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er i don't really understand what you mean by

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?

also , for

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

so its = L

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, csx

while the 'like' terms give me c

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'

but the question is something like this ,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?

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, [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

i know they are as such because my lecturer told us to memorize this relationshipBy 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 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

_{ij}L_{ik}= L^{T}_{ji}L_{ik}which is a matrix multiplicationso its = L

^{T}L = Ibut 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, csx

_{1}x_{2}are cancelled off.while the 'like' terms give me c

^{2}+ s^{2}= 1i 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?
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