Levi civita to knocker delta

  • Thread starter quietrain
  • Start date
  • #1
654
2
hi, i don't understand why levi civita symbols is equal to kronecker delta when i was reading the wikipedia on it http://en.wikipedia.org/wiki/Levi-Civita_symbol

namely,

1)why does EijkElmn = det [ 9 terms of kronecker delta]
how do we know its those 9 terms? like dil dim din etc

2)and since they use det, does it mean EijkElmn is a cross product?

also from this link http://folk.uio.no/patricg/teaching/a112/levi-civita/
on the proof of the vector triple product,
img29.png


3)why from (dmj dnk-dmk dnj)an bj ck it becomes bm ak ck-cm aj bj

i understand that the kronecker delta is something like a dirac delta function, which acts like a "sifting function" but if there are 2 kronecker delta, then how does it work?
so example instead of bm ak ck, why doesn't it become bmancn - ...

4)
and subsequently getting bm ak ck-cm aj bj
= [b(a.c)]m - [c(a.b)]m

what does the subscript m stands for? and how did they get to the bac-cab expression? if the j and k subscripts means dot products, then what are the other 2 components of the dot product? it has so many alphabets><
does it has anyhting to do with dummy index and free index?

also, from
img30.png

is it right to think that the subscript i is the x-plane in cartesian form?
so if i want the z-plane, then would it mean to use the k component? so in this equations, is the k used here equal to the k in cartesian? if it is the same, then how do i get the k-component since they will clash?

5)also what does the subscript i means? component?
6eff7c76572f03dcbe578a6469638d34.png


if i just want to have the triple product alone without the component i, what will it be?

thanks!
 
Last edited:

Answers and Replies

  • #2
654
2
erm i sort of understand point 3 already but the rest ><
 
  • #3
654
2
anybody?
 
  • #4
strangerep
Science Advisor
3,240
1,135
1)why does EijkElmn = det [ 9 terms of kronecker delta]
how do we know its those 9 terms? like dil dim din etc

Brute force, although you can get a little heuristic insight
by thinking about the symmetries of a determinant.
E.g., swapping two rows flips the sign of the det, as does
exchanging the corresponding indices in the eps's.

2)and since they use det, does it mean EijkElmn is a cross product?
No. It's a 6-th rank tensor (since there are no contracted indices).

3)[...] instead of bm ak ck, why doesn't it become bmancn - ...

Look up "summation convention", i.e., implicit summation over repeated indices.
(The latter are called dummy indices.)
Both these expressions are the same.


is it right to think that the subscript i is the x-plane in cartesian form?
Not exactly -- it's any particular component in a cartesian frame.
 
  • #5
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
I will only answer the questions about the very basic things here, because there's no point worrying about the difficult stuff before you know the basics. The m and the i you're asking about refers to the mth component of the vector and the ith component of the vector respectively. For example, when m=2, (5,6,7)m=6.

Summation convention: We can drop all the summation sigmas without ambiguity, because there's always a sum over indices that appear exactly twice, and there's never a sum over any of the others.

The ith component of x is of course the number [itex]x_i[/itex] in [itex]x=x_i e_i[/itex], where [itex]e_i[/itex] is the ith basis vector.

The dot product and cross product are defined by

[tex]x\cdot y=x_i y_i[/tex]

[tex](x\times y)_i=\varepsilon_{ijk}x_jy_k[/tex]

The Levi-Civita symbol is defined by [itex]\varepsilon_{123}=1[/itex] and the requirement that swapping any two indices changes the sign. (For example, [itex]\varepsilon_{ijk}=\varepsilon_{ikj}[/itex], regardless of the values of j and k).

The Kronecker delta is defined by [itex]\delta_{ij}=1[/itex] when i=j, and [itex]\delta_{ij}=0[/itex] when i≠j.

I suggest you do the following (easy) exercises:

Exercise 1: Use these definitions to find [itex](x\times y)_2[/itex].
Exercise 2: Show that if [itex]S_{ij}=S_{ji}[/itex] and [itex]A_{ij}=-A_{ji}[/itex], then [itex]S_{ij}A_{ij}=0[/itex]. (This result is extremely useful).
Exercise 3: Show that [itex]\nabla\cdot(\nabla\times F)=0[/itex].
 
Last edited:
  • #6
654
2
Brute force, although you can get a little heuristic insight
by thinking about the symmetries of a determinant.
E.g., swapping two rows flips the sign of the det, as does
exchanging the corresponding indices in the eps's.


No. It's a 6-th rank tensor (since there are no contracted indices).



Look up "summation convention", i.e., implicit summation over repeated indices.
(The latter are called dummy indices.)
Both these expressions are the same.



Not exactly -- it's any particular component in a cartesian frame.

i see thank you!
 
  • #7
654
2
I will only answer the questions about the very basic things here, because there's no point worrying about the difficult stuff before you know the basics. The m and the i you're asking about refers to the mth component of the vector and the ith component of the vector respectively. For example, when m=2, (5,6,7)m=6.

Summation convention: We can drop all the summation sigmas without ambiguity, because there's always a sum over indices that appear exactly twice, and there's never a sum over any of the others.

The ith component of x is of course the number [itex]x_i[/itex] in [itex]x=x_i e_i[/itex], where [itex]e_i[/itex] is the ith basis vector.

