Tensor equation [closed]
$begingroup$
If you have two antisymmetric tensors $A_{mu nu}$ and $B_{mu nu}$, and for every anti symmetric tensor $epsilon^{mu nu}$,
$epsilon^{mu nu} A_{mu nu} = epsilon^{mu nu} B_{mu nu}$
Is it true that
$A_{mu nu} = B_{mu nu}$ ?
Summation over the indices is assumed.
matrices mathematical-physics tensors index-notation
edited Jan 14 at 13:44
Sou
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
$endgroup$
closed as off-topic by Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste Jan 14 at 18:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If you have two antisymmetric tensors $A_{mu nu}$ and $B_{mu nu}$, and for every anti symmetric tensor $epsilon^{mu nu}$,
$epsilon^{mu nu} A_{mu nu} = epsilon^{mu nu} B_{mu nu}$
Is it true that
$A_{mu nu} = B_{mu nu}$ ?
Summation over the indices is assumed.
matrices mathematical-physics tensors index-notation
edited Jan 14 at 13:44
Sou
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
$endgroup$
closed as off-topic by Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste Jan 14 at 18:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44
add a comment |
$begingroup$
If you have two antisymmetric tensors $A_{mu nu}$ and $B_{mu nu}$, and for every anti symmetric tensor $epsilon^{mu nu}$,
$epsilon^{mu nu} A_{mu nu} = epsilon^{mu nu} B_{mu nu}$
Is it true that
$A_{mu nu} = B_{mu nu}$ ?
Summation over the indices is assumed.
matrices mathematical-physics tensors index-notation
edited Jan 14 at 13:44
Sou
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
$endgroup$
If you have two antisymmetric tensors $A_{mu nu}$ and $B_{mu nu}$, and for every anti symmetric tensor $epsilon^{mu nu}$,
$epsilon^{mu nu} A_{mu nu} = epsilon^{mu nu} B_{mu nu}$
Is it true that
$A_{mu nu} = B_{mu nu}$ ?
Summation over the indices is assumed.
matrices mathematical-physics tensors index-notation
matrices mathematical-physics tensors index-notation
edited Jan 14 at 13:44
Sou
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
edited Jan 14 at 13:44
Sou
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
edited Jan 14 at 13:44
Sou
3,2892923
edited Jan 14 at 13:44
Sou
3,2892923
edited Jan 14 at 13:44
Sou
3,2892923
3,2892923
asked Jan 14 at 8:27
HiggsinoHiggsino
595
asked Jan 14 at 8:27
HiggsinoHiggsino
595
asked Jan 14 at 8:27
HiggsinoHiggsino
595
595
closed as off-topic by Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste Jan 14 at 18:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste Jan 14 at 18:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Adrian Keister, Lee David Chung Lin, A. Pongrácz, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44
add a comment |
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For $n=2$, this is true. For the equality is
$$
epsilon^{12}A_{12}+epsilon^{21}A_{21} = epsilon^{12}B_{12}+epsilon^{21}B_{21} implies 2epsilon^{12}A_{12} = 2epsilon^{12}B_{12}.
$$
So $B_{12}=A_{12}$ and $A_{21}=-A_{12} = -B_{12} = B_{21}$ and $A_{ii}=B_{ii} = 0$ for all $i$. For $ngeq3$ this is false generally. The counterexample is
$$
[epsilon^{ij}] = begin{pmatrix} 0 & 1 & 1\ -1 & 0 & 1 \ -1 & -1 & 0 end{pmatrix},
[A_{ij}] = begin{pmatrix} 0 & 0 & 1\ 0 & 0 & 0 \ -1 & 0 & 0 end{pmatrix},
[B_{ij}] = begin{pmatrix} 0 & 1/3 & 1/3\ -1/3 & 0 & 1/3 \ -1/3 & -1/3 & 0 end{pmatrix}.
$$
answered Jan 14 at 9:06
SouSou
3,2892923
$endgroup$
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
add a comment |
$begingroup$
Reason I asked is because of this example found from the book "A first course in string theory".
