- #1

- 654

- 2

d

_{jm}d

_{kl}= d

_{km}

when j = l

thanks!

can i interchange it to d

_{jm}d

_{kl}= d

_{kl}d

_{jm}?

since l = j

do the "j" in d

_{kj}d

_{jm}signify summation?

so i get d

_{k1}d

_{1m}+ d

_{k2}d

_{2m}+ d

_{k3}d

_{3m}?

- Thread starter quietrain
- Start date

- #1

- 654

- 2

d

when j = l

thanks!

can i interchange it to d

since l = j

do the "j" in d

so i get d

- #2

Pengwuino

Gold Member

- 4,989

- 16

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

- #3

- 654

- 2

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

- #4

Pengwuino

Gold Member

- 4,989

- 16

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

- #6

- 74

- 0

More to this is to think of the kronecker delta as a symmetric unit matrix and that allows you to write;

[itex]\delta_{jm}=\delta_{mj}=\delta_{jl}\delta_{lm}=\delta_{lm}\delta_{jl}[/itex]

Last edited:

- #7

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

It's not. For example, [itex]\delta_{3m}\delta_{k3}[/itex] is not equal to [itex]\delta_{km}[/itex] in general. It is when m=k=3, but not when m=k=2.does anyone know why

d_{jm}d_{kl}= d_{km}

when j = l

On the other hand [itex]\delta_{jm}\delta_{kj}=\delta_{km}[/itex], 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, [itex]\delta_{ij}[/itex] is a real number, and real numbers commute.can i interchange it to d_{jm}d_{kl}= d_{kl}d_{jm}?

Yes.do the "j" in d_{kj}d_{jm}signify summation?

Yes.so i get d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}?

- #8

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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.More to this is to think of the kronecker delta as a symmetric unit matrix and that allows you to write;

[itex]\delta_{jm}=\delta_{mj}=\delta_{jl}\delta_{lm} =\delta_{lm}\delta_{jl}[/itex]

- #9

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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?)

*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.

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.

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 thatsince l = j

do the "j" in d_{kj}d_{jm}signify summation?

so i get d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}?

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}can i interchange it to d_{jm}d_{kl}= d_{kl}d_{jm}?

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}

Last edited:

- #10

- 654

- 2

i think hallsofivy knows what i am trying to say,

maybe i should rephrase it

i have this

d_{jm}d_{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_{jm}d_{kj} = d_{km} ,

but if i let j=l , issn't d_{jm}d_{kl} = d_{jm}d_{kj} ??

OR

OPTION 2)

d_{jm}d_{kl}

since l = j

d_{jm}d_{kj} where J is now a dummy variable? with einstein summation of repeated indices,

so i get d_{k1}d_{1m} + d_{k2}d_{2m} + d_{k3}d_{3m} ?

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

CONCLUSION

so is d_{km} = d_{k1}d_{1m} + d_{k2}d_{2m} + d_{k3}d_{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 d_{11}+d_{22}+d_{33} = 1+1+1=3?

so is d_{11}+d_{22}+d_{33} the same as d_{k1}d_{1m} + d_{k2}d_{2m} + d_{k3}d_{3m} ??? (they give same answer but do they mean the same thing?)

oh well, this is mind boggling ,,,

maybe i should rephrase it

i have this

d

when j = l

does it give me

OPTION 1)

d

but if i let j=l , issn't d

OR

OPTION 2)

d

since l = j

d

so i get d

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?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.

CONCLUSION

so is d

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

or are they 2 different entities?

but if i let k=m, d

so d

so is d

oh well, this is mind boggling ,,,

Last edited:

- #11

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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 differentbut if i let j=l , issn't d_{jm}d_{kl}= d_{jm}d_{kj}??

Halls and I said the same thing. Einstein's summation convention is that repeated indices are summed over. There's no repeated index in [itex]\delta_{jm}\delta_{kl}[/itex], but there is a repeated index in [itex]\delta_{jm}\delta_{kj}[/itex].hallsofivy says its not einstein summation? and fredrik says this is wrong?

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

Just a reminder of what we're talking about:

since l = j

do the "j" in d_{kj}d_{jm}signify summation?

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 is d_{km}= d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}??

Do you think a variable can represent different numbers at different placesso when k=m, i get 1+1+1 = 3?

You're not making sense here. However, both of the following statements are true:but if i let k=m, d_{kk}= 3 right? assuming k goes from 1 to 3

(a) [itex]\delta_{kk}=3[/itex].

(b) If [itex]k=m[/itex], then [itex]\delta_{km}=1[/itex].