The dot product and cross product are defined by

[tex]x\cdot y=x_i y_i[/tex]

[tex](x\times y)_i=\varepsilon_{ijk}x_jy_k[/tex]

The Levi-Civita symbol is defined by [itex]\varepsilon_{123}=1[/itex] and the requirement that swapping any two indices changes the sign. (For example, [itex]\varepsilon_{ijk}=\varepsilon_{ikj}[/itex], regardless of the values of j and k).

The Kronecker delta is defined by [itex]\delta_{ij}=1[/itex] when i=j, and [itex]\delta_{ij}=0[/itex] when i≠j.

I suggest you do the following (easy) exercises:

Exercise 1: Use these definitions to find [itex](x\times y)_2[/itex].
Exercise 2: Show that if [itex]S_{ij}=S_{ji}[/itex] and [itex]A_{ij}=-A_{ji}[/itex], then [itex]S_{ij}A_{ij}=0[/itex]. (This result is extremely useful).
Exercise 3: Show that [itex]\nabla\cdot(\nabla\times F)=0[/itex].

x X y = e123x2y3

so the 2nd component means 231-213?
so its x3y1 - x1y3?

for the 2nd exercise, i am not sure? are those matrix? or are they kronecker deltas? what does it mean by Sij = Sji? meaning S12 = S21? this is not levicivita symbols right?

for exercise 3

E123 ∇1∇2F3

so its 123-132 + 231-213 + 312 - 321 ?

so its ∇∇F - ∇F∇ + ∇F∇ - ∇∇F + F∇∇ - F∇∇

so we get 0 ?

btw on the last expression, is it right to swop ∇F∇ to ∇∇F? meaning are they equal? or do i have to introduce signs when not equal?

oh and with regards to one more thing

i notice that the triple vector product is given by the b(a.c) - c(a.b) rule

but i know a.c and a.b are the dot product. but what about the b()? is just just multiplication?

but since all are vectors, then is b() a cross or dot or ?

thanks!
 
  • #8
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
x X y = e123x2y3

so the 2nd component means 231-213?
so its x3y1 - x1y3?
That's the correct answer, but I can't tell if you used the definition of the Levi-Civita symbol or a definition of the cross product that doesn't use the Levi-Civita symbol. You should note that the sum [tex]\varepsilon_{2ij}x_iy_j[/tex] has nine terms, but only two that are non-zero. Do you see how the fact that the other seven are zero follows from my definition of the Levi-Civita symbol?

for the 2nd exercise, i am not sure? are those matrix? or are they kronecker deltas? what does it mean by Sij = Sji? meaning S12 = S21? this is not levicivita symbols right?
For each value of i and j, Sij and Aij are real numbers. I should perhaps have stated explicitly that the equalities I included in the problem statement hold for all values of i and j. So there are nine equalities involving the Sij and nine involving the Aij.

You can think of Sij and Aij as the components of two 3×3 matrices if you want to, but you can also choose not to.

This is just an exercise in using the summation convention, so it doesn't involve the Kronecker delta or the Levi-Civita symbol. (But the result allows you to simplify many expressions that do).

for exercise 3

E123 ∇1∇2F3
It's quite hard to read this notation. You should at least use "sub" tags for the indices. An alternative is to use LaTeX. If you click the quote button, you can see how I did the math in this post. (If you decide to try LaTeX you need to be aware that there's a bug that makes old images show up in previews. The workaround is to refresh and resend after each preview).

The solution I had in mind was to start with

[tex]\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k[/tex]

and then use (a simple generalization of) the result of exercise 2 to immediately conclude that the right-hand side is =0.

btw on the last expression, is it right to swop ∇F∇ to ∇∇F? meaning are they equal? or do i have to introduce signs when not equal?
What does the first expression mean?

i notice that the triple vector product is given by the b(a.c) - c(a.b) rule

but i know a.c and a.b are the dot product. but what about the b()? is just just multiplication?

but since all are vectors, then is b() a cross or dot or ?
It can't be a cross product or a dot product, because the thing in parentheses is a number, not a vector.
 
Last edited:
  • #9
654
2
That's the correct answer, but I can't tell if you used the definition of the Levi-Civita symbol or a definition of the cross product that doesn't use the Levi-Civita symbol. You should note that the sum [tex]\varepsilon_{2ij}x_iy_j[/tex] has nine terms, but only two that are non-zero. Do you see how the fact that the other seven are zero follows from my definition of the Levi-Civita symbol?


For each value of i and j, Sij and Aij are real numbers. I should perhaps have stated explicitly that the equalities I included in the problem statement hold for all values of i and j. So there are nine equalities involving the Sij and nine involving the Aij.

You can think of Sij and Aij as the components of two 3×3 matrices if you want to, but you can also choose not to.

This is just an exercise in using the summation convention, so it doesn't involve the Kronecker delta or the Levi-Civita symbol. (But the result allows you to simplify many expressions that do).


It's quite hard to read this notation. You should at least use "sub" tags for the indices. An alternative is to use LaTeX. If you click the quote button, you can see how I did the math in this post. (If you decide to try LaTeX you need to be aware that there's a bug that makes old images show up in previews. The workaround is to refresh and resend after each preview).

