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Levi civita symbol and kronecker delta identities in 4 dimensions

  1. Jul 17, 2014 #1
    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:
     
  2. jcsd
  3. Jul 17, 2014 #2

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
     
  4. Jul 18, 2014 #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. $$\varepsilon^{0 i j k} \varepsilon_{0 i j l} = - 2
    \delta^k_l$$
     
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