Why does the Kronecker delta interchange with itself when j equals l?

[quietrain]

does anyone know why

d_jmd_kl = d_km

when j = l

thanks!

can i interchange it to d_jmd_kl = d_kld_jm ?

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

so i get d_k1d_1m + d_k2d_2m + d_k3d_3m ?

 

[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}

 

[quietrain]

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

 

[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.

 

[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.

 

[de_brook]

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.

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}

 

[Fredrik]

does anyone know why

d_jmd_kl = d_km

when j = l

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.

can i interchange it to d_jmd_kl = d_kld_jm ?

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

do the "j" in d_kjd_jm signify summation?

Yes.

so i get d_k1d_1m + d_k2d_2m + d_k3d_3m ?

Yes.

 

[Fredrik]

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}

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.

 

[Fredrik]

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?)

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

so i get d_k1d_1m + d_k2d_2m + d_k3d_3m ?

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.

can i interchange it to d_jmd_kl = d_kld_jm ?

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.

 

[quietrain]

i think hallsofivy knows what i am trying to say,

maybe i should rephrase iti have this

d_jmd_kl

when j = l

does it give me

OPTION 1)

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

OR

OPTION 2)

d_jmd_kl

since l = j

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

so i get d_k1d_1m + d_k2d_2m + d_k3d_3m ?

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

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.

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 d_km = d_k1d_1m + d_k2d_2m + d_k3d_3m ?? 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, d_kk = 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 d_k1d_1m + d_k2d_2m + d_k3d_3m ? (they give same answer but do they mean the same thing?)

oh well, this is mind boggling ,,,

 

[Fredrik]

but if i let j=l , issn't d_jmd_kl = d_jmd_kj ??

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).

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

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}.

are you trying to say that since J is now a dummy variable, i cannot let l = j ?

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:

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

so is d_km = d_k1d_1m + d_k2d_2m + d_k3d_3m ??

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).

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

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.

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

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.

so is d₁₁+d₂₂+d₃₃ the same as d_k1d_1m + d_k2d_2m + d_k3d_3m ?

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

 

[quietrain]

QUESTION 1)
oh.. so ,d₁₁+d₂₂+d₃₃ is not the same as d_k1d_1m + d_k2d_2m + d_k3d_3m

because k or m cannot be 1,2,3 in the same expression?
Q 2)
for d_km = d_k1d_1m + d_k2d_2m + d_k3d_3m

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 d_km?

or does writing d_km 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 d_km - d_km ?

3 d_km came from this part : d_jld_km right? while

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

also, how did the 3 d_km part have a 3 popping out? shouldn't d_jld_km = 1d_km 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 }d_km which is 3d_km ?
Q4)
also, i really don't understand this part. d_kk is not equal to d_km when k=m as per your last post. you mean d_km where k=m, is 1, but i cannot write it as d_kk as the repeated k here would mean having repeated indices,which clashes with d_kk=3 when k goes from 1 to 3, so, only if the expression straight off the bat is d_kk, 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)= E_ijkA_jB_k=C_i 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!

 

[Fredrik]

QUESTION 1)
oh.. so ,d₁₁+d₂₂+d₃₃ is not the same as d_k1d_1m + d_k2d_2m + d_k3d_3m

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

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}.

Q 2)
for d_km = d_k1d_1m + d_k2d_2m + d_k3d_3m

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 d_km?

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

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

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

i was actually looking at this,

(weird the image code is not working?)

View attachment 38623

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

3 d_km came from this part : d_jld_km right? while

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

also, how did the 3 d_km part have a 3 popping out? shouldn't d_jld_km = 1d_km 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 }d_km which is 3d_km ?

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.

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

Yes, that's exactly right.

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)= E_ijkA_jB_k=C_i 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!

\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.

 

[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!

 

[Fredrik]

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.

 

[quietrain]

easier? perhaps

i found more questions though :(

 

[Fredrik]

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.

 

[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 X_i or X_ij 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?

 

[Fredrik]

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

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).

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

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

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.

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

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.

 

[quietrain]

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.

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

 

[Fredrik]

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.

If it's a combination of tensors, with no free indices, then it's always a scalar. If it's a combination of tensors, with at least one free index, then it's not a scalar. If it's not a combination of tensors at all, you will have to figure out how it associates numbers with coordinate systems.

Have you done the calculation that verifies that x_1^2+x_2^2 is invariant? If not, you should do it right away.

Matrix version: Since x'=Rx, we have x'^Tx'=\dots

Component version: Since x'_i=R_{ij}x_j, we have x'_i x'_i=\dots

Can you fill in the rest?