The solution I had in mind was to start with

[tex]\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k[/tex]

and then use (a simple generalization of) the result of exercise 2 to immediately conclude that the right-hand side is =0.


What does the first expression mean?


It can't be a cross product or a dot product, because the thing in parentheses is a number, not a vector.

for part 1 ,
the 7 other terms are 0 because of the repeated indices? but i don't know why a repeated indices means 0? definition?

for part 2, i made it a 2x2 matrix, and i realise wow it all cancels out if i let say,

S=
(1,1)
(1,1)

A=
(1,1)
(-1,-1)

am i right to let them have such values?

for part 3, how do you generalise the above to proof so?

for part 4, by cross product, it says like 132-123 is not interchangeable i assume?

but if the value example, 1i+2j+3k X 2i+3j+5k, then (2)(5) - 3(3) i + ...

if i swop 2and 5 it wouldn't matter right?

also, since ∇∇F are all vectors does it mean they are cross products or dot products? or just multiplication?

btw how do i quote here and quote there? breaking up the quotes like you did?

thanks!
 
  • #10
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
for part 1 ,
the 7 other terms are 0 because of the repeated indices? but i don't know why a repeated indices means 0? definition?
The 27 components of the Levi-Civita symbol are defined by the following two rules:

a) The 123 component is =1.
b) If you swap two indices, this changes the sign but not the absolute value.

Note that these two rules tell you the values of all 27 components. They tell you e.g. that [itex]\varepsilon_{312}=-\varepsilon_{132}=\varepsilon_{123}=1[/itex]. The second rule answers my question about why those seven terms are =0:

The fact that [itex]\varepsilon_{ijk}=-\varepsilon_{ikj}[/itex], regardless of the values of i,j,k, implies in particular that when j=k, we have [itex]\varepsilon_{ijk}=0[/itex]. So [itex]\varepsilon_{i11}=\varepsilon_{i22}=\varepsilon_{i33}=0[/itex] for all i.

for part 2, i made it a 2x2 matrix,
It has to be 3×3, since the indices go from 1 to 3.

and i realise wow it all cancels out if i let say,

S=
(1,1)
(1,1)

A=
(1,1)
(-1,-1)

am i right to let them have such values?
The idea was to prove the identity [itex]A_{ij}S_{ij}=0[/itex] for all values of the components that are consistent with the equalities I specified, so you can't choose specific values. The values you chose for A also contradict the condition [itex]A_{ij}=-A_{ji}[/itex].

I'll just do this one for you:

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}\quad\Rightarrow\quad A_{ij}S_{ij}=0[/tex]

Can you at least figure out why these equalities hold?

for part 3, how do you generalise the above to proof so?
The generalization I had in mind is that my proof works even if there are additional indices on A and/or S, for example

[tex]A_{ijkl}S_{ijkmn}=A_{jikl}S_{jikmn}=-A_{ijkl}S_{ijkmn}[/tex]

The conclusion is that if the entire expression is symmetric (doesn't change sign) under the exchange of two variables that are summed over, and antisymmetric (changes sign) under the exchange of the same two variables at the other place they appear, then the whole expression is zero. It doesn't even matter if the indices are distributed over one, two or more symbols.

This is an extremely useful result. A good way to remember it is "symmetric times antisymmetric equals zero". It solves exercise 3, because [itex]\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k[/itex] is symmetric under the exchange of i and j in the partial derivative, and antisymmetric under the exchange of i and j in the Levi-Civita symbol. So we immediately see that the whole expression is =0, without any more calculations.

for part 4, by cross product, it says like 132-123 is not interchangeable i assume?
I don't understand what you're asking here.

but if the value example, 1i+2j+3k X 2i+3j+5k, then (2)(5) - 3(3) i + ...

if i swop 2and 5 it wouldn't matter right?
You can swap them on the right, because real numbers commute. But 2i+3j+5k is obviously different from 5i+3j+2k.

also, since ∇∇F are all vectors does it mean they are cross products or dot products? or just multiplication?
I don't understand what you mean by ∇∇F. There are two different derivative operators used in exercise 3. They are often written as div and curl. The latter is defined by

[tex]\operatorname{curl} F=\varepsilon_{ijk}\frac{\partial F_k}{\partial x_j}[/tex]

Here F is a function from ℝ3 into ℝ3. The partial derivative operator is often abbreviated [itex]\partial_j[/itex]. If we define [itex]\nabla=(\partial_1,\partial_2,\partial_3)[/itex], we have

[tex]\operatorname{curl} F=\varepsilon_{ijk}\partial_j F_k=\nabla\times F[/tex]

I'll let you figure out "div" on your own.

btw how do i quote here and quote there? breaking up the quotes like you did?
I just copy and paste the quote tags that are created automatically.
 
Last edited:
  • #11
654
2
I'll just do this one for you:

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}\quad\Rightarrow\quad A_{ij}S_{ij}=0[/tex]

Can you at least figure out why these equalities hold?

oh so you mean Aij and Sij are levi civita symbols? so that when i change ij to ji i introduce minus sign. but since got 2, then it cancels to become the 2nd expression? for the 3rd expression we use the ij and ji defined by you . so 2AijSij = 0, so AijSij=0?


