Prove using the Levi-Civita Tensor/Kroenecker Delta that: (AxB)x(CxD) = (A.BxD).C-(A.BxC).D



To prove the vector identity \((\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{B} \times \mathbf{D}) \mathbf{C} - (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C}) \mathbf{D}\), we can use the **Levi-Civita tensor** \(\varepsilon_{ijk}\) and **Kronecker delta** \(\delta_{ij}\). Below is a step-by-step proof using these tools. 

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### 1. Expand each cross product in Levi-Civita notation

#### Cross product using the Levi-Civita symbol:
For two vectors \(\mathbf{A} \times \mathbf{B}\), the \(i\)-th component is given by:
\[
(\mathbf{A} \times \mathbf{B})_i = \varepsilon_{ijk} A_j B_k
\]
where the Levi-Civita symbol \(\varepsilon_{ijk}\) is 1 or -1 depending on the permutation of indices, and 0 if any two indices are repeated.

Similarly, for the nested cross product \((\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})\), the \(i\)-th component is:
\[
[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \varepsilon_{ilm} (\mathbf{A} \times \mathbf{B})_l (\mathbf{C} \times \mathbf{D})_m
\]
Using the definition of cross products for each term:
\[
(\mathbf{A} \times \mathbf{B})_l = \varepsilon_{lpq} A_p B_q, \quad (\mathbf{C} \times \mathbf{D})_m = \varepsilon_{mrs} C_r D_s
\]
Thus:
\[
[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \varepsilon_{ilm} \varepsilon_{lpq} A_p B_q \varepsilon_{mrs} C_r D_s
\]

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### 2. Use the Levi-Civita identity

We now use the **Levi-Civita contraction identity**:
\[
\varepsilon_{ilm} \varepsilon_{lpq} = \delta_{ip} \delta_{mq} - \delta_{iq} \delta_{mp}
\]
Substitute this into the expression:
\[
[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = (\delta_{ip} \delta_{mq} - \delta_{iq} \delta_{mp}) A_p B_q \varepsilon_{mrs} C_r D_s
\]
Distribute the terms:
\[
[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \delta_{ip} \delta_{mq} A_p B_q \varepsilon_{mrs} C_r D_s - \delta_{iq} \delta_{mp} A_p B_q \varepsilon_{mrs} C_r D_s
\]

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### 3. Simplify the expression

Using the properties of the Kronecker delta:
1. \(\delta_{ip} A_p = A_i\)
2. \(\delta_{mq} B_q = B_m\)
3. \(\delta_{iq} B_q = B_i\)
4. \(\delta_{mp} A_p = A_m\)

Thus, the first term becomes:
\[
A_i B_m \varepsilon_{mrs} C_r D_s = A_i (\mathbf{B} \times \mathbf{D})_s C_r
\]


Prove using the Levi-Civita Tensor/Kronecker Delta that: (AxB)x(CxD) = (A.BxD).C-(A.BxC).D

Explore a detailed proof of the vector identity (A × B) × (C × D) = (A · B × D) C - (A · B × C) D, utilizing the Levi-Civita tensor and Kronecker delta. This guide covers advanced vector calculus techniques, Levi-Civita tensor notation, and contraction identities, ideal for university-level physics or math students.

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