Proving Vector Identity Using Levi-Civita Symbol | Quantum Science Philippines
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Proving Vector Identity Using Levi-Civita Symbol

Roel N. Baybayon

MSPhysics1

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We are going to prove the following vector identity using Levi-Civita symbol:

\left(\vec{a}\times\vec{b}\right)\cdot\left(\vec{c}\times\vec{d}\right)=\left(\vec{a}\cdot\vec{c}\right)\left(\vec{b}\cdot\vec{d}\right)-\left(\vec{a}\cdot\vec{d}\right)\left(\vec{b}\cdot\vec{c}\right)

Solution:

Let   \vec{a}=a_i \hat{e_i} ,    \vec{b}=b_j \hat{e_j} ,   \vec{c}=c_k \hat{e_k} ,   \vec{d}=d_l\hat{e_l}.

Then,

\begin{array}{rcl} \left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{c} \times\vec{d}\right) & = & \left(a_i \hat{e_i} \times b_j \hat{e_j}\right) \cdot \left(c_k \hat{e_k} \times d_l\hat{e_l}\right) \\ & = & \left(\epsilon_{ijm}a_i b_j \hat{e_m}\right) \cdot \left(\epsilon_{kln} c_k d_l \hat{e_l}\right) \\ & = & \epsilon_{ijm}\epsilon_{kln} a_i b_j c_k d_l \hat{e_m}\cdot\hat{e_n} \\ & = & \epsilon_{ijm}\epsilon_{kln} a_i b_j c_k d_l \delta_{mn}\end{array}

By definition:

\delta_{mn} = \begin{cases} 1,  & \mbox{if } m=n \\ 0, & \mbox{if } m\neq n \end{cases}

We have to let m=n so that,

\left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{c} \times\vec{d}\right)= \epsilon_{ijm}\epsilon_{klm} a_i b_j c_k d_l

Levi-Civita symbol can be expressed in terms of Kronecker delta given by:

\epsilon_{ijm}\epsilon_{klm}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}

Thus,

\begin{array}{rcl} \left(\vec{a} \times \vec{b}\right) \cdot  \left(\vec{c} \times\vec{d}\right) & = & \left(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}\right)a_i b_j c_k d_l \\ & = & \delta_{ik}\delta_{jl}a_i b_j c_k d_l-\delta_{il}\delta_{jk}a_i b_j c_k d_l \\  & = & \left(a_ic_k\delta_{ik}\right)\left(b_jd_l\delta_{jl}\right)-\left(a_id_l\delta_{il}\right)\left(b_jc_k\delta_{jk}\right) \\ & = & \left(a_ic_k \hat{e_i}\cdot \hat{e_k}\right)\left(b_jd_l\hat{e_j}\cdot \hat{e_l}\right)-\left(a_id_l\hat{e_i}\cdot \hat{e_l}\right)\left(b_jc_k\hat{e_j}\cdot \hat{e_k}\right) \\ & = & \left(a_i\hat{e_i}\cdot c_k\hat{e_k}\right)\left(b_j\hat{e_j}\cdot d_l\hat{e_l}\right)-\left(a_i\hat{e_i}\cdot d_l\hat{e_l}\right)\left(b_j\hat{e_j}\cdot c_k\hat{e_k}\right) \\ & = & \left(\vec{a}\cdot\vec{c}\right)\left(\vec{b}\cdot\vec{d}\right)-\left(\vec{a}\cdot\vec{d}\right)\left(\vec{b}\cdot\vec{c}\right)\end{array}

 

 

One Response to “Proving Vector Identity Using Levi-Civita Symbol”

  1. Ome Says:

    Thanks a lot! Very helpful!

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