Vector Product and dot product identity: Levi-Civita symbols
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I want to prove that $vecacdot(vecbtimesvecc) = (vecc times veca)cdotvecb$ using the Levi-Civita symbols, however, I am not $100$% sure if my proof is correct.
Please see attached my proof, see the image
Or see the (using MathJax) equations below
$$vecacdot(vecbtimes vecc) = a_i(vecbtimesvecc)_i = a_iepsilon_ijkb_jc_ke_i = -epsilon_jika_ib_jc_ke_i = -(vecatimesvecc)cdotvecb = (vecctimesveca)cdotvecb$$
My main concern is that when I change the indices for epsilon from $(i,j,k)$ to $(j,i,k)$, should I also change the index for $e$ vector from $i$ to $j$ as well? It's just in my proof I assume that $b_je_i$ gives vector $b$ and I do not know if I can state that given the different indices.
Thank you in advance and I hope this all does not sound too confusing.
proof-verification vectors tensors
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up vote
0
down vote
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I want to prove that $vecacdot(vecbtimesvecc) = (vecc times veca)cdotvecb$ using the Levi-Civita symbols, however, I am not $100$% sure if my proof is correct.
Please see attached my proof, see the image
Or see the (using MathJax) equations below
$$vecacdot(vecbtimes vecc) = a_i(vecbtimesvecc)_i = a_iepsilon_ijkb_jc_ke_i = -epsilon_jika_ib_jc_ke_i = -(vecatimesvecc)cdotvecb = (vecctimesveca)cdotvecb$$
My main concern is that when I change the indices for epsilon from $(i,j,k)$ to $(j,i,k)$, should I also change the index for $e$ vector from $i$ to $j$ as well? It's just in my proof I assume that $b_je_i$ gives vector $b$ and I do not know if I can state that given the different indices.
Thank you in advance and I hope this all does not sound too confusing.
proof-verification vectors tensors
Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove that $vecacdot(vecbtimesvecc) = (vecc times veca)cdotvecb$ using the Levi-Civita symbols, however, I am not $100$% sure if my proof is correct.
Please see attached my proof, see the image
Or see the (using MathJax) equations below
$$vecacdot(vecbtimes vecc) = a_i(vecbtimesvecc)_i = a_iepsilon_ijkb_jc_ke_i = -epsilon_jika_ib_jc_ke_i = -(vecatimesvecc)cdotvecb = (vecctimesveca)cdotvecb$$
My main concern is that when I change the indices for epsilon from $(i,j,k)$ to $(j,i,k)$, should I also change the index for $e$ vector from $i$ to $j$ as well? It's just in my proof I assume that $b_je_i$ gives vector $b$ and I do not know if I can state that given the different indices.
Thank you in advance and I hope this all does not sound too confusing.
proof-verification vectors tensors
I want to prove that $vecacdot(vecbtimesvecc) = (vecc times veca)cdotvecb$ using the Levi-Civita symbols, however, I am not $100$% sure if my proof is correct.
Please see attached my proof, see the image
Or see the (using MathJax) equations below
$$vecacdot(vecbtimes vecc) = a_i(vecbtimesvecc)_i = a_iepsilon_ijkb_jc_ke_i = -epsilon_jika_ib_jc_ke_i = -(vecatimesvecc)cdotvecb = (vecctimesveca)cdotvecb$$
My main concern is that when I change the indices for epsilon from $(i,j,k)$ to $(j,i,k)$, should I also change the index for $e$ vector from $i$ to $j$ as well? It's just in my proof I assume that $b_je_i$ gives vector $b$ and I do not know if I can state that given the different indices.
Thank you in advance and I hope this all does not sound too confusing.
proof-verification vectors tensors
proof-verification vectors tensors
edited Feb 20 '17 at 2:19
Rafael Wagner
1,8122923
1,8122923
asked Oct 20 '16 at 17:05
Blondie
907
907
Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31
add a comment |Â
Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31
Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31
add a comment |Â
2 Answers
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For the Levi-Civita symbols we have that, for two vectors $veca$ and $vecb$,
$vecacdotvecb = sum_ia_ib_i$
and using Einstein notation convention we can just wright $vecacdotvecb = a_ib^i$, or, we can just use that $(vecacdotvecb )_i = a_ib_i$. We also have that
$(vecatimesvecb)_i = sum_j sum_k epsilon_ijka_jb_k$
Or just $(vecatimesvecb)_i = epsilon_ijka_jb_k$. So using that you can prove the relation as you did:
$$(vecacdot(vecbtimesvecc)) = sum_ia_i(vecbtimesvecc)_i = sum_isum_jsum_ka_iepsilon_ijkb_jc_k = sum_isum_jsum_kepsilon_jkic_ka_ib_j = sum_j(vecctimesveca)_jb_j = (vecbcdot(vecctimesveca))$$
Then this is what I think was your doubt.
add a comment |Â
up vote
0
down vote
That's almost right, but there are some inconsistencies in your notation.