Again, a variableso is d_{11}+d_{22}+d_{33}the same as d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}???

Last edited:

- #12

- 654

- 2

QUESTION 1)

oh.. so ,d_{11}+d_{22}+d_{33} is not the same as d_{k1}d_{1m} + d_{k2}d_{2m} + d_{k3}d_{3m}

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

Q 2)

for d_{km} = d_{k1}d_{1m} + d_{k2}d_{2m} + d_{k3}d_{3m}

so i only get d_{11} = d_{11}d_{11} +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_{jl}d_{km} right? while

- d_{km} came from the contraction of d_{jm}d_{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_{jl}d_{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 { d_{11}+ d_{22} +d_{33} }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, d_{12} 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_{ijk}A_{j}B_{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!

oh.. so ,d

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

Q 2)

for d

so i only get d

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

or does writing d

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

3 d

- d

also, how did the 3 d

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 { d

Q4)

also, i really don't understand this part. d

Q5)

so, what does say, d

since (A X B)= E

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

thanks!

Last edited:

- #13

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

Right. If k=m, the second expression can still represent three different calculations, but none of them is [itex]\delta_{11}\delta_{11} +\delta_{22}\delta_{22}+\delta_{33}\delta_{33}[/itex].QUESTION 1)

oh.. so ,d_{11}+d_{22}+d_{33}is not the same as d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}

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

Your calculation is correct. I don't know what it would mean to say that "k and m goes from 1 to 3 in dQ 2)

for d_{km}= d_{k1}d_{1m}+ d_{k2}d_{2m}+ d_{k3}d_{3m}

so i only get d_{11}= d_{11}d_{11}+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?

I don't like how the author uses the phrase "setting [itex]j=l[/itex]". 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 statementQ 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_{jl}d_{km}right? while

- d_{km}came from the contraction of d_{jm}d_{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_{jl}d_{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 { d_{11}+ d_{22}+d_{33}}d_{km}which is 3d_{km}?

[tex]\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}.[/tex]

Yes, that's exactly right.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?

[itex]\delta_{12}[/itex] is a symbol that that represents the number 0, nothing more, nothing less. There's nothing in the definition of [itex]\delta_{ij}[/itex] that says that i and j represents rows and columns, but for each value of i and j, [itex]\delta_{ij}[/itex] happens to be equal to the component on row i, column j, of the 3×3 identity matrix. So youQ5)

so, what does say, d_{12}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_{ijk}A_{j}B_{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!

Last edited:

- #14

- 654

- 2

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

thanks!

- #15

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

- #16

- 654

- 2

easier? perhaps

i found more questions though :(

i found more questions though :(

- #17

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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.

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:

- #18

- 654

- 2

for example, my prof says x

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

just purely X? not X

so since x

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

- #19

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

The expression [itex]x_1^2+x_2^2[/itex] is invariant under rotations in the 12 plane (the xy plane) in the sense that if you rotate the vector [itex](x_1,x_2)[/itex] in the 12 plane by some angle [itex]\theta[/itex] and the components of the new vector are [itex](x'_1,x'_2)[/itex], we have [itex]x'_1^2+x'_2^2=x_1^2+x_2^2[/itex], no matter what [itex]\theta[/itex] is.for example, my prof says x_{1}^{2}+ x_{2}^{2}is an invariant

This is how I prefer to think about these things (when I'm forced to think about it in terms of coordinates): The expression [itex]x_1^2+x_2^2[/itex] can be thought of as a formula that associates a real number with each coordinate system. This

(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 exactlyinvariant 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?

Because just like any other expression that's constructed from tensors but doesn't have anyso since x_{1}^{2}+ x_{2}^{2}has indexes, why is it invariant?

Last edited:

- #20

- 654

- 2

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

because my prof said [itex]x_1^2+x_2^2[/itex] is invariant and (i think he said this)[itex]x_1^3+x_2^3[/itex] 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

- #21

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

If it's a combination of tensors, with no free indices, then it's always a scalar. If it's a combination of tensors,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 [itex]x_1^2+x_2^2[/itex] is invariant and (i think he said this)[itex]x_1^3+x_2^3[/itex] is not invariant straight away.

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

Matrix version: Since [itex]x'=Rx[/itex], we have [itex]x'^Tx'=\dots[/itex]

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

Can you fill in the rest?

