Levi-Civita Symbol
📂Mathematical Physics Levi-Civita Symbol Definition The ϵ i j k \epsilon_{ijk} ϵ ijk Levi-Civita symbol .
ϵ i j k = { + 1 if ϵ 123 , ϵ 231 , ϵ 312 − 1 if ϵ 132 , ϵ 213 , ϵ 321 0 if i = j or j = k or k = i
\epsilon_{ijk} = \begin{cases} +1 & \text{if} \ \epsilon_{123}, \epsilon_{231}, \epsilon_{312}
\\ -1 & \text{if} \ \epsilon_{132}, \epsilon_{213}, \epsilon_{321}
\\ 0 & \text{if} \ i=j \ \text{or} \ j=k \ \text{or} \ k=i \end{cases}
ϵ ijk = ⎩ ⎨ ⎧ + 1 − 1 0 if ϵ 123 , ϵ 231 , ϵ 312 if ϵ 132 , ϵ 213 , ϵ 321 if i = j or j = k or k = i
Description While the Kronecker delta only considers whether the indices are equal, the Levi-Civita symbol, as shown in its definition, is also affected by the order of the indices. ϵ i j k \epsilon_{ijk} ϵ ijk + 1 +1 + 1 i i i j j j k k k ( 1 → 2 → 3 → 1 ) (1\to 2\to 3\to 1) ( 1 → 2 → 3 → 1 ) − 1 -1 − 1 ( 3 → 2 → 1 → 3 ) (3\to 2\to 1\to 3) ( 3 → 2 → 1 → 3 ) 0 0 0 3 × 3 × 3 = 27 3\times 3\times 3=27 3 × 3 × 3 = 27 6 6 6 0 0 0
i = 1 k = 1 k = 2 k = 3 j = 1 ϵ 111 = 0 ϵ 112 = 0 ϵ 113 = 0 j = 2 ϵ 121 = 0 ϵ 122 = 0 ϵ 123 = 1 j = 3 ϵ 131 = 0 ϵ 132 = − 1 ϵ 133 = 0 i = 2 k = 1 k = 2 k = 3 j = 1 ϵ 211 = 0 ϵ 212 = 0 ϵ 213 = − 1 j = 2 ϵ 221 = 0 ϵ 222 = 0 ϵ 223 = 0 j = 3 ϵ 231 = 1 ϵ 232 = 0 ϵ 233 = 0 i = 3 k = 1 k = 2 k = 3 j = 1 ϵ 311 = 0 ϵ 312 = 1 ϵ 313 = 0 j = 2 ϵ 321 = − 1 ϵ 322 = 0 ϵ 323 = 0 j = 3 ϵ 331 = 0 ϵ 332 = 0 ϵ 333 = 0
\begin{array}{|c|c|c|c|}\hline i=1 & k=1 & k=2 & k=3 \\
\hline j=1 & \epsilon_{111}=0 & \epsilon_{112}=0 & \epsilon_{113}=0 \\
j=2 & \epsilon_{121}=0 & \epsilon_{122}=0 & \epsilon_{123}=1 \\
j=3 & \epsilon_{131}=0 & \epsilon_{132}=-1 & \epsilon_{133}=0 \\
\hline \end{array}\quad \begin{array}{|c|c|c|c|}\hline i=2 & k=1 & k=2 & k=3 \\
\hline j=1 & \epsilon_{211}=0 & \epsilon_{212}=0 & \epsilon_{213}=-1 \\
j=2 & \epsilon_{221}=0 & \epsilon_{222}=0 & \epsilon_{223}=0 \\
j=3 & \epsilon_{231}=1 & \epsilon_{232}=0 & \epsilon_{233}=0 \\
\hline \end{array} \\
{} \\
\begin{array}{|c|c|c|c|}\hline i=3 & k=1 & k=2 & k=3 \\
\hline j=1 & \epsilon_{311}=0 & \epsilon_{312}=1 & \epsilon_{313}=0 \\
j=2 & \epsilon_{321}=-1 & \epsilon_{322}=0 & \epsilon_{323}=0 \\
j=3 & \epsilon_{331}=0 & \epsilon_{332}=0 & \epsilon_{333}=0 \\
\hline \end{array}
i = 1 j = 1 j = 2 j = 3 k = 1 ϵ 111 = 0 ϵ 121 = 0 ϵ 131 = 0 k = 2 ϵ 112 = 0 ϵ 122 = 0 ϵ 132 = − 1 k = 3 ϵ 113 = 0 ϵ 123 = 1 ϵ 133 = 0 i = 2 j = 1 j = 2 j = 3 k = 1 ϵ 211 = 0 ϵ 221 = 0 ϵ 231 = 1 k = 2 ϵ 212 = 0 ϵ 222 = 0 ϵ 232 = 0 k = 3 ϵ 213 = − 1 ϵ 223 = 0 ϵ 233 = 0 i = 3 j = 1 j = 2 j = 3 k = 1 ϵ 311 = 0 ϵ 321 = − 1 ϵ 331 = 0 k = 2 ϵ 312 = 1 ϵ 322 = 0 ϵ 332 = 0 k = 3 ϵ 313 = 0 ϵ 323 = 0 ϵ 333 = 0