- In the first step $acdot(btimes c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.
- In the second step $a_icdot(btimes c)_i$ you use a dot ($cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_i(btimes c)_i$
- Since $(btimes c)_i = epsilon_ijkb_jc_k$, in step three you should have just $a_iepsilon_ijkb_jc_k$, without the vectors $e_i$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.
- The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.
Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$acdot(btimes c) = a_i(btimes c)_i = a_iepsilon_ijkb_jc_k = -a_iepsilon_jikb_jc_k \= -epsilon_jika_ic_kb_j = -(atimes c)_jb_j = -(atimes c)cdot b = (ctimes a)cdot b$$
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the Levi-Civita symbols we have that, for two vectors $veca$ and $vecb$,
$vecacdotvecb = sum_ia_ib_i$
and using Einstein notation convention we can just wright $vecacdotvecb = a_ib^i$, or, we can just use that $(vecacdotvecb )_i = a_ib_i$. We also have that
$(vecatimesvecb)_i = sum_j sum_k epsilon_ijka_jb_k$
Or just $(vecatimesvecb)_i = epsilon_ijka_jb_k$. So using that you can prove the relation as you did:
$$(vecacdot(vecbtimesvecc)) = sum_ia_i(vecbtimesvecc)_i = sum_isum_jsum_ka_iepsilon_ijkb_jc_k = sum_isum_jsum_kepsilon_jkic_ka_ib_j = sum_j(vecctimesveca)_jb_j = (vecbcdot(vecctimesveca))$$
Then this is what I think was your doubt.
add a comment |Â
up vote
0
down vote
For the Levi-Civita symbols we have that, for two vectors $veca$ and $vecb$,
$vecacdotvecb = sum_ia_ib_i$
and using Einstein notation convention we can just wright $vecacdotvecb = a_ib^i$, or, we can just use that $(vecacdotvecb )_i = a_ib_i$. We also have that
$(vecatimesvecb)_i = sum_j sum_k epsilon_ijka_jb_k$
Or just $(vecatimesvecb)_i = epsilon_ijka_jb_k$. So using that you can prove the relation as you did:
$$(vecacdot(vecbtimesvecc)) = sum_ia_i(vecbtimesvecc)_i = sum_isum_jsum_ka_iepsilon_ijkb_jc_k = sum_isum_jsum_kepsilon_jkic_ka_ib_j = sum_j(vecctimesveca)_jb_j = (vecbcdot(vecctimesveca))$$
Then this is what I think was your doubt.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For the Levi-Civita symbols we have that, for two vectors $veca$ and $vecb$,
$vecacdotvecb = sum_ia_ib_i$
and using Einstein notation convention we can just wright $vecacdotvecb = a_ib^i$, or, we can just use that $(vecacdotvecb )_i = a_ib_i$. We also have that
$(vecatimesvecb)_i = sum_j sum_k epsilon_ijka_jb_k$
Or just $(vecatimesvecb)_i = epsilon_ijka_jb_k$. So using that you can prove the relation as you did:
$$(vecacdot(vecbtimesvecc)) = sum_ia_i(vecbtimesvecc)_i = sum_isum_jsum_ka_iepsilon_ijkb_jc_k = sum_isum_jsum_kepsilon_jkic_ka_ib_j = sum_j(vecctimesveca)_jb_j = (vecbcdot(vecctimesveca))$$
Then this is what I think was your doubt.
For the Levi-Civita symbols we have that, for two vectors $veca$ and $vecb$,
$vecacdotvecb = sum_ia_ib_i$
and using Einstein notation convention we can just wright $vecacdotvecb = a_ib^i$, or, we can just use that $(vecacdotvecb )_i = a_ib_i$. We also have that
$(vecatimesvecb)_i = sum_j sum_k epsilon_ijka_jb_k$
Or just $(vecatimesvecb)_i = epsilon_ijka_jb_k$. So using that you can prove the relation as you did:
$$(vecacdot(vecbtimesvecc)) = sum_ia_i(vecbtimesvecc)_i = sum_isum_jsum_ka_iepsilon_ijkb_jc_k = sum_isum_jsum_kepsilon_jkic_ka_ib_j = sum_j(vecctimesveca)_jb_j = (vecbcdot(vecctimesveca))$$
Then this is what I think was your doubt.
answered Feb 20 '17 at 2:29
Rafael Wagner
1,8122923
1,8122923
add a comment |Â
add a comment |Â
up vote
0
down vote
That's almost right, but there are some inconsistencies in your notation.
