term with j=m=k is =1 and the others =0.

Yes, because for each i and each j, $\delta_{ij}$ is a real number, and real numbers commute.

Yes.

Yes.

8. Sep 4, 2011

### Fredrik

Staff Emeritus
You need to make sure that your latex code contains a space at least once every 50 characters. If if it doesn't, vBulletin will insert a space and break the code.

9. Sep 5, 2011

### Fredrik

Staff Emeritus
Quietrain, do you understand the answers you've been given? If not, then please say something. (Would you rather start over from square one seven months from now?)

I didn't even notice what you said here, but you're making a serious error. It doesn't make sense to start this sentence with "since l=j". There's no l in the expression that follows, and the j that is a part of that expression is summed over, so it doesn't have a specific value. This means that no matter what l is, it wouldn't make any sense to say that l=j in that expression.

You are always extremely careless with your statements. I think you would find mathematics a lot easier if you made a habit of trying to express yourself as clearly as possible.

This question proves that you either think that one of the equalities \begin{align}
0\cdot 0=0\cdot 0\\
0\cdot 1=1\cdot 0\\
1\cdot 0=0\cdot 1\\
1\cdot 1=1\cdot 1
\end{align} is false, or that you don't understand the notation that puts one or more indices on a symbol. I don't believe that you would doubt those equalities, so I have to conclude that you don't understand what any symbol with an index on it represents. If you want to understand any of these things, you have to start by making sure that you understand the basics.

Last edited: Sep 5, 2011
10. Sep 6, 2011

### quietrain

i think hallsofivy knows what i am trying to say,

maybe i should rephrase it

i have this

djmdkl

when j = l

does it give me

OPTION 1)

dkm since Pengwuino said that its done by logic deduction. but fredrik says that its not? he says that djmdkj = dkm ,
but if i let j=l , issn't djmdkl = djmdkj ??

OR

OPTION 2)

djmdkl

since l = j

djmdkj where J is now a dummy variable? with einstein summation of repeated indices,

so i get dk1d1m + dk2d2m + dk3d3m ?

hallsofivy says its not einstein summation? and fredrik says this is wrong?

are you trying to say that since J is now a dummy variable, i cannot let l = j ? or else l is no longer a free index?

CONCLUSION

so is dkm = dk1d1m + dk2d2m + dk3d3m ?? debrook's post seems to suggest this is right?

so when k=m, i get 1+1+1 = 3?

or are they 2 different entities?

but if i let k=m, dkk = 3 right? assuming k goes from 1 to 3

so d11+d22+d33 = 1+1+1=3?

so is d11+d22+d33 the same as dk1d1m + dk2d2m + dk3d3m ??? (they give same answer but do they mean the same thing?)

oh well, this is mind boggling ,,,

Last edited: Sep 6, 2011
11. Sep 6, 2011

### Fredrik

Staff Emeritus
No. There's only one term on the left and three terms on the right. (Before we have actually checked, it's conceivable that the value of the expression on the left is the same as the value of the expression on the right for all values of the indices. But if you understand the notation at all, it's immediately obvious that the two expressions represent different calculations).

Halls and I said the same thing. Einstein's summation convention is that repeated indices are summed over. There's no repeated index in $\delta_{jm}\delta_{kl}$, but there is a repeated index in $\delta_{jm}\delta_{kj}$.

Maybe I should have just said that it doesn't make sense to say "since l=j" and then ask a question that doesn't involve l. But another problem with the question we're discussing (quoted again below) is that it's a yes/no question, and if the answer is yes, then the question fails to make sense in a second way (by setting a variable that's summed over to a fixed value).

Just a reminder of what we're talking about:

Why don't you check this yourself by plugging in all possible values of k and m? There are only 9 combos. (The answer is "yes", and if you read my first post again, you'll see that I said this there).

Do you think a variable can represent different numbers at different places in the same expression? This would mean that an equality like x+x=2x is usually false.

You're not making sense here. However, both of the following statements are true:
(a) $\delta_{kk}=3$.
(b) If $k=m$, then $\delta_{km}=1$.

Again, a variable never represents different numbers at different places in the same expression. That would make it impossible to do mathematics.

Last edited: Sep 6, 2011
12. Sep 7, 2011

### quietrain

QUESTION 1)
oh.. so ,d11+d22+d33 is not the same as dk1d1m + dk2d2m + dk3d3m

because k or m cannot be 1,2,3 in the same expression?

Q 2)
for dkm = dk1d1m + dk2d2m + dk3d3m

so i only get d11 = d11d11 +0 + 0 = 1
so its kind of redundant to write the other two terms? unless i specify that k and m goes from 1 to 3 in dkm?

or does writing dkm already implies that k and m must go from 1 to 3?

Q 3)
with regards to the l=j part,

i was actually looking at this,

(weird the image code is not working?)

so after j=l, how did it become 3 dkm - dkm ?

3 dkm came from this part : djldkm right? while

- dkm came from the contraction of djmdkl ? but you said this contraction is wrong in your first post?

also, how did the 3 dkm part have a 3 popping out? shouldn't djldkm = 1dkm where l=j ?

or is the crux of this problem lying with the levi -civita symbol on the left? where there is actually a summation symbol for repeated indices going from 1to 3?

so that i get { d11+ d22 +d33 }dkm which is 3dkm ?

Q4)
also, i really don't understand this part. dkk is not equal to dkm when k=m as per your last post. you mean dkm where k=m, is 1, but i cannot write it as dkk as the repeated k here would mean having repeated indices,which clashes with dkk=3 when k goes from 1 to 3, so, only if the expression straight off the bat is dkk, then its summation?