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).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

- #22

- 654

- 2

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 [Broken]

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...m}B_{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 learnt how to transform vectors like (x_{2} , -x_{1})

so that v_{1}' = L_{11}v_{1} + L_{12}v_{2}

so that v_{2}' = L_{21}v_{1} + L_{22}v_{2}

where my v_{1} takes the first vector x_{2}

and my v_{2} takes the 2nd vector -x_{1}

so with regards to the x_{1}^{2} + x_{2}^{2}

what is my v_{1} and v_{2} ??

am i suppose to take v_{1} = x_{1}^{2} + x_{2}^{2} ?

then what will my v_{2} be?

or am i suppose to take v_{1} = x_{1}

v_{2} = x_{2}

so that by the transformation R,

x_{1}^{2}' + x_{2}^{2}' = { L_{11}x_{1}+ L_{12}x_{2} } ^{2} +

L_{21}x_{1}+ L_{22}x_{2} } ^{2}

where c = cos, s = sin

= { cx_{1} + sx_{2} }^{2} + { -sx_{1} + cx_{2} }^{2} }

= x_{1}^{2} + x_{2}^{2}

so in the component form

is it suppose to read like this

[itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i= R_{ij}x_j R_{ik}x_k[/itex]

[itex]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][/itex]

middle of page 41

http://www.hereticalcosmology.com/articles/Tensor_Analysis-Chapter_1.pdf [Broken]

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

holds in all rotated coordinate frames where A, B, and C are of

Mth, Nth and (M+N-2)th order, then A

tensor.

with regards to the invariant question

i have only learnt how to transform vectors like (x

so that v

so that v

where my v

and my v

so with regards to the x

what is my v

am i suppose to take v

then what will my v

or am i suppose to take v

v

so that by the transformation R,

x

L

where c = cos, s = sin

= { cx

= x

so in the component form

is it suppose to read like this

[itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i= R_{ij}x_j R_{ik}x_k[/itex]

[itex]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][/itex]

Last edited by a moderator:

- #23

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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 forit 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 [Broken]

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.

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

i have only learnt how to transform vectors like (x_{2}, -x_{1})

so that v_{1}' = L_{11}v_{1}+ L_{12}v_{2}

so that v_{2}' = L_{21}v_{1}+ L_{22}v_{2}

where my v_{1}takes the first vector x_{2}

and my v_{2}takes the 2nd vector -x_{1}

No, you're supposed to "transform" the vector [itex](x_1,x_2)[/itex] the same way you transformed [itex](v_1,v_2)[/itex] above.so with regards to the x_{1}^{2}+ x_{2}^{2}

what is my v_{1}and v_{2}??

am i suppose to take v_{1}= x_{1}^{2}+ x_{2}^{2}?

Yes, that's one way of putting it.or am i suppose to take v_{1}= x_{1}

v_{2}= x_{2}

Correct.so that by the transformation R,

x_{1}^{2}' + x_{2}^{2}' = { L_{11}x_{1}+ L_{12}x_{2}}^{2}+

L_{21}x_{1}+ L_{22}x_{2}}^{2}

where c = cos, s = sin

= { cx_{1}+ sx_{2}}^{2}+ { -sx_{1}+ cx_{2}}^{2}}

= x_{1}^{2}+ x_{2}^{2}

Yes, but you can simplify it without explicitly performing the summation.so in the component form

is it suppose to read like this

[itex]x'_i=R_{ij}x_j[/itex], we have [itex]x'_i x'_i= R_{ij}x_j R_{ik}x_k[/itex]

[itex]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][/itex]

[tex]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[/tex]

It's actually much easier to do the whole thing with matrices instead of their components:

[tex]x'^Tx'=(Rx)^T(Rx)=x^TR^TRx=x^Tx[/tex]

Last edited by a moderator:

- #24

- 654

- 2

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)

- #25

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 412

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 [itex]A_{ij}[/itex] and their relationship with a scalar B is this:

What are you trying to do exactly? Are you trying to figure out if the numbers [itex]A_{ij}[/itex] 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 [itex]A_{ij}[/itex] in that coordinate system. So no matter what numbers the [itex]A_{ij}[/itex] 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 [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?

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.What are you trying to do exactly? Are you trying to figure out if the numbers [itex]A_{ij}[/itex] 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 [itex]A_{ij}[/itex] in that coordinate system. So no matter what numbers the [itex]A_{ij}[/itex] 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 [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?

Last edited:

- Last Post

- Replies
- 16

- Views
- 8K

- Last Post

- Replies
- 2

- Views
- 5K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 10

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 4

- Views
- 3K

- Replies
- 1

- Views
- 491

- Replies
- 2

- Views
- 4K