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

I'm actually not familiar with this rule. Can you state it exactly as it appears in your book or your lecture notes? (I assume that there are indices that you haven't included here).

 

[quietrain]

it was kind of hard to find anything on quotient law on the internet... weird.

middle of page 41
http://www.hereticalcosmology.com/articles/Tensor_Analysis-Chapter_1.pdf

this version is a lot more complicated though


below is the version that my lect gave me

if B and C are tensors and also that
A_pq...k...mB_ij...k...n = C_{pq...mij ...n}
holds in all rotated coordinate frames where A, B, and C are of
Mth, Nth and (M+N-2)th order, then A_pq...k...m is also a
tensor.




with regards to the invariant question

i have only learned how to transform vectors like (x₂ , -x₁)

so that v₁' = L₁₁v₁ + L₁₂v₂
so that v₂' = L₂₁v₁ + L₂₂v₂

where my v₁ takes the first vector x₂
and my v₂ takes the 2nd vector -x₁



so with regards to the x₁² + x₂²

what is my v₁ and v₂ ??

am i suppose to take v₁ = x₁² + x₂² ?

then what will my v₂ be?

or am i suppose to take v₁ = x₁
v₂ = x₂

so that by the transformation R,

x₁²' + x₂²' = { L₁₁x₁+ L₁₂x₂ } ² +
L₂₁x₁+ L₂₂x₂ } ²

where c = cos, s = sin

= { cx₁ + sx₂ }² + { -sx₁ + cx₂ }² }

= x₁² + x₂²


so in the component form

is it suppose to read like this

x'_i=R_{ij}x_j, we have x'_i x'_i= R_{ij}x_j R_{ik}x_k

x'_1 x'_1 + x'_2 x'_2 =[ R_{11}x_1 + R_{12} x_2 ][ R_{11}x_1 + R_{12}x_2] + [ R_{21}x_1 + R_{22} x_2 ][ R_{21}x_1 + R_{22}x_2]

 

[Fredrik]

it was kind of hard to find anything on quotient law on the internet... weird.

middle of page 41
http://www.hereticalcosmology.com/articles/Tensor_Analysis-Chapter_1.pdf

The proof on pages 41-42 is not too hard to understand, so you should work it through on your own. Ask if you get stuck on a detail. Note that it's essential that (13.5) holds for all tensors B, and in all the coordinate systems under consideration. If it just holds for one specific B, or in some coordinate systems, (13.5) doesn't really tell you anything.

I'm not sure what you need help with regarding the quotient law. You may have to post the specific problem along with your attempt to solve it.

with regards to the invariant question

i have only learned how to transform vectors like (x₂ , -x₁)

so that v₁' = L₁₁v₁ + L₁₂v₂
so that v₂' = L₂₁v₁ + L₂₂v₂

where my v₁ takes the first vector x₂
and my v₂ takes the 2nd vector -x₁

I don't understand what you're saying. I understand that L is a linear transformation that takes (v_1,v_2) to (v'_1,v'_2), but then you lost me. I can't make sense of the phrase "the first vector x₂".

so with regards to the x₁² + x₂²

what is my v₁ and v₂ ??

am i suppose to take v₁ = x₁² + x₂² ?

No, you're supposed to "transform" the vector (x_1,x_2) the same way you transformed (v_1,v_2) above.

or am i suppose to take v₁ = x₁
v₂ = x₂

Yes, that's one way of putting it.

so that by the transformation R,

x₁²' + x₂²' = { L₁₁x₁+ L₁₂x₂ } ² +
L₂₁x₁+ L₂₂x₂ } ²

where c = cos, s = sin

= { cx₁ + sx₂ }² + { -sx₁ + cx₂ }² }

= x₁² + x₂²

Correct.

so in the component form

is it suppose to read like this

x'_i=R_{ij}x_j, we have x'_i x'_i= R_{ij}x_j R_{ik}x_k

x'_1 x'_1 + x'_2 x'_2 =[ R_{11}x_1 + R_{12} x_2 ][ R_{11}x_1 + R_{12}x_2] + [ R_{21}x_1 + R_{22} x_2 ][ R_{21}x_1 + R_{22}x_2]

Yes, but you can simplify it without explicitly performing the summation.

x'_i x'_i=R_{ij}x_j R_{ik}x_k=x_j \underbrace{R_{ij}R_{ik}}_{=\delta_{jk}} x_k = x_j x_j
It's actually much easier to do the whole thing with matrices instead of their components:
x'^Tx'=(Rx)^T(Rx)=x^TR^TRx=x^Tx

 

[quietrain]

wow. so i did it through the brute force method :X , when i could have simplified it lol

with regards to the quotient law

if i have matrix A (3by3)

so if i mulitply it with a scalar B ( say 0) , then i get C = 0

then wouldn't it fulfill the quotient law?

since the orders of tensors are for A = 2 (M), B = 0 (N) , C = 0 (M+N-2)

 

[Fredrik]

What you're describing now has nothing to do with the quotient law, and it's not because of the orders of the tensors. Look at page 41 again. What the quotient law says about your components A_{ij} and their relationship with a scalar B is this:

Suppose that for all real numbers B, A_{ij}B are the components of a tensor. Then A_{ij} are the components of a tensor.​

This is of course an entirely trivial statement.