The conclusion is that if the entire expression is symmetric (doesn't change sign) under the exchange of two variables that are summed over, and antisymmetric (changes sign) under the exchange of the same two variables at the other place they appear, then the whole expression is zero. It doesn't even matter if the indices are distributed over one, two or more symbols.

This is an extremely useful result. A good way to remember it is "symmetric times antisymmetric equals zero". It solves exercise 3, because [itex]\nabla\cdot(\nabla\times F)=\partial_i\varepsilon_{ijk}\partial_j F_k[/itex] is symmetric under the exchange of i and j in the partial derivative, and antisymmetric under the exchange of i and j in the Levi-Civita symbol. So we immediately see that the whole expression is =0, without any more calculations.

but i and j are indices right? or are they variables? i thought variables are del, F etc?



I don't understand what you're asking here.


You can swap them on the right, because real numbers commute. But 2i+3j+5k is obviously different from 5i+3j+2k.


I don't understand what you mean by ∇∇F. There are two different derivative operators used in exercise 3. They are often written as div and curl. The latter is defined by

[tex]\operatorname{curl} F=\varepsilon_{ijk}\frac{\partial F_k}{\partial x_j}[/tex]

Here F is a function from ℝ3 into ℝ3. The partial derivative operator is often abbreviated [itex]\partial_j[/itex]. If we define [itex]\nabla=(\partial_1,\partial_2,\partial_3)[/itex], we have

[tex]\operatorname{curl} F=\varepsilon_{ijk}\partial_j F_k=\nabla\times F[/tex]

I'll let you figure out "div" on your own.

oh, so if del = d1+d2+d3

then del . (delXF)
= d1(d2F3-d3F2) + ... and so on? like cross product except the ijk unit vectors are replaced by the first del?

so del . (delXF) = div of the curl of F?

but is there a simple reason as to why the div of the curl is always 0? or the only way to proof is the levi civita way?

i saw an example from here
http://tutorial.math.lamar.edu/Classes/CalcIII/CurlDivergence.aspx

but it doesn't say why its 0.

i was thinking about total differtials, where if you differientiate wrt to x first, then y they will be the same for an expression if it is a total differiential. so does it mean that F is always a total differiential expression?
 
Last edited:
  • #12
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
oh so you mean Aij and Sij are levi civita symbols?
No, the Levi-Civita symbol is specifically [itex]\varepsilon_{ijk}[/itex], defined as described in my previous posts. A=[itex]\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}[/itex] is a set of 9 real numbers (not necessarily distinct). [itex]A_{ij}[/itex] is a member of that set, "the ij component". The nine equations [itex]A_{ij}=-A_{ji}[/itex] (one for each choice of i and j) describe how the nine members of the set are related to each other. You should be able to see that there are exactly three independent components of A, and exactly six independent components of S.

so that when i change ij to ji i introduce minus sign. but since got 2, then it cancels to become the 2nd expression? for the 3rd expression we use the ij and ji defined by you . so 2AijSij = 0, so AijSij=0?
I'm not sure I understand what you're saying, but it looks like you have correctly explained why the two equalities

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}[/tex]

together imply that [itex]A_{ij}S_{ij}=0[/itex], but failed to explain why the two equalities hold. You really shouldn't give up until you see it. I'll state the problem again:

Suppose that the equalities [itex]A_{ij}=-A_{ji}[/itex] and [itex]S_{ij}=S_{ji}[/itex] hold for all i and all j. Show that

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}[/tex]

and that

[tex]A_{ji}S_{ji}=-A_{ij}S_{ij}[/tex]


but i and j are indices right? or are they variables? i thought variables are del, F etc?
A variable is just a symbol that represents a member of a set. A constant is a variable that always represents the same member. So F and i are both variables, and the partial derivative operators can be thought of as constants.

oh, so if del = d1+d2+d3
It's not. You need to be more careful with the notation.

If [itex]f:\mathbb R^3\rightarrow\mathbb R[/itex], then

[tex]\operatorname{grad} f=(\partial_1 f,\partial_2f,\partial_3f)[/tex]

This is called the gradient of f. We're interested in the divergence of a vector field [itex]E:\mathbb R^3\rightarrow\mathbb R^3[/itex] defined by

[tex]\operatorname{div} E=\partial_i E_i[/tex].

The right-hand sides of these two definitions are often written as [itex]\nabla f[/itex] and [itex]\nabla\cdot E[/itex] respectively. To make sense of that, you need to think of [itex]\nabla[/itex] as a vector with 3 components: [itex]\nabla=(\partial_1,\partial_2,\partial_3)[/itex].

then del . (delXF)
= d1(d2F3-d3F2) + ... and so on? like cross product except the ijk unit vectors are replaced by the first del?

so del . (delXF) = div of the curl of F?
Yes.

but is there a simple reason as to why the div of the curl is always 0? or the only way to proof is the levi civita way?
The way I proved it is by far the easiest. That's one of the reasons you should make sure to understand exercise 2. But it's certainly not the only way. Just use any definitions of div and curl (that are equivalent to these), and you will see that every term cancels another term.

i was thinking about total differtials, where if you differientiate wrt to x first, then y they will be the same for an expression if it is a total differiential. so does it mean that F is always a total differiential expression?
No, F is an arbitrary vector field.
 
