Kronecker delta question

  • Thread starter quietrain
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  • #26
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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
tensor.jpg


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, [itex]A_{ij}B[/itex] are the components of a tensor. Then [itex]A_{ij}[/itex] are the components of a tensor.​
This is of course an entirely trivial statement.



also , for

By the way, do you understand why [itex]R^TR=I[/itex], or equivalently, why [itex]R_{ij}R_{ik}=\delta_{jk}[/itex]? 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 LijLik = LTjiLik which is a matrix multiplication

so its = LTL = 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 realise it is actually telling me the 'cross' terms, csx1x2 are cancelled 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 = Rij vj then they are a cartesian tensor?
 
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  • #27
Fredrik
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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
tensor.jpg
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 [itex]A^TA=I[/itex]. This equality is equivalent to [itex]A^{-1}=A^T[/itex]. So I would say that the answer to the one without the indices is that R is orthogonal.

so LijLik = LTjiLik which is a matrix multiplication

so its = LTL = I
OK, it's good that you recognize the matrix multiplication and the transpose operation. The definition of matrix multiplication is [itex](AB)_{ij}=A_{ik}B_{kj}[/itex], so [tex]R_{ij}R_{ik}=(R^T)_{ji}R_{ik}=(R^TR)_{jk}=I_{jk} =\delta_{jk}.[/tex]
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. [itex]\Leftrightarrow[/itex] The columns of R are orthonormal. [itex]\Leftrightarrow[/itex] 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 = Rij vj 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 [itex]v'_i=R_{ij}v_j[/itex], where R is the matrix that describes the coordinate change.
 
  • #28
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2
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-1


because 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
 
  • #29
Fredrik
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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.
 
  • #30
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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 thats 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
 
  • #31
Fredrik
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yes thats 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.
 
  • #32
654
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hmm its ok then , thats all the question says .

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

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