What are you trying to do exactly? Are you trying to figure out if the numbers A_{ij} are the components of a tensor? That wouldn't really make sense. You can always pick a coordinate system, and define A to be the tensor that has components A_{ij} in that coordinate system. So no matter what numbers the A_{ij} are, they are the components of infinitely many tensors. The question "Are these the components of a tensor?" only makes sense when what you've been given associates an indexed set of numbers with each coordinate system. Is there something about your matrix that gives you a reason to interpret it as an association of nine numbers with each coordinate system?

By the way, do you understand why R^TR=I, or equivalently, why R_{ij}R_{ik}=\delta_{jk}? Do you understand that those two statements are actually the same statement written in different notations?

 

[quietrain]

er i don't really understand what you mean by

The question "Are these the components of a tensor?" only makes sense when what you've been given associates an indexed set of numbers with each coordinate system. Is there something about your matrix that gives you a reason to interpret it as an association of nine numbers with each coordinate system?

but the question is something like this ,

show that the following is a 2nd order tensor

so am i suppose to find a scalar to multiply it with, so that it gives me back a scalar so that i satisfy the quotient law? issn't this what the law says?

Suppose that for all real numbers B, A_{ij}B are the components of a tensor. Then A_{ij} are the components of a tensor.​

This is of course an entirely trivial statement.



also , for

By the way, do you understand why R^TR=I, or equivalently, why R_{ij}R_{ik}=\delta_{jk}? Do you understand that those two statements are actually the same statement written in different notations?

i know they are as such because my lecturer told us to memorize this relationship
i understand the matrix one, because the inverse of a matrix X a matrix = identity matrix

but the index notation one? i think its because the L is orthogonal

so L_ijL_ik = L^T_jiL_ik which is a matrix multiplication

so its = L^TL = I

but to go straight to the kronecker delta? its telling me if j=\=k, then i have 0 , if j=k i have 1,

i realize it is actually telling me the 'cross' terms, csx₁x₂ are canceled off.
while the 'like' terms give me c² + s² = 1

i don't know the implicit reason being this? the creator who created this is amazing?


oh another quesiton

is the transformation R always
(c s 0)
(-s c 0)
(0 0 1)

? or is this only for rotation about an axis? so what other transformation R are there? i assume all the methodolog of understanding the indexes still hold?

so bascially there are other transformation matrices ? so as long as they satisfy the tensor transformation law of

v'_i = R_ij v_j then they are a cartesian tensor?

 

[Fredrik]

er i don't really understand what you mean by

I mean that a 3x3 matrix is just a set of nine numbers, but a tensor with two indices is something that associates nine numbers with each coordinate system (in a way that's consistent with the tensor transformation law).

but the question is something like this ,

show that the following is a 2nd order tensor

To make sense of this, we have to figure out a way that this associates nine numbers with each coordinate system. Can you do that? The only clue we have is the notation. (x,y,z) is a standard notation for the components of a vector in a coordinate system. (Hm, this actually looks pretty complicated).

so am i suppose to find a scalar to multiply it with, so that it gives me back a scalar so that i satisfy the quotient law? issn't this what the law says?

I told you what the quotient law says about this matrix and its relationship to a scalar B. It says something completely useless (that if A times a number is a tensor, then A is a tensor...this is useless because it's not easier to show that A times a number is a tensor, than to show that A is a tensor). What makes you think this problem has anything to do with the quotient law?

i understand the matrix one, because the inverse of a matrix X a matrix = identity matrix

but the index notation one? i think its because the L is orthogonal

You're on the right track, but your answer is still a bit odd. The definition of "orthogonal matrix" is that A is said to be orthogonal if A^TA=I. This equality is equivalent to A^{-1}=A^T. So I would say that the answer to the one without the indices is that R is orthogonal.

so L_ijL_ik = L^T_jiL_ik which is a matrix multiplication

so its = L^TL = I

OK, it's good that you recognize the matrix multiplication and the transpose operation. The definition of matrix multiplication is (AB)_{ij}=A_{ik}B_{kj}, so R_{ij}R_{ik}=(R^T)_{ji}R_{ik}=(R^TR)_{jk}=I_{jk} =\delta_{jk}.

is the transformation R always
(c s 0)
(-s c 0)
(0 0 1)

? or is this only for rotation about an axis? so what other transformation R are there? i assume all the methodolog of understanding the indexes still hold?