Last edited:
  • #13
654
2
No, the Levi-Civita symbol is specifically [itex]\varepsilon_{ijk}[/itex], defined as described in my previous posts. A=[itex]\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}[/itex] is a set of 9 real numbers (not necessarily distinct). [itex]A_{ij}[/itex] is a member of that set, "the ij component". The nine equations [itex]A_{ij}=-A_{ji}[/itex] (one for each choice of i and j) describe how the nine members of the set are related to each other. You should be able to see that there are exactly three independent components of A, and exactly six independent components of S.

ok, so i get A11,A22,A33,A12,A21 and so on and get 9 terms?
but what do you mean by 3 independent components of A? are you talking about the 11 22 and 33? so that they all have to be 0 to fulfil the equation Aij=-Aji?

so for S, i assume its the 12 ,21 13,31 etc that are the 6 terms of independent? again why for S are they independent? i mean more specifically independent from what?

I'm not sure I understand what you're saying, but it looks like you have correctly explained why the two equalities

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}=-A_{ij}S_{ij}[/tex]

together imply that [itex]A_{ij}S_{ij}=0[/itex], but failed to explain why the two equalities hold. You really shouldn't give up until you see it. I'll state the problem again:

Suppose that the equalities [itex]A_{ij}=-A_{ji}[/itex] and [itex]S_{ij}=S_{ji}[/itex] hold for all i and all j. Show that

[tex]A_{ij}S_{ij}=A_{ji}S_{ji}[/tex]

and that

[tex]A_{ji}S_{ji}=-A_{ij}S_{ij}[/tex]

oh , so for the first equation, its just that the set of 9 numbers determined by Aij is equal to the set of 9 numbers determined by Aji just that the ordering is different only? so they are all equal

so the 2nd equation follows from the condition you stated ?
 
  • #14
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
ok, so i get A11,A22,A33,A12,A21 and so on and get 9 terms?
but what do you mean by 3 independent components of A? are you talking about the 11 22 and 33? so that they all have to be 0 to fulfil the equation Aij=-Aji?

so for S, i assume its the 12 ,21 13,31 etc that are the 6 terms of independent? again why for S are they independent? i mean more specifically independent from what?
To specify A is to specify the values of its nine components. To specify an A that's consistent with the condition Aij=-Aji, we have to "choose" nine real numbers, but only the first three choices can be made independently (of each other and the other six "choices"). So we're really only choosing three. The other six are fixed by the condition Aij=-Aji.

I understand that it's a bit confusing to just say that "A has three independent components", but this terminology is pretty standard, at least in physics books.

Yes, one of the things I wanted you to realize was that the "diagonal" components A11, A22, A33 are all zero.

oh , so for the first equation, its just that the set of 9 numbers determined by Aij is equal to the set of 9 numbers determined by Aji just that the ordering is different only? so they are all equal

so the 2nd equation follows from the condition you stated ?
You explained the second equality correctly (except for leaving out the "s" at the end of "conditions"...there's one for A and one for S), and you're correct that exactly the same terms appear on both sides of the first equality. But I'm not sure how you came to that conclusion, because you still don't seem to have understood what this equality is about. The terms do not appear in different order on the two sides of the = sign. Both sides have exactly the same terms, in exactly the same order.

Now do you see that the same thing works no matter how many indices there are and no matter how many symbols they're attached to? For example, if T is symmetric (unchanged) under the exchange of its 2nd and 4th index, and antisymmetric (changes sign) under the exchange of its 1st and 6th index, then we can immediately conclude that

[tex]T_{ijkjmiknopq}=0[/tex]

for all values of m,n,o,p,q. When you understand this, you should have another look at the expression

[tex]\partial_i\varepsilon_{ijk}\partial_j F_k[/tex]
 
Last edited:
  • #15
654
2
To specify A is to specify the values of its nine components. To specify an A that's consistent with the condition Aij=-Aji, we have to "choose" nine real numbers, but only the first three choices can be made independently (of each other and the other six "choices"). So we're really only choosing three. The other six are fixed by the condition Aij=-Aji.

I understand that it's a bit confusing to just say that "A has three independent components", but this terminology is pretty standard, at least in physics books.

Yes, one of the things I wanted you to realize was that the "diagonal" components A11, A22, A33 are all zero.


You explained the second equality correctly (except for leaving out the "s" at the end of "conditions"...there's one for A and one for S), and you're correct that exactly the same terms appear on both sides of the first equality. But I'm not sure how you came to that conclusion, because you still don't seem to have understood what this equality is about. The terms do not appear in different order on the two sides of the = sign. Both sides have exactly the same terms, in exactly the same order.

Now do you see that the same thing works no matter how many indices there are and no matter how many symbols they're attached to? For example, if T is symmetric (unchanged) under the exchange of its 2nd and 4th index, and antisymmetric (changes sign) under the exchange of its 1st and 6th index, then we can immediately conclude that

[tex]T_{ijkjmiknopq}=0[/tex]

for all values of m,n,o,p,q. When you understand this, you should have another look at the expression

[tex]\partial_i\varepsilon_{ijk}\partial_j F_k[/tex]

the matrix i had in mind was
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)

so the 3 independent terms are 1,2 and 3? the 0s and the minus are all dude to the condition?

for S, it can be everything except the diagonal have to be 0, so 6 terms of independent?

on the part of proving the equations

do you mean that Aij = Aji = the matrix of A above?
so Aij and Aji are not really matrix on their own , but are conditions so to speak to relate to the matrix A?

oh so for T, if i index are symmetric, and j are anti, then its like

Tij.... = -Tji....

so i assume the rest of the index can be represented by Smnop....?

so its = 0 ? by the AS example you gave?

so the last expression of del , E and F will just mean all the same likewise as above = 0 ?
 
