Levi civita symbol and kronecker delta identities in 4 dimensions

In summary, the conversation discusses how to explicitly show that the expression $$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2 \delta^k_l$$ is true for all values of k and l. The conversation delves into the use of the convention $$\varepsilon^{\alpha \beta \gamma \delta} = - \varepsilon_{\alpha \beta \gamma \delta}$$ and discusses how to simplify the problem by making observations. Ultimately, the summary points out that the formula does not hold if k=l=0 and suggests a way to write the equation explicitly to understand it better.
  • #1
Emil
8
0
I'm trying to explicitly show that

[tex]\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2 \delta^k_l[/tex]

I sort of went off the deep end and tried to express everything instead of using snazzy tricks and ended up with

[tex]
\begin{eqnarray*}
\delta^{\mu \rho}_{\nu \sigma} & = & \delta^{\mu}_{\nu}
\delta^{\rho}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\rho}_{\nu}\\
& & \\
\delta^{\mu \rho_1 \rho_2}_{\nu \sigma_1 \sigma_2} & = & \delta^{\mu}_{\nu}
\delta^{\rho_1 \rho_2}_{\sigma_1 \sigma_2} - \delta^{\mu}_{\sigma_1}
\delta^{\rho_1 \rho_2}_{\nu \sigma_2} + \delta^{\mu}_{\sigma_1}
\delta^{\rho_1 \rho_2}_{\sigma_2 \nu}\\
& & \\
\delta^{\mu \rho_1 \rho_2 \rho_3}_{\nu \sigma_1 \sigma_2 \sigma_3} & = &
\delta^{\mu}_{\nu} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_1 \sigma_2
\sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\nu
\sigma_2 \sigma_3} + \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2
\rho_3}_{\sigma_2 \nu \sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1
\rho_2 \rho_3}_{\sigma_2 \sigma_3 \nu}\\
& & \\
\varepsilon^{0 i j k} \varepsilon_{0 i j l} = \delta^{0 i j k}_{0 i j l} & =
& \delta^0_0 \delta^{i j k}_{i j l} - \delta^0_i \delta^{i j k}_{0 j l} +
\delta^0_i \delta^{i j k}_{j 0 l} - \delta^0_i \delta^{i j k}_{j l 0}\\
& & \\
& = & \delta^0_0 \left( \delta^i_i \delta^{j k}_{j l} - \delta^i_j
\delta^{j k}_{i l} + \delta^i_i \delta^{j k}_{l j} \right) \ldots\\
& & - \delta^0_i \left( \delta^i_0 \delta^{j k}_{j l} - \delta^i_j
\delta^{j k}_{0 l} + \delta^0_j \delta^{j k}_{l 0} \right) \ldots\\
& & + \delta^0_i \left( \delta^i_j \delta^{j k}_{0 l} - \delta^i_0
\delta^{j k}_{j l} + \delta^0_0 \delta^{j k}_{l j} \right) \ldots\\
& & - \delta^0_i \left( \delta^i_j \delta^{j k}_{l 0} - \delta^i_l
\delta^{j k}_{j 0} + \delta^0_l \delta^{j k}_{0 j} \right)\\
& & \\
& = & \delta^0_0 \left( \delta^i_i \left( \delta^j_j \delta^k_l -
\delta^j_l \delta^k_j \right) - \delta^i_j \left( \delta^j_i \delta^k_l -
\delta^j_l \delta^k_i \right) + \delta^i_i \left( \delta^j_l \delta^k_j -
\delta^j_j \delta^k_l \right) \right) \ldots\\
& & - \delta^0_i \left( \delta^i_0 \left( \delta^j_j \delta^k_l -
\delta^j_l \delta^k_j \right) - \delta^i_j \left( \delta^j_0 \delta^k_l -
\delta^j_l \delta^k_0 \right) + \delta^0_j \left( \delta^j_l \delta^k_0 -
\delta^j_0 \delta^k_l \right) \right) \ldots\\
& & + \delta^0_i \left( \delta^i_j \left( \delta^j_0 \delta^k_l -
\delta^j_l \delta^k_0 \right) - \delta^i_0 \left( \delta^j_j \delta^k_l -
\delta^j_l \delta^k_j \right) + \delta^0_0 \left( \delta^j_l \delta^k_j -
\delta^j_j \delta^k_l \right) \right) \ldots\\
& & - \delta^0_i \left( \delta^i_j \left( \delta^j_l \delta^k_0 -
\delta^j_0 \delta^k_l \right) - \delta^i_l \left( \delta^j_j \delta^k_0 -
\delta^j_0 \delta^k_j \right) + \delta^0_l \left( \delta^j_0 \delta^k_j -
\delta^j_j \delta^k_0 \right) \right)\\
& & \\
& & 0 = i = j\\
& & \\
& = & \delta^0_0 \delta^i_i \delta^j_j \delta^k_l - \delta^0_0 \delta^i_j
\delta^j_i \delta^k_l - \delta^0_0 \delta^i_i \delta^j_j \delta^k_l \ldots\\
& & - \delta^0_i \delta^i_0 \delta^j_j \delta^k_l + \delta^0_i \delta^i_j
\delta^j_0 \delta^k_l + \delta^0_i \delta^0_j \delta^j_0 \delta^k_l \ldots\\
& & + \delta^0_i \delta^i_j \delta^j_0 \delta^k_l - \delta^0_i \delta^i_0
\delta^j_j \delta^k_l - \delta^0_0 \delta^j_j \delta^k_l \ldots\\
& & + \delta^0_i \delta^i_j \delta^j_0 \delta^k_l\\
& & \\
& = & \delta^k_l - \delta^k_l\\
\end{eqnarray*}
[/tex]