I don't understand the step when he cancels $epsilon$ on both sides because the dimension is generally bigger then 2 I think.
answered Jan 14 at 11:56
HiggsinoHiggsino
595
$endgroup$
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
add a comment |
$begingroup$
8.22) which you asked for is the following relation.
answered Jan 14 at 14:23
HiggsinoHiggsino
595
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $n=2$, this is true. For the equality is
$$
epsilon^{12}A_{12}+epsilon^{21}A_{21} = epsilon^{12}B_{12}+epsilon^{21}B_{21} implies 2epsilon^{12}A_{12} = 2epsilon^{12}B_{12}.
$$
So $B_{12}=A_{12}$ and $A_{21}=-A_{12} = -B_{12} = B_{21}$ and $A_{ii}=B_{ii} = 0$ for all $i$. For $ngeq3$ this is false generally. The counterexample is
$$
[epsilon^{ij}] = begin{pmatrix} 0 & 1 & 1\ -1 & 0 & 1 \ -1 & -1 & 0 end{pmatrix},
[A_{ij}] = begin{pmatrix} 0 & 0 & 1\ 0 & 0 & 0 \ -1 & 0 & 0 end{pmatrix},
[B_{ij}] = begin{pmatrix} 0 & 1/3 & 1/3\ -1/3 & 0 & 1/3 \ -1/3 & -1/3 & 0 end{pmatrix}.
$$
answered Jan 14 at 9:06
SouSou
3,2892923
$endgroup$
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
add a comment |
$begingroup$
For $n=2$, this is true. For the equality is
$$
epsilon^{12}A_{12}+epsilon^{21}A_{21} = epsilon^{12}B_{12}+epsilon^{21}B_{21} implies 2epsilon^{12}A_{12} = 2epsilon^{12}B_{12}.
$$
So $B_{12}=A_{12}$ and $A_{21}=-A_{12} = -B_{12} = B_{21}$ and $A_{ii}=B_{ii} = 0$ for all $i$. For $ngeq3$ this is false generally. The counterexample is
$$
[epsilon^{ij}] = begin{pmatrix} 0 & 1 & 1\ -1 & 0 & 1 \ -1 & -1 & 0 end{pmatrix},
[A_{ij}] = begin{pmatrix} 0 & 0 & 1\ 0 & 0 & 0 \ -1 & 0 & 0 end{pmatrix},
[B_{ij}] = begin{pmatrix} 0 & 1/3 & 1/3\ -1/3 & 0 & 1/3 \ -1/3 & -1/3 & 0 end{pmatrix}.
$$
answered Jan 14 at 9:06
SouSou
3,2892923
$endgroup$
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
add a comment |
$begingroup$
For $n=2$, this is true. For the equality is
$$
epsilon^{12}A_{12}+epsilon^{21}A_{21} = epsilon^{12}B_{12}+epsilon^{21}B_{21} implies 2epsilon^{12}A_{12} = 2epsilon^{12}B_{12}.
$$
So $B_{12}=A_{12}$ and $A_{21}=-A_{12} = -B_{12} = B_{21}$ and $A_{ii}=B_{ii} = 0$ for all $i$. For $ngeq3$ this is false generally. The counterexample is
$$
[epsilon^{ij}] = begin{pmatrix} 0 & 1 & 1\ -1 & 0 & 1 \ -1 & -1 & 0 end{pmatrix},
[A_{ij}] = begin{pmatrix} 0 & 0 & 1\ 0 & 0 & 0 \ -1 & 0 & 0 end{pmatrix},
[B_{ij}] = begin{pmatrix} 0 & 1/3 & 1/3\ -1/3 & 0 & 1/3 \ -1/3 & -1/3 & 0 end{pmatrix}.
$$
answered Jan 14 at 9:06
SouSou
3,2892923
$endgroup$
For $n=2$, this is true. For the equality is
$$
epsilon^{12}A_{12}+epsilon^{21}A_{21} = epsilon^{12}B_{12}+epsilon^{21}B_{21} implies 2epsilon^{12}A_{12} = 2epsilon^{12}B_{12}.
$$
So $B_{12}=A_{12}$ and $A_{21}=-A_{12} = -B_{12} = B_{21}$ and $A_{ii}=B_{ii} = 0$ for all $i$. For $ngeq3$ this is false generally. The counterexample is
$$
[epsilon^{ij}] = begin{pmatrix} 0 & 1 & 1\ -1 & 0 & 1 \ -1 & -1 & 0 end{pmatrix},
[A_{ij}] = begin{pmatrix} 0 & 0 & 1\ 0 & 0 & 0 \ -1 & 0 & 0 end{pmatrix},
[B_{ij}] = begin{pmatrix} 0 & 1/3 & 1/3\ -1/3 & 0 & 1/3 \ -1/3 & -1/3 & 0 end{pmatrix}.
$$
answered Jan 14 at 9:06
SouSou
3,2892923
answered Jan 14 at 9:06
SouSou
3,2892923
answered Jan 14 at 9:06
SouSou
3,2892923
answered Jan 14 at 9:06
SouSou
3,2892923
3,2892923
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
add a comment |
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
Wow, nice! Tensors :P !!!