- In the first step $acdot(btimes c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.
- In the second step $a_icdot(btimes c)_i$ you use a dot ($cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_i(btimes c)_i$
- Since $(btimes c)_i = epsilon_ijkb_jc_k$, in step three you should have just $a_iepsilon_ijkb_jc_k$, without the vectors $e_i$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.
- The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.
Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$acdot(btimes c) = a_i(btimes c)_i = a_iepsilon_ijkb_jc_k = -a_iepsilon_jikb_jc_k \= -epsilon_jika_ic_kb_j = -(atimes c)_jb_j = -(atimes c)cdot b = (ctimes a)cdot b$$
add a comment |Â
up vote
0
down vote
That's almost right, but there are some inconsistencies in your notation.
- In the first step $acdot(btimes c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.
- In the second step $a_icdot(btimes c)_i$ you use a dot ($cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_i(btimes c)_i$
- Since $(btimes c)_i = epsilon_ijkb_jc_k$, in step three you should have just $a_iepsilon_ijkb_jc_k$, without the vectors $e_i$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.
- The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.
Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$acdot(btimes c) = a_i(btimes c)_i = a_iepsilon_ijkb_jc_k = -a_iepsilon_jikb_jc_k \= -epsilon_jika_ic_kb_j = -(atimes c)_jb_j = -(atimes c)cdot b = (ctimes a)cdot b$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
That's almost right, but there are some inconsistencies in your notation.
- In the first step $acdot(btimes c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.
- In the second step $a_icdot(btimes c)_i$ you use a dot ($cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_i(btimes c)_i$
- Since $(btimes c)_i = epsilon_ijkb_jc_k$, in step three you should have just $a_iepsilon_ijkb_jc_k$, without the vectors $e_i$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.
- The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.
Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$acdot(btimes c) = a_i(btimes c)_i = a_iepsilon_ijkb_jc_k = -a_iepsilon_jikb_jc_k \= -epsilon_jika_ic_kb_j = -(atimes c)_jb_j = -(atimes c)cdot b = (ctimes a)cdot b$$
That's almost right, but there are some inconsistencies in your notation.
- In the first step $acdot(btimes c)$, you have a scalar. Nothing wrong here. But note that since you begin with a scalar, you should have scalars in all the next steps.
- In the second step $a_icdot(btimes c)_i$ you use a dot ($cdot$) between components. That is illegal. Components are numbers, and you can only use dot product between vectors. Hence, the second step should read just $a_i(btimes c)_i$
- Since $(btimes c)_i = epsilon_ijkb_jc_k$, in step three you should have just $a_iepsilon_ijkb_jc_k$, without the vectors $e_i$. This resonates with the note in (1), where I remarked you should have just scalars and not vector expressions. Also note that this goes against the summation convention, where it is only valid to sum over pairs of indices.
- The switch of indices and switch of sign is correct. Since, as mentioned in (3), you shouldn't write the vectors $e_i$, your concern about the index $i$ is just out of the question.
Steps 5 and 6 are indeed correct.
So, the correct derivation (with a pair of extra steps) is
$$acdot(btimes c) = a_i(btimes c)_i = a_iepsilon_ijkb_jc_k = -a_iepsilon_jikb_jc_k \= -epsilon_jika_ic_kb_j = -(atimes c)_jb_j = -(atimes c)cdot b = (ctimes a)cdot b$$
answered Jun 2 at 10:17
Jackozee Hakkiuz
710215
710215
add a comment |Â
add a comment |Â
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Well one issue I see is too many of the index i, there are three which makes the product ambiguous as which pair are summed over (since summing happens in pairs). You can fix this by omitting the unit vector as this is how the dot product works
â Triatticus
Oct 20 '16 at 17:13
Or you can represent the dot product part as a tensor operation through use of the Kronecker Delta, that is $acdot b = a_i b_j delta_ij$
â Triatticus
Oct 20 '16 at 17:16
Then you can write your product as $a cdot (btimes c) = a_i (b times c)_j delta_ij = a_i (epsilon_jklhate_j b_k c_l) delta_ij$
â Triatticus
Oct 20 '16 at 17:22
you should look at the problem of calculate the det[a.(bxc)] and compare
â janmarqz
Oct 20 '16 at 20:31