The bottom line is that all I want for christmas is to get [tex]- 2 \delta^k_l[/tex] from

[tex] \varepsilon^{0 i j k} \varepsilon_{0 i j l} = \delta^{0 i j k}_{0 i j l} =
\left|\begin{array}{cccc}
\delta^0_0 & \delta^0_i & \delta^0_j & \delta^0_l\\
\delta^i_0 & \delta^i_i & \delta^i_j & \delta^i_l\\
\delta^j_0 & \delta^j_i & \delta^j_j & \delta^j_l\\
\delta^k_0 & \delta^k_i & \delta^k_j & \delta^k_l
\end{array}\right| = [/tex]

in a way that doesn't involve 100000 kronecker deltas. THAAAAANKS :rofl:
 
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  • #2
The formula doesn't hold if k=l=0.

I have never tried to brute-force this sort of thing. It's so much easier to make observations that simplify the problem. Let k be an arbitrary element of {1,2,3}. ##\varepsilon^{0 i j k} \varepsilon_{0 i j l}## is a sum with 4×4=16 terms, but most of them are zero. Clearly all terms with i=j, all terms with i=0 or j=0, and all terms with i=k or j=k, are zero. This only leaves two terms!

Let a,b be the two elements of {1,2,3} that aren't equal to k. The only terms that we haven't proved are zero are (no summation) ##\varepsilon^{0 a b k}\varepsilon_{0 a b l}## and ##\varepsilon^{0 b a k}\varepsilon_{0 b a l}##. If ##l\neq k##, then ##l\in\{a,b\}##, and both terms are zero. If ##l=k##, then one of the terms is 1×1=1, and the other is (-1)×(-1)=1.

Hm, I didn't get a minus sign. I'm guessing that your convention isn't that ##\varepsilon^{0123}## and ##\varepsilon_{0123}## are both 1. One of them is defined to be -1, right?
 
  • #3
Not understanding how contra/covariance comes in, and what to sum over

Thanks for the quick reply!

I know that the convention in use is $$\varepsilon^{\alpha \beta \gamma \delta}
= - \varepsilon_{\alpha \beta \gamma \delta}$$ I'm not quite comfortable on how it produces the minus signs.
Does one of the terms become $$\varepsilon^{0
a b k} \varepsilon_{0 a b l} = \left( - 1 \right) \left( 1 \right) = -
1$$ and the other $$\varepsilon^{0 b a k} \varepsilon_{0 b a l} = \left( 1
\right) \left( - 1 \right) = - 1$$ If so, why explicitly? Then I want
to sum the two and stick a kronecker delta.

I'm shooting in the dark here but I think I need an equation explicitly written out to understand. I want to write
$$\varepsilon^{0 i j k} \varepsilon_{0 i j l} =EinsteinSummation?OverWhat?= \varepsilon^{0 a b
k} \varepsilon_{0 a b l} + \varepsilon^{0 b a k} \varepsilon_{0 b a l} =
\left( - 1 \right) \left( 1 \right) + \left( 1 \right)
\left( - 1 \right) = - 2 $$
for all $$k = l$$ i.e. $$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2
\delta^k_l$$
 

1. What is the Levi Civita symbol?

The Levi Civita symbol, also known as the permutation symbol, is a mathematical symbol used in vector calculus and differential geometry to express the orientation of a coordinate system. It is defined as a function that assigns a numerical value of either 1, -1, or 0 to each permutation of a set of indices.

2. What is the Kronecker delta?

The Kronecker delta, also known as the Kronecker symbol, is a mathematical symbol used to represent the identity matrix in linear algebra. It takes the value of 1 if the two indices are equal, and 0 if they are not equal.

3. How are the Levi Civita symbol and Kronecker delta related?

The Levi Civita symbol and Kronecker delta are related through the identity:

εijkδjl = δil

This means that the Kronecker delta can be expressed in terms of the Levi Civita symbol and vice versa.

4. How are these identities used in 4 dimensions?

In four dimensions, the Levi Civita symbol and Kronecker delta are used to manipulate vector and tensor equations. They are especially useful in expressing cross products and determinants. Additionally, the identities are used in the study of electromagnetism, general relativity, and fluid dynamics.

5. Are these identities only applicable in 4 dimensions?

No, the Levi Civita symbol and Kronecker delta can be used in any number of dimensions. However, in 4 dimensions, they are particularly useful and are often used in physics and engineering applications.

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