$endgroup$
– user2662833
Jan 14 at 11:42
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
I just double checked your counter example, I don't believe $e^{ij} A_{ij} = e^{ij} B_{ij}$ In this example. have I made a mistake?
$endgroup$
– user2662833
Jan 14 at 11:48
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
$begingroup$
@user2662833 Just compute lhs and rhs separately by using the antisymmetry property (or not, whatever). E.g. lhs is $epsilon^{12}A_{12}+epsilon^{13}A_{13}+epsilon^{21}A_{21}+epsilon^{23}A_{23}+epsilon^{31}A_{31}+epsilon^{32}A_{32} = 2epsilon^{12}A_{12}+2epsilon^{13}A_{13}+2epsilon^{23}A_{23}=2$
$endgroup$
– Sou
Jan 14 at 13:22
add a comment |
$begingroup$
Reason I asked is because of this example found from the book "A first course in string theory".
I don't understand the step when he cancels $epsilon$ on both sides because the dimension is generally bigger then 2 I think.
answered Jan 14 at 11:56
HiggsinoHiggsino
595
$endgroup$
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
add a comment |
$begingroup$
Reason I asked is because of this example found from the book "A first course in string theory".
I don't understand the step when he cancels $epsilon$ on both sides because the dimension is generally bigger then 2 I think.
answered Jan 14 at 11:56
HiggsinoHiggsino
595
$endgroup$
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
add a comment |
$begingroup$
Reason I asked is because of this example found from the book "A first course in string theory".
I don't understand the step when he cancels $epsilon$ on both sides because the dimension is generally bigger then 2 I think.
answered Jan 14 at 11:56
HiggsinoHiggsino
595
$endgroup$
Reason I asked is because of this example found from the book "A first course in string theory".
I don't understand the step when he cancels $epsilon$ on both sides because the dimension is generally bigger then 2 I think.
answered Jan 14 at 11:56
HiggsinoHiggsino
595
edited Jan 14 at 13:36
answered Jan 14 at 11:56
HiggsinoHiggsino
595
answered Jan 14 at 11:56
HiggsinoHiggsino
595
answered Jan 14 at 11:56
HiggsinoHiggsino
595
595
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
add a comment |
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
Here we must be careful that 8.55 follows from 8.22 which is a definition not an equality. So here we actually want to know what is the current looks like if we have such a Langrangian.
$endgroup$
– Sou
Jan 14 at 14:01
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
$begingroup$
I suppose the reason we can cancel $epsilon$ on both sides is because it consists of different parameters which are independent of each other. So we can vary one parameter on both sides keeping the other parameters constant. And therefore we can match terms even if we have a sum of terms on both sides. It's like the separation of variable method when solving partial differential equations.
$endgroup$
– Higgsino
Jan 14 at 14:19
add a comment |
$begingroup$
8.22) which you asked for is the following relation.
answered Jan 14 at 14:23
HiggsinoHiggsino
595
$endgroup$
add a comment |
$begingroup$
8.22) which you asked for is the following relation.
answered Jan 14 at 14:23
HiggsinoHiggsino
595
$endgroup$
add a comment |
$begingroup$
8.22) which you asked for is the following relation.
answered Jan 14 at 14:23
HiggsinoHiggsino
595
$endgroup$
8.22) which you asked for is the following relation.
answered Jan 14 at 14:23
HiggsinoHiggsino
595
answered Jan 14 at 14:23
HiggsinoHiggsino
595
answered Jan 14 at 14:23
HiggsinoHiggsino
595
answered Jan 14 at 14:23
HiggsinoHiggsino
595
595
add a comment |
add a comment |
$begingroup$
What is $epsilon^{munu}$ ?
$endgroup$
– Sou
Jan 14 at 8:39
$begingroup$
It's just a general tensor. But I changed the question now, supposing it is anti symmetric.
$endgroup$
– Higgsino
Jan 14 at 8:44