Q5)
so, what does say, d12 tell me? i know if index are equal, it is =1. the index are not really rows and columns indicators right? then what are they? is there something i can visualise or a simple example? since they are related to the levi-civita symbol, then it means that they have to have a meaning right? like for example, is my perception of levi-civita symbol correct as below?

since (A X B)= EijkAjBk=Ci means I must sum j and k(repeated index) for the ith component of C, thus if i goes from 1 to 3, then it has 3 components and thus is not a scalar but a vector like what the cross product gives? so i can imagine i to be say x-component, y, and z for 1,2,3

but for the kronecker delta, what do the index tell me?

thanks!

Last edited: Sep 7, 2011
13. Sep 7, 2011

### Fredrik

Staff Emeritus
Right. If k=m, the second expression can still represent three different calculations, but none of them is $\delta_{11}\delta_{11} +\delta_{22}\delta_{22}+\delta_{33}\delta_{33}$.

Your calculation is correct. I don't know what it would mean to say that "k and m goes from 1 to 3 in dkm".

I don't like how the author uses the phrase "setting $j=l$". I would interpret that as "when j and l represent the same number, we get..." That's not at all what he's trying to say. What he meant to say is that the statement

$$\text{For all j,k,l,m\in\{1,2,3\}, we have } \epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}.$$ implies the statement $$\text{For all k,m\in\{1,2,3\}, we have }\epsilon_{ijk}\epsilon_{ijm}=\delta_{jj}\delta_{km}-\delta_{jm}\delta_{kj}=3\delta_{km}-\delta_{km}=2\delta_{km}.$$ You might want to read my first post again to see what I'm actually saying.

Yes, that's exactly right.

$\delta_{12}$ is a symbol that that represents the number 0, nothing more, nothing less. There's nothing in the definition of $\delta_{ij}$ that says that i and j represents rows and columns, but for each value of i and j, $\delta_{ij}$ happens to be equal to the component on row i, column j, of the 3×3 identity matrix. So you can think of the indices as representing rows and columns if you want to. The indices are just variables that represent numbers in the set {1,2,3}. $\delta$ is a function from {1,2,3}×{1,2,3} into {0,1}. The fact that we write $\delta_{ij}$ instead of $\delta(i,j)$ is just a notational convention.

Last edited: Sep 7, 2011
14. Sep 7, 2011

### quietrain

alright i think i understand a lot now :O

its all about the basics.. like what my prof n you keep saying hahah

thanks!

15. Sep 7, 2011

### Fredrik

Staff Emeritus
I'm pleased that we've made progress. I think that now that we have found that one really important thing that you've been doing consistently wrong, you will find everything else much easier.

16. Sep 8, 2011

### quietrain

easier? perhaps

i found more questions though :(

17. Sep 8, 2011

### Fredrik

Staff Emeritus
Go ahead, ask them. Better to do it now than to wait until you've forgotten what you've learned in these threads.

I think it's best if you just ask one question at a time. I will be giving you hints rather than complete solutions from now on. When you ask a question, you should include your own attempt to answer it, up to the point where you get stuck.

I also have to request that when you write down a question, you think it through before you post it. You need to work harder to make sure that your questions make sense. It's too hard to try to figure out what you mean sometimes.

Last edited: Sep 8, 2011
18. Sep 8, 2011

### quietrain

ok this question is more about tensors

for example, my prof says x12 + x22 is an invariant

invariant means the tensor is order 0 right? means it has no index on it?

just purely X? not Xi or Xij etc , that is to say it has no components right?

so since x12 + x22 has indexes, why is it invariant?

also, what is an invariant? is it a scalar?

19. Sep 8, 2011

### Fredrik

Staff Emeritus
The expression $x_1^2+x_2^2$ is invariant under rotations in the 12 plane (the xy plane) in the sense that if you rotate the vector $(x_1,x_2)$ in the 12 plane by some angle $\theta$ and the components of the new vector are $(x'_1,x'_2)$, we have $x'_1^2+x'_2^2=x_1^2+x_2^2$, no matter what $\theta$ is.

This is how I prefer to think about these things (when I'm forced to think about it in terms of coordinates): The expression $x_1^2+x_2^2$ can be thought of as a formula that associates a real number with each coordinate system. This association is said to be invariant if it associates the same real number with each coordinate system. More generally, an association of an indexed set of numbers with each coordinate system is said to be a tensor if the indexed set associated with one coordinate system is related to the indexed set associated with another as described by the tensor transformation law.

(In this context, the only coordinate systems we're considering are related to each other by rotations).

Right. The indices indicate which component we're dealing with, and scalars (=invariants) don't have components. Edit: I mean that a scalar has exactly one component, so there's no need to use an index to distinguish between different components.

Because just like any other expression that's constructed from tensors but doesn't have any free indices, like e.g. $A_{ij}B_{ij}$, it associates the same real number with each coordinate system. By a "free" index, I mean one that isn't summed over. $x_1^2+x_2^2=x_i x_i$ clearly doesn't have any free indices.

Last edited: Sep 8, 2011
20. Sep 9, 2011

### quietrain

oh.. is there a way to "see" whether something is invariant or not? or do we have to manually work it out using the transformation law? i.e doing the matrix multiplication etc...

because my prof said $x_1^2+x_2^2$ is invariant and (i think he said this)$x_1^3+x_2^3$ is not invariant straight away.

also with regards to the quotient law of tensors that says

(A)(B) = C , if B,C are tensors, A is also a tensor ,if order of C = order of A + order of B - 2

so if i want to show that A is a 2nd order tensor, the idea to solving is just to mulitply it by a order 0 (scalar) so that i get order 0 (scalar)

so its something like this right

(A) (scalar) = (scalar , e.g 0 to make life easier)

so what is the idea to figuring out what (scalar) to mulitply A by?

because my prof gave me a 2nd order tensor (3x3 matrix) with crazy sin and cos terms everywhere