It is an example of a 3 3 3
Examples Cross Product Using the Levi-Civita symbol, the cross product of two vectors can be expressed very simply. In a 3-dimensional orthogonal coordinate system, the cross product of two vectors is as follows.
A × B = e ^ x ( A y B z − A z B y ) + e ^ y ( A z B x − A x B z ) + e ^ z ( A x B y − A y B x ) = ∣ e ^ x e ^ y e ^ z A x A y A z B x B y B z ∣
\begin{align*}
\mathbf{A} \times \mathbf{B} =&\ \hat{\mathbf{e}}_{x} (A_{y}B_{z}-A_{z}B_{y}) + \hat{\mathbf{e}}_{y} (A_{z}B_{x}-A_{x}B_{z}) + \hat{\mathbf{e}}_{z} (A_{x}B_{y}-A_{y}B_{x})
\\ =&\ \begin{vmatrix} \hat{\mathbf{e}}_{x} \quad \hat{\mathbf{e}}_{y} \quad \hat{\mathbf{e}}_{z}
\\ A_{x} \quad A_{y} \quad A_{z}
\\ B_{x} \quad B_{y} \quad B_{z} \end{vmatrix}
\end{align*}
A × B = = e ^ x ( A y B z − A z B y ) + e ^ y ( A z B x − A x B z ) + e ^ z ( A x B y − A y B x ) e ^ x e ^ y e ^ z A x A y A z B x B y B z
Here, if we set x = 1 x=1 x = 1 y = 2 y=2 y = 2 z = 3 z=3 z = 3
∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 ϵ i j k e ^ i A j B k
\sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} \hat{\mathbf{e}}_{i} A_{j}B_{k}}
i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk e ^ i A j B k
If we expand only the terms that are not 0 0 0
∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 ϵ i j k e ^ i A j B k = ϵ 123 e ^ 1 A 2 B 3 + ϵ 132 e ^ 1 A 3 B 2 + ϵ 231 e ^ 2 A 3 B 1 + ϵ 213 e ^ 2 A 1 B 3 + ϵ 312 e ^ 3 A 1 B 2 + ϵ 321 e ^ 3 A 2 B 1
\begin{align*}
& \sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} \hat{\mathbf{e}}_{i} A_{j}B_{k}}
\\ =&\ \epsilon_{123} \hat{\mathbf{e}}_{1}A_{2}B_{3} + \epsilon_{132} \hat{\mathbf{e}}_{1}A_{3}B_{2} + \epsilon_{ 231 }\hat{\mathbf{e}}_{2}A_{3}B_{1} + \epsilon_{213}\hat{\mathbf{e}}_{2}A_{1}B_{3} + \epsilon_{312}\hat{\mathbf{e}}_{3}A_{1}B_{2} + \epsilon_{321}\hat{\mathbf{e}}_{3}A_{2}B_{1}
\end{align*}
= i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk e ^ i A j B k ϵ 123 e ^ 1 A 2 B 3 + ϵ 132 e ^ 1 A 3 B 2 + ϵ 231 e ^ 2 A 3 B 1 + ϵ 213 e ^ 2 A 1 B 3 + ϵ 312 e ^ 3 A 1 B 2 + ϵ 321 e ^ 3 A 2 B 1
By substituting ϵ 123 = ϵ 231 = ϵ 312 = 1 \epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1 ϵ 123 = ϵ 231 = ϵ 312 = 1 ϵ 132 = ϵ 213 = ϵ 321 = − 1 \epsilon_{132}=\epsilon_{213}=\epsilon_{321}=-1 ϵ 132 = ϵ 213 = ϵ 321 = − 1
∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 ϵ i j k A i B j e ^ k = e ^ 1 ( A 2 B 3 − A 3 B 2 ) + e ^ 2 ( A 3 B 1 − A 1 B 3 ) + e ^ 3 ( A 1 B 2 − A 2 B 1 )
\begin{align*}
& \sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} A_{i}B_{j}}\hat{\mathbf{e}}_{k}
\\ =&\ \hat{\mathbf{e}}_{1}\left( A_{2}B_{3} - A_{3}B_{2} \right) + \hat{\mathbf{e}}_{2}\left( A_{3}B_{1} - A_{1}B_{3} \right) + \hat{\mathbf{e}}_{3}\left( A_{1}B_{2} - A_{2}B_{1} \right)
\end{align*}
= i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk A i B j e ^ k e ^ 1 ( A 2 B 3 − A 3 B 2 ) + e ^ 2 ( A 3 B 1 − A 1 B 3 ) + e ^ 3 ( A 1 B 2 − A 2 B 1 )
Finally, by substituting 1 1 1 2 2 2 3 3 3 x x x y y y z z z
e ^ x ( A y B z − A z B y ) + e ^ y ( A z B x − A x B z ) + e ^ z ( A x B y − A y B x )
\hat{\mathbf{e}}_{x}\left( A_{y}B_{z} - A_{z}B_{y} \right) + \hat{\mathbf{e}}_{y}\left( A_{z}B_{x} - A_{x}B_{z} \right) + \hat{\mathbf{e}}_{z}\left( A_{x}B_{y} - A_{y}B_{x} \right)
e ^ x ( A y B z − A z B y ) + e ^ y ( A z B x − A x B z ) + e ^ z ( A x B y − A y B x )
Therefore, we obtain the below result.
A × B = ∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 ϵ i j k e ^ i A j B k
\mathbf{A} \times \mathbf{ B } = \sum \limits_{i=1}^{3} \sum \limits_{j=1}^{3} \sum \limits_{k=1}^{3} \epsilon_{ijk}\hat{\mathbf{e}}_{i}A_{j}B_{k}
A × B = i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk e ^ i A j B k
Using Einstein notation , it can be expressed as follows.
A × B = ϵ i j k e ^ i A j B k
\mathbf{A} \times \mathbf{ B } = \epsilon_{ijk}\hat{\mathbf{e}}_{i}A_{j}B_{k}
A × B = ϵ ijk e ^ i A j B k
Each component of the cross product can be easily expressed as shown in the above equation, where the i i i ( A × B ) (\mathbf{A} \times \mathbf{B}) ( A × B )
( A × B ) i = ϵ i j k A j B k
(\mathbf{A} \times \mathbf {B})_{i}=\epsilon_{ijk}A_{j}B_{k}
( A × B ) i = ϵ ijk A j B k
Determinant The determinant of matrix 3 × 3 3 \times 3 3 × 3 A = [ a i j ] A = [a_{ij}] A = [ a ij ]
∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ = ∑ i , j , k = 1 3 ϵ i j k a 1 i a 2 j a 3 k
\begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{vmatrix} = \sum\limits_{i,j,k=1}^{3} \epsilon_{ijk}a_{1i}a_{2j}a_{3k}
a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 = i , j , k = 1 ∑ 3 ϵ ijk a 1 i a 2 j a 3 k
The demonstration is simple. By expanding the determinant and carefully observing the second index of each matrix component, one can see that the index’s Levi-Civita symbol corresponds to the sign of each term.