  • #16
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
the matrix i had in mind was
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)

so the 3 independent terms are 1,2 and 3? the 0s and the minus are all dude to the condition?

for S, it can be everything except the diagonal have to be 0, so 6 terms of independent?
If you interpret Aij as row i, column j of a matrix (which is 100% OK, but not necessary for what we've been doing), then the condition Aij=-Aji does indeed restrict it to the form

[tex]A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}[/tex]

But you're wrong about S. It would look like this:

[tex]A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}[/tex]

do you mean that Aij = Aji = the matrix of A above?
This is like asking if 100=-100=(100,-100,0). You seem to be extremely careless with the notation, and to have failed to understand that the expression "Aij" always refers to a number. Please read this again:
A=[itex]\{A_{ij}\in\mathbb R|i,j\in\{1,2,3\}\}[/itex] is a set of 9 real numbers (not necessarily distinct). [itex]A_{ij}[/itex] is a member of that set, "the ij component". The nine equations [itex]A_{ij}=-A_{ji}[/itex] (one for each choice of i and j) describe how the nine members of the set are related to each other.

so Aij and Aji are not really matrix on their own , but are conditions so to speak to relate to the matrix A?
How could a real number be a "condition"? If we define a matrix A by saying that for each i and each j, Aij is the number on row i, column j of A, then A23 is the number on row 2, column 3, and A32 is the number on row 3 column 2. If A23=5, then A32=-5, because Aij=-Aji holds for all values of i and j, including i=2 and j=3.

I don't think I want to discuss [itex]T_{ijkjmiknopq}[/itex] or [itex]\nabla\cdot(\nabla\times F)[/itex] until I'm 100% sure that you understand why [itex]A_{ij}S_{ij}=A_{ji}S_{ji}[/itex].
 
Last edited:
  • #17
654
2
If you interpret Aij as row i, column j of a matrix (which is 100% OK, but not necessary for what we've been doing), then the condition Aij=-Aji does indeed restrict it to the form

[tex]A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}[/tex]

But you're wrong about S. It would look like this:

[tex]A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}[/tex]


This is like asking if 100=-100=(100,-100,0). You seem to be extremely careless with the notation, and to have failed to understand that the expression "Aij" always refers to a number. Please read this again:


How could a real number be a "condition"? If we define a matrix A by saying that for each i and each j, Aij is the number on row i, column j of A, then A23 is the number on row 2, column 3, and A32 is the number on row 3 column 2. If A23=5, then A32=-5, because Aij=-Aji holds for all values of i and j, including i=2 and j=3.

I don't think I want to discuss [itex]T_{ijkjmiknopq}[/itex] or [itex]\nabla\cdot(\nabla\times F)[/itex] until I'm 100% sure that you understand why [itex]A_{ij}S_{ij}=A_{ji}S_{ji}[/itex].

i don't understand,

if i were to let
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
S=
(1 4 5)
(4 2 6)
(5 6 3)

then if i do a matrix multiplication of AxS, i get 9 numbers which has opposite signs to if i were to do -AxS.

but if i do the dot product
A11B11+A12B12+...
then i get 0 for both.

so why do i get 2 type of results ? the matrix multiplication is a cross product right?
so your notation of AijSij is a dot product?

but if i can visualize them as matrix, then how come i get different results?

in any case, if Aij = 5, Aji = -5,
how can 5S = -5S? unless you sum them all up using the dot product way and get 0 ?
 
  • #18
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
Edit: I wrote this post before I saw your post above. I will read it now.

OK, I'll just tell you the answer. The equality [itex]A_{ij}S_{ij}=A_{ji}S_{ji}[/itex] holds because both expressions are by definition exactly the same thing. There are no calculations involved, no properties of A or S, no properties of real numbers. These expressions are equal for the same reason that 5+3 is equal to 5+3.

Consider e.g. the equalities

[tex]\sum_{i=1}^2 x_i = x_1+x_2= \sum_{j=1}^2 x_i[/tex]

Earlier in the thread, you indicated that you understand that it never matters what symbol you're using as the summation index. But for some reason you never applied that principle to the expression [itex]A_{ij}S_{ij}[/itex].