This is a rotation around the z axis. A rotation is by definition an orthogonal matrix with determinant 1. (Some authors don't include the last requirement in the definition, and call a "rotation" with determinant 1 a "proper rotation"). It's useful to know, and easy to prove that

R is orthogonal. \Leftrightarrow The columns of R are orthonormal. \Leftrightarrow The rows of R are orthonormal.​

so bascially there are other transformation matrices ? so as long as they satisfy the tensor transformation law of

v'_i = R_ij v_j then they are a cartesian tensor?

This is not a condition on R. It's a condition on v. R defines a coordinate change (if and only if it's orthogonal and its determinant is 1). The association of 3 numbers with each coordinate system that you're dealing with is a tensor (specifically a vector) if the relationship between the three numbers associated with an arbitrary coordinate system and the three numbers associated with another arbitrary coordinate system is v'_i=R_{ij}v_j, where R is the matrix that describes the coordinate change.

 

[quietrain]

i see...

with regards to the quotient law,

the question asks me to use the quotient law to show that that matrix is a 2nd order tensor. but anyway, from what i see, the matrix is symmetric. so if i diagonalize the matrix

P_-1AP = D

is this a form of the quotient law?

also, is this right?

P^-1APP^-1 = DP^-1

P^-1A = DP^-1because in the other thread, https://www.physicsforums.com/showthread.php?p=3497945#post3497945
hallsofivy told me

P^-1AP= D

AP= P^-1D

i haven't got a clue how she got AP= P^-1D

 

[Fredrik]

with regards to the quotient law,

the question asks me to use the quotient law to show that that matrix is a 2nd order tensor.

Have you really told me everything that this problem says?

but anyway, from what i see, the matrix is symmetric. so if i diagonalize the matrix

P_-1AP = D

is this a form of the quotient law?

What makes you think that it might be?

The fact that the matrix is symmetric is a good observation. Since symmetric matrices can be diagonalized by orthogonal matrices, we know that if the components of the matrix are the components of a tensor, there's a coordinate system in which it's diagonal.

also, is this right?

P^-1APP^-1 = DP^-1

P^-1A = DP^-1

Of course. That's what you get when you multiply P^-1AP = D by P^-1 from the right.

hallsofivy told me

P^-1AP= D

AP= P^-1D

i haven't got a clue how she got AP= P^-1D

The right-hand side should be PD, not P^-1D.

 

[quietrain]

Have you really told me everything that this problem says? What makes you think that it might be?

The fact that the matrix is symmetric is a good observation. Since symmetric matrices can be diagonalized by orthogonal matrices, we know that if the components of the matrix are the components of a tensor, there's a coordinate system in which it's diagonal.

.

yes that's all the question wants. use quotient law to show that matrix is a 2nd order tensor.

so how do i proceed? is diagonalizing it the right way?

i think my prof also said that by the quotient law, if the matrix

(A) (scalar) = invariant, it shows that A is tensor. or something along the lines of these.

anyway, i still don't really have any idea how to solve this quesiton

i don't seem to be able to draw the link between diagonal matrix and quotient law

 

[Fredrik]

yes that's all the question wants. use quotient law to show that matrix is a 2nd order tensor.

I don't see how the quotient law can have anything to do with this.

so how do i proceed? is diagonalizing it the right way?

I have no idea. What I do know is that it's pointless to do that unless you have a way to proceed once you have found the diagonal matrix.

i think my prof also said that by the quotient law, if the matrix

(A) (scalar) = invariant, it shows that A is tensor. or something along the lines of these.

If A times a scalar is an invariant, then A is obviously invariant too. So I'm pretty sure that's not what your professor told you.

When you have a statement that you think might be useful, that may or may not be what you have been told by a professor or seen in a book, you should at least think about what it would mean if the statement is true. If you had done this here, it would have saved us both some time.

anyway, i still don't really have any idea how to solve this quesiton

I don't either, and to be honest, I still doubt that you have stated the problem correctly.

Do you have any solved examples from your book or your lecture notes that you can show me? This could at least give us some idea about how the quotient law is relevant.

 

[quietrain]

hmm its ok then , that's all the question says .

thanks for everything though, i finally understood what those indexes meant