det A = a 11 ( a 22 a 33 − a 23 a 32 ) + a 12 ( a 23 a 31 − a 21 a 33 ) + a 13 ( a 21 a 32 − a 22 a 31 ) = a 11 a 22 a 33 − a 11 a 23 a 32 + a 12 a 23 a 31 − a 12 a 21 a 33 + a 13 a 21 a 32 − a 13 a 22 a 31 = a 1 1 a 2 2 a 3 3 − a 1 1 a 2 3 a 3 2 + a 1 2 a 2 3 a 3 1 − a 1 2 a 2 1 a 3 3 + a 1 3 a 2 1 a 3 2 − a 1 3 a 2 2 a 3 1 = ϵ 123 a 1 1 a 2 2 a 3 3 + ϵ 132 a 1 1 a 2 3 a 3 2 + ϵ 231 a 1 2 a 2 3 a 3 1 + ϵ 213 a 1 2 a 2 1 a 3 3 + ϵ 312 a 1 3 a 2 1 a 3 2 + ϵ 321 a 1 3 a 2 2 a 3 1 = ∑ i , j , k = 1 3 ϵ i j k a 1 i a 2 j a 3 k
\begin{align*}
& \det A \\
&= a_{11}(a_{22}a_{33} - a_{23}a_{32}) + a_{12}(a_{23}a_{31} - a_{21}a_{33}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \\
&= a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} + a_{12}a_{23}a_{31} - a_{12}a_{21}a_{33} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \\
&= a_{1\textcolor{red}{1}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{3}} - a_{1\textcolor{red}{1}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{2}} + a_{1\textcolor{red}{2}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{1}} - a_{1\textcolor{red}{2}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{3}} + a_{1\textcolor{red}{3}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{2}} - a_{1\textcolor{red}{3}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{1}} \\
&= \epsilon_{\textcolor{red}{123}}a_{1\textcolor{red}{1}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{3}} + \epsilon_{\textcolor{red}{132}}a_{1\textcolor{red}{1}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{2}} + \epsilon_{\textcolor{red}{231}}a_{1\textcolor{red}{2}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{1}} + \epsilon_{\textcolor{red}{213}}a_{1\textcolor{red}{2}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{3}} + \epsilon_{\textcolor{red}{312}}a_{1\textcolor{red}{3}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{2}} + \epsilon_{\textcolor{red}{321}}a_{1\textcolor{red}{3}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{1}} \\
&= \sum\limits_{i,j,k=1}^{3} \epsilon_{ijk}a_{1i}a_{2j}a_{3k}
\end{align*}
det A = a 11 ( a 22 a 33 − a 23 a 32 ) + a 12 ( a 23 a 31 − a 21 a 33 ) + a 13 ( a 21 a 32 − a 22 a 31 ) = a 11 a 22 a 33 − a 11 a 23 a 32 + a 12 a 23 a 31 − a 12 a 21 a 33 + a 13 a 21 a 32 − a 13 a 22 a 31 = a 1 1 a 2 2 a 3 3 − a 1 1 a 2 3 a 3 2 + a 1 2 a 2 3 a 3 1 − a 1 2 a 2 1 a 3 3 + a 1 3 a 2 1 a 3 2 − a 1 3 a 2 2 a 3 1 = ϵ 123 a 1 1 a 2 2 a 3 3 + ϵ 132 a 1 1 a 2 3 a 3 2 + ϵ 231 a 1 2 a 2 3 a 3 1 + ϵ 213 a 1 2 a 2 1 a 3 3 + ϵ 312 a 1 3 a 2 1 a 3 2 + ϵ 321 a 1 3 a 2 2 a 3 1 = i , j , k = 1 ∑ 3 ϵ ijk a 1 i a 2 j a 3 k
(a) When one index is the same: ϵ i j k ϵ l m k = δ i l δ j m − δ i m δ j l \epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl} ϵ ijk ϵ l mk = δ i l δ jm − δ im δ j l
(b) When two indices are the same: ϵ i j k ϵ l j k = 2 δ i l \epsilon_{ijk}\epsilon_{ljk}=2\delta_{il} ϵ ijk ϵ l jk = 2 δ i l
(c) When three indices are the same: ϵ i j k ϵ i j k = 6 \epsilon_{ijk}\epsilon_{ijk}=6 ϵ ijk ϵ ijk = 6