[tex]\sum_{i=1}^3\sum_{j=1}^3A_{ij}S_{ij}=\sum_{k=1}^3\sum_{l=1}^3A_{kl}S_{kl}=\sum_{j=1}^3\sum_{i=1}^3A_{ji}S_{ji}[/tex]
 
  • #19
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
i don't understand,

if i were to let
A=
(0 1 2)
(-1 0 3)
(-2 -3 0)
S=
(1 4 5)
(4 2 6)
(5 6 3)

then if i do a matrix multiplication of AxS, i get 9 numbers which has opposite signs to if i were to do -AxS.
Aren't you saying something entirely trivial here, i.e. that with these specific choices of A and S, we have -(-AS)=AS? This identity is obviously true for all matrices A and S. (By the way, you shouldn't write the product as A×S. A product of matrices is always written without a product symbol).

but if i do the dot product
A11B11+A12B12+...
then i get 0 for both.
The dot product of what? The dot product isn't defined for matrices. The sum you typed is equal to the dot product of the first row of A and the first row of S. Why are you calculating that? If you're doing it as a part of a calculation of AS, then you're doing it wrong. AS is defined by [itex](AS)_{ij}=A_{ik}S_{kj}[/itex], so the entry on row i, column j of AS is the dot product of the ith row of A and the jth column of S.

the matrix multiplication is a cross product right?
No, it has nothing to do with the cross product.

so your notation of AijSij is a dot product?
No. (What two vectors would it be a dot product of?)

but if i can visualize them as matrix, then how come i get different results?
I can't answer that, because I don't understand what you're doing, or why you're doing it.

in any case, if Aij = 5, Aji = -5,
how can 5S = -5S?
:confused: I have no idea what you're talking about here. I haven't said anything that suggests that 5S=-5S, and it certainly isn't implied by Aij=-Aji.



the condition Aij=-Aji does indeed restrict it to the form

[tex]A=\begin{pmatrix}0 & a & b\\ -a & 0 & c\\ -b & -c & 0\end{pmatrix}[/tex]

But you're wrong about S. It would look like this:

[tex]A=\begin{pmatrix}d & a & b\\ a & e & c\\ b & c & f\end{pmatrix}[/tex]
The second equality here is a typo, but you probably understood that. I meant to type "S=", not "A=". Note that Aij=-Aji implies that the diagonal entries of A are =0, but Sij=Sji says nothing at all about the diagonal entries of S.
 
  • #20
654
2
ok this is what i have in mind

AijSij =
(0 1 2)(1 4 5)
(-10 3)(4 2 6)
(-2-30)(5 6 3)

which gives me
(14 14 12)
(14 14 4)
(-14-14-28)

now AjiSji =
(0-1-2)(1 4 5)
(1 0 -3)(4 2 6)
(2 3 0)(5 6 3)

which gives me
(-14-14-12)
(-14-14-4)
(14 14 28)

so i see that the AijSij = - AjiSji
but is not equal to AjiSji ?

this is because of the condition Aij = -Aji ?

but you are telling me they are all dummy variables? but from the matrix it shows that AijSij is different from AjiSji

><
 
  • #21
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
We can choose to interpret the Aij and Sij for i,j=1,2,3 as the components of two matrices A and S, but even if we do, there's no way AijSij can be a matrix. It's clearly just a sum of real numbers.

You're still not thinking about what definitions you need to use. If you want to treat this expression as something involving a product of matrices, this is what you need:

Definition of matrix multiplication: (XY)ij=XikYkj
Definition of transpose: (XT)ij=Xji
Definition of trace: Tr X=Xii

These definitions tell us that

[tex]A_{ij}S_{ij}=(A^T)_{ji}S_{ij}=(A^TS)_{jj}=\operatorname{Tr}(A^TS)[/tex]

Alternatively,

[tex]A_{ij}S_{ij}=A_{ij}(S^T)_{ji}=(AS^T)_{ii}=\operatorname{Tr}(AS^T)[/tex]

You can also reverse the order of the matrices in the trace, because Tr(XY)=Tr(YX) for all matrices X and Y. (This is very easy to prove using the definitions above).

You computed AS and ATST. The conditions I put on the components of A and S translate to AT=-A and ST=S, so ATST=(-A)S=-AS.
 
Last edited:
  • #22
654
2
so, in AiSj , theres no summation cos the indices do not repeat? so is there any meaning to this notation? or does it mean A1S1 + A2S1 +A3S1 +A1S2+ .. .

but in AiSi it is a dot product where A1S1 + A2S2 + A3S3 ?

so the way to compute AijSij = A11S11 + A11S12 + A11S13 + A21S11 +A21S12+ AA21S13 +. .. . .

so i get 81 terms? 9terms of A x 9 terms of S? or does it follow the kronecker delta where the non-identical indices = 0 and so we are left with the 11 22 and 33 ? so i assume AijSij is not equal to the matrix AS i computed since you said its just a number summed?

on 2nd thought,

if AiSi = A1S1 +A2S2 + ...

then does it mean AijSij = A11S11 +A12S12 + A13S13+A21S21+...

that means i cannot have a A11S12 where the indices of A, 11, is not equal to S, 12 ? so i get 9 terms instead of 81?

in any case, if i so if AijSij = AjiSji
(this is just dummy variables right? its not meaning that i swop the Aij=A12 to a Aji=A21 right? )

then it means AijSij = -AijSij from the condtion you set,

so 2AijSij=0
so AijSij = 0 ?

thus
[PLAIN]https://www.physicsforums.com/latex_images/31/3102234-8.png [Broken]
so, the deltas are the symmetric S and the levi civita are the antisymmetric A, and F is the other components that you generalized? am i right to say that there is summation here due to the levi civita symbol? or else like at the top of this post, AiSj would have no meaning without a summation symbol "[tex]\sum[/tex]" placed to the left of the expression ?
 
Last edited by a moderator:
  • #23
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
so, in AiSj , theres no summation cos the indices do not repeat? so is there any meaning to this notation? or does it mean A1S1 + A2S1 +A3S1 +A1S2+ .. .
If i=1 and j=1, then it means A1S1.
If i=2 and j=1, then it means A2S1
etc.

I have to say, it's pretty frustrating that you are still stuck on the meaning of (Something)something this late in the thread. I have spent many hours trying to explain this to you. One last time: We're dealing with finite sets whose members are identified by the values of one or more indices that take integer values from 1 to 3. The members of the set are usually real numbers, but they can also be e.g. basis vectors or partial derivative operators.

but in AiSi it is a dot product where A1S1 + A2S2 + A3S3 ?
Yes. A person who understands that (Something)something refers to a member of an indexed set of real numbers, and who understands the summation convention, can take the following as the definition of the dot product:

[tex](A_1,A_2,A_3)\cdot (S_1,S_2,S_3)=A_i S_i[/tex]

so the way to compute AijSij = A11S11 + A11S12 + A11S13 + A21S11 +A21S12+ AA21S13 +. .. . .

so i get 81 terms? 9terms of A x 9 terms of S? or does it follow the kronecker delta where the non-identical indices = 0 and so we are left with the 11 22 and 33 ? so i assume AijSij is not equal to the matrix AS i computed since you said its just a number summed?

on 2nd thought,

if AiSi = A1S1 +A2S2 + ...

then does it mean AijSij = A11S11 +A12S12 + A13S13+A21S21+...

that means i cannot have a A11S12 where the indices of A, 11, is not equal to S, 12 ? so i get 9 terms instead of 81?
It's also frustrating to see this, because it means that you still haven't understood that the summation convention is nothing more than the practice to not bother to write the summation sigmas when an index occurs twice. Or is the problem that you don't understand the sigma notation for sums? Do you not see that you just asked me if

[tex]\sum_{i=1}^3\sum_{j=1}^3 X_{ij}[/tex]

has 81 terms?

in any case, if i so if AijSij = AjiSji
(this is just dummy variables right? its not meaning that i swop the Aij=A12 to a Aji=A21 right? )
Right. Dummy variables. No swapping.

then it means AijSij = -AijSij from the condtion you set,

so 2AijSij=0
so AijSij = 0 ?
Correct.

thus
[PLAIN]https://www.physicsforums.com/latex_images/31/3102234-8.png [Broken]
so, the deltas are the symmetric S and the levi civita are the antisymmetric A, and F is the other components that you generalized?
Yes.

[tex]\partial_i\varepsilon_{ijk}\partial_j F_k=\partial_j\varepsilon_{jik}\partial_i F_k=-\partial_i\varepsilon_{ijk}\partial_j F_k\quad\Rightarrow\quad \partial_i\varepsilon_{ijk}\partial_j F_k=0[/tex]

am i right to say that there is summation here due to the levi civita symbol? or else like at the top of this post, AiSj would have no meaning without a summation symbol "[tex]\sum[/tex]" placed to the left of the expression ?
There's a summation over i because i occurs twice (and for no other reason).
There's a summation over j because j occurs twice (and for no other reason).
There's a summation over k because k occurs twice (and for no other reason).

Without the summation convention, the right-hand side is written as

[tex]\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\partial_i\varepsilon_{ijk}\partial_j F_k[/tex]
 
Last edited by a moderator:
  • #24
654
2
ok i better start from scratch

if (must i put in a summation here?since they are not repeated indices?)[tex]\sum[/tex]Aij , i,j from 1 to 3
means A11+A12+A13+21+22+23.... 9 terms

if i impose a condition of Aij=-Aji (why not just put Aij=-Aij since they are just dummy variable?)

then the value of Aij = 0 summed since they all cancel each other out right?

so now if i multiply in a Sij term which is governed by the condition = Sji
then (i drop summation symbol due to convention due to repeated indices)AijSij = A11S11+A12S12+A13S13+...
so i get 9 terms now and they all add up to a number. and so this number = 0 because of the equations AijSij=AjiSji=-AijSij=0

and the rest follows...

well i thank you for your patience :))
 
  • #25
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
413
if (must i put in a summation here?since they are not repeated indices?)[tex]\sum[/tex]Aij , i,j from 1 to 3
means A11+A12+A13+21+22+23.... 9 terms
Correct.

if i impose a condition of Aij=-Aji (why not just put Aij=-Aij since they are just dummy variable?)

then the value of Aij = 0 summed since they all cancel each other out right?
Yes, the sum of all the components of A add up to zero. As for your other question, the alternative condition that you're suggesting clearly just says that all the components are zero. Your question shows that you have completely misunderstood the concept of dummy variables.

so now if i multiply in a Sij term which is governed by the condition = Sji
then (i drop summation symbol due to convention due to repeated indices)AijSij = A11S11+A12S12+A13S13+...
so i get 9 terms now and they all add up to a number. and so this number = 0 because of the equations AijSij=AjiSji=-AijSij=0
Correct.
 

Related Threads on Levi civita to knocker delta

  • Last Post
Replies
9
Views
512
  • Last Post
Replies
2
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
34K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
1
Views
2K
Top